General Chemistry
Core Principles

Noble Trex

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Contents

1 Foundations of Chemistry
 1.1 The Scientific Method and Classification of Matter
 1.2 SI Units and Measurement Scales
 1.3 Uncertainty, Precision, and Significant Figures
 1.4 Dimensional Analysis and Unit Conversions
 1.5 Atomic Theory and Subatomic Particles
 1.6 Atomic Number, Mass Number, and Isotopes
 1.7 Problems
2 Stoichiometry and the Mole
 2.1 The Mole Concept and Avogadro’s Constant
 2.2 Molar Mass and Chemical Formulas
 2.3 Percent Composition and Empirical Formulas
 2.4 Chemical Equations and Balancing
 2.5 Reaction Stoichiometry: Mass Relations
 2.6 Limiting Reagents and Percent Yield
 2.7 Combustion Analysis
 2.8 Problems
3 Reactions in Aqueous Solution
 3.1 General Properties of Aqueous Solutions
 3.2 Precipitation Reactions and Solubility Rules
 3.3 Acid-Base Reactions
 3.4 Oxidation-Reduction Reactions
 3.5 Concentration of Solutions: Molarity
 3.6 Solution Stoichiometry and Titrations
 3.7 Problems
4 Electronic Structure and Periodic Trends
 4.1 The Wave Nature of Light and Energy Quantization
 4.2 Bohr Model and Atomic Spectra
 4.3 Wave Mechanics and Quantum Numbers
 4.4 Atomic Orbitals and Their Shapes
 4.5 Electron Configurations
 4.6 Periodic Trends: Radius and Ionization Energy
 4.7 Electron Affinity and Metallic Character
 4.8 Problems
5 Chemical Bonding and Molecular Structure
 5.1 Types of Chemical Bonds
 5.2 Electronegativity and Bond Polarity
 5.3 Lewis Structures and the Octet Rule
 5.4 Formal Charge and Resonance
 5.5 Exceptions to the Octet Rule
 5.6 VSEPR Theory: Molecular Shapes
 5.7 Molecular Polarity
 5.8 Problems
6 Advanced Bonding Theories
 6.1 Valence Bond Theory
 6.2 Hybridization of Atomic Orbitals
 6.3 Multiple Bonds and Delocalization
 6.4 Introduction to Molecular Orbital Theory
 6.5 Homonuclear Diatomic Molecules
 6.6 Problems
7 Gases and the Kinetic Molecular Theory
 7.1 Pressure and Gas Laws
 7.2 The Ideal Gas Equation
 7.3 Gas Mixtures and Partial Pressures
 7.4 Stoichiometry of Gas Reactions
 7.5 Kinetic Molecular Theory
 7.6 Real Gases and Deviation from Ideality
 7.7 Problems
8 Thermochemistry
 8.1 Energy, Work, and Heat
 8.2 The First Law of Thermodynamics
 8.3 Enthalpy and Enthalpy of Reaction
 8.4 Calorimetry
 8.5 Hess’s Law
 8.6 Standard Enthalpies of Formation
 8.7 Problems
9 Intermolecular Forces, Liquids, and Solids
 9.1 Intermolecular Forces
 9.2 Properties of Liquids
 9.3 Phase Changes and Heating Curves
 9.4 Vapor Pressure and Boiling Point
 9.5 Phase Diagrams
 9.6 Structure of Solids
 9.7 Problems
10 Properties of Solutions
 10.1 The Solution Process
 10.2 Concentration Units
 10.3 Factors Affecting Solubility
 10.4 Colligative Properties: Vapor Pressure
 10.5 Boiling Point Elevation and Freezing Point Depression
 10.6 Osmotic Pressure
 10.7 Problems
11 Chemical Kinetics
 11.1 Reaction Rates
 11.2 Rate Laws and Reaction Order
 11.3 Integrated Rate Laws
 11.4 Reaction Half-Life
 11.5 Temperature and the Arrhenius Equation
 11.6 Reaction Mechanisms and Catalysis
 11.7 Problems
12 Chemical Equilibrium
 12.1 Dynamic Equilibrium
 12.2 The Equilibrium Constant
 12.3 The Reaction Quotient
 12.4 Calculating Equilibrium Concentrations
 12.5 Le Châtelier’s Principle
 12.6 Problems
13 Acids and Bases
 13.1 Acid-Base Definitions and Molecular Structure
 13.2 The pH Scale and Water Autoionization
 13.3 Strong Acids and Bases
 13.4 Weak Acids and Ionization Constants
 13.5 Weak Bases and Salt Properties
 13.6 Problems
14 Aqueous Ionic Equilibria
 14.1 The Common Ion Effect
 14.2 Buffered Solutions
 14.3 Acid-Base Titrations
 14.4 Solubility Equilibria (Ksp)
 14.5 Precipitation and Complex Ions
 14.6 Problems
15 Thermodynamics and Electrochemistry
 15.1 Entropy and the Second Law
 15.2 Gibbs Free Energy
 15.3 Free Energy and Equilibrium
 15.4 Galvanic Cells and Standard Potentials
 15.5 Cell Potential and Free Energy
 15.6 The Nernst Equation
 15.7 Electrolysis
 15.8 Problems

Introduction

Chemistry is often famously termed the "central science," a designation that acknowledges its unique position bridging the abstract laws of physics and the complex, systemic realities of biology and geology. To study chemistry is to learn the language of matter itself. It is a discipline that challenges us to look beyond the macroscopic world we perceive with our senses-the rusting of iron, the burning of wood, the dissolving of salt-and to find explanations in the invisible realm of atoms and molecules.

This textbook, General Chemistry: Core Principles, is designed to guide you through this intellectual transformation. The journey we undertake is not merely a collection of facts or a catalog of formulas; it is a systematic development of a worldview. The chemist’s perspective is one that toggles constantly between two realities: the observable world of bulk matter and the microscopic world of particles. The coherence of chemistry lies in the fact that the behavior of the former is entirely dictated by the properties of the latter.

The structure of this book reflects the logical progression required to master this dual perspective. We begin, in the opening chapters, by laying the foundations of the discipline. Chemistry is an experimental science, and as such, it relies on a rigorous framework of measurement, uncertainty, and dimensional analysis. Here, we introduce the concept of the mole-the chemist’s bridge between the atomic scale and the laboratory scale. By mastering stoichiometry and the types of reactions that occur in aqueous solution, you acquire the essential toolkit required for quantitative analysis.

Once these tools are in hand, we turn our gaze inward to the architecture of the atom. To understand how elements combine, we must first understand the electron. We will explore the quantum mechanical model of the atom, a theory that explains why the periodic table is shaped the way it is and why elements exhibit specific periodic trends. This understanding of electronic structure serves as the basis for our exploration of chemical bonding. From simple Lewis structures to the more

advanced theories of orbital hybridization and molecular orbitals, we will learn how nature constructs complex geometries from simple atomic building blocks.

With a firm grasp of molecular structure, the text expands to examine the states of matter. We move from the idealized behavior of gases, described by the kinetic molecular theory, to the condensed phases of liquids and solids, where intermolecular forces reign supreme. This section elucidates how the physical properties of a substance-its boiling point, viscosity, and crystal lattice-are direct consequences of the strength of forces acting between its constituent particles.

Finally, we address the dynamic nature of chemical change. It is not enough to know how matter is built; we must understand why and how fast it changes. We explore chemical kinetics to understand the rates of reactions and the mechanisms by which they proceed. We then delve into the elegant concept of equilibrium, encompassing gaseous, acid-base, and solubility systems, where forward and reverse processes achieve a delicate balance. The book culminates in a study of thermodynamics and electrochemistry, the powerful theoretical frameworks that allow us to predict the spontaneity of the universe and harness chemical energy to do work.

As you navigate these chapters, you will encounter the same concepts revisited in increasing layers of complexity. This is intentional. Chemistry is cumulative. The quantum numbers learned in Chapter 4 explain the bonding in Chapter 5; the intermolecular forces defined in Chapter 9 dictate the solubility rules in Chapter 10; and the equilibrium constants calculated in Chapter 12 are fundamentally linked to the free energy changes calculated in Chapter 15.

We invite you to approach this material with curiosity and patience. General chemistry is a rigorous subject that demands precision in calculation and imagination in visualization. However, the reward is a profound understanding of the material universe, providing you with the intellectual tools to analyze the world around you, from the energy powering your devices to the biological processes sustaining life itself. Welcome to the study of chemistry.

Chapter 1
Foundations of Chemistry

Chemistry acts as the bridge between the macroscopic world we observe and the microscopic realm that dictates behavior. To cross this bridge, we must first master the universal language of measurement and the architecture of matter itself. This chapter builds the chemist’s toolkit: the precision required to quantify nature through the International System of Units, the statistical handling of experimental data, and the foundational atomic theories that transformed alchemy into a predictive science.

1.1 The Scientific Method and Classification of Matter

Chemistry is the study of matter—its composition, structure, properties, and the transformations it can undergo. Because chemical systems are often complex, chemistry relies on a disciplined way of asking questions and testing answers. This section introduces (1) the logic of scientific inquiry and (2) the way chemists classify matter, which provides a common language for describing substances and designing experiments.

1.1.1 Scientific inquiry: from observation to explanation

Observations and the role of measurement

Scientific investigation begins with observations: descriptions of the natural world obtained by the senses and, crucially, by instruments. Chemistry depends especially on quantitative observations—measurements of mass, volume, temperature, pressure, time, and electrical quantities—because numerical patterns can reveal underlying regularities.

It is useful to distinguish:

Qualitative observations: non-numerical descriptions (“the solution is blue,” “a solid precipitate forms”).

Quantitative observations: numerical measurements with units (“the

sample has a mass of 12.43 g”).

A key feature of chemistry is that many chemical ideas are supported not by a single dramatic observation but by reproducible patterns across many measurements. For that reason, good experimental practice emphasizes careful record-keeping, appropriate controls, and awareness of uncertainty (developed later in the chapter).

Hypotheses, models, and predictions

After observations, scientists propose a hypothesis: a tentative, testable explanation of a pattern. In chemistry, hypotheses are often embedded in models—simplified representations of reality that capture essential features while omitting distracting detail. For example, the particulate model of matter treats substances as composed of discrete particles (atoms, molecules, ions), allowing one to explain macroscopic properties such as pressure or density.

A scientifically useful hypothesis must be falsifiable: it must make predictions that could, in principle, be shown wrong by experiment. The hypothesis gains credibility when its predictions agree with data over many independent tests, and it must be revised or rejected when results consistently contradict it.

Experiments: testing cause-and-effect

An experiment is a controlled procedure designed to test a hypothesis. Control is essential because chemical outcomes often depend on multiple variables.

A typical experimental design identifies:

Independent variable: deliberately changed (e.g., temperature).

Dependent variable: measured response (e.g., reaction rate).

Controlled variables: held constant (e.g., concentrations, volume, catalysts).

For example, to test whether temperature affects the rate of a reaction producing a gas, one would vary temperature while keeping the amounts of reactants and the vessel volume constant, then measure the rate of gas formation. If multiple factors change at once, cause-and-effect becomes ambiguous.

Theory vs. law: different roles in science

As evidence accumulates, a successful explanatory framework becomes a theory. In science, the word “theory” does not mean a guess; it means an explanation supported by extensive evidence and capable of making accurate predictions. Atomic theory, for instance, explains why elements combine in fixed ratios and why matter is conserved in ordinary chemical processes.

A scientific law is a concise description (often mathematical) of a consistent natural relationship. A law summarizes what happens; a theory explains why it happens. These are not rungs on a ladder where hypotheses “graduate” into theories and then into laws. They are different kinds of scientific knowledge.

As an analogy:

A law is like a reliable rulebook statement: “Under these conditions, the relationship is this.”

A theory is like a coherent story of mechanisms: “Here is the model that explains why the rule holds and predicts new outcomes.”

Iteration and community verification

Scientific knowledge is iterative. New data refine old models; improved instruments reveal new phenomena; and unexpected results can motivate better hypotheses. Chemistry also depends on community verification: independent replication, peer review, and the requirement that methods and data be communicated in sufficient detail for others to evaluate and reproduce.

1.1.2 Matter and its physical states

Defining matter

Matter is anything that has mass and occupies space. Chemistry focuses on matter at the macroscopic scale (what we can weigh, measure, and observe) and connects those properties to the microscopic scale (atoms, molecules, and ions).

Physical state: solid, liquid, gas, and plasma

The physical state of matter is determined largely by particle arrangement and motion, which are influenced by temperature and pressure.

Solid. Solids have a definite shape and volume. Their particles are closely packed and typically vibrate about fixed positions. The strong attractions between particles give solids rigidity and low compressibility.

Liquid. Liquids have a definite volume but take the shape of their container. Particles remain close together but can move past one another, allowing flow. Liquids are only slightly compressible.

Gas. Gases have neither definite shape nor definite volume; they expand to fill their container. Particles are far apart and move freely, making gases highly compressible and low in density compared with liquids and solids.

Plasma. A plasma is a high-energy state in which many particles are ionized (electrons separated from nuclei). Plasmas conduct electricity and respond strongly to electromagnetic fields. Plasmas occur in flames, lightning, and stars.

1.1.3 Classification of matter by composition

Physical state describes how matter appears under given conditions. Classification by composition addresses what matter is made of. This distinction is essential: a substance can change state (ice → liquid water → steam) without changing its chemical identity (H2O remains H2O).

Pure substances vs. mixtures

Matter is classified by whether its composition is fixed or variable.

Pure substance. A pure substance has a constant composition and characteristic properties. It cannot be separated into different components by physical means

without changing its chemical identity. Pure substances are either elements or compounds.

Mixture. A mixture is a physical combination of two or more substances in variable proportions. Components of a mixture retain their chemical identities and can be separated by physical methods (e.g., filtration, distillation). Mixtures are either homogeneous or heterogeneous.

Elements

An element is a pure substance that cannot be broken down into simpler substances by chemical means. Each element is defined by its atoms (developed later through atomic number), and each has characteristic properties. Examples include Cu (copper), O2 (oxygen gas, an elemental form of oxygen), and Ne (neon).

A key point: an element may exist as single atoms (e.g., He) or as molecules containing only one type of atom (e.g., O2, S8). The defining feature is not “single atom” but “single type of atom.”

Compounds

A compound is a pure substance composed of two or more elements chemically combined in fixed proportions. Compounds have properties that can differ dramatically from their constituent elements. For example, elemental sodium is a reactive metal and elemental chlorine is a toxic gas, yet NaCl (sodium chloride) is a stable ionic solid.

Because compounds have fixed ratios, their composition can be represented by a chemical formula (e.g., H2O, CO2, CaCO3). Breaking a compound into its elements requires a chemical change (a reaction), not a physical separation.

Homogeneous mixtures (solutions)

A homogeneous mixture has a uniform composition throughout: any small sample has the same composition as the whole. Homogeneous mixtures are often called solutions. Examples include air (a mixture primarily of N2, O2, and other gases), saltwater, and many metal alloys (e.g., brass as a copper–zinc mixture).

Uniformity is a macroscopic criterion. Microscopically, the different particles are intermingled, but they are not chemically bonded into a single repeating unit as in a compound.

Heterogeneous mixtures

A heterogeneous mixture has a nonuniform composition; it contains distinct regions (phases) with different properties. Examples include oil and water, granite, and a suspension such as sand in water.

The boundary between homogeneous and heterogeneous is sometimes a matter of scale and observation. For example, milk appears uniform to the eye but is a colloid (a heterogeneous mixture at the microscopic scale). Chemistry addresses such gray areas by specifying the observational scale and the separation behavior (e.g., filtration, centrifugation).

1.1.4 Separating mixtures: physical methods

Because mixtures are combinations of substances without chemical bonding between components, they can often be separated by exploiting differences in physical properties such as particle size, boiling point, solubility, and affinity for a surface.

Filtration

Filtration separates a solid from a fluid (liquid or gas) by passing the mixture through a porous barrier. It is effective when the solid particles are large enough (or aggregated enough) to be retained by the filter.

Example: separating sand from water using filter paper. The sand is the residue; the liquid that passes through is the filtrate.

Distillation

Distillation separates components of a mixture based on differences in boiling points (or, more precisely, differences in volatility). In simple distillation, the more volatile component vaporizes preferentially, is condensed, and collected.

Example: separating ethanol from water in an ethanol–water mixture (though because ethanol and water have intermolecular interactions and comparable boiling points, more effective separation often uses fractional distillation).

Chromatography

Chromatography separates components based on their differing affinities for a stationary phase (a solid or liquid on a solid support) and a mobile phase (a liquid or gas). Components that interact strongly with the stationary phase move more slowly.

Example: separating dyes in ink on paper chromatography as solvent rises by capillary action.

1.1.5 Why classification matters in chemistry

The scientific method tells us how to build reliable knowledge; classification tells us what kind of system we are studying. These ideas work together in practice:

If a sample is a pure substance, its properties (melting point, density, spectroscopic signatures) should be reproducible and characteristic. This supports identification and quality control.

If a sample is a mixture, measured properties often depend on composition. Scientific questions shift toward determining the components and their proportions, and separation techniques become part of the experimental plan.

Distinguishing physical change (state changes, dissolving, mixing) from chemical change (reactions forming new substances) helps interpret observations correctly and choose appropriate tests.

As the chapter progresses, the classification of matter by composition will connect naturally to the atomic theory of matter: elements correspond to different kinds of atoms, compounds correspond to specific combinations of atoms, and mixtures correspond to physical combinations of particles. This particle-level viewpoint will allow chemistry to move from description to explanation and prediction.

1.2 SI Units and Measurement Scales

The previous section emphasized that chemistry advances through quantitative observation: measurements that are reproducible, interpretable, and communicable. Measurements only become scientifically useful when they are expressed in a shared language of units. The International System of Units (SI, from the French Système International d’Unités) provides that language by defining a small set of base units from which essentially all other scientific units can be constructed.

1.2.1 Measurement as a comparison to a standard

A measurement answers the question: “How much?” by comparing an unknown quantity to a defined standard unit. Every reported measurement therefore has two essential parts:

a number (how many units), and

a unit (which standard is being used).

For example, “12.43” is not a complete measurement; “12.43 g” communicates a mass using grams as the unit.

Because chemistry routinely combines multiple measured quantities in calculations (density, concentration, energy, gas pressure), consistent units are critical. Using SI units makes dimensional relationships explicit and reduces conversion errors.

1.2.2 The seven SI base units

SI is built on seven base quantities, each with a base unit. All other SI units are derived from these.

Base quantity | Unit name | Unit symbol | What it measures in chemistry
Length | meter | m | dimensions of apparatus; wavelengths; diffusion distances
Mass | kilogram | kg | amount of material (macroscopic); gravimetric analysis
Time | second | s | reaction times; half-lives; instrument sampling intervals
Thermodynamic temperature | kelvin | K | temperature for thermodynamics and kinetics
Amount of substance | mole | mol | chemical “counting unit” for atoms/molecules/ions
Electric current | ampere | A | electrochemical measurements; conductivity
Luminous intensity | candela | cd | optical measurements (less central in general chemistry)

Two points deserve special emphasis early in general chemistry: mass, temperature, and the mole.

Mass: kilogram and gram

The SI base unit of mass is the kilogram (kg). In laboratory chemistry, masses are commonly recorded in grams (g):

Balances measure mass, not weight; however, near Earth’s surface the distinction is often neglected in introductory contexts because gravitational acceleration is approximately constant for most laboratory settings.

Temperature: kelvin as the thermodynamic scale

The SI base unit of temperature is the kelvin (K). Kelvin is an absolute scale: zero kelvin corresponds to the lowest possible thermal energy state (absolute zero). Many physical laws (gas laws, thermodynamic relationships) require absolute temperature, so chemistry calculations conventionally use kelvin even when temperatures are measured in ∘C.

Amount of substance: the mole

The mole (mol) is the SI unit for amount of substance. It is a counting unit that links microscopic particles to macroscopic measurements. Conceptually, it plays the same role for particles that a “dozen” plays for eggs, but on a vastly larger scale.

A key relationship (developed more fully in later chapters) is that the mole connects mass to number of entities through the molar mass. At this stage, the essential idea is: chemical equations describe reactions in terms of numbers of particles, while experiments measure masses and volumes; the mole bridges these descriptions.

1.2.3 SI prefixes and scientific notation

Chemical measurements span many orders of magnitude. Atomic-scale lengths are on the order of 10⁻¹⁰ m, while laboratory volumes may be 10⁻³ m³ (liters) or larger. SI addresses this range with prefixes that denote powers of ten.

Prefix | Symbol | Factor | Typical chemistry context
tera | T | 10¹² | (rare in gen chem) large data rates, some spectroscopy scales
giga | G | 10⁹ | (rare) large counts in instrumentation
mega | M | 10⁶ | (occasionally) large molar quantities in industry
kilo | k | 10³ | kilograms, kilometers
deci | d | 10⁻¹ | (rare) deciliter (dL) in some contexts
centi | c | 10⁻² | centimeters; 1 cm³ as a volume unit
milli | m | 10⁻³ | milliliters (mL); milligrams (mg)
micro | μ | 10⁻⁶ | microliters (μL); micromolar (μM)
nano | n | 10⁻⁹ | nanometers (nm) for wavelengths/particles
pico | p | 10⁻¹² | picometers (pm) for atomic radii
femto | f | 10⁻¹⁵ | femtoseconds (fs) for ultrafast processes (advanced)

Scientific notation as a unit-handling tool

Scientific notation expresses numbers as

which helps track significant figures and powers of ten during calculations. For example:

The value is unchanged; only the representation differs.

1.2.4 Derived SI units: constructing new units from base units

A derived unit is formed by multiplying and/or dividing base units (and sometimes raising them to powers). Derived units are central in chemistry because they encode physical relationships.

Area and volume

Area and volume follow directly from length:

Chemistry frequently uses the liter:

Also common is the milliliter:

The equivalence 1 mL = 1 cm³ is especially useful in density and displacement measurements.

Density

Density relates mass to volume:

In SI base units, density is

In laboratory practice, density is often expressed as g∕mL or g∕cm³. These are consistent because 1 mL = 1 cm³.

Pressure

Pressure is force per unit area:

The derived SI unit is the pascal (Pa):

So in base units,

Chemistry also uses non-SI pressure units (atm, bar, mmHg or torr). Conversions are handled systematically by dimensional analysis (developed in a later section), but the SI framework clarifies what pressure is physically.

Energy

The SI unit of energy is the joule (J):

Energy units appear throughout chemistry: heat transfer, bond energies, reaction energetics, and spectroscopy.

Concentration

A central chemical derived quantity is amount concentration (molarity):

Its common unit is

Although the liter is not an SI base unit, mol∕L is accepted and pervasive in chemistry. In strict SI base units, concentration is mol∕m³, and

1.2.5 Temperature scales: Kelvin, Celsius, and Fahrenheit

Temperature is measured on several scales. Chemistry uses three commonly encountered ones: Kelvin (K), Celsius (∘C), and Fahrenheit (∘F). The key to avoiding mistakes is to distinguish temperature differences (intervals) from temperature values (points on a scale).

Celsius and Kelvin: same size degree, different zero

The Celsius and Kelvin scales have the same size unit increment:

They differ only in where zero is placed. The relationship is

Therefore:

Why Kelvin is required in many equations. Gas behavior provides a clear reason. In the ideal gas law,

temperature must be in kelvin. Using Celsius would incorrectly imply that V = 0 at 0 ∘C, which is physically false. Kelvin avoids this because its zero corresponds (in an idealized extrapolation) to the point where thermal motion approaches a minimum.

Fahrenheit: different increment and different zero

Fahrenheit uses a different sized degree and a different zero point. The conversions are

Combining with the Celsius–Kelvin relation gives a direct Fahrenheit-to-kelvin conversion:

Reference points and physical meaning

Three reference points anchor intuition:

Absolute zero: 0 K = −273.15 ∘C = −459.67 ∘F.

Freezing point of water (at 1 atm): 273.15 K = 0 ∘C = 32 ∘F.

Boiling point of water (at 1 atm): 373.15 K = 100 ∘C = 212 ∘F.

In later chapters, the pressure dependence of boiling and freezing will be revisited; for now, these familiar points provide a practical framework for conversions.

1.2.6 Dimensional consistency: the “unit grammar” of chemistry

Even before learning systematic conversion methods, one powerful habit improves nearly every calculation: check dimensional consistency. Physical equations must balance in units the same way chemical equations balance in atoms.

For example, density must have units of mass divided by volume. If you compute

and obtain units of g ⋅ mL (mass times volume), then a division or unit conversion was set up incorrectly.

Thinking of units as algebraic objects that multiply and cancel is a foundational skill. The next sections will formalize this idea through uncertainty analysis and then dimensional analysis (factor-label method), allowing you to convert units and propagate measurement information with confidence.

1.3 Uncertainty, Precision, and Significant Figures

The previous section established SI units as the common language for quantitative chemistry. However, a measurement is never “exact” in the everyday sense. Every reported value is an estimate limited by instrument resolution, experimental technique, environmental fluctuations, and (often) assumptions built into the procedure. This section develops three linked ideas:

uncertainty is inherent to measurement,

accuracy and precision describe different aspects of measurement quality, and

significant figures provide a practical convention for communicating precision in calculated results.

1.3.1 Why uncertainty is unavoidable

To measure is to compare an unknown quantity to a standard unit with a finite resolution. Consider reading the volume in a graduated cylinder. The scale might have marks every 1.0 mL. You can often estimate between marks (perhaps to 0.1 mL), but you cannot justify reporting to 0.001 mL from that instrument.

This leads to a fundamental principle:

A measurement consists of certain digits (read directly) plus one estimated digit (the last reported digit).

The last digit is uncertain because it depends on judgment and small variations. The goal of good laboratory practice is not to eliminate uncertainty (impossible), but to minimize, quantify, and correctly communicate it.

Uncertainty vs. error: a useful distinction

In introductory chemistry, the terms uncertainty and error are often used together, but they emphasize different ideas:

Uncertainty is the expected range of variation in a measurement due to limitations in the measurement process (instrument resolution, noise, uncontrolled influences).

Error is the difference between a measured value and a true (or accepted) value.

You can have uncertainty even when you do not know the true value. In many real chemical problems (new compounds, unknown samples), the true value is not available, so uncertainty is the more general concept.

1.3.2 Sources of measurement error: systematic and random

Measurement error is commonly classified as systematic or random. The classification matters because the strategies for improvement differ.

Systematic error (bias)

A systematic error is a consistent, repeatable shift that pushes results in the same direction away from the true value. It affects accuracy.

Common sources include:

Instrument calibration error: a balance that reads 0.020 g for all

objects.

Zero offset: not taring a container; a miszeroed pH meter.

Method bias: an analytical method that systematically loses a product (e.g., incomplete precipitation, volatile loss on heating).

Parallax or reading bias: consistently reading a meniscus from above or below eye level.

Systematic error is reduced by calibration, blanks, standards, and improved experimental design. Importantly, repeated trials do not necessarily remove systematic error; you can repeatedly get the same wrong answer.

Random error (scatter)

A random error varies unpredictably in sign and magnitude from trial to trial. It affects precision.

Examples include:

small fluctuations in temperature that affect gas volume,

minor differences in how a sample is positioned on a balance pan,

electronic noise in sensors,

human reaction time when starting/stopping a timer.

Random error can often be reduced by repeated measurements and averaging. Because random errors partly cancel, the mean value tends to stabilize as the number of measurements increases.

1.3.3 Accuracy and precision: what each word means

Definitions

Two words that are often confused in everyday language have distinct technical meanings in measurement science:

Accuracy describes how close a measured value is to the true or accepted value.

Precision describes how closely repeated measurements agree with one another (reproducibility).

In chemistry, both matter. A method can be precise but inaccurate (repeatable but biased), or accurate on average but imprecise (centered but scattered).

Target analogy (conceptual model)

The following diagram uses a bullseye target to visualize the four combinations of accuracy and precision. Think of each “X” as one repeated measurement of the same quantity.

Quantifying accuracy: absolute and percent error

When an accepted value xtrue is known (e.g., a certified standard), accuracy can be quantified.

Absolute error.

Percent error.

The absolute value is often used because the sign (high vs. low) is less important than the magnitude when discussing accuracy.

Quantifying precision: spread and standard deviation

Precision is about the spread of repeated measurements.

Suppose a quantity is measured n times, giving values x₁,x₂,…,xn. The mean (average) is

A widely used measure of precision is the sample standard deviation:

A smaller s indicates a tighter clustering of values (greater precision). The n − 1 in the denominator accounts for the fact that x is estimated from the same data (a correction appropriate for samples).

Relative standard deviation (RSD). To compare precision across different magnitudes, the standard deviation is often scaled by the mean:

An RSD of 0.20% indicates much tighter precision than an RSD of 5%, regardless of the units.

1.3.4 Significant figures: communicating precision

In chemistry, we frequently perform calculations with measured values. The result cannot be more precise than the least precise information that went into it. Significant figures (sig figs) are a practical convention for reporting a number with an appropriate level of precision.

What a significant figure represents

A reported measurement implies that all digits shown are meaningful. In general:

All certain digits are significant.

The final (estimated) digit is significant.

For example, a mass recorded as 12.43 g implies uncertainty in the hundredths place; the balance resolution and reading justify digits through the hundredths place but not beyond.

Counting significant figures

Use the following rules to determine the number of significant figures in a value.

Rule 1: Nonzero digits are significant.

Rule 2: Zeros between nonzero digits are significant (captive zeros).

Rule 3: Leading zeros are not significant. Leading zeros only locate the decimal point.

Rule 4: Trailing zeros are significant only if a decimal point is present.

But

The number “2500” could mean 2.5 × 10³ (2 sig figs), 2.50 × 10³ (3 sig figs), or 2.500 × 10³ (4 sig figs). Scientific notation removes ambiguity.

Exact numbers and defined conversions Some values have no uncertainty and therefore do not limit significant figures:

Counted quantities: 12 test tubes, 3 trials.

Defined relationships: 1 L = 1000 mL (definition), 1 in = 2.54 cm (definition).

Such numbers can be treated as having infinitely many significant figures in calculations.

Rounding: how to drop digits correctly

When rounding to a specified number of significant figures:

Identify the last digit to keep.

Look at the next digit (the first to drop).

If the next digit is 0–4, round down (leave the last kept digit unchanged).

If the next digit is 5–9, round up (increase the last kept digit by 1).

Example (3 significant figures):

In multi-step calculations, it is good practice to carry extra guard digits and round only at the end, to avoid compounding rounding error.

1.3.5 Significant figures in calculations

Significant-figure rules approximate how uncertainty propagates. They are not a substitute for full statistical propagation, but they are effective for routine introductory chemistry calculations.

Multiplication and division: least number of significant figures

For multiplication or division, the result should have the same number of significant figures as the factor with the fewest significant figures.

Example. Compute density from m = 12.43g and V = 3.1mL.

Since 3.1 mL has 2 significant figures, report the result with 2 significant figures:

Addition and subtraction: least number of decimal places

For addition or subtraction, the limiting factor is the decimal place (not the total number of significant figures). The result should be reported to the least precise decimal place among the inputs.

Example. Add 12.43 mL and 0.6 mL:

Because 0.6 mL is precise only to the tenths place, round the sum to the tenths place:

Mixed operations: apply rules stepwise

For expressions involving multiple operations, apply the appropriate rule at each stage, but avoid excessive intermediate rounding.

Example. Suppose a sample mass is found by difference:

For subtraction, the limiting decimal place is hundredths (from 23.10 g), so

If this mass is then used in a multiplication/division, its 3 significant figures and hundredths-place precision constrain subsequent results.

1.3.6 Practical reporting guidelines in the laboratory

Record all instrument digits

When reading an instrument, record all certain digits plus one estimated digit. Do not round during data collection; record the measurement as read.

Use units as part of the measurement

Units are not decorations; they encode meaning and prevent errors. Write units at every step of a calculation whenever feasible. This habit sets up the factor-label method developed in the next section.

Do not confuse precision with accuracy

A digital display can encourage false confidence. A pH meter reading of 7.003 may be precise (many digits) but not accurate if the electrode is improperly calibrated. Always connect reported digits to instrument specifications and calibration status.

1.3.7 Looking ahead: uncertainty as part of scientific reasoning

The scientific method relies on comparing predictions to measurements. That comparison is only meaningful when uncertainty is accounted for:

A model can appear to “fail” if measurement uncertainty is ignored.

Two results that differ numerically may still agree within experimental

uncertainty.

Careful uncertainty reporting allows other scientists to replicate, validate, and build on results.

In the next section, dimensional analysis will provide a systematic method for converting units and checking the internal consistency of calculations—a complementary tool to significant figures for producing results that are both numerically and conceptually trustworthy.

1.4 Dimensional Analysis and Unit Conversions

The previous sections established two foundational ideas: (1) chemistry depends on quantitative measurements expressed in a common unit system (SI), and (2) every measurement carries limited precision, which must be communicated through appropriate significant figures. The next essential skill is dimensional analysis: a method for organizing calculations so that units behave like algebraic quantities and the final numerical answer must have the correct units.

Dimensional analysis is not merely a conversion trick. It is a general strategy for:

verifying that an equation is physically sensible (dimensional consistency),

converting quantities from one unit system to another (factor-label method),

reducing complex multi-step problems to a systematic chain of unit-canceling factors.

1.4.1 Dimensions vs. units: what you are really checking

Physical dimensions

A dimension is the physical type of a quantity (length, mass, time, temperature, amount of substance, etc.). For example, distance and wavelength have the same dimension: length.

Units as human-chosen standards

A unit is a standard used to measure a dimension (meters, centimeters, inches are all units of length). Units can change without changing the underlying physical dimension.

Dimensional analysis works because it respects the idea that the same physical quantity can be expressed in different units, and those expressions are related by conversion factors.

1.4.2 Dimensional consistency: the “grammar” of equations

An equation is physically meaningful only if both sides have the same dimensions (and therefore the same units, up to conversion factors).

Example: checking density

Density is defined as

If m is in kilograms (kg) and V is in cubic meters (m³), then ρ must be in kg∕m³. Any result with units like kg m³ indicates a setup error.

Example: checking pressure

Pressure is

Force has units of newtons (N) and area has units of m², so pressure has units of N∕m², defined as the pascal (Pa). This unit structure is not decorative; it encodes the relationship that pressure increases if the same force acts over a smaller area.

Dimensional consistency is a fast way to catch mistakes before you worry about arithmetic.

1.4.3 The factor-label method (unit factor method)

Conversion factors as ratios equal to 1

A conversion factor is a ratio of two equivalent quantities. Because the two quantities are equal, the ratio has the numerical value 1, even though it contains units.

For example,

is a defined relationship, so both ratios

are valid conversion factors.

Multiplying a measured quantity by a conversion factor changes units without changing the underlying physical amount.

Core procedure

To convert units using factor-label:

Write the given quantity with its units.

Multiply by conversion factors arranged so that unwanted units cancel.

Confirm that the remaining unit is the desired unit.

Perform the arithmetic and apply significant-figure rules at the end.

1.4.4 Single-step conversions

Example: converting grams to kilograms

Convert 45.2g to kilograms.

Use 1 kg = 1000 g:

The g cancels, leaving kilograms. The result has 3 significant figures, matching the given measurement.

Example: converting milliliters to cubic meters

Convert 250 mL to m³.

Use the chain 1 mL=1 × 10⁻³ L and 1 L=1 × 10⁻³ m³:

This illustrates a common pattern: small laboratory volumes correspond to very small m³ values.

1.4.5 Multi-step (chain) conversions

Many chemistry problems require multiple conversions in one calculation (e.g., converting a laboratory-scale unit into an SI-derived unit). Dimensional analysis turns these into a single line where units cancel sequentially.

Example: miles per hour to meters per second

Convert 55.0 mi∕h to m∕s.

Use:

Set up the chain:

All unwanted units cancel, leaving m∕s. The result is reported to 3 significant figures, limited by 55.0 mi∕h.

Practical advice: write units on every factor

Errors in chain conversions almost always trace back to missing units or using a reciprocal conversion factor by mistake. Writing units on every numerator and denominator is slower at first but dramatically reduces errors.

1.4.6 Converting units raised to powers

A common source of mistakes is converting squared and cubed units. The conversion factor must be applied the same number of times as the power indicates.

Example: converting cm³ to m³

Because

cubing both sides gives

So, for instance,

Example: converting cm² to m²

Similarly,

1.4.7 Derived units as built-in dimensional analysis

Many SI derived units are compact names for combinations of base units. Recognizing these combinations makes conversions and checks easier.

Example: the pascal

So if a calculation yields kg m⁻¹ s⁻², you may (when appropriate) rewrite it as pascals.

Example: molarity

If a problem requires SI base units, you can convert liters to cubic meters:

The dimensional structure (amount per volume) remains the same.

1.4.8 Exact vs. measured conversion factors and significant figures

Not all numbers in conversions have the same status.

Exact conversions

Some conversions are defined and therefore exact; they do not limit significant figures:

1 m = 100 cm (definition of prefix centi-),

1 L = 1000 mL (definition),

1 in = 2.54 cm (definition).

Measured (non-exact) conversions

Other values are measured or conventional and may carry limited precision depending on context. In most introductory chemistry problems, commonly used conversion values (e.g., 1 mi=1609 m) are provided with sufficient digits and can be treated as exact for the purpose of significant figures unless otherwise stated by the problem.

Sig figs in conversions: the key rule

In a calculation, the final answer should generally be rounded according to the least precise measured quantity in the problem (as described in the significant figures section). Exact conversion factors do not reduce the number of significant figures.

1.4.9 A structured problem-solving template

When you encounter a conversion or a derived-unit calculation, use this checklist:

Identify the target unit. Write it explicitly (e.g., kg∕m³).

Write the given quantity with units.

Choose conversion factors that connect the given units to the desired units.

Arrange factors so units cancel. If a unit is in the numerator and you want it gone, it must appear in the denominator of a conversion factor (and vice versa).

Check the final unit before computing the final number.

Compute and round using significant figures at the end.

1.4.10 Common pitfalls (and how to avoid them)

Pitfall 1: converting only the number, not the unit

Writing 55 mi∕h and then converting “55” without tracking units often leads to missing factors of 60 or 1000. Units must be carried through the calculation.

Pitfall 2: forgetting powers on units

The most frequent exponent mistake is treating cm³ as if it were cm. Always cube (or square) the conversion factor consistently:

Pitfall 3: rounding too early

Rounding intermediate results can introduce noticeable error in multi-step calculations. Keep extra digits during the work and round once at the end, consistent with significant-figure rules.

1.4.11 Dimensional analysis as a bridge to later chemistry

Dimensional analysis will appear repeatedly throughout the course:

converting between mass and moles using molar mass,

relating gas volume, pressure, and temperature in gas laws,

converting energy units and comparing thermodynamic quantities,

working with concentrations in solution stoichiometry.

In each case, the core logic is the same: choose conversion factors that act like “unit-canceling gears” and let the units guide the mathematics. This habit builds calculations that are not only correct, but also self-checking.

1.5 Atomic Theory and Subatomic Particles

The preceding sections developed the measurement tools used to study matter: SI units, significant figures, and dimensional analysis. Those tools are essential because atomic theory is not a philosophical claim; it is an evidence-based model built from quantitative experiments. In this section we trace how the modern picture of the atom emerged and summarize the basic properties of the three principal subatomic particles: the electron, proton, and neutron.

1.5.1 From macroscopic matter to microscopic particles

Classifying matter as elements, compounds, and mixtures is powerful, but it naturally raises deeper questions:

Why do elements combine in fixed ratios to form compounds?

Why do some substances conduct electricity while others do not?

Why do atoms of different elements have different masses and chemical behaviors?

Atomic theory addresses these questions by proposing that matter is composed of discrete particles whose structure controls chemical properties.

1.5.2 Dalton’s atomic theory: a starting point

In the early 1800s, John Dalton proposed an atomic theory to explain several experimental laws of chemical combination. A modernized summary of Dalton’s key ideas is:

Matter is composed of atoms.

Atoms of a given element share characteristic properties and mass (modern correction: isotopes exist, so atoms of the same element can have different masses).

Compounds form when atoms combine in simple whole-number ratios.

Chemical reactions rearrange atoms; atoms are not created or destroyed in ordinary chemical changes (modern correction: nuclear reactions can change atoms).

Dalton’s theory was successful because it explained quantitative patterns such as constant composition and multiple proportions. However, Dalton treated atoms as indivisible. The next set of experiments demonstrated internal structure.

1.5.3 Discovery of the electron: Thomson’s cathode rays

Cathode ray tube observations

In the late 1800s, experiments with evacuated glass tubes containing electrodes showed a beam emitted from the negative electrode (cathode) when a high voltage was applied. J. J. Thomson studied these cathode rays and found that:

the beam was deflected by electric and magnetic fields (so it carried charge),

it traveled from the cathode toward the anode (indicating negative charge),

its behavior was independent of the electrode material and the gas in the tube (suggesting a universal component of matter).

Thomson measured the charge-to-mass ratio (e∕m) of the particles in cathode rays. The large magnitude of e∕m implied either a very large charge or a very small mass. Subsequent work established the latter: the particles were extremely light compared with atoms.

The electron

Thomson’s work led to the identification of the electron, a negatively charged subatomic particle present in all atoms. This discovery overturned the idea that atoms are indivisible and demanded a new model of atomic structure.

Thomson’s “plum pudding” model (historical)

To account for electrical neutrality of bulk matter, Thomson proposed that negative electrons were embedded in a diffuse positive charge. While historically important, this model could not explain later scattering experiments and was replaced by the nuclear model.

1.5.4 Quantization of charge: Millikan’s oil-drop experiment

Thomson determined e∕m for the electron, but to find the electron mass one must know its charge. Robert Millikan’s oil-drop experiment measured the electron’s charge by balancing gravitational and electric forces on tiny charged oil droplets.

Central result: charge comes in discrete packets

Millikan found that the charges on droplets were always integer multiples of a smallest value, demonstrating that electric charge is quantized. The smallest unit of charge corresponds to the magnitude of the electron’s charge.

Once e (electron charge) was known, the electron mass could be obtained from Thomson’s e∕m ratio:

This combination of experiments illustrates a recurring scientific pattern: one experiment constrains a ratio, another gives an absolute scale, and together they yield fundamental constants.

1.5.5 Rutherford’s gold foil experiment: discovery of the nucleus

By the early 1900s, it was clear atoms contained electrons and positive charge, but the distribution of that positive charge was unknown. Ernest Rutherford (with Geiger and Marsden) used scattering to probe atomic structure.

Experimental design: scattering as a “microscope”

Rutherford directed a beam of alpha particles (helium nuclei, 4He2+) at a very thin sheet of gold foil. Alpha particles are:

massive compared with electrons,

positively charged,

energetic enough to pass through thin matter.

A fluorescent screen surrounding the foil detected alpha particles by tiny flashes of light, revealing scattering angles.

Key observations and interpretation

Most alpha particles passed through with little or no deflection. A small fraction were deflected at large angles, and a very tiny number scattered nearly backward.

These results are inconsistent with the plum pudding model (diffuse positive

charge), which would produce only small deflections. Rutherford concluded:

The atom is mostly empty space.

Nearly all the mass and all positive charge are concentrated in a very small central region: the nucleus.

Electrons occupy the surrounding volume, forming an electron cloud that accounts for most of the atom’s size.

Physical reasoning: why large-angle scattering implies a small nucleus

Alpha particles are positively charged. A large deflection requires a strong repulsive electric force, which (by Coulomb’s law) becomes large only when the alpha particle approaches very close to the positive charge. If the positive charge were spread throughout the atom, the repulsion would be weaker and more gradual, producing only small deflections. Therefore, large-angle and backscattered events imply a highly concentrated positive region: the nucleus.

1.5.6 The nuclear atom model and the proton

Rutherford’s model required that the nucleus contain positive charge. The proton is the positively charged subatomic particle located in the nucleus. While the detailed discovery pathway includes multiple experiments and naming conventions, for our purposes the essential outcome is:

The nucleus contains protons, carrying +1 elementary charge each.

Electrons outside the nucleus carry −1 elementary charge each.

Neutral atoms have equal numbers of protons and electrons.

This immediately explains why atoms are electrically neutral overall, while still containing charged components.

1.5.7 The neutron: completing the picture of nuclear mass

Early nuclear models faced a puzzle: nuclear charge (number of protons) could not account for the full nuclear mass. For example, a nucleus with charge +2 (two protons) is much more massive than two protons alone would suggest. The resolution is the neutron, a neutral particle in the nucleus with a mass similar to that of the proton.

The neutron accounts for:

the extra mass in nuclei beyond what protons provide,

the existence of isotopes: atoms of the same element (same protons) but different masses (different neutrons),

aspects of nuclear stability (some proton–neutron combinations are stable; others are radioactive).

1.5.8 Properties of subatomic particles

Atomic structure can be summarized in terms of charge, mass, and location.

Charge

Charges are expressed in multiples of the elementary charge magnitude, e:

In coulombs, e = 1.602 × 10⁻¹⁹ C (useful in electrochemistry and physics contexts).

Mass and relative scale

It is helpful to know both absolute masses (in kg) and relative masses. Chemistry commonly uses the atomic mass unit (amu or u), defined so that 1 u is approximately the mass scale of nucleons.

Particle | Charge (e) | Mass (u) | Mass (kg) | Location and chemical role
Electron (e⁻) | −1 | 5.4858×10⁻⁴ | 9.109×10⁻³¹ | Outside nucleus; determines bonding, ion formation, and most chemical properties
Proton (p⁺) | +1 | 1.0073 | 1.673×10⁻²⁷ | In nucleus; determines element identity (atomic number) and contributes to mass
Neutron (n⁰) | 0 | 1.0087 | 1.675×10⁻²⁷ | In nucleus; affects isotopic mass and nuclear stability

Two comparisons are especially important:

Protons and neutrons are about the same mass (≈ 1 u each).

A proton or neutron is about 1800 times more massive than an electron.

Therefore, almost all the mass of an atom is in the nucleus, while most of the volume is the electron cloud.

1.5.9 What the nuclear model explains (and what it does not)

Explains

The nuclear atom model explains several core observations:

Scattering results: rare large deflections require a compact nucleus.

Atomic mass: dominated by nucleons (protons + neutrons).

Ions: atoms gain or lose electrons, changing charge with little change in mass.

Does not yet explain

However, the nuclear model alone does not explain:

why electrons do not spiral into the nucleus,

why atoms emit and absorb light at discrete wavelengths (line spectra),

the detailed patterns of chemical bonding.

Those questions require quantum theory and electron structure (addressed later). For now, the nuclear model provides the essential structural framework: a tiny, massive nucleus containing protons and neutrons, surrounded by electrons whose behavior governs chemistry.

1.5.10 Summary: the atom as an evidence-based model

Atomic theory developed through a sequence of experiments guided by the scientific method:

Dalton linked atoms to quantitative laws of chemical combination.

Thomson demonstrated universal negatively charged particles (electrons).

Millikan measured the fundamental unit of charge, enabling the electron mass to be determined.

Rutherford revealed a small, dense, positively charged nucleus through scattering.

The proton and neutron completed the nuclear picture of mass and charge.

In the next section, we will formalize how protons and neutrons define atomic identity and isotopic variation, introducing atomic number, mass number, and isotope notation.

1.6 Atomic Number, Mass Number, and Isotopes

The previous section introduced the nuclear model of the atom: a tiny, dense nucleus containing protons and neutrons, surrounded by electrons. That model immediately suggests a rigorous way to define what an element is and why atoms of the same element can nevertheless differ in mass. This section formalizes three core ideas:

atomic number (Z) determines elemental identity,

mass number (A) counts nucleons (protons + neutrons),

isotopes are atoms of the same element with different numbers of neutrons, leading to average atomic mass as a weighted average over natural abundances.

1.6.1 Atomic number: the definition of an element

What makes an element an element?

In chemical classification, an element is a pure substance that cannot be decomposed by chemical means. At the microscopic level, an element is defined by the number of protons in the nucleus.

The integer Z is called the atomic number. Every atom with Z = 6 is carbon; every atom with Z = 8 is oxygen. Changing Z changes the element itself.

Why Z determines chemistry

Chemical behavior is governed primarily by electrons. In a neutral atom, the number of electrons equals the number of protons:

Because Z determines how many electrons an atom has (and therefore how those electrons are arranged), Z ultimately controls the patterns of bonding and reactivity studied throughout chemistry.

1.6.2 Mass number: counting nucleons in the nucleus

Definition

The mass number A is the total number of nucleons (protons + neutrons) in the nucleus:

where N is sometimes used to denote the number of neutrons.

Finding the number of neutrons

If A and Z are known, the number of neutrons is

Example (counting neutrons). For 31P (phosphorus-31), Z = 15 (phosphorus) and A = 31, so

Mass number vs. atomic mass

It is important to distinguish:

A (mass number): a count of nucleons, always an integer.

atomic mass: a measured mass (in u), generally not an integer.

Mass number correlates with mass but is not equal to it, because actual nuclear masses are affected by binding energy and because atomic masses include electrons (a small but measurable contribution).

1.6.3 Isotopes: same element, different mass

Definition and key idea

Isotopes are atoms of the same element (same Z) with different numbers of neutrons (different N), and therefore different mass numbers A.

For example, carbon has several isotopes:

All are carbon because all have Z = 6. They differ in neutron count:

Chemical similarity and physical differences

Because isotopes share the same atomic number, their electron configurations (for neutral atoms) are essentially the same, so they have very similar chemical behavior. However, isotopic mass differences can produce measurable physical differences, such as:

slightly different densities,

slightly different vibrational frequencies (important in spectroscopy),

different nuclear stability (some isotopes are radioactive).

In most introductory chemical reactions, isotopes of an element are treated as chemically interchangeable unless isotopic labeling or kinetic isotope effects are specifically under study.

1.6.4 Isotope notation

Nuclide symbol

A compact notation for a specific isotope (nuclide) is

where X is the element symbol.

Example.

denotes chlorine with 17 protons and mass number 35 (so 18 neutrons).

Hyphen notation

In many chemical contexts, isotopes are written using hyphen notation:

Thus, 35Cl is “chlorine-35,” and 14C is “carbon-14.”

Ions and electron count

The nucleus determines Z and A; the electron count depends on charge. For an ion with charge q (in units of e),

Here q is positive for cations and negative for anions.

Example (electrons in an ion). For 35Cl– , Z = 17 and q = −1, so

1.6.5 Atomic mass and the atomic mass unit

The atomic mass unit (u)

Chemistry commonly uses the atomic mass unit (u or amu) to express atomic and molecular masses on a convenient scale. The unit is defined by convention so that the mass of a 12C atom is exactly 12 u:

This definition anchors the mass scale and makes typical atomic masses on the order of 1–200 u rather than extremely small numbers in kilograms.

Mass spectrometry (conceptual overview)

How are isotopic masses and abundances determined? The central experimental technique is mass spectrometry, which measures the mass-to-charge ratio (m∕z) of ions. Conceptually, a mass spectrometer:

converts atoms or molecules into ions,

accelerates them using an electric potential,

separates them based on how their trajectories respond to electric and/or magnetic fields (which depends on m∕z),

detects the number of ions arriving at each m∕z value.

Each isotope typically produces a distinct peak (or group of peaks) at a characteristic m∕z. Peak intensities are proportional to relative abundance.

At this stage, the important point is not instrument design but the logic: isotopes are distinguished because different masses produce different responses in the analyzer, allowing both isotopic masses and abundances to be measured.

1.6.6 Average atomic mass: a weighted mean over isotopes

Why the periodic table values are not integers

The atomic masses listed on the periodic table are typically not whole numbers because naturally occurring elements are mixtures of isotopes. A sample of elemental chlorine, for example, contains mostly 35Cl and 37Cl. The periodic table mass is the average over that mixture.

Weighted-average formula

If an element has isotopes with masses m₁,m₂,…,mk (in u) and fractional abundances f₁,f₂,…,fk (where each fi is between 0 and 1 and ∑ fi = 1), then the average atomic mass m is

If abundances are given as percent, convert to fractions by dividing by 100.

Example: two-isotope element (chlorine-like)

An element X has two isotopes:

Compute the average atomic mass:

This illustrates why a periodic-table atomic mass can lie between two isotope masses and why it can be closer to the more abundant isotope.

Significant figures in average atomic mass

When calculating an average atomic mass:

isotopic masses are measured values with stated precision,

abundances may be given with limited precision,

the final reported average should reflect the precision of the input data.

In many textbook problems, enough digits are supplied to support reporting the final mass to two decimal places, but the correct rounding should follow the significant-figure conventions established earlier.

1.6.7 Interpreting the periodic table: what the atomic mass means

The periodic-table atomic mass is best interpreted as:

the mass (in u) of a statistically typical atom of that element, averaged over natural isotopic composition.

It is not the mass number of any one atom. For elements with one overwhelmingly abundant isotope, the atomic mass may appear close to an integer (e.g., fluorine near 19). For elements with multiple comparable isotopic abundances, the atomic mass can be far from an integer (e.g., chlorine near 35.45).

1.6.8 Summary: nucleus-based definitions that support chemical calculation

The nuclear model becomes quantitatively useful once we define:

Atomic number Z: number of protons; determines the element.

Mass number A: number of protons + neutrons; identifies a specific nuclide.

Isotopes: same Z, different A; explain non-integer atomic masses and many nuclear phenomena.

Average atomic mass: weighted average over isotopic masses using natural abundances.

These concepts will be used repeatedly: in counting subatomic particles in ions, in understanding periodic trends (through electron count tied to Z), and in stoichiometry (where atomic masses connect measured mass to moles of atoms and molecules).

1.7 Problems

1.7.1 Dimensional Analysis: Converting a Compound Unit to SI

A runner’s pace is reported as 6.50 furlongs/fortnight. Using dimensional analysis (factor-label method), convert this speed to ms⁻¹. You may use the exact conversions 1 furlong = 201.168m, 1 fortnight = 14 days, 1 day = 24 h, and 1 h = 3600 s. Report your final answer with the correct number of significant figures and show a clear cancellation of units.

1.7.2 Density from Water Displacement with Significant Figures

An irregular metal nugget is placed into a graduated cylinder containing 35.0mL of water. The new water level reads 47.6mL. The mass of the nugget is measured as 98.27g.

Determine the volume of the nugget.

Calculate the density in g mL⁻¹.

Convert the density to kg m⁻³.

For each part, apply appropriate significant-figure rules and explicitly state how many significant figures your final answers contain.

1.7.3 Counting Subatomic Particles in an Ion

For the ion 204Pb2+:

Determine the atomic number Z of lead and use it to find the number of protons.

Determine the number of neutrons in the nucleus.

Determine the number of electrons in the ion.

Compute the total nuclear charge (in units of e) and the net ionic charge (in units of e).

Present your work clearly using A, Z, and N, and justify each step with the relevant definitions.

1.7.4 Temperature Conversion: Cryogenic Conditions

A cryogenic bath is reported to be at −320.0 °F.

Convert this temperature to degrees Celsius.

Convert this temperature to kelvin.

State whether the result is above or below the boiling point of liquid nitrogen (use 77.36 K as the boiling point at 1 atm).

Show the conversion formulas and keep guard digits until the final rounding step.

1.7.5 Average Atomic Mass of a Fictional Element

A fictional element X has three naturally occurring isotopes:

Calculate the average atomic mass of X in u.

Suppose the percent abundances were measured with an uncertainty of ±0.05% (absolute). Discuss qualitatively which input values most strongly influence the uncertainty in the calculated average atomic mass and why.

Report the numerical result with appropriate significant figures.

1.7.6 Accuracy vs. Precision from Repeated Density Measurements

A student measures the density of a solid and obtains the following values (in g mL⁻¹):

The accepted density is 2.85g mL⁻¹.

Compute the mean density x.

Determine the absolute error and percent error of the mean relative to the accepted value.

Calculate the sample standard deviation s and the relative standard deviation (RSD, in %).

Based on your results, decide whether the measurements are accurate, precise, both, or neither. Support your conclusion quantitatively.

1.7.7 Chain Calculation: From Mass and Geometry to Number of Atoms

A pure copper sphere has a radius of 1.250cm. Assume the density of copper is 8.96g cm⁻³ and the molar mass of copper is 63.546 g mol⁻¹. Use NA = 6.022 × 10²³ mol⁻¹.

Calculate the volume of the sphere in cm³.

Calculate the mass of the sphere in grams.

Convert the mass to moles of copper atoms.

Calculate the number of copper atoms in the sphere.

Throughout, track units carefully and report the final number of atoms with appropriate significant figures.

1.7.8 Significant Figures: Zeros and Scientific Notation

For each of the following values, determine the number of significant figures

and rewrite the value in scientific notation (without changing the implied precision):

Then, explain in one or two sentences why 1200 (without a decimal point) is ambiguous in precision, while 1200. is not.

1.7.9 Classifying Matter: Elements, Compounds, and Mixtures

Classify each sample as an element, a compound, a homogeneous mixture, or a heterogeneous mixture. For mixtures, indicate whether separation by a physical method is plausible and name one appropriate technique (filtration, distillation, chromatography, or another physical method) that would be suitable.

CO2(g)

Air in a sealed container (assume dry air)

Granite

NaCl(s)

A beaker containing ice water in equilibrium at 0∘C

Brass (copper–zinc alloy)

Oil and vinegar salad dressing (not emulsified)

Pure neon gas, Ne(g)

Sweetened tea in which sugar has fully dissolved

A sample labeled “distilled water” that contains dissolved Na+ and Cl– at 1.0 × 10⁻⁵ moldm⁻³

Your classifications should explicitly reference the definitions of pure substances vs. mixtures and homogeneous vs. heterogeneous systems.

1.7.10 Atomic Radius from Molar Volume and Packing Efficiency

A crystalline metal has a molar volume of 7.10cm³ mol⁻¹. Assume:

Each atom can be approximated as a sphere of radius r.

The atoms pack with a packing efficiency of 0.740 (fraction of space actually occupied by spheres).

Avogadro’s number is NA = 6.022 × 10²³ mol⁻¹.

Determine the total volume occupied by the atoms (spherical volume only) in 1.00mol of the metal.

Use this to find the volume per atom.

Assuming spherical atoms, solve for the atomic radius r in centimeters.

Convert the radius to picometers (pm).

Use dimensional analysis explicitly for all conversions (including cm³ to m³ if you choose to pass through SI base units), and report the final radius with appropriate significant figures.

Chapter 2
Stoichiometry and the Mole

Chemistry explains the behavior of matter, but without a standard method to count the invisible components of that matter, explanation cannot yield prediction. How do we weigh a handful of powder and know exactly how many atoms it contains? This chapter constructs the essential bridge between the microscopic world of atomic mass units and the macroscopic reality of laboratory balances. By defining the mole and applying the principles of conservation, we transform chemical equations from simple qualitative descriptions into precise mathematical tools capable of predicting the outcome of any reaction.

2.1 The Mole Concept and Avogadro’s Constant

Chemistry connects two radically different “worlds.” In the laboratory we can readily measure macroscopic quantities such as mass (grams), volume (liters), pressure (kilopascals), and temperature (kelvin). However, chemical change is governed by microscopic entities: atoms, molecules, ions, and electrons. A central challenge in chemistry is therefore a counting problem:

How can we describe, measure, and compare numbers of particles when those numbers are far too large to count directly?

The answer is the mole, a counting unit designed specifically for particles.

2.1.1 Counting by grouping: why chemists need a special unit

In everyday life, we rarely count individual small objects when there are many of them. We use grouping words such as:

dozen = 12 objects,

ream = 500 sheets of paper,

pair = 2 objects.

These words are convenient because they let us specify how many without listing each object. Chemistry uses the same idea, but the group size must be extraordinarily large because particles are extraordinarily small.

For example, a small grain of table salt contains on the order of 10¹⁸ ions. A few grams of water contain on the order of 10²³ molecules. The mole is defined so that these huge particle numbers can be expressed with manageable numerical values.

2.1.2 The mole as an SI base unit (2019 definition)

The amount of substance is one of the seven base quantities in the International System of Units (SI). Its SI base unit is the mole (symbol: mol).

Since the 2019 revision of the SI, the mole is defined by fixing the numerical value of the Avogadro constant NA:

Therefore,

Several points in this definition deserve careful attention:

The mole is a counting unit. It is fundamentally like “dozen,” except the number per mole is much larger.

The entity must be specified. One mole can mean one mole of atoms, molecules, ions, electrons, or formula units, but you must say which:

NA is exact by definition. The number 6.02214076 × 10²³ is not a measured approximation; it is the defining value of the mole in the modern SI. (In calculations, we often use rounded values such as 6.022 × 10²³ mol⁻¹ when high precision is not required.)

The mole does not inherently mean a certain mass. Mass relationships enter through molar mass, developed in the next section. Here, the essential idea is counting.

2.1.3 Why Avogadro’s constant is needed: bridging atomic mass and laboratory mass

Atoms and molecules are so small that their masses are impractical to express in grams. Instead, chemistry uses the atomic mass unit (also called the unified atomic mass unit), symbol u or Da (dalton). Atomic and molecular masses listed on the periodic table are based on this scale.

The key conceptual bridge is this:

Atomic masses in u and molar masses in g mol⁻¹ share the same numerical values (within standard rounding), because the mole links the microscopic mass scale to the macroscopic one.

Carbon-12 provides the reference point. A single atom of carbon-12 has a mass of exactly 12 u on the atomic mass scale. When we speak of “12 grams of carbon-12,” we are describing a macroscopic sample containing a mole of carbon-12 atoms. The numerical equivalence (12) is the same, but the units and the meaning are different:

12 u refers to the mass of one 12C atom.

12 g refers (approximately) to the mass of a mole of 12C atoms.

This is not coincidence; it is precisely what makes the mole useful. Because the periodic table gives relative masses on an atomic scale, the mole allows those relative masses to be used as practical laboratory conversion factors once we scale up to NA particles.

2.1.4 Dual-scale meaning of “12”: atomic-scale vs laboratory-scale

The first statement is a microscopic mass statement; the second is a macroscopic mass statement. What connects them is how many atoms are in a mole.

2.1.5 Interpreting Avogadro’s constant as a conversion factor

Because the mole is a counting unit, Avogadro’s constant functions as a conversion factor between amount of substance (moles) and number of entities (particles). In dimensional analysis form,

Thus, for any specified entity,

Example 2.1: Converting moles of atoms to number of atoms

How many carbon atoms are present in 0.250 mol of C?

Example 2.2: Converting number of molecules to moles

A sample contains 3.00 × 10²² molecules of H2O. What amount of substance is this?

These conversions are conceptually simple but foundational: much of stoichiometry can be viewed as repeated use of unit factors that convert between measurable macroscopic quantities and particle counts.

2.1.6 What “amount of substance” means (and what it does not mean)

The SI quantity “amount of substance” can be misunderstood because the word “amount” is used casually in everyday language. In chemistry and SI metrology, amount of substance has a specific meaning:

Amount of substance (n) measures how many specified entities are present, expressed in mol.

It is not the same as mass (m), which measures how much matter is present in grams or kilograms.

It is not the same as volume (V ), which measures space occupied (e.g., mL, L).

In a laboratory context, the mole is powerful precisely because it allows you to connect:

However, the mass and volume connections require additional ideas (molar mass, density, gas laws). At this stage, the essential point is that the mole gives a precise way to count particles by using NA.

2.1.7 Choosing the entity: atoms, molecules, ions, and formula units

Chemistry frequently uses the mole to count different kinds of entities. The correct interpretation depends on what the chemical formula represents:

Elements in atomic form: For Fe, C, Ne, “1 mol” usually means 1 mole of atoms.

Molecular elements: For H2, O2, N2, Cl2, “1 mol” usually means 1 mole of molecules. (Counting atoms would require an extra factor of 2.)

Molecular compounds: For H2O, CO2, C6H12O6, “1 mol” means 1 mole of molecules.

Ionic compounds: For NaCl, CaF2, “1 mol” refers to 1 mole of formula units (the simplest electrically neutral ratio of ions). If you need moles of individual ions, you must use the subscripts:

This explicit attention to entities prevents common mistakes in stoichiometry, such as treating “moles of molecules” and “moles of atoms” as interchangeable. They are related, but only through the chemical formula.

2.1.8 Summary: the mole as the central bridge in chemical measurement

The mole is the chemist’s bridge between the particulate and macroscopic descriptions of matter.

The mole is an SI base unit for amount of substance.

It is defined by a fixed value of Avogadro’s constant:

It functions as a conversion factor between number of entities and amount of substance.

By scaling up microscopic masses to macroscopic samples, the mole enables the numerical correspondence between atomic/molecular masses in u and molar masses in g mol⁻¹ (developed formally in the next section).

With the mole established as a precise counting unit, we are prepared to connect amount of substance to mass through molar mass and to use dimensional analysis as a systematic tool for chemical calculations.

2.2 Molar Mass and Chemical Formulas

The previous section established the mole as a precise counting unit:

To use the mole in the laboratory, we must connect counting to a quantity we can measure directly. The most common laboratory measurement is mass. The connection between mass and amount of substance is the molar mass.

2.2.1 Molar mass: definition and units

The molar mass of a substance, denoted ℳ, is the mass per mole:

with SI-compatible units of g mol⁻¹ (or kg mol⁻¹ in strict SI base units, but g mol⁻¹ is standard in general chemistry).

Rearranging gives the two most frequently used conversion equations:

n | = |  |
m | = nℳ. |  |

2.2.2 Why molar mass numerically matches atomic and formula masses

A powerful and often confusing fact is that atomic masses in u have the same numerical values as molar masses in g mol⁻¹. For example,

The units differ because the physical meaning differs (one atom versus a mole of atoms), but the numerical correspondence is what allows periodic table masses to become practical laboratory conversion factors once we interpret them on a per-mole basis.

2.2.3 Chemical formulas as counting instructions

A chemical formula communicates more than composition; it also encodes a particle-level ratio. For example:

H2O indicates 2 H atoms per 1 O atom in each molecule.

CaCl2 indicates 1 Ca ion per 2 Cl ions in each formula unit.

C6H12O6 indicates 6 C, 12 H, and 6 O atoms per molecule.

When we speak of one mole of a substance, the formula subscripts become mole ratios among atoms or ions inside that mole:

This idea underlies later stoichiometry, but here it is essential for correctly interpreting “number of entities” when converting among moles, particles, and mass.

2.2.4 Formula mass vs. molecular mass vs. molar mass

The terminology depends on the type of substance.

Molecular mass (molecular weight) for covalent substances

For a molecular compound, the molecular mass is the sum of the atomic masses in

a single molecule, usually expressed in u. For water:

The corresponding molar mass is then:

Formula mass for ionic substances

Ionic solids do not exist as discrete molecules; they form extended lattices. For ionic compounds we use the formula mass: the sum of atomic masses in one formula unit. For sodium chloride:

so

Even though NaCl is not molecular, the molar mass is still well-defined because the mole counts formula units just as it can count molecules.

Practical note on precision

Periodic table atomic masses are weighted averages of naturally occurring isotopes (not usually integers). Consequently, molar masses are rarely whole numbers. In calculations, keep enough significant figures to avoid rounding errors, and round only at the end according to the limiting measurement.

2.2.5 Computing molar mass from a chemical formula

To find the molar mass of a compound:

Identify each element and its subscript (including any parentheses multipliers).

Multiply each element’s atomic mass by the number of that atom in the formula.

Sum the contributions.

Example 2.3: Molar mass of a molecular compound (C2H5OH)

Ethanol can be written as C2H6O (equivalent to C2H5OH in atom counts).

ℳ(C2H6O) | = 2(12.011) + 6(1.008) + 1(15.999) g mol⁻¹ |  |
| = 24.022 + 6.048 + 15.999 |  |
| = 46.069 g mol⁻¹ ≈ 46.07 g mol⁻¹. |  |

Example 2.4: Molar mass with parentheses (Al2(SO4)3)

For aluminum sulfate:

ℳ(Al2(SO4)3) | = 2(26.982) + 3(32.06) + 12(15.999) g mol⁻¹ |  |
| = 53.964 + 96.18 + 191.988 |  |
| = 342.132 g mol⁻¹ ≈ 342.13 g mol⁻¹. |  |

2.2.6 Interconverting mass and moles

Because molar mass is a conversion factor, it can be used in dimensional analysis in either direction:

Example 2.5: Mass → moles

How many moles of NaCl are in a 15.0 g sample?

n(NaCl) | = 15.0 g × |  |
| = 0.2567 mol ≈ 0.257 mol (3 sig figs). |  |

Example 2.6: Moles → mass

What mass of ethanol corresponds to 0.125 mol of C2H6O?

m(C2H6O) | = 0.125 mol × 46.07 g mol⁻¹ |  |
| = 5.75875 g ≈ 5.76 g (3 sig figs). |  |

2.2.7 Interconverting moles and number of entities

Avogadro’s constant NA connects moles to particle counts. The key is to specify the entity correctly (molecules vs. atoms vs. ions vs. formula units).

Example 2.7: Moles of molecules → molecules (CO2)

How many CO2 molecules are in 0.750 mol of CO2?

N(CO2 molecules) | = 0.750 mol × 6.022 × 10²³  |  |
| = 4.52 × 10²³ molecules (3 sig figs). |  |

Example 2.8: Molecules → atoms (careful interpretation of the formula, CO2)

How many oxygen atoms are present in 0.750 mol of CO2? Each CO2 molecule contains 2 oxygen atoms. Therefore:

n(O atoms) | = 0.750 mol CO2 × |  |
| = 1.50 mol O atoms. |  |

Then convert moles of O atoms to number of O atoms:

N(O atoms) | = 1.50 mol × 6.022 × 10²³  |  |
| = 9.03 × 10²³ atoms. |  |

This two-step structure is common: use the formula to convert between moles of compound and moles of constituent atoms (or ions), then use NA if a particle count is required.

2.2.8 The full set of core conversions: mass, moles, particles

Many problems in this chapter reduce to selecting the appropriate sequence of conversion factors. Conceptually, the mole is the central bridge:

The two fundamental conversion factors are:

Mass ↔ Moles: | ℳ  |  |
Moles ↔ Particles: | NA  |  |

Example 2.9: Mass → particles (a complete conversion chain, H2O)

Step 1: Convert mass to moles.

n(H2O) | = 9.00 g × |  |
| = 0.4996 mol ≈ 0.500 mol (3 sig figs). |  |

Step 2: Convert moles to molecules.

N(H2O molecules) | = 0.4996 mol × 6.022 × 10²³  |  |
| = 3.01 × 10²³ molecules (3 sig figs). |  |

Optional extension: molecules → atoms. If the question instead asked for hydrogen atoms, you would multiply by 2 after finding moles of H2O (or after finding molecules) because each molecule contains two H atoms.

2.2.9 Common pitfalls and how to avoid them

Pitfall 1: Confusing the entity being counted

Writing “0.10 mol of oxygen” is ambiguous: it could mean O atoms, O2 molecules, or O2– ions depending on context. Prefer explicit statements such as:

Pitfall 2: Misreading subscripts as charges or coefficients

In CaCl2, the “2” is part of the formula (composition). In a chemical equation, coefficients (e.g., 2CaCl2) represent how many formula units or moles participate. Do not interchange them.

Pitfall 3: Dropping units during conversion

Stoichiometry is more reliable when treated as dimensional analysis. Carry units through every step; the correct pathway is the one that cancels unwanted units and leaves the desired unit.

Molar mass ℳ links laboratory mass to amount of substance:

Molar masses are computed from chemical formulas by summing atomic masses (accounting for subscripts and parentheses).

Avogadro’s constant links moles to particle counts:

Chemical formulas act as counting instructions, enabling conversions between moles of compounds and moles of constituent atoms or ions.

2.3 Percent Composition and Empirical Formulas

The previous section established how chemical formulas function as counting instructions and how molar mass connects formulas to laboratory measurements of mass. In this section we use those ideas in two directions:

Forward direction (formula → composition): Given a chemical formula, determine the percent composition by mass.

Reverse direction (composition →formula): Given mass data (often as mass percentages), determine the empirical formula and, when additional molar-mass information is available, the molecular formula.

These calculations are among the most common ways chemists convert between experimental measurements and symbolic representations of matter.

2.3.1 Percent composition by mass

Definition

The mass percent (percent by mass) of element E in a compound is

When the compound is pure and its chemical formula is known, the calculation is most conveniently performed on a per-mole basis:

Here the numerator and denominator are both masses per mole of compound, so the ratio is dimensionless.

Example 2.10: Percent composition of H2O

For water, ℳ(H2O) = 18.015 g mol⁻¹.

% H | = × 100% = 11.19% |  |
% O | = × 100% = 88.81%. |  |

The results sum to 100.00% (within rounding), as expected for a compound consisting only of H and O.

Example 2.11: Percent composition of an ionic compound, CaCl2

First compute the molar mass:

Then compute mass percent:

% Ca | = × 100% = 36.11% |  |
% Cl | = × 100% = 63.89%. |  |

Hydrates and mixtures: a note on interpretation

Percent composition describes a single pure compound with a well-defined formula. If a sample is a mixture (for example, sand contaminated with salt) there is no single

chemical formula, so percent composition must be interpreted with care. Hydrated salts are pure compounds with formulas such as CuSO4 ⋅5H2O; in that case, water is part of the compound, and percent composition can be computed from the full formula.

2.3.2 From mass data to empirical formula

Empirical formula: meaning and limitations

The empirical formula is the simplest whole-number ratio of atoms in a compound. It captures composition but not necessarily the true molecular size. For example:

Both formulas imply the same atom ratio C:H:O = 1 : 2 : 1, but the second contains six times as many atoms per molecule.

Core strategy (mass → moles → ratio)

The empirical-formula procedure is a direct application of molar mass conversions:

Obtain the mass of each element. Data may be given in grams directly or as percent by mass.

Convert each element’s mass to moles. Use n = m∕ℳ for each element.

Form mole ratios. Divide all mole values by the smallest to normalize.

Convert to whole numbers. If ratios are close to integers, round. If ratios are close to simple fractions (e.g., 1.5, 1.33, 1.25), multiply all ratios by a small integer to clear the fraction.

Why the "assume 100 g" shortcut works for percent data. If composition is given as mass percent, one convenient choice is to imagine a 100.0 g sample. Then each percent value becomes a mass in grams (e.g., 40.0% becomes 40.0 g). This does not change mole ratios because scaling the sample mass scales all component masses proportionally.

Example 2.12: Empirical formula from percent composition (CHO compound)

A compound contains 40.00% C, 6.71% H, and 53.29% O by mass. Determine the empirical formula.

Step 1: Assume a 100.00 g sample.

Step 2: Convert each mass to moles.

nC | = = 3.330 mol |  |
nH | = = 6.656 mol |  |
nO | = = 3.331 mol. |  |

Step 3: Normalize by dividing by the smallest number of moles. The smallest is approximately 3.330.

[picture] | = 1.000 |  |
[picture] | = 1.999 ≈ 2.000 |  |
[picture] | = 1.000. |  |

Step 4: Write the empirical formula. The ratio C:H:O ≈ 1 : 2 : 1, so the empirical formula is

(These percentages are consistent with the empirical formula of carbohydrate units.)

Example 2.13: Empirical formula requiring a multiplier (fractional ratios)

A compound is found to be 52.14% C and 47.86% O by mass. Determine the empirical formula.

Step 1: Assume a 100.00 g sample.

Step 2: Convert to moles.

nC | = = 4.342 mol |  |
nO | = = 2.991 mol. |  |

Step 3: Normalize by the smaller value.

[picture] | = 1.452 |  |
[picture] | = 1.000. |  |

The carbon ratio is close to 1.5, which suggests a fraction of .

Step 4: Clear the fraction by multiplying all ratios by 2.

Thus the simplest whole-number ratio is approximately C:O = 3 : 2:

Interpreting near-integers. Experimental data and atomic masses rarely produce exact integers. The goal is to identify the smallest reasonable integers consistent with the data. Ratios such as 1.49 or 1.51 often indicate 1.5; ratios such as 1.33 indicate ; ratios such as 1.25 indicate . The multipliers 2, 3, or 4 are most common.

2.3.3 From empirical formula to molecular formula

Molecular formula: definition

The molecular formula gives the actual number of each atom in a molecule. It is always an integer multiple of the empirical formula:

where k is a positive integer.

Determining the multiplier k

If the compound’s molar mass ℳmolecular is known (from experiment), compute the empirical formula mass ℳemp and take the ratio:

Because the molecular formula must be a whole-number multiple of the empirical formula, k should be very close to an integer.

Example 2.14: Determining a molecular formula from empirical formula and molar mass

A compound has empirical formula CH2O. Its molar mass is found to be 180.16 g mol⁻¹. Determine the molecular formula.

Step 1: Compute the empirical formula mass.

ℳemp(CH2O) | = 12.011 + 2(1.008) + 15.999 |  |
| = 30.026 g mol⁻¹. |  |

Step 2: Compute the integer multiplier.

k | = = 5.999 ≈ 6. |  |

Step 3: Multiply subscripts by k.

Thus the molecular formula is

2.3.4 Working from direct mass data (not percentages)

Sometimes elemental masses are given directly (for example, from synthesis or decomposition). The method is identical: convert each mass to moles, normalize, and convert to integers.

Example 2.15: Empirical formula from gram amounts

A sample contains 2.80 g Fe and 1.20 g O. Determine the empirical formula.

Step 1: Convert to moles.

nFe | = = 0.0501 mol |  |
nO | = = 0.0750 mol. |  |

Step 2: Normalize by the smaller.

[picture] | = 1.00 |  |
[picture] | = 1.50. |  |

Step 3: Clear the fraction. Multiply both by 2:

So the empirical formula is

2.3.5 Consistency checks and common pitfalls

Check 1: Percentages should sum to 100% (within rounding)

If mass percentages are given, they should add to 100% to within rounding uncertainty. A noticeable mismatch may signal missing elements (for example, oxygen by difference) or a transcription error.

Check 2: Empirical formula must reflect the simplest ratio

If all subscripts share a common factor (e.g., C2H4, N2O2), reduce them (CH2, NO) for the empirical formula.

Pitfall 1: Rounding too early

Carry at least 3–4 significant figures through the mole calculations and normalize before rounding. Premature rounding can distort ratios, especially when one element is present in small amounts.

Pitfall 2: Guessing an integer without considering simple fractions

Ratios such as 1.49, 1.51, 2.49, and 2.51 often indicate half-integers. Ratios near 1.33 or 1.67 often indicate thirds. Use a small-integer multiplier only after you have normalized by the smallest mole value.

Pitfall 3: Confusing empirical and molecular formulas

Percent composition determines the empirical formula. Determining the molecular formula additionally requires the molar mass (or some other information specifying molecular size).

Percent composition is found from a known chemical formula using molar masses:

Empirical formulas are determined by converting element masses (or mass percentages) to moles, normalizing to the simplest whole-number ratio.

Molecular formulas are integer multiples of empirical formulas:

These techniques turn composition measurements into chemical formulas, preparing us to write and balance chemical equations and to use those equations for quantitative reaction calculations.

2.4 Chemical Equations and Balancing

Chemical equations are the symbolic language that connects observable chemical change to the particle-level rearrangement of atoms. The earlier sections developed the mole and chemical formulas as counting instructions. A balanced chemical equation is the next essential step: it tells us the stoichiometric ratios in which substances react and form products, and it does so in a way that respects a fundamental physical constraint: matter is conserved.

2.4.1 The Law of Conservation of Mass

In ordinary chemical reactions (not nuclear reactions), atoms are neither created nor destroyed. They are rearranged into new combinations. This is the Law of Conservation of Mass:

In a chemical reaction, the total mass of reactants equals the total mass of products (within experimental uncertainty).

Because mass is conserved, the number of atoms of each element must be the same on both sides of a chemical equation. Balancing an equation is the process of choosing coefficients so that this condition is met.

2.4.2 Anatomy of a chemical equation

A chemical equation has three core components:

Reactants (left): substances consumed.

Products (right): substances formed.

Arrow (→): indicates direction of the chemical change under stated conditions.

For example, methane combusts in oxygen to form carbon dioxide and water:

Several conventions add important information:

Physical states

States of matter may be included as subscripts in parentheses:

The state symbols mean:

(s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).

State symbols describe the macroscopic form of substances under reaction conditions; they do not change the balancing rules, but they are essential for interpreting the chemistry (for example, whether ions are present in solution).

Reaction conditions

Conditions such as heat, catalysts, or light may be written above or below the arrow:

Here Δ indicates heating.

Reversible reactions

Some reactions proceed appreciably in both directions and are written with a double arrow:

Balancing still requires equal atom counts on both sides.

2.4.3 What balancing means: coefficients vs. subscripts

Balancing is accomplished by adjusting stoichiometric coefficients—the numbers placed in front of formulas. Never change subscripts in chemical formulas to balance an equation.

Why subscripts must not be changed

A subscript is part of a substance’s identity. For example:

Changing subscripts changes the chemical species itself. When we balance, we are not changing what reacts; we are changing how many units of each substance are involved.

Interpretation of coefficients

A coefficient applies to the entire formula unit:

At the mole level, coefficients represent mole ratios:

This is the critical link to stoichiometry: once an equation is balanced, its coefficients define the proportional amounts of reactants and products.

2.4.4 Writing chemical equations from words

A word description identifies reactants and products; writing the equation requires correct formulas and often state symbols. The general workflow is:

1. Write correct formulas for all reactants and products.

2. Add state symbols if known.

3. Balance using coefficients.

Example 2.16: Translating a word equation

Word description: Aqueous sodium carbonate reacts with aqueous calcium chloride to form solid calcium carbonate and aqueous sodium chloride.

Step 1: Write formulas (unbalanced).

Step 2: Balance. Count atoms:

Left: Na 2, C 1, O 3, Ca 1, Cl 2

Right: Na 1, C 1, O 3, Ca 1, Cl 1

Sodium and chlorine are short on the right; place a 2 before NaCl:

2.4.5 Systematic balancing by inspection

Many equations can be balanced by inspection using a consistent strategy. The key idea is to treat the coefficients as unknown multipliers and enforce conservation of each element.

A practical balancing strategy

List each element and count atoms on both sides.

Balance elements that appear in the fewest formulas first.

Leave H and O for last in reactions where they appear in multiple compounds (common in combustion and aqueous reactions).

Treat polyatomic ions as intact units if they appear unchanged on both sides (for example, SO42– in many reactions). This simplifies bookkeeping.

Reduce coefficients to the smallest whole-number set at the end.

Example 2.17: Balancing a combustion reaction

Balance the combustion of propane:

Step 1: Balance carbon. There are 3 C in C3H8, so put 3 in front of CO2:

Step 2: Balance hydrogen. There are 8 H in C3H8, so put 4 in front of H2O:

Step 3: Balance oxygen. On the right: 3CO2 contains 6 O atoms; 4H2O contains 4 O atoms; total 10 O atoms ⇒ 5 O2 molecules.

2.4.6 Treating polyatomic ions as units

When a polyatomic ion appears on both sides unchanged, count it as a group.

Example 2.18: Balancing with an intact polyatomic ion

Balance:

Here SO4 appears on both sides.

Step 1: Balance sulfate groups. There are 3 SO4 on the left, so put 3 before CaSO4:

Step 2: Balance calcium. Now there are 3 Ca on the right, so put 3 before Ca(OH)2:

Step 3: Balance aluminum. There are 2 Al on the left, so put 2 before Al(OH)3:

Step 4: Check hydroxide. Left has 3 × 2 = 6 OH groups; right has 2 × 3 = 6 OH groups. Balanced.

2.4.7 The algebraic method for balancing complex equations

Some equations are difficult to balance by inspection, especially when many species contain the same elements. In such cases, treat coefficients as unknowns and write conservation equations for each element.

General approach

For an unbalanced equation

write one equation per element by equating atom counts on both sides. Solve the linear system (often with one free variable), then scale to the smallest whole numbers.

Example 2.19: Algebraic balancing of a combustion-type equation

Balance:

Let coefficients be a,b,c,d:

Conservation equations:

C: | 2a = c |  |
H: | 6a = 2d  ⇒  3a = d |  |
O: | 2b = 2c + d |  |

Choose a = 1 (free choice for scaling). Then c = 2, d = 3. Oxygen:

Multiply all coefficients by 2 to clear the fraction:

This method guarantees a correct balance when applied carefully.

2.4.8 Final checks: what a balanced equation guarantees (and what it does not)

What balancing guarantees

A balanced chemical equation guarantees:

Atom conservation: each element has the same atom count on both sides.

Mass conservation: total mass is conserved (chemical reactions only).

Stoichiometric ratios: coefficients give mole ratios among substances.

What balancing does not guarantee

A balanced equation does not, by itself, guarantee:

that the reaction will occur (thermodynamics and kinetics matter),

the extent of reaction or yield,

the mechanism (sequence of steps),

the correct physical states (those require chemical context).

Chemical equations symbolize chemical change and must satisfy the Law of Conservation of Mass.

Balance by adjusting coefficients only; never change subscripts in formulas.

Coefficients represent counts of particles and, at the macroscopic level, mole ratios.

Many equations can be balanced by inspection using a systematic strategy; complex cases can be balanced reliably using an algebraic method.

With balanced equations in hand, we are prepared to perform reaction stoichiometry: converting among masses, moles, and particles of reactants and products using the coefficients as the central chemical conversion factors.

2.5 Reaction Stoichiometry: Mass Relations

The previous section showed how to write and balance chemical equations so that atom counts are conserved. A balanced equation does more than satisfy conservation of mass: it provides quantitative equivalences among reactants and products. These equivalences are expressed through stoichiometric factors (mole ratios) derived from the coefficients in the balanced equation.

This section develops a systematic method for using balanced equations to convert between measurable laboratory quantities (especially mass) of one substance and the required or produced amount of another.

2.5.1 Stoichiometric factors: the mole ratio as a conversion factor

Consider the balanced reaction

The coefficients provide mole relationships:

From this, we can build stoichiometric factors, which are conversion factors between moles of different species in the same reaction:

Each ratio is valid only for this balanced equation. Stoichiometry is therefore an application of dimensional analysis in which the balanced equation provides the bridge between substances.

2.5.2 The stoichiometry roadmap (the “mole bridge” idea)

Most mass-relation problems follow the same logical path:

Mass of A →Moles of A →Moles of B →Mass of B

The first and last steps use molar mass; the middle step uses a stoichiometric factor from the balanced equation. When particles or volume are involved, additional conversion factors may appear (Avogadro’s constant, density, molar volume, gas laws), but the central bridge remains the mole ratio from the balanced equation.

2.5.3 The standard mass–mass stoichiometry procedure

A reliable problem-solving template is:

Write and balance the chemical equation.

Identify the given substance (A) and the target substance (B).

Convert the given quantity to moles of A.

Use the balanced equation to convert moles of A to moles of B.

Convert moles of B to the requested unit (often grams).

Each step should be written as a unit-canceling calculation. This approach makes errors visible (units that do not cancel signal a wrong conversion factor).

2.5.4 Mass of product from a given mass of reactant

Example 2.20: Mass of CO2 produced from methane combustion

Methane burns in oxygen:

What mass of CO2 is produced when 16.0 g of CH4 reacts completely (with excess O2)?

Step 1: Convert mass of CH4 to moles.

n(CH4) | = 16.0 g × = 0.997 mol. |  |

Step 2: Convert moles of CH4 to moles of CO2 using the balanced equation. The coefficients show 1 mol CH4 produces 1 mol CO2:

n(CO2) | = 0.997 mol CH4 × = 0.997 mol CO2. |  |

Step 3: Convert moles of CO2 to mass.

m(CO2) | = 0.997 mol × 44.009 g mol⁻¹ = 43.9 g. |  |

(rounded to three significant figures to match the given mass).

2.5.5 Mass of reactant required to produce a desired mass of product

Stoichiometry often runs “backward”: given a target production mass, how much reactant must be supplied?

Example 2.21: Mass of CaCO3 needed to produce a given mass of CO2

Carbon dioxide can be produced by thermal decomposition of calcium carbonate:

What mass of CaCO3 is required to produce 10.0 g of CO2?

Step 1: Convert CO2 mass to moles.

Step 2: Use the stoichiometric factor to get moles of CaCO3. From the balanced equation, 1 mol CaCO3 produces 1 mol CO2:

Step 3: Convert moles of CaCO3 to mass.

2.5.6 Using stoichiometry with particles (entities)

Because the mole connects to particles through Avogadro’s constant, a stoichiometry problem can begin or end with particle counts. The balanced equation still provides the mole ratio.

Example 2.22: Molecules of product from molecules of reactant

Nitrogen and hydrogen form ammonia:

If 9.03 × 10²³ molecules of H2 react completely with excess N2, how many molecules of NH3 are formed?

Step 1: Convert molecules of H2 to moles.

Step 2: Use the stoichiometric factor to convert to moles of NH3.

Step 3: Convert moles of NH3 to molecules.

2.5.7 Stoichiometry with volume via density (pure liquids and solids)

Laboratory quantities are sometimes measured by volume rather than by mass,

especially for liquids. If the density ρ is known, volume can be converted to mass:

Then proceed with standard stoichiometry.

Example 2.23: Using density to convert a liquid volume to moles

Ethanol combusts according to:

A sample of ethanol has volume 25.0 mL and density 0.789 g mL⁻¹. Assuming excess oxygen, what mass of CO2 can be produced?

Step 1: Convert volume to mass of ethanol.

Step 2: Convert mass to moles.

Step 3: Use the balanced equation to convert to moles of CO2.

Step 4: Convert moles of CO2 to mass.

(three significant figures).

2.5.8 Interpreting coefficients: mass is not conserved species-by-species

A frequent conceptual mistake is to assume that if coefficients match, masses must match, or that mass is conserved for each substance separately. Conservation of mass applies to the total system, not to each reactant individually. Coefficients represent mole ratios, not mass ratios.

For example, in

the mole ratio H2:O2 is 2:1, but the mass ratio is

very different from 2:1. This is why stoichiometry must be performed through moles.

2.5.9 A checklist for reliable stoichiometric calculations

Is the equation balanced? If not, stop and balance first.

Are formulas correct? Incorrect formulas ruin the calculation even if the equation is balanced.

Is the stoichiometric factor oriented correctly? The desired unit must be in the numerator.

Are you converting through moles? Mass-to-mass problems must pass through moles.

Do units cancel cleanly? If not, a conversion factor is missing or inverted.

Is the final answer reasonable? Compare to the given amount and the chemistry (e.g., a small reactant mass cannot produce an enormous product mass without additional reactant mass from elsewhere).

A balanced equation provides stoichiometric factors (mole ratios) that relate reactants and products.

The standard mass-relation pathway is:

Particle counts and volumes can be incorporated by additional conversion factors (Avogadro’s constant, density), but the mole bridge from the balanced equation remains central.

Stoichiometry is most reliable when treated as dimensional analysis with units carried throughout.

The next section extends these ideas to non-stoichiometric mixtures, where one reactant may limit the amount of product formed, and to the comparison of theoretical and actual yields in real chemical processes.

2.6 Limiting Reagents and Percent Yield

The previous section treated reaction stoichiometry as an idealized conversion problem: given an amount of one reactant, the balanced equation provides the mole ratio needed to compute the amount of product. That approach implicitly assumes that all other reactants are present in excess. In real laboratory and industrial settings, reactants are often mixed in non-stoichiometric proportions, so one reactant runs out first and stops the reaction.

This section introduces two closely related ideas:

Limiting reagent: the reactant that is consumed first and therefore limits the amount of product formed.

Percent yield: a measure of how efficiently a reaction produces the desired product compared with the theoretical maximum.

2.6.1 Conceptual model: the limiting reagent as the “reactant that runs out”

Consider a simple reaction written in terms of particles:

If we start with 5 particles of A and 6 particles of B, B will run out first because each product unit requires 2 B particles. Only 3 product units can form (consuming 6 B), leaving 2 A particles unreacted. The same logic applies to moles and grams; the only complication is that we must convert to moles to compare amounts properly.

2.6.2 Definitions

Limiting reagent (limiting reactant)

The limiting reagent is the reactant that is completely consumed when the reaction goes to completion. It determines the maximum amount of product that can form.

Excess reagent

An excess reagent is any reactant that remains after the limiting reagent is consumed. Excess reagents do not determine the amount of product; they merely ensure that the limiting reagent can be used up completely.

Theoretical yield

The theoretical yield is the maximum amount of product predicted by stoichiometry (based on the limiting reagent), assuming the reaction proceeds to completion with no losses and no side reactions.

Actual yield

The actual yield is the amount of product actually obtained experimentally.

Percent yield

The percent yield quantifies efficiency:

2.6.3 How to identify the limiting reagent

For a reaction involving two or more reactants, there are two standard stoichiometric approaches. Both are valid; choose the one that is clearest for you.

Method 1: Compute product from each reactant (and choose the smaller)

Convert each reactant amount to moles.

For each reactant in turn, assume it is limiting and compute the moles of product it could form using the balanced equation.

The reactant that produces the least product is the limiting reagent.

Method 2: Compare available-to-required mole ratios

Convert reactant amounts to moles.

Use the balanced equation to compute the stoichiometric ratio required:

where ν₁ and ν₂ are coefficients.

Compare to the ratio actually available. If the available ratio is smaller than required, Reactant 1 is limiting; if larger, Reactant 2 is limiting.

Method 1 is more general and tends to reduce sign errors because the “smaller product” criterion is unambiguous. We will use Method 1 in the main example, then compute excess remaining as an additional step.

2.6.4 Worked example: limiting reagent and excess remaining

Example 2.24: Formation of ammonia

Ammonia is formed by:

A mixture contains 10.0 g of N2 and 2.00 g of H2. Identify the limiting reagent and calculate:

the theoretical yield of NH3 (in grams),

the mass of the excess reagent remaining (in grams).

Step 1: Convert each reactant mass to moles.

n(N2) | = 10.0g × = 0.357mol |  |
n(H2) | = 2.00g × = 0.992mol. |  |

Step 2: Compute moles of NH3 producible from each reactant.

If N2 were limiting:

If H2 were limiting:

The smaller value is 0.661 mol, so H2 is the limiting reagent and N2 is in excess.

Step 3: Theoretical yield of NH3.

Step 4: Amount of excess N2 remaining. First compute how much N2 is consumed when 0.992 mol of H2 reacts completely. From Eq. (??), 3 mol H2 require 1 mol N2:

Initial n(N2) = 0.357mol, so remaining:

Convert to grams:

2.6.5 Understanding the particle diagram (and its limitations)

The diagram emphasizes the key qualitative outcome: the limiting reagent is exhausted and the excess reagent remains. In real calculations we do not count molecules one by one; we count in moles. The underlying logic, however, is identical: the reaction can only proceed as long as all required reactants are available in the proper ratio.

2.6.6 Theoretical yield and percent yield

Theoretical yield comes from the limiting reagent

Once the limiting reagent is known, the theoretical yield of any product is found by standard stoichiometry:

No matter how much excess reagent is present, it cannot increase the theoretical yield beyond what the limiting reagent allows.

Percent yield quantifies real-world performance

Percent yield compares experimental results to the theoretical maximum (Eq. (??)).

Example 2.25: Percent yield for ammonia synthesis

In Example 2.24, the theoretical yield of ammonia was 11.3 g. Suppose the experiment actually produced 9.20 g of NH3. Then:

2.6.7 Why yields are often less than 100%

Percent yields below 100 % are common and do not necessarily indicate “bad chemistry.” Typical causes include:

Incomplete reaction (equilibrium): Some reactions do not go to completion; reactants and products coexist at equilibrium.

Side reactions: Reactants may form undesired byproducts, reducing formation of the desired product.

Loss during processing: Transfers, filtration, washing, drying, and purification steps often lose material mechanically.

Product decomposition or volatility: The product may decompose, evaporate, or react with air or moisture before it is measured.

Measurement and purity issues: Impure reactants, wet solids, or systematic measurement errors can alter the apparent yield.

2.6.8 Yields greater than 100%: what they usually mean

A calculated percent yield above 100 % is a warning sign. It most commonly results from:

Impure product: the measured “product” contains solvent, water, salts, or other contaminants.

Incomplete drying: residual water increases mass.

Side products counted as product: a mixture is weighed as if it were pure desired compound.

In such cases, the actual chemical production of the desired product is not truly above the theoretical maximum; rather, the measured mass does not represent pure product.

2.6.9 A compact algorithm for limiting reagent problems

For reactions with multiple reactants:

Balance the equation.

Convert each reactant amount to moles.

Compute moles of product from each reactant (using stoichiometric factors).

The smallest product amount identifies the limiting reagent.

Use the limiting reagent to compute theoretical yields of all products of interest.

If requested, compute excess remaining by subtracting moles consumed from moles initially present.

If actual yield is given, compute percent yield.

In non-stoichiometric mixtures, the limiting reagent is consumed first and determines the maximum (theoretical) amount of product.

The theoretical yield is calculated using stoichiometry based on the limiting reagent.

The percent yield measures efficiency:

Percent yields below 100 % commonly arise from incomplete reactions, side reactions, and mechanical losses; yields above 100 % usually indicate impurities or measurement issues.

Having developed limiting-reagent logic and yield analysis, we can now apply stoichiometry to specialized experimental data sets. The next section focuses on combustion analysis, where measured masses of CO2 and H2O are used to infer the empirical formula of an unknown compound.

2.7 Combustion Analysis

The previous sections developed the core stoichiometric toolkit: the mole, molar mass, balanced equations, and limiting-reagent logic. Combustion analysis is a specialized application of these ideas. In combustion analysis, we determine the empirical formula of an unknown compound (typically an organic compound) by burning it completely and measuring how much CO2 and H2O form.

The method is powerful because carbon and hydrogen are converted into products whose amounts can be measured accurately. From those product amounts we back-calculate the amounts of C and H in the original sample, then infer the empirical formula.

2.7.1 What combustion analysis measures

In a typical combustion experiment, a known mass of the unknown compound is reacted with excess oxygen so that combustion is complete. The products CO2 and H2O are collected (often by absorption) and their masses are measured. The key experimental data are therefore:

mass of the unknown sample, msample,

mass of CO2 produced, m(CO2),

mass of H2O produced, m(H2O).

Depending on the compound, other products may also form. In many introductory problems the unknown contains C and H, and possibly O (a CHO compound). More advanced cases include N, S, or halogens; those require additional measurements or assumptions.

2.7.2 Underlying chemical idea: carbon → CO2, hydrogen → H2O

Complete combustion of a generic hydrocarbon CxHy produces carbon dioxide and water:

If oxygen is present in the original compound (CxHyOz), the products are still CO2 and H2O under complete combustion; the oxygen in products comes partly from O2 and partly from the compound itself:

For combustion analysis, the crucial point is that:

Every mole of CO2 contains exactly 1 mole of C atoms.

Every mole of H2O contains exactly 2 moles of H atoms.

These fixed relationships allow us to count atoms in the original sample by counting molecules of combustion products.

2.7.3 The combustion analysis algorithm (CHO compounds)

For a compound containing only C, H, and O, the standard procedure is:

1. Convert measured masses m(CO2) and m(H2O) to moles.

2. Determine moles of C from moles of CO2:

3. Determine moles of H from moles of H2O:

4. Convert n(C) and n(H) to masses of C and H in the original sample.

5. Find the mass of oxygen by difference:

6. Convert m(O) to moles n(O).

7. Form the mole ratio C:H:O by dividing by the smallest mole value, then convert to small whole numbers to obtain the empirical formula.

Why oxygen is found “by difference.” In a CHO combustion experiment, oxygen in CO2 and H2O comes partly from the sample and partly from the O2 supplied. Therefore, the oxygen content of the original sample cannot be determined directly from the amounts of CO2 and H2O alone. However, if the sample contains only C, H, and O, then the sample mass must equal the sum of masses of C, H, and O in the compound. Subtracting the masses of C and H (which are determined from the products) reveals the oxygen mass.

2.7.4 Worked example: empirical formula of a CHO compound

Example 2.26: Combustion analysis with oxygen by difference

A 1.500 g sample of an unknown compound containing only C, H, and O is completely combusted. The reaction produces 2.200 g of CO2 and 0.900 g of H2O. Determine the empirical formula of the compound.

Step 1: Convert CO2 and H2O masses to moles.

n(CO2) | = 2.200 g × = 0.04999 mol |  |
n(H2O) | = 0.900 g × = 0.04996 mol. |  |

Step 2: Determine moles of C and H. Each CO2 contains 1 C:

Each H2O contains 2 H:

Step 3: Convert moles of C and H to masses.

m(C) | = 0.04999 mol × 12.011 g mol⁻¹ = 0.600 g |  |
m(H) | = 0.09992 mol × 1.008 g mol⁻¹ = 0.1007 g. |  |

Step 4: Determine oxygen mass by difference.

Step 5: Convert oxygen mass to moles.

Step 6: Form the simplest mole ratio. Divide all by the smallest (approximately 0.04994):

[picture] | = = 1.001 ≈ 1 |  |
[picture] | = = 2.001 ≈ 2 |  |
[picture] | = = 1.000 ≈ 1. |  |

Thus the empirical formula is

2.7.5 Connecting combustion analysis to percent composition and molecular formulas

Combustion analysis yields moles (and thus mass fractions) of each element. Once the empirical formula is known, you can:

compute percent composition directly from the empirical formula (as in the previous section), and

if the molar mass of the compound is known, determine the molecular formula by multiplying the empirical formula by the integer

2.7.6 Compounds containing nitrogen (CHN): a common extension

Many organic compounds contain nitrogen. In complete combustion:

carbon still becomes CO2,

hydrogen still becomes H2O,

nitrogen often becomes N2 (under many combustion-analysis conditions) rather than NOx.

If the mass (or moles) of N2 produced is measured in addition to CO2 and H2O, then moles of N in the original sample can be found:

If N2 is not measured, nitrogen content cannot generally be determined from CO2 and H2O alone without additional data (such as percent composition from elemental analysis).

2.7.7 Precision and error propagation: why careful bookkeeping matters

Combustion analysis is indirect: we infer the composition of the original sample from measured product masses. Small measurement errors can shift mole ratios, especially when one element is present in a small fraction. Several practical guidelines improve reliability:

Carry extra significant figures through intermediate calculations and round at the end.

Check for physical plausibility: computed m(C) and m(H) must be less than the sample mass; oxygen by difference must be nonnegative.

Treat near-integers thoughtfully: mole ratios such as 1.49 or 1.51 often indicate ; ratios near 1.33 indicate . Multiply all ratios by a small integer to obtain whole numbers, as in empirical-formula work.

Be explicit about assumptions: oxygen-by-difference is only valid if the compound contains only C, H, and O.

2.7.8 Common pitfalls

Pitfall 1: Confusing oxygen in products with oxygen in the sample

The oxygen atoms in CO2 and H2O are not a direct measure of oxygen in the original compound because O2 supplies oxygen to the products. Oxygen must be found by difference (CHO case) or by independent measurement.

Pitfall 2: Forgetting the factor of 2 for hydrogen

Because each mole of water contains 2 moles of H atoms, the conversion is

not n(H) = n(H2O).

Pitfall 3: Switching from atoms to molecules midstream

Combustion analysis counts atoms in the original compound, but the measured products are molecules. Always make the conversion explicitly through moles and the fixed atom-per-molecule relationships.

Combustion analysis determines empirical formulas by combusting a known mass of an unknown compound and measuring the masses of CO2 and H2O produced.

Moles of C are found directly from CO2:

Moles of H are found from H2O:

For CHO compounds, oxygen is determined by mass difference:

Convert element moles to the simplest whole-number ratio to obtain the empirical formula; use molar mass (if available) to determine the molecular formula.

The next step in mastering stoichiometry is extensive practice. The problem set at the end of the chapter integrates mole concepts, molar mass, equation balancing, reaction stoichiometry, limiting reagents, yield, and combustion analysis.

2.8 Problems

2.8.1 Mole–particle conversions on a macroscopic scale

A backyard swimming pool is filled with 6.50 × 10⁴ L of water at room temperature. Assume the density of water is 1.00 g mL⁻¹ and the molar mass of water is 18.015 g mol⁻¹.

Calculate the mass of water in the pool in kilograms.

Calculate the amount of substance of water in moles.

Calculate the number of water molecules in the pool.

Calculate the total number of hydrogen atoms present in the pool.

Report all answers with appropriate significant figures.

2.8.2 Molar mass of a complex hydrated salt

A laboratory reagent is labeled (NH4)2Fe(SO4)2 ⋅6H2O (Mohr’s salt).

Determine the molar mass of (NH4)2Fe(SO4)2 ⋅6H2O.

How many moles of formula units are present in a 12.50 g sample?

How many moles of Fe2+ ions are present in that sample?

How many water molecules of hydration are present in that sample?

2.8.3 Empirical formula from percent composition (mineral analysis)

A mineral sample is found (by mass percent) to contain 43.64% P and 56.36% O. Assume the mineral contains only phosphorus and oxygen.

Determine the empirical formula of the mineral.

From your result, compute the mass percent of P in your empirical formula and compare to the given data (quantify the discrepancy, if any).

If the molar mass of the mineral is determined to be 283.89 g mol⁻¹, determine the molecular formula (or formula unit) consistent with this molar mass.

2.8.4 Balancing a complex combustion reaction with fractional coefficients

An unknown hydrocarbon is represented as C7H16. Write and balance the complete combustion reaction of C7H16 in O2 to form CO2 and H2O.

First balance using inspection but allow a fractional coefficient for O2.

Then multiply through to obtain the smallest set of whole-number coefficients.

Verify conservation of C, H, and O atoms explicitly by counting atoms on each side.

2.8.5 Stoichiometry involving density of a liquid reactant

Nitric acid can be produced by the reaction:

Balance the equation.

A reaction vessel contains 35.0 g of NO2 and 25.0 mL of liquid water. Assume the density of water is 0.997 g mL⁻¹ at the reaction temperature. Determine the limiting reagent.

Calculate the theoretical yield (in grams) of HNO3.

Calculate the mass (in grams) of the excess reactant remaining after completion.

2.8.6 Limiting reagent in a three-reactant industrial process

Titanium(IV) chloride is produced by the high-temperature reaction:

A feed mixture contains 5.00 mol TiO2, 8.00 mol Cl2, and 12.0 mol C.

Identify the limiting reagent.

Calculate the theoretical amount (in moles) of TiCl4 that can be produced.

Determine how many moles of each excess reactant remain after the reaction goes to completion.

If TiCl4 is condensed and collected with an 87.0% yield (based on limiting reagent), calculate the actual amount (in moles) collected.

2.8.7 Theoretical and percent yield in a multi-step synthesis

A two-step synthesis is carried out:

Step 1: A → B is 1:1 in moles. You start with 25.0 g of A (ℳA = 125.0 g mol⁻¹) and obtain 16.8 g of B (ℳB = 140.0 g mol⁻¹).

Step 2: B + C → D is 1:1:1 in moles. You react all the B produced with 12.0 g of C (ℳC = 60.0 g mol⁻¹) and collect 18.5 g of D (ℳD = 200.0 g mol⁻¹).

Calculate the percent yield of Step 1.

Determine the limiting reagent in Step 2.

Calculate the theoretical yield (in grams) of D for Step 2.

Calculate the percent yield of Step 2.

Calculate the overall percent yield of the two-step synthesis, defined as (moles of D actually obtained)/(moles of D theoretically obtainable from initial A) ×100%.

2.8.8 Combustion analysis of a compound containing C, H, and N

A 1.250 g sample of an unknown compound containing only C, H, and N is completely combusted. The products are collected and the following masses are measured:

In a separate measurement, nitrogen from the sample is converted quantitatively to N2(g), and 0.224 L of N2 is collected at 1.00 atm and 273.15 K.

Determine the moles of C and H in the original sample from the combustion data.

Determine the moles of N in the original sample from the N2 data (assume ideal gas behavior and use 22.414 Lmol⁻¹ at STP).

Determine the empirical formula of the compound.

If the molar mass of the compound is 108.14 g mol⁻¹, determine the molecular formula.

2.8.9 Reverse stoichiometry: required reactant mass for a target yield

Aluminum reacts with chlorine gas to form aluminum chloride:

You wish to produce 50.0 g of AlCl3 as actual product. The reaction is known (from past runs) to proceed with a 78.0% yield, based on the limiting reagent. Chlorine gas is provided in excess.

Calculate the theoretical yield of AlCl3 required to obtain 50.0 g actual yield.

Calculate the moles of AlCl3 corresponding to that theoretical yield.

Determine the moles of Al required.

Determine the mass of Al required (in grams).

2.8.10 Conceptual problem: atomic scale vs. molecular scale vs. laboratory scale

Answer the following conceptual questions clearly and concisely. Use dimensional analysis and explicit statements of “entity” (atoms, molecules, formula units) where

appropriate.

Explain why the statement “one mole of C has a mass of 12 g” is approximately true, and identify what is exact and what is not in that statement.

A student claims: “Because H2O has two hydrogen atoms, 1 mol of water contains 2NA molecules.” Identify the error and write the correct relationship.

Two sealed flasks each contain 1.00 mol of gas at the same temperature and pressure: Flask A contains He; Flask B contains CO2. Compare (with justification) the following quantities between the flasks: (i) number of molecules, (ii) total mass, (iii) total number of atoms.

In your own words, explain why balanced equation coefficients correspond to mole ratios rather than mass ratios, and give one concrete example.

Chapter 3
Reactions in Aqueous Solution

Water is the universal solvent, a medium whose unique polarity and hydrogen-bonding capabilities create the stage for the vast majority of biochemical and environmental chemistry. In this chapter, we move beyond the static description of atoms and molecules to the dynamic interactions of species in solution. We will rigorously define the thermodynamic and electrostatic principles that dictate whether substances dissolve, dissociate, or precipitate, and we will categorize the driving forces behind the three major classes of aqueous reactions. Finally, we bridge the gap between the molecular scale and the laboratory bench by introducing molarity and volumetric analysis, essential tools for quantifying matter in the liquid phase.

3.1 General Properties of Aqueous Solutions

Many of the reactions studied in general chemistry occur in aqueous solution, meaning that reactants are dissolved in water and the reacting species move and collide while surrounded by water molecules. To understand why certain substances dissolve, why some dissolved substances conduct electricity while others do not, and why solution reactions often proceed rapidly, we need a molecular-level picture of water as a solvent and the energetic changes that accompany dissolution.

In earlier discussions of bonding and intermolecular forces, two ideas were central:

Electrostatic attraction (between full charges in ionic compounds, and between partial charges in polar molecules).

Competition between interactions: separating particles costs energy, while forming new attractions releases energy.

Dissolving a substance in water is a direct application of both ideas. Water can stabilize ions and polar molecules by forming strong solute–solvent attractions, and the balance of energetic and entropic contributions determines whether the dissolution process is thermodynamically favorable.

3.1.1 Water as a Solvent: Polarity and Hydrogen Bonding

Molecular structure and polarity

A water molecule has two polar O–H bonds and a bent geometry (bond angle ≈ 104.5∘). Oxygen is more electronegative than hydrogen, so electron density is pulled toward oxygen. As a result, water has:

a partial negative charge on oxygen (δ⁻),

partial positive charges on the hydrogens (δ⁺),

a substantial permanent dipole moment.

This polarity enables water to interact strongly with both:

ions (via ion–dipole attractions), and

polar molecules (via dipole–dipole interactions and hydrogen bonding).

Hydrogen bonding network

In liquid water, each molecule participates in a dynamic, three-dimensional hydrogen-bond network. A hydrogen bond forms when a hydrogen atom covalently bound to an electronegative atom (here, oxygen) is attracted to a lone pair on another electronegative atom. Hydrogen bonding in water is strong enough to influence boiling point, viscosity, and heat capacity—but it is also labile, breaking and reforming on very short timescales. This fluid network is crucial: it allows water to reorganize around solute particles and create hydration shells.

3.1.2 Hydration, Solvation, and Dissociation

When a solute dissolves, it becomes surrounded by solvent molecules. The general term for this process is solvation. When the solvent is water, solvation is called hydration.

Hydration of ions: ion–dipole interactions and hydration shells

If an ionic compound dissolves, individual cations and anions separate from the crystal and become hydrated. Water molecules orient themselves to maximize electrostatic attraction:

Around a cation (positive ion), the δ⁻ oxygen ends point inward.

Around an anion (negative ion), the δ⁺ hydrogen ends point inward.

The structured set of water molecules closest to the ion is the primary hydration shell. Beyond this region, water is less ordered but still influenced by the ion’s electric field.

Solvation of molecular solutes: dissolution without dissociation

Many molecular compounds dissolve in water without forming ions. For example, sucrose (C12H22O11) dissolves because it contains many polar O–H and C–O bonds. Water can hydrogen-bond to these sites and stabilize sucrose molecules separated from the solid. Importantly, the sucrose molecules remain intact in solution; dissolution occurs without dissociation into ions.

Molecular-level comparison: NaCl vs. sucrose in water

The distinction between ionic dissociation and molecular dissolution is foundational for predicting solution behavior, especially electrical conductivity and reaction pathways.

The key takeaway from this diagram is that the identity of the dissolved particles differs:

NaCl(aq) consists primarily of Na+ and Cl− ions dispersed in water.

Sucrose(aq) consists of neutral sucrose molecules dispersed in water.

This difference controls both the thermodynamics of dissolution and the electrical properties of the resulting solution.

3.1.3 The Energetics of Dissolution: ΔH, ΔS, and ΔG

A three-step energy model

At the molecular level, dissolving a solute can be viewed as three conceptual steps:

Separate solute particles (break solute–solute attractions).

Separate solvent molecules locally (disrupt solvent–solvent attractions to make space).

Form solute–solvent attractions (hydrate/solvate the solute particles).

Each step has an associated enthalpy change. A useful bookkeeping expression is

Here:

ΔHsolute is typically positive (energy required to separate ions or molecules from the solid/liquid solute),

ΔHsolvent is typically positive (energy required to disrupt water–water interactions),

ΔHmix is typically negative (energy released when water forms stabilizing interactions with the solute).

Ionic solutes. For an ionic crystal, separating ions corresponds to overcoming the lattice energy (a large, stabilizing energy associated with the ionic solid). Hydration of ions is also strong. Whether ΔHsolution is negative (exothermic) or positive (endothermic) depends on the competition between lattice energy and hydration energy.

Molecular solutes. For molecular solids, ΔHsolute reflects intermolecular forces (dispersion, dipole–dipole, hydrogen bonding) holding the solid together. Solvation may be strong if the solute is polar and can hydrogen-bond; it is weak if the solute is nonpolar.

Entropy and the “ordering” effect of hydration

Enthalpy alone does not determine whether dissolution occurs. The thermodynamic criterion for spontaneity at constant T and P is the sign of Gibbs free energy:

Dissolving a solid often increases disorder because particles become dispersed throughout the solvent, suggesting ΔS > 0. However, hydration can also order nearby water molecules, especially around small, highly charged ions (high charge density). This local ordering can reduce entropy and partially offset the dispersal effect.

Therefore, ΔS for dissolution is not universally positive or negative; it depends on a balance between:

dispersal of solute particles (tends to increase entropy),

structuring of solvent in hydration shells (tends to decrease entropy).

Why some substances dissolve and others do not

Dissolution is favorable when ΔG < 0, which can occur in several ways:

ΔH < 0 and ΔS > 0: strongly favorable (many salts and polar solutes).

ΔH > 0 but ΔS sufficiently positive: favorable at higher T (some endothermic dissolutions).

ΔH < 0 but ΔS < 0: may still be favorable at lower T, but can become less favorable as T increases.

This framework explains why solubility can depend strongly on temperature and on the nature of both solute and solvent.

3.1.4 Electrolytes and Nonelectrolytes: Dissolved Particles Determine Conductivity

Electrical conductivity in solution

Pure water conducts electricity very poorly because it contains extremely low concentrations of ions. When dissolved species include ions, they act as charge carriers. Under an applied electric field, cations migrate toward the negative electrode (cathode) and anions migrate toward the positive electrode (anode), producing an electric current through the solution.

Thus, the conductivity of an aqueous solution depends primarily on:

the concentration of ions present, and

the mobility of ions (how easily they move through the solvent).

Classification by degree of ion formation

Solutes are classified according to how completely they produce ions in water.

Strong electrolytes. A strong electrolyte produces (essentially) only ions in aqueous solution. This category includes:

most soluble ionic compounds (e.g., NaCl, KNO3),

strong acids (e.g., HCl, HNO3),

strong bases (e.g., NaOH, KOH).

For a soluble ionic compound, we represent dissociation as

The single arrow emphasizes that, for a strong electrolyte, the dissolved solute exists predominantly as separated ions (not as intact formula units).

Weak electrolytes. A weak electrolyte produces a mixture of ions and intact molecules in water. This behavior occurs when ion formation is thermodynamically allowed but not strongly favored. A classic example is acetic acid:

The double arrow indicates an equilibrium: both reactants and products are present, and the extent of ion formation is limited.

Nonelectrolytes. A nonelectrolyte dissolves without forming ions and therefore conducts electricity poorly. Many molecular compounds fall into this category, including sugars (such as sucrose) and many alcohols. Their dissolution can be represented simply as a physical dispersal:

A molecular-level summary: why conductivity differs

It is tempting to think that “more soluble” always means “more conductive,” but conductivity is not controlled by solubility alone; it is controlled by the presence of ions.

NaCl(aq): ions form; solution conducts well.

sucrose(aq): molecules remain neutral; solution conducts poorly even though sucrose is quite soluble.

This distinction is central for writing ionic equations in later sections. In precipitation, acid–base, and redox reactions in water, it is often the ions (not the undissociated compounds) that are the reacting species.

3.1.5 Weak Electrolytes and Chemical Equilibrium (Conceptual Preview)

Some aqueous reactions depend on species that are present only in small amounts because they are generated by equilibria. Weak acids and weak bases are important examples. Their key features are:

they partially ionize in water,

the extent of ionization depends on an equilibrium constant,

the solution contains both molecular and ionic species simultaneously.

At this stage, the essential point is qualitative: equilibrium in solution is a natural consequence of competing tendencies—forming ions can be stabilized by hydration, but creating charge separation has energetic and entropic costs. Later chapters will quantify these ideas using equilibrium constants and pH, but the molecular picture developed here is the foundation.

3.1.6 Common Pitfalls and Productive Habits

Do not confuse “dissolves” with “dissociates.” Many substances dissolve without forming ions.

Use the correct representation. Strong electrolytes are written as producing ions; weak electrolytes are written with equilibrium arrows.

Ask: what particles are actually in solution? This question guides conductivity predictions and sets up net ionic equations.

3.1.7 Section Summary

Water’s polarity and hydrogen bonding make it an unusually effective solvent for ions and polar molecules.

Dissolution involves breaking solute–solute and solvent–solvent interactions and forming solute–solvent interactions; the sign of ΔHsolution depends on their balance.

Whether dissolution is thermodynamically favorable is determined by ΔG = ΔH − TΔS, so both enthalpy and entropy matter.

Electrical conductivity in aqueous solution arises from mobile ions. Strong electrolytes produce many ions, weak electrolytes produce some ions via equilibrium, and nonelectrolytes produce essentially no ions.

3.2 Precipitation Reactions and Solubility Rules

In the previous section, we established a molecular picture of aqueous solutions: ionic compounds that dissolve produce mobile ions (strong electrolytes), whereas many molecular solutes dissolve without ion formation (nonelectrolytes). This distinction is the key to understanding precipitation reactions, which occur when ions in solution combine to form an insoluble ionic compound that separates as a solid.

Precipitation reactions are among the most visually striking aqueous reactions: two clear solutions are mixed and a cloudy solid appears. Chemically, the driving force is the formation of a stable solid lattice that removes ions from solution.

3.2.1 What Is a Precipitation Reaction?

A precipitation reaction is an aqueous reaction in which a solid ionic compound forms from ions that were initially dissolved. The solid is called a precipitate.

Most precipitation reactions are metathesis (double-displacement) reactions, in which cations and anions exchange partners:

Here A+ and B+ are cations, and X− and Y− are anions. Whether products remain aqueous or form solids depends on their solubilities in water.

3.2.2 Solubility: A Macroscopic Observation with a Molecular Cause

Meaning of “soluble” and “insoluble”

In general chemistry, a salt is described as:

soluble if it dissolves to an appreciable extent in water, producing ions (an aqueous solution),

insoluble if it dissolves only to a very small extent, so that a solid phase persists.

At the molecular level, solubility reflects competition between:

the lattice energy of the solid (favoring the intact crystal), and

the hydration (solvation) energy of the ions (favoring separation into hydrated ions).

If hydration can compensate for the energetic cost of separating ions from the lattice (and the overall ΔG is favorable), the salt tends to be soluble. If not, the solid tends to persist and the salt is classified as insoluble.

Solubility is not all-or-nothing

Even “insoluble” salts dissolve to some extent, establishing an equilibrium between solid and dissolved ions. In this chapter, we use solubility rules as a practical predictive tool and defer quantitative treatment (solubility products, Ksp) to later discussions.

3.2.3 A Systematic Strategy for Predicting Precipitates

When two aqueous ionic solutions are mixed, follow this workflow:

Step 1: Identify the ions present (strong electrolytes dissociate)

Write each soluble ionic compound as separated ions. For example:

This step uses the earlier concept of strong electrolytes: if a compound is soluble and ionic, it exists primarily as ions in solution.

Step 2: Predict the possible products (metathesis exchange)

Swap anions between cations to form two candidate products, keeping charge neutrality and correct formulas:

Step 3: Use solubility rules to determine physical states

Decide whether each product is soluble (aq) or insoluble (s). If either product is insoluble, it will precipitate (assuming sufficient ion concentrations).

Step 4: Write balanced molecular, total ionic, and net ionic equations

The molecular equation shows intact compounds with phase labels. The total ionic equation shows strong electrolytes as ions. The net ionic equation removes spectator ions to emphasize the actual chemical change.

This process does more than provide bookkeeping: it reveals the driving force of the reaction—the formation of a solid phase.

3.2.4 Solubility Rules (Empirical Patterns)

Solubility rules are experimentally derived generalizations about common ionic

compounds in water at room temperature. They are not perfect, but they are reliable for most introductory problems.

Rules that predict high solubility

The following are generally soluble:

Group 1 (alkali metal) and ammonium salts: Li+, Na+, K+, Rb+, Cs+, and NH 4+ salts are soluble with almost all anions.

Nitrates, acetates, and perchlorates: NO3−, CH 3CO2− (or C2H3O2−), and ClO 4− salts are soluble.

Most chlorides, bromides, iodides: Cl−, Br−, I− salts are soluble except with certain cations (notably Ag+, Pb2+, Hg 22+).

Most sulfates: SO42− salts are soluble except with Ba2+, Sr2+, Pb2+ (and often Ca2+ is treated as low solubility in many courses).

Rules that predict low solubility

The following are generally insoluble unless paired with Group 1 cations or NH4+:

Carbonates: CO32−

Phosphates: PO43−

Sulfides: S2−

Hydroxides: OH− (with notable soluble/partly soluble exceptions such as Ba(OH)2, Sr(OH)2, and Ca(OH)2)

How to use the rules efficiently

A productive habit is to test for “always soluble” categories first. If a compound contains NO3− or a Group 1 cation, you can immediately label it (aq). If not, check whether it falls into a mostly soluble category with exceptions (halides, sulfates), and then whether it is in a mostly insoluble class (carbonates, phosphates, hydroxides, sulfides).

3.2.5 Spectator Ions and Ionic Equations

Why ionic equations matter

In aqueous solution, ions move independently. When a reaction occurs, typically only a subset of ions actually undergo a chemical change. The ions that remain unchanged are spectator ions. Removing them from the equation clarifies what is truly happening.

Definitions

Molecular equation: compounds written as formula units, with phase labels.

Total ionic equation: soluble strong electrolytes written as dissociated ions.

Net ionic equation: only species that undergo change, with spectators removed.

3.2.6 Worked Example: BaCl2(aq) Mixed with AgNO3(aq)

1. Predict products and precipitation

Consider mixing aqueous BaCl2 and AgNO3. The metathesis products are Ba(NO3)2 and AgCl.

Using solubility rules:

All nitrates are soluble ⇒ Ba(NO3)2(aq).

Most chlorides are soluble, but AgCl is an important exception ⇒ AgCl(s).

Therefore, a precipitate of AgCl(s) forms.

2. Molecular equation

Balance atoms and charge at the formula-unit level:

3. Total ionic equation

Dissociate strong electrolytes (aqueous ionic compounds), but do not dissociate

solids:

Ba2+(aq) + 2Cl−(aq) + 2Ag+(aq) + 2NO 3−(aq) → Ba2+(aq) + 2NO 3−(aq) + 2AgCl(s) |  |

4. Net ionic equation

Cancel spectator ions: Ba2+ and NO 3− appear unchanged on both sides.

This net ionic equation captures the essence of the precipitation event: hydrated ions collide, and if the combination corresponds to an insoluble lattice, a solid forms and removes ions from solution.

3.2.7 Recognizing “No Reaction” Cases

Sometimes metathesis predicts two products that are both soluble. In that case, no precipitate forms, and there is no net ionic equation beyond the ions remaining dispersed.

Example

Mixing NaNO3(aq) and KCl(aq) predicts NaCl(aq) and KNO3(aq). All are soluble (Group 1 salts; nitrates), so no precipitation occurs:

The “reaction” is simply mixing; the solution composition changes, but no new phase forms.

3.2.8 What Drives Precipitation? A Particle-Level View

When an insoluble salt forms, ions transition from a dispersed, hydrated state to a highly ordered ionic lattice. This is favorable when the lattice formation is energetically strong enough (and overall ΔG becomes negative) to overcome the stabilizing hydration of separate ions.

Two practical consequences follow:

Precipitation removes ions from solution, often shifting related equilibria (a theme that becomes important in equilibrium chapters).

The net ionic equation shows the driving force: the formation of (s) is the chemical change.

3.2.9 A Checklist for Success

Before concluding that a precipitate forms, verify each item:

The reactants are present as ions (they are soluble strong electrolytes).

The product formulas are written correctly (charge-balanced).

At least one product is insoluble by solubility rules.

The equation is balanced before converting to ionic form.

Spectator ions are removed correctly in the net ionic equation.

3.2.10 Section Summary

Precipitation reactions are typically metathesis reactions in which ions exchange partners to form an insoluble ionic solid.

Solubility rules provide a systematic, empirical method to predict whether a proposed ionic compound will be aq or s.

Writing molecular, total ionic, and net ionic equations isolates the chemical change and identifies spectator ions.

The appearance of a solid phase is a strong driving force in aqueous reactions because it removes ions from solution and forms a stable lattice.

3.3 Acid-Base Reactions

The previous sections developed two essential tools for understanding reactions in water: (1) a molecular picture of dissolved species (strong electrolytes dissociate into ions; weak electrolytes partially ionize), and (2) a systematic method for predicting precipitation reactions and writing net ionic equations. Acid–base reactions are the next major class of aqueous reactions. Like precipitation, they are often driven by the formation of a particularly stable product—but here the key event is proton transfer and, in many cases, the formation of liquid water or a gas.

3.3.1 Arrhenius Acids and Bases: The First Operational Definition

Historically, acids and bases were classified by how they change the composition of water.

Arrhenius acid

An Arrhenius acid is a substance that increases the concentration of H+(aq) when dissolved in water. Because a bare proton cannot exist as an isolated species in water (it is far too small and strongly attracted to electron density), H+ is better understood as being associated with water molecules to form hydronium:

. In introductory work, the symbols H+(aq) and H 3O+(aq) are often used interchangeably to represent acidic solutions, but H3O+ reflects the more realistic molecular picture.

Common Arrhenius acids include strong acids such as HCl, HBr, and HNO3, which produce essentially complete ionization in water:

Arrhenius base

An Arrhenius base is a substance that increases the concentration of OH−(aq) when dissolved in water. Many bases are ionic hydroxides that dissociate completely (strong electrolytes), for example:

The Arrhenius definition is practical and works well for many aqueous reactions, especially neutralization, but it is limited: it applies only to water as solvent and to substances that directly produce H+ or OH−.

3.3.2 Brønsted–Lowry Perspective: Proton Transfer as the Central Event

A more general and conceptually powerful framework is the Brønsted–Lowry definition:

A Brønsted–Lowry acid is a proton donor.

A Brønsted–Lowry base is a proton acceptor.

This definition emphasizes what actually happens in an acid–base reaction: a proton is transferred from one species to another. In aqueous solution, water often participates directly as the proton acceptor (base) or proton donor (acid).

Example: weak acid ionization is an equilibrium

Acetic acid is a weak electrolyte; in water it establishes an equilibrium:

Here, CH3CO2H donates a proton (acid) and H2O accepts a proton (base), forming H3O+. The reverse reaction also occurs: CH 3CO2− can accept a proton (base behavior), and H3O+ can donate a proton (acid behavior). Acid–base chemistry therefore connects naturally to equilibrium concepts introduced earlier for weak electrolytes.

Conjugate acid–base pairs (introductory)

When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid. In the acetic acid example:

This pairing helps track proton transfer consistently and becomes essential in later quantitative acid–base treatments.

3.3.3 Strong vs. Weak Acids and Bases (Qualitative)

Because aqueous reactions depend on the particles present in solution, it is important to distinguish strong and weak acids/bases in the same way we distinguish strong and weak electrolytes.

Strong acids are strong electrolytes: in water they produce nearly complete formation of H3O+ and the conjugate base anion. Examples commonly treated as strong in general chemistry include HCl, HBr, HI, HNO3, HClO4, and (for the first proton) H2SO4.

Weak acids ionize only partially: both the molecular acid and ions are present (equilibrium).

Strong bases (such as Group 1 hydroxides and some Group 2 hydroxides) dissociate to produce OH− to a large extent.

Weak bases react with water to produce OH− only partially (equilibrium).

In this chapter’s qualitative treatment, the most important consequence is:

Strong acids and strong bases are written as ions in total ionic equations; weak acids and weak bases are usually written as molecules (because they do not fully ionize).

3.3.4 Neutralization Reactions: Formation of Water as a Driving Force

Molecular description

A neutralization reaction is an acid–base reaction in which H+ (or H 3O+) and OH− combine to form water:

Because H2O(l) is a stable molecular product and is not an ion, the reaction removes charge carriers from solution. This strong tendency to form water is a primary driving force for many acid–base reactions in water.

Strong acid–strong base neutralization: a canonical net ionic equation

Consider HCl(aq) mixed with NaOH(aq).

Molecular equation

Total ionic equation Because HCl, NaOH, and NaCl are strong electrolytes in water:

Net ionic equation Cancel spectators Na+ and Cl−:

This result is remarkably general: every strong acid–strong base neutralization has the same net ionic equation. The particular salt formed (NaCl, KNO3, etc.) depends on spectator ions.

Stoichiometry: monoprotic vs. polyprotic acids

Acids differ in how many protons they can donate. For example:

Monoprotic acids donate one proton per molecule (e.g., HCl, HNO3).

Diprotic acids can donate two protons (e.g., H2SO4, H2CO3).

The stoichiometric neutralization requirement depends on the number of acidic protons and hydroxide ions:

Even though the net ionic core still involves H+ and OH−, the balanced molecular equation must reflect the correct mole ratio.

Neutralization when the acid or base is weak

If an acid is weak (e.g., CH3CO2H), it is written largely as a molecule in the total ionic equation. Yet neutralization can still proceed because OH− consumes H 3O+ and effectively pulls the acid dissociation equilibrium forward.

For example, acetic acid reacting with hydroxide:

This net ionic equation shows a proton transfer from the weak acid to hydroxide. The key point is procedural:

Dissociate strong electrolytes.

Keep weak electrolytes largely undissociated.

Form the net ionic equation by canceling spectators.

3.3.5 Gas-Forming Acid–Base Reactions

Not all acid–base reactions are driven primarily by water formation. Another powerful driving force is the formation of a gas that escapes from solution. Gas evolution removes product from the reaction mixture, promoting completion and

producing observable bubbling (effervescence).

Two important families in aqueous chemistry involve carbonates and sulfites.

Carbonates and bicarbonates: formation of CO2(g)

Carbonate ions act as bases in water and react with acids. The overall reaction typically produces carbon dioxide gas and liquid water.

Example with carbonate Consider mixing Na2CO3(aq) with HCl(aq).

Molecular equation

Net ionic equation Dissociate strong electrolytes and cancel spectators (Na+, Cl−):

Chemical logic This net ionic equation can be interpreted as a sequence of proton transfers leading to carbonic acid, followed by decomposition:

Even if these intermediate steps are not written explicitly in introductory problems, they explain why CO2 is produced and why the reaction is strongly product-favored.

Bicarbonate Bicarbonate (HCO3−) requires only one proton to form CO 2 and water:

This stoichiometric difference is important in quantitative work.

Sulfites: formation of SO2(g)

Sulfite ions also react with acids to form a gas, sulfur dioxide:

In the laboratory, SO2 has a sharp, pungent odor and is toxic; such reactions should be carried out with appropriate ventilation and safety practices.

Recognizing gas-forming patterns

A practical recognition strategy:

If an acid is mixed with a carbonate or bicarbonate, expect CO2(g).

If an acid is mixed with a sulfite or bisulfite, expect SO2(g).

In each case, gas evolution is part of the reaction driving force, analogous to how precipitate formation drives precipitation reactions.

3.3.6 Acid–Base Reactions That Also Involve Precipitation

In real mixtures, more than one driving force can occur simultaneously. For example, an acid might react with a basic anion to form water and a gas, while spectator ions form an insoluble salt (or vice versa). The recommended approach is still systematic:

Identify ions present (strong electrolytes dissociate).

Consider likely driving forces: formation of H2O(l), formation of a gas, formation of a precipitate.

Use solubility rules to check for solids among potential products.

Write the net ionic equation that reflects the driving force(s).

3.3.7 Writing Net Ionic Equations for Acid–Base Reactions: A Procedure

Write the balanced molecular equation with phase labels.

Write the total ionic equation by dissociating strong electrolytes ((aq) ionic compounds, strong acids, strong bases).

Do not dissociate weak acids, weak bases, or insoluble solids.

Cancel spectator ions to obtain the net ionic equation.

Check: atoms balanced? charge balanced? phases reasonable?

3.3.8 Section Summary

Acid–base reactions in water are fundamentally proton-transfer processes.

Neutralization reactions are often driven by the formation of H2O(l), with the central net ionic equation H+ + OH− → H 2O for strong acid–strong base reactions.

Weak acids and weak bases participate in neutralization even though they do not fully dissociate; they are typically written as molecules in total ionic equations.

Many acid–base reactions are driven by gas formation, especially acids reacting with carbonates/bicarbonates (CO2(g)) and sulfites (SO2(g)).

The same systematic method used for precipitation reactions—identify dissolved species, predict products, and write net ionic equations—applies to acid–base chemistry, with an emphasis on H+, OH−, water formation, and gas evolution.

3.4 Oxidation-Reduction Reactions

The previous sections focused on two major classes of aqueous reactions: precipitation (driven by formation of an insoluble solid) and acid–base reactions (driven by proton transfer, often producing H2O(l) or a gas). A third broad class is oxidation–reduction (redox) reactions, in which the central event is electron transfer. Redox chemistry connects macroscopic observations (color changes, metal deposition, gas evolution, heat release) to microscopic changes in electron distribution and is foundational for topics such as corrosion, batteries, and metabolism.

This section develops three complementary tools:

the conceptual definition of oxidation and reduction,

a rigorous method for assigning oxidation numbers (oxidation states),

the activity series of metals for predicting spontaneous single-displacement reactions in water.

3.4.1 Oxidation and Reduction: Electron Transfer and Charge

Definitions

A redox reaction involves transfer of electrons from one species to another. Two processes occur simultaneously:

Oxidation: loss of electrons.

Reduction: gain of electrons.

The mnemonic OIL RIG (Oxidation Is Loss, Reduction Is Gain) is useful, but it is even better to connect these ideas to observable charge changes:

If a species loses electrons, its charge becomes more positive (or its oxidation state increases).

If a species gains electrons, its charge becomes more negative (or its oxidation state decreases).

Oxidizing and reducing agents

Because electron transfer always has a donor and an acceptor:

The reducing agent (reductant) donates electrons and is oxidized.

The oxidizing agent (oxidant) accepts electrons and is reduced.

Students often confuse these roles; the key is to focus on what the agent does to the other species. An oxidizing agent causes oxidation (by taking electrons), and a reducing agent causes reduction (by giving electrons).

Half-reactions

It is often helpful to split a redox reaction conceptually into two half-reactions:

The overall reaction must conserve both mass and charge, so electrons lost in oxidation must equal electrons gained in reduction.

3.4.2 Oxidation Numbers: A Bookkeeping System for Electron Flow

In covalent compounds and polyatomic ions, electrons are shared rather than fully transferred. Nevertheless, we can track electron redistribution using oxidation numbers (also called oxidation states). An oxidation number is the hypothetical charge an atom would have if bonding electrons were assigned to the more electronegative atom in each bond.

Oxidation numbers are a bookkeeping tool, not a claim that the atom literally has that integer charge in the real electron density distribution. Their power lies in enabling you to identify which atoms are oxidized and reduced, even in complex molecular reactions.

Rules for assigning oxidation numbers

The following rules are sufficient for most introductory redox problems:

Elements in their standard state have oxidation number 0. Examples: Zn(s), O2(g), Cl2(g), S8(s).

Monatomic ions have oxidation numbers equal to their ionic charges. Examples: Na+ is +1; Cl− is −1; Fe3+ is +3.

The sum of oxidation numbers equals the overall charge: for a neutral compound the sum is 0, and for a polyatomic ion the sum equals the ion charge.

Group 1 metals are almost always +1 in compounds; Group 2 metals are almost always +2.

Fluorine is always −1 in its compounds.

Hydrogen is usually +1 when bonded to nonmetals, and −1 in metal hydrides (e.g., NaH).

Oxygen is usually −2 in compounds, with important exceptions: peroxides (oxygen is −1, e.g., H2O2); superoxides (oxygen is − on average, e.g., KO2); oxygen bonded to fluorine (oxygen is positive, e.g., OF2, where O is +2).

Procedure: identifying what is oxidized and what is reduced

A systematic approach:

Assign oxidation numbers to each element in reactants and products.

Identify which element’s oxidation number increases (oxidation).

Identify which element’s oxidation number decreases (reduction).

The species containing the oxidized element is the reducing agent.

The species containing the reduced element is the oxidizing agent.

3.4.3 Common Categories of Redox Reactions

Redox reactions appear in several recurring patterns.

Combination (synthesis)

Two or more reactants form one product. Often, elements combine with oxygen:

Here Mg is oxidized (0 to +2) and O is reduced (0 to −2).

Decomposition

One reactant breaks into simpler products. A classic example:

This is a disproportionation reaction: the same element (oxygen) is both reduced and oxidized.

Combustion

A substance reacts with oxygen, often producing CO2 and H2O for hydrocarbons:

Carbon is oxidized and oxygen is reduced.

Single-displacement (metal displacement)

A more active metal displaces a less active metal ion from solution:

These reactions are especially important in aqueous chemistry and are predicted using the activity series.

3.4.4 The Activity Series of Metals

Meaning of the activity series

The activity series ranks metals by their tendency to be oxidized (to lose electrons) in aqueous solution. Metals higher in the series:

are more easily oxidized (stronger reducing agents),

can displace ions of metals lower in the series from solution.

For example, because Zn is above Cu in the activity series, Zn(s) can reduce Cu2+(aq) to Cu(s), while Zn(s) is oxidized to Zn2+(aq). Conversely, Cu(s) will not reduce Zn2+(aq) under typical conditions.

Predicting single-displacement reactions using the activity series

To predict whether a reaction such as

is spontaneous, compare M and N in the activity series:

If M is above N: reaction tends to occur (spontaneous).

If M is below N: reaction tends not to occur (no reaction).

The activity series is an empirical summary of many redox experiments and provides a practical decision tool in aqueous systems.

3.4.5 Aqueous Redox Example: Zn(s) in CuSO4(aq)

The reaction between a zinc strip and copper(II) sulfate solution is a classic demonstration because the macroscopic changes are dramatic and directly tied to microscopic electron transfer.

Macroscopic observations

When a clean zinc strip is placed in a blue CuSO4(aq) solution:

the blue color (due to Cu2+(aq)) fades as Cu2+ is consumed,

reddish-brown copper metal deposits on the zinc surface,

the zinc strip gradually dissolves as Zn2+(aq) forms.

Molecular and ionic description

The sulfate ion SO42− is a spectator ion in this process. The net ionic equation is:

This equation emphasizes the species undergoing change and parallels the net ionic equations used for precipitation and acid–base reactions.

Half-reaction analysis

Write the two half-reactions:

Oxidation (zinc loses electrons)

Reduction (copper(II) gains electrons)

Adding them cancels electrons and yields the net ionic equation. This makes the electron transfer explicit and shows conservation of charge.

Oxidation number changes

Zn: 0 → +2 (oxidation)

Cu: +2 → 0 (reduction)

Therefore:

Zn(s) is the reducing agent.

Cu2+(aq) is the oxidizing agent.

3.4.6 Visualizing Macro and Micro Views of Electron Transfer (Activity Series Context)

This split view emphasizes an important general principle in aqueous redox chemistry: what you see in the beaker (metal dissolving, another metal depositing, color change) is a direct consequence of electron transfer between a solid metal and dissolved metal ions. The activity series provides a shortcut for predicting the direction of this electron transfer without calculating energies explicitly.

3.4.7 Redox Reactions in Aqueous Solution: Spectator Ions and Net Ionic Form

Many redox reactions in water involve ionic compounds, so the same approach used earlier remains essential:

Write the balanced molecular equation (if given reactants as compounds).

Dissociate strong electrolytes to write the total ionic equation.

Cancel spectator ions to obtain the net ionic equation.

Identify oxidation and reduction either by oxidation numbers or half-reactions.

Example (spectator ion cancellation).

Total ionic:

Net ionic:

The sulfate ion does not change and is therefore a spectator ion.

3.4.8 Section Summary

Redox reactions involve electron transfer: oxidation is electron loss and reduction is electron gain.

Oxidation numbers provide a systematic way to identify which atoms are oxidized and reduced, even in covalent compounds.

The oxidizing agent is reduced (accepts electrons); the reducing agent is oxidized (donates electrons).

The activity series ranks metals by their tendency to be oxidized and allows prediction of spontaneous single-displacement reactions in aqueous solution.

The reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) illustrates how macroscopic observations (metal deposition, color change) arise from microscopic electron transfer and how spectator ions are removed in net ionic form.

3.5 Concentration of Solutions: Molarity

The previous sections treated aqueous reactions primarily at the level of which

species are present (ions vs. molecules) and what drives change (formation of a precipitate, water, a gas, or electron transfer). To make these ideas quantitative, we must describe how much of each dissolved species is present. In aqueous chemistry, concentration determines reaction direction, reaction extent, and the amounts of products that can form.

This section introduces molarity, the most common concentration unit for solution stoichiometry, and develops the practical calculations and laboratory practices used to prepare solutions of known molarity.

3.5.1 Concentration as an Intensive Property

Extensive vs. intensive quantities

A quantity is extensive if it depends on the amount of material (mass, volume, moles). A quantity is intensive if it does not depend on the total amount (density, temperature, concentration).

Concentration is intensive because it is defined as a ratio of two extensive quantities. If you double both the amount of solute and the volume of solution, the concentration remains unchanged.

Why concentration matters in aqueous reactions

In the reactions studied so far, “aq” indicates dissolved species, but the extent of precipitation, neutralization, or redox change depends on how many moles of the relevant ions are present. For example:

A precipitate forms only if enough ions are present to exceed the solubility limit (treated qualitatively here).

Neutralization depends on the mole ratio of H+ to OH−.

Metal displacement depends on the availability of oxidizing/reducing species in solution.

Molarity provides a direct bridge from measured solution volumes to moles of

dissolved solute.

3.5.2 Definition of Molarity

Molarity (M)

Molarity M is defined as moles of solute per liter of solution:

with units:

Interpreting molarity

A 1.00molL⁻¹ NaCl(aq) solution contains 1.00mol of NaCl formula units (and therefore approximately 1.00mol Na+ and 1.00mol Cl−, because it is a strong electrolyte) in every 1.00L of solution.

Key distinction: liters of solution, not solvent. The denominator is the final volume of the solution after dissolution and mixing, not the volume of water used. Dissolving solute can change volume slightly, so laboratory preparation targets a final volume mark (e.g., in a volumetric flask).

Unit consistency

Because M is in molL⁻¹:

volumes must be in liters (convert from mL: 1mL = 1 × 10⁻³ L),

moles come from mass using molar mass when preparing solutions from solids.

3.5.3 From Molarity to Moles (and Back)

The definition of molarity can be rearranged to the two conversions used constantly in solution stoichiometry:

Example: moles from volume and molarity

How many moles of KNO3 are present in 35.0mL of 0.200molL⁻¹ KNO3(aq)?

This conversion will later serve as the entry point to stoichiometric calculations in solution: measured volumes become moles.

3.5.4 Preparing Solutions from Solids

In laboratory work, you often prepare a solution of specified molarity by dissolving a measured mass of solid solute and diluting to a final volume.

Calculation strategy

Given desired molarity M and final volume V , compute required moles:

Convert moles to mass using molar mass Mm (here Mm is molar mass, not molarity):

Example: preparing a sodium carbonate solution

To prepare 250.0mL of 0.500molL⁻¹ Na2CO3(aq):

Using Mm(Na2CO3) ≈ 105.99g mol⁻¹,

A mass of 13.2g (to appropriate significant figures for the context) is weighed and diluted to a final volume of 250.0mL.

Laboratory technique: volumetric flask method

The conceptually correct preparation emphasizes the final volume:

Accurately weigh the solute (using an analytical balance when appropriate).

Transfer the solid to a clean volumetric flask (using a funnel if needed).

Add a portion of water and dissolve completely (swirl until no solid remains).

Add water until the bottom of the meniscus is exactly at the calibration mark.

Stopper and invert the flask multiple times to ensure uniform concentration.

Why mixing matters before final adjustment. If undissolved solute remains or the solution is not mixed, concentration gradients can exist and the meniscus reading may not correspond to the intended final composition.

Common sources of error (conceptual)

Using solvent volume instead of solution volume: measuring “250mL of water” and adding solute does not guarantee a 250mL solution.

Incomplete transfer: solute left on weighing paper or funnel reduces actual moles in solution.

Incomplete dissolution: undissolved solid means fewer moles are actually in solution.

Not mixing: concentration near the top may differ from concentration near the bottom.

3.5.5 Dilution and the Conservation of Moles

Solutions are frequently prepared by diluting a more concentrated stock solution. In a dilution, the number of moles of solute remains constant; only the volume changes.

Derivation of the dilution equation

Let:

M₁ and V ₁ be the molarity and volume of the initial (concentrated) solution portion used,

M₂ and V ₂ be the molarity and volume after dilution.

Moles before dilution:

Moles after dilution (same solute, no reaction):

Set them equal:

Example: diluting concentrated HCl(aq)

A student takes 15.0mL of 12.0molL⁻¹ HCl(aq) and dilutes it to a total volume of 500.0mL. Find the final molarity.

Notice that mL can be used directly here because the units cancel, as long as both volumes use the same unit.

Serial dilutions

When very low concentrations are needed, a single dilution may be impractical because V ₁ would be too small to measure accurately. A serial dilution uses multiple dilution steps, each with a measurable V ₁, and multiplies their dilution factors.

If step i dilutes by a factor Di = , then the overall dilution factor is the product:

and

3.5.6 Molarity and Ionic Species: Connecting to Reaction Chemistry

Earlier, we emphasized that many solutes are strong electrolytes and exist in solution as ions. Molarity is typically defined for the formula units dissolved, but stoichiometry often depends on ion amounts.

Example: ion concentrations from a strong electrolyte

For 0.150molL⁻¹ CaCl2(aq), assuming complete dissociation:

Therefore:

This relationship is crucial in precipitation and acid–base stoichiometry, where ionic mole ratios follow directly from the balanced dissociation equation.

Weak electrolytes: molarity is not the same as ion concentration

For weak acids and bases, the analytical molarity (amount dissolved per liter) does not equal the concentration of ions produced because ionization is incomplete. In this chapter, we treat weak electrolytes qualitatively; later equilibrium chapters will quantify the fraction ionized.

3.5.7 Section Summary

Molarity is an intensive concentration unit defined by M = with units molL⁻¹.

Molarity allows direct conversion between measured volumes and moles using n = MV , providing the entry point for solution stoichiometry.

Solutions of specified molarity are commonly prepared by dissolving a calculated mass of solute and diluting to a final volume in a volumetric flask.

Dilutions conserve moles of solute, leading to the relationship M₁V ₁ = M₂V ₂; serial dilutions multiply dilution factors.

For strong electrolytes, molarity of the dissolved compound can be related to ion concentrations using the dissociation stoichiometry.

3.6 Solution Stoichiometry and Titrations

In the previous section, we introduced molarity as a practical concentration unit that connects measurable solution volumes to moles of dissolved solute. We now combine

molarity with balanced chemical equations to solve quantitative problems involving reactions in aqueous solution. This integration is called solution stoichiometry. A central application is titration, a volumetric method in which the volume of a standardized solution required to react completely with an analyte is measured to determine an unknown concentration or amount.

This section emphasizes a consistent strategy:

1. Identify the relevant reacting species (often ions in solution).

2. Write and balance the appropriate chemical equation (molecular or net ionic).

3. Convert volumes and molarities to moles.

4. Use stoichiometric mole ratios to relate reactants and products.

5. Convert back to the desired quantity (volume, molarity, mass, or moles).

3.6.1 Solution Stoichiometry: The Core Idea

From measured volume to moles

For a solution of molarity M, a measured volume V contains:

where V must be in liters.

This is the key “bridge” between laboratory measurements (mL delivered from a buret or pipet) and chemical stoichiometry (moles reacting according to a balanced equation).

The equation must match the chemistry in solution

In aqueous systems, it is often most efficient to use the net ionic equation, because it reflects the actual species undergoing change and avoids carrying spectator ions through the mole-ratio step.

For example:

Strong acid–strong base neutralization: H+(aq) + OH−(aq) → H 2O(l).

Precipitation of silver chloride: Ag+(aq) + Cl−(aq) → AgCl(s).

Metal displacement: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).

Using a molecular equation can also work, but it can hide the key reacting species. Regardless of form, the equation must be balanced before any quantitative work begins.

3.6.2 Limiting Reactants in Solution

Why limiting reactants matter in aqueous mixtures

When two solutions are mixed, each contributes a certain number of moles of reactive species. Typically one reactant is present in a smaller stoichiometric amount and is therefore limiting; it is consumed first and determines how much product can form. The other reactant is in excess and remains partly

unreacted.

The procedure mirrors limiting-reactant problems for solids and gases, but initial amounts are found from n = MV .

General procedure

Convert each solution volume and molarity to moles of reactive species.

Use the balanced equation to determine which reactant runs out first.

Use the limiting reactant to compute moles of product (or moles remaining).

Convert to requested units (mass, concentration, volume, etc.).

Illustrative example (precipitation)

Suppose 25.0mL of 0.100molL⁻¹ AgNO3(aq) is mixed with 10.0mL of 0.200molL⁻¹ NaCl(aq). Determine the limiting reactant and the moles of AgCl(s) that can form.

Step 1: net ionic equation

Step 2: moles of reacting ions

Step 3: limiting reactant The stoichiometric ratio is 1 : 1. Since Cl− has fewer moles, Cl− is limiting.

Step 4: moles of precipitate

This example demonstrates the standard pattern: compute moles from molarity and volume, then apply the mole ratio from the balanced (net ionic) equation.

3.6.3 Titrations: Volumetric Analysis

Definition and purpose

A titration is an analytical technique in which a solution of known concentration (the titrant) is delivered from a buret into a solution containing the analyte (the analyte) until the reaction reaches completion, as indicated by an endpoint signal.

Common goals include:

determining the concentration of an unknown acid or base,

determining the amount (moles or mass) of a substance in a sample,

standardizing a solution whose concentration is not known with sufficient precision.

Equivalence point vs. endpoint

The equivalence point is the theoretical point at which stoichiometrically equivalent amounts of titrant and analyte have reacted (according to the balanced reaction).

The endpoint is the experimentally observed point (often a color change) used to estimate the equivalence point.

A good indicator produces an endpoint very close to the equivalence point. The difference between them is called the titration error (or indicator error).

Indicators (introductory)

An indicator is a substance that changes color over a certain chemical condition range. In acid–base titrations, indicators are typically weak acids or bases whose conjugate forms have different colors. Selection of an appropriate indicator depends on the expected pH change near equivalence (treated quantitatively in later chapters). For now, the essential operational point is:

Choose an indicator whose color change occurs sharply near the equivalence condition for the reaction being used.

3.6.4 Titration Apparatus and Technique

Accurate titration depends on both correct stoichiometry and careful measurement of delivered volumes.

Reading the buret

The delivered volume is found by difference:

where both readings are taken at the bottom of the meniscus. Consistent eye level and proper meniscus reading are essential for accuracy.

3.6.5 Stoichiometric Relationships at the Equivalence Point

At the equivalence point, the moles of titrant delivered and the moles of analyte present satisfy the stoichiometric ratio in the balanced equation.

General equivalence-point relationship

For a reaction:

the equivalence-point condition is:

with n = MV for each solution species.

Acid–base example: monoprotic acid with strong base

For a monoprotic acid HA titrated with NaOH:

The stoichiometric ratio is 1 : 1, so at equivalence:

which becomes:

This equation resembles the dilution relation M₁V ₁ = M₂V ₂ but has a different meaning: it comes from reaction stoichiometry, not conservation of solute during dilution.

Polyprotic acids and stoichiometric factors

If an acid can donate more than one proton, the mole ratio changes. For example, sulfuric acid provides two acidic protons in complete neutralization:

At equivalence:

or, in terms of molarity and volume,

The factor of 2 arises from the balanced equation and must be included.

3.6.6 Worked Titration Calculation (Acid–Base)

Example: determining an unknown acid concentration

A 25.00mL sample of HCl(aq) is titrated with 0.1000molL⁻¹ NaOH(aq). The endpoint occurs after 18.46mL of base is delivered. Find the molarity of the acid.

Step 1: net ionic equation and mole ratio Strong acid–strong base:

Step 2: moles of base delivered

Step 3: moles of acid At equivalence, n(H+) = n(OH−), so moles of HCl in the aliquot are 1.846 × 10⁻³ mol.

Step 4: acid molarity

Significant figures are determined by measurement precision in the data provided.

3.6.7 Standardization: Making a Solution’s Concentration Reliable

A titrant concentration must be known accurately. If it was prepared by weighing a reagent that is impure, hygroscopic, or otherwise unreliable, the nominal molarity may differ from the true value. Standardization uses a titration against a primary standard to determine the titrant’s actual concentration.

Primary standard (conceptual criteria)

A primary standard is a substance suitable for preparing solutions of accurately known concentration by direct mass measurement. Desirable characteristics include:

high purity,

stability in air (non-hygroscopic, non-reactive with CO2 or H2O vapor),

relatively large molar mass (reduces relative weighing error),

well-defined stoichiometry in reaction.

Standardization workflow

Weigh a known mass of primary standard and dissolve to make an analyte solution (or dissolve in the titration flask).

Titrate with the titrant solution to an endpoint.

Use stoichiometry to compute moles of titrant delivered at equivalence.

Compute Mtitrant = .

3.6.8 Common Pitfalls and Best Practices

Pitfall 1: confusing dilution with reaction stoichiometry

M₁V ₁ = M₂V ₂ applies to dilution (moles conserved).

Titration relationships come from the balanced reaction and may require stoichiometric coefficients.

Pitfall 2: using the wrong reacting species

In aqueous reactions, the reacting species may be ions rather than the written compound. For strong electrolytes this is straightforward, but for weak acids/bases, the appropriate net ionic equation may involve the molecular form:

Choosing the correct equation ensures correct mole ratios.

Pitfall 3: ignoring units or forgetting to convert mL to L

Always convert volumes to liters when using n = MV unless you are using a ratio where volume units cancel consistently.

Pitfall 4: not balancing the equation

Stoichiometric coefficients are not optional. They determine the mole ratios that connect titrant and analyte.

3.6.9 Section Summary

Solution stoichiometry combines molarity with balanced chemical equations to relate measured volumes to moles and reaction amounts.

Limiting-reactant analysis in solution begins by converting each solution’s M and V into moles of reactive species.

Titration is a volumetric method in which a titrant of known concentration reacts with an analyte of unknown amount or concentration.

The equivalence point is the stoichiometric completion point; the endpoint is the observed indicator signal used to approximate it.

At equivalence, moles satisfy the stoichiometric ratio from the balanced equation; for acid–base titrations this often reduces to simple relationships, but polyprotic systems require coefficient factors.

Standardization uses a primary standard to determine the true concentration of a titrant solution.

3.7 Problems

3.7.1 Electrolytes in Water: Strong, Weak, or None?

Classify each of the following substances as a strong electrolyte, weak electrolyte, or nonelectrolyte when dissolved in water at room temperature. For each, briefly justify your classification in terms of the particles present in solution (ions vs. molecules). Assume solubility unless the substance is well known to be insoluble.

1. Na2SO4

2. NH4NO3

3. C2H5OH

4. CH3CO2H

5. Ca(OH)2

6. HClO4

7. K2CO3

8. CO(NH2)2 (urea)

9. H3PO4

10. BaCl2

Then, write the predominant dissolution/ionization equation in water for each substance you classified as a strong or weak electrolyte (use → for strong, ↔ for weak).

3.7.2 Precipitation and Ionic Equations: BaCl2(aq) with AgNO3(aq)

Aqueous solutions of BaCl2 and AgNO3 are mixed.

1. Using solubility rules, predict whether a precipitate forms. If so, identify it and write its formula.

2. Write the balanced molecular equation with correct phase labels.

3. Write the total ionic equation.

4. Write the net ionic equation.

5. Identify the spectator ion(s).

Finally, state explicitly which observation(s) you would expect in the lab (e.g., formation of a solid, color change, gas evolution) and connect the observation to the net ionic equation.

3.7.3 Predicting Precipitation Upon Mixing: Pb(NO3)2 and KI

You mix 50.0 mL of 0.100 moldm⁻³ Pb(NO3)2(aq) with 25.0 mL of 0.300 moldm⁻³ KI(aq).

1. Predict whether a precipitate forms, and write the balanced molecular equation with phase labels.

2. Write the net ionic equation.

3. Determine the limiting reactant using moles (compute n from M and V ).

4. Calculate the moles of precipitate that can form.

5. Determine which ion(s) remain in excess in the final solution and calculate their moles remaining.

6. Compute the final concentrations (in moldm⁻³) of the major remaining ions after reaction, assuming additive volumes.

3.7.4 Neutralization Stoichiometry: H2SO4 and KOH

A 25.0 mL sample of 0.200 moldm⁻³ H2SO4(aq) is completely neutralized by 0.150 moldm⁻³ KOH(aq).

1. Write the balanced molecular equation and the net ionic equation for complete neutralization.

2. Calculate the volume of KOH required (in mL).

3. After mixing at equivalence, identify the major dissolved species in solution (ignore autoionization of water). Which ions are spectators in the net ionic equation?

3.7.5 Oxidation Numbers and Redox Identification

For the reaction

1. Assign oxidation numbers to every element in every reactant and product.

2. Identify which element(s) are oxidized and which are reduced.

3. Identify the oxidizing agent and the reducing agent.

4. Determine the total number of electrons transferred per balanced reaction as written.

Your work must show how oxidation number changes lead to your conclusions.

3.7.6 Redox Concept Check: Mg Reacting with O2

Magnesium metal reacts with oxygen gas to form magnesium oxide:

1. Assign oxidation numbers to Mg and O in reactants and products.

2. Identify the species oxidized, the species reduced, the oxidizing agent, and the reducing agent.

3. Write separate oxidation and reduction half-reactions showing electrons explicitly, and then combine them to recover the overall balanced equation.

4. Explain (in one to three sentences) why this is classified as a redox reaction even though there are no aqueous ions.

3.7.7 Preparing a Solution from a Solid: Na2CO3

You need 250.0 mL of 0.500 moldm⁻³ Na2CO3(aq) for an experiment.

1. Calculate the mass of anhydrous Na2CO3(s) required.

2. Describe a correct volumetric-flask preparation procedure in 5–7 steps, emphasizing actions that ensure the final molarity is accurate.

3. If you accidentally dilute to 255.0 mL instead of 250.0 mL, calculate the actual molarity obtained.

3.7.8 Dilution of Concentrated Acid

A student prepares dilute hydrochloric acid by transferring 15.0 mL of 12.0 moldm⁻³ HCl(aq) into a volumetric flask and diluting to a total volume of 500.0 mL.

1. Calculate the final molarity of HCl(aq).

2. Calculate the moles of H+ present in 20.0 mL of the diluted solution, assuming complete dissociation.

3. The student instead adds water to the flask first, then adds the acid, then dilutes to the mark. Does this change the final molarity (assume perfect mixing and no volume change upon mixing)? Explain briefly.

3.7.9 Gravimetric Analysis: Determining an Unknown Chloride Salt

A 0.3250 g sample of an unknown soluble chloride salt (containing a single cation and Cl– ) is dissolved in water. Excess AgNO3(aq) is added, forming 0.7432 g of AgCl(s) which is filtered, washed, and dried.

1. Write the net ionic equation for the precipitation.

2. Calculate the moles of AgCl formed and the moles of Cl− in the original sample.

3. Determine the mass of chlorine in the original sample.

4. Calculate the mass of the cation portion in the original sample.

5. If the salt is known to be a chloride of a Group 1 metal (Li+, Na+, K+, Rb+, or Cs+), determine which one it most likely is by calculating the molar mass of the salt and comparing to plausible candidates.

3.7.10 Acid–Base Titration: Molar Mass of an Unknown Weak Acid

A 0.500 g sample of an unknown monoprotic weak acid HA is dissolved in water and titrated with 0.100 moldm⁻³ NaOH(aq). The endpoint occurs after 34.0 mL of base is delivered.

1. Write the balanced net ionic equation for the titration reaction.

2. Calculate the moles of OH− delivered at the endpoint and the corresponding moles of HA originally present.

3. Determine the molar mass (in g mol⁻¹) of the unknown acid.

4. If the measured endpoint systematically occurs after the true equivalence point (indicator changes color slightly too late), state qualitatively whether the reported molar mass would be too high or too low, and explain why in one or two sentences.

Chapter 4
Electronic Structure and Periodic Trends

The periodic table is not merely a chart of elements but a map of the quantum mechanical landscape. This chapter bridges the gap between the abstract wave mechanics of the electron and the observable chemical properties of the elements. By exploring the quantization of energy and the probabilistic nature of orbitals, we uncover the deep structural rules that govern electron configurations, ultimately revealing why elements behave as they do and how periodic trends emerge directly from the geometry of the atom.

4.1 The Wave Nature of Light and Energy Quantization

Chemistry is ultimately an account of how matter exchanges and redistributes energy. In earlier chapters, we treated energy as if it can change continuously (for example, heat flow and work in thermochemistry). At the atomic scale, however, experiments show that energy exchange between matter and electromagnetic radiation is not continuous. This section develops (1) a wave description of light that successfully explains interference and diffraction, and then (2) the experimental evidence that requires a particle description and leads to quantized energy.

4.1.1 Electromagnetic Radiation as a Wave

Electromagnetic (EM) radiation is a form of energy that can propagate through space, carrying energy and momentum. It can be described as a coupled oscillation of electric and magnetic fields that travels as a wave. For introductory chemistry, it is sufficient to focus on the measurable wave properties: wavelength, frequency, amplitude, and speed.

Wavelength and frequency

A periodic wave is characterized by:

Wavelength λ: the distance between successive equivalent points on a wave (e.g., crest-to-crest). SI unit: m.

Frequency ν (Greek letter nu): the number of wave cycles that pass a point per second. SI unit: s⁻¹, also called hertz (Hz).

Amplitude: related to intensity (brightness for visible light). In many chemical applications, intensity is proportional to the square of amplitude.

All electromagnetic radiation in vacuum travels at the same speed, the speed of light:

Because one wavelength corresponds to one cycle, wave speed is the product of wavelength and frequency:

Thus, wavelength and frequency are inversely related: higher-frequency radiation has shorter wavelength.

The electromagnetic spectrum

The electromagnetic spectrum is the continuous range of wavelengths (or frequencies) of EM radiation. Chemistry routinely uses multiple regions:

Radiofrequency and microwave: probe nuclear spins and molecular rotations.

Infrared (IR): excites molecular vibrations; used to identify functional groups.

Visible: excites valence electrons; responsible for color.

Ultraviolet (UV): higher-energy electronic transitions; photochemistry.

X-rays: core-electron processes and crystal structure (diffraction).

Energy carried by a wave

Classically, the energy carried by an EM wave increases with intensity (amplitude). Classical wave theory does not impose restrictions on how small an energy transfer can be; in principle, an arbitrarily weak wave could transfer arbitrarily small energy to matter. As we will see, specific experiments contradict this expectation.

4.1.2 Wave Behavior: Interference and Diffraction

Wave descriptions are strongly supported by phenomena such as:

Interference: When two waves overlap, their electric fields add. Constructive interference increases intensity; destructive interference reduces it.

Diffraction: Waves spread when passing through an opening or around an obstacle, producing interference patterns.

These are not merely qualitative curiosities: X-ray diffraction, for example, depends on interpreting X-rays as waves with wavelengths comparable to atomic spacings (∼1 × 10⁻¹⁰ m). Thus, any complete model of light must preserve its wave character.

4.1.3 A Crisis for Classical Waves: The Photoelectric Effect

Despite its success, the classical wave model fails to explain how light transfers energy to electrons in metals. The decisive evidence comes from the photoelectric effect: when light shines on a clean metal surface in vacuum, electrons may be emitted from the metal into the vacuum (photoelectrons).

Key experimental observations

Experiments reveal several robust facts:

1. Threshold frequency. For a given metal, there exists a minimum frequency ν₀ below which no electrons are emitted, no matter how intense the light is.

2. Kinetic energy depends on frequency, not intensity. For frequencies above threshold, the maximum kinetic energy of emitted electrons increases as the light frequency increases. Increasing intensity (at fixed frequency above threshold) increases the number of electrons emitted, but not their maximum kinetic energy.

3. No measurable time lag. When light above threshold frequency strikes the metal, electrons are emitted essentially immediately, even at low intensity.

Classical wave theory predicts the opposite behavior in two important ways:

If energy transfer depended only on intensity, then sufficiently intense low-frequency light should eventually eject electrons (contradicting the threshold).

At low intensity, electrons should require time to “accumulate” enough energy before emission (contradicting the absence of time lag).

These contradictions force a new model of how light carries and delivers energy.

Photon model and quantized energy

Albert Einstein explained the photoelectric effect by proposing that light energy is quantized and delivered in discrete packets called photons. Each photon has energy proportional to the frequency of the radiation:

where h is Planck’s constant:

This relation was consistent with Planck’s earlier work on blackbody radiation and, crucially, it explains the photoelectric observations with a simple energy bookkeeping argument.

Work function and the photoelectric equation

Electrons in a metal are not free to leave the surface without an energy input. The minimum energy required to remove an electron from a particular metal surface is called the work function, denoted ϕ (phi). It depends on the identity of the metal and the condition of its surface.

In the photon model, a single photon transfers its energy to a single electron (in the simplest description). The electron must “pay” the work function to escape; any remaining energy becomes kinetic energy of the emitted electron:

where K is the kinetic energy of the photoelectron. Because emitted electrons come from a distribution of initial energies and may suffer collisions, experiments commonly report the maximum kinetic energy Kmax:

The threshold frequency ν₀ occurs when electrons are emitted with essentially zero kinetic energy (Kmax = 0):

These equations immediately account for the experimental facts:

Below ν₀, photons do not carry enough energy to overcome ϕ, so no electrons are emitted regardless of intensity.

Increasing ν increases photon energy hν, so Kmax increases linearly with frequency.

Increasing intensity at fixed ν increases the number of photons arriving per second, so more electrons are emitted per second, but each photon still carries energy hν, so Kmax does not increase.

No time lag occurs because an electron that absorbs a photon above threshold can be emitted essentially immediately.

A measurable consequence: stopping potential

In a typical experiment, emitted electrons are collected by a second electrode. By applying a reverse (retarding) electric potential, one can reduce the current and find the potential V s that just stops even the fastest electrons. Then:

where e = 1.602176634 × 10⁻¹⁹ C is the elementary charge. A plot of V s (or Kmax) versus frequency is a straight line with slope h∕e and intercept related to −ϕ. This provides direct experimental access to Planck’s constant and the work function.

4.1.4 Energy Quantization: From Continuous to Discrete Exchange

The photon model introduces a critical shift in scientific thinking: even if the wave can have a continuum of intensities, the exchange of energy between radiation and matter occurs in discrete amounts of size hν.

Quantization as a rule for energy transfer

The central idea is not that light loses all wave properties; rather, the energy of light at frequency ν is packaged such that absorption and emission occur in integer numbers of photons:

This is the first appearance of quantization in our study of atomic structure. In subsequent sections, quantization will reappear in a deeper way: atoms and molecules possess discrete allowed energy levels, and photons are absorbed or emitted precisely when their energy matches the energy difference between these levels.

4.1.5 Wave–Particle Duality: A Unified View

The photoelectric effect compels a particle-like model (photons) for explaining energy transfer, while interference and diffraction compel a wave-like model for explaining propagation and spatial patterns. The modern description is wave–particle duality:

Light propagates in a way accurately described by waves (wavelength, interference, diffraction).

Light interacts with matter in discrete packets of energy (photons), with E = hν.

This duality is not a matter of choosing the “right” classical picture; it is a signal that classical categories (wave versus particle) are incomplete for microscopic phenomena. In the next section, we will leverage quantization to interpret atomic spectra, where atoms emit and absorb light at discrete wavelengths—direct evidence that electron energies in atoms are quantized.

4.2 Bohr Model and Atomic Spectra

The photoelectric effect showed that electromagnetic radiation exchanges energy with matter in discrete packets, photons, with energy E = hν. We now encounter the complementary statement: atoms themselves possess only certain allowed energies. When an atom changes from one allowed energy to another, it must emit or absorb a photon whose energy matches the energy difference. This idea is revealed most clearly by the atomic line spectrum of hydrogen.

4.2.1 Continuous (Thermal) Spectra vs. Line Spectra

Many hot objects (a filament, molten metal, the Sun) emit a continuous spectrum: all visible wavelengths appear, with intensity varying smoothly with wavelength. By contrast, a low-pressure gas excited by an electrical discharge emits a line spectrum: discrete, sharply defined wavelengths.

Hydrogen is the simplest atom and its emission spectrum is particularly important historically and conceptually. When excited hydrogen atoms relax back to lower-energy states, the emitted light is not continuous; it consists of a set of lines. The existence of discrete lines implies that:

Hydrogen atoms can exist only in certain allowed energy states.

Energy changes during emission occur in discrete steps, not continuously.

The photon energy is tied to the atomic energy change:

Negative ΔEatom indicates emission (the atom loses energy).

4.2.2 The Hydrogen Emission Spectrum and Series

Hydrogen emission lines are not random. They group into series characterized by a common lower (final) energy level.

Lyman series: transitions ending at n = 1 (ultraviolet).

Balmer series: transitions ending at n = 2 (visible and near-UV; historically crucial).

Paschen series: transitions ending at n = 3 (infrared).

Here n is an integer label that will soon be identified with an allowed energy level. The regularity of the lines suggests a simple mathematical pattern.

4.2.3 The Rydberg Equation

The wavelengths of hydrogen emission lines are well described by the Rydberg equation:

where

λ is the emitted wavelength,

ni is the initial principal quantum number (higher energy),

nf is the final principal quantum number (lower energy),

RH is the Rydberg constant for hydrogen:

Several features of this equation are conceptually important:

Only integer values of ni and nf appear. This integer structure is a signature of quantization.

For a fixed nf, taking ni = nf + 1,nf + 2,… generates a series of lines.

As ni →∞ (very high initial level), 1∕ni² → 0, so each series has a series limit:

This corresponds to the smallest photon energy (longest wavelength) that can still end at nf from extremely high levels.

Example: A Balmer line

For the Balmer series, nf = 2. The line with ni = 3 → nf = 2 is:

This produces a wavelength in the visible region (the well-known red Balmer line).

The Rydberg equation describes what the wavelengths are, but it does not, by itself, explain why hydrogen behaves this way. The Bohr model provides a historically important explanation by postulating quantized electron energies.

4.2.4 Bohr’s Model: Quantized Orbits and Quantized Energy

In the early 20th century, the atom was often pictured (classically) as electrons orbiting a nucleus, somewhat like planets orbiting the Sun. Classical electrodynamics predicts that an accelerating charge (an orbiting electron) should radiate energy continuously and spiral into the nucleus—atoms would be unstable. Hydrogen is stable, so the classical picture is incomplete.

Bohr proposed a model with two key postulates that intentionally break with classical expectations:

The electron in hydrogen can occupy only certain stationary states (allowed orbits) with specific energies. While in a stationary state, it does not radiate energy.

Radiation occurs only when the electron transitions between stationary states. The photon energy equals the energy difference:

To select the allowed states, Bohr quantized the electron’s angular momentum:

where me is electron mass, v is orbital speed, r is orbital radius, and ℏ = h∕(2π).

Although this angular-momentum postulate is not how modern quantum mechanics ultimately treats electrons, it leads to energy levels that match hydrogen spectra remarkably well.

4.2.5 Energy Levels of Hydrogen in the Bohr Model

Bohr’s model predicts that the energy of hydrogen depends only on the integer n:

It is common to express this in joules per atom or in electronvolts (eV). Using RHhc ≈ 2.18 × 10⁻¹⁸ J and 1 eV = 1.602 × 10⁻¹⁹ J, we obtain:

Interpretation of the energy expression

The negative sign indicates that the electron is in a bound state (its energy is lower than that of a free electron infinitely far from the nucleus, defined as E = 0).

The ground state is the lowest-energy state, n = 1:

States with n ≥ 2 are excited states with higher (less negative) energy.

The spacing between adjacent energy levels decreases as n increases because En ∝ −1∕n². This explains why spectral lines crowd together near the series limit.

Connection to the Rydberg equation

For an emission transition from ni to nf (ni > nf):

ΔE | = Enf − Eni |  |
| = −RHhc. |  |

The emitted photon has energy hν = −ΔE, so

hν | = RHhc |  |
ν | = cRH. |  |

Using c = λν, we recover:

Thus, Bohr’s quantized energies provide a physical interpretation: spectral lines are fingerprints of energy-level differences.

4.2.6 Absorption Spectra and Excitation

Emission occurs when an atom loses energy by moving to a lower level. The reverse process is absorption: a photon is absorbed if and only if its energy matches the gap between two allowed levels:

This requirement has an important chemical implication: atoms and molecules interact selectively with radiation. Not every wavelength is absorbed; only those that correspond to allowed energy transitions.

4.2.7 Ionization Energy and the Meaning of the Series Limit

In the Bohr model, the reference point E = 0 corresponds to a free electron infinitely far from the nucleus. Ionization from level n means raising the electron from En to 0:

For the ground state (n = 1), the ionization energy is 13.6 eV per atom (about 2.18 × 10⁻¹⁸ J per atom). The series limit in each spectral series corresponds to transitions from very high ni down to a fixed nf; physically, this is closely related to

the onset of ionization from the level nf.

This connects spectra to a major periodic property discussed later: ionization energy—the energy required to remove an electron from an atom.

4.2.8 Successes and Limitations of the Bohr Model

Why the Bohr model matters

The Bohr model is historically pivotal and remains pedagogically valuable because it:

predicts the hydrogen energy levels and spectra accurately,

introduces the idea of quantized energy states in atoms,

provides a concrete link between photon energy (hν) and atomic energy differences (ΔE),

gives a useful framework for discussing ground and excited states, emission vs. absorption, and ionization.

Where the Bohr model fails

Despite its success for hydrogen, the Bohr model is not a general theory of atoms.

It does not correctly predict spectra of multi-electron atoms without additional assumptions and corrections.

It treats electrons as orbiting particles with definite radii and speeds, which conflicts with the more accurate quantum-mechanical view developed in the next section.

It cannot naturally explain line intensities or fine spectral details (splitting due to electron spin, relativistic effects, and interactions).

The key idea that survives and becomes foundational is quantization: atoms have discrete allowed energies. In the next section, we replace classical orbits with wave mechanics and introduce the quantum numbers that describe electron states in a way that extends beyond hydrogen.

4.3 Wave Mechanics and Quantum Numbers

The Bohr model accounts for the hydrogen spectrum by assigning the electron a discrete set of energies labeled by an integer n. However, it does so by postulating special “allowed orbits,” a concept that does not extend cleanly to multielectron atoms and conflicts with the modern understanding of measurement at microscopic scales. Quantum mechanics replaces the idea of a precise electron trajectory with a wave function that encodes what can be known and predicted about the electron. The allowed energies emerge not from assumed orbits, but from the mathematical requirement that the wave function be physically acceptable.

4.3.1 Matter Waves: The de Broglie Hypothesis

Wave–particle duality applies not only to light but also to matter. Louis de Broglie proposed that any particle of momentum p has an associated wavelength:

where m is mass and v is speed (nonrelativistic).

This proposal provides an immediate conceptual bridge from the Bohr model to wave mechanics: a stable electron state in an atom can be pictured (qualitatively) as a standing wave around the nucleus. Standing waves occur only for certain wavelengths, which suggests only certain energies are allowed.

Standing-wave intuition (qualitative)

For a standing wave on a circular path, the circumference must contain an integer number of wavelengths:

Substituting λ = h∕(mv) gives mvr = n(h∕2π) = nℏ, which resembles Bohr’s angular momentum rule. In modern quantum mechanics, this “standing wave” picture is only an analogy, but it correctly hints that quantization is a wave condition rather than an arbitrary restriction.

4.3.2 The Heisenberg Uncertainty Principle

A decisive break from classical trajectories comes from the Heisenberg uncertainty principle. For position x and momentum px,

Here Δx and Δpx represent the inherent uncertainties (more precisely, standard deviations) in repeated measurements on identically prepared systems.

Why uncertainty matters for atomic electrons

An electron confined to the small space of an atom must have a comparatively uncertain (and typically large) momentum. Thus, it is not physically meaningful to imagine an electron moving in a sharply defined orbit with both a precise radius and a precise momentum. Quantum mechanics therefore focuses on predicting probabilities for measurement outcomes rather than deterministic paths.

4.3.3 The Schrödinger Equation and the Wave Function

The wave function ψ

In quantum mechanics, the state of an electron in an atom is described by a wave function ψ(r), where r is the position vector. The wave function itself is not directly observable, but it is the central mathematical object from which observable predictions are extracted.

The most important physical quantity derived from ψ is the probability density:

Interpreting |ψ|²:

|ψ(r)|² dτ is proportional to the probability of finding the electron in a small volume element dτ around r.

Because total probability must be 1, acceptable wave functions must be normalizable:

The time-independent Schrödinger equation

For a particle of mass m moving in a potential energy function V (r), the time-independent Schrödinger equation is:

where Ĥ is the Hamiltonian operator,

For the hydrogen atom, the electron experiences the Coulomb attraction to the proton:

where r is the electron–nucleus distance and 𝜀₀ is the permittivity of vacuum.

Solving Ĥψ = Eψ yields:

a set of allowed energies E (quantized),

corresponding wave functions ψ (often called atomic orbitals for hydrogen-like atoms).

Boundary conditions and quantization

Quantization in quantum mechanics is not imposed by hand; it arises because only certain solutions satisfy physical requirements:

ψ must be single-valued (one value at each point),

ψ must be finite everywhere,

ψ must be continuous (and typically smoothly varying),

ψ must be normalizable.

These constraints restrict the allowed solutions to a discrete set, producing quantized energies and discrete quantum numbers.

4.3.4 Quantum Numbers from the Hydrogen-Atom Solution

The Schrödinger equation for hydrogen is most naturally solved in spherical coordinates because the potential depends only on r. The solutions separate into a radial part and an angular part:

where Y ℓmℓ are spherical harmonics. Each allowed wave function is labeled by a set of quantum numbers. In addition, electrons possess an intrinsic form of angular momentum called spin, described by a fourth quantum number.

1. Principal quantum number n

The principal quantum number n is a positive integer:

For hydrogen, n determines the energy:

More generally, in multielectron atoms n remains central for describing shells and (approximately) size and energy.

Physical meaning. Larger n corresponds to orbitals that, on average, extend farther from the nucleus and are higher in energy (less negative for hydrogen).

2. Angular momentum (azimuthal) quantum number ℓ

For each n, the angular momentum quantum number ℓ can take integer values:

The value of ℓ determines the electron’s orbital angular momentum magnitude:

Connection to subshells. In chemistry, ℓ labels subshell type:

Thus, n labels the shell (1, 2, 3, …) and ℓ labels the subshell within that shell.

3. Magnetic quantum number mℓ

For a given ℓ, the magnetic quantum number mℓ takes integer values:

There are 2ℓ + 1 allowed values of mℓ for each ℓ.

Physical meaning. The quantum number mℓ specifies the allowed orientations of the orbital angular momentum relative to a chosen axis (commonly an external magnetic field direction). While |L| is fixed by ℓ, the component along an axis is quantized:

In chemical language, the different mℓ values correspond to different orbitals within a subshell (e.g., the three 2p orbitals correspond to mℓ = −1,0,+1).

4. Spin quantum number ms

Electrons possess intrinsic angular momentum called spin. Spin is not the electron literally spinning like a tiny sphere; it is a quantum property with measurable consequences (e.g., splitting of spectral lines in magnetic fields and the structure of the periodic table).

The electron spin quantum number can take two values:

Chemical significance. The two allowed spin states are essential for electron configurations because they permit two electrons to occupy the same spatial orbital (same n, ℓ, mℓ) provided they have opposite spins. This point will be formalized later as the Pauli exclusion principle.

4.3.5 Counting Orbitals and Electron Capacity (Preview for Electron Configurations)

Although a full discussion of electron configurations comes later, it is useful now to connect the quantum numbers to how many electrons “fit” in shells and subshells.

Orbitals in a subshell

For a fixed (n,ℓ), the allowed mℓ values are −ℓ to +ℓ, giving:

Each orbital can hold two electrons (spin ±), so the electron capacity of a subshell is:

Thus:

Orbitals in a shell

For a fixed n, the allowed ℓ values are 0 to n− 1. The total number of orbitals in shell n is:

so the maximum number of electrons in shell n is:

These counting rules are direct consequences of the allowed quantum numbers, and they will become the bookkeeping foundation for electron configurations and periodic trends.

4.3.6 What Quantum Numbers Do Not Tell Us (Yet)

At this stage, it is important to separate what has been established from what will be refined later:

For hydrogen, energy depends only on n; all orbitals with the same n are degenerate (same energy), regardless of ℓ and mℓ.

For multielectron atoms, electron–electron repulsions and shielding break this degeneracy; energies depend on both n and ℓ (and sometimes more subtle effects). This is why 3s, 3p, and 3d are not equal in energy in real atoms, even though they share n = 3.

The Schrödinger equation thus provides a framework that (1) reproduces hydrogen quantization and (2) generalizes to complex atoms through additional physical effects. In the next section, we use these quantum numbers to describe the shapes and nodal structures of atomic orbitals, transforming ψ from an abstract function into a chemically meaningful picture of electron distribution in space.

4.4 Atomic Orbitals and Their Shapes

In the previous section, we introduced the wave function ψ(r) as the central description of an electron in an atom, and we learned that allowable states are labeled by quantum numbers (n,ℓ,mℓ,ms). We now translate those abstract labels

into spatial structure: what an orbital “looks like,” where the electron is likely to be found, and where it is never found (nodes).

4.4.1 Orbitals as Wave Functions in Three Dimensions

For hydrogen-like atoms (one electron moving in a spherically symmetric Coulomb potential), the Schrödinger equation separates in spherical coordinates:

where:

Rnℓ(r) is the radial part, controlling how ψ varies with distance from the nucleus.

Y ℓmℓ(𝜃,ϕ) is the angular part (a spherical harmonic), controlling how ψ varies with direction.

Two related but distinct quantities are used to describe “where the electron is”:

Probability density (in space):

so ρdτ gives the probability of finding the electron in a small volume element dτ around r.

Radial probability distribution (in distance):

which is the probability of finding the electron between r and r + dr, regardless of direction. The factor 4πr² is crucial: even if |R|² is largest near the nucleus, the volume of space increases with r², often shifting the most probable radius outward.

4.4.2 Probability Surfaces and Orbital Phase

Orbitals are often visualized using probability (electron-density) surfaces. A common convention is to draw the surface that encloses a fixed fraction of the total probability (e.g., 90% or 95%). Inside the surface, the electron is likely to be found; outside it, less likely.

Phase (sign) of ψ and why it is shown

A wave function can be positive or negative. The probability density depends on |ψ|² and is therefore always nonnegative, but the sign (phase) of ψ matters profoundly when orbitals combine to form molecular orbitals and chemical bonds. For this reason, orbital pictures often use two colors to represent regions where ψ has opposite sign (phase). In this section:

one color represents ψ > 0,

the other color represents ψ < 0.

4.4.3 Nodes: Where the Electron Cannot Be Found

A node is a location where ψ = 0, so the probability density is zero. Nodes are not “low probability” regions; they are zero probability surfaces or points imposed by the mathematical structure of the wave function.

There are two main kinds of nodes:

Angular nodes arise from the angular part Y ℓmℓ(𝜃,ϕ) and depend on ℓ. They form nodal planes or cones (surfaces passing through the nucleus).

Radial nodes arise from the radial part Rnℓ(r) and depend on n and ℓ. They form nodal spheres (spherical shells at certain radii where ψ = 0).

For hydrogen-like orbitals, the node counts are:

number of angular nodes | = ℓ, |  |
number of radial nodes | = n − ℓ − 1, |  |
total number of nodes | = (n − ℓ − 1) + ℓ = n − 1. |  |

These relationships provide a powerful consistency check: if you know (n,ℓ), you know how many nodal surfaces the orbital must have.

4.4.4 s Orbitals (ℓ = 0): Spherical Symmetry

If ℓ = 0, then mℓ = 0 and the angular function Y ₀₀ is a constant. This means the orbital has no directional preference: it is spherically symmetric. The simplest example is the 1s orbital (n = 1,ℓ = 0).

Shape and nodes of 1s

Angular nodes: ℓ = 0 ⇒ 0 angular nodes.

Radial nodes: n − ℓ − 1 = 1 − 0 − 1 = 0 radial nodes.

Thus, 1s has no nodes at all (other than the limit ψ → 0 as r →∞). Its probability density decreases smoothly with distance from the nucleus.

Higher s orbitals and radial nodes

For 2s (n = 2,ℓ = 0), there is 2 − 0 − 1 = 1 radial node: a spherical surface where ψ = 0. This produces an “inner” and “outer” region of opposite phase. As n increases (e.g., 3s,4s), the number of radial nodes increases, reflecting increasingly complex radial structure.

4.4.5 p Orbitals (ℓ = 1): Directional Dumbbells and a Nodal Plane

For ℓ = 1, there are 2ℓ + 1 = 3 orbitals corresponding to mℓ = −1,0,+1. In introductory chemistry, we commonly label these three orbitals as px, py, and pz to emphasize their orientations along mutually perpendicular axes.

Shape and nodes of 2p

Consider 2p (n = 2,ℓ = 1):

Angular nodes: ℓ = 1 ⇒ 1 angular node (a nodal plane passing through the nucleus).

Radial nodes: n − ℓ − 1 = 2 − 1 − 1 = 0 radial nodes.

A p orbital therefore has two lobes of opposite phase separated by a nodal plane. The electron probability density is zero on that plane because ψ = 0 there.

Orientation versus energy (important distinction)

In the absence of external fields (and for hydrogen), the different 2p orbitals have the same energy; they are degenerate. Their orientations are different, but their energies are identical because the Coulomb potential is spherically symmetric. In multielectron atoms, the p subshell remains threefold in orbital count, but its energy ordering relative to other subshells depends on shielding and penetration (developed later).

4.4.6 d Orbitals (ℓ = 2): Five Orbitals with Characteristic Lobes

For ℓ = 2, there are 2ℓ + 1 = 5 orbitals. These are commonly described as:

Four of these have a four-lobed “cloverleaf” appearance; dz² is distinct, with two lobes along z and a torus (donut) of opposite phase in the equatorial region.

Nodes in a 3d orbital

For 3d (n = 3,ℓ = 2):

Angular nodes: ℓ = 2 ⇒ 2 angular nodes (planes or cones, depending on the orbital).

Radial nodes: n − ℓ − 1 = 3 − 2 − 1 = 0 radial nodes.

Thus, the complexity of d orbitals arises primarily from angular structure rather than radial nodes (for 3d specifically).

4.4.7 f Orbitals (ℓ = 3): Seven Highly Directional Orbitals

For ℓ = 3, there are 2ℓ + 1 = 7 orbitals. Their shapes are more intricate (often 6–8 lobes with additional nodal surfaces). In general chemistry, the most important facts about f orbitals are:

They exist starting at n = 4 (because ℓ ≤ n − 1).

They contribute to the electron configurations and chemical behavior of the lanthanides and actinides.

They have three angular nodes (ℓ = 3), making their angular patterns especially complex.

4.4.8 Common Misconceptions About Orbital Pictures

An orbital is not a path. The electron is not traveling along the surface of a drawn orbital. The picture represents a probability distribution in space.

Orbital boundaries are conventions. A 90% probability surface is useful for visualization, but the probability density does not abruptly drop to zero at the boundary.

Lobes represent phase, not charge. Oppositely colored lobes do not mean positive and negative charge; they indicate the sign of ψ.

4.4.9 Representative Shapes of 1s, 2p, and 3d Orbitals

The following schematic illustrates the standard qualitative shapes and phase coloring for the 1s orbital, the three 2p orbitals, and the five 3d orbitals. Colors distinguish opposite phases of ψ.

4.4.10 From Orbital Shapes to Chemical Insight

Orbital shapes and nodes are not merely geometric facts; they predict patterns in bonding, spectroscopy, and periodic trends:

Directionality: p, d, and f orbitals have preferred directions, which strongly influences molecular geometry and bond angles.

Penetration and shielding (preview): Even among orbitals with the same n, different ℓ values produce different electron-density distributions near the nucleus, affecting effective nuclear charge and energy ordering in multielectron atoms.

Phase: The sign pattern of ψ governs constructive or destructive overlap, distinguishing bonding from antibonding interactions in molecular orbital theory.

In the next section, we use these orbitals as the “seats” that electrons occupy in atoms, introducing the rules that determine electron configurations across the periodic table.

4.5 Electron Configurations

The shapes and quantum numbers of atomic orbitals provide the “seats” available to electrons. Chemistry now asks a practical, predictive question: how are the electrons distributed among these orbitals in the ground state of an atom or ion? The answer is the electron configuration. Electron configurations connect quantum mechanics to periodic properties (size, ionization energy, bonding patterns) because they determine which electrons are valence electrons and how strongly those electrons are held.

4.5.1 What an Electron Configuration Represents

An electron configuration specifies the occupied orbitals of an atom (or ion), consistent with the lowest possible total energy subject to quantum-mechanical restrictions. It can be written:

Spectroscopic notation (compact, text-based):

Orbital diagram (graphical): boxes (orbitals) and arrows (electron spins).

The configuration is a model of the ground state; it does not depict electron motion, and it does not imply that an electron can be assigned a precise path. It is a bookkeeping scheme rooted in the quantum numbers and energy ordering of orbitals in multielectron atoms.

4.5.2 The Three Foundational Rules

Determining configurations relies on three principles: Aufbau, Pauli, and Hund. Each is an operational rule that summarizes deeper quantum-mechanical facts.

Aufbau principle (build-up)

The Aufbau principle states that electrons occupy the available orbitals of lowest energy first, producing the lowest-energy (ground-state) electron arrangement.

For hydrogen, all orbitals with the same n are degenerate, but in multielectron atoms orbital energies depend on both n and ℓ because of electron–electron repulsion, shielding, and penetration. As a result, the energy ordering is not simply n = 1 then n = 2 then n = 3; instead, subshells interleave (e.g., 4s is typically lower in energy than 3d for neutral atoms).

Pauli exclusion principle

The Pauli exclusion principle states:

No two electrons in the same atom can have the same set of four quantum numbers.

Because an orbital is defined by (n,ℓ,mℓ), the Pauli principle implies that each orbital can contain at most two electrons, and if two electrons share an orbital, they must have opposite spins:

Hund’s rule (maximum multiplicity)

Hund’s rule applies within a set of degenerate orbitals (such as the three p orbitals or five d orbitals in a given subshell):

Electrons occupy degenerate orbitals singly with parallel spins before pairing occurs.

This rule minimizes electron–electron repulsion within a subshell and is consistent with the stabilizing effects of exchange interactions (beyond the scope of general chemistry, but the empirical rule is essential for correct configurations and for predicting magnetism).

4.5.3 Energy Ordering and the n + ℓ (Madelung) Guideline

For many ground-state neutral atoms, the approximate subshell energy ordering follows the n + ℓ rule:

Lower n + ℓ fills first.

If two subshells have the same n + ℓ, the one with lower n fills first.

This guideline produces the commonly used filling order:

The ordering reflects increasing energy in typical neutral atoms, but it is not a universal law; exceptions and ion formation require additional care.

4.5.4 Writing Electron Configurations: A Systematic Procedure

A reliable method for any main-group or transition-metal atom is:

Determine the electron count. For a neutral atom, electrons = Z (atomic number). For ions, adjust: add electrons for anions, subtract for cations.

Fill orbitals in energy order (Aufbau), using superscripts to count electrons in each subshell.

Enforce Pauli (max 2 per orbital, opposite spins if paired).

Apply Hund within each subshell to distribute electrons across degenerate orbitals.

Check the total number of electrons matches the intended species.

Example: Oxygen, O (Z = 8)

Fill in order:

So:

Orbital diagram for the 2p electrons (three degenerate orbitals):

after Hund’s rule places one electron in each orbital before pairing the fourth.

4.5.5 Orbital Diagrams and Spin Pairing

Orbital diagrams make Hund’s rule and Pauli’s principle visible.

Box-and-arrow conventions

Each box represents one orbital (a specific mℓ value within a subshell).

Each arrow represents an electron.

Arrow direction indicates spin: ↑ corresponds to ms = +, ↓ corresponds to ms = −.

A paired orbital must be ↑↓ (Pauli).

Counting orbitals per subshell

Recall: a subshell with angular momentum quantum number ℓ has 2ℓ + 1 orbitals. Thus:

These counts determine how electrons distribute under Hund’s rule.

4.5.6 Condensed (Noble-Gas) Configurations

For heavy atoms, writing every occupied subshell is cumbersome. A condensed configuration uses the previous noble gas as a core:

Why noble-gas cores are chemically meaningful

Noble gases have filled valence shells and are relatively unreactive. Writing [noble gas] highlights the valence electrons—the electrons outside the noble-gas core—which largely control chemical bonding and reactivity.

4.5.7 Electron Configurations for Ions

Ions form by losing or gaining electrons. Electron configurations for ions must be determined carefully because the order of filling is not always the same as the order of electron removal.

General rules for ions

Anions (negative ions): add electrons to the next available lowest-energy orbitals, following Hund and Pauli.

Cations (positive ions): remove electrons from the highest principal quantum number n first (outermost shell), which often corresponds to valence orbitals.

Main-group cations: remove from the highest n

Example: Mg (Z = 12):

Then:

The two 3s electrons are removed.

Transition-metal cations: why 4s electrons are removed before 3d

In neutral atoms, 4s typically fills before 3d (Aufbau ordering). However, once the 3d subshell begins to fill, the relative energies shift: in many transition-metal atoms and their ions, the 3d electrons are stabilized (lowered in energy) relative to 4s. As a result, when forming cations, electrons are usually removed from 4s before 3d.

Example: iron:

Then:

The 4s electrons are removed first, then one 3d electron for Fe3+.

4.5.8 The Special Stability of Half-Filled and Filled Subshells

Some configurations predicted by a simple Aufbau filling order are not the true ground states because electron correlation and exchange effects stabilize certain patterns, especially:

Half-filled subshells: p³, d⁵, f⁷

Filled subshells: p⁶, d¹⁰, f¹⁴

These patterns often reduce electron–electron repulsion and increase exchange stabilization, making them slightly lower in energy than nearby alternatives.

4.5.9 Anomalous Configurations: Chromium and Copper

Two classic, chemically important exceptions occur in the first-row transition metals: chromium and copper. They are not arbitrary anomalies; they reflect the competition between subshell energies that are extremely close.

Chromium

A naive Aufbau prediction might give:

Experimentally, the ground state is:

By promoting one electron from 4s to 3d, chromium achieves a half-filled 3d⁵ subshell, which is especially stable.

Copper

A naive prediction:

Observed:

Here, the promotion yields a filled 3d¹⁰ subshell, also especially stable.

What these exceptions teach

Subshell energies can be close enough that electron–electron interactions decide the winner.

The Aufbau ordering is a highly effective guideline, but the true ground state is determined by the lowest total energy, not by a rigid fill sequence.

4.5.10 Periodic Table Blocks and Configuration Patterns

Electron configurations explain the block structure of the periodic table:

s-block: groups 1–2 (and helium by configuration), filling ns.

p-block: groups 13–18, filling np.

d-block: transition metals, filling (n − 1)d.

f-block: lanthanides and actinides, filling (n − 2)f.

This block pattern is not merely organizational; it encodes which subshell contains the valence electrons and thus predicts broad similarities in chemistry.

4.5.11 Valence Electrons and Chemical Reactivity (Preview)

In general chemistry, valence electrons are the electrons outside the noble-gas core, especially those in the highest n shell for main-group elements. These electrons participate most directly in bonding and reactions. For transition metals, valence behavior can involve both ns and (n − 1)d electrons, which helps explain variable oxidation states and coordination chemistry.

Electron configurations therefore set the stage for periodic trends: atomic size, ionization energy, electron affinity, and metallic character all follow from how strongly the nucleus attracts valence electrons and how effectively inner electrons shield them. We turn next to these trends and their underlying cause: effective nuclear charge.

4.6 Periodic Trends: Radius and Ionization Energy

Electron configurations tell us which electrons are valence electrons and which orbitals they occupy. Periodic trends describe how strongly those electrons are held and how far, on average, they reside from the nucleus. Two of the most foundational trends are atomic size (atomic/ionic radius) and ionization energy. Both are governed primarily by the competition between (1) the attractive force from the nucleus and (2) the screening (shielding) of that attraction by other electrons.

4.6.1 Effective Nuclear Charge and Shielding

Nuclear charge versus effective nuclear charge

The nucleus contains Z protons, so the total nuclear charge is +Ze. A valence electron, however, does not feel the full nuclear charge because inner electrons partially screen the nucleus.

We therefore define the effective nuclear charge felt by a given electron as:

where S is a shielding constant that represents the reduction in nuclear attraction due to other electrons. This is a model (not a directly measurable quantity), but it is extremely useful for interpreting trends.

Penetration and why orbitals with the same n differ

Electrons in different subshells penetrate toward the nucleus to different extents. An electron with significant electron density close to the nucleus is less shielded and experiences larger Zeff. A standard qualitative ordering of penetration (and therefore typical stabilization in multielectron atoms) within a given shell is:

Thus, within the same principal shell n:

ns electrons feel a larger Zeff than np electrons,

which feel a larger Zeff than (n − 1)d electrons, etc.

This penetration/shielding concept helps explain both the order of orbital energies and the detailed structure of periodic trends.

4.6.2 Atomic Radius: What It Means and How It Varies

Defining atomic radius

An isolated atom has no sharp boundary, so “atomic radius” must be defined operationally. A common chemical definition is based on bonding distances:

Covalent radius: half the distance between nuclei in a homonuclear diatomic molecule (e.g., Cl2).

Metallic radius: half the distance between adjacent nuclei in a metallic crystal.

Although the numerical values depend on the definition, the periodic trends are robust.

Trend across a period (left to right): radius decreases

Moving left to right across a period:

Z increases by 1 each element.

Electrons are added to the same principal shell (same n for main-group elements).

Shielding by electrons in the same shell is incomplete; valence electrons do not perfectly shield one another.

As a result, Zeff increases across a period, pulling the valence electron density closer to the nucleus. Therefore, atomic radius generally decreases from left to right across a period.

Trend down a group: radius increases

Moving down a group:

The principal quantum number n of the valence shell increases (new shells are added).

Inner shells provide substantial shielding.

Even though Z increases significantly, the increase in distance (larger n) and shielding dominate, so the outer electron cloud extends farther out. Therefore, atomic radius increases down a group.

Important refinement: transition metals and lanthanide contraction (qualitative)

Across the transition series, radius changes are more gradual because added electrons enter (n− 1)d orbitals, which provide imperfect shielding. In heavier elements, poor shielding by 4f electrons contributes to the lanthanide contraction, which influences sizes of 5d and 6p elements. A detailed treatment is beyond the scope of this section, but the key idea remains: trends reflect how effectively added electrons shield the nucleus.

4.6.3 Ionic Radius: Cations, Anions, and Isoelectronic Series

Cations are smaller than their parent atoms

When an atom forms a cation, electrons are removed. This produces two effects:

Reduced electron–electron repulsion: fewer electrons means less mutual repulsion, allowing the remaining electrons to contract inward.

Higher Zeff per electron: the same nuclear charge attracts fewer electrons.

Thus, cation radius is smaller than atomic radius. The effect is especially large when ionization removes the entire valence shell (e.g., Na → Na+, losing the 3s electron leaves the filled n = 2 shell as the outermost shell).

Anions are larger than their parent atoms

When an atom forms an anion, electrons are added to the valence shell:

Electron–electron repulsion increases.

Shielding among valence electrons increases, reducing the net attraction felt by each.

Therefore, anion radius is larger than atomic radius.

Isoelectronic series: the cleanest test of nuclear attraction

In an isoelectronic series, species have the same number of electrons but different Z. Example:

(all have 10 electrons). Within an isoelectronic series, the radius decreases as Z increases because electrons are pulled in more strongly:

This illustrates the central role of nuclear attraction when electron count is held constant.

4.6.4 Ionization Energy: Definition and Physical Meaning

First ionization energy

The first ionization energy (IE₁) is the minimum energy required to remove the most weakly bound electron from a gaseous atom:

Ionization energies are always positive because energy must be supplied to separate an electron from the attractive force of the nucleus.

Successive ionization energies

Second and later ionizations remove additional electrons:

X+(g) → X2+(g) + e− | IE 2, |  |
X2+(g) → X3+(g) + e− | IE 3, etc. |  |

In general:

because removing electrons makes the ion more positive and the remaining electrons feel stronger attraction.

Large jumps as evidence for valence shells

A dramatic increase between successive ionization energies indicates that the next electron to be removed is a core electron rather than a valence electron. For example, after removing all valence electrons from a main-group element, the next ionization must disrupt a filled inner shell, requiring much more energy. These “jumps” are a powerful diagnostic for valence electron counts and thus connect directly to electron configurations.

4.6.5 Periodic Trend in First Ionization Energy

General trend across a period: increases

Across a period, Zeff increases and atomic radius decreases, so the valence electron is held more tightly. Therefore, first ionization energy generally increases left to right across a period.

General trend down a group: decreases

Down a group, the valence electron resides in a shell with larger n and is more shielded by inner electrons. Even though Z increases, the increased distance and shielding dominate. Therefore, first ionization energy generally decreases down a group.

Why the trend is periodic

Electron configurations repeat in a regular pattern across periods:

Alkali metals (group 1) have a single ns¹ valence electron: relatively easy to remove ⇒ low IE₁.

Noble gases (group 18) have filled valence shells (ns²np⁶ for n ≥ 2): electrons are tightly held ⇒ high IE₁.

This repetition creates the “sawtooth” periodicity seen in IE₁ versus atomic number.

4.6.6 Explaining Common Exceptions in Ionization-Energy Trends

While the overall periodic pattern is strong, there are predictable, instructive deviations. These deviations are not random; they reflect subshell energies and electron pairing.

Exception type 1: entering a new subshell (group 2 → group 13)

Across a period, IE₁ generally increases, but there is often a drop when moving from group 2 to group 13 (e.g., Be to B, Mg to Al). Reason:

Group 2 ends with ns².

Group 13 begins with ns²np¹.

A p electron is typically higher in energy (less penetrating, more shielded) than an s electron in the same shell, so it is easier to remove, lowering IE₁ relative to the simple trend.

Exception type 2: electron pairing within a subshell (group 15 → group 16)

Another common drop occurs from group 15 to group 16 (e.g., N to O, P to S). Reason:

Group 15 has a half-filled p subshell: np³ (one electron in each p orbital), which is relatively stable.

Group 16 has np⁴, forcing one p orbital to contain a paired set of electrons, increasing electron–electron repulsion.

The paired electron is slightly easier to remove, reducing IE₁ compared to the expected monotonic increase.

4.6.7 Connecting Radius and Ionization Energy

Atomic radius and first ionization energy are closely linked but not identical:

Smaller radius usually implies a more tightly held valence electron and thus larger IE₁.

Both trends reflect increasing Zeff across a period and increasing shielding/downward expansion down a group.

However, electron configuration details (subshell type, penetration, pairing) can produce local differences. This is why interpreting the periodic table requires both broad principles (Zeff, shielding) and the orbital-level structure introduced earlier.

In the next section, we extend the same framework to electron affinity and metallic character, completing the set of core periodic trends that predict how atoms gain, lose, and share electrons in chemical reactions.

4.7 Electron Affinity and Metallic Character

Atomic radius and ionization energy quantify two complementary aspects of electron-nucleus attraction: how far valence electrons extend (size) and how much energy is required to remove an electron (ionization). A third property, electron affinity, addresses the energetic favorability of adding an electron. Together with metallic character—the tendency of an element to lose electrons and form cations—these ideas unify the main periodic trends into a coherent predictive framework.

4.7.1 Electron Affinity: Definition and Sign Convention

Thermochemical definition

The electron affinity (EA) of an atom is the energy change when an electron is added to a gaseous atom to form a gaseous anion:

Electron affinity is an energy change for a process involving an isolated atom in the gas phase. Like ionization energy, it is defined for atoms, not for bulk solids or molecules (although related concepts appear there as well).

Important note on sign conventions

Two sign conventions are used in chemistry texts:

Thermochemical convention (used here): EA is the energy change ΔE. If energy is released upon electron addition, then EA < 0 (exothermic).

Magnitude convention: some tables report electron affinity as the magnitude of energy released (a positive number for favorable additions).

To avoid confusion, always read the definition attached to a numerical table. In this

chapter, we interpret trends using the thermochemical convention: more negative EA means a stronger tendency to accept an electron.

First versus second electron affinity

The definition above refers to the first electron affinity. Adding a second electron to an anion,

is always energetically unfavorable in the gas phase because the incoming electron is repelled by the negatively charged ion. Thus, the second electron affinity is positive (endothermic). In real ionic compounds, lattice energy can compensate for this repulsion; therefore, stable dianions can exist in solids even if the gas-phase second electron affinity is unfavorable.

4.7.2 Electron Affinity vs. Ionization Energy: Similarities and Differences

Ionization energy and electron affinity are sometimes conflated because both involve electrons and energy changes, but they correspond to different processes:

| Ionization Energy (IE) | Electron Affinity (EA)
Process | X(g) → X+(g) + e– | X(g) + e– → X– (g)
Meaning | Difficulty of removing an electron | Favorability of gaining an electron
Sign (typical) | IE > 0 (always endothermic) | often EA < 0 (often exothermic)
Most relevant for | Metals, cation formation | Nonmetals, anion formation

Despite these differences, both are governed by the same underlying physics: effective nuclear charge, shielding, and the stability of particular electron configurations.

4.7.3 Periodic Trend in Electron Affinity

General trend across a period

Across a period (left → right), Zeff increases and atomic radius decreases. A smaller atom with higher Zeff attracts an added electron more strongly. Therefore, electron affinity generally becomes more negative across a period.

However, the trend shows significant structure because adding an electron places it into specific subshells, and subshell stability matters.

General trend down a group

Down a group, the added electron goes into a higher-n shell, farther from the nucleus and more shielded. Therefore, electron affinity generally becomes less negative (less exothermic) down a group. In practice, this trend can have exceptions due to subtle electron-electron repulsion and subshell effects.

4.7.4 Subshell Effects and Notable Exceptions

Electron affinity does not vary smoothly because adding an electron can:

complete a subshell (stabilizing),

begin a new subshell (less stabilizing),

force electron pairing in a compact orbital (destabilizing due to repulsion).

Noble gases: electron affinity near zero or positive

Noble gases have closed-shell configurations (ns2np6 for n ≥ 2). Adding an electron would place it into a new, higher-energy subshell (e.g., (n + 1)s), and the closed shell is already very stable. As a result, noble gases have electron affinities that are small in magnitude and often positive (unfavorable) in the gas phase.

Alkaline earth metals and group 12: low tendency to gain electrons

Group 2 elements (ns2) and group 12 (d10s2) do not strongly favor gaining an electron because the next electron would enter a higher-energy p orbital, creating an electron configuration that is not especially stabilized. Their electron affinities are therefore less negative than neighboring elements.

Group 15: reduced electron affinity due to half-filled p subshells

Group 15 elements have np3, a half-filled p subshell with one electron in each p orbital (Hund’s rule). Adding one electron forces pairing in one p orbital, increasing electron-electron repulsion. This can make electron affinity less negative than expected from the general across-period trend.

Halogens: strongly negative electron affinities

Halogens (group 17) have ns2np5 and are one electron short of a filled valence shell. Adding an electron yields the stable noble-gas configuration ns2np6, so electron affinity is strongly negative. This helps explain their strong oxidizing behavior and the prevalence of X– ions in ionic compounds.

4.7.5 Metallic Character

Operational definition

Metallic character refers to the tendency of an element to behave as a metal: to lose electrons easily, form cations, conduct electricity in the solid state, and form metallic bonds. From an electronic-structure viewpoint, metallic character correlates with:

low first ionization energy (electrons are easily removed),

large atomic radius (valence electrons are farther from the nucleus),

less negative electron affinity (weaker tendency to accept electrons).

Periodic trend in metallic character

Across a period (left →right): metallic character decreases. Increasing Zeff pulls valence electrons closer, increasing ionization energies and favoring electron gain rather than loss.

Down a group: metallic character increases. Valence electrons occupy higher-n shells, are more shielded, and are removed more easily.

These trends explain the broad placement of metals (left and center), nonmetals (upper right), and metalloids (along the dividing staircase) on the periodic table.

4.7.6 Synthesis: How Radius, Ionization Energy, and Electron Affinity Work Together

The most chemically useful insight is that these properties are different windows on the same underlying cause: effective nuclear charge and the electronic structure of valence shells.

Predicting tendency to form ions

A simple, powerful qualitative rule is:

Atoms that form cations (metals): large radius, low IE₁, weakly negative or small-magnitude EA.

Atoms that form anions (nonmetals): smaller radius, high IE₁, strongly negative EA (especially halogens).

Atoms that resist ion formation (noble gases): high IE₁ and near-zero/positive EA; they rarely form stable monatomic ions under ordinary conditions.

Predicting bond type and polarity (preview)

Although bonding is treated in later chapters, periodic trends already suggest bonding behavior:

Large differences in metallic character (typically metal + nonmetal) favor ionic bonding.

Similar metallic character (nonmetal + nonmetal) favors covalent bonding.

Intermediate differences (metalloid behavior or polar covalent interactions) produce mixed character and partial charge separation.

A unified mental model

It is helpful to summarize the direction of change for the core properties:

Increasing Zeff (across a period) ⇒ radius decreases, IE₁ increases, EA becomes more negative, metallic character decreases.

Increasing n (down a group) with increased shielding ⇒ radius increases, IE₁ decreases, EA becomes less negative, metallic character increases.

This synthesis is the conceptual endpoint of the chapter’s development: wave mechanics and quantization lead to orbitals; orbitals lead to electron configurations; electron configurations, combined with Zeff and shielding, generate the repeating chemical patterns encoded by the periodic table.

4.8 Problems

4.8.1 Photon Energy from Frequency (Quantization)

Monochromatic ultraviolet light of frequency ν = 9.20 × 10¹⁴ s⁻¹ is used in a photoelectric experiment.

1. Calculate the energy of a single photon in joules.

2. Convert the photon energy to electronvolts (eV).

3. Calculate the wavelength λ of the radiation in nanometers.

4. A metal has a work function ϕ = 2.10eV. Determine whether photoelectrons are emitted, and if so, calculate the maximum kinetic energy Kmax (in eV) predicted by the photon model.

4.8.2 de Broglie Wavelength of a Moving Electron

An electron is accelerated from rest through a potential difference of 150V in vacuum.

1. Assuming nonrelativistic motion, compute the electron’s speed.

2. Compute the de Broglie wavelength of the electron.

3. Compare this wavelength to a typical atomic diameter (∼ 1.0×10⁻¹⁰ m). State whether diffraction by atomic spacings is plausible and justify briefly.

4.8.3 Bohr Energy Levels and Hydrogen Transitions

Consider the hydrogen atom and the Bohr energy-level expression En = −13.6eV∕n².

1. Calculate ΔE (in eV and J) for the emission transition ni = 5 → nf = 2.

2. Determine the wavelength of the emitted photon in nanometers.

3. Identify the spectral series (Lyman, Balmer, Paschen, etc.) for this transition.

4. Determine the series limit wavelength for this series (i.e., as ni → ∞ with the same nf).

4.8.4 Valid and Invalid Sets of Quantum Numbers

For each proposed set of quantum numbers (n,ℓ,mℓ,ms), decide whether it is allowed for an electron in an atom. If it is invalid, identify exactly which rule(s) it violates.

1. (3,2,−2,+)

2. (2,2,0,−)

3. (4,0,1,+)

4. (5,4,−5,−)

5. (1,0,0,−)

6. (3,1,0,+)

4.8.5 Radial Probability and Nodes of the 3s Orbital

The hydrogen-like orbital node rules state: number of radial nodes = n − ℓ − 1 and number of angular nodes = ℓ.

1. For a 3s orbital, determine the number of radial nodes and angular nodes.

2. Sketch a qualitative graph of the radial probability distribution P(r) versus r for a 3s electron. Your sketch must indicate the location of radial nodes as points where P(r) = 0; the relative number of peaks (maxima); and the behavior as r → 0 and as r →∞.

3. Explain (in one or two sentences) why P(r) can be small at r = 0 even if |ψ(0)|² is nonzero for an s orbital.

4.8.6 Electron Configurations of Fe and Fe3+

1. Write the full ground-state electron configuration of neutral iron, Fe (Z = 26).

2. Write the condensed (noble-gas) electron configuration of Fe.

3. Form the Fe3+ ion. Write the condensed electron configuration of Fe3+ and explicitly state which subshell(s) lose electrons.

4. Draw an orbital diagram (box-and-arrow) for the 3d subshell only for Fe and for Fe3+ consistent with Hund’s rule.

5. Based solely on your orbital diagrams, predict which species has more unpaired electrons and briefly justify.

4.8.7 Explaining the Anomalous Configuration of Copper

Copper has an experimentally observed ground-state configuration of [Ar]4s¹ 3d¹⁰ rather than the naive Aufbau expectation [Ar]4s² 3d⁹.

1. Using the ideas of subshell stability and electron–electron repulsion, give a concise explanation (3–6 sentences) for why the observed configuration is favored.

2. Write the condensed electron configuration of Cu+ and Cu2+ and specify from which subshell(s) electrons are removed.

3. Compare the 3d occupancy of Cu, Cu+, and Cu2+ and comment on which is expected to be especially stable and why.

4.8.8 Ranking Atomic Radius and First Ionization Energy

Without using tabulated numerical values, rank each set in order of increasing atomic radius and then in order of increasing first ionization energy. For each ranking, provide a short justification based on effective nuclear charge, shielding, and electron configuration.

1. Na, Mg, Al, Si

2. O, S, Se

3. K, Ca, Ga, Br

4.8.9 Estimating Zeff for Valence Electrons: Carbon vs. Silicon

Using Slater’s rules (state any assumptions clearly), estimate Zeff for a valence 2p electron in carbon and for a valence 3p electron in silicon.

1. Write the electron configuration of C and Si.

2. Apply Slater’s rules to compute an approximate shielding constant S for each target electron.

3. Compute Zeff = Z − S for each and compare the results.

4. Explain how your computed Zeff values help rationalize the trends in atomic radius and ionization energy between C and Si.

4.8.10 Predicting Ionic Compound Formulas from Periodic Trends

For each pair of elements below, predict the most likely formula of the binary ionic compound formed (if any). Justify your prediction by identifying the most likely ion charges based on periodic trends, electron configurations, and the drive toward noble-gas-like configurations.

1. Mg and N

2. Al and S

3. Ba and F

4. K and O

5. Ga and Cl

In one additional sentence for each, state whether you expect the compound to be primarily ionic or to have significant covalent character, and briefly explain in terms of metallic character and electronegativity trends (qualitative only).

Chapter 5
Chemical Bonding and Molecular Structure

Why does table salt form brittle crystals while water flows as a liquid? The answer lies in the invisible architecture of chemical bonds. In this chapter, we move beyond the isolated atom to explore the forces that hold matter together. By mastering the localized electron model, Lewis structures, and the geometric predictions of VSEPR theory, students will learn to translate a chemical formula into a three-dimensional object, predicting its physical properties and reactivity based on the subtle interplay of electrons and nuclear geometry.

5.1 Types of Chemical Bonds

Chemical bonds are the forces that hold atoms together in stable arrangements. From an energy perspective, bonding occurs because the total potential energy of a collection of atoms can be lowered when attractive electrostatic interactions outweigh repulsive interactions at an appropriate separation distance. From an electronic perspective, bonding models are distinguished by what happens to valence electrons (the outermost electrons involved most directly in chemical change). This section introduces three foundational bonding models—ionic, covalent, and metallic—as idealized limits that describe how valence electrons are distributed in matter.

Although real substances may show mixed character, these three models provide a powerful organizing framework:

Ionic bonding: valence electrons are effectively transferred, producing discrete cations and anions that assemble into a lattice.

Covalent bonding: valence electrons are shared between specific pairs of atoms, producing directional bonds and discrete molecules or network solids.

Metallic bonding: valence electrons are delocalized over many atoms,

producing a “sea” (or band) of mobile electrons surrounding metal cations.

5.1.1 Bonding as a Potential Energy Minimum

Two neutral atoms far apart interact weakly; as they approach, several electrostatic interactions change simultaneously:

Attractive interactions (between positively charged nuclei and negatively charged electrons of the other atom) tend to lower the potential energy.

Repulsive interactions (nucleus–nucleus and electron–electron) tend to raise the potential energy.

A stable bond corresponds to a separation distance where the net effect is a minimum in potential energy. This distance is the equilibrium bond length re. If atoms move closer than re, repulsions rise sharply; if they move farther apart than re, attractions weaken and the potential energy increases toward the energy of separated atoms.

A useful quantity is the bond energy (bond dissociation energy), the energy required to break a bond in the gas phase:

The fact that D(A−B) is positive reflects that the bonded state has lower potential energy than separated atoms. Later sections will quantify how electron distribution (localized vs. shared vs. delocalized) controls both bond strength and molecular properties.

5.1.2 The Ionic Bond Model: Electron Transfer and Lattice Formation

What “electron transfer” means physically

In the ionic model, bonding is described as a transfer of one or more electrons from one atom (typically a metal) to another (typically a nonmetal), producing ions:

The driving force is the formation of strong electrostatic attractions between oppositely charged ions. Importantly, a single isolated ion pair is not the full story for most ionic compounds. In the condensed phase, ions arrange into an extended, repeating three-dimensional crystal lattice. Each cation is surrounded by several anions and vice versa, maximizing attractions while balancing repulsions.

Energetics: why lattices matter

Forming ions costs energy (e.g., ionization energy for producing cations), but placing ions into a lattice releases energy due to many ion–ion attractions. This energy release is often described by the lattice energy (the energy change associated with forming the ionic solid from gaseous ions). Strong ionic solids typically involve:

Higher charges (2+/2− stronger than 1+/1−),

Smaller ionic radii (shorter separations increase attraction).

These trends reflect Coulomb’s law: the magnitude of electrostatic attraction increases with charge magnitude and decreases with distance.

Characteristic properties explained by the ionic model

The ionic model connects electron localization (electrons associated primarily with anions) to macroscopic behavior:

High melting and boiling points: separating ions in a lattice requires overcoming many strong electrostatic attractions.

Brittleness: shifting lattice planes can bring like charges near each other, causing strong repulsion and fracture.

Electrical conductivity in melts/solutions but not as solids: in a solid lattice, ions are fixed in place; in a melt or aqueous solution, ions can move and carry charge.

5.1.3 The Covalent Bond Model: Electron Sharing and Directional Bonds

Shared electron density between nuclei

In the covalent model, a bond forms when two atoms share one or more pairs of valence electrons. A simple representation is a bonding electron pair located between two nuclei. Electron density between nuclei is stabilizing because each electron is attracted to both nuclei, increasing nucleus–electron attractions while allowing the nuclei to remain at a separation where repulsions are not overwhelming.

A single covalent bond corresponds (in the simplest Lewis picture) to one shared pair; multiple bonds correspond to multiple shared pairs:

As the number of shared pairs increases, bonds tend to become shorter and stronger because electron density between nuclei increases (though real trends depend on the specific atoms and orbitals involved).

Localized bonding leads naturally to molecular shapes

Because shared electron density tends to occupy specific regions between specific nuclei, covalent bonds are directional. Directionality underlies two crucial ideas developed later:

Lewis structures as a bookkeeping tool for shared pairs and lone pairs,

Molecular geometry (VSEPR theory), where electron domains arrange in 3D to minimize repulsions.

Thus, the covalent model is the natural starting point for explaining the existence of discrete molecules (like H2O and CH4) and the three-dimensional structures that control reactivity and physical properties.

Network covalent solids

Not all covalent substances exist as small molecules. In network covalent solids (e.g., diamond, SiO2 in quartz), covalent bonds extend throughout the entire crystal. In such cases:

The bonding is covalent (electron sharing), but there are no discrete “molecules”.

Strong covalent bonds throughout the lattice often give very high melting points.

This highlights a key theme: the bonding model describes electron distribution, while the material’s structure (molecular vs. extended network) shapes properties.

5.1.4 The Metallic Bond Model: Delocalization and Collective Behavior

From localized pairs to an electron sea

Metals are composed of atoms with relatively low ionization energies and valence electrons that are not strongly held to a single nucleus in the solid. In the metallic model:

Metal atoms contribute valence electrons to a delocalized pool spread over many atoms.

The remaining positively charged metal cores (often idealized as cations) occupy lattice positions.

The delocalized electrons provide the attractive force that holds the lattice together.

Unlike covalent bonding, where electron density is concentrated between particular pairs of atoms, metallic bonding is collective: each electron can be shared among many nuclei simultaneously.

Characteristic properties explained by electron delocalization

The metallic model gives direct, intuitive explanations for hallmark metallic properties:

Electrical conductivity: mobile electrons respond to an applied electric field, producing current.

Thermal conductivity: mobile electrons transport kinetic energy efficiently through the solid.

Malleability and ductility: metal cations can slide past one another while the electron sea maintains attraction, so the solid deforms rather than fractures.

Luster: delocalized electrons interact strongly with light, leading to reflection.

5.1.5 Bonding Models as Ideal Limits (and Why This Matters)

The ionic, covalent, and metallic models are best viewed as idealized endpoints describing electron distribution:

localized on one atom   ↔   shared between two atoms   ↔   delocalized over many atoms

Real substances can deviate from these ideals:

Many bonds are neither purely ionic nor purely covalent; they show polar covalent character.

Some solids contain both metallic and covalent features (e.g., semimetals; alloys with complex bonding).

Rather than weakening the models, this motivates a more refined question: How unevenly are electrons shared? The next section introduces electronegativity as a quantitative tool for predicting bond polarity and placing real bonds along the ionic–covalent continuum.

5.1.6 Summary: Electron Behavior as the Unifying Theme

Chemical bonding is stabilized by a balance of attractive and repulsive electrostatic interactions, producing a potential-energy minimum at the equilibrium bond length.

Ionic bonding is characterized by electron localization primarily on anions and a stabilizing ion lattice.

Covalent bonding is characterized by shared electron density between specific nuclei, producing directional bonds and molecular (or network) structures.

Metallic bonding is characterized by delocalized valence electrons shared among many atoms, producing conductivity and malleability.

5.2 Electronegativity and Bond Polarity

The bonding models introduced in the previous section (ionic, covalent, and metallic) are best understood as ideal limits describing how valence electrons are distributed. Many real bonds fall between the ionic and covalent limits: electrons are shared, but not shared equally. This unequal sharing produces bond polarity, which in turn influences molecular structure, intermolecular forces, solubility, boiling point, and chemical reactivity.

This section develops three closely connected ideas:

Electronegativity as a predictive measure of an atom’s tendency to attract shared electrons.

Bond dipoles and dipole moments as quantitative descriptors of bond polarity.

The bonding continuum: ionic and covalent bonding are not separate categories but endpoints of a spectrum.

5.2.1 Electronegativity: A Measure of Electron-Attracting Power

Definition and physical meaning

Electronegativity (EN) is a dimensionless quantity that describes how strongly an atom attracts electron density toward itself when it is bonded to another atom. In a covalent bond, electrons occupy a region of space shared by two nuclei. If one nucleus exerts a stronger attraction on that shared electron density, the electron cloud shifts toward that atom.

Electronegativity is not the same as:

Ionization energy (energy required to remove an electron from a gaseous atom),

Electron affinity (energy change when a gaseous atom gains an electron),

Atomic charge (which depends on the bonding situation).

However, electronegativity trends are broadly consistent with the underlying physics captured by ionization energy and electron affinity: atoms that hold valence electrons tightly and stabilize added electron density tend to be more electronegative.

The Pauling scale and periodic trends

In general chemistry, electronegativity is commonly discussed using the Pauling

scale. The Pauling values allow useful comparisons among elements even though electronegativity is not directly measurable as a single isolated property of an atom.

Two periodic trends are especially important:

Across a period (left → right): electronegativity generally increases as effective nuclear charge increases and atomic radius decreases.

Down a group (top → bottom): electronegativity generally decreases as valence electrons are farther from the nucleus and more shielded by inner electrons.

Fluorine is the most electronegative element on the Pauling scale; cesium and francium are among the least electronegative (values may vary slightly depending on the dataset used).

Electronegativity difference

For a bond between atoms A and B, the electronegativity difference is

As ΔEN increases, electron sharing becomes more uneven. This does not mean that a sharp boundary exists where “covalent becomes ionic.” Instead, increasing ΔEN gradually increases the bond’s ionic character.

5.2.2 From Unequal Sharing to Bond Dipoles

Partial charges and bond polarity

When two different atoms form a covalent bond, the more electronegative atom draws electron density toward itself. The resulting separation of charge is commonly represented using partial charges:

if B is more electronegative than A. The symbol δ emphasizes that the charges are typically fractional rather than full integer charges (±1,±2,…) found in ionic compounds.

A bond with no charge separation (equal sharing) is nonpolar covalent. A bond with a significant but incomplete separation is polar covalent. In the limiting case of nearly complete transfer, the bond is described as ionic. These terms are qualitative labels for different regions on a continuum.

Bond dipole moment as a quantitative measure

A bond dipole is a vector quantity describing both:

the magnitude of charge separation, and

the direction of separation (from δ⁺ toward δ⁻).

The magnitude of the dipole moment is modeled as

where:

Q is the magnitude of the separated charge (in coulombs, C),

r is the distance between the centers of positive and negative charge (in meters, m).

Because molecular-scale dipoles are small, dipole moments are commonly reported in Debye (D):

Interpreting μ and separating its contributors

The expression μ = Qr highlights two independent contributors to bond polarity:

How much charge is separated (Q), which generally increases as ΔEN increases.

How far the charge is separated (r), which depends on bond length (often increasing down a group as atoms get larger).

Thus, a bond with a modest electronegativity difference can still have a sizable dipole moment if the bond is long, while a short bond can have a smaller dipole even if Q is substantial.

5.2.3 The Ionic–Covalent Continuum

Why a continuum is necessary

The ionic model pictures electrons localized on ions; the covalent model pictures electrons shared between nuclei. In reality, the electron density in a bond is a continuous distribution in space, not a discrete object that is either fully transferred or perfectly shared. The idea of a continuum reflects this physical reality:

At very small ΔEN, the electron density is symmetrically distributed (nonpolar covalent).

At moderate ΔEN, the density is skewed (polar covalent).

At large ΔEN, the density is strongly localized near one atom and the material is well described as ionic in many contexts.

Percent ionic character (conceptual overview)

Chemists sometimes describe bond character using percent ionic character, a quantity that compares the observed dipole moment of a bond to the dipole moment expected for complete charge separation:

Here, μionic is the dipole moment predicted if charges were fully separated by one elementary charge (e) over the internuclear distance r:

This ratio is a useful interpretive tool, but it should be used with care:

Real charge distributions are not point charges located exactly on nuclei.

The “ionic” reference assumes full ±e charges and a specific separation distance.

Despite these limitations, percent ionic character strongly correlates with ΔEN and reinforces the key message: bonding character changes gradually, not abruptly.

5.2.4 Practical Use: Predicting Bond Polarity from Electronegativity

A working procedure

To predict and represent bond polarity in typical general chemistry problems:

Identify the two bonded atoms.

Compare their electronegativities to determine which is more electronegative.

Assign partial charges: the more electronegative atom is δ⁻; the other is δ⁺.

Draw a bond dipole arrow (if requested), pointing toward δ⁻ with a “cross” (or plus-like mark) at the δ⁺ end.

This procedure is qualitative but highly reliable for recognizing polar bonds and anticipating molecular polarity (developed later).

Examples of increasing polarity

Example 1: Homonuclear bond (nonpolar covalent). In F2 or Cl2, the atoms are identical, so ΔEN = 0 and the bond is nonpolar. Electron density is shared symmetrically:

Example 2: Moderately polar covalent bond. In hydrogen fluoride, fluorine is much more electronegative than hydrogen, so electron density shifts toward F:

This bond is polar covalent with a substantial bond dipole.

Example 3: Largely ionic bond. In a bond such as LiF (in the solid, an ionic lattice), the electronegativity difference is large enough that the ionic model is highly effective:

Even here, the term “ionic” does not require perfect electron transfer at the quantum level; it signals that electron density is strongly localized and that the crystal is composed of ions.

5.2.5 Bond Polarity and Chemical Consequences (Preview)

Bond polarity is not merely a labeling exercise; it has measurable consequences:

Polar bonds can create attractive interactions between molecules (dipole–dipole forces) and can enable hydrogen bonding when H is bonded to highly electronegative atoms (notably N, O, F).

Polar bonds often indicate reactive sites: the δ⁺ end is more susceptible to nucleophilic attack, while the δ⁻ end can attract electrophiles (language that becomes more precise in organic and advanced chemistry).

In polyatomic species, bond dipoles combine as vectors; geometry can cause them to cancel or reinforce, determining molecular polarity (treated in a later section).

5.2.6 Key Takeaways

Electronegativity (Pauling scale) describes an atom’s tendency to attract shared electrons.

Differences in electronegativity produce unequal electron sharing, giving polar covalent bonds with partial charges.

The dipole moment magnitude is modeled by μ = Qr and is commonly reported in Debye (D).

Ionic and covalent bonding form a continuum; increasing ΔEN generally increases ionic character, but no sharp boundary exists.

5.3 Lewis Structures and the Octet Rule

The previous sections established two ideas that make Lewis structures useful: (1) in many compounds, bonding can be described as localized electron pairs (covalent bonding) or localized charge on ions (ionic bonding), and (2) when electrons are shared unequally, bond polarity can be predicted from electronegativity differences. Lewis structures build on these ideas by providing a systematic, two-dimensional bookkeeping method for valence electrons: which atoms are connected, how many electrons are involved in bonds, and how many remain as lone pairs.

A Lewis structure is not a photograph of electron density. Instead, it is a model that answers three practical questions:

Which atoms are bonded to which (connectivity)?

How many electron pairs are shared (single, double, triple bonds)?

How many nonbonding electrons remain on each atom (lone pairs)?

This information provides the foundation for later topics such as formal charge, resonance, molecular shape (VSEPR), and molecular polarity.

5.3.1 The Octet Rule and Its Scope

Statement of the rule

The octet rule states that many main-group atoms form bonds so that each atom achieves a valence-shell electron count of eight (an octet), corresponding to the electron configuration of the nearest noble gas. For hydrogen, the analogous goal is a duet (two electrons), corresponding to helium.

In Lewis structures:

A bonding pair (a line) counts as two electrons shared by two atoms.

A lone pair (two dots) counts as two electrons belonging to a single atom.

Why the octet rule works (and when it does not)

For many elements in Period 2 (C, N, O, F) the octet rule is especially reliable because their valence shell consists of the 2s and 2p orbitals, which can accommodate at most eight electrons. However, there are important exceptions (treated later): electron-deficient compounds (e.g., BF3), odd-electron species (radicals), and expanded-octet species (common for Period 3 and below, e.g., SF6). In this section we focus on the standard use of the octet rule as an organizing principle for typical molecules and polyatomic ions.

5.3.2 Valence Electron Accounting: The Core Skill

Lewis structures are based on conservation of electrons: the total number of valence electrons in the structure must equal the sum contributed by each atom, adjusted for charge.

Step 1: Count total valence electrons

To determine the total number of valence electrons available:

Sum the valence electrons of each atom (using group numbers for main-group elements).

Add one electron for each negative charge, or subtract one electron for each positive charge.

For example:

Example: NF3. Nitrogen (Group 15) contributes 5 valence electrons; each fluorine (Group 17) contributes 7.

Step 2: Choose a skeletal structure (connectivity)

Lewis structures require a plausible arrangement of atoms:

The least electronegative atom is often central (exceptions: hydrogen is never central; halogens are usually terminal).

In molecules with one obvious central atom (e.g., C in many organics; N in NF3), choose that as the center.

For NF3, nitrogen is central with three N–F single bonds.

Step 3: Form single bonds and subtract bonding electrons

Each single bond uses 2 electrons. After drawing the skeletal structure with single

bonds, subtract 2× (number of bonds) from the total.

For NF3:

Step 4: Complete octets on terminal atoms first

Distribute remaining electrons as lone pairs to terminal atoms until each reaches an octet (or duet for H). This step works well because terminal atoms typically cannot form additional bonds without changing connectivity.

For NF3, each fluorine already has 2 electrons in the N–F bond; each needs 6 more as three lone pairs:

Remaining electrons:

Step 5: Place remaining electrons on the central atom

Any electrons still unassigned are placed on the central atom as lone pairs.

For NF3, the remaining 2 electrons form one lone pair on nitrogen. Nitrogen now has:

so the octet is satisfied.

5.3.3 Algorithmic Flowchart for Lewis Structures

The following flowchart summarizes a reliable procedure. It includes decision points for the most common complication encountered at this stage: the central atom lacks an octet after all terminal atoms have been satisfied. In that case, one or more lone pairs on terminal atoms can be converted into multiple bonds.

5.3.4 Reading and Writing Lewis Structures

Conventions

Lewis structures are drawn using standard conventions:

Single bond: one line.

Double bond: two lines.

Triple bond: three lines.

Lone pairs: pairs of dots (or sometimes short lines) on atoms.

For polyatomic ions, the entire structure is enclosed in brackets with the net charge written outside:

Electron-count check (a required final step)

After drawing a structure, always verify:

The total number of electrons shown equals the total valence electrons counted.

Hydrogen atoms have 2 electrons (duet).

Most Period 2 atoms have 8 electrons (octet), unless the species is an exception (later).

This check prevents the most common errors: missing lone pairs, incorrect charge adjustment, and incorrect number of bonds.

5.3.5 Typical Patterns That Reduce Guesswork

As you gain experience, common valence patterns help you choose reasonable skeletal structures and spot mistakes quickly:

Halogens (F,Cl,Br,I) typically form one bond and have three lone pairs in neutral molecules.

Oxygen often forms two bonds and has two lone pairs (neutral).

Nitrogen often forms three bonds and has one lone pair (neutral).

Carbon often forms four bonds and has no lone pairs (neutral).

Hydrogen forms one bond and has no lone pairs.

These are not inviolable rules, but they are consistent with the octet rule and provide a useful “sanity check.”

5.3.6 Worked Illustrations (No Formal Charge Yet)

The next section introduces formal charge as a systematic way to compare multiple valid Lewis structures. For now, we focus on generating structures that satisfy electron counting and octets.

Example: CO2

Count valence electrons:

Skeleton: O−C−O. Single bonds use 4 e⁻, leaving 12 e⁻. Completing octets on both oxygens uses all 12 electrons, but then carbon has only 4 electrons around it (two single bonds). Carbon lacks an octet, so convert one lone pair from each oxygen into a bonding pair to form double bonds:

Now carbon has 8 electrons (two double bonds), each oxygen has an octet, and the total electron count remains 16.

Example: CN–

Count valence electrons:

Skeleton: C−N. A triple bond (6 e⁻) plus one lone pair on each atom (4 e⁻) uses 10 electrons:

(orientation and placement of the negative charge will be clarified using formal charge in the next section). Both atoms have octets (6 bonding electrons + 2 lone-pair electrons on each).

5.3.7 Limitations of Lewis Structures (Preview)

Lewis structures are intentionally simplified:

They depict electron pairs as localized, even though real electron density can be delocalized.

They are two-dimensional; they do not directly give 3D shape (VSEPR does).

They may not be unique: some species require multiple contributing structures (resonance).

Nevertheless, they remain one of the most effective first tools for predicting connectivity, bond order, and many trends in molecular behavior.

5.3.8 Key Takeaways

Lewis structures are a valence-electron bookkeeping model representing bonds and lone pairs.

The octet rule (and duet rule for H) guides typical Lewis structures for many main-group compounds.

A reliable algorithm is: count electrons →choose skeleton →single bonds → terminal octets → central atom → multiple bonds if needed → final electron-count check.

When multiple structures seem possible or when charges appear ambiguous, formal charge and resonance provide the next level of evaluation.

5.4 Formal Charge and Resonance

Lewis structures provide a systematic way to account for valence electrons and satisfy octets, but they do not always produce a unique picture of bonding. Often, more than one Lewis structure can be drawn for the same connectivity and total electron count. In such cases, we need additional criteria to evaluate which structure (or combination of structures) best represents the electron distribution.

Two ideas refine the Lewis model:

Formal charge is a bookkeeping method that assigns electron ownership in a bond as if the electrons were shared equally. Formal charges help identify the most reasonable Lewis structure among alternatives.

Resonance acknowledges that when electrons are delocalized, no single Lewis structure can represent the real electron distribution; instead, the best description is a resonance hybrid.

These tools build directly on earlier themes: bond polarity arises from unequal sharing, and some electron density is shared between atoms rather than localized. Resonance extends this further: in certain molecules and ions, electron density is shared among three or more atoms in a way that cannot be captured by a single arrangement of double and single bonds.

5.4.1 Formal Charge: Definition and Calculation

What formal charge is (and is not)

Formal charge (FC) is the charge an atom would have if:

all lone-pair electrons are assigned to the atom on which they are drawn, and

bonding electrons are divided equally between the two bonded atoms,

regardless of electronegativity.

Formal charge is therefore not the same as the real partial charges (δ⁺, δ⁻) discussed in bond polarity. Partial charges arise from unequal sharing; formal charges arise from an equal-sharing bookkeeping rule. Despite this simplification, formal charge is extremely useful for selecting good Lewis structures.

The formal charge formula

For an atom in a Lewis structure,

where:

V = number of valence electrons in the free (neutral) atom,

N = number of nonbonding (lone-pair) electrons assigned to that atom in the structure,

B = number of bonding electrons (2 per bond line) shared by that atom in the structure.

A common equivalent form uses bond lines directly:

This works because each bond line corresponds to two bonding electrons, and the atom “owns” half (one electron per bond line) under the formal-charge convention.

A required check: formal charges must sum to the net charge

For a correct Lewis structure of an ion, the sum of formal charges must equal the overall ionic charge:

For a neutral molecule, the sum must be zero.

5.4.2 Using Formal Charge to Evaluate Lewis Structures

Formal charge does not guarantee a correct structure, but it provides strong guidance. The following rules are widely effective for main-group compounds under the octet rule.

Rules of thumb for “best” Lewis structures

Among plausible Lewis structures with the same connectivity and total electron count, the most important criteria are:

Minimize the magnitude of formal charges. Structures with many atoms bearing large + or − charges are typically less favorable.

Minimize charge separation. If charges must exist, structures with fewer separated + and − centers are often preferred.

Place negative formal charge on more electronegative atoms. This aligns the bookkeeping assignment with the physical tendency of electronegative atoms to stabilize electron density.

Place positive formal charge on less electronegative atoms. Less electronegative atoms more readily tolerate electron deficiency in many contexts.

These criteria connect to earlier ideas: electronegativity reflects an atom’s ability to stabilize electron density, so it is chemically reasonable when formal charge assignments follow electronegativity trends.

Example: CO (multiple valid octet structures)

Carbon monoxide has 10 valence electrons:

A triple bond with one lone pair on each atom satisfies octets:

Formal charges:

This result may feel counterintuitive because oxygen is more electronegative, yet carries +1 formal charge. This highlights a key point: formal charge is not a direct map of electron density. In CO, the bonding picture is better refined by molecular orbital theory in later courses; nonetheless, the Lewis structure is still valuable for electron counting and predicting many reaction patterns.

Example: SO2 (formal charge guides multiple-bond placement)

Sulfur dioxide has 18 valence electrons:

Several Lewis structures satisfy the electron count, including forms with different placements of double bonds. Formal charge helps identify contributors that minimize charge separation and place negative charge on oxygen (more electronegative), though SO2 also illustrates that Period 3 elements can show bonding patterns not perfectly captured by simple octet-based Lewis structures. The main message is methodological: formal charge is the tool used to compare candidate Lewis structures.

5.4.3 Resonance: When One Lewis Structure Is Not Enough

Definition of resonance structures

Resonance structures (also called resonance contributors) are two or more valid Lewis structures for the same species that:

have the same arrangement of atoms (same connectivity),

differ only in the placement of electrons (locations of multiple bonds and/or lone pairs),

each satisfy the total valence electron count.

Resonance structures are typically written with a double-headed arrow:

What resonance does not mean

Resonance does not mean that the molecule or ion rapidly switches back and forth between structures. The real species is a single, stable arrangement with a single electron distribution. Resonance indicates that our Lewis model cannot capture that distribution with only one structure.

The resonance hybrid

The real electron distribution is described as a resonance hybrid: a weighted combination of resonance contributors. Contributors that:

minimize formal charges,

minimize charge separation, and

place negative charge on electronegative atoms

tend to contribute more strongly to the hybrid.

A practical consequence is bond equalization: if several resonance contributors place a double bond in different positions, the actual bonds often have lengths and strengths intermediate between single and double bonds.

5.4.4 Formal Charge Rules for Resonance Contributors

When evaluating resonance contributors, apply the same formal charge criteria, with

two additional points:

All resonance contributors must be valid Lewis structures (correct electron count; octets satisfied where applicable).

Equivalent contributors contribute equally. If resonance structures are related by symmetry and have identical formal charges, they have equal weight in the hybrid.

5.4.5 Resonance in a Central Example: The Nitrate Ion, NO3 –

The nitrate ion is a canonical example of resonance and bond equalization. It has 24 valence electrons:

A typical Lewis structure under the octet rule places nitrogen in the center with one N−−O double bond and two N−O single bonds; the negative charge appears on the singly bonded oxygens. However, there are three equivalent ways to place the double bond (on any one of the three oxygens), producing three equivalent resonance contributors.

Formal charges in a nitrate resonance contributor

Consider one contributor with one double bond and two single bonds:

Nitrogen has four bond lines (one double, two single) and no lone pairs. Using FC = V − (nonbonding e⁻) − (bond lines):

Double-bonded oxygen has two lone pairs (4 nonbonding electrons) and two bond lines:

Each single-bonded oxygen has three lone pairs (6 nonbonding electrons) and one bond line:

The sum of formal charges is +1 + 0 + (−1) + (−1) = −1, matching the ion’s charge.

Because the three contributors are equivalent by symmetry, they contribute equally. Therefore, the three N−O bonds in the actual ion are equivalent, each with bond order between 1 and 2 (often described as approximately 1).

Visualizing resonance contributors and the hybrid

The resonance hybrid is often drawn with:

three equivalent N−O bonds shown as dotted (partial double-bond) lines, and

the negative charge distributed evenly over the three oxygens.

In a simplified equal-weight picture, each oxygen carries an average formal charge of − in the hybrid: across the three contributors, each oxygen is single-bonded (and thus −1) in two contributors and double-bonded (0) in one contributor, giving an average of = −.

5.4.6 Nitrate Resonance Diagram (Contributors and Hybrid)

5.4.7 Implications of Resonance for Bonding and Reactivity

Bond lengths and bond orders

Resonance often leads to bond lengths that are intermediate between those expected for single and double bonds. In NO3−, all three N−O bonds are experimentally equivalent, consistent with a delocalized bonding picture rather than one localized double bond.

A useful qualitative rule is:

If resonance contributors redistribute π bonding among several positions, the affected bonds tend to become more equal in length and strength.

Charge delocalization and stability

Delocalization of negative charge over multiple electronegative atoms typically stabilizes an ion. This is one reason many polyatomic anions (e.g., nitrate, carbonate, sulfate) are relatively stable species. The Lewis/resonance framework captures this stabilization qualitatively: spreading charge reduces charge density and lowers electrostatic energy.

5.4.8 Key Takeaways

Formal charge is calculated by FC = V − N − and must sum to the net charge of the species.

Formal charge helps identify the most reasonable Lewis structures by minimizing charge magnitude and placing negative charge on more electronegative atoms.

Resonance arises when multiple valid Lewis structures differ only in electron placement; the real structure is a resonance hybrid, not an oscillation.

Resonance often produces bond equalization (intermediate bond order and equal bond lengths) and stabilization through charge delocalization.

5.5 Exceptions to the Octet Rule

The octet rule and Lewis structures provide a powerful framework for predicting connectivity, bond order, and electron distribution in many main-group compounds, especially those involving Period 2 elements such as C, N, O, and F. However, the octet rule is a model with a domain of usefulness rather than a universal law. Some stable molecules and ions cannot be represented with every atom having exactly eight valence electrons (and hydrogen two). In these cases, forcing an octet-counting approach produces incorrect structures, impossible electron counts, or formal-charge patterns that do not match known chemical behavior.

This section organizes octet-rule exceptions into three major classes:

Odd-electron species (radicals): an odd total number of valence electrons prevents all electrons from pairing.

Incomplete octets (electron-deficient compounds): some atoms are stable with fewer than eight electrons, most commonly Be and B.

Expanded octets (hypervalent compounds): certain central atoms (typically Period 3 and below) accommodate more than eight electrons in their valence shell in Lewis structures.

Throughout, we emphasize how to recognize these species from electron counting and how to draw consistent Lewis structures without violating basic bookkeeping rules.

5.5.1 A Strategy for Recognizing Octet Exceptions

When constructing Lewis structures using the algorithm from the previous section, an exception is usually signaled by one of these red flags:

The species has an odd total number of valence electrons (radical likely).

After completing terminal octets and placing remaining electrons on the central atom, the central atom still has fewer than eight electrons and there is no reasonable way to create multiple bonds without producing implausible formal charges (electron-deficient central atom likely).

After completing octets, there are still extra electrons that must be placed on the central atom (expanded octet likely), and the central atom is Period 3 or below.

Because formal charge provides a diagnostic tool, it should be used whenever multiple structures are possible. However, the existence of an octet exception is first and foremost determined by electron counting, not by preference.

5.5.2 Odd-Electron Species: Free Radicals

What makes radicals unavoidable

A radical is a species with one or more unpaired electrons. If the total number of valence electrons is odd, at least one unpaired electron must exist because electrons are discrete particles and Lewis structures place electrons in pairs (bonds or lone pairs). Therefore, odd electron count ⇒ radical is a simple and reliable diagnostic.

Radicals are important in combustion, atmospheric chemistry, polymerization, and many reaction mechanisms. Although some radicals are highly reactive, many are sufficiently stable to be detected and studied.

How to draw Lewis structures for radicals

The Lewis-structure algorithm remains the same, with one modification: one atom will end up with a single electron (a dot) in addition to any lone pairs and bonds. A radical electron is typically placed:

on the atom that minimizes formal charge magnitudes, and/or

on the less electronegative atom when alternatives are comparable (because very electronegative atoms tend to stabilize electron pairs rather than unpaired electron density in simple Lewis pictures).

This is a guideline rather than a strict rule; in practice, radical electron placement is best understood with molecular orbital ideas. In general chemistry, the goal is to produce a structure with the correct electron count and plausible bonding.

Example: nitric oxide, NO

Count valence electrons:

A common Lewis structure uses an N−−O double bond, with one unpaired electron and lone pairs arranged to give reasonable formal charges:

One can verify the electron count: a double bond contributes 4 bonding electrons; the remaining 7 electrons appear as lone pairs plus the unpaired electron.

Example: nitrogen dioxide, NO2

Count valence electrons:

A representative Lewis structure has one N– –O double bond, one N–O single bond, and one unpaired electron on nitrogen. Because there are two equivalent positions for the double bond, NO2 is also a resonance problem:

The unpaired electron is delocalized in the resonance hybrid, and the two N–O bonds become more similar than a pure single/double description suggests.

5.5.3 Incomplete Octets: Electron-Deficient Molecules

Common electron-deficient centers: Be and B

Some atoms form stable compounds in which the central atom has fewer than eight electrons in its valence shell. The most common cases in general chemistry are:

Beryllium (Group 2) compounds such as BeCl2 (often covalent in the gas phase),

Boron (Group 13) compounds such as BF3 and BCl3.

These atoms have relatively few valence electrons and can form stable structures without reaching an octet in simple Lewis terms.

Example: boron trifluoride, BF3

Count valence electrons:

Skeleton: B central with three single bonds to F uses 6 e₋, leaving 18 e₋. Completing octets on the three fluorines uses exactly 18 electrons (each F needs 6 as lone pairs). No electrons remain for boron. Boron has only three bonds (6 electrons) around it:

This is a legitimate Lewis structure even though B has only six valence electrons.

Why not force an octet by adding a double bond?

One might attempt to draw a B−−F double bond to give boron an octet. However, formal charge analysis typically reveals a less favorable charge separation (placing positive charge on F or negative charge on B in ways inconsistent with electronegativity), and experimentally BF3 behaves as an electron-pair acceptor (a Lewis acid), consistent with an electron-deficient boron center. Thus, the incomplete octet is chemically meaningful: it helps predict reactivity.

Lewis acidity as a consequence of electron deficiency

An atom with fewer than eight electrons in its valence shell can often accept a lone pair from a donor (Lewis base) to form an adduct. For example:

Here, NH3 donates a lone pair to boron. In Lewis notation, the coordinate covalent bond is often drawn with an arrow pointing from the donor to the acceptor:

This bonding picture remains consistent with earlier themes: electron distribution and electron-pair sharing determine bonding and properties.

Example: beryllium chloride, BeCl2

Count valence electrons:

Skeleton Cl−Be−Cl uses 4 electrons; completing both chlorine octets uses the remaining 12 electrons. Beryllium is left with two single bonds (4 electrons), far short of an octet:

Again, this is a valid electron-counting Lewis structure. (In the condensed phase, BeCl2 forms polymeric structures, and bonding is better described with more advanced models; the key point here is recognizing the electron-deficient pattern.)

5.5.4 Expanded Octets: Hypervalent Compounds

When expanded octets appear in Lewis structures

Many stable compounds contain central atoms surrounded by more than four electron pairs (more than eight electrons) in a Lewis structure. These are called expanded-octet or hypervalent species.

In general chemistry, expanded octets are most commonly drawn for:

Period 3 and heavier central atoms: P, S, Cl, Br, I, Xe.

Species with highly electronegative terminal atoms (often F or O) that stabilize bonding patterns and reduce unfavorable formal charges.

A simple diagnostic is:

Example: sulfur hexafluoride, SF6

Count valence electrons:

Skeleton with six S−F single bonds uses 12 e₋, leaving 36 e₋. Completing octets on six fluorines requires 6 × 6 = 36 e₋, leaving none. Sulfur is surrounded by six bonding pairs:

Thus the Lewis structure has sulfur with 12 electrons in its valence shell.

Example: phosphorus pentachloride, PCl5

Count valence electrons:

With five single bonds, P has 10 electrons around it. The Lewis structure is consistent with an expanded octet on phosphorus:

(geometry is treated later by VSEPR; here we focus only on electron counting and correct Lewis representation).

Example: sulfate ion, SO42– (expanded octet and resonance)

Count valence electrons:

A structure with four single bonds uses 8 electrons, leaving 24. Completing oxygen octets uses all 24. This gives sulfur 8 electrons (an octet) but produces formal charges of −1 on each oxygen and +2 on sulfur. Alternative structures introduce one or more S––O double bonds, reducing charge separation. These alternatives are related by resonance and are often used in general chemistry to represent the resonance hybrid of SO42– .

This example illustrates a theme: expanded octets often appear as part of a formal-charge minimization strategy in resonance-capable oxyanions.

5.5.5 Why Expanded Octets Are Allowed (Conceptual Explanation)

Size, electron density, and bonding models

For Period 2 elements, the octet rule is strict in the Lewis model because only the 2s and 2p valence orbitals are available, limiting the number of electrons that can be accommodated close to the nucleus. For Period 3 and heavier elements, the situation is more flexible:

These atoms are larger, so electron density can be distributed over a larger volume, reducing electron-electron repulsions for a given electron count.

Bonding in hypervalent molecules is often better described using delocalized bonding models (molecular orbitals), which can distribute electron density over several atoms and bonds.

In introductory Lewis terms, we treat expanded octets as acceptable for Period 3 and beyond when they provide a structure with correct electron count and more reasonable formal charges.

A practical rule set

In this course-level framework:

Do not give expanded octets to Period 2 elements (C, N, O, F).

Expanded octets are commonly used for Period 3+ elements (P, S, Cl, Br, I, Xe) when electron counting requires it or when it significantly improves formal charges.

5.5.6 Summary Table: Three Classes of Octet Exceptions

Exception type | Diagnostic | Representative examples
Odd-electron species (radicals) | Total valence electron count is odd; at least one unpaired electron must appear in the Lewis structure | NO, NO2, ClO2
Incomplete octet (electron-deficient) | Central atom has fewer than 8 electrons after all terminal octets are completed; multiple bonds do not resolve the deficiency without implausible formal charges | BF3, BCl3, BeCl2
Expanded octet (hypervalent) | Central atom (Period 3+) has more than 8 electrons in the Lewis structure; often occurs with many terminal atoms or when reducing formal charge separation | PCl5, SF6, SO42–

5.5.7 Key Takeaways

The octet rule is a highly useful guideline, especially for Period 2 elements, but it has systematic exceptions.

Odd-electron counts necessarily produce radicals; Lewis structures must show an unpaired electron.

Electron-deficient compounds (especially boron and beryllium) can be stable with incomplete octets and often act as Lewis acids.

Expanded-octet compounds are common for Period 3 and heavier central atoms; Lewis structures may show more than eight electrons around the central atom, particularly when this reduces formal charge separation.

When octet exceptions arise, prioritize: correct electron count, plausible connectivity, and formal-charge patterns consistent with electronegativity and known behavior.

5.6 VSEPR Theory: Molecular Shapes

Lewis structures describe connectivity (which atoms are bonded) and the distribution of valence electrons into bonding pairs and lone pairs. However, Lewis structures are two-dimensional. Real molecules are three-dimensional, and their shapes strongly influence polarity, physical properties (boiling point, solubility), and chemical reactivity (which directions bonds can form or break).

Valence Shell Electron Pair Repulsion (VSEPR) theory provides a systematic method to predict three-dimensional molecular geometry from a Lewis structure. VSEPR is based on a simple idea: electron domains around a central atom arrange themselves to minimize repulsion, producing a set of characteristic geometries.

5.6.1 Core Idea: Electron Domains Repel

Electron domains

In VSEPR, we count regions of electron density around a central atom. These regions

are called electron domains (or electron groups). An electron domain can be:

a single bond (one sigma bond),

a multiple bond (double or triple bond) counted as one domain,

a lone pair on the central atom,

(less commonly at this level) a single unpaired electron in a radical.

The key rule is:

VSEPR domain-counting rule: Each bond—single, double, or triple—counts as one electron domain. Each lone pair counts as one electron domain.

This rule is a practical approximation: multiple bonds contain more electron density than single bonds and can cause stronger repulsions, but for determining the basic geometry, counting them as one domain is usually sufficient.

Why domains arrange in specific geometries

Electron domains repel because electron clouds carry negative charge and occupy space. Domains spread out to maximize separation, which lowers electron–electron repulsion and stabilizes the molecule. For a given number of domains, there is a characteristic arrangement that maximizes separation:

2 domains: linear (180∘)

3 domains: trigonal planar (120∘)

4 domains: tetrahedral (109.5∘)

5 domains: trigonal bipyramidal (90∘, 120∘, 180∘)

6 domains: octahedral (90∘, 180∘)

5.6.2 Electron Geometry vs. Molecular Geometry

Electron geometry

The electron geometry (also called electron-domain geometry) describes the arrangement of all electron domains (bonding and nonbonding) around the central atom.

Molecular geometry

The molecular geometry (shape) describes the arrangement of atoms (bonding domains only) around the central atom. Lone pairs affect shape because they occupy positions in space, but they are not seen as atoms in the molecular geometry.

For example, NH3 has four electron domains (three bonds and one lone pair), so its electron geometry is tetrahedral. But because one domain is a lone pair, the molecular geometry is trigonal pyramidal.

5.6.3 Predicting Molecular Shape: A Step-by-Step Procedure

Given a correct Lewis structure:

Identify the central atom (the atom bonded to multiple atoms).

Count electron domains around the central atom: number of bonded atoms attached to the central atom (each bond counts once), plus the number of lone pairs on the central atom.

Assign the electron geometry based on the domain count (2–6).

Determine the molecular geometry by “removing” lone-pair positions and naming the remaining arrangement of atoms.

Estimate bond angles, remembering that lone pairs typically compress bond angles.

5.6.4 Relative Repulsions and Bond Angle Distortions

VSEPR predicts ideal angles when all domains are equivalent. In practice, domains repel with different strengths:

Lone pairs are more spatially diffuse (less constrained between nuclei) and therefore occupy more space, causing them to repel more strongly. As a result:

Lone pairs decrease bond angles between bonding pairs compared to ideal values.

Multiple bonds can also slightly increase repulsion relative to single bonds, sometimes causing measurable distortions.

Example: H2O. Oxygen has four electron domains (2 bonds + 2 lone pairs): tetrahedral electron geometry. The molecular geometry is bent. The ideal tetrahedral angle is 109.5∘, but in water the ∠H−O−H is about 104.5∘ because two lone pairs compress the bond angle.

5.6.5 Comprehensive VSEPR Reference (Electron Domains 2–6)

The table below provides a primary reference mapping electron-domain count and lone-pair count to electron geometry and molecular shape. At this level, we use the common AXE notation:

where A is the central atom, X is a bonded atom (or group), m is the number of bonded atoms, and E is a lone pair on the central atom.

5.6.6 Applying the Reference: Worked Shape Determinations

Linear electron geometry (2 domains)

CO2. Lewis structure: O−−C−−O. Central carbon has two electron domains (two double bonds, each counts as one domain):

Ideal bond angle: 180∘.

Trigonal planar electron geometry (3 domains)

BF3. Electron-deficient but still analyzed by VSEPR: boron has three bonding domains and no lone pairs:

Even though boron has only six electrons, VSEPR geometry prediction is still effective because the shape depends on electron-domain arrangement, not on whether the central atom has an octet.

SO2. A common resonance-aware Lewis description has sulfur with two bonding domains (to O) and one lone pair: AX₂E.

Bond angle is less than 120∘ due to lone-pair compression.

Tetrahedral electron geometry (4 domains)

CH4. Central carbon has four bonding domains, AX₄:

NH3. Nitrogen has three bonds and one lone pair, AX₃E:

Bond angles are slightly less than 109.5∘.

H2O. Oxygen has two bonds and two lone pairs, AX₂E₂:

Bond angle is significantly compressed relative to 109.5∘ because two lone pairs exert strong repulsions.

Trigonal bipyramidal electron geometry (5 domains)

Axial vs. equatorial positions. Trigonal bipyramidal geometry has two axial positions and three equatorial positions. Repulsions differ:

An axial domain experiences three 90∘ interactions with equatorial domains.

An equatorial domain experiences two 90∘ interactions (with axial domains) and two 120∘ interactions (with equatorial domains).

Because 90∘ interactions are more repulsive than 120∘, lone pairs (which repel more strongly) preferentially occupy equatorial positions to minimize the number of 90∘ interactions. This rule explains the standard shapes derived from AX₅.

PCl5. Five bonding domains, AX₅:

SF4. Four bonds and one lone pair on sulfur, AX₄E:

The lone pair occupies an equatorial position, leading to bond angles that deviate from the ideal 90∘ and 120∘.

ClF3. Three bonds and two lone pairs, AX₃E₂:

Both lone pairs occupy equatorial positions, leaving one equatorial and two axial bonding positions.

XeF2. Two bonds and three lone pairs, AX₂E₃:

All three lone pairs occupy equatorial positions, leaving two axial bonds opposite each other.

Octahedral electron geometry (6 domains)

SF6. Six bonding domains, AX₆:

BrF5. Five bonds and one lone pair, AX₅E:

XeF4. Four bonds and two lone pairs, AX₄E₂:

The two lone pairs occupy positions opposite each other (trans) to minimize lone pair–lone pair repulsion.

5.6.7 Common Pitfalls and How to Avoid Them

Mistake: counting double bonds as two domains. Correct: a double bond counts as one electron domain for basic VSEPR geometry.

Mistake: using the wrong central atom. The central atom is usually the least electronegative (except H) and is the atom with multiple connections in the Lewis structure.

Mistake: ignoring lone pairs on the central atom. Lone pairs determine electron geometry and strongly affect molecular geometry and bond angles.

Mistake: forcing octets before confirming the Lewis structure. VSEPR depends on a correct Lewis structure (including resonance and octet exceptions when relevant).

5.6.8 Key Takeaways

VSEPR predicts molecular shape by arranging electron domains around a central atom to minimize repulsion.

Electron geometry counts all domains (bonds and lone pairs); molecular geometry describes only the arrangement of bonded atoms.

Lone pairs repel more strongly than bonding pairs and typically compress bond angles.

For 5-domain (trigonal bipyramidal) structures, lone pairs preferentially occupy equatorial positions; for 6-domain (octahedral) structures, lone pairs often arrange trans when possible.

A correct Lewis structure is the required starting point: VSEPR is a geometry model built on electron-domain counting.

5.7 Molecular Polarity

In earlier sections, bond polarity was introduced as a consequence of unequal

electron sharing, quantified by a bond dipole moment μ = Qr. VSEPR theory then provided the three-dimensional geometry needed to understand how multiple bond dipoles combine in a molecule. Molecular polarity is the synthesis of these two ideas: a molecule is polar only if it has both (1) polar bonds or charge separation and (2) a geometry that prevents those dipoles from canceling.

This section develops a systematic method for determining whether a molecule has a net dipole moment (a nonzero molecular dipole), and therefore is polar.

5.7.1 Bond Dipoles as Vectors

Direction and magnitude

A bond dipole is a vector: it has both magnitude and direction. By convention, the bond dipole vector points from the partial positive end (δ⁺) toward the partial negative end (δ⁻). The magnitude depends on both the charge separation and bond length:

In a polyatomic molecule, each polar bond contributes a dipole vector. The molecule’s overall dipole moment is the vector sum:

If μnet≠0, the molecule is polar. If μnet = 0, the molecule is nonpolar (even if individual bonds are polar).

A frequent misconception

It is not enough to ask “Does the molecule contain polar bonds?” Many molecules contain polar bonds but are nonpolar overall because symmetry causes cancellation. Conversely, a molecule can be polar even if it contains only moderately polar bonds if the geometry produces an asymmetric vector sum.

5.7.2 A Procedure for Determining Molecular Polarity

Given a molecular formula, determine molecular polarity using the following workflow:

1. Draw a correct Lewis structure (including resonance if relevant) to determine connectivity.

2. Use VSEPR to determine the molecular geometry around the central atom(s).

3. Identify which bonds are polar using electronegativity differences, and assign approximate dipole directions.

4. Add the bond dipoles as vectors using symmetry and geometry. Decide whether cancellation occurs.

This workflow is reliable for most main-group molecules studied in general chemistry.

5.7.3 Symmetry and Cancellation: The Essential Criterion

When do polar bonds cancel?

Bond dipoles cancel when:

they have equal magnitudes (often true for identical bonds in a symmetric environment), and

they are arranged symmetrically so that vectors sum to zero.

Common symmetric shapes that frequently lead to μnet = 0 (when all surrounding atoms are identical) include:

linear AX₂: CO2, BeCl2 (gas phase),

trigonal planar AX₃: BF3,

tetrahedral AX₄: CCl4, CH4,

trigonal bipyramidal AX₅: PCl5,

octahedral AX₆: SF6.

The key phrase is when all surrounding atoms are identical. If different atoms occupy different positions (e.g., CH3Cl vs. CCl4), symmetry is broken and the molecule may be polar.

Geometries that often produce polarity

Molecular shapes with lone pairs typically reduce symmetry and therefore often yield a net dipole:

bent (AX₂E, AX₂E₂): H2O, SO2,

trigonal pyramidal (AX₃E): NH3,

seesaw (AX₄E), T-shaped (AX₃E₂), square pyramidal (AX₅E).

These shapes are not guaranteed to be polar in every case, but they are strong indicators.

5.7.4 Vector Addition Examples: CO2 vs. H2O

The contrast between CO2 and H2O illustrates the central idea: both contain polar bonds, but only one has a net dipole due to geometry.

Interpreting the diagrams

CO2. Each C– –O bond is polar (oxygen is more electronegative), so two equal bond dipoles point from carbon toward each oxygen. Because CO2 is linear, the dipoles are equal in magnitude and opposite in direction, so they cancel:

H2O. Each O–H bond is polar, with dipoles pointing toward oxygen. Because H2O is bent, the vectors do not oppose each other directly; their components add to produce a net dipole pointing roughly toward oxygen:

5.7.5 More Examples: Polar Bonds vs. Polar Molecules

Nonpolar molecules with polar bonds

CCl4 (tetrahedral). Each C–Cl bond is polar, but the tetrahedral arrangement is symmetric. The four bond dipoles cancel:

BF3 (trigonal planar). Each B–F bond is polar, but the trigonal planar symmetry cancels the dipoles:

Polar molecules due to asymmetry

CH3Cl (tetrahedral, but substituted). The tetrahedral arrangement is not symmetric because one substituent is Cl and the others are H. The C–Cl bond dipole is not canceled by the C–H dipoles:

NH3 (trigonal pyramidal). The N–H bonds are polar and the lone pair produces a trigonal pyramidal shape. The dipoles add to a net dipole pointing toward nitrogen:

5.7.6 Molecules with Multiple Centers: Local Dipoles and Net Dipoles

Many molecules have more than one central atom. In such cases:

determine the geometry around each central atom,

identify the major polar bonds,

consider the overall symmetry of the entire molecule.

For example, in ethanol CH3CH2OH, the O–H and C–O bonds are strongly polar and their dipoles do not cancel due to the molecule’s overall asymmetry, leading to a net dipole.

5.7.7 Connection to Intermolecular Forces (Preview)

Molecular polarity strongly influences intermolecular attractions:

Polar molecules exhibit dipole–dipole attractions.

Nonpolar molecules rely primarily on London dispersion forces.

Highly polar bonds involving H bonded to N, O, or F can enable hydrogen bonding, provided molecular geometry exposes appropriate donors and acceptors.

These effects are developed in detail when intermolecular forces are studied; here, the critical point is that polarity is a molecular property derived from both electronegativity and geometry.

5.7.8 Key Takeaways

Bond dipoles are vectors. Molecular polarity depends on the vector sum μnet = ∑ μi.

A molecule can have polar bonds and still be nonpolar if its geometry causes dipole cancellation (e.g., CO2, CCl4).

Lone pairs often reduce symmetry and make molecules polar (e.g., H2O, NH3).

The reliable workflow is: Lewis structure → VSEPR geometry → bond polarity → vector addition/symmetry test.

5.8 Problems

5.8.1 Problem 1: Bonding Model Classification and Properties

For each substance below, classify the dominant bonding model as ionic, covalent (molecular), covalent (network), or metallic. Then, for each, predict (with a brief justification) whether it is expected to be (i) a good electrical conductor as a

solid, (ii) a good electrical conductor when molten, and (iii) high-melting or low-melting relative to typical molecular substances.

Your justifications should explicitly reference electron localization/delocalization and the presence or absence of mobile charge carriers.

5.8.2 Problem 2: Electronegativity, Bond Polarity, and Dipole Moments

Consider the bonds H−F, H−Cl, C−O, C−H, and N−H.

Rank these bonds from least polar to most polar based on electronegativity differences (use qualitative periodic trends if exact values are not provided).

For H−Cl and C−O, indicate the direction of the bond dipole arrow and label the partial charges (δ⁺, δ⁻).

A particular A−B bond has a dipole moment of 1.50 D and a bond length of 1.00 × 10⁻¹⁰ m. Estimate the magnitude of charge separation Q in coulombs and express it as a fraction of the elementary charge e. (Use 1 D = 3.33564 × 10⁻³⁰ Cm and e = 1.602 × 10⁻¹⁹ C.)

5.8.3 Problem 3: Lewis Structures Under the Octet Rule (Electron Counting Emphasis)

Draw valid Lewis structures (including lone pairs) for the following species. Show your total valence-electron count for each before drawing.

For each structure, clearly indicate which atom is central and state whether all atoms satisfy the octet rule (or duet rule for H).

5.8.4 Problem 4: Formal Charge Analysis and Selecting the Best Structure

For each of the following, more than one Lewis structure can be drawn (with the same atom connectivity). For each species:

draw two distinct valid Lewis structures (different placement of multiple bonds and/or charges),

compute the formal charge on every atom in each structure,

identify which structure is the best contributor based on formal-charge criteria (minimize charge magnitude and separation; place negative charge on more electronegative atoms).

Species:

In your justification, explicitly state which formal-charge rule(s) you used.

5.8.5 Problem 5: Resonance and Resonance Hybrids

Draw all major resonance contributors for NO3− and CO 32−. Use double-headed arrows between contributors.

For each ion, compute formal charges for one contributor and use symmetry to infer the others.

For each ion, state the predicted average bond order for each central-atom–oxygen bond in the resonance hybrid.

Explain (conceptually) why resonance contributors do not represent a molecule “flipping” between structures.

5.8.6 Problem 6: Exceptions to the Octet Rule (Classification and Structure Drawing)

For each species below:

classify it as an odd-electron radical, incomplete-octet (electron-deficient), or expanded-octet (hypervalent) case (or “no exception” if applicable),

draw a Lewis structure consistent with your classification (including the unpaired electron for radicals),

state the total valence-electron count used.

Species:

5.8.7 Problem 7: VSEPR Domain Counting and Molecular Geometry (Including Lone-Pair Placement)

For each species, draw a Lewis structure, then determine:

the number of electron domains around the central atom,

the electron geometry,

the molecular geometry (shape),

approximate bond angles (ideal and whether angles are compressed/expanded relative to ideal),

the AXmEn designation.

Species:

For trigonal bipyramidal cases, explicitly state whether lone pairs occupy axial or equatorial positions and justify using VSEPR repulsion arguments.

5.8.8 Problem 8: Molecular Polarity via Vector Addition and Symmetry

For each molecule, determine whether it is polar or nonpolar. Your answer must explicitly reference both (i) bond polarity and (ii) molecular geometry/symmetry and the vector sum of bond dipoles.

For CH2Cl2, provide a brief sketch indicating the relative directions of the C−Cl and C−H bond dipoles and explain why cancellation is incomplete.

5.8.9 Problem 9: Integrated Structural Comparison (Isomers, Formal Charge, and Polarity)

Two different Lewis structures (same formula) are proposed for a species with formula N2O. One has connectivity N−N−O; the other has connectivity N−O−N.

Draw both connectivities as valid Lewis structures (multiple bonds as needed), ensuring correct electron count and octets where possible.

Compute formal charges for each structure and determine which connectivity is more plausible based on formal-charge placement and electronegativity considerations.

For the more plausible connectivity, use VSEPR to predict the molecular geometry about the central atom and determine whether the molecule is polar.

Your reasoning should be explicit at each step: electron count → Lewis structure → formal charge → geometry → polarity.

5.8.10 Problem 10: Multi-Step Synthesis of Concepts (From Bonding to Shape to Polarity)

A neutral molecule has the formula C2H2Cl2. Consider two structural possibilities:

Draw complete Lewis structures for (A) and (B), including all lone pairs.

For each structure, identify the bonding model most appropriate for describing the bonding within the molecule (ionic vs. covalent vs. metallic) and justify.

For each carbon atom in each structure, determine the electron-domain count and predict the local geometry (use VSEPR). State whether the local geometry is trigonal planar or tetrahedral.

Using electronegativity and bond dipoles, determine whether (A) and (B) are polar or nonpolar molecules. For (A), consider the possibility of different spatial arrangements around the C−−C bond and explain how that could affect the net dipole.

Rank (A) and (B) in expected boiling point, assuming comparable molar masses, and justify your ranking primarily using polarity and intermolecular forces (do not use external data).

Chapter 6
Advanced Bonding Theories

Why is liquid oxygen magnetic while nitrogen is not? Why does methane form a perfect tetrahedron despite carbon’s ground-state electron configuration suggesting otherwise? While Lewis structures provide a convenient shorthand for connectivity, they fail to explain the nuance of molecular geometry, magnetism, and excited states. This chapter pulls back the curtain on the quantum mechanical nature of the chemical bond. We will explore how atomic wavefunctions mix to create hybrid orbitals, how electrons are shared across entire molecules in molecular orbitals, and how constructive and destructive interference patterns dictate the stability of the matter around us.

6.1 Valence Bond Theory

Valence Bond Theory (VBT) is a quantum-mechanical model of covalent bonding that treats a bond as a localized region of enhanced electron density between two atomic nuclei. In this view, a covalent bond forms when two atoms each contribute one electron and the corresponding half-filled atomic orbitals overlap in space. The overlap allows the electrons to be concentrated between the nuclei, lowering the potential energy of the system relative to separated atoms.

This section builds directly on two ideas established earlier in the text:

Electron configurations and atomic orbitals: electrons occupy orbitals (quantum states) characterized by shapes and orientations in space (e.g., s and p orbitals).

Lewis structures and valence electrons: covalent bonds involve valence electrons, but Lewis theory does not specify how electron density becomes concentrated between nuclei or why bonding has directionality.

VBT provides a bridge: it retains the chemical intuition of localized electron pairs (like Lewis structures) while grounding bonding in orbital wavefunctions and their overlap.

6.1.1 Orbitals as wavefunctions: bonding as interference

An atomic orbital is not a physical “path” but a wavefunction ψ(r) describing the quantum state of an electron. The square magnitude, |ψ(r)|², is a probability density (and, for many-electron systems, relates to electron density). A crucial feature is that a wavefunction has phase (sign): in sketches of orbitals, lobes may be labeled “+” and “−” to indicate the sign of ψ in that region.

When two atoms approach, their atomic wavefunctions overlap. Two limiting combinations (linear combinations) are possible:

ψconstructive | ∝ ψA + ψB |  |
ψdestructive | ∝ ψA − ψB |  |

In the constructive case, the wavefunctions reinforce one another in the internuclear region, increasing electron density between the nuclei. In the destructive case, cancellation produces a node (a surface where ψ = 0) between the nuclei, depleting electron density from that region.

Valence Bond Theory emphasizes that a stable covalent bond corresponds to constructive interference in the region between the nuclei, because electron density between positively charged nuclei provides electrostatic stabilization and reduces repulsion between the nuclei by “screening” their charges.

6.1.2 Energy changes as two atoms approach

Consider two atoms A and B far apart. Their electrons occupy atomic orbitals on separate centers, and the total energy is approximately the sum of the isolated-atom energies. As the atoms move closer, several interactions change:

Attraction: each nucleus attracts the other atom’s electrons. If electron density accumulates between the nuclei, this attraction is enhanced.

Repulsion: nucleus–nucleus repulsion increases strongly at short distance; electron–electron repulsion also increases as electron clouds overlap.

Quantum mechanical stabilization: constructive overlap allows electron density to concentrate in the bonding region without violating the Pauli principle (provided appropriate spin pairing), lowering kinetic and potential energy relative to separated atoms.

These competing effects produce a characteristic potential energy curve as a function of internuclear distance r:

At large r, interaction is negligible and the energy approaches that of separated atoms.

At intermediate r, attraction dominates and energy decreases.

At very small r, strong repulsions dominate and energy rises steeply.

The minimum of this curve defines:

the bond length (equilibrium internuclear distance, re),

and the bond energy (the energy required to separate the atoms from re to infinite separation, in the gas phase).

Thus, in VBT a covalent bond is not merely “sharing” electrons; it is the result of a net energetic minimum created by electron density buildup between nuclei from orbital overlap.

6.1.3 Conditions for effective covalent bonding in VBT

Not all overlaps produce strong bonds. VBT highlights three practical conditions:

(1) Half-filled orbitals (or available electron pairs)

The simplest covalent bond forms when each atom provides a half-filled valence orbital. The two electrons can pair with opposite spins in the overlap region, producing a localized electron pair bond.

More generally, bonding can involve:

half-filled–half-filled overlap (typical single bond formation),

filled–empty overlap (coordinate or dative bonding; treated later),

or interactions involving hybrid orbitals (next section).

For the introductory VBT picture, the key case is two half-filled orbitals producing a shared electron pair.

(2) Proper symmetry and orientation

Orbitals must overlap with compatible symmetry relative to the internuclear axis. The overlap integral (a measure of how strongly the wavefunctions overlap)

depends strongly on direction. This is the origin of bond directionality: p orbitals bond strongly in some orientations and weakly (or not at all) in others.

(3) Comparable orbital energies

Overlap is most effective when the interacting orbitals have similar energies. This is why bonds between atoms of similar electronegativity often involve substantial covalent character: their valence orbitals can mix and overlap effectively. (A deeper treatment of energy matching and mixing appears in Molecular Orbital Theory later in the chapter.)

6.1.4 Sigma (σ) bonds: head-on overlap along the internuclear axis

A sigma bond (σ bond) is a covalent bond in which electron density is concentrated along the line connecting the nuclei (the internuclear axis). In wavefunction terms, a σ bond results from head-on (end-to-end) overlap of orbitals.

Key characteristics of σ bonds:

Cylindrical symmetry: the electron density is symmetric upon rotation about the bond axis.

Strong overlap: head-on overlap generally maximizes overlap for a given pair of orbitals, making σ bonds typically stronger than π bonds (introduced later).

Single bonds are σ bonds: every single bond is a σ bond; multiple bonds contain one σ bond plus additional π bond(s).

Common types of σ-bond overlap

Three foundational overlap patterns occur frequently in main-group chemistry:

s–s overlap: two spherical s orbitals overlap directly (e.g., H2).

s–p overlap: a spherical s orbital overlaps with one lobe of a directional p orbital (e.g., HCl treated in a simplified picture).

p–p end-to-end overlap: two p orbitals aligned on the internuclear axis overlap head-on (e.g., the σ component in F2 or the σ framework of many bonds involving second-row atoms).

6.1.5 Spin pairing and the role of the Pauli principle

Two electrons can occupy the same region of space (the bonding region) only if their spins are paired (opposite). This constraint is a consequence of the Pauli exclusion principle: no two electrons in an atom (or molecule) can have the same set of quantum numbers.

In VBT language, a covalent bond corresponds to pairing of two electrons with opposite spin in an overlapping region of two atomic orbitals. This explains an important feature of bonding capacity:

Atoms with unpaired valence electrons tend to form covalent bonds readily (e.g., H with one unpaired electron, Cl with one unpaired electron in a p orbital).

Atoms with all electrons paired in their valence shell are typically less reactive toward covalent bond formation under ordinary conditions (e.g., noble gases), because forming a bond would require promoting electrons to create half-filled orbitals, which is energetically costly.

6.1.6 What VBT explains well (and where it struggles)

Valence Bond Theory captures several central chemical observations:

Bond formation is directional: overlap depends on orbital orientation; this naturally leads to specific bond angles in many cases.

Single bonds are localized electron pairs: consistent with Lewis structures, but now with a wavefunction basis.

σ bonds are strong and lie along the internuclear axis: consistent with observed bond strengths and the structural role of single bonds as the “framework” of molecules.

However, the most basic VBT picture—using only unmodified atomic s and p orbitals on a central atom—has difficulty reproducing the observed equivalence and geometry of many polyatomic molecules. For example, the four C–H bonds in methane are experimentally identical and arranged tetrahedrally, yet one 2s and three 2p orbitals are not equivalent in shape and energy. Addressing such discrepancies motivates the next section: hybridization, in which atomic orbitals are mathematically mixed to form new, equivalent orbitals oriented to match observed molecular geometries.

In Valence Bond Theory, a covalent bond forms when half-filled atomic orbitals overlap and the two electrons pair with opposite spins.

Bonding corresponds to constructive interference of wavefunctions in the internuclear region, producing enhanced electron density between nuclei.

σ bonds result from head-on overlap (s–s, s–p, or end-to-end p–p), yielding electron density concentrated along the bond axis.

The energetic stability of a bond arises from a balance of attractions and repulsions, leading to an equilibrium bond length and a bond energy.

6.2 Hybridization of Atomic Orbitals

Valence Bond Theory (VBT) explains covalent bonding as the overlap of half-filled atomic orbitals to form localized electron density between nuclei. A serious limitation appears immediately when we apply that simple picture to polyatomic molecules: unmodified (“pure”) atomic s and p orbitals often predict the wrong number of equivalent bonds and the wrong bond angles.

A classic example is methane, CH4. Experimentally, methane has:

four identical C–H bonds (same length and bond energy),

a tetrahedral geometry with bond angles near 109.5∘.

However, the ground-state valence electron configuration of carbon is

which (in a naive VBT interpretation) contains only two unpaired electrons in the 2p subshell and therefore seems capable of forming only two covalent bonds.

Hybridization is VBT’s solution: it is a mathematical mixing of atomic orbitals on the same atom to generate a new set of equivalent, directionally oriented orbitals that match observed molecular geometries (as summarized by VSEPR theory). Hybridization does not change the number of orbitals; it changes their shapes and orientations so that bonding overlap is maximized in the correct directions.

6.2.1 Why hybridization is needed: equivalence of bonds and molecular geometry

In many molecules, a central atom forms multiple equivalent σ bonds. Equivalent bonds imply that the bonding orbitals on the central atom are also equivalent (same energy and same extent of overlap). But pure 2s and 2p orbitals are not equivalent:

A 2s orbital is spherically symmetric and typically lower in energy.

A 2p orbital is directional and higher in energy, with two lobes of opposite phase separated by a nodal plane.

If carbon used one 2s and three 2p orbitals to form four C–H bonds, VBT would predict one C–H bond different from the other three. Methane does not behave this way. Therefore, the bonding model must provide four identical carbon orbitals directed toward the corners of a tetrahedron.

Hybridization achieves exactly this by constructing new orbitals (linear combinations of 2s and 2p) that are:

degenerate (equal energy within the model),

oriented in space to match electron-domain geometry (VSEPR),

well-suited to strong head-on overlap (strong σ bonds).

6.2.2 Hybrid orbitals as linear combinations (the mathematical idea)

A hybrid orbital is a linear combination of atomic orbitals on the same atom. In the simplest conceptual form,

where the coefficients ci are chosen so that:

the resulting hybrid orbitals are orthonormal (independent and properly normalized),

their electron density is directed along specific axes,

they are energetically suitable for bonding in the observed geometry.

Two points prevent common misconceptions:

Hybridization is not a physical “process” that occurs before bonding. It is a modeling choice within VBT: we choose an orbital basis on the central atom that best represents bonding.

Mixing orbitals does not violate quantum mechanics. Any set of orbitals spanning the same subspace is an equally valid description of the electron distribution; hybrid orbitals are chosen because they provide a chemically intuitive, geometry-aligned basis.

6.2.3 Carbon in methane: promotion and mixing into sp³ hybrids

To describe four C–H σ bonds in methane, VBT employs two linked ideas:

Electron “promotion” (conceptual step): one electron is promoted from 2s to 2p to create four unpaired electrons.

Hybridization (mixing step): one 2s and three 2p orbitals mix to yield four equivalent sp³ hybrid orbitals directed tetrahedrally.

Promotion: creating four half-filled orbitals

In the ground state, carbon has two unpaired electrons (in 2p). In a promoted configuration (often written for bonding discussions),

carbon now has four unpaired electrons, one in each of four valence orbitals. Promotion costs energy, but methane formation releases substantial energy via formation of four strong C–H bonds; the net result is highly favorable. The promoted configuration should be viewed as part of the bonding model, not as an isolated excited atom that must exist independently.

sp³ hybridization: four equivalent orbitals

Mixing 2s with all three 2p orbitals produces four equivalent sp³ hybrids. Each sp³ hybrid has:

one large lobe (high electron density) and one small lobe (opposite phase),

a direction in space aligned toward a tetrahedral vertex,

energy intermediate between 2s and 2p (in the model).

Each half-filled sp³ orbital overlaps head-on with a half-filled H(1s) orbital to form a σ(C–H) bond.

Figure 6.1: Energy-level correlation for carbon bonding in methane (VBT model). Left: carbon valence orbitals in the ground state showing 2s² 2p² with two unpaired electrons; Right: promotion to 2s¹ 2p³ and mixing into four degenerate sp³ hybrids.

6.2.4 Hybridization and electron-domain geometry (VSEPR connection)

VSEPR theory predicts electron-domain geometry based on the number of electron domains (bonding regions and lone pairs) around a central atom. Hybridization provides a VBT orbital picture consistent with the same count.

Define the steric number (SN) as:

Within the hybridization model (for main-group compounds), the steric number correlates with the type of hybridization and the idealized electron-domain geometry:

SN | Hybrid set | Electron-domain geometry | Typical examples (central atom)
2 | sp | linear (180∘) | BeCl 2(g) (Be), CO2 (C), HC#CH (each C)
3 | sp² | trigonal planar (120∘) | BF 3 (B), C2H4 (each C), SO3 (S)
4 | sp³ | tetrahedral (109.5∘) | CH 4 (C), NH3 (N, 1 lone pair), H2O (O, 2 lone pairs)
5 | sp³d | trigonal bipyramidal | PCl 5 (P), SF4 (S, 1 lone pair)
6 | sp³d² | octahedral | SF 6 (S), BrF5 (Br, 1 lone pair)

Two cautions are important for correct use:

Hybridization tracks electron domains, not the molecular shape directly. For instance, NH3 and H2O are commonly described as sp³ because the electron-domain geometry is tetrahedral, even though the molecular shapes are trigonal pyramidal and bent, respectively, due to lone pairs.

Hybridization is a localized model. It is often very effective for sigma-bond frameworks, but delocalized π systems (conjugation, aromaticity) may require resonance descriptions or Molecular Orbital Theory for a more complete picture.

6.2.5 sp hybridization: two hybrid orbitals and linear geometry

When the steric number is 2, one s orbital mixes with one p orbital to form two equivalent sp hybrids oriented 180∘ apart. Two p orbitals remain unhybridized and

perpendicular to the bond axis; these leftover p orbitals are crucial for forming π bonds in triple bonds and in some cumulated systems (discussed in the next section).

Example: carbon in CO2.

A simplified localized picture treats carbon as sp hybridized:

two sp orbitals form two σ bonds to the oxygens in a linear arrangement,

two unhybridized p orbitals on carbon can participate in π bonding (one π bond to each oxygen in a resonance description).

The key point here is geometric: sp hybrids naturally align with the observed linear arrangement of electron density.

6.2.6 sp² hybridization: three hybrid orbitals and trigonal planar geometry

For steric number 3, one s mixes with two p orbitals to form three equivalent sp² hybrids in a plane, 120∘ apart. One p orbital remains unhybridized and perpendicular to that plane; it is positioned ideally for side-by-side overlap to make a π bond.

Example: carbon in ethene, C2H4.

Each carbon is commonly described as sp² hybridized:

three sp² orbitals form σ bonds (two C–H and one C–C),

the remaining unhybridized p orbitals overlap to form the π bond of the double bond.

Here again, the hybrid model matches both equivalent bonding within the plane and the planar geometry around each carbon.

6.2.7 sp³ hybridization: four hybrid orbitals and tetrahedral electron-domain geometry

For steric number 4, sp³ hybridization gives four equivalent hybrids oriented toward a tetrahedron. This model applies to:

molecules with four σ bonds (e.g., CH4),

molecules with three σ bonds and one lone pair (e.g., NH3),

molecules with two σ bonds and two lone pairs (e.g., H2O).

In each case, the electron-domain geometry remains tetrahedral, consistent with VSEPR, while the observed bond angles decrease as lone pairs occupy more space and repel bonding pairs more strongly.

6.2.8 Expanded-octet hybridizations: sp³d and sp³d²

For steric numbers 5 and 6, introductory VBT often describes bonding using sp³d (trigonal bipyramidal) and sp³d² (octahedral) hybrid sets. These labels provide a useful geometry-aligned bookkeeping scheme that matches VSEPR electron-domain geometries and helps organize molecular shapes (seesaw, T-shaped, square pyramidal, square planar in some cases).

However, it is important to understand the status of these descriptions:

The hybridization labels for expanded octets are models used to rationalize geometry and σ bonding patterns.

More advanced treatments (including Molecular Orbital Theory and modern bonding analyses) often describe bonding in hypervalent molecules without requiring substantial participation from valence d orbitals on third-period elements.

In this chapter, we use these hybridization schemes primarily as a consistent way to connect electron-domain count (VSEPR) to a localized orbital picture (VBT).

6.2.9 Using hybridization correctly: a practical workflow

To apply hybridization as a predictive tool, proceed in this order:

Draw a plausible Lewis structure (including lone pairs and formal charges if needed).

Count electron domains around the central atom (each single, double, or triple bond counts as one domain; each lone pair counts as one domain).

Assign steric number and map to sp, sp², sp³, sp³d, or sp³d².

Predict the electron-domain geometry (linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral).

Infer the molecular shape by removing positions occupied by lone pairs (VSEPR shape classification).

6.2.10 Limitations and the transition to multiple bonding

Hybridization is extremely effective for describing σ bonding frameworks and

rationalizing many molecular geometries. Nevertheless, two limitations foreshadow the next section:

Multiple bonds require unhybridized p orbitals. Double and triple bonds cannot be described by hybrid orbitals alone; π bonding relies on side-by-side overlap of remaining p orbitals.

Delocalization challenges purely localized pictures. Conjugated and aromatic systems distribute electron density over multiple atoms; localized hybrid orbitals plus localized π bonds must be supplemented by resonance or by Molecular Orbital Theory.

Hybridization is a mathematical mixing of atomic orbitals on the same atom that produces equivalent, directional hybrid orbitals suited for strong σ bonding.

The methane problem motivates hybridization: carbon’s observed four identical bonds and tetrahedral geometry are modeled by four equivalent sp³ hybrids (often preceded conceptually by promotion to create four unpaired electrons).

Hybridization correlates with VSEPR electron-domain geometry via the steric number:

Hybridization sets the stage for multiple bonding: sp and sp² schemes leave unhybridized p orbitals available for π bond formation.

6.3 Multiple Bonds and Delocalization

The previous sections developed Valence Bond Theory (VBT) as a localized bonding model based on overlap of half-filled orbitals and introduced hybridization as a way to generate directional, equivalent orbitals aligned with VSEPR geometries. That framework explains many single bonds as σ bonds formed by head-on overlap of hybrid orbitals (or s∕p orbitals in simpler cases).

Multiple bonds require one additional ingredient: unhybridized p orbitals. A double bond cannot be described as “two σ bonds” between the same atoms, because only one orbital pair can overlap head-on along a given internuclear axis. Instead, VBT represents multiple bonds with the σ–π framework:

A double bond consists of one σ bond and one π bond.

A triple bond consists of one σ bond and two π bonds.

In this section we develop the orbital origin of π bonding, connect it to observable molecular properties (bond lengths, bond strengths, restricted rotation), and then show why delocalization and resonance are necessary extensions when localized bonds cannot adequately represent the electron distribution.

6.3.1 From hybridization to multiple bonding: leaving unhybridized p orbitals

Hybridization is best viewed as a geometry-aligned partition of valence orbitals into:

hybrid orbitals that form the σ framework (localized head-on overlaps), and

unhybridized p orbitals (when present) that can form π bonds by side-by-side overlap.

The key cases are:

sp³: no unhybridized p orbitals remain ⇒ typically only σ bonds.

sp²: one unhybridized p orbital remains ⇒ up to one π bond per atom involved.

sp: two unhybridized p orbitals remain (mutually perpendicular) ⇒ up to two π bonds per atom involved.

Thus, multiple bonding is enabled by hybridization schemes that preserve unhybridized p orbitals.

6.3.2 The π bond: side-by-side overlap and a nodal plane

A π bond forms by side-by-side (lateral) overlap of two parallel unhybridized p orbitals on adjacent atoms. Because each p orbital has two lobes of opposite phase, constructive overlap occurs when lobes of the same phase overlap above and below the internuclear axis.

Two defining features distinguish π from σ bonds:

Electron density is concentrated in two lobes above and below (or in front of and behind) the bond axis, not directly on the axis.

There is a nodal plane that contains the internuclear axis in which the π-bonding wavefunction changes sign and the probability density is reduced. (This nodal plane is a direct consequence of the sign change of the p orbitals.)

Because lateral overlap is less extensive than head-on overlap, π bonds are generally weaker than σ bonds, although the total strength of a double or triple bond exceeds that of a single bond.

6.3.3 Double bonds in VBT: ethene as the prototype

Ethene, C2H4, is the simplest molecule that cleanly illustrates the σ–π framework. Experimental observations to keep in mind are:

each carbon is approximately trigonal planar (bond angles near 120∘),

the C– –C bond is shorter and stronger than a C–C single bond,

rotation about the C– –C bond is strongly restricted.

Orbital picture of ethene

In the localized VBT description:

Each carbon is sp² hybridized. Three sp² orbitals lie in a plane: two form σ bonds to hydrogen (C–H); one forms the σ bond between the carbons (C–C σ).

Each carbon retains one unhybridized 2p orbital perpendicular to the sp² plane.

These two unhybridized p orbitals overlap side-by-side to form the π bond, creating electron density above and below the molecular plane.

Consequences of π bonding: restricted rotation

A single C–C σ bond allows nearly free rotation because rotating one carbon relative to the other does not significantly change the head-on overlap along the bond axis.

A π bond is different: it requires the two unhybridized p orbitals to remain parallel for effective side-by-side overlap. Rotation about the C– –C axis would twist the p orbitals out of alignment and destroy the π overlap, which costs energy. Therefore:

double bonds exhibit restricted rotation,

and this restriction underlies cis/trans (E/Z) isomerism in substituted alkenes.

From the VBT perspective, the barrier to rotation is the energetic penalty for losing constructive lateral overlap.

Bond length and bond strength trends in the σ–π framework

Adding a π bond increases electron density between the nuclei and draws the nuclei closer together. Thus, for a given pair of atoms:

Bond length decreases as bond order increases (single > double > triple).

Bond enthalpy increases as bond order increases (single < double < triple), though the increase is not linear because π bonds contribute less than σ bonds.

VBT rationalizes these trends by noting that π bonding adds additional constructive

overlap and additional electron density in the internuclear region, strengthening attraction and shortening equilibrium distance.

6.3.4 Triple bonds: one σ and two π bonds

A triple bond requires two independent π overlaps, which in turn requires two unhybridized p orbitals on each bonded atom. This naturally corresponds to sp hybridization.

Orbital picture of acetylene, C2H2

In ethyne (HC– – –CH):

Each carbon is sp hybridized: two sp orbitals lie along the internuclear axis in a linear arrangement.

One sp orbital on each carbon forms the C–C σ bond (head-on).

The other sp orbital on each carbon forms a C–H σ bond.

Two unhybridized p orbitals remain on each carbon; the pairs overlap side-by-side in two perpendicular planes to form two π bonds.

This produces a linear geometry and a very short, strong C– – –C bond. The two π bonds are orthogonal, reflecting the orthogonality of the two remaining p orbitals.

6.3.5 Delocalization: when localized π bonds are not enough

The σ–π framework describes many isolated multiple bonds well. Difficulties arise when a molecule can be drawn with multiple Lewis structures that differ only in the placement of π bonds and lone pairs. In such cases, experimental data often show:

bond lengths that are intermediate between single and double bonds,

electron density that is spread out over three or more atoms,

enhanced stability relative to a single localized-bond structure.

These observations indicate delocalization: electrons (especially π electrons and sometimes lone pairs) are not confined to one bond between two atoms but are distributed over a larger region of the molecule.

In a strict localized VBT picture, electrons belong to specific bonds; delocalization therefore forces an extension of the model. The standard extension at this level is resonance.

6.3.6 Resonance in Valence Bond Theory: the resonance hybrid

Resonance structures are not real molecules

A resonance structure is one of two or more valid Lewis structures that:

have the same arrangement of nuclei,

differ only in the placement of electrons (typically π bonds and lone pairs),

and together describe a bonding situation that cannot be represented adequately by a single localized structure.

The actual molecule is a resonance hybrid: a single real structure whose electron distribution is best approximated as a weighted combination of the resonance structures.

Example: the allyl system (three-atom π delocalization)

Consider the allyl cation, C3H5+. Two resonance structures can be drawn with the positive charge on either terminal carbon and the double bond shifted:

Experimentally and conceptually, the π electron density is delocalized over all three carbons. In localized VBT terms, neither single structure alone captures the symmetry and intermediate bond lengths; the resonance hybrid does.

Example: benzene and aromatic delocalization

Benzene, C6H6, can be drawn with alternating double bonds in two equivalent Kekulé structures:

However, all six C–C bonds in benzene are experimentally equivalent and have a bond length intermediate between typical single and double bonds. The resonance hybrid description expresses that the π electrons are delocalized around the ring.

Within the VBT framework, resonance is therefore a necessary tool: it preserves the localized σ bond skeleton while acknowledging that π electrons may be shared across multiple adjacent atoms.

6.3.7 How to recognize when resonance is required

A practical set of indicators that a single localized structure is insufficient:

Multiple valid Lewis structures exist without moving nuclei, differing only by shifting π bonds/lone pairs.

The system contains a conjugated sequence of p orbitals (e.g., C––C–C––C, C– –C–C+, C– –C–N:).

Measured bond lengths suggest partial bond order (often close to 1.5 in aromatic rings).

Charge appears “shared” over more than one atom (common in polyatomic ions like NO3 – , CO32– , SO32– ).

In each case, resonance is a way of stating that the most appropriate electron-pair description involves delocalization.

Multiple bonds are described in VBT by the σ–π framework: double bonds are σ + π, triple bonds are σ + 2π.

A π bond forms from side-by-side overlap of unhybridized parallel p orbitals and contains a nodal plane that includes the internuclear axis.

In ethene, each carbon is sp² hybridized: an sp² σ framework lies in a plane and a perpendicular p–p overlap forms the π bond, leading to restricted rotation.

Triple bonds typically involve sp hybridization with two unhybridized p orbitals available for two perpendicular π bonds.

Delocalization occurs when π electrons (and sometimes lone pairs) are spread over several atoms; in localized VBT this is treated using resonance structures and the resonance hybrid.

6.4 Introduction to Molecular Orbital Theory

Valence Bond Theory (VBT) and hybridization provide a powerful localized picture of covalent bonding: bonds arise from overlap of atomic orbitals, and electron pairs are assigned to specific regions between specific pairs of atoms. This view explains bond directionality, σ and π bonding patterns, and many molecular geometries.

However, the previous section also exposed a structural limitation: when electrons are delocalized (as in conjugated systems and aromatic rings), the localized electron-pair picture must be supplemented by resonance. Molecular Orbital (MO) Theory takes a different approach from the start: it treats electrons as occupying orbitals that can be delocalized over the entire molecule. This delocalization is not a “patch”; it is built into the model.

In this section, we develop the core ideas of MO theory that will be used throughout the remainder of the chapter:

construction of molecular orbitals by the Linear Combination of Atomic Orbitals (LCAO) method,

interpretation of bonding and antibonding orbitals in terms of wave interference and nodes,

electron filling rules for MOs,

calculation and meaning of bond order, and how bond order correlates with bond length, bond strength, and magnetism.

6.4.1 The central idea: electrons in molecular orbitals

In MO theory, a molecule is treated as a multi-nuclear quantum system. Electrons are assigned to molecular orbitals ψi that belong to the entire molecule rather than to a particular atom or bond. The electron density is then obtained by summing contributions from all occupied MOs:

where ni is the number of electrons in MO i (0, 1, or 2).

This viewpoint has two immediate conceptual consequences:

A single MO can contribute electron density to multiple internuclear regions, naturally describing delocalization.

Bonding is not assigned to a single electron pair in a single region; it is inferred from the net effect of all occupied orbitals on the distribution of electron density and energy.

6.4.2 LCAO: building molecular orbitals from atomic orbitals

To make MO theory practical, we need a method to construct molecular orbitals. The most common introductory approach is the Linear Combination of Atomic Orbitals (LCAO) approximation:

Here, the ψA,ψB,… are atomic orbitals (AOs) on the constituent atoms, and the coefficients ci describe how strongly each AO contributes to the MO.

When can two atomic orbitals combine effectively?

Two AOs combine to form MOs most effectively when they:

have compatible symmetry with respect to the internuclear axis (they must overlap constructively in some combination),

have significant spatial overlap,

have similar energies (large energy mismatches yield little mixing).

These conditions parallel VBT’s qualitative requirements for strong overlap, but MO theory uses them to decide which AOs mix to form the actual orbitals of the molecule.

6.4.3 Bonding and antibonding orbitals as constructive and destructive interference

Because orbitals are wavefunctions, adding and subtracting them leads to constructive or destructive interference. For a simple diatomic molecule formed from two atomic orbitals ψA and ψB, two combinations are possible:

Bonding MO (constructive combination)

Features:

increased electron density in the internuclear region,

no node between nuclei for the simplest σ bonding combinations,

lower energy than the original AOs (stabilizing).

Antibonding MO (destructive combination)

Features:

a node between the nuclei (a region where ψ = 0),

reduced electron density in the internuclear region,

higher energy than the original AOs (destabilizing).

Why a node raises energy

A node means the wavefunction changes sign over a short distance. Rapid spatial variation corresponds to higher kinetic energy (roughly, more “curvature” in the wavefunction). In addition, removing electron density from between nuclei weakens electrostatic attraction that would otherwise stabilize the system. Together these effects make antibonding orbitals higher in energy.

6.4.4 Notation for molecular orbitals: σ and π labels

MOs are labeled by their symmetry around the internuclear axis:

σ orbitals: cylindrically symmetric about the internuclear axis (electron density on the axis).

π orbitals: have one nodal plane containing the internuclear axis (electron density above and below the axis).

An asterisk denotes antibonding:

For homonuclear diatomics, additional labels g (gerade, even under inversion) and u (ungerade, odd under inversion) are often used. These become important in spectroscopy and detailed MO diagrams, but the most chemically essential distinction at this stage is bonding vs. antibonding.

6.4.5 Filling molecular orbitals: Aufbau, Pauli, and Hund rules

Electrons occupy MOs according to the same fundamental rules used for atomic orbitals:

Aufbau principle: fill lower-energy MOs before higher-energy MOs.

Pauli exclusion principle: each MO holds at most two electrons, which must have opposite spins.

Hund’s rule (for degenerate MOs): when two or more MOs have the same energy (e.g., the pair of π orbitals), electrons occupy them singly with parallel spins before pairing occurs.

These rules allow MO theory to predict not only bond strength trends but also magnetic properties. Any molecule with one or more unpaired electrons is paramagnetic; if all electrons are paired, it is diamagnetic. (This becomes decisive in explaining O2 later.)

6.4.6 Bond order in MO theory

In MO theory, bond order is defined by comparing electrons in bonding and antibonding orbitals. For diatomic molecules (and many simple cases), the MO bond order is:

where Nbonding is the total number of electrons in bonding MOs and Nantibonding is the total number of electrons in antibonding MOs.

Physical meaning of the bond order expression

The factor of 1∕2 reflects that two electrons in a bonding MO correspond, roughly, to one “bond unit” of stabilization. Antibonding electrons cancel bonding stabilization electron-for-electron within this bookkeeping scheme. While real bonding is more nuanced than a single number, MO bond order is remarkably predictive for:

whether a bond exists,

relative bond strengths,

relative bond lengths,

trends across series of related molecules and ions.

Bond order and stability

If bond order is positive, the molecule is predicted to have net bonding stabilization (a bound state may exist).

If bond order is zero, bonding and antibonding contributions cancel; a stable bond is not expected under ordinary conditions.

This idea will later explain why He2 does not exist as a stable molecule in the ground state.

6.4.7 How bond order correlates with bond length and bond enthalpy

MO bond order provides a quantitative link between electronic structure and measurable properties:

Higher bond order ⇒shorter bond length (nuclei can approach more closely when electron density stabilizes the internuclear region).

Higher bond order ⇒larger bond enthalpy (more energy required to separate atoms).

These are trends, not exact equations, because bond length and bond enthalpy also depend on atomic sizes, electronegativity, and repulsion terms. Still, within a family of related species (e.g., O2, O2+, O2 – ), bond order trends are strongly diagnostic.

6.4.8 A minimal worked example: H2 vs. He2

Even without a full MO diagram, the bond-order concept can be illustrated with the simplest 1s-based combinations.

H2

H2 has 2 electrons. The 1s AOs combine to form σ1s (bonding) and σ1s∗ (antibonding). Both electrons occupy the lower-energy σ1s:

A single bond is predicted, consistent with experiment.

He2

He2 would have 4 electrons. Filling gives two electrons in σ1s and two in σ1s∗:

No net bond is predicted, consistent with the absence of a stable He2 molecule in the ground state.

This comparison shows why MO theory is not merely a different language: it makes stability judgments by explicitly accounting for antibonding occupancy, which is not naturally represented in Lewis structures and is only indirectly handled in localized VBT.

6.4.9 MO theory versus VBT: complementary strengths

It is useful to keep a balanced perspective:

VBT/hybridization excels at predicting local bonding patterns and geometries in many main-group molecules, especially for σ-bond frameworks.

MO theory excels at describing delocalization and explaining properties that depend sensitively on electron occupancy patterns, such as magnetism and trends across isoelectronic series.

In the next section, we will construct explicit MO diagrams for second-period homonuclear diatomic molecules and use them to explain bond orders and the paramagnetism of O2, a central success of MO theory.

Molecular Orbital Theory treats electrons as occupying molecular orbitals delocalized over the entire molecule.

The LCAO method constructs MOs as linear combinations of atomic orbitals; orbitals mix effectively when they have compatible symmetry, significant overlap, and similar energies.

Two key combinations arise from AO mixing: bonding (constructive, lower energy, enhanced internuclear density) and antibonding (destructive, node, higher energy).

MO filling follows Aufbau, Pauli, and Hund rules, enabling predictions of bond order and magnetism.

MO bond order is given by

and correlates with bond length and bond enthalpy trends.

6.5 Homonuclear Diatomic Molecules

The previous section introduced the central machinery of Molecular Orbital (MO) Theory: molecular orbitals form as linear combinations of atomic orbitals (LCAO), producing bonding and antibonding MOs; electrons fill MOs by the Aufbau principle, Pauli exclusion principle, and Hund’s rule; and bond order is computed from the occupancy of bonding vs. antibonding orbitals.

We now apply these principles to a particularly important family of systems: second-period homonuclear diatomic molecules, Li2 through Ne2. This set provides a complete, internally consistent set of examples that reveal how MO theory:

predicts bond orders and relative bond strengths,

explains magnetic behavior (diamagnetism vs. paramagnetism),

and rationalizes a key change in MO energy ordering caused by s–p mixing.

6.5.1 What an MO diagram represents

An MO diagram is an energy-level correlation sketch showing:

Atomic orbitals (AOs) of each atom on the left and right,

Molecular orbitals (MOs) in the center formed by mixing compatible AOs,

electron occupancy indicated by arrows (spin).

For second-period diatomics, we focus on valence MOs derived from the 2s and 2p AOs. Core 1s electrons are usually omitted from bond order discussions because they occupy both bonding and antibonding combinations and cancel out (and because they lie far lower in energy and are chemically inert).

6.5.2 Valence MOs for second-period diatomics

When two second-period atoms combine, their 2s and 2p orbitals generate the following types of MOs:

From 2s AOs: a bonding σ2s and an antibonding σ2s∗.

From 2p AOs: one bonding σ2p (from end-to-end overlap of the p orbitals aligned along the internuclear axis), two degenerate bonding π2p orbitals (from side-by-side overlap of the two p orbitals perpendicular to the axis), two degenerate antibonding π2p∗ orbitals, and one antibonding σ2p∗.

The labels σ and π describe symmetry about the internuclear axis, as in the previous section. What matters most for chemical predictions is the relative energy ordering of these MOs and their electron occupancy.

6.5.3 Two energy orderings: the role of s–p mixing

A key subtlety distinguishes early and late second-period diatomics: the ordering of σ2p relative to π2p.

Why ordering can change

The σ2s and σ2p orbitals have the same symmetry (both are σ), so they can mix (interact) with each other. This interaction is called s–p mixing (more precisely, mixing between σ2s and σ2p MOs). When the energy gap between the 2s and 2p AOs is relatively small, the interaction is significant and can shift MO energies:

σ2s is pushed down somewhat (stabilized),

σ2p is pushed up (destabilized),

while π2p (different symmetry) is largely unaffected by this particular mixing.

As a result, in molecules where s–p mixing is strong, σ2p can end up higher in energy than π2p.

As nuclear charge increases across the period, the 2s orbital becomes more stabilized relative to 2p (greater penetration and poorer shielding), increasing the 2s–2p energy gap and reducing s–p mixing. Consequently:

Strong s–p mixing: B2, C2, N2 (and often Li2, Be2 in a simplified picture).

Weak s–p mixing: O2, F2, Ne2.

The two standard orderings (valence MOs)

We will use two orderings for second-period homonuclear diatomics:

Ordering A (with significant s–p mixing): B2, C2, N2

Ordering B (“normal” ordering, weak s–p mixing): O2, F2, Ne2

This single change—σ2p crossing relative to π2p—is enough to change predicted electron configurations and magnetic properties in chemically meaningful ways.

6.5.4 Step-by-step procedure for constructing and using an MO diagram

For second-period homonuclear diatomics, a reliable workflow is:

1. Count valence electrons. For X2, this is 2× (valence electrons of X).

2. Choose the correct MO ordering. Use Ordering A for B2–N2, Ordering B for O2–Ne2.

3. Fill MOs from lowest to highest using Aufbau, Pauli, and Hund rules.

4. Compute bond order:

5. Determine magnetism. Any unpaired electrons ⇒ paramagnetic; all paired ⇒ diamagnetic.

6.5.5 Comparison of N2 and O2: ordering, bond order, and magnetism

Nitrogen and oxygen provide the most famous contrast in second-period MO theory:

N2 has a very strong triple bond and is diamagnetic.

O2 has a double bond by Lewis ideas but is paramagnetic, an observation that localized electron-pair theories cannot explain.

The difference emerges naturally from MO occupancy and the change in ordering caused by s–p mixing.

Electron configurations and bond orders

N2: diamagnetic, strong bond Each nitrogen contributes 5 valence electrons, so N2 has 10 valence electrons. Using Ordering A (with s–p mixing):

Fill σ2s², then σ2s∗2 (these largely cancel in bond order),

then fill π2p⁴,

then fill σ2p².

Counting bonding and antibonding electrons:

Thus MO theory predicts a triple bond, consistent with the short, strong N– – –N bond. All electrons are paired, so N2 is diamagnetic.

O2: paramagnetic with two unpaired electrons Each oxygen contributes 6 valence electrons, so O2 has 12 valence electrons. Using Ordering B (normal ordering):

Fill σ2s², then σ2s∗2,

then fill σ2p²,

then fill π2p⁴,

the remaining 2 electrons enter the degenerate π2p∗ orbitals singly (Hund’s rule).

Bonding/antibonding counts:

MO theory predicts a double bond overall, consistent with O– –O being shorter and stronger than a single bond but weaker than N– – –N.

Most importantly, the two electrons in π2p∗ remain unpaired (one in each degenerate antibonding π∗ orbital), so O2 is paramagnetic. This explains the well-known observation that liquid oxygen is attracted into a magnetic field—a phenomenon that cannot be reproduced by Lewis structures or a simple localized electron-pair picture.

6.5.6 Bond order and periodic trends across Li2 to Ne2

MO theory predicts systematic trends as we add electrons across the period, because added electrons occupy progressively higher-energy MOs and may eventually populate antibonding orbitals.

Key bond order outcomes (valence-only view)

Without drawing every diagram, the main results are:

Li2: 2 valence electrons → σ2s² ⇒ BO = 1 (bond exists).

Be2: 4 valence electrons → σ2s²σ2s∗2 ⇒ BO = 0 (no stable bond predicted in the simplest model).

B2: 6 valence electrons ⇒ BO = 1 and (critically) paramagnetic due to unpaired electrons in π2p (Ordering A).

C2: 8 valence electrons ⇒ BO = 2 (diamagnetic).

N2: 10 valence electrons ⇒ BO = 3 (diamagnetic).

O2: 12 valence electrons ⇒ BO = 2 (paramagnetic with two unpaired electrons in π2p∗).

F2: 14 valence electrons ⇒ BO = 1 (diamagnetic; antibonding π2p∗ largely filled).

Ne2: 16 valence electrons ⇒ BO = 0 (no stable bond predicted).

Two broad trends are worth emphasizing:

Bond strength generally increases with bond order. The strongest bond in this series is predicted for N2 (BO = 3).

Antibonding occupancy weakens bonds. From O2 to F2, electrons are added primarily into antibonding π2p∗ orbitals, lowering bond order from 2 to 1.

6.5.7 Magnetism as a diagnostic of MO occupancy

Magnetism provides an experimental test that is unusually direct for bonding theories:

Diamagnetic species have all electrons paired; they are weakly repelled by a magnetic field.

Paramagnetic species have one or more unpaired electrons; they are attracted into a magnetic field.

MO theory predicts magnetism by explicit electron placement. Two especially important predictions (both experimentally correct) are:

B2 is paramagnetic (unpaired electrons in π2p when ordering places π2p below σ2p).

O2 is paramagnetic (two unpaired electrons in π2p∗).

This is a major conceptual advantage over localized bonding models, where electron pairing is often implicitly assumed in bonds and lone pairs.

6.5.8 Interpreting bond length and bond enthalpy using MO theory

Within a related set of diatomics, MO bond order correlates with measurable properties:

Higher BO ⇒ shorter bond length and larger bond enthalpy.

Lower BO ⇒ longer bond length and smaller bond enthalpy.

Thus, MO theory rationalizes why N2 has an exceptionally short, strong bond, while F2 has a comparatively weak single bond: in F2, antibonding π2p∗ orbitals are heavily occupied, reducing net bonding.

Second-period homonuclear diatomics are described using valence MOs derived from 2s and 2p atomic orbitals: σ2s, σ2s∗, σ2p, π2p, π2p∗, σ2p∗.

A key periodic effect is s–p mixing, which can raise σ2p above π2p for B2, C2, and N2, but becomes weak for O2, F2, and Ne2.

MO diagrams, filled using Aufbau/Pauli/Hund rules, allow direct prediction of bond order via BO = and magnetism from unpaired electrons.

The paramagnetism of O2 arises from two unpaired electrons in the degenerate π2p∗ orbitals, a hallmark success of MO theory.

6.6 Problems

6.6.1 Hybridization and σ∕π counting in aspirin

Aspirin (acetylsalicylic acid) has the connectivity shown by the condensed structural description: a benzene ring bearing adjacent substituents –CO2H and –OCOCH3.

Draw a valid Lewis structure for aspirin, clearly indicating all double bonds in one Kekulé form of the aromatic ring.

For each non-hydrogen atom, assign the hybridization state (sp, sp², or sp³) consistent with the best localized VBT description.

Count the total number of σ bonds and the total number of π bonds in your structure.

Identify which π bonds are part of a delocalized system (aromatic or conjugated) and briefly state why a single localized Lewis structure is insufficient for the aromatic ring.

6.6.2 Hybridization and electron-domain geometry in a polyfunctional drug-like fragment

Consider the fragment (written with explicit multiple bonds and lone-pair-bearing heteroatoms):

Assign the hybridization of each heavy atom and the electron-domain geometry at each heavy atom (use VSEPR electron-domain language).

Identify every atom that must possess an unhybridized p orbital to support the given π bonding pattern.

Determine whether the C– –C–C(– –O) segment can be conjugated. If yes, state which orbitals overlap to produce conjugation and what geometric requirement must be met.

Predict (qualitatively) whether the C–C single bond between the alkene and the carbonyl carbon should have more, less, or similar rotational freedom compared with an isolated C–C single bond, and justify using the σ∕π framework.

6.6.3 Orbital-overlap sketches for σ bonds

For each molecule below, provide a clear orbital-overlap description of every unique σ bond to the central atom. Your answer must name the overlapping orbitals (e.g., “C(sp²)–O(sp²) head-on overlap”) and indicate the approximate orientation in space.

CH3Cl

BF3

NH3

HCN

Additionally, for HCN identify which orbitals account for the π bonding and how many π bonds are present.

6.6.4 Ethene and restricted rotation: quantitative reasoning without numbers

Ethene, C2H4, exhibits restricted rotation about the C– –C bond.

Using the VBT sp² + p description, state precisely what must happen to the p–p overlap during a 90∘ twist about the C– –C axis.

Without using numerical data, rank the expected rotational barrier (highest to lowest) for: H2C– –CH2, H3C–CH3, and HC– – –CH. Justify each comparison in terms of σ overlap and π overlap.

Explain why the C–C σ bond remains intact during twisting in H2C– –CH2 even though the molecule becomes much less stable.

6.6.5 Conjugation versus isolation: drawing and evaluating resonance

Consider the three molecules:

For each molecule, identify which carbon atoms are sp² hybridized and which are sp³ hybridized.

For (B) and (C), determine whether there exists a continuous array of overlapping p orbitals across four carbons. Explain your reasoning using hybridization and geometry.

For (B), draw at least two valid resonance structures and describe (qualitatively) how resonance changes the predicted bond lengths compared with a purely localized structure.

For (A) and (C), explain why resonance of the same type as in (B) is not applicable (or is much less significant).

6.6.6 Aromaticity as a delocalization test case: benzene versus cyclohexene

Draw a Kekulé structure for benzene, C6H6, and a Lewis structure for cyclohexene, C6H10.

For each carbon in each molecule, assign the hybridization.

Using VBT language, identify which orbitals provide the σ framework and which provide the π system (if present).

Explain why benzene requires a resonance-hybrid description (delocalized π electrons), whereas cyclohexene can be reasonably described with one localized π bond.

Predict which molecule should have more uniform C–C bond lengths and justify.

6.6.7 MO diagram construction for a heteronuclear diatomic ion: CN–

Treat CN– using a second-period diatomic MO approach (valence MOs derived from 2s and 2p orbitals), and assume the overall ordering is analogous to N2 (i.e., the ordering associated with significant s–p mixing).

Count the total number of valence electrons in CN– .

Construct the MO electron configuration (list occupied MOs with electron counts).

Determine the bond order and predict whether CN– is diamagnetic or paramagnetic.

Compare the predicted bond order of CN– to N2. Based on bond order alone, which would you expect to have the shorter bond? State your reasoning.

6.6.8 MO diagram construction for a heteronuclear cation: NO+

Treat NO+ using a second-period diatomic MO approach (valence MOs derived from 2s and 2p orbitals). For the purpose of this problem, treat it as isoelectronic with N2 and use the same valence MO ordering as N2.

Count the total number of valence electrons in NO+.

Construct the MO electron configuration and compute the bond order.

Predict the magnetic behavior (dia- or paramagnetic) and justify explicitly from your MO occupancy.

Compare NO+ with neutral NO (which has one additional electron). Without drawing a full second diagram, state how adding one electron would change bond order and magnetism, and identify which type of MO the added electron would enter.

6.6.9 Bond-length comparison using MO bond order: O2, O2+, and O2 –

Using the MO framework for O2 (normal ordering with σ2p below π2p):

Determine the bond order of O2, O2+, and O2 – . Show your electron-count logic clearly.

Rank the three species by expected bond length (shortest to longest) and separately by expected bond enthalpy (largest to smallest). Justify solely using bond order trends.

For each species, state whether it is paramagnetic or diamagnetic, and give the number of unpaired electrons predicted.

6.6.10 Non-existence (or extreme weakness) of He2: an energetic argument in MO terms

Construct the simplest MO description of He2 using only 1s orbitals: identify the bonding and antibonding MOs, fill with electrons, and compute bond order.

Explain, in words, why occupying the antibonding MO cancels the stabilization of the bonding MO in terms of electron density between nuclei and nodal structure.

Contrast He2 with He2+. Compute the bond order of He2+ and predict whether it should be more stable than He2. State whether you expect He2+ to be paramagnetic or diamagnetic.

Provide a brief discussion of why a bond order greater than zero does not automatically guarantee a long-lived stable molecule under all conditions (address at least one additional factor beyond bond order).

Chapter 7
Gases and the Kinetic Molecular Theory

We live submerged at the bottom of an ocean of air, yet the fluid that sustains us remains largely invisible. This chapter demystifies the chaotic dance of gas particles that creates the pressure we feel and the volume we measure. By transitioning from historical experiments to statistical mechanics, we reveal how the simple assumption of ’elastic collisions’ allows us to predict the behavior of everything from hot air balloons to deep-sea diving tanks, while also understanding exactly when and why these simple rules break down.

7.1 Pressure and Gas Laws

Gases are the simplest state of matter to model quantitatively because their macroscopic behavior is strongly constrained by a small set of measurable variables. In earlier chapters, we emphasized that the mole is the bridge between laboratory-scale measurements and particulate-scale reasoning. In this chapter we extend that bridge to gases: by relating pressure, volume, temperature, and amount of substance, we develop a compact set of laws that allow prediction and interpretation of gas behavior across a wide range of conditions.

7.1.1 State variables for gases: P, V , T, and n

A sample of gas is described (to a very good approximation in many situations) by four macroscopic variables:

Pressure, P: the force per unit area exerted by gas molecules on the walls of a container.

Volume, V : the space the gas occupies (often the container volume).

Temperature, T: a measure tied to molecular kinetic energy; gas laws require an absolute temperature scale.

Amount of substance, n: the number of moles of gas particles.

A state of a gas is a specific set (P,V,T,n). A process is a change from one state to another. When we write a gas law, we must always state which quantities are held constant (e.g., “at constant T and n”).

Pressure as momentum transfer

At the microscopic level, gas pressure results from collisions of molecules with the container walls. Each collision changes molecular momentum; Newton’s laws imply the wall experiences an equal and opposite impulse. The average force from many collisions divided by the wall area is the pressure. This microscopic picture will be formalized later using the kinetic molecular theory; for now it motivates why pressure depends on (i) how many particles are present in a given volume and (ii) how fast they move (temperature dependence).

Units of pressure

The SI unit of pressure is the pascal:

Because atmospheric-scale pressures are large in pascals, several non-SI units are common in chemistry:

bar: 1bar = 1 × 10⁵ Pa.

atmosphere (atm): defined historically from average sea-level pressure.

torr and mmHg: based on the height of a mercury column supported by a pressure difference.

Conversion relationships used throughout general chemistry:

Temperature: absolute vs. Celsius scales

Gas laws involve temperature because thermal energy affects molecular speeds. The relevant measure is absolute temperature:

A common error is to substitute Celsius values directly into gas-law equations. This fails because ratios like V∕T only make physical sense on a scale whose zero corresponds to vanishing thermal motion; that is the Kelvin scale.

7.1.2 Measuring pressure: barometers and manometers

Pressure is not directly visible, but it can be compared to a well-understood mechanical effect. The classic devices used in chemistry exploit hydrostatic balance: a pressure difference supports a column of liquid.

The mercury barometer (atmospheric pressure)

A mercury barometer consists of a glass tube closed at one end, filled completely with mercury, and inverted into a mercury reservoir. A column of mercury remains suspended, leaving an evacuated space (Torricellian vacuum) at the top. The atmosphere pushes on the reservoir surface and supports the mercury column.

Hydrostatic balance gives:

where ρ is the mercury density, g is gravitational acceleration, and h is the height difference between the mercury levels in the tube and reservoir.

Because h is often measured in mm, it is convenient to express pressure as mmHg (or torr). Under typical sea-level conditions, h ≈ 760mm, motivating the historical definition 1atm = 760mmHg.

Why mercury? Mercury has a high density, so the required column height is manageable (tens of centimeters rather than many meters). It also has low vapor pressure at room temperature, so the space above the column is close to a vacuum, simplifying interpretation.

Open-end manometer (gas pressure relative to atmosphere)

To measure the pressure of a gas sample, a manometer uses a U-shaped tube partially filled with a liquid (often mercury or oil). One arm is connected to the gas; the other is open to the atmosphere. The gas pressure is determined by comparing liquid heights.

Let Δh be the vertical height difference between the liquid columns. Then:

If the liquid level is higher on the gas side, the gas pressure is lower than atmospheric:

If the liquid level is lower on the gas side, the gas pressure is higher than atmospheric:

When mercury is the manometer fluid and Δh is measured in mm, the term ρgΔh is often directly treated as Δh in units of mmHg. In that case,

Closed-end manometer (gas pressure in absolute units)

In a closed-end manometer, one arm is sealed and contains (ideally) a vacuum above the liquid. Then the gas pressure is absolute:

independent of atmospheric pressure. This configuration conceptually mirrors the barometer and is especially useful when Patm is unknown or variable.

7.1.3 Empirical gas laws: relationships among P, V , T, and n

The foundational gas laws were established empirically: chemists and physicists measured how one variable changed as others were held constant. These laws are not arbitrary curve-fits; they reflect deep constraints of molecular motion and will later be derived (approximately) from the kinetic molecular

theory.

Boyle’s Law: pressure–volume relationship at constant T and n

Boyle’s Law states that for a fixed amount of gas at constant temperature, pressure is inversely proportional to volume:

Equivalently, the product PV is constant:

Physical interpretation. If the temperature is constant, the average molecular kinetic energy is unchanged. Compressing the gas into a smaller volume increases the frequency of wall collisions (molecules traverse shorter distances between collisions with the walls), raising pressure. Doubling the volume halves the collision frequency, lowering pressure by a factor of two.

Isothermal curves. A plot of P versus V at constant T produces a hyperbola (an isotherm). Different temperatures correspond to different isotherms: at higher T, the pressure is higher at the same volume.

Charles’s Law: volume–temperature relationship at constant P and n

Charles’s Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to absolute temperature:

Thus:

Why absolute temperature is required. If V is linear in T, extrapolating experimental V vs. T(∘C) data to V = 0 yields approximately −273.15∘C, motivating the Kelvin scale. The intercept indicates that the proportionality is fundamentally tied to an absolute zero of thermal motion.

Physical interpretation. At constant pressure, heating a gas increases molecular speeds; to keep the same pressure, the gas must expand (larger V ) so that collision frequency per unit area does not increase too much. Cooling has the opposite effect.

Avogadro’s Law: volume–amount relationship at constant P and T

Avogadro’s Law states that at constant pressure and temperature, the volume is directly proportional to the amount of gas:

So:

Physical interpretation. At fixed P and T, the average kinetic energy and the required collision rate with the walls are constrained. Adding more gas particles (increasing n) requires a larger volume so that particle density decreases enough to keep the pressure constant.

7.1.4 Graphical summary of Boyle’s and Charles’s Laws

The empirical laws become much more memorable when tied to their characteristic plots: hyperbolic P vs. V at constant T (Boyle) and linear V vs. T at constant P (Charles), with extrapolation to absolute zero.

7.1.5 From separate laws to a combined relationship

Boyle’s, Charles’s, and Avogadro’s laws each describe how one variable responds when two others are held constant. In practice, laboratory processes often involve changes in multiple variables at once. We therefore seek a single relationship that captures the simultaneous dependence of P, V , T, and n.

Start with the proportionalities:

Combining them suggests that for a given gas sample,

Rearranging gives a useful constant for a fixed amount of gas:

This expression is the Combined Gas Law for a constant amount of gas:

Using the combined gas law correctly

To apply the combined gas law, ensure:

Temperature is in kelvins.

Pressure units are consistent on both sides (atm with atm, kPa with kPa, etc.).

Amount of gas is constant (no leaks, no reactions consuming/producing gas, no dissolution into liquids, etc.).

Conceptual checkpoint. The combined gas law is not yet the most general gas equation because it assumes constant n. The next step in the chapter is to incorporate Avogadro’s insight explicitly into a single equation relating P, V , T, and n with a universal proportionality constant. That equation of state will be the ideal gas law.

7.1.6 Common experimental contexts and practical cautions

Empirical gas laws are simple, but experimental conditions can complicate their use. Keep these issues in mind:

Pressure measurements may be relative. An open-end manometer gives Pgas relative to Patm; a closed-end manometer gives absolute pressure directly.

Temperature equilibration matters. During rapid compression or expansion, temperature can change transiently (non-isothermal behavior). Boyle’s law applies to isothermal changes; in real experiments, allow time for thermal equilibration.

Gases can deviate from ideal behavior. At high pressures and low temperatures, intermolecular forces and finite molecular volume become significant. The empirical laws are excellent at moderate conditions, but later sections quantify deviations.

7.1.7 Summary of key relationships

For quick reference, the three classical empirical laws and the combined gas law are:

Law | Conditions held constant | Mathematical form
Boyle’s Law | T,n | PV = constant; P₁V ₁ = P₂V ₂
Charles’s Law | P,n | = constant; =
Avogadro’s Law | P,T | = constant; =
Combined Gas Law | n | = constant; =

7.2 The Ideal Gas Equation

In the previous section, Boyle’s, Charles’s, and Avogadro’s laws were unified into the combined gas law,

showing that PV scales linearly with T when the amount of gas does not change. The next conceptual step is to remove the restriction that n is constant. Avogadro’s law (V ∝ n at constant P and T) implies that the “constant” in PV∕T must be proportional to the amount of substance. This leads directly to the central equation of state for an ideal gas.

7.2.1 From empirical laws to the ideal gas equation

From the combined dependence established earlier,

we introduce a proportionality constant R such that the relationship becomes an equality:

Equation (??) is the ideal gas law. It is called an equation of state because it relates state variables P, V , and T for a given amount n of gas.

Meaning of “ideal”

An ideal gas is a model system in which gas particles:

occupy negligible volume compared with the container, and

exert no intermolecular attractions or repulsions except during collisions.

Real gases approach this limit at low pressure (molecules far apart) and high temperature (kinetic energy dominates attractions). Later sections will quantify deviations from ideality; for now, the ideal gas law provides a remarkably accurate and versatile framework for typical laboratory and atmospheric conditions.

7.2.2 The universal gas constant R: value, units, and physical significance

How R is determined

The gas constant R is the proportionality constant connecting the macroscopic variables of a gas. It can be determined by choosing a reference state where P, V , n, and T are known, then solving

Historically, one convenient reference uses approximately atmospheric pressure and temperatures near room temperature. In modern chemistry, R is treated as a universal constant whose numerical value depends only on the units chosen for P and V .

Common numerical values of R

Two unit systems dominate general chemistry:

Pressure in atm and volume in liters

Pressure in pascals and volume in cubic meters (SI form)

Since 1Pam³ = 1J,

Other useful forms

Because 1kPaL = 1J, another convenient value is

Dimensional interpretation of R

In the SI form, R has units of energy per mole per kelvin. This hints at its deeper meaning: R sets the energy scale that links temperature to mechanical variables. In later sections, we will connect R to the Boltzmann constant kB and to average molecular kinetic energy.

Unit consistency and a practical warning

The ideal gas law is algebraically simple but unforgiving of inconsistent units. The most common errors are:

using T in ∘C instead of K;

using a value of R that does not match the units of P and V ;

mixing pressure units (e.g., P in kPa but using R in Latmmol⁻¹ K⁻¹).

A reliable strategy is to decide first whether you will work in SI (Pa,m³) or in the common laboratory set (atm,L), and then stay consistent.

7.2.3 Solving problems with the ideal gas law

Equation (??) can be rearranged to isolate any variable:

The conceptual content is the same in each form: at fixed n, pressure increases with temperature and decreases with volume; at fixed P and T, volume is proportional to moles.

State changes at constant amount: the combined gas law as a special case

If n is constant in two states, the ideal gas law implies

which is exactly the combined gas law derived empirically. Thus, the ideal gas law subsumes the earlier laws as special cases, and also handles changes in n.

7.2.4 Density of a gas from the ideal gas law

One of the most powerful applications of the ideal gas law is converting between measurable macroscopic properties (pressure and temperature) and composition-dependent properties (molar mass, density).

Derivation of the gas density equation

Density is ρ = m∕V . The ideal gas law provides n∕V because

Relate moles to mass via n = m∕M, where M is molar mass:

Rearrange to obtain the ideal-gas density equation:

Interpretation

At fixed T, density increases with pressure (more moles per volume). At fixed P, density decreases as temperature increases (gas expands). At fixed P and T, heavier gases (larger M) are denser because each mole contains more mass.

Unit check (recommended habit)

If P is in Pa, M in kg mol⁻¹, R in Jmol⁻¹ K⁻¹, and T in K, then ρ emerges in kg m⁻³, because

7.2.5 Molar mass from gas measurements (vapor density methods)

If the mass of a gas sample is measured along with P, V , and T, the ideal gas law can determine the molar mass M. Start from n = PV∕RT and M = m∕n:

Equation (??) is the basis for classical vapor density measurements (often used for volatile liquids that can be vaporized into a known volume at a known T and P).

Why this works

At a given P, V , and T, the ideal gas law fixes the number of moles present:

If you can measure the mass m that occupies that state, the ratio m∕n is the molar mass. This method is conceptually straightforward: it compares how much mass corresponds to one mole under the measured conditions.

Common experimental cautions

When applying Eq. (??) to real measurements, systematic errors often come from:

Non-ideal behavior (especially for polarizable vapors near condensation).

Incorrect pressure if the gas is collected over water and water vapor pressure is not subtracted (covered in the next section on partial pressures).

Temperature gradients (gas not actually at the thermometer reading).

Incomplete vaporization or presence of air (changes n without changing measured mass in the intended way).

7.2.6 Standard Temperature and Pressure (STP) and molar volume

Because gases are compressible, comparisons are simplest when all samples are referenced to standard conditions. However, the meaning of “standard” depends on context, so it is important to use a definition consistently.

STP and why the definition matters

In general chemistry problem solving, STP is commonly taken as:

Under these conditions, one mole of an ideal gas occupies the molar volume

This value is frequently rounded to 22.4 Lmol⁻¹.

A caution on alternative “standard” definitions

In some scientific and engineering contexts, a “standard pressure” of 1 bar is used instead of 1 atm, which changes the molar volume slightly. In this text, unless explicitly stated otherwise, you should interpret STP as 273.15 K and 1.00 atm when using the shortcut 22.4 Lmol⁻¹. When the problem provides different P and T, do not force STP; use PV = nRT directly.

When the molar volume shortcut is appropriate

Equation (??) is a shortcut for a specific set of conditions. It is valid only when:

the gas behaves ideally (usually reasonable near 1 atm for many gases), and

the sample is at STP as defined for the problem.

If T or P differs significantly from STP, you should compute n from PV∕RT rather than scaling by 22.4 Lmol⁻¹ unless the scaling is explicitly derived from the ideal gas law.

7.2.7 Non-standard conditions: scaling and strategy

Many problems involve comparing gas samples at different conditions. A consistent strategy is:

Identify what is constant (often n, sometimes V , sometimes P).

Choose the simplest form of the ideal gas law or use ratios.

Convert temperatures to kelvins and ensure unit consistency for R.

Two-state form with changing moles (generalized combined relationship)

If a gas changes from state 1 to state 2, and n may change, then from PV = nRT,

This is simply the ideal gas law written to emphasize proportionality across two states.

7.2.8 Summary: key equations and when to use them

The ideal gas law is the core tool:

From it follow two especially useful derived relationships:

ρ | = (gas density at given P,T), |  |
M | = (molar mass from mass and gas state). |  |

At STP (as defined here: 1.00 atm, 273.15 K), the molar volume of an ideal gas is:

These results will be extended in the next section to gas mixtures, where the key additional concept is that total pressure is shared among components through partial pressures.

7.3 Gas Mixtures and Partial Pressures

The ideal gas equation PV = nRT describes a single gas sample in terms of macroscopic state variables. Many real systems, however, involve mixtures of gases: air (mostly N2 and O2), natural gas (often CH4 with other hydrocarbons), and laboratory gas collections (a product gas mixed with water vapor). When the gases in a mixture do not react chemically, their behavior is described to excellent approximation by extending the ideal gas model to multiple components.

This section develops the core quantitative idea: in an ideal-gas mixture, each component contributes independently to the total pressure, and the contribution is proportional to its mole fraction. We then apply this to a ubiquitous laboratory technique: collecting gases over water.

7.3.1 Dalton’s Law of Partial Pressures

Consider a mixture of k non-reacting gases in the same container (same V and T). Let the components be labeled 1,2,…,k, with amounts n₁,n₂,…,nk. The total amount is

Definition of partial pressure

The partial pressure Pi of component i is defined as the pressure that component i would exert if it alone occupied the container volume V at the same temperature T.

For an ideal gas, applying PV = nRT to component i gives

Statement and derivation of Dalton’s Law

The total pressure of the mixture is measured at the container walls and (for an ideal mixture) equals the sum of the component contributions:

This is Dalton’s Law of Partial Pressures.

Derivation from the ideal gas law. Apply the ideal gas law to the entire mixture:

Now substitute ntot = ∑ ini:

Divide both sides by V :

Using Eq. (??), each term is Pi, so

which is Eq. (??).

Interpretation. In an ideal mixture, the gases do not “compete” for space by exerting special interactions. At fixed V and T, each component’s pressure is determined only by how many moles of that component are present. Total pressure is additive because total collision rate with the container walls is the sum of collision rates from each component.

7.3.2 Mole fraction and its connection to partial pressure

To use Dalton’s law efficiently, we need a compact way to specify mixture composition.

Mole fraction

The mole fraction xi of component i is

Mole fractions are dimensionless, satisfy 0 ≤ xi ≤ 1, and sum to unity:

Deriving Pi = xiPtot

From Eq. (??) and the ideal gas law for the total mixture:

Divide the first equation by the second:

Therefore,

This equation is one of the most useful forms of Dalton’s law: partial pressure equals mole fraction times total pressure.

Consequences.

If you know the mixture composition in moles (or mole fractions) and the total pressure, you can compute each partial pressure directly.

If you know partial pressures, you can infer mole fractions:

Composition in mole fraction is the same as composition in partial-pressure fraction for ideal mixtures.

7.3.3 Partial pressures in chemical contexts

Gas-phase stoichiometry often occurs in mixtures. In many applications, a reacting species is present in a background gas (e.g., O2 in air), and it is the partial pressure of the reacting component that matters for collision frequency and reaction rate.

Even when reactions are not considered, partial pressures are essential for:

determining the amount of a component gas in a mixture (ni = PiV∕RT);

correcting laboratory gas collections (gas + water vapor);

interpreting atmospheric composition (e.g., PO2 rather than total atmospheric pressure).

7.3.4 Collecting gases over water: the pneumatic trough

A common laboratory method for collecting a gas is to displace water in an inverted container (graduated cylinder or gas jar) submerged in a water trough. The gas produced in a reaction is bubbled into the inverted container, where it accumulates above the water. This method is convenient, but it introduces a crucial complication: the collected gas is not pure. It is a mixture of the desired gas and water vapor.

What pressure is measured?

The gas space at the top of the inverted container contains:

the desired gas (e.g., H2, O2, CO2 if sufficiently insoluble), and

water vapor from the liquid water surface.

If the system is at equilibrium, the water vapor has a partial pressure equal to the vapor pressure of water at the experiment temperature:

This vapor pressure depends strongly on temperature and is tabulated.

If the water levels inside and outside the container are equalized (or corrected for any height difference), the total pressure of the trapped gas mixture equals atmospheric pressure:

Dalton’s law then gives:

Therefore, the partial pressure of the dry collected gas is

Hydrostatic correction when water levels are not equal

If the water level inside the collection vessel is higher than outside by a vertical difference Δh, then the trapped gas must support an additional hydrostatic pressure. In that case:

and if the inside level is lower:

Then Dalton’s law is applied to Ptot, not directly to Patm:

In routine introductory experiments, the water levels are typically equalized to eliminate Δh, but the correction is conceptually important and sometimes experimentally unavoidable.

7.3.5 Diagram: pneumatic trough gas collection and pressure balance

7.3.6 Example framework: collecting H2 over water (no numerical solution)

The following workflow illustrates how Dalton’s law enters a typical calculation. Suppose hydrogen is collected over water, and you measure:

volume of gas collected V ,

temperature T (of the water bath / gas),

atmospheric pressure Patm,

water levels (to decide whether a hydrostatic correction is needed).

Step 1: Determine the total pressure of the trapped mixture

If water levels are equal: Ptot = Patm.

If not equal: apply the hydrostatic correction to obtain Ptot.

Step 2: Subtract water vapor pressure to get the dry gas partial pressure

Look up PH2O∗(T) from a vapor pressure table, then compute:

Step 3: Use the ideal gas law on the dry gas

Now apply PV = nRT to the dry hydrogen alone:

This nH2 can then be connected to reaction stoichiometry (mass of reactant consumed, theoretical yield, percent yield, etc.) in later problem types.

7.3.7 Summary of key equations

For a non-reacting ideal-gas mixture in a single container at T and V :

Ptot | = ∑ iPi (Dalton’s law), |  |
xi | = , |  |
Pi | = xiPtot, |  |
ni | = . |  |

For gas collected over water (after any hydrostatic correction to obtain Ptot):

These relationships allow you to move cleanly between measured pressures and chemically meaningful amounts of substance, preparing for gas-phase stoichiometry and kinetic theory in subsequent sections.

7.4 Stoichiometry of Gas Reactions

The previous sections established (1) how gas state variables P, V , T, and n are related by the ideal gas law and (2) how to treat mixtures using Dalton’s law and water-vapor corrections. We now combine these tools with chemical stoichiometry. The key idea is simple but powerful:

Balanced equations relate moles of substances. The ideal gas law relates moles of a gas to measurable P, V , and T.

Once those two links are in place, you can convert between masses, volumes, and pressures of gaseous reactants and products under a wide variety of conditions.

7.4.1 Stoichiometry review: coefficients are mole ratios

For a balanced chemical equation

the coefficients give mole ratios:

These ratios are exact consequences of atom conservation.

When gases are involved, the ideal gas law provides the additional bridge:

Thus, in gas stoichiometry problems, you will repeatedly move between:

7.4.2 Gay-Lussac’s Law of Combining Volumes (volume–volume ratios)

Empirical statement and its meaning

Gay-Lussac’s Law of Combining Volumes states that, for reacting gases measured at the same temperature and pressure, the volumes of reactants and products (if gases) are in simple whole-number ratios.

This observation is a direct consequence of the ideal gas law. If two gas samples are at the same T and P, then V ∝ n (Avogadro’s law). Because stoichiometric coefficients relate moles, they also relate volumes under the same T and P.

Derivation from the ideal gas law

For any gas at a fixed T and P,

Therefore, for two gases 1 and 2 at the same T and P,

In a reaction, if n₁ : n₂ is fixed by the balanced equation, then V ₁ : V ₂ must be the same ratio provided both volumes are measured under identical conditions of T and P.

Example: synthesis of hydrogen chloride (gas phase)

At a common T and P, the stoichiometric mole ratio is 1 : 1 : 2, so the volume ratio is also

Thus, 3.0 L of H2 reacts with 3.0 L of Cl2 to form 6.0 L of HCl if all gases are compared at the same T and P.

What Gay-Lussac’s law does not say

It does not mean that the reaction necessarily occurs at constant T and P, nor that volume ratios hold when conditions differ. It is a statement about measured volumes of reacting gases at the same T and P, not about what happens in a rigid vessel or under changing conditions.

7.4.3 The ideal gas law as a stoichiometric bridge

In most laboratory problems, gases are not conveniently presented as “liters at the same T and P” for each reactant and product. Instead, you might know a mass of a solid reactant that produces a gas, or you might measure a gas volume at non-standard conditions. The ideal gas law allows you to compute moles in these non-idealized circumstances.

General workflow for gas stoichiometry

A robust workflow is:

Write and balance the chemical equation.

Convert given quantities to moles. If given mass: n = m∕M. If given gas state: n = PV∕RT (use dry gas partial pressure if needed).

Use mole ratios from the balanced equation to find moles of the desired species.

Convert moles to the requested form. To mass: m = nM. To volume at specified P,T: V = nRT∕P. To partial pressure in a mixture: Pi = niRT∕V or Pi = xiPtot.

7.4.4 Molar volume at STP: a useful shortcut with strict limits

At STP (as defined in earlier sections: T = 273.15K, P = 1.00atm), an ideal gas has molar volume

This gives a shortcut:

When the shortcut is appropriate

The molar-volume shortcut is appropriate when:

the gas is near-ideal,

the problem explicitly states STP (or the exact P and T match the definition being used), and

you only need modest precision consistent with the rounding.

When to avoid it

Avoid the shortcut when:

P and/or T are not STP (even if “close”),

high precision is required,

the gas is likely non-ideal (high pressure, low temperature), or

the gas is collected over water and the total pressure is not the dry gas pressure.

In these cases, use PV = nRT directly.

7.4.5 Mass–volume stoichiometry: producing a gas from a measured mass

A common experimental scenario is a solid or liquid reactant that generates a gas. The stoichiometric question becomes: how much gas (in moles or in volume at given conditions) is produced?

Example framework: CaCO3 reacting with acid

Consider

From the equation, 1 mol CaCO3 produces 1 mol CO2, assuming HCl is in excess.

If m grams of CaCO3 are consumed:

If the CO2 is collected at some P and T, the predicted volume is

If collected over water, PCO2 must be the dry gas partial pressure:

This explicitly links gas stoichiometry to the partial-pressure corrections developed earlier.

7.4.6 Volume–mass stoichiometry: using a measured gas volume to infer mass

The reverse situation is also common: you measure a gas volume at known conditions and use it to infer how much reactant was consumed or product was

formed.

Example framework: metal + acid producing H2

For a typical reaction

a measured amount of H2 corresponds directly to moles of Zn consumed (1:1 ratio). If the experiment collects H2 over water:

PH2 | = Ptot − PH2O∗(T), |  |
nH2 | = , |  |
nZn | = nH2, |  |
mZn | = nZnMZn. |  |

The gas measurement becomes a quantitative probe of how much solid reacted.

7.4.7 Stoichiometry with gaseous reactants: limiting reactants in the gas phase

Limiting-reactant analysis becomes especially transparent for gases because amounts can be computed from P, V , and T. The underlying logic is unchanged: the limiting reactant is the one that would produce the least amount of product after accounting for stoichiometric ratios.

Limiting-reactant strategy for gases

Suppose gases A(g) and B(g) react to form C(g):

Given initial states for A and B, the strategy is:

Compute nA and nB using n = PV∕RT for each (or use volume ratios only if both are at the same T and P).

Compare the available mole ratio to the stoichiometric requirement using

Identify the limiting reactant and compute the moles of product formed.

If desired, compute leftover moles of the excess reactant and then translate to a final pressure or composition using the ideal gas law and Dalton’s law.

Special case: volume ratios at common T and P

If A and B are in separate containers but both measured at the same T and P, then n ∝ V and limiting-reactant comparisons can be made using volumes directly:

This is essentially Gay-Lussac’s law applied in a stoichiometric context. If T and P are not common, you must compute moles.

7.4.8 Products and final states: connecting stoichiometry to final P, V , and composition

Stoichiometry predicts how many moles of each species remain after reaction. The gas laws then predict what that implies for measurable properties.

Constant V and T: pressure changes reflect changes in total moles

In a rigid container (V constant) at constant temperature, the ideal gas law implies

Thus, at fixed V and T, any reaction that changes the total number of moles of gas changes the pressure proportionally:

For example, in the gas-phase reaction

the stoichiometry converts 3 mol gas to 2 mol gas (for complete reaction in the correct proportions), so pressure decreases in a sealed container at constant T.

Mixtures after reaction: partial pressures from mole amounts

If multiple gases remain, Dalton’s law and mole fractions connect stoichiometry to partial pressures:

This is often the cleanest way to answer questions such as “What is the partial pressure of the excess reactant after the reaction?”

7.4.9 Summary: the core toolbox for gas-reaction stoichiometry

Gas-reaction stoichiometry relies on three linked ideas:

Balanced equations give mole ratios.

The ideal gas law relates moles to P, V , and T: n = PV∕RT.

Dalton’s law relates mixture composition to pressure: Ptot = ∑ Pi, and Pi = xiPtot.

Gay-Lussac’s volume relationships emerge as a special case when comparing gases at common T and P, and the molar volume at STP provides a limited but useful shortcut. These methods will remain central as we move to a microscopic interpretation of gas behavior in the kinetic molecular theory.

7.5 Kinetic Molecular Theory

The gas laws developed earlier in this chapter—Boyle’s law, Charles’s law, Avogadro’s law, Dalton’s law, and the ideal gas equation PV = nRT—are macroscopic relationships. They describe what gases do without explicitly stating why. The kinetic molecular theory (KMT) provides the microscopic explanation: it connects the measurable state variables P, V , and T to the motion of molecules.

KMT is not merely a qualitative story. With a small set of idealizing assumptions, it produces quantitative predictions: the ideal gas law itself, the dependence of molecular speeds on temperature and molar mass, and the forms of diffusion and effusion laws.

7.5.1 Postulates of the kinetic molecular theory (ideal gas model)

KMT models an ideal gas as a large number of microscopic particles (atoms or molecules) with the following properties:

Particles are point masses. The volume of individual gas particles is negligible compared with the container volume.

Constant random motion. Particles move continuously and randomly in straight lines between collisions.

No intermolecular forces (except during collisions). Particles neither attract nor repel one another appreciably.

Elastic collisions. Collisions among particles and with container walls conserve total kinetic energy (and momentum).

Large numbers and statistical behavior. Macroscopic properties are averages over enormous numbers of particles.

These assumptions become increasingly accurate at low pressure and high temperature, where particles are far apart and move rapidly.

7.5.2 Pressure from molecular collisions

A microscopic definition of pressure

Pressure is force per unit area. In KMT, the force arises from molecules striking the container walls and rebounding. Each collision changes a molecule’s momentum; the wall experiences an equal and opposite momentum change per unit time. The average over many collisions yields a steady macroscopic pressure.

For a single molecule of mass m moving with x-component of velocity ux in a cubic container of side L, an elastic collision with a wall perpendicular to x reverses ux and changes momentum by Δpx = −2mux (the sign indicates direction). The frequency of collisions with that wall is proportional to ux∕L. Summing over all molecules and averaging leads to a central KMT result:

where

N is the number of molecules,

V is the container volume,

u is the molecular speed (magnitude of velocity),

⟨u²⟩ is the mean of the squared speeds (an average over all molecules).

The factor 1∕3 arises from isotropy: in random motion, average kinetic contributions are equally distributed among the three spatial directions, so ⟨ux²⟩ = ⟨uy²⟩ = ⟨uz²⟩ = ⟨u²⟩.

Interpretation. Equation (??) shows two independent ways to increase pressure:

Increase number density N∕V (more molecules per volume ⇒ more impacts).

Increase mean squared speed ⟨u²⟩ (harder impacts).

The first is tied to compression (Boyle’s law); the second is tied to heating (Charles’s law).

7.5.3 Temperature and kinetic energy

Average translational kinetic energy

The translational kinetic energy of one molecule is

Taking an average over all molecules,

A deep result of statistical mechanics (consistent with KMT assumptions) is that the average translational kinetic energy depends only on absolute temperature:

where kB is the Boltzmann constant. For one mole of molecules, multiply by Avogadro’s number NA, and use R = NAkB:

Consequences.

is proportional to T, not to ∘C. This provides the microscopic justification for using kelvins.

At the same T, all ideal gases have the same average translational kinetic energy per molecule, regardless of molar mass. Heavier molecules move

more slowly on average, but their greater mass compensates in Ek = mu².

7.5.4 Deriving the ideal gas law from KMT

Combine Eq. (??) with Eq. (??).

From Eq. (??):

Substitute into Eq. (??):

Rearrange:

Convert from molecules N to moles n using N = nNA, and substitute R = NAkB:

which is exactly the ideal gas law derived previously from empirical evidence. KMT therefore explains not only that PV scales with nT, but why: pressure is the macroscopic manifestation of molecular impacts, and temperature is the measure of average kinetic energy.

7.5.5 Molecular speed: most probable, average, and root-mean-square

Because gas molecules move randomly, they do not all have the same speed at a given temperature. Instead, speeds are distributed over a range. Three characteristic speeds are commonly used:

Most probable speed, ump: the speed at the peak of the distribution (the mode).

Average speed, uavg: the mean value of the speed.

Root-mean-square speed, urms: defined by

For an ideal gas (Maxwell–Boltzmann statistics), these are

ump | = , |  |
uavg | = , |  |
urms | = , |  |

where M is the molar mass in kg mol⁻¹ when R is in Jmol⁻¹ K⁻¹, yielding speeds in ms⁻¹.

Ordering and interpretation

Because of how each average weights high-speed molecules,

All three increase with temperature (∝) and decrease with molar mass (∝ 1∕). This captures an important microscopic point:

At the same temperature, lighter gases move faster on average than heavier gases.

7.5.6 Maxwell–Boltzmann speed distributions

The Maxwell–Boltzmann distribution gives the probability density f(u) that a molecule has speed between u and u + du. Its general form is

While you will not often need to use this full expression in introductory calculations, it explains the characteristic shape of speed distributions:

f(u) = 0 at u = 0 (the u² factor),

a peak at u = ump,

a long tail toward high speeds.

Two qualitative trends are essential:

For the same gas, higher T broadens and shifts the distribution to higher speeds.

At the same T, lighter gases have distributions shifted to higher speeds than heavier gases.

7.5.7 Diffusion and effusion

Molecular motion leads to spontaneous mixing and escape processes. Although diffusion and effusion are distinct, both reflect the same underlying speed distributions.

Diffusion

Diffusion is the net movement of molecules from regions of higher concentration to regions of lower concentration due to random thermal motion. In gases, diffusion tends to be rapid because molecules are far apart and move quickly.

From a molecular viewpoint:

Individual molecules move randomly in all directions.

If there is a concentration gradient, more molecules cross a reference plane from the high-concentration side than from the low-concentration side, producing a net flux.

A full quantitative treatment involves transport theory (Fick’s laws) and is beyond the scope of this section, but an important qualitative prediction is immediate: higher temperature and lower molar mass generally increase diffusion rates because typical molecular speeds are higher.

Effusion

Effusion is the escape of gas molecules through a tiny hole (small compared with the mean free path) into a vacuum or low-pressure region. If the hole is sufficiently small, molecules pass through without significant collisions at the opening, and the effusion rate depends mainly on molecular speeds.

Graham’s Law of Effusion

Graham’s law states that, at the same temperature and pressure, effusion rates are inversely proportional to the square root of molar mass:

This follows from the speed scaling in Eqs. (??)–(??), which imply typical speeds u ∝ 1∕ at fixed T. Effusion rate is proportional to the number of molecules reaching the hole per unit time, which scales with their average speed for comparable conditions.

Using Graham’s law carefully.

The gases must be at the same temperature. Temperature controls the speed distribution.

The hole must be small enough for true effusion (molecules pass one-by-one without many collisions at the hole).

Graham’s law is most accurate when both gases behave ideally and do not interact strongly with the wall material.

7.5.8 Mean free path and collision frequency

KMT assumes molecules travel in straight lines between collisions. The average distance a molecule travels between collisions is the mean free path, denoted λ.

Definition and qualitative dependence

Mean free path depends on how many molecules are present per volume and on the effective molecular collision diameter:

At higher pressure (greater number density), collisions are more frequent, so λ decreases.

At higher temperature (for fixed pressure), number density decreases (because n∕V = P∕RT), so λ increases.

A common theoretical expression is

where d is an effective collision diameter and P is pressure. The factor arises from relative motion of molecules in a gas (collisions depend on relative speed, not just a single molecule’s speed).

Why λ matters.

Diffusion and viscosity: transport properties depend on how far molecules travel before colliding.

Effusion criterion: effusion assumes the hole diameter is much smaller than λ.

Deviations from ideality: when molecules are close enough that collisions are strongly influenced by intermolecular forces and finite size, ideal assumptions can fail (explored in the next major section on real gases).

7.5.9 Key results and conceptual connections

Kinetic molecular theory provides a unified microscopic foundation for the macroscopic gas laws:

Pressure arises from momentum transfer in molecular collisions.

Temperature measures average translational kinetic energy:

These lead to the ideal gas law:

Molecular speeds are distributed (Maxwell–Boltzmann), with characteristic speeds scaling as

Effusion rates follow Graham’s law:

The mean free path λ connects microscopic collisions to transport behavior and helps define when effusion and ideal assumptions are valid.

These results set the stage for understanding real gases: when particles have finite size and intermolecular forces, the idealized KMT assumptions break down, and measurable deviations from PV = nRT appear.

7.6 Real Gases and Deviation from Ideality

The ideal gas law PV = nRT is extraordinarily useful, but it is built on the kinetic molecular theory assumptions that gas particles have negligible volume and exert no intermolecular forces. Real molecules are neither point-like nor noninteracting. As a result, all real gases deviate from ideal behavior to some extent. These deviations become most pronounced under conditions that amplify the neglected effects:

High pressure: molecules are forced close together, making their finite size and short-range repulsions important.

Low temperature: molecular speeds decrease, so intermolecular attractions have more influence on motion and pressure.

This section introduces a compact quantitative measure of deviation (the compressibility factor Z) and then develops a more realistic equation of state (the van der Waals equation) that explicitly corrects for finite molecular volume and intermolecular attractions.

7.6.1 When and why the ideal gas model fails

Finite molecular volume (excluded volume)

In the ideal gas model, molecules occupy no volume; the entire container volume V is available for translational motion. In reality, each molecule excludes a region of space from being occupied by the centers of other molecules. At high pressures, the fraction of the container volume occupied (or effectively excluded) becomes non-negligible. The immediate consequence is that the effective free volume is smaller than the measured container volume, which tends to make real gases less compressible than predicted at sufficiently high P.

Intermolecular attractions

Real molecules attract each other at moderate separation due to dispersion forces and (for polar molecules) dipole interactions and other specific forces. Attractions reduce the frequency and intensity of collisions with the container walls: near a wall, molecules are pulled back toward the bulk by neighbors. This tends to reduce the measured pressure relative to the ideal prediction, especially at moderate pressures where attractions act without being overwhelmed by excluded-volume effects.

Competing effects and the typical pattern

As pressure increases from near zero:

At low to moderate P, attractions often dominate ⇒ Preal < Pideal.

At higher P, excluded volume and short-range repulsions dominate ⇒ Preal > Pideal.

Whether a gas shows a pronounced dip below ideality depends on its intermolecular forces (e.g., more polarizable gases typically show stronger attraction effects).

7.6.2 Compressibility factor Z

Definition

The compressibility factor Z is defined by comparing the real equation of state to the ideal gas law:

If a gas behaves ideally, then PV = nRT and therefore

Interpreting values of Z

The sign of deviation carries physical information:

Z < 1: the gas is more compressible than ideal; attractions effectively lower the pressure (or reduce PV ).

Z > 1: the gas is less compressible than ideal; excluded volume/repulsions effectively raise the pressure (or increase PV ).

At very low pressure, most gases approach ideal behavior and

This reflects the fact that when molecules are far apart, finite size and intermolecular forces become negligible.

Why Z is so useful

The compressibility factor packages complex real-gas behavior into a single dimensionless number. Once Z is known (from experiment or from an equation of state), the ideal gas law can be corrected in a minimal way:

Equation (??) does not explain why deviations occur, but it provides a direct computational correction.

7.6.3 Typical Z vs. P behavior for real gases

At a fixed temperature (especially one not too high above the condensation region), many gases show the characteristic pattern: Z ≈ 1 at low P, then a dip below 1 (attraction-dominated), and then a rise above 1 (repulsion/excluded-volume dominated). Different gases show different magnitudes of these effects.

7.6.4 The van der Waals equation: a corrected equation of state

The compressibility factor describes deviations, but it does not explain them mechanistically. A more informative approach is to modify the ideal gas law to incorporate the two primary non-ideal effects:

finite molecular volume → reduces free volume;

intermolecular attractions → reduce measured pressure.

This leads to the van der Waals equation.

Correcting for excluded volume: the b term

In the ideal gas law, molecules are assumed to have access to the full container volume V . If molecules occupy a finite volume, the free volume is smaller. Van der Waals proposed replacing V by (V − nb), where b is an empirical constant characteristic of the gas:

Here, b has units of volume per mole (e.g., Lmol⁻¹). The combination nb represents an approximate excluded volume for n moles.

Physical meaning of b. Larger molecules generally have larger b values because they exclude more volume. Thus, b is connected to effective molecular size.

Correcting for attractions: the a term

Intermolecular attractions reduce the measured pressure because molecules approaching a wall are pulled inward by neighboring molecules, decreasing the net momentum transfer to the wall. This effect grows with molecular crowding: stronger when more molecules are packed into a given volume.

A simple proportionality captures this:

The square arises because attractions depend on pairs of molecules. Van der Waals proposed adding a term to the measured pressure to estimate the “idealized” pressure that would exist without attractions:

where a is another gas-specific constant (with units such as L²atm∕mol² when using L and atm).

Physical meaning of a. Gases with stronger intermolecular attractions (greater polarizability, permanent dipoles, etc.) generally have larger a values.

Assembling the van der Waals equation

Replace P by P + a(n∕V )² and V by V − nb in the ideal gas law:

Substitute Eqs. (??) and (??):

This is the van der Waals equation for n moles. For one mole (n = 1), it is often written:

where V m = V∕n is the molar volume.

Recovering ideal behavior as a limiting case

At low pressure (large V ) and/or high temperature, the correction terms become small:

Then Eq. (??) reduces to PV = nRT. The ideal gas law is therefore a limiting approximation of the more general real-gas description.

7.6.5 Connecting van der Waals corrections to the sign of Z − 1

The van der Waals equation predicts both Z < 1 and Z > 1 regimes, consistent with experimental trends:

Attraction-dominated regime (often moderate P, lower T). The a(n∕V )² term is significant, meaning the measured pressure P is smaller than the pressure expected without attractions. This drives PV below nRT, so Z < 1.

Excluded-volume-dominated regime (high P). The reduction V −nb becomes significant, meaning the effective free volume is smaller. To satisfy the equation of state, the measured pressure must be higher than ideal, driving PV above nRT, so Z > 1.

The important conceptual point is that Z is not merely an error metric; it encodes which physical effect dominates under the conditions of interest.

7.6.6 Using real-gas corrections in calculations

Two common approaches

In practical problem solving, real-gas behavior is typically handled in one of two ways:

Approach 1: Compressibility factor correction. If Z is known (from a table, chart, or equation of state),

This is algebraically simple and is often accurate when Z is reliable for the given conditions.

Approach 2: Use an equation of state (e.g., van der Waals). When a and b are known (tabulated for many gases), the van der Waals equation can be solved for the desired variable. This typically requires algebraic manipulation and sometimes numerical solving because the equation can be nonlinear in V .

Estimating error in the ideal assumption

A useful way to quantify the error of using the ideal gas law at given conditions is:

For example, Z = 0.94 corresponds to an ideal-gas deviation of about 6% in PV at those conditions.

7.6.7 Summary: what to remember

Real gases deviate from ideality most strongly at high pressure and low temperature.

The compressibility factor quantifies deviation:

The van der Waals equation incorporates two physical corrections:

where a measures attractions and b measures excluded volume (molecular size).

In the low-pressure/high-temperature limit, real gases approach ideal behavior and both Z → 1 and the van der Waals equation reduces to PV = nRT.

With these tools, the ideal gas law becomes not just a formula to apply, but a benchmark model: real-gas behavior is understood in terms of systematic, physically interpretable deviations from ideality.

7.7 Problems

7.7.1 Multi-step gas stoichiometry with collection over water

Zinc metal reacts with hydrochloric acid according to

A student reacts 0.412g of Zn with excess HCl and collects the H2 over water in a eudiometer. The volume of gas collected is 148.2mL at 24.0∘C. The atmospheric pressure is 755.6Torr. The water levels are equal inside and outside the eudiometer. The vapor pressure of water at 24.0∘C is 22.4Torr.

Determine PH2 (dry hydrogen pressure).

Calculate the moles of H2 collected.

Determine the theoretical moles of H2 expected from the given mass of Zn.

Compute the percent yield of H2.

Assuming the only systematic error is a leak that lost H2, estimate the volume of H2 lost (in mL) at the collection conditions.

7.7.2 Dalton’s law in connected vessels (two bulbs, stopcock opened)

Two rigid bulbs are connected by a stopcock and maintained at constant temperature T = 298.15K. Bulb A has volume V A = 2.00L and contains N2(g) at PA = 1.20atm. Bulb B has volume V B = 3.50L and contains CO2(g) at PB = 0.650atm. The stopcock is opened and the gases mix without reacting.

Calculate the final total pressure Ptot,final.

Determine the final partial pressures PN2 and PCO2.

Compute the mole fractions xN2 and xCO2.

If, after mixing, the system is compressed isothermally into a single rigid 3.00L vessel, determine the new total pressure and each partial pressure.

7.7.3 Manometer reasoning and gas-law state change

A sample of gas is confined in a rigid container connected to an open-end mercury manometer. Initially, the mercury level is 84mm higher on the gas side than on the atmospheric side. Atmospheric pressure is 742mmHg. The container is then heated from 21.0∘C to 125.0∘C with no change in volume and no gas loss.

Determine the initial gas pressure in mmHg and atm.

Assuming ideal behavior, calculate the final gas pressure.

Predict the final manometer height difference Δh (state whether the gas-side mercury is higher or lower than the atmospheric side).

7.7.4 Gas density and molar mass from non-standard conditions

A gas has density ρ = 1.87g L⁻¹ at T = 35.0∘C and P = 0.955atm.

Assuming ideal behavior, determine the molar mass M of the gas.

The gas is suspected to be a mixture of CO2 and N2. Assuming ideal mixing and that both gases are present, determine the mole fraction of CO2 that would produce the measured molar mass.

Using your composition, compute the partial pressures of CO2 and N2 at the stated conditions.

7.7.5 Limiting reactant with gaseous reactants at different states

Ammonia is produced by the Haber process:

A reactor is charged with N2 and H2 from two separate tanks. Tank 1 contains N2 at V ₁ = 12.0L, P₁ = 2.20atm, T₁ = 310K. Tank 2 contains H2 at V ₂ = 18.0L, P₂ = 1.85atm, T₂ = 295K. The contents are transferred into a single rigid 25.0L reactor and brought to a uniform temperature of 300K. Assume complete reaction to equilibrium that consumes all of the limiting reactant (ignore real equilibrium limitations) and that NH3 remains gaseous.

Determine the initial moles of N2 and H2 transferred.

Identify the limiting reactant and calculate the moles of NH3 formed.

Determine the moles of any excess reactant remaining.

Calculate the final total pressure in the reactor at 300K.

Calculate the final partial pressures of NH3, N2, and H2.

7.7.6 Graham’s law: comparative effusion and time-to-effuse

A rigid container with a small pinhole effuses gas into a vacuum at constant temperature. In one trial, Ne requires 125s to effuse a certain fraction of the gas. In a second trial under identical conditions, the container is filled with Xe.

Use Graham’s law to predict the time required for Xe to effuse the same fraction.

If instead the container holds an unknown gas and it effuses in 96.0s under the same conditions as neon, determine the molar mass of the unknown.

Discuss briefly (one or two sentences) whether Graham’s law should be expected to be more accurate at low pressure or high pressure, and why, in terms of the mean free path.

7.7.7 Maxwell–Boltzmann speeds and temperature scaling

For O2(g), compute the following at T = 298.15K:

ump, uavg, and urms.

The ratio urms(600K)∕urms(300K) and interpret the result.

At T = 298.15K, compute urms for He and compare it to O2 by calculating the ratio urms(He)∕urms(O2).

7.7.8 Real-gas deviation using compressibility factor

A 2.50mol sample of CH4(g) is placed in a 3.00L container at T = 300.0K.

Compute the pressure predicted by the ideal gas law.

Suppose experimental data at this T and approximate pressure range indicate Z = 0.88 for methane. Compute the corrected pressure using PV = ZnRT.

Determine the percent error in pressure if the ideal gas law is used instead of the Z-corrected value.

State whether the sign of Z − 1 suggests attractions or excluded-volume effects dominate under these conditions.

7.7.9 van der Waals calculation and percent error relative to ideal

A 1.000mol sample of N2(g) is confined to a volume V = 1.000L at T = 300.0K. For nitrogen, use van der Waals constants a = 1.39L² atmmol⁻² and b = 0.0391Lmol⁻¹.

Compute the pressure predicted by the ideal gas law.

Compute the pressure predicted by the van der Waals equation:

Determine the percent error made by using the ideal gas law (take the van der Waals result as the best estimate).

Using your computed values, determine the compressibility factor Z implied by the van der Waals pressure at these conditions.

7.7.10 Multi-step partial pressures with reaction, then mixing

A sealed rigid vessel of volume V = 10.00L at T = 298.15K initially contains a mixture of CO(g) and O2(g) with partial pressures PCO = 0.800atm and PO2 = 0.500atm. The mixture is sparked and reacts according to

Assume ideal-gas behavior and complete consumption of the limiting reactant.

Determine which reactant is limiting and the moles of CO2 produced.

Determine the final moles of each gas remaining in the vessel.

Compute the final total pressure and each partial pressure in the vessel.

The vessel is then connected to an evacuated rigid vessel of volume 5.00L at the same temperature and allowed to equilibrate (no further reaction). Compute the new total pressure and new partial pressures in the combined

volume.

Chapter 8
Thermochemistry

Chemical reactions are not merely the reshuffling of atoms; they are profound transactions of energy that dictate the direction and feasibility of natural processes. By bridging the microscopic world of bond energies with the macroscopic observations of heat and work, thermochemistry provides the rigorous accounting system of the chemical universe. This chapter establishes the fundamental principles of the First Law of Thermodynamics, enabling precise prediction of energy flow in chemical transformations.

8.1 Energy, Work, and Heat

Thermochemistry applies the First Law of Thermodynamics to chemical and physical processes. Before we can quantify energy changes during reactions, we must define what we mean by energy, how energy is exchanged between a chemical system and its environment, and why some quantities (such as temperature and pressure) depend only on the current state of the system while others (such as heat and work) depend on the process by which the system changes.

8.1.1 The thermodynamic universe: system, surroundings, and boundary

In thermodynamics, the universe is defined as

The system is the part of the universe we choose to study. In chemistry, the system is often the reacting mixture (for example, an aqueous acid–base reaction in a beaker,

or a gaseous reaction in a cylinder). Everything outside the system is the surroundings. The system and surroundings are separated by a boundary, which may be real (walls of a vessel) or conceptual (an imaginary surface around a reaction mixture).

Common types of systems include:

Open system: exchanges both matter and energy with surroundings (e.g., a reaction in an uncovered beaker).

Closed system: exchanges energy but not matter (e.g., gas in a sealed piston–cylinder).

Isolated system: exchanges neither energy nor matter (an idealization; real systems can only approximate isolation).

Because chemical reactions occur through molecular rearrangements, it is essential to keep the “bookkeeping” straight: whenever the system gains energy, the surroundings must lose the same amount, and vice versa. Thermochemistry formalizes this exchange through two mechanisms: heat and work.

8.1.2 Energy in chemical systems: kinetic and potential contributions

Energy is the capacity to do work or to produce heat flow. At the microscopic level, the energy of a chemical system can be conceptually grouped into kinetic and potential contributions.

Kinetic energy at the molecular scale

Kinetic energy arises from motion. In a chemical sample, particles possess several kinds of motion:

Translational motion: molecules moving through space (dominant in gases).

Rotational motion: molecules rotating about axes.

Vibrational motion: atoms oscillating within molecules.

Temperature is an experimentally accessible indicator of how the system’s energy is distributed among these motions, especially translational motion in gases. However, temperature is not itself “the energy” of the system; it is a state variable that correlates with average molecular kinetic energy. Two systems at the same temperature can still have very different total energies because they may contain different amounts of substance or different forms of potential energy.

Potential energy in chemical bonding and interactions

Potential energy arises from positions and interactions. In chemistry, the most important forms include:

Chemical potential energy: energy stored in chemical bonds and electronic structure. Breaking bonds typically requires energy; forming bonds typically releases energy.

Intermolecular potential energy: attractions/repulsions between molecules (e.g., hydrogen bonding, dipole–dipole interactions, dispersion forces), crucial in phase changes and solution processes.

Electrostatic potential energy: interactions among ions in ionic solids or between charged species in solution.

Macroscopic potential energy: for example, gravitational potential energy of a raised piston (important when discussing expansion work in a piston–cylinder assembly).

A central idea of thermochemistry is that a reaction may convert energy between these forms while conserving the total energy of the universe. For example, combustion converts chemical potential energy (bonds in fuel and O2) into thermal energy in the surroundings (increased kinetic energy of surrounding particles) and sometimes into work (expansion of hot gases).

8.1.3 Heat and work: two modes of energy transfer

Thermodynamics distinguishes between energy a system has and energy a system transfers. The internal energy U (introduced formally in the next section) is a property of the system. In contrast, heat q and work w describe energy transfer across the boundary.

Heat q

Heat is energy transfer driven by a temperature difference between system and surroundings. Heat flows spontaneously from the higher-temperature region to the lower-temperature region until thermal equilibrium is reached.

Heat is not a substance contained in an object; it is a process variable describing energy in transit. Once the transfer is complete, we say the system’s internal energy has changed, but we do not say the system “contains heat.”

Work w

Work is energy transfer associated with a force acting through a distance. Many forms of work exist (electrical work in electrochemical cells, surface work in creating new surface area, etc.), but in general chemistry the most common is pressure–volume work when a gas expands or is compressed against an external pressure.

As with heat, work is not a state property stored in a system; it is an energy transfer that depends on the details of the process.

8.1.4 Sign conventions for q and w

Thermochemistry relies on a consistent sign convention that connects our calculations to physical intuition.

q > 0: heat is absorbed by the system from the surroundings (the system gains energy thermally).

q < 0: heat is released by the system to the surroundings (the system loses energy thermally).

w > 0: work is done on the system by the surroundings (the system gains energy mechanically or electrically).

w < 0: work is done by the system on the surroundings (the system expends energy to move the surroundings).

These conventions are not arbitrary: they ensure that when the system gains energy by either mechanism, the corresponding term in the energy balance is positive. The following schematic summarizes the system–surroundings picture and the sign conventions.

8.1.5 State functions versus path-dependent quantities

A major conceptual step in thermochemistry is distinguishing state functions from path-dependent (process) quantities.

State functions

A state function depends only on the current state of the system, not

on how the system arrived there. If the state is specified (for example, by giving T, P, composition, and amount), then a state function has a definite value.

Common state functions include:

Temperature T

Pressure P

Volume V

Internal energy U (defined in the next section)

Enthalpy H (introduced later in this chapter)

Heat and work are not state functions

Heat q and work w are not state functions because they describe how energy crosses the boundary during a change. Different processes connecting the same initial and final states can involve different amounts of heat and work.

An everyday example makes this intuitive: suppose you want to raise the temperature of a sample of water from 20 ∘C to 30 ∘C. You could:

warm it gently in a water bath (energy transfer primarily as heat), or

use an electric immersion heater (energy transfer as electrical work, which is then dissipated internally and appears as heating).

The initial and final temperatures are the same (a state description), but the mechanism of energy transfer differs, so the bookkeeping in terms of q and w differs.

8.1.6 Units and measurement language in thermochemistry

Because thermochemistry is quantitative, units matter:

Energy is measured in joules (J); 1 J = 1 kg m² s⁻².

In chemistry, kilojoules (kJ) are commonly used for reaction energies.

Heat and work are measured in the same units as energy because they are energy transfers.

In experimental practice, we often measure a temperature change of the surroundings (for example, a water bath in calorimetry) and infer the heat flow q associated with a process. This is possible because temperature is a state variable of the surroundings, even though heat is not a state function of the system.

8.1.7 A microscopic picture: why chemical reactions exchange heat

Why does a reaction sometimes heat its surroundings (exothermic behavior) and sometimes cool them (endothermic behavior)? At the molecular level:

When product bonding and interactions are overall lower in potential energy than those of reactants, the difference must leave the system as energy transfer—often as heat to the surroundings.

When products are overall higher in potential energy, the system must acquire energy from the surroundings to reach that higher-energy arrangement.

Thermochemistry provides the macroscopic accounting framework: instead of tracking every molecular collision and energy redistribution, we track net energy transfers (q and w) and net changes in state functions (especially U and H).

8.1.8 Preview: connecting these ideas to the First Law

The concepts in this section establish the language needed for the First Law of Thermodynamics, which will formalize energy conservation for chemical systems. The key bridge is:

The system’s energy change can occur only through heat and work crossing the boundary.

The quantities q and w depend on the process, but their combined effect produces a definite change in the system’s internal energy between the initial and final states.

In the next section, we will define internal energy U and express the First Law in its working mathematical form,

then develop the especially important case of pressure–volume work relevant to gases and many reaction systems.

8.2 The First Law of Thermodynamics

The previous section established the thermodynamic bookkeeping framework: a system exchanges energy with its surroundings through two process variables, heat q and work w, using the sign conventions q > 0 and w > 0 for energy entering the system. We now formalize these ideas into the First Law of Thermodynamics and develop the most important type of work in chemical applications: pressure–volume work.

8.2.1 Conservation of energy and the First Law

The First Law of Thermodynamics is the statement of conservation of energy for thermodynamic systems:

Energy cannot be created or destroyed; it can only be transferred and transformed.

Because the universe is defined as system plus surroundings, conservation of energy implies

or equivalently,

In thermochemistry, we track how energy crosses the system boundary. The First Law is written in terms of the system’s internal energy U:

Equation (??) is the working form used throughout this chapter.

8.2.2 Internal energy U: what it is and what it is not

Definition and physical meaning

The internal energy U of a system is the total microscopic energy contained in the system: the sum of the kinetic and potential energies of its particles (molecules, ions, atoms, electrons) associated with their motions and interactions. Contributions include:

kinetic energy from translational, rotational, and vibrational motion,

potential energy from chemical bonds and electronic structure,

potential energy from intermolecular and electrostatic interactions.

Internal energy is a state function: if the system is brought from an initial equilibrium state to a final equilibrium state, ΔU = Ufinal −Uinitial depends only on those states, not on the path taken between them.

Internal energy is not directly measurable in absolute terms

Absolute values of U are not directly measurable. Thermochemistry focuses on changes in internal energy, ΔU, because these changes are linked to measurable energy transfers (q and w) during controlled processes. This is typical of state functions: differences are accessible even when absolute values are not.

8.2.3 Deriving ΔU = q + w from energy conservation

Consider a process in which the system changes from state 1 to state 2 while exchanging energy with the surroundings. By definition:

q is the energy transferred as heat into the system,

w is the energy transferred as work into the system.

If the system gains heat (q > 0), the surroundings must lose the same amount of energy as heat. Likewise, if work is done on the system (w > 0), the surroundings lose energy by doing that work. Therefore, from conservation of energy,

which is Equation (??).

A key conceptual point is that q and w are path dependent, but their sum for a given initial and final state is constrained by ΔU, which is path independent. Different paths can trade off heat and work while still producing the same ΔU.

8.2.4 Implications of the First Law for chemical processes

Energy changes in reactions

Chemical reactions rearrange electrons and nuclei, changing the system’s potential energy (bond energies and intermolecular interactions). The resulting ΔU must be balanced by energy transfer:

If products have lower internal energy than reactants, then ΔU < 0 and energy must leave the system as heat, work, or a combination.

If products have higher internal energy than reactants, then ΔU > 0 and energy must enter the system.

Why we care about separating q and w

In laboratory chemistry, energy transfer often appears as a temperature change (heat transfer) and/or a volume change against atmospheric pressure (expansion/compression work). By treating heat and work as distinct channels, we can connect:

measured temperature changes (calorimetry) to q,

measured pressure–volume changes to w,

and then infer ΔU using the First Law.

8.2.5 Pressure–volume work (PV work)

Mechanical origin of PV work

A gas expanding in a piston–cylinder assembly pushes against an external pressure Pext. If the piston moves outward by a distance Δx while the gas exerts a force F, the work done by the gas on the surroundings is

Pressure is force per area, Pext = F∕A, and the volume change is ΔV = AΔx. Substituting gives

By the thermodynamic sign convention used in chemistry (positive work means work on the system), the work on the system is the negative of the work by the system:

Interpreting the sign in w = −PextΔV

Equation (??) encodes two physically distinct cases:

Expansion: ΔV > 0. Then w < 0, meaning the system does work on the surroundings and loses energy through the work channel.

Compression: ΔV < 0. Then w > 0, meaning the surroundings do work on the system and the system gains energy through the work channel.

These conclusions match intuition: expanding gas can lift a piston or displace the atmosphere (energy leaves the system as work), whereas compressing a gas requires energy input (energy enters as work).

Constant external pressure

Equation (??) is exact when the external pressure opposing expansion/compression is constant during the process. This is a common idealization for introductory thermochemistry and is appropriate for many processes against constant atmospheric pressure or constant applied pressure (e.g., a piston with a fixed mass on it).

When Pext is not constant, the general expression for PV work is written using an integral,

but for now we emphasize the constant-Pext form because it leads directly to practical calculations in chemistry.

8.2.6 Combining PV work with the First Law

Substituting Equation (??) into the First Law gives a particularly useful relation for processes where PV work dominates:

Equation (??) makes an important point explicit: even if a process releases heat (q < 0), the internal energy change depends also on whether the system expands or contracts. For example:

A reaction that produces gas may expand against the atmosphere (ΔV > 0), making −PextΔV negative and more energy leave the system through work.

A reaction that consumes gas may reduce volume (ΔV < 0), making −PextΔV positive and partially offsetting heat loss by gaining energy as work done on the system.

8.2.7 Units and typical magnitudes for PV work

In SI units, pressure is in pascals (Pa) and volume is in cubic meters (m³). Then work is in joules (J) because

In chemistry, pressures and volumes are often given in bar or atm and liters. Useful conversions include:

Thus,

This provides a quick estimate: PV work in typical laboratory reactions (volume changes of a few liters against ∼ bar) is often on the order of 1 × 10² to 1 × 10³ J, whereas reaction heats are frequently reported in kJmol⁻¹. The comparison foreshadows why constant-pressure heat flow (enthalpy) becomes especially convenient for chemistry: it incorporates PV work in a way that aligns with common experimental conditions.

8.2.8 Common pitfalls and conceptual checkpoints

Do not confuse internal energy with heat

Heat q is a transfer; internal energy U is a state property. After a process, the system has a new value of U, but it does not “contain heat.”

Sign conventions must match the First Law

Equation (??) assumes the chemistry convention w > 0 for work done on the system. If a different sign convention is used (common in some physics texts), the equation changes form. In this text, we will consistently use ΔU = q + w.

Path dependence of q and w

Even though ΔU is fixed by initial and final states, q and w can vary with the process conditions. This is why experimental design matters: a reaction carried out in a rigid container (constant volume, w ≈ 0 for PV work) can have a different heat flow than the same reaction carried out in an open beaker (approximately constant pressure, nonzero PV work).

8.2.9 Preview: why enthalpy is introduced

Most reactions in the laboratory are performed at approximately constant external pressure (open to the atmosphere). Under these conditions, the heat measured by calorimetry is commonly qp (heat at constant pressure), not directly ΔU. The next section introduces a new state function, enthalpy H, designed so that at constant pressure the heat flow is equal to a state-function change:

This will allow us to tabulate reaction energy changes and use them predictively across many chemical processes.

8.3 Enthalpy and Enthalpy of Reaction

The First Law of Thermodynamics established the fundamental energy balance for a chemical system:

and for processes where pressure–volume work dominates,

Many reactions in general chemistry are carried out in open containers (beakers,

flasks) exposed to the atmosphere, so the external pressure is approximately constant at Pext ≈ Patm. Under these constant-pressure conditions, we would like a state function whose change directly equals the heat flow measured in the laboratory. That state function is enthalpy.

8.3.1 Why introduce enthalpy? Constant pressure is the common laboratory condition

In an open vessel, a reaction may release or consume heat and may also expand or contract, doing PV work against the atmosphere. If we measure the heat exchange with the surroundings at constant pressure, we are measuring qp (the heat at constant pressure). Equation (??) shows that q is not generally equal to ΔU when ΔV ≠0.

This motivates the following idea: incorporate the PV term into a new state function so that, at constant pressure,

Enthalpy provides exactly this simplification.

8.3.2 Definition of enthalpy H

Enthalpy H is defined as

where U is internal energy, P is the system pressure, and V is the system volume.

Several important points follow immediately:

H is a state function because it is defined in terms of state functions (U, P, and V ).

H has units of energy (J), because PV has units of energy.

We typically care about changes in enthalpy, ΔH, rather than absolute values.

8.3.3 Derivation: ΔH = qp for processes at constant pressure

Starting from the definition in Equation (??), take the change between an initial state and a final state:

For many introductory applications, we consider processes occurring at constant pressure, so P is constant and

Substituting into Equation (??) gives

Now use the First Law written for PV work against a constant external pressure. Under common laboratory conditions, the system pressure during a mechanically balanced expansion/compression is effectively the external pressure, so PV work is

Insert this into ΔU = q + w:

Substitute into Equation (??):

Therefore, for a process carried out at constant pressure where PV work is the only significant work,

Interpretation. Equation (??) is a major practical result: at constant pressure, the heat absorbed or released by the system is equal to the enthalpy change. Because ΔH is a state function, this means constant-pressure heats of reaction can be treated as properties of the reaction itself (for specified physical states), not of the particular experimental path.

8.3.4 Physical meaning of the PV term

The term PV is sometimes described as “flow energy” in contexts like gases moving in and out of containers, but in general chemistry its most useful interpretation is bookkeeping for expansion/compression at constant pressure.

If a reaction produces more moles of gas than it consumes (at fixed T and P), the system tends to expand and do PV work on the surroundings. Enthalpy packages this PV work together with internal energy so that the measured constant-pressure heat qp directly corresponds to ΔH.

A helpful way to think about this is:

At constant pressure, ΔH counts the energy change of the system including the energy needed to make room for the system in the surroundings.

8.3.5 Enthalpy changes and chemical reactions

Enthalpy of reaction

For a chemical reaction carried out at constant pressure, the enthalpy change of reaction is defined as

where the enthalpies refer to the substances in their specified physical states (e.g., H2O(l) versus H2O(g)).

When the reaction occurs at constant pressure with PV work as the only work,

Thus, ΔHrxn is the heat exchanged with the surroundings (at constant pressure) for the reaction as written.

Extensive nature and stoichiometric scaling

Enthalpy is an extensive property: it scales with the amount of material. Consequently, ΔHrxn scales with the stoichiometric amount of reaction:

If you multiply a balanced chemical equation by a factor n, then ΔHrxn is multiplied by n.

If you reverse a reaction, the sign of ΔHrxn changes.

These algebraic properties will be crucial later when applying Hess’s law.

8.3.6 Endothermic and exothermic processes

Because ΔH = qp at constant pressure, the sign of ΔH provides a direct classification:

Exothermic process: ΔH < 0. The system releases heat to the surroundings.

Endothermic process: ΔH > 0. The system absorbs heat from the surroundings.

Common observational cues (qualitative). In an exothermic reaction run in an open container, the surroundings often warm up (the container feels hot). In an endothermic process, the surroundings may cool (the container feels cold). These observations reflect the direction of heat flow, but calorimetry is required for quantitative measurement.

8.3.7 Energy-level diagrams for ΔH

The sign and magnitude of ΔH are often communicated using an energy-level

diagram (also called an enthalpy diagram). The vertical axis represents enthalpy, and the horizontal axis represents reaction progress (a qualitative coordinate indicating the transformation from reactants to products). In such diagrams:

Products lower than reactants correspond to ΔH < 0 (exothermic).

Products higher than reactants correspond to ΔH > 0 (endothermic).

8.3.8 Thermochemical equations and sign conventions

A thermochemical equation is a balanced chemical equation that includes an enthalpy change. For example, the combustion of methane under specified conditions might be written as

The negative sign indicates that heat is released at constant pressure.

Two equivalent ways to present thermochemical information are common:

Attach ΔH to the equation. This emphasizes that ΔH corresponds to the reaction as written.

Include heat as a reactant or product. For an exothermic reaction, heat can be written on the product side; for an endothermic reaction, heat on the reactant side. This convention is useful when reasoning qualitatively about temperature changes and Le Châtelier-type responses, but in thermochemistry we typically prefer explicit ΔH notation.

State matters. The enthalpy change depends on physical state. For example, forming H2O(l) versus H2O(g) differs by the enthalpy of vaporization of water. Therefore, thermochemical equations must specify states, such as (s), (l), (g), or (aq).

8.3.9 Connection between ΔH and ΔU for reactions involving gases

Although ΔH is often the most convenient quantity at constant pressure, it is useful to see how it relates to ΔU. Begin with

For reactions carried out at constant pressure,

If the system behaves approximately as an ideal gas during the volume change, then PV = nRT, so at constant T,

where Δngas is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants) for the reaction as written.

Thus, under common approximations (ideal gas behavior; temperature not changing significantly during the comparison),

Why this matters. Equation (??) explains why ΔH and ΔU are often similar for reactions that do not change the total moles of gas, and why they differ more for reactions that produce or consume gas. This distinction becomes especially important when comparing constant-pressure calorimetry (measuring ΔH) with constant-volume calorimetry (measuring ΔU).

8.3.10 Summary of core ideas

Enthalpy is defined by H = U + PV and is a state function.

For constant pressure with PV work as the only work, ΔH = qp.

Exothermic processes have ΔH < 0; endothermic processes have ΔH > 0.

Thermochemical equations must specify physical states; ΔH scales with reaction stoichiometry and changes sign when a reaction is reversed.

For reactions involving gases, ΔH and ΔU are related approximately by ΔH ≈ ΔU + RT Δngas under ideal-gas/constant-T assumptions.

In the next section, we develop the experimental methods used to measure heat flow (and thus ΔH or ΔU) using calorimetry.

8.4 Calorimetry

Thermochemistry connects measurable quantities (temperature, mass, pressure, volume) to the energy transfers q and w introduced earlier. In practice, heat flow is rarely measured directly. Instead, we observe a temperature change in a material of known heat capacity and infer the heat exchanged. This approach is called calorimetry.

A calorimetric measurement rests on two ideas already developed:

Energy conservation (First Law): the energy change of the universe is zero.

Heat flow at specified conditions: at constant pressure, qp = ΔH; at constant volume (with only PV work possible), qv = ΔU.

Calorimeters are designed to approximate either constant-pressure or constant-volume conditions and to minimize unwanted heat exchange with the external environment.

8.4.1 The calorimetric energy balance: system and surroundings

In calorimetry we explicitly define:

System: the chemical or physical process of interest (reaction, dissolution, phase change).

Surroundings: the calorimeter hardware and any water (or other medium) whose temperature change is measured.

If the calorimeter is well insulated, the calorimeter plus its immediate contents are approximately isolated from the outside world, so energy conservation for system and calorimeter gives

or

The measured temperature change is used to determine qsurr, then Equation (??) gives the heat associated with the system.

8.4.2 Heat capacity and the temperature–heat relationship

Heat capacity

The heat capacity C of an object is the amount of heat required to raise its temperature by 1 K (or 1 ∘C):

Heat capacity depends on the amount of material (it is extensive).

Specific heat capacity and molar heat capacity

To compare materials independent of sample size, we define:

Specific heat capacity cs (often written c): heat required to raise the temperature of 1 g of a substance by 1 K:

Molar heat capacity Cm: heat required to raise the temperature of 1 mol by 1 K:

Sign convention reminder. In Equations (??) and (??), ΔT = Tfinal − Tinitial carries the sign of the temperature change. Thus if the surroundings warm (ΔT > 0), then qsurr > 0 and the system released heat (qsys < 0). If the surroundings cool (ΔT < 0), then qsurr < 0 and the system absorbed heat (qsys > 0).

Additivity of heat capacities

When multiple components experience the same temperature change (for example, water plus the calorimeter body), their heat capacities add:

This additivity underlies most calorimetry calculations.

8.4.3 Constant-pressure calorimetry (coffee-cup calorimetry)

Concept and typical applications

Coffee-cup calorimetry is performed in an insulated container (often nested polystyrene cups with a lid) open to the atmosphere. Because the pressure remains approximately constant at 1 atm, the heat measured corresponds to qp, and therefore to enthalpy change:

Coffee-cup calorimetry is widely used for:

reactions in aqueous solution (neutralization, precipitation, redox),

dissolution processes (e.g., salts dissolving in water),

mixing processes where temperature change is measurable.

Energy balance for a solution reaction

In a typical setup, the surroundings include the solution (water plus dissolved species) and the calorimeter apparatus. If the solution mass is msoln with effective specific heat capacity csoln (often approximated as that of water, 4.184 Jg⁻¹ K⁻¹, when solutions are dilute), then

Combining with qsys = −qsurr,

At constant pressure,

From measured heat to molar enthalpy of reaction

Calorimetry yields the heat for the amount reacted. To report a molar enthalpy change, divide by the moles of limiting reactant consumed (or by the “moles of reaction” as written):

Because ΔH scales with the stoichiometric equation, it is essential to state the balanced reaction and the physical states (especially for aqueous species and water).

8.4.4 Constant-volume calorimetry (bomb calorimetry)

Concept and what is measured

In bomb calorimetry, the reaction occurs inside a rigid, sealed steel vessel (the “bomb”) immersed in a known mass of water. The bomb is designed to withstand high pressures generated by combustion reactions.

Because the reaction vessel volume is constant, PV work is effectively zero:

If we neglect other forms of work, the First Law gives

Thus, a bomb calorimeter directly measures the internal energy change (for the reaction as carried out at constant volume), not the enthalpy change.

Bomb calorimetry is especially important for:

heats of combustion of fuels,

energy content of foods (dietary “Calories” are based on combustion heat),

reactions that are too vigorous or produce gases in ways that make open-vessel calorimetry impractical.

How the temperature signal is related to heat

In a bomb calorimeter, the surroundings include:

the water bath,

the steel bomb and internal hardware (ignition leads, sample cup),

often the thermometer and stirrer (depending on calibration convention).

All of these experience (approximately) the same temperature change ΔT due to stirring and thermal equilibration. Therefore,

where Ccal is the total heat capacity of the calorimeter assembly (water + hardware), typically determined by calibration.

Then the heat of reaction at constant volume is

Calibration of a bomb calorimeter

Because Ccal includes many components and is difficult to compute from first principles, it is usually determined experimentally by combusting a standard substance with a known energy of combustion (commonly benzoic acid) under the same conditions. If the standard releases a known heat qstd for the amount burned and produces a measured temperature rise ΔTstd, then

(The negative sign appears because qstd < 0 for an exothermic combustion while ΔTstd > 0.)

Once Ccal is known, Equation (??) converts any measured ΔT into qv for an unknown sample.

From ΔU to ΔH for combustion reactions

Bomb calorimetry yields qv = ΔU (Equation (??)). However, many tables and chemical engineering applications report ΔH (enthalpy change) at constant pressure. For reactions involving gases, a useful connection is

where Δngas is the change in moles of gaseous species (products minus reactants) for the balanced reaction as written and T is the absolute temperature at which the comparison is made.

Interpretation. If a combustion reaction produces fewer moles of gas than it consumes (often true when O2(g) is converted to CO2(g) and liquid water), then Δngas may be negative, making ΔH slightly more negative than ΔU. The correction is frequently modest compared with the magnitude of combustion energies, but it can be important for high-accuracy work.

8.4.5 Bomb calorimeter schematic (constant-volume combustion)

The essential features of a bomb calorimeter are shown in the following cross-sectional schematic: a sealed steel reaction vessel immersed in a water bath with stirring to ensure a uniform temperature, ignition wires to initiate combustion, and insulation to approximate an adiabatic boundary with respect to the external environment.

8.4.6 Practical considerations and sources of error

Real calorimeters are not perfectly isolated. Accurate measurements require accounting for, or minimizing, several effects:

Heat loss/gain to the external environment: Even well-insulated calorimeters exchange some heat with the lab air. Modern bomb calorimeters often use jacketed designs and computational corrections to approximate adiabatic behavior.

Incomplete reaction: If combustion is incomplete (soot formation, CO formation), the measured heat is smaller in magnitude than the true complete-combustion value.

Heat capacity assumptions: In coffee-cup calorimetry, approximating csoln as water’s c can introduce error for concentrated solutions. The calorimeter constant Ccal can also be significant and must be included or calibrated.

Temperature measurement and mixing: Poor stirring produces temperature gradients so the thermometer does not reflect the true average temperature. Proper mixing is essential for reliable ΔT.

8.4.7 Summary: what calorimetry provides

Calorimetry converts a temperature change into a heat flow:

using heat capacities to compute qsurr.

Coffee-cup calorimetry (constant P): measures qp = ΔH for the process.

Bomb calorimetry (constant V ): measures qv = ΔU for the process, which can be related to ΔH when needed.

These measurements provide the experimental foundation for thermochemical quantities used later in the chapter, including Hess’s law calculations and standard enthalpies of formation.

8.5 Hess’s Law

Calorimetry provides a direct experimental route to reaction enthalpies ΔHrxn for processes we can conveniently run in a calorimeter. However, many reactions of interest are difficult to measure directly: they may be too slow, too dangerous, or not proceed cleanly in a single step. The key to extending thermochemistry beyond direct measurement is the fact that enthalpy is a state function. This property leads to Hess’s law, which allows us to compute ΔHrxn for a target reaction by algebraically combining other reactions whose enthalpy changes are known.

8.5.1 Enthalpy as a state function: the core idea

From earlier sections:

Enthalpy H is defined by H = U + PV .

For constant-pressure processes (typical of many reactions), ΔH = qp.

Most importantly here, H is a state function: the change ΔH depends only on the initial and final states, not on the path taken between them.

This means that if we can get from the same reactants to the same products by different sequences of chemical steps, the total enthalpy change must be the same for all sequences.

8.5.2 Statement of Hess’s law

Hess’s Law of Constant Heat Summation states:

If a reaction can be expressed as the sum of two or more reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the component reactions.

Symbolically, if Reaction (target) is obtained by adding reactions (1), (2), …, then

provided that the equations are combined to yield the target stoichiometry and physical states.

8.5.3 Why Hess’s law works (conceptual derivation)

Consider an initial state consisting of reactants R and a final state consisting of products P. Because enthalpy is a state function,

is uniquely determined.

Now imagine two different paths:

Path A: the reaction proceeds directly R → P with enthalpy change ΔHA.

Path B: the reaction proceeds through intermediates, e.g., R → I₁ → I₂ →⋯ → P, with enthalpy changes ΔHB1,ΔHB2,….

Along Path B, additivity of changes in a state function gives

Since both paths connect the same initial and final states, ΔHA = ΔHB. This equality is Hess’s law.

8.5.4 Thermochemical equations: algebra rules for ΔH

Using Hess’s law is primarily an exercise in manipulating balanced chemical equations while keeping track of how ΔH changes. The rules follow from the extensive nature of enthalpy.

Rule 1: Reversing a reaction changes the sign of ΔH

If

then

Rule 2: Multiplying a reaction multiplies ΔH

If

then

This includes fractional multipliers, which are sometimes used to match stoichiometric coefficients. Fractional coefficients are mathematically allowed, but the final reported target reaction is usually written with integer coefficients.

Rule 3: Adding reactions adds ΔH

If

and

then adding gives

Species that appear on both sides of the summed equation cancel just like algebraic terms.

Rule 4: Physical states must match

Because enthalpy depends on physical state, the following are not interchangeable:

If your target reaction requires H2O(l) but a source equation contains H2O(g), you must include an additional step (e.g., condensation) with its own enthalpy change to correct the state.

8.5.5 A structured method for Hess’s law problems

Complex Hess’s law problems become manageable if you follow a consistent workflow.

Step 1: Write the target equation clearly

Include:

correct balanced stoichiometry,

physical states (s, l, g, aq),

the unknown ΔHtarget.

Step 2: List the given equations with their ΔH

Keep each reaction and its enthalpy paired. Using a table can help prevent sign and scaling mistakes.

Step 3: Decide how to manipulate each given equation

For each provided reaction:

reverse it if needed (flip sign of ΔH),

multiply by a factor to match coefficients (scale ΔH),

aim for cancellation of intermediates upon addition.

Step 4: Add manipulated equations and cancel species

Add the left-hand sides together and the right-hand sides together, then cancel any species appearing on both sides.

Step 5: Add enthalpy changes to get the result

Sum the correspondingly manipulated ΔH values.

Step 6: Sanity-check the result

Useful checks include:

Does the final equation exactly match the target (including states)?

Are all coefficients correct and balanced?

Does the sign of ΔH make qualitative sense (e.g., many combustions should be strongly exothermic)?

Are the magnitudes reasonable for the scale of reaction as written (per mole of reaction)?

8.5.6 Worked example 1 (conceptual): cancellation of intermediates

Suppose you want the enthalpy change for:

and you are given:

C(graphite) + O2(g) → CO2(g) | ΔH₁ |  |
CO(g) + O2(g) → CO2(g) | ΔH₂ |  |

Reverse the second equation to make CO(g) a product:

Now add to the first equation:

Cancel O2(g) on both sides to obtain the target:

Then,

This example highlights the main mechanism of Hess’s law: construct the target by cancellation of intermediates and track enthalpy changes with the same algebra.

8.5.7 Worked example 2 (with states): formation of liquid water from gaseous steps

Target:

If the available data (from a source table or prior measurements) include the reaction producing water vapor,

then to obtain liquid water you must also include condensation:

Adding gives the target, and

This illustrates that Hess’s law is not just about chemical rearrangement; it also accounts for physical processes (phase changes, dissolution) so long as each step is written as a thermochemical equation with a known enthalpy change.

8.5.8 Common pitfalls and how to avoid them

Forgetting that ΔH follows the equation as written

A reported ΔH value is tied to the specific balanced equation and the specified physical states. Changing coefficients without scaling ΔH, or changing states without adding a correction step, is a frequent source of errors.

Cancelling species incorrectly

Only species that appear identically on both sides (same formula and same physical state) may be cancelled. For example, you cannot cancel H2O(l) with H2O (g).

Mixing “per mole” interpretations

A thermochemical equation has an implicit scale: ΔH corresponds to the reaction as written. If a table lists “ΔH per mole of CO2 formed,” that is a different convention. Always rewrite the reaction so that the meaning is unambiguous.

8.5.9 Summary: why Hess’s law is central in thermochemistry

Hess’s law is a practical consequence of enthalpy being a state function:

It allows indirect calculation of ΔHrxn for reactions that cannot be measured directly.

It turns thermochemistry into a form of algebra: manipulate and add thermochemical equations while applying consistent rules to ΔH.

It lays the groundwork for using tabulated standard enthalpies of formation to compute ΔHrxn for essentially any reaction, which is the focus of the next section.

8.6 Standard Enthalpies of Formation

Hess’s law showed how reaction enthalpies can be obtained by adding and scaling thermochemical equations, relying on the fact that ΔH is a state-function change. The remaining practical challenge is data availability: to use Hess’s law broadly, we need a small set of tabulated enthalpy values that can be combined to obtain ΔHrxn for essentially any reaction. Standard enthalpies of formation provide that standardized data set.

8.6.1 Standard states and the meaning of the superscript ∘

Thermochemical quantities depend on conditions (pressure, temperature) and on physical state (s, l, g, aq). To make tabulated data comparable and widely usable, thermodynamics defines a standard state for each substance. In general chemistry, the most common convention is:

Standard pressure: P∘ = 1bar.

Standard temperature for tabulations: usually T = 298.15K (25 ∘C), unless otherwise noted.

Standard state of a pure substance: the most stable form of that substance at 1 bar and the specified temperature.

Standard state of a gas: the gas behaving ideally at 1 bar.

Standard state of a solute in solution: typically 1 molL⁻¹ (1 M) for aqueous species, with the solution behaving ideally (an approximation in many real solutions).

A superscript ∘ on a thermodynamic quantity (such as ΔH∘) indicates the process occurs with reactants and products in their standard states. Thus, ΔHrxn∘ means the enthalpy change for the reaction as written when all reactants and products are in their standard states.

8.6.2 Definition of the standard enthalpy of formation ΔHf∘

The standard enthalpy of formation of a substance, ΔHf∘, is defined as:

The enthalpy change for the formation of one mole of a compound from its constituent elements, with all reactants and products in their standard states.

This definition has several consequences that are easy to miss but essential for correct calculations:

“Formation” means the product is one mole of the compound, not necessarily an integer-coefficient equation.

The reactants are the elements in their reference (standard) forms, not arbitrary elemental forms.

The physical states must be specified; ΔHf∘ for H2O(l) is different from that for H2O(g).

8.6.3 Reference values: elements in their standard states have ΔHf∘ = 0

By convention,

for any element in its standard state at the specified temperature.

Examples at 298.15 K and 1 bar include:

O2(g), N2(g), H2(g), Cl2(g) have ΔHf∘ = 0.

C(graphite) has ΔHf∘ = 0 (graphite is the standard state of carbon at 298.15 K and 1 bar).

Br2(l) has ΔHf∘ = 0 (liquid bromine is the standard state at 298.15 K).

Hg(l) has ΔHf∘ = 0 (liquid mercury is the standard state at 298.15 K).

Common pitfall: nonstandard elemental forms. O(g) is not the standard state of oxygen at 298.15 K and 1 bar; O2(g) is. Likewise, C(diamond) is not the standard state of carbon at 298.15 K; C(graphite) is. Those nonstandard forms have nonzero enthalpies of formation because forming them from the standard state involves an enthalpy change.

8.6.4 Using ΔHf∘ with Hess’s law to calculate ΔHrxn∘

Key idea: any reaction can be built from formation reactions

Because formation reactions connect compounds to elements, and because Hess’s law allows us to add reactions and add enthalpy changes, we can construct a target

reaction by:

forming the products from the elements (using ΔHf∘ values),

forming the reactants from the elements (using ΔHf∘ values),

subtracting the reactant-formation contributions from the product-formation contributions.

This leads to the central working equation:

ΔHrxn∘ | = ∑ iνi ΔHf,i∘(products) −∑ jνj ΔHf,j∘(reactants), |  |

where ν are stoichiometric coefficients from the balanced reaction (positive for products and reactants in their respective sums).

Why the coefficients matter. ΔHf∘ is defined per one mole of compound formed. If a balanced equation produces 2 moles of a compound, its ΔHf∘ contribution is multiplied by 2.

Units and interpretation

Tabulated ΔHf∘ values are commonly given in kJmol⁻¹. When using Equation (??), the result ΔHrxn∘ corresponds to the reaction as written (i.e., per “mole of reaction” with the given stoichiometric coefficients).

8.6.5 Worked example (symbolic): combustion of a hydrocarbon

Consider the combustion of methane, written with physical states:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) |  |

Using Equation (??),

ΔHrxn∘ | = ΔHf∘(CO 2(g)) + 2ΔHf∘(H 2O(l))−ΔHf∘(CH 4(g)) + 2ΔHf∘(O 2(g)). |  |

Because O2(g) is an element in its standard state, ΔHf∘(O2(g)) = 0, so

This illustrates the practical power of ΔHf∘: once you have tabulated values for common compounds, you can compute standard reaction enthalpies without measuring each reaction calorimetrically.

8.6.6 Writing and recognizing formation reactions

Formation reactions are defined by the product being one mole of the compound and reactants being elements in their standard states.

Examples (all at standard conditions):

C(graphite) + O2(g) → CO2(g) | ΔHf∘(CO 2(g)), |  |
H2(g) + O2(g) → H2O(l) | ΔHf∘(H 2O(l)), |  |
Na(s) + Cl2(g) → NaCl(s) | ΔHf∘(NaCl(s)), |  |
N2(g) + 3H2(g) →2 NH3(g) | 2ΔHf∘(NH 3(g)). |  |

The last example emphasizes a subtlety: the balanced equation shown forms 2 moles of NH3(g). A strict formation reaction for NH3(g) would be

Both are acceptable as long as you interpret the enthalpy correctly (scaling by stoichiometry).

8.6.7 Tabulated data and careful bookkeeping

Physical states are part of the identity

When using a table of ΔHf∘, treat each state as a distinct entry:

Similarly, aqueous ions have their own standard formation values, typically defined relative to conventions for aqueous solutions at 1 molL⁻¹.

A compact bookkeeping format

A table format can reduce errors when applying Equation (??). For a generic reaction,

you can tabulate coefficients and ΔHf∘ values:

Species (state) | Role | ν | ν ΔHf∘
Product 1 | product | ν₁ | ν₁ ΔHf∘
Product 2 | product | ν₂ | ν₂ ΔHf∘
Reactant 1 | reactant | ν₁ | ν₁ ΔHf∘
Reactant 2 | reactant | ν₂ | ν₂ ΔHf∘

Then compute:

This mirrors Hess’s-law logic and helps ensure coefficients and signs are handled consistently.

8.6.8 Bond enthalpies as an estimation method (when formation data are unavailable)

Sometimes ΔHf∘ values are not available for a particular compound, especially for unstable intermediates or exotic species. In such cases, a rough estimate of ΔHrxn can be obtained from average bond enthalpies.

Conceptual basis

Bond enthalpy is the enthalpy change associated with breaking a particular bond in the gas phase (typically defined for homolytic cleavage). Using average bond enthalpies, we estimate:

ΔHrxn ≈∑ (bond enthalpies of bonds broken) −∑ (bond enthalpies of bonds formed). |  |

This method is approximate because:

bond enthalpies are averages over many compounds,

the method usually assumes gas-phase species,

it neglects solvent effects and detailed molecular environment.

Nevertheless, it provides useful order-of-magnitude insight and a quick check on whether a reaction is likely to be strongly exothermic or endothermic.

8.6.9 Limits and interpretation of standard reaction enthalpies

“Standard” does not mean “always”

ΔHrxn∘ applies to the reaction with all species in their standard states. Real laboratory conditions may differ in temperature, pressure, and concentration, so the actual enthalpy change can differ from the standard value. However, ΔHrxn∘ is a well-defined reference point and is the basis for many thermodynamic calculations.

Temperature dependence

Enthalpy changes vary with temperature because heat capacities vary with temperature. Introductory thermochemistry often treats ΔH as approximately

constant near 298 K; more advanced treatments incorporate temperature corrections using heat capacity data.

8.6.10 Summary of core results

Standard states (often 1 bar and 298.15 K) allow consistent tabulation of thermochemical data.

The standard enthalpy of formation, ΔHf∘, is the enthalpy change for forming 1 mol of a compound from its elements in their standard states.

Elements in their standard states have ΔHf∘ = 0 by convention.

Standard reaction enthalpies can be computed from formation enthalpies using:

When formation data are unavailable, bond enthalpies can provide approximate estimates via “broken minus formed” (Equation (??)).

With these tools, thermochemistry becomes broadly predictive: once a database of ΔHf∘ values is available, enthalpy changes for a wide range of reactions can be computed systematically and consistently.

8.7 Problems

8.7.1 PV work during expansion against constant external pressure

A sample of N2(g) in a frictionless piston–cylinder expands irreversibly against a constant external pressure of 0.950 bar from V ₁ = 2.50L to V ₂ = 9.80L at T = 298K. Assume the only work is PV work.

Calculate the work w (in J) for the gas (use the chemistry sign convention).

If, during the expansion, the gas absorbs q = +1.20kJ of heat, calculate ΔU for the gas.

Explain in one or two sentences why the sign of w is consistent with the physical direction of energy transfer.

8.7.2 Specific heat determination by metal–water calorimetry

A 125.0 g sample of an unknown metal is heated to 98.0 ∘C and then quickly transferred into 250.0 g of water initially at 22.5 ∘C in a coffee-cup calorimeter. The final temperature is 27.8 ∘C. The calorimeter has a heat capacity of Ccal = 45.0JK⁻¹. Assume no heat loss to the external environment and take cwater = 4.184Jg⁻¹ K⁻¹.

Determine the specific heat capacity cs of the metal in Jg⁻¹ K⁻¹.

State clearly which quantities you treat as the “system” and which as the “surroundings” for the energy balance.

8.7.3 Coffee-cup calorimetry: enthalpy of neutralization

In a coffee-cup calorimeter at 1 bar, 50.00 mL of 1.000 moldm⁻³ HCl(aq) at 24.20 ∘C is mixed with 50.00 mL of 1.000 moldm⁻³ NaOH(aq) at 24.20 ∘C. The maximum temperature observed is 30.58 ∘C. Assume the total solution mass is 100.00 g, csoln = 4.184Jg⁻¹ K⁻¹, and the calorimeter heat capacity is negligible.

Calculate qsurr and qrxn.

Determine ΔHrxn (in kJmol⁻¹) for the reaction

using the limiting reagent and the stoichiometry of the neutralization.

Briefly explain why this experiment yields ΔH rather than ΔU.

8.7.4 Hess’s law: synthesis enthalpy from multiple steps (I)

Use the following thermochemical equations to determine ΔH∘ for the target reaction:

Given:

N2(g) + 2O2(g) →2 NO2(g) | ΔH₁∘ = +66.4 kJ |  |
2NO(g) + O2(g) →2 NO2(g) | ΔH₂∘ = −114.1 kJ |  |

Show, by explicitly manipulating the equations, how to obtain the target reaction.

Compute ΔH∘ for the target reaction.

8.7.5 Hess’s law: reaction enthalpy with phase correction (II)

Compute ΔH∘ for

from the following data:

CO(g) + 2H2(g) → CH3OH(g) | ΔH₁∘ = −90.5kJ |  |
CH3OH(l) → CH3OH(g) | ΔH₂∘ = 37.4kJ |  |

Identify which equation must be reversed and why.

Determine ΔH∘ for forming liquid methanol as written.

Explain the physical meaning of the phase-change correction in one or two sentences.

8.7.6 Standard enthalpies of formation: compute ΔHrxn∘ from tabulated data (I)

Using standard enthalpies of formation at 298.15 K (values provided below), calculate ΔHrxn∘ for:

Data:

Show the full products-minus-reactants calculation with stoichiometric coefficients.

Report ΔHrxn∘ in kJ for the reaction as written.

8.7.7 Standard enthalpies of formation: combustion enthalpy and mass-based energy (II)

Given the following standard enthalpies of formation at 298.15 K:

calculate:

ΔHrxn∘ for the complete combustion of ethanol:

The heat released (kJ) when 35.0 g of ethanol is burned under standard conditions (assume complete combustion and that the reaction occurs at constant pressure).

The heat released per gram (kJ/g) for ethanol based on your result.

8.7.8 Bond enthalpy estimate of reaction enthalpy

Estimate ΔHrxn (in kJ/mol of reaction) for:

using average bond enthalpies:

Clearly identify which bonds are broken and which are formed.

Use the “bonds broken minus bonds formed” approach to compute ΔHrxn.

Briefly state two reasons why this estimate may differ from a tabulated ΔHrxn∘ based on formation enthalpies.

8.7.9 Integrative problem: relating ΔU and ΔH for a combustion reaction

A hydrocarbon combusts at 298.15 K according to:

A bomb calorimeter experiment (constant volume) gives qv = −2219.0kJ per mole of C3H8 combusted (assume this equals ΔU for the reaction as written, per mole of propane).

Determine Δngas for the reaction.

Using the approximation

calculate ΔH at 298.15 K (use R = 8.314Jmol⁻¹ K⁻¹ and keep careful track of units).

Explain, in a few sentences, why ΔH and ΔU differ for this reaction, referring explicitly to PV work and the change in gaseous moles.

Chapter 9
Intermolecular Forces, Liquids, and Solids

From the intricate lattice of a snowflake to the high surface tension of water, the physical state of matter is dictated not merely by the bonds within molecules, but by the subtle yet powerful forces acting between them. This chapter bridges the gap between microscopic electrostatic attractions and macroscopic material properties, exploring how intermolecular forces govern phase stability, fluid dynamics, and the ordered architecture of solids.

9.1 Intermolecular Forces

In earlier chapters, we treated chemical bonds (ionic, covalent, and metallic) as the forces that hold atoms together within a substance. Those intramolecular forces largely determine chemical identity (which atoms are connected and how). In this chapter we shift to intermolecular forces (IMFs): the noncovalent electrostatic interactions between atoms, ions, and molecules. Although IMFs are much weaker than chemical bonds, they dominate many physical properties: melting point, boiling point, vapor pressure, viscosity, and solubility.

9.1.1 What intermolecular forces are (and why they matter)

Intermolecular forces arise from Coulombic attraction between positive and negative charges. The “charges” can be:

permanent (formal ionic charge, or a molecular dipole with partial charges δ⁺ and δ⁻), or

instantaneous/induced (temporary charge separation created by electron motion).

A useful organizing idea is that the potential energy of a collection of particles decreases (becomes more negative) when attractive interactions are formed. When a substance melts or boils, energy must be supplied to partially overcome these

attractions. Thus, stronger IMFs generally imply:

9.1.2 A hierarchy of common intermolecular forces

Four types of interactions are especially important in general chemistry:

London dispersion forces (LDF): attractions between instantaneous and induced dipoles; present in all substances.

Dipole–dipole interactions: attractions between permanent molecular dipoles.

Hydrogen bonding: a particularly strong and directional subtype of dipole–dipole interaction involving H bonded to N, O, or F.

Ion–dipole forces: attractions between an ion and a polar molecule; central to solvation of ionic compounds.

In real substances these often act together; for example, liquid water exhibits dispersion, dipole–dipole interactions, and hydrogen bonding simultaneously. The art in prediction is to identify which interaction dominates and how molecular structure

alters its magnitude.

9.1.3 London dispersion forces (instantaneous-induced dipoles)

Origin: electron motion creates transient dipoles

Even in a nonpolar atom or molecule, electrons are in constant motion. At any instant, the electron cloud may be slightly more concentrated on one side, creating an instantaneous dipole. This instantaneous dipole can polarize (distort) the electron cloud of a neighboring particle, producing an induced dipole. The two dipoles attract each other.

Because the process depends only on electron motion and polarizability, London dispersion forces occur in:

noble gases (e.g., Ar(l)),

nonpolar molecules (e.g., CH4, CO2),

polar molecules as well (dispersion is always present, even if not dominant).

Polarizability: why bigger often means stronger

Polarizability is the ease with which a particle’s electron cloud can be distorted. It increases when:

the electron cloud is larger (more electrons, larger atomic/molecular size),

outer electrons are held less tightly (more diffuse, higher-energy valence shells).

Consequences:

Down a group (e.g., He, Ne, Ar, Kr, Xe), dispersion forces increase and boiling points rise.

For similar shapes, larger molar mass often correlates with higher Tb because of increased polarizability (though shape can compete strongly with mass).

Molecular shape and surface area: contact matters

Dispersion forces depend not only on how polarizable a molecule is, but also on how effectively two molecules can approach each other. Extended shapes with larger surface area allow more points of contact and stronger overall attraction.

A classic comparison is between structural isomers:

n-pentane (more extended) has a higher boiling point than neopentane (more compact), even though both are C5H12.

The difference is not in bonding (intramolecular structure is the same formula) but in how the molecular surfaces pack and interact.

9.1.4 Dipole–dipole interactions (permanent dipoles)

Permanent dipoles arise from polar bonds and geometry

A molecule has a dipole moment μ when it has an uneven distribution of electron density that does not cancel by symmetry. Two requirements are needed:

one or more polar bonds (electronegativity differences), and

a molecular geometry that does not cancel bond dipoles.

Examples:

HCl is polar (one bond dipole).

CH3Cl is polar (tetrahedral, dipoles do not cancel).

CO2 has polar C– –O bonds but is nonpolar overall (linear cancellation).

CCl4 has polar C–Cl bonds but is nonpolar overall (tetrahedral cancellation).

Orientation and temperature

Dipole–dipole attraction is strongest when molecules align so that δ⁺ regions approach δ⁻ regions. Thermal motion disrupts alignment; therefore:

Dipole–dipole effects are often more evident at lower temperatures (where alignment persists longer).

At higher temperatures, increased kinetic energy reduces the time-averaged orientation alignment.

In condensed phases, dipole alignment is still partial and dynamic; the liquid structure is constantly rearranging.

How dipole–dipole compares with dispersion

For molecules of similar size and shape, a polar molecule generally has a higher boiling point than a comparable nonpolar molecule because dipole–dipole attraction

adds to dispersion. However, dispersion can dominate when a nonpolar molecule is much larger or more polarizable. For example, a heavy nonpolar halogenated compound can have a higher boiling point than a smaller polar molecule due to strong dispersion.

9.1.5 Hydrogen bonding (a special, strong dipole–dipole interaction)

Definition: the donor–hydrogen–acceptor pattern

A hydrogen bond forms when:

Donor: a hydrogen atom is covalently bonded to N, O, or F (written X–H where X = N, O, F),

Acceptor: a lone pair on N, O, or F of a neighboring molecule interacts with that hydrogen.

It is commonly represented as

where X–H is the donor and Y is the acceptor atom bearing the lone pair.

Why only N, O, and F? These atoms are very electronegative and small:

The X–H bond is highly polar, creating a strongly positive δ⁺ hydrogen.

The acceptor lone pairs are relatively concentrated (high charge density), strengthening the electrostatic attraction.

Directionality and structure

Hydrogen bonds are directional: the interaction is strongest when the donor bond and acceptor lone pair align approximately linearly (often discussed as a preference for near-linear X–H⋅..Y arrangements). This directionality helps explain the structured networks in substances such as water and many biological molecules (proteins, DNA), where hydrogen bonds influence shape and recognition.

Consequences: unusually high boiling points and unique properties

Hydrogen bonding substantially increases cohesion in liquids and raises boiling points. A striking example is water:

H2O has a much higher normal boiling point than H2S, H2Se, and H2Te, even though those heavier molecules have stronger dispersion forces.

The hydrogen-bonding network also contributes to water’s high surface tension and heat capacity (topics developed in later sections of this chapter).

Hydrogen bonding also influences solubility: molecules capable of donating/accepting hydrogen bonds tend to be more soluble in water, because they can form favorable interactions with water molecules.

9.1.6 Ion–dipole forces (ions interacting with polar molecules)

Origin: charge interacting with a permanent dipole

An ion–dipole interaction occurs between an ion (full charge, e.g., Na+, Cl– ) and the partial charges of a polar molecule (e.g., δ⁻ oxygen in H2O). The electrostatic attraction is often stronger than typical dipole–dipole or hydrogen-bond interactions because one partner carries a full ionic charge.

Solvation and hydration: why salts dissolve

When an ionic solid dissolves in a polar solvent, the solvent molecules orient around the ions:

For cations, the δ⁻ end of the dipole points toward the ion (oxygen toward Na+ in water).

For anions, the δ⁺ end points toward the ion (hydrogen toward Cl– in water).

This organized shell of solvent is called a solvation shell; in water it is specifically hydration.

Dissolution of an ionic solid can be viewed as a competition between:

Ion–ion attractions in the crystal lattice (lattice energy), and

Ion–dipole attractions between ions and solvent molecules (solvation energy).

If the energetic and entropic changes overall favor separation of ions into solution, the solid dissolves.

Dependence on ion charge density

Ion–dipole strength increases with:

larger ionic charge (e.g., Mg2+ interacts more strongly with water than Na+ under comparable conditions),

smaller ionic radius (higher charge density leads to stronger electric fields).

Thus, small highly charged ions tend to be strongly solvated and can significantly structure nearby solvent molecules.

9.1.7 Predicting dominant intermolecular forces from structure

Step 1: Identify particle types

Neutral molecules: consider dispersion, dipole–dipole, hydrogen bonding.

Ions present with polar molecules: include ion–dipole (often dominant in solution contexts).

Step 2: Determine polarity and hydrogen-bonding capability

Is the molecule polar overall (geometry + bond polarity)?

Does it contain N–H, O–H, or F–H (donor)?

Does it contain N, O, or F with lone pairs (acceptor)?

Note: molecules like alcohols (ROH) both donate and accept hydrogen bonds; ethers (R–O–R′) accept but do not donate.

Step 3: Assess dispersion contributions (size and shape)

Larger/heavier and more polarizable → stronger dispersion.

More extended shape → greater surface contact → stronger dispersion.

Branching often lowers boiling point by reducing surface area contact, even at constant molar mass.

Step 4: Combine effects (additive but not equal)

Intermolecular forces are not “either/or”; they add. But some are typically decisive:

Hydrogen bonding often explains unexpectedly high boiling points and high cohesion.

Ion–dipole interactions often control solubility of ionic compounds in polar solvents.

Dispersion dominates in large nonpolar molecules and noble gases, and can rival dipole effects in heavy molecules.

9.1.8 A caution about language: “van der Waals forces”

The term van der Waals forces is used inconsistently. In some texts it refers to all intermolecular attractions between neutral molecules (dispersion + dipole–dipole, sometimes including induced-dipole interactions). In others it is used as a synonym for dispersion only. In this chapter we will use the specific terms (dispersion, dipole–dipole, hydrogen bonding, ion–dipole) to avoid ambiguity.

9.1.9 Summary table: key features of the four major interactions

Interaction | What interacts? | When is it important? | Typical qualitative strength
London dispersion | Instantaneous dipole with induced dipole | Always present; dominates in nonpolar and/or large, polarizable species; sensitive to shape/surface area | Weak individually; can be substantial in large molecules
Dipole–dipole | Permanent dipoles of polar molecules | Polar molecules; orientation-dependent; adds to dispersion | Moderate
Hydrogen bonding | N–H, O–H, or F–H donor with N, O, or F lone-pair acceptor | Molecules containing N–H, O–H, F–H; strongly affects boiling points and liquid structure | Strong (among neutral molecules)
Ion–dipole | An ion with the partial charges of a polar molecule | Dissolution/solvation of ions in polar solvents; strongest driver of hydration shells | Often very strong; increases with ionic charge density

9.2 Properties of Liquids

In the previous section we described intermolecular forces (IMFs) as electrostatic attractions between particles: dispersion forces, dipole–dipole interactions, hydrogen bonding, and ion–dipole forces. In a liquid, these attractions are strong enough to keep molecules close together (condensed phase) yet weak enough to allow continual rearrangement. The result is a state of matter with a distinctive combination of properties:

Definite volume (low compressibility),

No fixed shape (the liquid flows to adopt the container shape),

A surface (interface) with measurable energy (surface tension),

Transport properties (viscosity) that strongly reflect molecular interactions.

This section connects three macroscopic liquid behaviors—viscosity, surface tension, and capillary action—to two microscopic ideas:

cohesion (attraction between like molecules) and adhesion (attraction between unlike molecules).

9.2.1 Cohesion, adhesion, and the “tug-of-war” at interfaces

Within the bulk of a liquid, each molecule is attracted to many neighbors in all directions. These attractions largely balance, so there is no net force on a molecule due solely to its surroundings. At an interface (liquid–gas or liquid–solid), that symmetry is broken:

At a liquid–gas interface, molecules at the surface have fewer neighboring molecules above them, so they experience a net inward attraction. This creates surface tension.

At a liquid–solid interface, molecules in the liquid interact with the surface of the solid. If adhesive forces to the solid are strong relative to cohesive forces within the liquid, the liquid tends to spread (wetting). If adhesive forces are weak, the liquid tends to bead up (non-wetting).

These same intermolecular attractions also resist relative motion of layers of liquid, producing viscosity.

9.2.2 Viscosity: resistance to flow

Definition and molecular picture

Viscosity is a measure of a fluid’s resistance to flow. Qualitatively, it answers: How difficult is it for one layer of liquid to slide past another?

At the molecular level, flow requires continual breaking and reforming of intermolecular attractions as molecules move to new neighbors. Stronger cohesion generally increases the energy barrier for rearrangement, raising viscosity.

Two everyday comparisons illustrate the role of IMFs and molecular structure:

Water vs. glycerol: Glycerol (C3H8O3) has multiple O–H groups and forms extensive hydrogen bonding, making it much more viscous than water at room temperature.

Branching in hydrocarbons: For isomeric alkanes, more compact, highly branched molecules often have lower viscosity because they experience less surface contact and can slide past one another more easily than long, flexible chains.

Trends with intermolecular forces and molecular shape

Viscosity increases with:

Stronger intermolecular forces: hydrogen bonding and strong dipole–dipole interactions tend to raise viscosity compared with purely dispersion-dominated liquids of similar size.

Greater molecular surface area and entanglement: long-chain or planar molecules have greater contact and can become physically entangled, increasing resistance to flow.

It is helpful to separate two contributions:

Energetic: stronger attractions increase the work required to rearrange neighbors.

Geometric: shape and flexibility affect how easily molecules can move past one another.

Temperature dependence

For liquids, viscosity decreases as temperature increases. Increased thermal energy gives molecules more kinetic energy, helping them overcome cohesive attractions and move between configurations more readily.

This is the opposite of the typical trend for gases (gas viscosity often increases with temperature due to faster molecular motion and momentum transfer), emphasizing that liquid viscosity is dominated by cohesive interactions rather than by free-flight collisions.

Dynamic vs. kinematic viscosity (brief note)

Two related quantities appear in science and engineering:

Dynamic viscosity η (units: Pa ⋅ s) characterizes internal friction.

Kinematic viscosity ν = η∕ρ (units: m²∕s) accounts for fluid density ρ.

In chemistry, dynamic viscosity is most commonly discussed because it connects directly to intermolecular cohesion.

9.2.3 Surface tension: energy cost of creating surface area

Why liquids minimize surface area

A surface is energetically unfavorable because surface molecules are less stabilized by intermolecular attractions than interior molecules. A liquid

therefore tends to minimize its surface area, which is why small droplets become nearly spherical: the sphere has the smallest surface area for a given volume.

Definition and units

Surface tension γ is defined as:

the force per unit length acting along a line on the surface, or equivalently,

the energy per unit area required to create new surface.

These are equivalent because N∕m = J∕m².

Conceptually, if you increase the surface area by ΔA, the minimum reversible work required is

A large γ means it is “costly” (energetically) to create surface area, reflecting strong cohesion.

Microscopic origin: unbalanced attractions at the surface

In the bulk, a molecule experiences attractions in all directions. At the surface, there are fewer neighbors above, so the net cohesive force points inward. To expand the surface, molecules from the interior must be brought to the surface, which reduces their average intermolecular stabilization. The required energy shows up macroscopically as surface tension.

Factors affecting surface tension

Surface tension increases with stronger cohesive forces:

Liquids with extensive hydrogen bonding (e.g., water) generally have high surface tension.

Nonpolar liquids with only dispersion forces tend to have lower surface tension, though large polarizable molecules can still exhibit moderate values.

Surface tension decreases with temperature, because higher thermal energy disrupts intermolecular ordering and makes it easier for molecules to occupy surface positions.

Surfactants: reducing surface tension by disrupting cohesion

A surfactant (surface-active agent) is a molecule that preferentially adsorbs at the liquid surface and reduces γ. Many surfactants have:

a hydrophilic (polar/ionic) head group and

a hydrophobic (nonpolar) tail.

By accumulating at the interface, surfactants replace strong liquid–liquid interactions at the surface with weaker or differently structured interactions, lowering the energy cost of surface area. This is why detergents help water spread across greasy surfaces and penetrate fabrics.

9.2.4 Capillary action: competition between cohesion and adhesion

What capillary action is

Capillary action is the spontaneous rise or fall of a liquid in a narrow tube or porous material. It is responsible for phenomena such as:

water moving upward through plant xylem,

absorption of ink in paper,

wicking of liquids in thin channels and porous solids.

Capillary behavior is driven by surface tension and the balance between adhesive and cohesive forces at the solid boundary.

Meniscus shape and wetting

When a liquid contacts a solid surface (such as glass), the liquid surface typically curves near the wall, forming a meniscus. The curvature depends on relative strengths:

Adhesion > cohesion: the liquid wets the surface and climbs the wall, producing a concave meniscus. Water in clean glass is a standard example because polar Si–O groups on glass interact strongly with water via hydrogen bonding and dipole interactions.

Cohesion > adhesion: the liquid does not wet the surface and pulls away from the wall, producing a convex meniscus. Mercury in glass is a standard example because Hg–Hg metallic cohesion is very strong, while Hg–glass adhesion is comparatively weak.

The degree of wetting is quantified by the contact angle 𝜃, measured through the liquid at the solid surface:

𝜃 < 90∘: wetting, concave meniscus (capillary rise).

𝜃 > 90∘: non-wetting, convex meniscus (capillary depression).

Why narrow tubes amplify the effect

The surface tension force acts along the contact line where liquid meets solid. In a narrow tube, the circumference-to-area ratio is large, so surface effects (which scale with perimeter) compete effectively against gravity (which scales with volume). Thus, capillary rise is greater in smaller-radius tubes.

Quantitative description: Jurin’s law

For a capillary tube of radius r, the height h a liquid rises (or falls) relative to the bulk liquid level is described by Jurin’s law:

where:

γ is the surface tension,

𝜃 is the contact angle,

ρ is the liquid density,

g is gravitational acceleration,

r is the tube radius.

This equation encodes the core physical ideas:

Stronger surface tension (γ larger) increases the magnitude of h.

Better wetting (𝜃 smaller, so cos𝜃 closer to 1) increases capillary rise.

Denser liquids (ρ larger) rise less because gravity is stronger for the same volume.

Narrower tubes (smaller r) produce larger rise or depression.

A sign change is built into cos𝜃: if 𝜃 > 90∘, then cos𝜃 < 0, and h becomes negative (capillary depression).

Connecting back to intermolecular forces

Capillary action is a macroscopic outcome of two microscopic interactions:

Cohesion within the liquid (liquid–liquid IMFs) determines γ and the tendency to minimize surface area.

Adhesion to the solid (liquid–solid attractions, often dipole–dipole, hydrogen bonding, or ion–dipole depending on surface chemistry) determines 𝜃, i.e., whether wetting occurs.

Thus, by identifying the dominant intermolecular interactions in each part of the system (liquid–liquid and liquid–solid), we can predict whether a meniscus will be concave or convex and whether the liquid will rise or fall in a capillary.

9.2.5 Summary: linking microscopic forces to macroscopic behavior

Property | Macroscopic description | Microscopic (IMF-based) interpretation
Viscosity | Resistance to flow; “thickness” of a liquid | Stronger cohesion and greater molecular contact/entanglement hinder molecular rearrangement; viscosity decreases with temperature
Surface tension | Energy cost to create surface area; tendency to minimize surface | Surface molecules have fewer neighbors and are less stabilized; stronger cohesion increases γ; γ decreases with temperature and surfactants
Capillary action | Rise or fall in narrow tubes/pores; meniscus curvature | Competition between adhesion (liquid–solid) and cohesion (liquid–liquid) sets contact angle; surface tension and tube radius determine magnitude via h =

9.3 Phase Changes and Heating Curves

The previous sections connected intermolecular forces (IMFs) to liquid properties such as viscosity, surface tension, and capillary action. The same IMFs also determine the energy required to separate particles enough to change phase. In this section we treat phase transitions using thermochemistry: how heat flow relates to temperature change, enthalpy changes, and the shape of a heating curve.

9.3.1 Energy, temperature, and phase: separating kinetic and potential

contributions

A common misconception is that adding heat always increases temperature. In reality, when a substance changes phase at constant pressure, heat can be absorbed or released with no temperature change. The key is to distinguish two roles for energy:

Kinetic energy change: manifests as a change in temperature. When a single phase is heated, added energy increases the average molecular speed.

Potential energy change: manifests as a change in intermolecular arrangement. During a phase change, added (or removed) energy changes the degree of intermolecular attraction (how strongly particles are held together), not the average speed.

Because IMFs are electrostatic attractions, converting solid → liquid → gas generally requires energy to overcome these attractions. Stronger IMFs usually imply larger enthalpies of phase change and higher melting/boiling points.

9.3.2 Enthalpy changes for phase transitions

At constant pressure, the heat absorbed or released by a system equals the enthalpy change:

For phase changes, we define three standard enthalpy quantities (per mole of substance):

Enthalpy of fusion, ΔHfus

Fusion is melting:

ΔHfus is the enthalpy change to melt 1 mol of solid at its melting point. The reverse process, freezing, has the same magnitude but opposite sign:

Enthalpy of vaporization, ΔHvap

Vaporization is boiling/evaporation:

ΔHvap is the enthalpy change to convert 1 mol of liquid to gas at its boiling point. Condensation is the reverse:

Because molecules in a gas are much farther apart than in a liquid, vaporization requires far more energy than fusion for most substances:

Enthalpy of sublimation, ΔHsub

Sublimation is:

The enthalpy of sublimation satisfies (Hess’s law):

provided all quantities refer to the same temperature and appropriate states (often approximated using values at the melting and boiling points, respectively).

9.3.3 Heating within a single phase: specific heat capacity

When a substance remains in a single phase, heat is related to the temperature change by:

where m is mass, c is specific heat capacity, and ΔT = Tfinal − Tinitial.

It is often convenient to use molar heat capacity Cm (units Jmol⁻¹ K⁻¹):

The slope of a heating curve segment (temperature vs. heat added) depends on the heat capacity:

Larger heat capacity ⇒ more heat needed per degree ⇒ smaller slope.

Different phases have different heat capacities because particle motion and degrees of freedom differ (vibrations in solids, rotational/translational freedom in liquids and gases).

9.3.4 Heating curves: a map of temperature vs. heat added

A heating curve is a plot of temperature as a function of heat added. It provides an integrated view of:

warming within phases (sloped segments), and

phase changes (flat plateaus).

The canonical example is water at 1 atm, heated from below 0∘C to above 100∘C. The curve contains five segments:

Segment | Process | Heat expression
1 | Warm ice (−20∘C → 0∘C) | q 1 = mciceΔT
2 | Melt at 0∘C | q 2 = nΔHfus
3 | Warm liquid water (0∘C → 100∘C) | q₃ = mcliqΔT
4 | Vaporize at 100∘C | q 4 = nΔHvap
5 | Warm steam (100∘C → 120∘C) | q₅ = mcsteamΔT

9.3.5 Quantitative heating curve for water (1 atm)

The heating curve below is drawn using commonly used thermochemical values for water:

cice ≈ 2.09Jg⁻¹ ∘C⁻¹

cliq ≈ 4.18Jg⁻¹ ∘C⁻¹

csteam ≈ 2.01Jg⁻¹ ∘C⁻¹

ΔHfus ≈ 6.01kJmol⁻¹

ΔHvap ≈ 40.7kJmol⁻¹

The vaporization plateau is much longer than the fusion plateau, reflecting that much more energy is required to separate molecules into the gas phase than to loosen them into a liquid.

9.3.6 How to compute energy for multi-step heating or cooling

A heating curve is more than a picture: it is a recipe for energy accounting. The general strategy is:

Step 1: Identify the initial and final states

Determine:

initial phase and temperature,

final phase and temperature,

whether any phase boundaries are crossed (melting point, boiling point at the given pressure).

Step 2: Break the process into segments

Each time you cross a phase boundary or change phase, a new segment begins. Typical segments include:

warming/cooling within one phase: q = mcΔT,

melting/freezing: q = nΔHfus,

vaporization/condensation: q = nΔHvap,

sublimation/deposition (less common in basic heating curves): q = nΔHsub.

Step 3: Track signs carefully

Use a consistent sign convention:

If the system absorbs heat (endothermic), q > 0.

If the system releases heat (exothermic), q < 0.

Thus:

Step 4: Sum the segment heats

The total heat is the algebraic sum:

9.3.7 Why plateaus occur: phase equilibrium at constant temperature

During a phase change at constant pressure, the two phases coexist in equilibrium (e.g., ice + liquid water at 0∘C and 1 atm). As heat is added:

energy goes into disrupting the intermolecular arrangement of the old phase,

the temperature stays fixed because the average kinetic energy does not increase until the phase change is complete.

Once the last portion of the old phase disappears, additional heat again raises temperature.

9.3.8 Supercooling and superheating (metastable states)

In ideal equilibrium, freezing occurs at the melting point and boiling occurs at the boiling point. Real systems can deviate if nucleation is delayed:

Supercooling

A liquid can be cooled below its normal freezing point without solidifying if crystal nucleation does not occur. The liquid is metastable; once nucleation begins (often triggered by an impurity or disturbance), freezing can proceed rapidly and release heat.

Superheating

A liquid can be heated above its normal boiling point without boiling if vapor bubble formation is suppressed (for example, in a very smooth container). When boiling finally initiates, it may be sudden and vigorous (“bumping”), which is why boiling chips or stirring are used in the laboratory to provide nucleation sites.

9.3.9 Key takeaways

Temperature changes reflect changes in average kinetic energy; phase changes reflect changes in intermolecular potential energy.

At constant pressure, heat for phase change is computed with enthalpies: q = nΔHfus, q = nΔHvap, q = nΔHsub.

Heating curves combine q = mcΔT segments (sloped) with phase-change plateaus (flat).

For most substances, ΔHvap is far larger than ΔHfus because forming a gas requires much greater separation of molecules.

In the next section, we will connect vaporization enthalpy and intermolecular forces to vapor pressure and boiling point, and we will develop a quantitative relationship between vapor pressure and temperature.

9.4 Vapor Pressure and Boiling Point

The heating-curve analysis in the previous section emphasized that vaporization requires substantial energy (ΔHvap) to overcome intermolecular attractions and separate molecules into the gas phase. Here we connect that energy requirement to vapor pressure and boiling, two ideas that explain why some liquids evaporate

quickly at room temperature while others do not, and why boiling points change with external pressure.

9.4.1 Evaporation and condensation: a dynamic molecular process

Even below its boiling point, a liquid continuously exchanges molecules with its vapor:

Evaporation occurs when a surface molecule has enough kinetic energy and an appropriate direction of motion to escape the liquid’s intermolecular attractions and enter the gas phase.

Condensation occurs when a vapor molecule collides with the liquid surface and becomes trapped by intermolecular attractions, returning to the liquid.

Because kinetic energies are distributed (Maxwell–Boltzmann distribution), at any temperature there is always a fraction of molecules with unusually high kinetic energy. These energetic molecules are most likely to evaporate. This explains why evaporation is observed at temperatures far below the normal boiling point.

9.4.2 Vapor pressure: definition and equilibrium meaning

Consider a closed container containing a pure liquid and its vapor at a fixed temperature. Initially, evaporation dominates and the amount of vapor increases. As the vapor concentration rises, condensation becomes more frequent. Eventually, a dynamic equilibrium is reached:

At equilibrium, the pressure exerted by the vapor is the vapor pressure, Pvap, of the liquid at that temperature.

Two important points follow immediately:

Vapor pressure depends strongly on temperature.

Vapor pressure depends on the strength of intermolecular forces in the liquid (through ΔHvap).

What vapor pressure is not

Vapor pressure is not the pressure needed to “push” vapor out of the liquid. It is an equilibrium property: the pressure of vapor that results when evaporation and condensation balance at a given temperature.

9.4.3 How intermolecular forces influence vapor pressure

A molecule must overcome cohesive attractions to leave the liquid. Stronger intermolecular forces (especially hydrogen bonding and strong dipole–dipole interactions) increase the energy barrier to escape, decreasing the fraction of molecules that can evaporate at a given temperature. Therefore:

Stronger IMFs ⇒lower vapor pressure at the same T.

This relationship is the mirror image of boiling-point trends:

Lower vapor pressure at a given T ⇒higher boiling point.

9.4.4 Boiling vs. evaporation

Evaporation: surface phenomenon

Evaporation occurs only at the surface because molecules must escape into the gas phase. It can occur at any temperature.

Boiling: bulk phenomenon

Boiling occurs when vapor bubbles can form within the bulk of the liquid and rise without collapsing. For a bubble to persist, the pressure inside the bubble (the liquid’s vapor pressure at that temperature) must be at least as large as the external pressure (plus small additional contributions from surface tension and hydrostatic pressure).

In general chemistry we capture the central condition for boiling as:

Thus, the boiling point depends on external pressure.

Normal boiling point

The normal boiling point is defined as the temperature at which:

Why boiling point decreases at high altitude

At high altitude, atmospheric pressure is lower, so the condition Pvap = Pext is met at a lower temperature. Water boils below 100 ∘C, which is why cooking times can increase: the maximum temperature of boiling water is reduced.

9.4.5 Temperature dependence of vapor pressure: the Clausius–Clapeyron equation

Vapor pressure increases rapidly (nonlinearly) with temperature because higher T shifts the kinetic energy distribution to higher energies, increasing the fraction of molecules able to escape.

A quantitative relationship between Pvap and T can be obtained from the Clausius–Clapeyron equation. In its commonly used two-point form:

where:

P₁, P₂ are vapor pressures at temperatures T₁, T₂ (in kelvin),

ΔHvap is the enthalpy of vaporization (assumed approximately constant over the temperature range),

R is the gas constant (8.314 Jmol⁻¹ K⁻¹).

Linear form and interpretation

Rearranging shows that a plot of lnP vs. 1∕T is approximately linear:

with slope −ΔHvap∕R. Thus:

a larger ΔHvap produces a more negative slope,

meaning vapor pressure rises more slowly with temperature when vaporization is energetically costly.

This directly encodes the role of intermolecular forces: stronger IMFs ⇒ larger ΔHvap ⇒ lower vapor pressures and higher boiling points.

9.4.6 Comparing liquids: ether, ethanol, and water

The following comparison illustrates how intermolecular forces shape vapor pressure curves:

Diethyl ether (C4H10O) has relatively weak intermolecular attractions (no O–H bond, so it cannot donate hydrogen bonds). It has a high vapor pressure at room temperature and a low normal boiling point.

Ethanol (C2H5OH) can hydrogen-bond (donor and acceptor), so it has lower vapor pressure than ether at the same T and a higher normal boiling point.

Water (H2O) forms an extensive hydrogen-bonding network, giving it the lowest vapor pressure among the three at a given T and the highest normal boiling point.

9.4.7 Using Clausius–Clapeyron calculations in practice

The two-point Clausius–Clapeyron equation is used in two common ways.

Predicting vapor pressure at a new temperature

Given P₁ at T₁ and ΔHvap, compute P₂ at T₂:

Practical reminders:

Use kelvin for temperature.

Keep pressure units consistent (mmHg, atm, kPa) because only the ratio P₂∕P₁ appears.

Use ΔHvap in Jmol⁻¹ if R is 8.314 Jmol⁻¹ K⁻¹.

Predicting boiling point at a nonstandard pressure

Because boiling occurs when Pvap = Pext, you can treat the external pressure as the target vapor pressure:

If P₁ = 760mmHg at the normal boiling point T₁ = Tb,normal, then solving for T₂ gives the boiling point at the new pressure.

9.4.8 Key takeaways

Vapor pressure is the equilibrium pressure of a vapor above its liquid in a closed system at fixed T.

Stronger intermolecular forces lower vapor pressure and raise boiling point.

Boiling occurs when Pvap(T) = Pext; the normal boiling point is defined at 760 mmHg.

The Clausius–Clapeyron equation provides a quantitative link between vapor pressure and temperature and relates the slope of lnP vs. 1∕T to ΔHvap.

9.5 Phase Diagrams

The previous section introduced vapor pressure and boiling as equilibrium phenomena and emphasized that phase stability depends on both temperature and pressure. A phase diagram is the natural extension of that idea: it is a compact

map showing which phase (solid, liquid, or gas) is thermodynamically stable at each combination of temperature and pressure for a pure substance.

9.5.1 What a phase diagram shows

A phase diagram plots pressure P versus temperature T. Each point on the diagram represents a condition under which a pure substance will exist in the phase with the lowest Gibbs free energy G (at that T and P).

The diagram is divided into regions and boundary lines:

Regions: single-phase stability fields (solid, liquid, gas).

Boundary lines (coexistence curves): conditions where two phases coexist in equilibrium (their molar Gibbs free energies are equal).

Crossing a boundary line changes the stable phase. For example, at 1 atm increasing T typically moves a substance from solid → liquid → gas, but the path and the temperatures at which transitions occur depend on the substance and pressure.

9.5.2 The three coexistence curves

A typical phase diagram contains three equilibrium lines:

Solid–liquid line (fusion curve)

Points on this line represent equilibrium between solid and liquid:

where μ is chemical potential (molar Gibbs free energy). At a given pressure, the temperature on this line is the melting point (or freezing point).

Liquid–gas line (vaporization curve)

Points on this line represent equilibrium between liquid and vapor:

At a given pressure, the temperature on this line is the boiling point. The earlier condition for boiling,

can be interpreted directly on a phase diagram: boiling occurs when the system crosses from the liquid region into the gas region at the chosen external pressure.

Solid–gas line (sublimation curve)

Points on this line represent equilibrium between solid and vapor:

At pressures below the triple point pressure, a solid heated sufficiently will cross this line and sublime directly to the gas phase without forming a liquid.

9.5.3 Special points: triple point and critical point

Triple point

The triple point is the unique (T,P) at which all three phases coexist in equilibrium:

It is the intersection of the sublimation, fusion, and vaporization curves. Because it is a fixed point for a pure substance, triple points are used in temperature calibration (the water triple point is historically important in metrology).

Critical point and supercritical fluid

The liquid–gas coexistence curve terminates at the critical point (Tc,Pc). At temperatures and pressures above this point, there is no sharp distinction between liquid and gas; the substance becomes a supercritical fluid:

It can have liquid-like density (good solvent power) and gas-like transport properties (low viscosity, high diffusivity).

Crossing from “liquid-like” to “gas-like” behavior can occur continuously without crossing a phase boundary.

9.5.4 How to read a phase diagram

A phase diagram answers three common types of questions.

1. What phase is stable at a given (T,P)?

Locate the point (T,P) and identify which region contains it (solid, liquid, or gas). If the point lies on a boundary line, two phases coexist; if at the triple point, three phases coexist.

2. At a fixed pressure, what happens as temperature changes?

Draw a horizontal line at that pressure and move along it as T increases:

Crossing the fusion curve: melting/freezing occurs.

Crossing the vaporization curve: boiling/condensation occurs.

If the pressure is below the triple point: the horizontal line never enters the liquid region; the solid sublimes upon heating.

3. At a fixed temperature, what happens as pressure changes?

Draw a vertical line at that temperature:

Increasing pressure generally favors denser phases (often solid or liquid over gas).

Crossing boundaries corresponds to phase transitions driven by compression or decompression.

9.5.5 The slope of coexistence curves: qualitative meaning

The slope of a phase boundary indicates how the transition temperature changes with pressure. While a full derivation involves the Clapeyron equation, the qualitative idea is accessible:

Increasing pressure favors the phase with smaller molar volume (higher density).

If the solid is denser than the liquid (the usual case), increasing pressure tends to favor the solid and raises the melting point; the solid–liquid line has a positive slope.

If the liquid is denser than the solid (an anomaly), increasing pressure favors the liquid and lowers the melting point; the solid–liquid line has a negative slope.

This is precisely the key difference between H2O and CO2 phase diagrams.

9.5.6 Comparison: phase diagrams of water and carbon dioxide

Water exhibits an important structural anomaly: ice is less dense than liquid water near 0 ∘C. Carbon dioxide behaves more “typically”: its solid is denser than its liquid. These density relationships appear in the slope of the solid–liquid line.

Water: negative fusion-curve slope and the density anomaly

For H2O near 0 ∘C:

Ice has an open hydrogen-bonded structure, making it less dense (larger molar volume) than liquid water.

Increasing pressure favors the denser phase (liquid), so the melting point decreases with pressure.

This is why, in principle, applying pressure can promote melting of ice at temperatures slightly below 0 ∘C. The phenomenon is subtle in everyday contexts but is conceptually important and unique among common substances.

Carbon dioxide: positive fusion-curve slope and no liquid at 1 atm

For CO2:

The solid is denser than the liquid, so the fusion curve has a positive slope.

The triple point pressure for CO2 is above 1 atm. Therefore, at 1 atm, CO2 cannot exist as a liquid; heating solid CO2 (dry ice) causes it to sublime directly to gas.

This illustrates a practical reading rule: if your pressure lies below the triple point, the liquid region is inaccessible and phase changes occur via sublimation/deposition rather than melting/boiling.

9.5.7 Phase-diagram reasoning patterns (worked qualitatively)

Although phase diagrams can include numerical scales, many problems are qualitative and require only consistent reasoning.

Following a path at constant pressure

At fixed pressure P∗, increase temperature:

If P∗ intersects the fusion curve and then the vaporization curve, you observe melting and then boiling.

If P∗ is below the triple point, you intersect only the sublimation curve: solid → gas.

Following a path at constant temperature

At fixed temperature T∗, increase pressure:

You may condense a gas to a liquid, and then freeze the liquid to a solid, depending on whether your vertical line crosses those coexistence curves.

Above the critical temperature Tc, you cannot liquefy the gas by pressure alone; there is no liquid–gas boundary to cross.

9.5.8 Key takeaways

A phase diagram summarizes the stable phase of a pure substance as a function of T and P, with boundary lines marking equilibria between phases.

The triple point is where solid, liquid, and gas coexist; below the triple-point pressure, a liquid phase cannot exist.

The critical point ends the liquid–gas boundary; above it, the substance is a supercritical fluid.

The slope of the solid–liquid line reflects density differences between solid and liquid: water has a negative slope because ice is less dense than liquid water.

CO2 has a positive slope because its solid is denser than its liquid.

9.6 Structure of Solids

Earlier in this chapter we developed a molecular-level view of condensed phases: intermolecular forces determine cohesion in liquids, the energetics of phase changes, and the boundaries on phase diagrams. A solid is a condensed phase in which particles (atoms, ions, or molecules) occupy positions that are fixed on average; motion is primarily vibrational. The strength and directionality of forces between particles determine not only whether a substance is solid, but also how it is arranged in the solid state.

This section classifies solids by structure (crystalline vs. amorphous) and by bonding/constituent particles (ionic, molecular, covalent-network, metallic). We then focus on crystalline solids, introducing the unit cell concept, three common cubic unit cells (SC, BCC, FCC), and quantitative measures of packing efficiency. Finally, we briefly introduce X-ray diffraction as the experimental route from diffraction patterns to structure (Bragg’s law).

9.6.1 Crystalline vs. amorphous solids

Crystalline solids: long-range order

A crystalline solid has particles arranged in a repeating three-dimensional pattern

called a crystal lattice. The defining feature is long-range order: knowing the arrangement in one small region allows you to predict the arrangement far away by translation.

Typical characteristics:

Sharp melting point: melting occurs over a narrow temperature range because a well-defined lattice must be disrupted.

Anisotropy: properties such as hardness, electrical conductivity, and refractive index can depend on direction within the crystal.

Cleavage planes: crystals often break along specific planes where bonding or packing is weaker.

Amorphous solids: short-range order only

An amorphous solid lacks long-range periodicity. Particles can have local (short-range) organization, but there is no repeating lattice extending through the material. Common examples include glass and many polymers.

Typical characteristics:

No sharp melting point: amorphous materials soften over a temperature range (glass transition behavior).

Isotropy: properties are often similar in all directions because there is no long-range directional order.

9.6.2 Classes of crystalline solids by bonding and constituent particles

A useful chemical classification focuses on what the lattice points represent and what forces hold them together.

Ionic solids

Constituents: cations and anions.
Forces: ion–ion electrostatic attractions (Coulombic).
Typical properties: high melting points, brittle, poor electrical conductors as solids but conduct when molten or dissolved.

Example: NaCl(s) consists of a 3D lattice of Na+ and Cl– . Breaking or shifting layers can bring like charges close together, producing strong repulsion and brittleness.

Molecular solids

Constituents: neutral molecules.
Forces: intermolecular forces (dispersion, dipole–dipole, hydrogen bonding).
Typical properties: relatively low melting points, often soft, generally poor conductors.

Examples:

CO2(s) (dry ice): dispersion forces dominate; sublimation occurs readily at 1 atm.

H2O(s): hydrogen bonding creates a strong but open network.

Covalent-network (atomic) solids

Constituents: atoms linked by covalent bonds in an extended network.
Forces: covalent bonds throughout the structure.
Typical properties: very high melting points, hard, often brittle; electrical behavior varies.

Examples:

diamond: each C is tetrahedrally bonded to 4 neighbors; extremely hard.

graphite: layered covalent sheets with delocalized electrons; conducts electricity along planes.

SiO2 (quartz): covalent network; high melting point.

Metallic solids

Constituents: metal atoms/ions in a lattice.
Forces: metallic bonding (positive cores in a “sea” of delocalized electrons).
Typical properties: good electrical and thermal conductors, malleable, ductile, variable melting points.

Because metallic bonding is non-directional, many metals adopt high-symmetry, efficiently packed crystal structures such as BCC, FCC, or HCP (hexagonal close-packed).

9.6.3 The crystal lattice and the unit cell

Lattice points and motifs

A crystal lattice is a periodic array of lattice points. Each lattice point represents an identical environment in the crystal. The actual arrangement of atoms, ions, or molecules associated with each lattice point is called the motif (or basis). Repeating the motif at every lattice point generates the entire crystal structure.

Unit cell: the repeating building block

A unit cell is the smallest repeating volume that, when translated in three dimensions, reproduces the whole lattice. Unit cells are described by:

edge lengths a,b,c,

interaxial angles α,β,γ.

In many introductory problems, we focus on cubic unit cells where a = b = c and all

angles are 90∘. Cubic cells are especially important for metals and provide a clear context for counting atoms and computing packing efficiency.

9.6.4 Counting atoms in cubic unit cells

In a unit cell representation, atoms are often shown at corners, faces, and/or the body center. The crucial idea is that atoms at the boundaries of a unit cell are shared with neighboring unit cells. The fraction of a boundary atom that “belongs” to a given unit cell is determined by how many unit cells share it:

Corner atom: shared by 8 unit cells ⇒ contribution = .

Edge-centered atom: shared by 4 unit cells ⇒ contribution = . (Not used in SC/BCC/FCC, but important conceptually.)

Face-centered atom: shared by 2 unit cells ⇒ contribution = .

Body-centered atom: entirely inside one unit cell ⇒ contribution = 1.

Simple cubic (SC)

Atom positions: 8 corners.
Atoms per unit cell:

Body-centered cubic (BCC)

Atom positions: 8 corners + 1 body center.
Atoms per unit cell:

Face-centered cubic (FCC)

Atom positions: 8 corners + 6 face centers.
Atoms per unit cell:

9.6.5 Geometry of cubic unit cells: relating atomic radius to edge length

To compute density or packing efficiency, we often relate the unit-cell edge length a to the atomic radius r (assuming atoms are hard spheres that touch along certain

directions).

Simple cubic

Atoms touch along the edge:

Body-centered cubic

Atoms touch along the body diagonal. The body diagonal has length a. Along that diagonal, the sequence corner–center–corner corresponds to 4r:

Face-centered cubic

Atoms touch along the face diagonal. The face diagonal has length a. Along that diagonal, corner–face center–corner corresponds to 4r:

These relationships are essential for density calculations because the unit-cell volume is a³.

9.6.6 Packing efficiency and density

Packing efficiency (atomic packing factor)

The packing efficiency (also called the atomic packing factor, APF) is the fraction of unit-cell volume occupied by atoms (treated as hard spheres):

For a cubic unit cell containing N atoms:

Using the a–r relations above:

SC: APF = ≈ 0.52.

BCC: APF = ≈ 0.68.

FCC: APF = ≈ 0.74.

Thus FCC (and HCP) are close-packed structures with the highest packing efficiency for equal spheres.

Density of a crystalline metal from its unit cell

If a metal crystallizes in a known structure, its density can be computed from atomic mass and unit-cell geometry. The mass of one unit cell is:

where:

N is atoms per unit cell (1, 2, or 4 for SC/BCC/FCC),

M is molar mass,

NA is Avogadro’s number.

The volume is V cell = a³. Therefore,

If the atomic radius r is given, a can be computed using the geometry relationships above.

9.6.7 Unit-cell visualizations (SC, BCC, FCC)

The diagrams below show, for each cubic unit cell, both a “ball-and-stick” style view emphasizing lattice positions and a “space-filling” view emphasizing packing. The fractional contributions of atoms at corners and faces are indicated to reinforce atom-counting rules.

9.6.8 X-ray diffraction and Bragg’s law (structure determination)

Visual models of unit cells are useful, but real crystal structures are determined experimentally. The primary method is X-ray diffraction (XRD).

Why X-rays?

Typical interatomic distances in crystals are on the order of 10⁻¹⁰ m (1 Å), comparable to X-ray wavelengths. When X-rays strike a crystal, the periodic arrangement of atoms scatters the radiation. Under certain angles, scattered waves interfere constructively, producing diffraction peaks.

Bragg’s law

For a set of crystal planes separated by distance d, constructive interference occurs when:

where:

n is an integer (order of diffraction),

λ is the X-ray wavelength,

𝜃 is the incident angle relative to the planes.

Measuring diffraction angles allows determination of interplanar spacings d, and from these one can infer unit-cell dimensions and ultimately the crystal structure.

9.6.9 Summary table: solids, bonding, and structure

Solid type | Constituents | Dominant forces | Common properties
Ionic | cations + anions | ion–ion (Coulombic) | high Tm, brittle, conducts when molten/aqueous
Molecular | molecules | IMFs (dispersion/dipole/H-bond) | low–moderate Tm, soft, poor conductors
Covalent-network | atoms in extended network | covalent bonds | very high Tm, hard, brittle; conductivity varies
Metallic | metal atoms/ions | metallic bonding | conductive, malleable, ductile; many are BCC/FCC/HCP

9.6.10 Key takeaways

Crystalline solids exhibit long-range periodic order; amorphous solids do not and typically soften over a range.

Solids can be classified by constituent particles and bonding: ionic, molecular, covalent-network, metallic.

Unit cells are repeating building blocks; boundary atoms are shared, so counting atoms per cell uses fractional contributions.

SC, BCC, and FCC cubic unit cells contain 1, 2, and 4 atoms per unit cell, respectively, and have increasing packing efficiencies (≈ 0.52 < 0.68 < 0.74).

X-ray diffraction (Bragg’s law) links diffraction angles to interplanar spacings and unit-cell geometry.

In the final section of this chapter, you will apply these ideas—from intermolecular forces to phase behavior and crystal structure—to multi-step quantitative calculations and conceptual reasoning problems.

9.7 Problems

9.7.1 Boiling Point Ranking from Intermolecular Forces

Rank the following substances in order of increasing normal boiling point. For each adjacent pair in your ranking, provide a brief structural argument identifying the dominant intermolecular forces (dispersion, dipole–dipole, hydrogen bonding, ion–dipole) and any important effects of molecular size/shape (e.g., branching vs. extended surface area).

CH4, CH3OH, CH3OCH3, C3H8, CH3Cl, CCl4

Your explanation should explicitly address:

why CH3OH differs from CH3OCH3 despite identical molecular formulas,

why CCl4 behaves differently from smaller polar molecules,

why nonpolar molecules can still show substantial cohesion.

9.7.2 Multi-Step Heating Curve Calculation: Ice at −10∘C to Steam at 110∘C

A 45.0 g sample of ice at −10.0 ∘C is converted to steam at 110.0 ∘C at 1 atm.

Use the following data:

cice = 2.09Jg⁻¹ ∘C⁻¹

cliq = 4.18Jg⁻¹ ∘C⁻¹

csteam = 2.01Jg⁻¹ ∘C⁻¹

ΔHfus = 6.01kJmol⁻¹

ΔHvap = 40.7kJmol⁻¹

M(H2O) = 18.015g mol⁻¹

Draw and label the five conceptual segments of the heating curve relevant to this process.

Calculate q for each segment and then the total heat qtotal in kJ.

Identify which segment contributes the largest fraction of the total energy and explain why in terms of intermolecular forces and molecular separation.

9.7.3 Clausius–Clapeyron: Vapor Pressure from ΔHvap and a Normal Boiling Point

A liquid has a normal boiling point of 82.0 ∘C and an enthalpy of vaporization ΔHvap = 33.5kJmol⁻¹ (assume it is constant over the temperature range of interest).

Use the Clausius–Clapeyron equation to estimate the vapor pressure at 25.0 ∘C.

At what temperature (in ∘C) will the vapor pressure reach 200 mmHg?

Briefly state how your answers would change (increase/decrease) if the true ΔHvap increases as temperature decreases, and justify qualitatively.

9.7.4 Phase Diagram Reasoning: Identifying Phase Along a Path

A hypothetical substance has a phase diagram with a triple point at moderate pressure and a critical point at high temperature. Consider the following process paths (each begins at the stated condition and proceeds as described). For each path, list the sequence of phases encountered and identify any phase transition(s) (melting, freezing, vaporization, condensation, sublimation, deposition). Assume equilibrium is maintained and ignore superheating/supercooling.

Start at low T and low P in the solid region. Increase T at constant P where P is below the triple-point pressure.

Start in the liquid region at a pressure above the triple-point pressure but below the critical pressure. Increase T at constant P until the gas region is reached.

Start in the gas region at a temperature above the critical temperature. Increase P at constant T to a very high pressure.

For the third path, explicitly state whether a liquid phase boundary is crossed and what phase description applies at the end.

9.7.5 FCC Density from Atomic Radius

A metallic element crystallizes in a face-centered cubic (FCC) structure. Its atomic radius is r = 143pm, and its molar mass is M = 63.55g mol⁻¹.

Determine the unit-cell edge length a in pm using the FCC geometry relation.

Calculate the unit-cell volume in cm³.

Compute the density ρ of the metal in g cm⁻³.

The measured density is 8.9 g cm⁻³. Comment on whether the result is consistent within typical rounding/ideal-sphere approximations, and list at least two physical reasons (beyond arithmetic) why the calculated value might differ from experiment.

9.7.6 Viscosity and Isomerism: Shape vs. Intermolecular Forces

Consider two structural isomers: n-butanol (CH3CH2CH2CH2OH) and tert-butanol ((CH3)3COH).

Predict which has the higher viscosity at 25 ∘C, and justify your choice using both intermolecular-force arguments and molecular-shape/surface-area arguments.

Predict which has the higher normal boiling point, and justify.

Suppose both are dissolved separately in water at the same mole fraction. Predict which solution has the higher viscosity and justify in terms of hydrogen bonding and disruption of water’s structure.

9.7.7 X-ray Diffraction and Bragg’s Law: Identifying a Metal

A monochromatic X-ray beam with wavelength λ = 154pm (Cu Kα) is used to analyze a crystalline metal. A strong diffraction peak is observed at 2𝜃 = 44.7∘ corresponding to the (110) planes of a body-centered cubic (BCC) structure. The molar mass of the metal is 55.85 g mol⁻¹.

Use Bragg’s law (take n = 1) to determine the interplanar spacing d₁₁₀.

For a cubic lattice, dhkl = . Use this to calculate the unit-cell edge length a.

Determine the density of the metal and compare it to known values to identify the most likely element.

Briefly describe (one paragraph) why identifying crystal structure from diffraction peaks requires indexing (hkl) planes rather than using a single peak.

9.7.8 Meniscus Shape: Water vs. Mercury in Glass

Explain, using a molecular-level argument, why water forms a concave meniscus in clean glass whereas mercury forms a convex meniscus.

Your response must include:

clear definitions of cohesion and adhesion,

the relationship between adhesion/cohesion balance and the contact angle 𝜃,

a discussion of which interactions (hydrogen bonding, dipole interactions, metallic bonding, etc.) dominate in each case.

9.7.9 Enthalpy of Sublimation from ΔHfus and ΔHvap

A substance has ΔHfus = 12.4kJmol⁻¹ at its melting point and ΔHvap = 45.2kJmol⁻¹ at its boiling point. Assume (for the purpose of this problem) that these values can be combined directly.

Calculate ΔHsub for the process solid → gas.

If 7.50 g of the solid sublimes completely, calculate the heat absorbed in kJ.

Provide a brief critique of the assumption that ΔHsub = ΔHfus + ΔHvap can be formed by directly adding values measured at different temperatures. What additional information would be needed to refine the result?

9.7.10 Qualitative Phase-Diagram Analysis: Does It Sublime at 1 atm?

A hypothetical substance has the following qualitative phase-diagram features:

The triple point pressure is 2.5 atm.

The vaporization curve exists only for P > 2.5atm.

At 1 atm, the solid–gas boundary (sublimation curve) is crossed at T = 45∘C.

Answer the following:

If the substance is a solid at 20 ∘C and 1 atm, what phase change occurs upon heating at constant 1 atm? At what temperature does it occur?

If the substance is held at 60 ∘C and compressed from 1 atm to 5 atm, describe the likely sequence of phase changes (if any). State whether a liquid phase must appear at some point and justify using the triple point condition.

Explain, in one paragraph, how the location of the triple point determines whether “boiling” is a meaningful concept at 1 atm for this substance.

Chapter 10
Properties of Solutions

From the saline composition of blood plasma to the antifreeze in an engine, solutions are ubiquitous in both biological systems and industrial applications. This chapter transitions from pure substances to mixtures, exploring how the introduction of a solute fundamentally alters the physical properties of a solvent. By applying the principles of intermolecular forces and thermodynamics, we uncover the quantitative laws governing solubility and the colligative behaviors that depend solely on the number of solute particles present.

10.1 The Solution Process

A solution is a homogeneous mixture at the molecular (or ionic) scale. Forming a solution from a pure solute and a pure solvent is not merely “mixing” in the everyday sense; it is a thermodynamic process governed by (1) the energetic cost of disrupting existing intermolecular attractions and (2) the energetic benefit (and entropy gain) of creating a new, more disordered arrangement of particles.

In earlier chapters, we used thermodynamics to predict spontaneity through the Gibbs free energy:

This same criterion determines whether dissolution is thermodynamically favorable under specified conditions. In this section, we analyze solution formation by separating it into physically meaningful steps, then connecting these steps to ΔHsoln, ΔS, and ultimately ΔG. We then use intermolecular-force ideas to rationalize the empirical rule “like dissolves like” and to distinguish ideal and non-ideal solutions.

10.1.1 What Does It Mean to “Dissolve”? A Molecular Picture

At the particle level, dissolution involves converting two ordered phases (pure solute and pure solvent) into one mixed phase where solute particles are dispersed and stabilized by interactions with solvent particles. The key microscopic event is solvation: the surrounding of solute particles by solvent molecules.

If the solvent is water, solvation is often called hydration.

For ionic solutes, solvation usually involves strong ion–dipole attractions.

For molecular solutes, solvation can involve dipole–dipole, hydrogen bonding, and dispersion forces.

A solution forms spontaneously when the overall free-energy change for moving from the separated state to the mixed state is negative:

10.1.2 A Three-Step Thermodynamic Cycle for Solution Formation

It is useful to decompose the enthalpy change for making a solution, ΔHsoln, into three conceptual steps. These steps are not necessarily separate events in time; rather, they form a thermodynamic cycle that helps us track energy changes.

Step 1: Separate solute particles (endothermic)

To dissolve, the solute must be “pulled apart” from itself:

Ionic solids: disruption of the ionic lattice (related to lattice energy).

Molecular solids/liquids: disruption of intermolecular attractions (dispersion, dipole–dipole, hydrogen bonding).

This step requires energy:

Step 2: Separate solvent particles (endothermic)

Solvent molecules must create cavities or spaces to accommodate the solute. This requires partially overcoming solvent–solvent attractions:

Step 3: Mix and form solute–solvent interactions (often exothermic)

When the solute becomes surrounded by solvent molecules, new attractive interactions form (ion–dipole, hydrogen bonding, etc.). This releases energy:

Net enthalpy of solution

Summing the steps gives:

The sign and magnitude of ΔHsoln therefore depend on competition between:

Energy required to disrupt solute–solute and solvent–solvent attractions (ΔH₁ + ΔH₂),

Energy released when solute–solvent attractions form (ΔH₃).

10.1.3 Enthalpy Is Not the Whole Story: The Central Role of Entropy

A common misconception is that dissolution is favorable only when ΔHsoln < 0. In reality, many dissolutions occur even when ΔHsoln > 0 because the entropy term can dominate:

Why does ΔSsoln often increase?

Mixing typically increases the number of accessible microstates (ways to distribute particles in space and energy). When solute particles disperse among many solvent particles, the mixed state is usually more disordered than the separated state:

However, ΔSsoln is not guaranteed to be positive. Two important countereffects can reduce entropy:

Solvent structuring: If solvent molecules become more ordered around the solute (e.g., strong hydration shells, hydrophobic hydration), entropy can decrease locally and sometimes overall.

Association/ion pairing: If dissolved species cluster (e.g., contact ion pairs), the effective number of independent particles decreases.

Thus, predicting the sign of ΔSsoln requires a molecular view of both solute dispersion and solvent reorganization.

Temperature as a “lever”

Because ΔG = ΔH − TΔS, temperature changes the balance:

If ΔHsoln > 0 and ΔSsoln > 0, dissolution becomes more favorable at higher T.

If ΔHsoln < 0 and ΔSsoln < 0, dissolution becomes less favorable at higher T.

This thermodynamic reasoning anticipates the temperature dependence of solubility developed later in the chapter.

10.1.4 Intermolecular Forces and “Like Dissolves Like”

The phrase “like dissolves like” summarizes a physically precise idea: dissolution is favored when the new solute–solvent interactions (Step 3) are comparable in strength and type to those disrupted in Steps 1 and 2.

Polar and ionic solutes in polar solvents

Water is highly polar and forms strong hydrogen bonds. It dissolves:

Ionic compounds when ion–dipole interactions are strong enough to compete with lattice energy.

Polar molecular solutes capable of strong dipole–dipole interactions and/or hydrogen bonding with water.

Example: hydration of ions. For an ionic solute such as NaCl(s), dissolution can be represented as:

The notation (aq) emphasizes that ions are stabilized by hydration shells. Oxygen atoms in water orient toward Na+, while hydrogen atoms orient toward Cl– , reflecting electrostatic attraction.

Nonpolar solutes in nonpolar solvents

Nonpolar solvents (e.g., hexane) cannot provide strong ion–dipole or hydrogen-bond interactions. They dissolve solutes primarily through dispersion forces. Nonpolar solutes (oils, many hydrocarbons) dissolve well because:

Steps 1 and 2 involve disrupting dispersion-dominated attractions,

Step 3 replaces them with similar dispersion interactions between solute and solvent.

The thermodynamic meaning of poor solubility

Poor solubility often corresponds to one (or both) of these conditions:

ΔHsoln is too positive because solute–solvent attractions are much weaker than solute–solute and solvent–solvent attractions.

ΔSsoln is not sufficiently positive (or is negative), often due to strong solvent ordering around the solute.

10.1.5 Ideal versus Non-Ideal Solution Formation

To connect molecular interactions to measurable behavior, it is useful to define an ideal solution as a limiting model.

Ideal solutions: a reference model

In an ideal solution, the average strength of solute–solvent interactions is essentially the same as the average of solute–solute and solvent–solvent interactions. As a consequence:

Mixing causes minimal enthalpy change:

The driving force for solution formation is primarily the entropy of mixing:

Ideal behavior is most closely approached by mixtures of chemically similar, nonpolar liquids (e.g., two hydrocarbons of comparable size), where interactions are all dispersion-like and similar in magnitude.

Non-ideal solutions: when interaction strengths differ

When solute–solvent interactions differ significantly from the original interactions, ΔHmix can be noticeably nonzero:

Stronger solute–solvent attractions than in the pure components ⇒ ΔHmix < 0 (exothermic mixing).

Weaker solute–solvent attractions ⇒ ΔHmix > 0 (endothermic mixing).

Non-ideality has measurable consequences. Later in this chapter, deviations from ideality will appear in vapor-pressure behavior (Raoult’s law deviations) and in colligative-property measurements when solute particles associate or dissociate.

10.1.6 Connecting the Molecular Picture to Calorimetry: ΔHsoln and Heat Effects

Because ΔHsoln refers to an enthalpy change at (approximately) constant pressure, it directly relates to heat flow:

for dissolution carried out under conditions where the only significant process is solution formation.

If ΔHsoln < 0, dissolution is exothermic; the solution warms and heat flows to the surroundings.

If ΔHsoln > 0, dissolution is endothermic; the solution cools and heat flows from the surroundings into the system.

Importantly, an endothermic dissolution can still be spontaneous if the entropy gain is large enough that ΔGsoln < 0.

10.1.7 Summary: A Predictive Framework for Dissolution

The formation of a solution can be analyzed by combining thermodynamics with molecular interactions:

Three-step enthalpy cycle:

Solute separation and solvent separation cost energy; solute–solvent attraction formation often releases energy.

Spontaneity criterion:

Even when ΔHsoln > 0, dissolution can proceed if TΔS is sufficiently positive.

Intermolecular-force matching: “Like dissolves like” reflects whether Step 3 can energetically compensate Steps 1 and 2 without imposing excessive ordering (entropy penalty) on the solvent.

Ideal vs non-ideal behavior: Ideal solutions have ΔHmix ≈ 0; non-ideal solutions have significant enthalpic contributions due to interaction-strength differences.

These ideas provide the foundation for quantifying solutions. Next, we develop concentration units and then examine how temperature, pressure, and molecular structure determine solubility limits and the colligative properties of solutions.

10.2 Concentration Units

The thermodynamic analysis of dissolution developed in the previous section explains why solutions form. To use solutions quantitatively—in stoichiometry, equilibrium, kinetics, and colligative-property calculations—we must also describe how much solute is present relative to the solvent or the total solution. This section develops the most common concentration units, emphasizes

what each unit physically represents, and highlights which units change with temperature.

10.2.1 Basic Definitions: Solute, Solvent, and Solution Composition

For a solution containing one solute and one solvent:

Solute is the component present in the smaller amount (not necessarily by mass).

Solvent is the component present in the larger amount (and often the phase that defines the solution state).

Two distinct “reference bases” appear repeatedly:

Per volume of solution (depends on solution volume).

Per mass of solvent or per total mass (depends only on mass).

Because volume changes with temperature (thermal expansion), any concentration unit based on volume is typically temperature dependent.

10.2.2 Molarity (M): Amount of Solute per Volume of Solution

Molarity is defined as moles of solute per liter of solution:

Interpretation and common use

Molarity answers the question: How many moles of solute are present in a given volume of solution? It is the standard unit for:

preparing solutions in volumetric glassware,

reaction stoichiometry in solution (e.g., titrations),

many equilibrium expressions that use molar concentrations.

Temperature dependence

Because V solution changes with temperature, so does M:

In most general-chemistry laboratory conditions, this effect is small but conceptually important. When high precision is required (or temperature varies widely), mass-based units are preferred.

Example: preparing a molar solution (conceptual)

To prepare 250.0 mL of 0.1000 molL⁻¹ NaCl(aq), compute moles required:

Then convert moles to mass using molar mass and dilute to the mark in a volumetric flask. The defining feature is that the final volume of the solution is fixed.

10.2.3 Molality (m): Amount of Solute per Mass of Solvent

Molality is defined as moles of solute per kilogram of solvent:

Interpretation and common use

Molality answers: How many moles of solute are present per kilogram of solvent? It is especially important when concentration must remain valid as temperature changes, and in applications where the solvent mass is the natural reference:

colligative properties (ΔTb, ΔTf, osmotic pressure treatments often specify m or depend on mass),

thermodynamic work where temperature varies,

solutions at elevated temperatures.

Temperature independence

Molality is temperature independent because it uses mass, not volume:

Important distinction: solvent vs solution

Molality uses mass of solvent, not mass of solution. This distinction matters most for concentrated solutions.

10.2.4 Mole Fraction (X): A Dimensionless Composition Variable

The mole fraction of component i is:

For a binary solution (solute s and solvent v):

Interpretation and common use

Mole fraction answers: What fraction of all particles (in the mole-count sense) are of a given component? It is widely used in:

thermodynamics of mixtures and phase equilibria,

vapor pressure relationships (e.g., Raoult’s law uses Xsolvent),

expressing composition independent of total amount.

Temperature independence

Mole fraction is temperature independent for a closed system of fixed composition, because it depends only on mole ratios.

10.2.5 Mass Percent: Solute Mass as a Fraction of Total Mass

Mass percent (or percent by mass) is:

Interpretation and common use

Mass percent answers: What percentage of the solution’s mass comes from solute? It is common in:

commercial formulations (acids, bleach, saline solutions),

laboratory stock solutions (especially when density is not fixed or not known),

situations where mass measurements are easier and more accurate than volume.

Temperature independence

Because it is mass-based, mass percent is temperature independent (for a system with fixed composition).

10.2.6 Parts per Million (ppm) and Parts per Billion (ppb): Very Dilute Solutions

For very dilute solutions, it is convenient to use parts per million:

and parts per billion:

Mass-based definition (most common in general chemistry)

In environmental and aqueous contexts, ppm is commonly interpreted as a mass ratio:

Thus:

For dilute aqueous solutions, 1 kg solution ≈ 1 L solution, so a frequently used approximation is:

Similarly, for very dilute aqueous solutions:

These approximations are widely used but should be justified by density and dilution assumptions when precision matters.

Why ppm/ppb are useful

ppm and ppb provide readable numbers for trace concentrations (e.g., contaminants in drinking water) where molarity would be inconveniently small.

10.2.7 Converting Among Concentration Units: What Information Is Required?

Conversions are easiest when you track what the denominator represents (volume of solution, mass of solvent, total moles, total mass). A key practical point is that some conversions require density.

Conversions that typically do not require density

If you start with explicit masses (or moles) of solute and solvent, you can compute:

molality m (need nsolute and msolvent),

mass percent (need msolute and msolution),

mole fractions Xi (need moles of each component).

Conversions that require density

Any conversion involving molarity requires the solution volume. If you do not directly know volume but you know solution mass, then you need density ρ:

Typical examples:

mass percent → molarity (requires ρ),

molality → molarity (requires ρ and molar mass),

ppm → molarity (requires a clear ppm definition and often ρ).

10.2.8 Common Pitfalls and Best Practices

Do not confuse molarity and molality. M uses liters of solution; m uses kilograms of solvent.

Keep track of “solution” vs “solvent”. Mass percent and ppm are typically based on solution mass; molality is based on solvent mass.

State assumptions for ppm/ppb. The shortcut 1 ppm ≈ 1mg L⁻¹ is not universal; it relies on density and dilution.

Use temperature-independent units when temperature varies. For colligative properties and many thermodynamic applications, choose m or Xi rather than M.

10.2.9 Summary of Major Concentration Units

Molarity, M: nsolute∕V solution. Temperature dependent.

Molality, m: nsolute∕msolvent(kg). Temperature independent.

Mole fraction, Xi: ni∕∑ nj. Temperature independent.

Mass percent: msolute∕msolution × 100%. Temperature independent.

ppm, ppb: very dilute ratios; commonly ppm by mass = mg kg⁻¹, and for dilute aqueous solutions approximately mg L⁻¹.

These concentration measures are not interchangeable without additional information (most often density). In subsequent sections, concentration units become the quantitative language for solubility limits, vapor-pressure lowering, and other colligative properties.

10.3 Factors Affecting Solubility

The previous sections established (1) a molecular and thermodynamic picture of why dissolution can be favorable and (2) quantitative ways to express composition. We now focus on solubility, the maximum amount of solute that can dissolve in a given amount of solvent at a specified temperature and (when relevant) pressure. Solubility is an equilibrium property: it reflects a balance between the tendency of solute particles to leave their original phase and the tendency of solvated particles to return (by crystallization, precipitation, or gas escape).

10.3.1 Solubility as an Equilibrium Condition

Consider a solid solute in contact with its solution. At equilibrium, two opposing processes occur at equal rates:

dissolution: solute particles enter solution,

crystallization (precipitation): solute particles leave solution and join the solid.

When the solution reaches this dynamic equilibrium, it is saturated. If less solute is present than the equilibrium amount, the solution is unsaturated. Under special conditions, a solution can temporarily contain more solute than the equilibrium solubility; this unstable state is supersaturated and tends to crystallize when disturbed (seed crystal, scratching glass, etc.).

Because solubility is an equilibrium property, it depends strongly on variables that shift equilibria: molecular structure (intermolecular forces), temperature, and pressure.

10.3.2 Molecular Structure: Intermolecular Forces and the Energetics of Solvation

The “like dissolves like” principle developed in the solution-process section becomes predictive when we tie it to the energetics of separating and solvating particles.

The competition that sets solubility

For a solute to be soluble, solute–solvent interactions must be sufficiently favorable to compensate for:

breaking solute–solute attractions (e.g., lattice energy in ionic solids, hydrogen bonding networks in molecular solids),

disrupting solvent–solvent attractions (e.g., hydrogen bonding in water).

If the energetic penalty is too large and/or the entropy gain is too small, ΔGsoln is not negative enough to permit substantial dissolution, and the solubility is low.

Ionic solutes: lattice energy versus hydration energy

For ionic solids, solubility reflects a balance between:

lattice energy (energy required to separate ions from the crystal),

hydration/solvation energy (energy released when ions are stabilized by solvent).

Small, highly charged ions tend to have:

high lattice energies (strong ion–ion attractions),

also high hydration energies (strong ion–dipole attractions).

Thus, trends in ionic solubility can be subtle: increasing charge or decreasing ionic

radius can either increase or decrease solubility depending on which effect dominates.

Molecular solutes: polarity, hydrogen bonding, and dispersion

For molecular substances, the key structural features are:

Polarity: polar solutes are generally more soluble in polar solvents.

Hydrogen bonding capability: solutes that can donate/accept hydrogen bonds often dissolve well in water (e.g., alcohols, small amines).

Molecular size and shape: for a homologous series, increasing nonpolar surface area usually decreases water solubility because more solvent structuring is required to accommodate hydrophobic regions.

Example (qualitative).

Methanol and ethanol are completely miscible with water because their polar O–H group forms strong hydrogen bonds with water. As hydrocarbon chains lengthen (e.g., butanol, hexanol), the nonpolar region grows and water solubility decreases markedly.

The hydrophobic effect (conceptual)

Nonpolar solutes in water often cause water molecules to adopt more ordered arrangements near the solute surface. This can reduce ΔSsoln and thereby reduce solubility. The term hydrophobic does not mean “afraid of water”; it means the system lowers free energy by minimizing the contact area between water and nonpolar surfaces, often leading to aggregation of nonpolar solute molecules.

10.3.3 Temperature and Solubility of Solids and Liquids

Temperature alters solubility because it changes the relative free energies of the

dissolved and undissolved states. A practical approach is to connect solubility trends to the enthalpy change associated with the dissolution equilibrium.

Dissolution as an equilibrium with a heat term

For a solid solute A:

If dissolving A is endothermic overall (ΔHsoln > 0), adding heat favors the dissolved state, so solubility tends to increase with temperature. If dissolving is exothermic (ΔHsoln < 0), increasing temperature tends to decrease solubility.

This statement is an application of Le Châtelier’s principle in thermal form:

endothermic dissolution: heat is a reactant ⇒ higher T shifts toward products (more dissolved solute),

exothermic dissolution: heat is a product ⇒ higher T shifts toward reactants (less dissolved solute).

Why many solids become more soluble as T increases

Many ionic solids have ΔHsoln > 0 because disrupting the crystal lattice costs substantial energy; thus solubility often increases with temperature. However, this is not a universal rule: some solids dissolve exothermically and become less soluble as temperature rises.

Solubility curves

A solubility curve plots equilibrium solubility versus temperature. Its slope is an empirical fingerprint of the thermodynamics:

positive slope: commonly indicates endothermic dissolution,

negative slope: commonly indicates exothermic dissolution.

Such curves are used to design crystallization and purification steps: a compound with steeply increasing solubility with T can often be purified by dissolving at high T and recrystallizing upon cooling.

10.3.4 Temperature and Solubility of Gases in Liquids

Gas solubility shows a different and often opposite temperature trend compared with many solids.

Why gases usually become less soluble as T increases

Dissolving a gas in a liquid typically:

releases energy (often ΔHsoln < 0) due to attractive interactions between gas molecules and the solvent,

decreases entropy because gas molecules lose translational freedom when confined to the liquid phase (ΔSsoln < 0 is common).

When T increases, the term −TΔS becomes less favorable if ΔS < 0, and the equilibrium shifts toward the gas phase. As a result, gas solubility generally decreases as temperature increases.

Everyday manifestation.

Warm soda goes “flat” faster than cold soda because CO2 is less soluble at higher temperatures.

Practical consequences

Aquatic life: warmer water holds less dissolved O2, stressing fish and other organisms.

Industrial processes: gas absorption (scrubbing) is often more effective at lower temperatures.

10.3.5 Pressure and Gas Solubility: Henry’s Law

For gases, pressure plays a dominant role. Increasing the partial pressure of a gas above a liquid increases the amount of that gas dissolved at equilibrium.

Statement of Henry’s law

At constant temperature, the concentration C of a dissolved gas is proportional to its partial pressure P above the solution:

where k is Henry’s law constant. The value and units of k depend on:

the identity of the gas,

the solvent,

the temperature,

the choice of concentration unit (e.g., molarity, molality, mole fraction).

Interpretation

Henry’s law describes an equilibrium between gas molecules in the vapor phase and gas molecules solvated in the liquid. Raising P increases the frequency with which gas molecules strike the liquid surface and enter solution; equilibrium is restored at a higher dissolved concentration.

Limitations and conditions for validity

Henry’s law is most accurate when:

the gas does not react chemically with the solvent,

the solution is dilute in the gas,

pressures are not so high that non-ideal gas behavior dominates.

If the gas reacts (e.g., CO2 forming carbonic acid in water, or NH3 reacting with water), the apparent solubility can be higher and may deviate from simple proportionality.

10.3.6 Applications: Decompression Sickness and Carbonation

Decompression sickness (“the bends”)

Divers breathe air (or gas mixtures) under elevated pressure. At depth, the partial pressure of N2 is higher than at the surface, increasing dissolved N2 in blood and tissues by Henry’s law. If ascent is too rapid:

ambient pressure drops quickly,

dissolved N2 becomes supersaturated,

N2 bubbles can form in tissues and bloodstream, causing pain and potentially severe injury.

Controlled ascent and decompression stops allow N2 to be eliminated through respiration before bubbles form.

Carbonated beverages

Soft drinks are bottled under elevated CO2 pressure. When the container is opened:

the CO2 partial pressure above the liquid drops abruptly,

equilibrium solubility decreases,

CO2 escapes as bubbles until a new equilibrium is reached.

Cooling increases CO2 solubility, which is why carbonated beverages retain carbonation better at low temperature.

10.3.7 Summary: Predicting Solubility Trends

Solubility is governed by both molecular interactions and thermodynamic driving forces:

Structure and intermolecular forces: solubility increases when solute–solvent interactions effectively replace solute–solute and solvent–solvent interactions without imposing excessive ordering.

Temperature: many solids: solubility often increases with T (commonly endothermic dissolution); gases: solubility generally decreases with T.

Pressure: gas solubility increases with gas partial pressure according to Henry’s law, C = kP, under appropriate conditions.

10.4 Colligative Properties: Vapor Pressure

The previous sections established how solutions form (enthalpy–entropy balance) and how to quantify composition (mole fraction, molality, etc.). We now begin the study of colligative properties: changes in physical properties of a solvent caused by the presence of solute particles. The distinguishing feature is that, in the simplest (ideal) treatment, these changes depend primarily on the number of dissolved particles rather than their chemical identity.

The first colligative property we consider is vapor-pressure lowering. The vapor pressure of a liquid reflects the tendency of molecules to escape into the gas phase; dissolving a solute modifies that tendency by changing the composition of the liquid phase and, in non-ideal solutions, by changing intermolecular interactions.

10.4.1 Vapor Pressure and Dynamic Equilibrium

For a pure liquid solvent, a dynamic equilibrium exists at a fixed temperature:

The equilibrium vapor pressure Psolvent 0 depends strongly on temperature and on the strength of solvent–solvent attractions. A higher vapor pressure indicates that solvent molecules can more readily escape from the liquid surface.

When a solute is added, the system becomes a solution, and we ask: How does the equilibrium vapor pressure of the solvent above the solution compare to that above the pure solvent?

10.4.2 Non-Volatile Solutes: Why Vapor Pressure Decreases

A non-volatile solute has negligible vapor pressure compared to the solvent under the conditions of interest (examples: NaCl, sugar in water). Such a solute contributes essentially no molecules to the vapor phase.

A surface and probability argument (molecular picture)

Evaporation requires that a solvent molecule at the surface escape into the gas phase. In a solution, the surface is shared by solvent and solute particles (or, more precisely, by solvent molecules and solvated species). The presence of solute reduces the fraction of surface sites occupied by solvent molecules, thereby reducing the number of solvent molecules available to evaporate at any instant. Condensation from vapor back to liquid is largely unaffected by the solute because the vapor phase is still primarily solvent molecules.

At equilibrium, the reduction in evaporation rate must be matched by a lower vapor pressure (fewer solvent molecules in the vapor), so that the condensation rate

decreases to match the reduced evaporation rate. The thermodynamic statement of this idea is Raoult’s law.

10.4.3 Raoult’s Law for Ideal Solutions

Statement of Raoult’s law

For an ideal solution, the partial vapor pressure of component A above the solution is proportional to its mole fraction in the liquid:

where:

PA is the partial vapor pressure of A over the solution,

XA is the mole fraction of A in the liquid solution,

PA⁰ is the vapor pressure of pure A at the same temperature.

Vapor-pressure lowering by a non-volatile solute

If the solute is non-volatile, only the solvent contributes significantly to the vapor

pressure. Denoting the solvent by “solv”,

The vapor-pressure lowering is

ΔP | ≡ Psolv 0 − P solv |  |
| = Psolv 0 − X solvPsolv 0 |  |
| = (1 − Xsolv)Psolv 0 |  |

For a solution with one solute s and solvent v,

Thus, vapor-pressure lowering increases as the solute mole fraction increases. This is the first clear example of a colligative dependence: the effect depends on how many solute particles are present relative to solvent particles.

Why mole fraction is the natural variable here

Raoult’s law uses XA rather than molarity M or molality m because vapor pressure is an equilibrium property tied to chemical potential, and the chemical potential of a component in an ideal solution has a logarithmic dependence on mole fraction. In practice, mole fraction is also the composition measure that directly expresses the fraction of particles that are solvent molecules.

10.4.4 Volatile Components: Raoult’s Law for Each Component

A volatile solute has a non-negligible vapor pressure. Many liquid–liquid mixtures fall into this category (e.g., ethanol and water; benzene and toluene). In an ideal binary solution of volatile liquids A and B,

and by Dalton’s law of partial pressures the total vapor pressure is

Because XA + XB = 1, Ptotal is a linear function of composition in the ideal case.

Composition of the vapor phase

The vapor phase is usually enriched in the more volatile component (the one with larger P⁰). The mole fraction of A in the vapor, yA, is:

Even if XA is modest in the liquid, yA can be large if PA⁰ is large. This principle underlies fractional distillation, where repeated vaporization/condensation steps exploit differences in volatility.

10.4.5 Ideal Versus Non-Ideal Solutions: The Role of Intermolecular Forces

Raoult’s law is exact only for ideal solutions, where the interactions are “balanced”:

When solute–solvent interactions differ significantly from the pure-component interactions, the solution is non-ideal and deviates from Raoult’s law.

Activity (conceptual preview)

A rigorous thermodynamic treatment replaces XA with an effective concentration called activity, aA:

where γA is the activity coefficient. In general chemistry, we often focus on the qualitative meaning: deviations arise from non-ideal intermolecular interactions.

10.4.6 Positive and Negative Deviations from Raoult’s Law

Deviations are classified by whether the measured vapor pressure is higher or lower than the ideal prediction.

Positive deviation: weaker A−B interactions

If the attractions between unlike molecules (A−B) are weaker than those between like molecules (A−A, B−B), molecules escape more readily from the liquid:

mixing is often endothermic (ΔHmix > 0),

vapor pressures are higher than Raoult’s law predicts,

the total vapor pressure curve lies above the ideal straight line.

Negative deviation: stronger A−B interactions

If A−B attractions are stronger, molecules are held more tightly in the liquid:

mixing is often exothermic (ΔHmix < 0),

vapor pressures are lower than Raoult’s law predicts,

the total vapor pressure curve lies below the ideal straight line.

Physical interpretation

The sign of deviation directly reflects the molecular forces emphasized earlier in the chapter:

Stronger solute–solvent attractions stabilize the liquid phase and suppress vaporization (negative deviation).

Weaker solute–solvent attractions destabilize the mixed liquid relative to the pure liquids, promoting escape to the vapor (positive deviation).

10.4.7 Colligative-Property Emphasis and the “Particle Count” Idea

For non-volatile solutes in sufficiently dilute, nearly ideal solutions, vapor-pressure lowering is largely controlled by Xsolv, which in turn is determined primarily by the ratio of solute particles to solvent molecules. In this regime:

The chemical identity of the solute matters mainly through how many particles it produces in solution (electrolytes vs nonelectrolytes), a theme we develop more fully for freezing-point depression, boiling-point elevation, and osmotic pressure.

Deviations from ideality become more important at higher concentrations,

where intermolecular-force differences and ion pairing change the effective particle count.

Dissolving a non-volatile solute lowers the solvent vapor pressure because the solvent’s mole fraction in the liquid decreases.

Raoult’s law for an ideal solution: PA = XAPA⁰. For a non-volatile solute, Psolution = XsolvPsolv 0.

For volatile mixtures, each component obeys Raoult’s law in the ideal case and Ptotal = ∑ iXiPi⁰.

Non-ideal solutions deviate from Raoult’s law: positive deviation when unlike interactions are weaker; negative deviation when unlike interactions are stronger.

Vapor-pressure lowering sets the stage for understanding why solutions boil at higher temperatures and freeze at lower temperatures than the pure solvent, which we quantify next.

10.5 Boiling Point Elevation and Freezing Point Depression

In the previous section, we showed that adding a (non-volatile) solute lowers the solvent vapor pressure. That single result has two important consequences for phase changes:

A solution must be heated to a higher temperature than the pure solvent to reach the condition for boiling.

A solution must be cooled to a lower temperature than the pure solvent to reach the condition for freezing.

These effects are colligative: in dilute solutions they depend mainly on the number of dissolved particles, not on their chemical identity. This section develops the quantitative relationships

and explains the meaning and limitations of the van’t Hoff factor i.

10.5.1 From Vapor-Pressure Lowering to Boiling and Freezing Shifts

Boiling point: a vapor-pressure criterion

A liquid boils when its vapor pressure equals the external pressure:

For a solution containing a non-volatile solute, the solvent vapor pressure is lowered at every temperature:

Therefore, to reach the same external pressure Pext, the solution must be heated to a higher temperature than the pure solvent:

The difference is the boiling point elevation:

Freezing point: a chemical-potential criterion

Freezing occurs when the chemical potentials of the solid and liquid phases of the solvent are equal. In the presence of solute, the liquid solvent is stabilized (its chemical potential is lowered) relative to the pure liquid because mixing increases entropy. The solid solvent phase (pure solid solvent) contains essentially no solute, so it is comparatively unaffected. The equality condition is reached only at a lower temperature:

The difference is the freezing point depression:

10.5.2 Quantitative Laws for Dilute Solutions

For sufficiently dilute solutions, experiment shows that both temperature shifts are proportional to solute concentration expressed as molality m (not molarity). Molality is used for two reasons:

it is temperature independent (mass-based),

it naturally expresses “solute particles per fixed amount of solvent,” which aligns with the colligative origin.

Boiling point elevation

For a nonelectrolyte solute (does not dissociate),

where:

m is the molality of the solute, molkg⁻¹,

Kb is the ebullioscopic constant of the solvent, with units ∘Ckg mol⁻¹ (or Kkg mol⁻¹).

The constant Kb depends only on the solvent (and pressure condition, typically 1atm), not on the solute.

Freezing point depression

Similarly, for a nonelectrolyte,

where Kf is the cryoscopic constant of the solvent.

Sign conventions and careful definitions

The two ΔT expressions are written to produce ΔTb > 0 and ΔTf > 0 by definition:

Thus, when applying the formulas, compute ΔT as a positive magnitude and then adjust the final temperature accordingly.

10.5.3 Electrolytes and the van’t Hoff Factor i

Colligative properties depend on the number of dissolved particles. Electrolytes produce more particles by dissociating into ions.

Ideal particle-count expectation

If one formula unit yields ν ions in solution, an idealized treatment predicts ν times the effect. For example:

To incorporate this, we define the van’t Hoff factor i:

For ideal (complete) dissociation, i ≈ ν. Then the colligative equations become:

Real solutions: why i is often smaller than ν

In real electrolyte solutions, especially at moderate to high concentrations, ions do not behave as completely independent particles. The effective particle count can be reduced by:

ion pairing (temporary association of oppositely charged ions),

electrostatic screening and non-ideal interactions that cause departures from ideal-solution assumptions.

As a result, measured values of i are frequently less than the “stoichiometric” value ν, and i may vary with concentration.

Weak electrolytes

For weak acids and bases (e.g., CH3CO2H), dissociation is incomplete, so i may be only slightly greater than 1 at typical concentrations. In such cases, i is closely tied to the dissociation equilibrium, linking colligative properties with chemical equilibrium concepts.

10.5.4 Phase Diagram Interpretation: Why Freezing Decreases and Boiling Increases

A phase diagram provides a unified picture of both effects. In a pure solvent, the solid–liquid boundary intersects 1atm at the normal freezing point, and the liquid–vapor boundary intersects 1atm at the normal boiling point. Introducing a non-volatile solute lowers the chemical potential of the liquid solvent, shifting the equilibrium boundaries such that:

the solid–liquid boundary shifts to lower temperature (freezing point depression),

the liquid–vapor boundary shifts to higher temperature (boiling point elevation).

10.5.5 How to Use the Equations in Calculations

Step-by-step strategy

For either boiling point elevation or freezing point depression in a dilute solution:

Convert solute amount to moles, nsolute.

Convert solvent mass to kilograms, msolvent(kg).

Compute molality:

Determine whether the solute is a nonelectrolyte (i = 1) or electrolyte (use an appropriate i, ideally measured or specified).

Apply ΔTb = iKbm or ΔTf = iKfm.

Convert ΔT to the final transition temperature using the definitions of ΔTb or ΔTf.

Units and temperature scale

Because ΔT is a difference in temperature, it has the same numerical value in K and ∘ C:

However, when reporting absolute temperatures, always use a consistent scale and be careful about converting between ∘C and K.

10.5.6 Limits of the Simple Colligative Formulas

The relationships ΔT = iKm are accurate under these typical conditions:

solutions are dilute,

the solute is non-volatile (for the simplest derivations),

the solution behaves nearly ideally with respect to solvent chemical potential,

for electrolytes, the effective particle count is well represented by the chosen i.

At higher concentrations, non-ideality, ion pairing, and changes in solvent structure can cause measurable deviations. In advanced treatments, activities replace concentrations and i is not constant.

Vapor-pressure lowering causes a solution to boil at a higher temperature and freeze at a lower temperature than the pure solvent.

For dilute solutions:

The van’t Hoff factor i corrects for the effective number of dissolved particles; i may differ from the stoichiometric ion count due to incomplete dissociation and ion pairing.

The phase-diagram view unifies both effects: the presence of solute shifts equilibrium boundaries, producing ΔTf and ΔTb.

10.6 Osmotic Pressure

The colligative properties discussed so far—vapor-pressure lowering, boiling point elevation, and freezing point depression—all arise because dissolved solute particles lower the chemical potential of the solvent in the liquid phase. Osmosis is another manifestation of the same driving force, but it is expressed mechanically: a difference in solvent chemical potential across a semipermeable boundary produces a pressure that can be measured and used quantitatively.

10.6.1 Osmosis: The Physical Situation

Semipermeable membranes

A semipermeable membrane allows some species to pass through while blocking others. In the classical osmotic experiment:

the membrane is permeable to the solvent,

the membrane is impermeable to the solute.

Common examples include biological membranes (selectively permeable lipid bilayers with transport proteins) and synthetic membranes used in dialysis and reverse osmosis.

Definition of osmosis

Osmosis is the net flow of solvent through a semipermeable membrane from a region of higher solvent chemical potential to a region of lower solvent chemical potential. For an idealized case of pure solvent on one side and solution on the other, solvent tends to flow into the solution side because the solvent chemical potential is lower there.

Chemical-potential interpretation

The key thermodynamic statement is:

for a solution containing a non-volatile solute at the same T and P. Mixing increases entropy; the solvent in the solution has more available microstates (mixed with solute) and thus a reduced chemical potential.

10.6.2 Osmotic Pressure Π: Definition and Meaning

If solvent is allowed to flow, the solution level (or pressure) on the solution side rises until the driving force for further solvent flow is eliminated. The required pressure difference is the osmotic pressure, denoted Π.

Operational definition

Osmotic pressure Π is the external pressure that must be applied to the solution side to prevent net solvent flow through the membrane at a given temperature:

Thus, Π is not a pressure exerted by the solute alone; it is the pressure required to counterbalance the chemical potential difference caused by solute particles.

Direction of solvent flow

Without applied pressure, solvent flows from lower solute concentration to higher solute concentration (from higher μ to lower μ).

If a pressure Π is applied to the solution, net flow stops.

If a pressure greater than Π is applied, solvent can be forced out of the solution (the basis of reverse osmosis).

10.6.3 The van’t Hoff Equation for Osmotic Pressure

For dilute solutions, osmotic pressure follows an equation formally analogous to the ideal gas law.

Nonelectrolytes

For a dilute solution of a nonelectrolyte (no dissociation),

where:

Π is osmotic pressure (commonly in atm, bar, or Pa),

M is the molarity of the solute (molL⁻¹),

R is the gas constant (choose units consistent with Π and M),

T is absolute temperature (K).

Electrolytes and the van’t Hoff factor

If the solute produces multiple dissolved particles (electrolytes), the effective particle concentration is higher, and we write:

where i is the van’t Hoff factor introduced previously. In ideal dilute solutions, i approaches the number of ions formed per formula unit; in real solutions i can be smaller due to ion pairing and other non-ideal effects.

Why Π behaves like a “gas pressure” (conceptual)

In an ideal gas, pressure reflects the number density of particles and temperature. In dilute solutions, osmotic pressure reflects the tendency of solvent to dilute solute particles, and the magnitude of that tendency scales with:

the number density of solute particles (captured by M and i),

the thermal energy scale RT.

The analogy is not that solute particles are a gas, but that both phenomena depend on particle number and thermal motion in a way that produces a linear, ideal-law relationship in dilute regimes.

10.6.4 Units and Practical Use of Π = iMRT

Unit consistency

A common and convenient unit set is:

which gives Π in atm when M is in molL⁻¹ and T is in K. If Π is measured in Pa, use

and convert M to molm⁻³.

Magnitude of osmotic pressures

Even modest molarities can produce large osmotic pressures. For example, a 0.10 molL⁻¹ nonelectrolyte solution at room temperature has an osmotic pressure on the order of a few atm. This is why osmotic phenomena are biologically and technologically significant (cell swelling/shrinking; desalination).

10.6.5 Osmosis in Biological Contexts: Isotonic, Hypertonic, Hypotonic

Biological systems are particularly sensitive to osmotic pressure because cell membranes act as semipermeable barriers and because many solutes (ions, sugars, proteins) are unevenly distributed between intracellular and extracellular fluids.

Tonicity definitions

Tonicity compares the effective osmotic pressure of a solution to that of another reference solution across a membrane (often the cell membrane). For a cell placed in an external solution:

Isotonic: same effective osmotic pressure as the cell interior ⇒ no net water movement; cell volume remains stable.

Hypertonic: higher effective osmotic pressure outside ⇒ water leaves the cell; the cell shrinks (crenation in animal cells).

Hypotonic: lower effective osmotic pressure outside ⇒ water enters the cell; the cell swells and may lyse (animal cells) or become turgid (plant cells with cell walls).

Important nuance: permeating versus non-permeating solutes

Tonicity depends on solutes that cannot cross the membrane on the relevant timescale. A solute that can permeate will not sustain a long-term osmotic pressure difference by itself, even if it initially contributes to Π. Thus, in physiology, the relevant “iM” is an effective concentration of non-permeating particles.

10.6.6 Determining Molar Mass from Osmotic Pressure

One of the most powerful uses of osmotic pressure is determining the molar mass of large solutes (polymers, proteins) that are difficult to analyze by other colligative methods. Freezing point depression and boiling point elevation often become too small to measure accurately for macromolecules because the required solute molality is limited by solubility and viscosity; osmotic pressure, however, can remain measurable even at low solute concentrations.

Derivation of the working equation

Suppose a mass ms of a nonelectrolyte solute is dissolved to make a solution volume V (in L), and the osmotic pressure Π is measured at temperature T. Then:

Solve for moles:

The molar mass Ms (here using Ms for molar mass to avoid confusion with molarity M) is:

Electrolytes and association

If the solute dissociates or associates, the measured osmotic pressure corresponds to an effective particle concentration. For electrolytes in dilute solution:

and the apparent molar mass derived assuming i = 1 will be too small if dissociation occurs (more particles ⇒ larger Π) and too large if association occurs (fewer particles ⇒ smaller Π). For macromolecules, association (dimerization, aggregation) can be diagnosed by concentration dependence in Π∕M analyses (beyond the scope of this chapter).

Procedure outline (conceptual)

To determine molar mass of a protein by osmometry:

Prepare a dilute protein solution with known mass concentration (e.g., g L⁻¹) and measure solution volume V .

Measure Π at a known temperature T using an osmometer with a membrane that retains the protein but allows water and small ions to pass.

Use Ms = (or an appropriately corrected relation if non-ideal behavior is significant).

10.6.7 Limitations of the van’t Hoff Equation

The relation Π = iMRT is most accurate when:

the solution is dilute (ideal behavior),

the solute does not react with the solvent in a way that changes particle count beyond what i captures,

the membrane behaves as ideally semipermeable (solvent passes freely; solute is fully retained).

At higher solute concentrations, Π often exceeds the ideal prediction due to solute–solute interactions and non-idealities. In more advanced treatments, activities replace concentrations and osmotic coefficients are introduced.

Osmosis is solvent flow across a semipermeable membrane driven by a solvent chemical potential difference.

Osmotic pressure Π is the pressure required to stop net osmosis.

For dilute solutions:

Osmotic pressure provides a practical route to molar mass determination of large solutes:

Biological effects of osmosis are described by isotonic, hypertonic, and hypotonic environments, with cell volume changes determined by effective (non-permeating) particle concentrations.

10.7 Problems

10.7.1 Thermodynamic cycle for dissolution and sign of ΔHsoln

A student dissolves three different solutes in water at 25 ∘C: NH4NO3(s), NaOH(s), and sucrose(s). For each solute, the student observes either a temperature decrease, a temperature increase, or essentially no temperature change (within experimental uncertainty).

For each solute, predict whether dissolution is likely endothermic, exothermic, or near-thermoneutral. Justify your prediction using the three-step enthalpy cycle (ΔH₁, ΔH₂, ΔH₃) and dominant intermolecular interactions (ion–dipole, hydrogen bonding).

Explain how it is possible for a dissolution with ΔHsoln > 0 to be spontaneous at 25 ∘C. In your explanation, state the sign you expect for ΔSsoln and connect it to the molecular picture of mixing and solvation.

Suppose NH4NO3 dissolves with ΔHsoln > 0 and ΔSsoln > 0. Predict qualitatively how its solubility should change with temperature, and explain why.

10.7.2 Concentration-unit conversions with density

A laboratory stock solution of hydrochloric acid is labeled “37.0% HCl by mass, density = 1.19 g mL⁻¹ at 25 ∘C.”

Calculate the molarity M of HCl in this solution at 25 ∘C.

Calculate the molality m of HCl in this solution.

Determine the mole fractions XHCl and XH2O.

If the same solution warms to 35 ∘C and the density decreases to 1.17 g mL⁻¹ (assume mass percent remains 37.0%), compute the new molarity. Explain why molality and mole fraction do or do not change.

10.7.3 Gas solubility, Henry’s law, and decompression

At 20 ∘C, a diver’s tissues can be modeled as an aqueous phase in equilibrium with breathing gas. Assume Henry’s law in the form C = kP applies to dissolved N2 with k = 6.0 × 10⁻⁴ mol L⁻¹atm⁻¹ at 20 ∘C.

Estimate the dissolved N2 concentration in tissues when the diver breathes air at 1.00 atm total pressure (take PN2 = 0.78atm).

Estimate the dissolved N2 concentration at a depth where the total pressure is 4.00 atm (take PN2 = 0.78 × 4.00atm).

If the diver ascends rapidly from 4.00 atm to 1.00 atm, compute the change in equilibrium dissolved N2 concentration predicted by Henry’s law. Discuss why this change can lead to bubble formation (supersaturation).

Qualitatively predict how these concentrations would change if the water temperature increased, and explain your reasoning in terms of gas solubility versus temperature.

10.7.4 Raoult’s law for a non-volatile solute and determination of mole fraction

At 25 ∘C, the vapor pressure of pure benzene is PC6H6 0 = 95.1Torr. A non-volatile solute is dissolved in benzene to form a solution whose measured vapor pressure is 90.3 Torr at the same temperature.

Assuming ideal behavior and a non-volatile solute, calculate the mole fraction of benzene in the solution.

If the solution was prepared by dissolving 2.50 g of the solute in 75.0 g of benzene, determine the molar mass of the solute.

Explain two experimental or conceptual reasons why the molar mass determined in part (2) might be inaccurate (consider non-ideality and possible association/dissociation of solute).

10.7.5 Raoult’s law for two volatile liquids and vapor composition

A binary mixture of two volatile liquids A and B behaves ideally at 30 ∘C.

The vapor pressures of the pure liquids at 30 ∘C are PA⁰ = 120Torr and PB⁰ = 60Torr.

For a liquid mixture with XA = 0.35, calculate PA, PB, and Ptotal.

Calculate the mole fraction of A in the vapor, yA, above this solution.

Determine whether the vapor phase is enriched in A or B relative to the liquid phase, and explain why.

Suppose a single equilibrium vaporization/condensation step is performed (a simple distillation stage). Qualitatively describe how the distillate composition compares to the original liquid composition.

10.7.6 Diagnosing non-ideality: positive vs negative deviations from Raoult’s law

A mixture of liquids A and B shows a measured total vapor pressure at 25 ∘C that is consistently higher than the Raoult’s-law prediction across intermediate compositions. Separately, calorimetry shows that mixing A and B is endothermic.

Identify whether this mixture exhibits a positive or negative deviation from Raoult’s law. Justify using both vapor-pressure behavior and the sign of ΔHmix.

Sketch (conceptually) how the actual Ptotal vs XB curve compares to the ideal straight line.

Provide a molecular-level explanation in terms of relative strengths of A−A, B−B, and A−B interactions.

Predict whether such a mixture is more likely to have a minimum-boiling or maximum-boiling azeotrope, and justify your reasoning qualitatively (no numerical calculation required).

10.7.7 Freezing point depression with electrolytes and ion pairing

A solution is prepared by dissolving 15.0 g of CaCl2 in 200.0 g of water. The freezing point of the solution is measured to be −3.95 ∘C. For water, Kf = 1.86∘Ckg mol⁻¹.

Calculate the molality of CaCl2 in the solution.

Using the measured freezing point, determine the experimental van’t Hoff factor i.

Compare your experimental i to the ideal dissociation value for CaCl2. Propose a molecular explanation for any discrepancy.

If ion pairing reduces the effective number of particles, describe qualitatively how i would be expected to change as the solution becomes more concentrated.

10.7.8 Boiling point elevation and solvent identification

An unknown, non-volatile nonelectrolyte solute is dissolved in an unknown solvent. A

solution is made by dissolving 3.25 g of solute in 75.0 g of solvent. The boiling point of the pure solvent is 78.20 ∘C, and the boiling point of the solution is 78.84 ∘C.

Calculate ΔTb for the solution.

If the molar mass of the solute is known to be 180.16 g mol⁻¹, determine the ebullioscopic constant Kb of the solvent.

Compare your calculated Kb to typical values you may know for common solvents and suggest a plausible identity for the solvent (state your assumptions).

Explain why molality (not molarity) is used in the boiling point elevation equation.

10.7.9 Osmotic pressure and molar mass of a macromolecule

A biochemist dissolves 0.250 g of a protein in water to make 25.0 mL of solution at 298 K. The osmotic pressure is measured as 7.65 mmHg. Assume the protein is a nonelectrolyte and the solution is sufficiently dilute to use the van’t Hoff equation.

Convert the osmotic pressure to atm and calculate the molarity M of the protein.

Determine the molar mass of the protein.

If the protein dimerizes in solution (two monomers associate to form a dimer), describe qualitatively how the measured osmotic pressure would change at the same mass concentration, and how that would bias the molar mass calculation if dimerization were ignored.

10.7.10 Integrated multipart: designing a colligative-property experiment

You are tasked with determining whether an unknown solute S is a nonelectrolyte, a strong electrolyte, or a weak electrolyte in water. You may use freezing point depression, boiling point elevation, and osmotic pressure measurements. You have access to a balance, thermometer (precision ±0.02∘C), an osmometer, and standard volumetric glassware.

Propose an experimental plan that uses at least two different colligative properties to classify S. Specify what measurements you will take and which equations you will apply.

Explain how you would extract an experimental van’t Hoff factor i from each chosen method, and how you would compare those i values to decide whether S is strong, weak, or non-electrolytic.

Identify at least three sources of systematic error or non-ideality (e.g., ion pairing, volatility, concentration dependence, temperature gradients) and explain how each would affect the inferred i.

If S is found to be a weak electrolyte, describe how you could use concentration dependence of i to infer qualitative information about its dissociation equilibrium.

Chapter 11
Chemical Kinetics

While thermodynamics determines the direction of chemical change, it remains silent on the dimension of time. Chemical kinetics bridges this gap, quantifying the speed at which reactants transform into products-from the femtosecond breakage of bonds to the geological timescales of radioactive decay. This chapter provides a rigorous mathematical framework for analyzing reaction rates, deducing rate laws from experimental data, and using temperature dependence to probe the energy barriers of chemical transformations. By connecting macroscopic measurements to microscopic collision theory and reaction mechanisms, we gain the ability to control and optimize chemical processes.

11.1 Reaction Rates

In previous chapters, we emphasized stoichiometry: chemical equations encode the amount relationships between reactants and products. Chemical kinetics asks a different but complementary question: how fast do those stoichiometric changes occur? The central quantitative idea in kinetics is the reaction rate, defined in terms of how concentrations change with time.

11.1.1 Concentration as a Function of Time

Most kinetic measurements follow how the concentration of a species varies during the course of a reaction. For a simple reaction,

the reactant concentration [A] decreases with time, while the product concentration [B] increases with time.

A typical concentration–time curve is not a straight line. Often, the reaction is fastest at the beginning (when reactant concentrations are highest) and slows as reactants are consumed. This curvature is the reason we distinguish average rates (over a finite time interval) from instantaneous rates (at a specific moment).

11.1.2 Average Rate of Reaction

Over a finite time interval from t₁ to t₂, the average rate of disappearance of reactant A is defined as

where Δ[A] = [A]₂ − [A]₁ and Δt = t₂ − t₁.

The minus sign is included because for a reactant, [A]₂ < [A]₁ during forward reaction progress, so Δ[A] is negative. The rate is reported as a positive quantity.

Similarly, the average rate of appearance of product B is

Units. Because concentration is typically in molL⁻¹ (M) and time is in seconds, the most common rate units are molL⁻¹ s⁻¹.

11.1.3 Instantaneous Rate and the Derivative

Average rates are useful but depend on the chosen interval. To characterize the speed of reaction at a particular moment, we define the instantaneous rate. Conceptually, it is obtained by shrinking the time interval until it becomes infinitesimally small.

Mathematically, the instantaneous rate of disappearance of A at time t is the derivative

Likewise, the instantaneous rate of appearance of B is

Geometric meaning: slope of a tangent line. On a plot of concentration versus time, the derivative is the slope of the tangent line at the chosen time. For a reactant, the curve slopes downward, so d[A]∕dt < 0 and −d[A]∕dt > 0. For a product, the curve slopes upward, so d[B]∕dt > 0.

The initial rate. A particularly important instantaneous rate is the initial rate, the rate at t = 0:

Experimentally, initial rates are valuable because early in a reaction the reverse process is often negligible, side reactions may be minimal, and concentrations are known most accurately.

11.1.4 A Unique Reaction Rate and Stoichiometric Normalization

For A → B (with 1:1 stoichiometry), the magnitudes of the rates of disappearance of A and appearance of B are equal:

However, many reactions have coefficients other than 1, and then these derivatives are not numerically equal unless we account for stoichiometry.

Consider the general reaction

The concentration changes are constrained by stoichiometry: when the reaction advances by one “stoichiometric event,” a moles of A are consumed per liter, b moles of B are consumed per liter, and so on. To define a single, consistent reaction rate (sometimes called the rate of reaction), we normalize by the stoichiometric coefficients:

This definition ensures that no matter which species you monitor, you obtain the same reaction rate (within experimental error).

Why the normalization matters (an explicit example). For

stoichiometry requires that NO2 disappears twice as fast (in concentration units) as O2 appears:

The unique reaction rate is therefore

Without the coefficient-based factors, different species would appear to give different “rates” simply because the balanced equation demands different numerical changes in concentration.

11.1.5 Interpreting Rate Data in Practice

Rates are time-dependent. Because slopes of the concentration–time curves typically become less steep as reactants are consumed, both average and

instantaneous rates generally decrease over time for many reactions under fixed conditions. Thus, any reported rate must specify the time interval (for an average rate) or the time point (for an instantaneous rate).

Experimental reality: measured signals and calibration. In practice, one often measures a signal (absorbance, pressure, conductivity, mass loss) that can be related to concentration. Once a calibration or model relates signal to [A] or [B], the definitions above apply unchanged: reaction rates are still expressed as slopes (finite differences for average rates; derivatives for instantaneous rates) of concentration versus time.

A preview of what comes next. This section has defined what a rate is. The next step in kinetics is to determine how the rate depends on concentration (the rate law) and how that dependence reveals information about the reaction process.

11.2 Rate Laws and Reaction Order

The previous section defined the reaction rate as a concentration change per unit time and showed how stoichiometric normalization produces a unique rate for the overall reaction. The next step in kinetics is to determine how that rate depends on the concentrations of reactants. This concentration dependence is summarized by the rate law. Rate laws are not obtained from the balanced chemical equation (except for elementary steps); instead, they are determined experimentally and reflect the underlying molecular mechanism.

11.2.1 From Measured Rates to a Differential Rate Law

Consider a reaction written in overall stoichiometric form,

From Section 11.1, the (stoichiometrically normalized) instantaneous rate is

This expression is a definition of rate in terms of derivatives. Kinetics now posits that, under fixed conditions (especially temperature), the rate is often well described by a product of concentrations raised to powers:

Equation (??) is called the differential rate law (or simply the rate law). The constant k is the rate constant, and the exponents m,n,… are the reaction orders with respect to each reactant.

What the rate law does (and does not) say.

The rate law is an empirical relationship: it summarizes how rate responds to concentration for a given reaction under specified conditions.

The exponents m,n,… are not generally equal to the stoichiometric coefficients a,b,… in the overall balanced equation.

The rate constant k is constant only when conditions such as temperature and solvent are fixed; k changes strongly with temperature (Chapter 11.5).

11.2.2 Reaction Order: Meaning and Interpretation

For a rate law of the form

m is the order with respect to A.

n is the order with respect to B.

The overall order is m + n (for this two-reactant example).

The order quantifies sensitivity. For example, if m = 2, doubling [A] (with everything else fixed) increases the rate by a factor of 2² = 4. If m = 0, changing [A] does not change the rate at all (zero-order in A).

Orders can be non-integers. Reaction orders are often integers (0, 1, 2), but fractional or even negative orders can occur, especially in complex mechanisms (e.g., chain reactions or reactions involving adsorption/inhibition). The rate law captures what is observed; the mechanism explains why the observed orders arise (Chapter 11.6).

11.2.3 Units of the Rate Constant

Because rates have units of concentration per time, the units of k depend on the overall order. For

the overall order is m + n, and the units of k are

Examples:

Overall first order (m + n = 1): [k] = s⁻¹.

Overall second order (m + n = 2): [k] = mol∕d³m⁻¹∕s.

Overall zero order (m + n = 0): [k] = moldm⁻³ s⁻¹.

This dependence is a practical diagnostic: once orders are known, the units of k are fixed.

11.2.4 Rate Laws vs. Stoichiometry: A Crucial Warning

The balanced equation tells us net consumption and formation. It does not, by itself, specify how molecules react. Only for an elementary step (a single molecular event) do stoichiometric coefficients directly determine the rate law. For example:

Unimolecular elementary step: A → products implies rate = k[A].

Bimolecular elementary step: A + B → products implies rate = k[A][B].

Termolecular elementary step (rare): A + B + C → products implies rate = k[A][B][C].

Most overall reactions proceed through multiple steps; their rate laws can be more complicated than the overall stoichiometry suggests. Thus, in general chemistry, you should treat the rate law as experimental input unless the problem explicitly states that a step is elementary or provides a mechanism.

11.2.5 Determining Reaction Order Experimentally: The Method of Initial Rates

One common and conceptually clean way to find orders is the Method of Initial Rates. The strategy is to measure the initial rate (the instantaneous rate at t ≈ 0) for a set of experiments in which initial concentrations are systematically varied.

Why initial rates are especially useful

At very early times:

Reactant concentrations are essentially their initial values [A]₀,[B]₀.

Product concentrations are near zero, making reverse reactions or product inhibition less likely.

The rate is often easiest to approximate by the initial slope of a concentration–time curve.

Therefore, initial-rate data tend to reflect the forward reaction rate law with minimal complications.

Core idea: isolate one variable at a time

Suppose the rate law is hypothesized to be

To determine m, compare two experiments in which [B]₀ is held constant while [A]₀ is changed. Taking the ratio of initial rates eliminates k and [B]₀n:

[picture] | = = m. |  |

Then,

A completely analogous comparison (holding [A]₀ constant and varying [B]₀) yields n.

The “doubling” shortcut. If [A]₀ is doubled while all else is constant, then

Thus:

If the rate doubles, m = 1.

If the rate quadruples, m = 2.

If the rate is unchanged, m = 0.

If the rate increases by a factor of , m = .

Real data often deviate slightly because of experimental uncertainty; one typically chooses the simplest order consistent with the data.

A worked experimental pattern (illustrative, not a proof)

Imagine the following initial-rate data at fixed temperature:

Experiment | [A]₀ (mol dm⁻³) | [B] 0 (mol dm⁻³) | Initial rate (mol dm⁻³ s⁻¹)
1 | 0.10 | 0.10 | 2.0 × 10⁻³
2 | 0.20 | 0.10 | 4.0 × 10⁻³
3 | 0.10 | 0.20 | 8.0 × 10⁻³

Compare Experiments 1 and 2: [B]₀ is constant, [A]₀ doubles, and the rate doubles. This indicates first order in A (m = 1).

Compare Experiments 1 and 3: [A]₀ is constant, [B]₀ doubles, and the rate increases by a factor of 4. This indicates second order in B (n = 2).

Thus the empirical rate law consistent with these data is

and the overall order is 1 + 2 = 3.

Finding the rate constant once orders are known

Once m and n are established, solve for k using any experiment:

Using multiple experiments and averaging k (or using regression methods) improves robustness against experimental noise.

A note on consistency. If the calculated k values from different experiments differ substantially beyond expected experimental uncertainty, it may indicate that:

the assumed orders are incorrect,

the measured “initial” rates are not truly initial (e.g., induction periods, mixing delays),

side reactions or reverse reaction contributions are significant,

temperature or other conditions are not constant.

11.2.6 Connecting Rate Laws to the Rate Definition

Because the rate law is a relationship for the reaction rate, it must be consistent with the stoichiometric normalization used to define rate. For example, for

the rate is uniquely

If the experimentally determined rate law is

then the corresponding differential equations for concentration changes are

− | = k[A]²[B], |  |
− | = k[A]²[B], |  |
[picture] | = k[A]²[B]. |  |

These expressions show two complementary viewpoints:

Stoichiometry constrains the relative rates of concentration change of different species.

The rate law specifies how fast the overall reaction proceeds as a function of concentrations.

11.2.7 What You Can Do Once You Have a Rate Law

A differential rate law is the starting point for two major kinetic tasks developed in later sections:

Predicting how concentrations change with time by integrating the rate law (Integrated Rate Laws, Section 11.3).

Interpreting molecular behavior by comparing the observed rate law to proposed mechanisms (Reaction Mechanisms, Section 11.6).

At this stage, the key outcomes are:

Reaction order is an experimentally determined exponent that measures sensitivity of rate to concentration.

The Method of Initial Rates uses controlled concentration changes and rate ratios to extract those exponents cleanly.

Once orders are known, k follows directly from measured rates and carries order-dependent units.

11.3 Integrated Rate Laws

The differential rate laws developed in the previous section relate reaction rate to concentration at an instant:

This form is powerful, but it answers the question How fast is the reaction right now? Many experimental and practical questions instead ask What will the concentration be after some time? Integrated rate laws provide that connection by linking [A] directly to time t.

In this section we derive the integrated rate laws for three common cases—zero, first, and second order in a single reactant—and then show how to use linear plots to determine reaction order from concentration–time data.

11.3.1 General Strategy: Separating Variables

For a reaction in which the rate depends only on the concentration of one reactant A,

we can rearrange to separate the variables:

Integrating from the initial condition [A] = [A]₀ at t = 0 to the condition [A] = [A]t at time t yields an equation relating [A]t and t. The results depend on the value of n.

Important note on scope. These derivations assume that:

the rate law is valid over the time interval considered,

k is constant (fixed temperature and conditions),

the reaction is effectively irreversible over the time scale of the measurement (or that the measured rate law already accounts for any reverse contribution).

More complex situations (multiple reactants, reversible reactions, changing conditions) require additional techniques, but the cases below are foundational and widely applicable.

11.3.2 Zero-Order Integrated Rate Law (n = 0)

A reaction is zero order in A if the rate is independent of [A]:

Separate and integrate:

− d[A] | = k dt, |  |
∫ [A]₀[A]t (−d[A]) | = ∫ ₀tk dt. |  |

This gives

Key features.

A plot of [A]t versus t is a straight line with slope −k and intercept [A]₀.

Zero-order behavior often arises when a catalyst surface is saturated or when a limiting factor other than reactant concentration controls the rate (ideas expanded in the catalysis section).

Because Eq. (??) predicts that [A]t decreases linearly, it will eventually reach zero at t = [A]₀∕k. Physically, the rate cannot remain constant past depletion of reactant, so the model applies only while A is present and the conditions supporting zero order persist.

11.3.3 First-Order Integrated Rate Law (n = 1)

A reaction is first order in A if

Separate variables:

Integrate:

∫ [A]₀[A]t − | = ∫ ₀tk dt, |  |
−[A]₀[A]t | = kt, |  |

so

An equivalent and often-used exponential form is obtained by exponentiating both sides:

Key features.

A plot of ln[A]t versus t is linear with slope −k and intercept ln[A]₀.

First-order decay is characterized by a constant fraction consumed per unit time (not a constant amount). This makes the concentration–time curve exponential rather than linear.

Many unimolecular processes and radioactive decays exhibit first-order kinetics (radioactive decay is explored further in the half-life section and problem set).

11.3.4 Second-Order Integrated Rate Law (n = 2)

A reaction is second order in A (single-reactant form) if

Separate variables:

Integrate:

∫ [A]₀[A]t − | = ∫ ₀tk dt, |  |
−[A]₀[A]t | = kt, |  |

which simplifies to

Key features.

A plot of 1∕[A]t versus t is linear with slope +k and intercept 1∕[A]₀.

Second-order kinetics often arise from bimolecular processes such as dimerization, 2A → products, or from A+B → products when [A] = [B] throughout (a special case discussed in more advanced kinetics).

11.3.5 Diagnosing Reaction Order Using Linear Forms

In many experiments, one collects [A] versus t data and must decide which integrated rate law describes it. The integrated rate laws can be written in linear forms:

Order | Linear form | Diagnostic plot (should be linear)
Zero | [A]t = [A]₀ − kt | [A] vs t (slope = −k)
First | ln[A]t = ln[A]₀ − kt | ln[A] vs t (slope = −k)
Second | = + kt | 1∕[A] vs t (slope = +k)

Practical procedure.

Measure [A] at several times t.

Construct the three candidate plots: [A] vs t, ln[A] vs t, and 1∕[A] vs t.

The plot that is most nearly a straight line indicates the reaction order (within the scope of these models).

Extract k from the slope (with sign and units appropriate to the order).

What "most nearly linear" means. Real data contain noise. In practice, linear regression is used and one compares measures of goodness-of-fit (e.g., R²) and, equally importantly, inspects residuals for systematic curvature. A visually straight plot with random residuals is a strong indicator that the corresponding integrated rate law is appropriate.

11.3.6 Using the Integrated Rate Laws for Prediction

Once the order and k are known, the integrated laws allow direct prediction of concentration at a future time (or the time needed to reach a target concentration).

Solving for time. Each integrated law can be rearranged to solve for t:

Zero order: | t = , |  |
First order: | t = ln, |  |
Second order: | t = . |  |

These forms are particularly useful in environmental persistence calculations, shelf-life estimation, and analysis of reaction completion times.

11.3.7 Conceptual Summary: How Order Shapes the Concentration–Time Curve

Although all three cases describe decreasing [A] with time, the shape encodes the kinetics:

Zero order: [A] decreases by equal amounts in equal times (linear decrease).

First order: [A] decreases by equal fractions in equal times (exponential decay).

Second order: [A] decreases rapidly at first but slows strongly as [A] becomes small; 1∕[A] increases linearly.

These integrated rate laws set the stage for the next major kinetic concept: half-life and how it depends on reaction order.

11.4 Reaction Half-Life

The integrated rate laws derived in the previous section connect concentration directly to time. A particularly useful time scale that can be extracted from an integrated rate law is the half-life, denoted t1∕2. Half-life provides an intuitive measure of reaction speed and, importantly, reveals how kinetic behavior depends on concentration and reaction order.

11.4.1 Definition of Half-Life

For a reaction in which the concentration of reactant A decreases with time, the half-life t1∕2 is defined as the time required for [A] to fall to half its initial

value:

Half-life is not defined by stoichiometry; it is defined by the time dependence encoded in the integrated rate law. Therefore, how t1∕2 depends on [A]₀ varies with reaction order.

A conceptual warning. It is tempting to treat half-life as an intrinsic “constant” of a reaction. That is only true for first-order kinetics. For zero- and second-order reactions, the half-life depends on the initial concentration, so two experiments at different [A]₀ will have different half-lives even at the same temperature.

11.4.2 Half-Life for a Zero-Order Reaction

For a zero-order reaction in A, the integrated rate law is

Set [A]t = [A]₀∕2 and t = t1∕2:

[picture] | = [A]₀ − kt1∕2, |  |
kt1∕2 | = [A]₀ − = . |  |

Thus,

Interpretation.

The half-life is directly proportional to the initial concentration [A]₀.

If you start with twice as much reactant, it takes twice as long to consume half of it, because the reaction consumes a constant amount per unit time.

Units check: for zero order, k has units Ms⁻¹, so [A]₀∕k has units of time.

11.4.3 Half-Life for a First-Order Reaction

For a first-order reaction in A, the integrated rate law can be written as

At t = t1∕2, [A]t = [A]₀∕2:

ln | = ln[A]₀ − kt1∕2. |  |

Use logarithm properties:

ln[A]₀ − ln2 | = ln[A]₀ − kt1∕2. |  |

Cancel ln[A]₀ from both sides:

− ln2 | = −kt1∕2. |  |

Therefore,

Interpretation: why first-order half-life is constant.

The half-life does not depend on [A]₀.

First-order kinetics describe exponential decay: a constant fraction of A is consumed per unit time. Halving is a fractional change, so it always takes the same time.

Units check: for first order, k has units s⁻¹, so 1∕k is a time.

Successive half-lives. Because first-order half-life is constant, the concentration decreases by factors of 2 over equal time intervals:

After n half-lives,

This relationship is widely used to estimate the remaining fraction of a substance after a given time, especially in radioactive decay and in pharmacokinetics.

11.4.4 Half-Life for a Second-Order Reaction

For a second-order reaction in A (single-reactant form), the integrated rate law is

At half-life, [A]t = [A]₀∕2:

[picture] | = + kt1∕2, |  |
[picture] | = + kt1∕2. |  |

Subtract 1∕[A]₀ from both sides:

Therefore,

Interpretation.

The half-life is inversely proportional to the initial concentration.

If you start with a larger [A]₀, the reaction is much faster initially because the rate depends on [A]²; as a result, it reaches half of [A]₀ sooner.

Units check: for second order, k has units M⁻¹ s⁻¹, so 1∕(k[A]₀) has units of time.

11.4.5 Comparing Half-Life Behavior Across Orders

The dependence of t1∕2 on [A]₀ provides a simple diagnostic for reaction order:

Order in A | Half-life t1∕2 | Dependence on [A]₀
Zero | [picture] | Increases linearly with [A]₀
First | [picture] | Independent of [A]₀
Second | [picture] | Decreases as [A]₀ increases

How to use this in experiments. If you can measure half-life at several different initial concentrations (while keeping temperature and other conditions constant), the pattern suggests the order:

Constant t1∕2 across experiments ⇒ first order.

t1∕2 proportional to [A]₀ ⇒ zero order.

t1∕2 inversely proportional to [A]₀ ⇒ second order.

In practice, determining order from half-life alone is less common than using linear plots (Section 11.3), but half-life trends are an excellent conceptual cross-check.

11.4.6 Half-Life and the Meaning of “Fast”

Half-life is a time scale, not a rate. Two reactions can have the same initial rate under certain conditions yet different half-lives if their rates change differently as concentration decreases. The order determines how strongly the rate slows as reactants are depleted:

Zero order: rate is constant until reactant depletion or a regime change.

First order: rate decreases proportionally with [A].

Second order: rate decreases sharply with [A]², so the reaction slows dramatically as [A] becomes small.

Thus, half-life is especially meaningful when paired with an identified rate law: it tells you how the reaction “ages” as it proceeds.

11.4.7 Looking Ahead

Half-life relationships are convenient tools, but they are ultimately consequences of the rate constant k. The next section explains why k is so sensitive to temperature and introduces the Arrhenius equation, which connects kinetic measurements to the concept of an activation energy barrier.

11.5 Temperature and the Arrhenius Equation

The preceding sections showed how reaction rates depend on concentrations (rate laws) and how concentrations change with time (integrated rate laws and half-life). In all of those treatments, the rate constant k was assumed constant. In reality, k is extremely sensitive to temperature: a modest increase of 10 ∘C–20 ∘C can increase the rate of many reactions by a factor of 2 or more. This section explains why temperature affects rates using collision theory and introduces the Arrhenius equation, which provides a quantitative link between k, temperature T, and activation energy Ea.

11.5.1 Why Temperature Matters: A Molecular View

At the molecular level, reactions occur when particles collide and then successfully reorganize their bonds to form products. Temperature affects reaction rates primarily because it changes the distribution of molecular kinetic energies. Higher temperature means higher average kinetic energy and, more importantly, a larger fraction of molecules in the high-energy “tail” of the distribution.

Two key ideas follow:

Increasing T generally increases the frequency of collisions (molecules move faster).

Increasing T increases the fraction of collisions that are effective (more collisions have enough energy to overcome the energy barrier to reaction).

For most reactions, the second effect dominates; this is why rates can be very temperature-sensitive.

11.5.2 Collision Theory: Necessary (but Not Sufficient) Conditions

Collision theory provides a conceptual model for bimolecular reactions (especially in the gas phase) and motivates the factors that appear in more sophisticated

theories.

Collision frequency

For a bimolecular reaction such as

the rate of collision between A and B increases as molecules move faster. As temperature rises, molecular speeds increase, and so does collision frequency. If all collisions caused reaction, then increasing temperature would produce only a modest increase in rate.

Orientation (steric) factor

Even when A and B collide, they may not be oriented in a way that allows bond rearrangement. For many reactions, only certain orientations lead to product formation. This is commonly represented by an orientation factor (or steric factor) p (with 0 < p ≤ 1). Complex molecules often have small p values because correct alignment is rare.

Energy barrier and activation energy

Most crucially, collisions must have sufficient energy to pass an energy barrier associated with breaking old bonds and forming new ones. This barrier is quantified by the activation energy Ea. A collision with kinetic energy smaller than Ea is unlikely to produce products, even if the orientation is favorable.

Effective collisions. A useful summary is:

The last term increases rapidly with temperature, leading to the pronounced temperature dependence of k.

11.5.3 Reaction Coordinate Diagrams and Activation Energy

A reaction coordinate diagram (also called a potential energy profile) depicts how the potential energy of a reacting system changes as reactants transform into products. Although it does not show time, it clarifies the existence of an energy barrier and relates kinetics to energetics.

Activation energy and the transition state

The peak of the curve corresponds to the transition state (also called the activated complex), a high-energy configuration in which bonds are partially broken and partially formed. The activation energy for the forward reaction is

Likewise, for the reverse reaction,

Enthalpy change vs. activation energy

The overall enthalpy change (approximately the energy difference between products and reactants in the diagram) is

It is crucial not to confuse ΔH (a thermodynamic quantity describing energetic favorability) with Ea (a kinetic quantity describing the barrier to reaction). A reaction may be strongly exothermic (ΔH < 0) yet still proceed slowly if Ea is large.

11.5.4 The Arrhenius Equation

Collision theory suggests that rates depend on both how often molecules collide and how many of those collisions have enough energy to overcome the barrier. The Arrhenius equation captures this relationship empirically and is consistent with more advanced theories (transition state theory) in its basic form:

Here:

k is the rate constant for a specific rate law at temperature T.

A is the pre-exponential factor (or frequency factor).

Ea is the activation energy.

R is the gas constant (R = 8.314Jmol⁻¹ K⁻¹).

T is the absolute temperature in kelvins (K).

Interpreting the parameters A and Ea

Activation energy Ea. The exponential term e−Ea∕(RT) strongly controls the magnitude of k. Because it depends on 1∕T, increasing temperature makes the exponent less negative, increasing k (often dramatically). Larger Ea means a stronger temperature dependence.

Pre-exponential factor A. The factor A reflects, in a simplified way, the frequency of collisions and the probability that collisions have favorable orientations and lead toward products. While A can vary with temperature, it typically changes much more slowly with T than the exponential term, so the exponential term usually dominates temperature dependence.

11.5.5 Linear (Logarithmic) Form and the Arrhenius Plot

Taking the natural logarithm of Eq. (??) gives a linear relationship:

This has the form y = b + mx with

Therefore, a plot of lnk versus 1∕T (an Arrhenius plot) should be approximately linear over a temperature range where the mechanism does not change. From the slope, one can determine Ea:

Why the slope is negative. As T increases, 1∕T decreases. Since higher T generally increases k, lnk increases while 1∕T decreases, producing a negative slope.

11.5.6 Two-Point Arrhenius Equation

Often, only two rate constants are available (or two temperatures are measured). Subtract Eq. (??) written at two temperatures T₁ and T₂:

lnk₁ | = lnA −, |  |
lnk₂ | = lnA −. |  |

Subtracting eliminates lnA:

Solving for Ea:

Common pitfalls.

Use kelvins. Temperatures must be in K, not ∘C.

Use consistent units for Ea. If R = 8.314Jmol⁻¹ K⁻¹, then Ea comes out in Jmol⁻¹ (often converted to kJmol⁻¹).

Make sure k₁ and k₂ correspond to the same rate law. The reaction order and rate law form should not change across the temperature range. If the mechanism changes with temperature, the Arrhenius plot may curve.

11.5.7 Connecting Temperature Effects to Rate Laws and Half-Life

The Arrhenius equation describes how k changes with temperature, and k controls both instantaneous rates and integrated time scales:

For a given set of concentrations, increasing k increases the rate because rate = k[A]m[B]n⋯.

Half-life depends on k in all orders, for example:

Thus, as temperature increases and k increases, half-life decreases.

A qualitative summary. Temperature changes typically do not change the reaction order (which is mechanistic in origin) over modest temperature ranges, but temperature often changes k substantially. Therefore, temperature is one of the most effective practical levers for speeding up (or slowing down) a reaction without changing initial concentrations.

11.5.8 Looking Ahead: Mechanisms and Catalysis

Activation energy is not only a parameter in an equation; it represents the barrier that must be crossed along the reaction pathway. The next section uses this perspective to explain catalysis: catalysts increase reaction rates primarily by providing an alternative pathway with a lower activation energy, thereby increasing k at a given temperature without being consumed.

11.6 Reaction Mechanisms and Catalysis

The rate laws and temperature dependence developed in the previous sections describe what a reaction does kinetically: how the rate depends on concentration (reaction order), how concentrations evolve with time (integrated rate laws), and how the rate constant k changes with temperature (Arrhenius behavior and activation energy). In this section we address the underlying question: what molecular sequence of events produces the observed rate law?

A reaction mechanism is a proposed step-by-step molecular pathway by which reactants become products. Mechanisms link macroscopic measurements (rate laws, activation energies) to microscopic events (collisions, bond breaking/forming), and they provide the framework for understanding catalysis, in which reaction rates are increased by providing an alternative pathway with a lower activation energy.

11.6.1 Elementary Steps and the Overall Reaction

Most reactions written as a single balanced equation actually occur through multiple molecular events. Each individual event is an elementary step. The sum of the elementary steps (after canceling species that appear on both sides) must reproduce the overall balanced reaction.

Example: overall reaction vs. steps. Suppose the overall reaction is

A plausible two-step mechanism might be

NO + O2 → NO3 | (step 1) |  |
NO3 + NO →2NO2 | (step 2) |  |

Adding the two steps cancels NO3 (produced in step 1 and consumed in step 2), yielding the overall reaction. Species that cancel are not present in the overall equation, which motivates the concept of an intermediate (Section ??).

11.6.2 Molecularity: Counting Reactant Particles in a Step

The molecularity of an elementary step is the number of reacting particles (molecules, atoms, ions) that must collide (or rearrange) in that step.

Unimolecular: one reactant particle transforms, e.g. A → products.

Bimolecular: two particles collide, e.g. A + B → products or 2A → products.

Termolecular: three particles collide simultaneously, e.g. A + B + C → products (rare in solution and gas phase because three-body collisions are unlikely).

Why molecularity matters. Molecularity controls the form of the rate law for that step: because an elementary step corresponds to a single molecular event, its rate is proportional to the probability that the required reactant particles meet.

11.6.3 Rate Laws for Elementary Steps

A central kinetic rule is:

For an elementary step, the rate law follows directly from the step’s molecularity (stoichiometric coefficients of that step).

Examples:

A → products | : rate = k[A] |  |
A + B → products | : rate = k[A][B] |  |
2A → products | : rate = k[A]² |  |

Contrast with overall reactions. This rule does not apply to the overall balanced equation unless the overall reaction is known to be elementary (uncommon for multi-reactant balanced equations). This explains why experimentally determined orders often differ from stoichiometric coefficients of the overall reaction.

11.6.4 Intermediates and the Mechanistic “Bookkeeping”

An intermediate is a species that is:

produced in one elementary step,

consumed in a subsequent step,

and therefore does not appear in the net (overall) reaction.

Intermediates are not catalysts: they are transient species formed from reactants and ultimately converted into products. In mechanisms, intermediates are essential because they provide a way to break a complex transformation into simpler molecular events.

Detectability. Some intermediates can be observed experimentally (spectroscopy, trapping experiments), while others are too short-lived. Even when an intermediate is not directly observed, its inclusion in a mechanism may be justified if it explains the observed rate law and other experimental facts.

11.6.5 The Rate-Determining Step

In a multi-step mechanism, the overall rate is often controlled by the slowest elementary step. This step is called the rate-determining step (RDS). The RDS functions as a bottleneck: faster steps can occur only as rapidly as material flows through the slow step.

A simple (but common) case

Consider a two-step mechanism:

A + B → I | (slow) |  |
I + C → products | (fast) |  |

If step (??) is slow and therefore rate-determining, then the overall rate is approximately the rate of that step:

This is powerful because the observed rate law can sometimes reveal which step is slow and which reactants participate in it.

Important limitation. The RDS approximation is most straightforward when the slow step is the first step and its rate law contains only stable reactants (not intermediates). When the slow step involves an intermediate, or when a fast pre-equilibrium precedes the RDS, additional analysis is needed to eliminate intermediate concentrations (Section ??).

11.6.6 Mechanism Validity: Necessary Conditions and Common Tests

A proposed mechanism must satisfy multiple constraints. Two are especially

important in general chemistry:

(1) Stoichiometric consistency

When elementary steps are added, intermediates cancel and the overall balanced equation must be recovered exactly. If not, the mechanism cannot be correct.

(2) Kinetic consistency

The mechanism must predict a rate law compatible with experiment. In practice:

Identify the slow (rate-determining) step (or use steady-state/pre-equilibrium methods).

Write the rate law for that step using molecularity.

If the resulting expression contains intermediates, use additional relationships (pre-equilibrium or steady-state) to rewrite the rate law in terms of measurable reactant concentrations.

A mechanism that cannot reproduce the observed orders (within experimental uncertainty) is rejected.

Non-uniqueness. Kinetic data alone do not always yield a unique mechanism; different mechanisms can sometimes produce the same overall rate law. Mechanistic proposals are strengthened by additional evidence (intermediate detection, isotope effects, stereochemistry, product distributions).

11.6.7 Eliminating Intermediates: Pre-Equilibrium and Steady-State Ideas

Many realistic mechanisms require one additional conceptual tool: intermediate concentrations are usually not directly measurable. To connect the mechanism to an experimentally testable rate law, we must eliminate intermediates from the rate expression.

Fast pre-equilibrium preceding a slow step

A common pattern is:

A + B ↔ I | (fast equilibrium) |  |
I → products | (slow) |  |

The rate is controlled by the slow step:

But [I] is an intermediate concentration. If step (??) is at equilibrium, then

so [I] = K[A][B]. Substitution yields

where kobs = k₂K is an effective rate constant under these conditions. The observed rate law appears bimolecular in A and B even though the slow step involves only the intermediate.

Steady-state approximation (conceptual introduction)

Another widely used idea is the steady-state approximation: the concentration of an intermediate remains small and approximately constant because it is formed and consumed at nearly the same rate:

Solving this condition can express [I] in terms of reactant concentrations. While full steady-state derivations can be algebraically intensive, the physical meaning is simple: intermediates do not accumulate; they act as transient carriers of matter along the pathway.

Why this matters for reaction order. Both pre-equilibrium and steady-state reasoning can generate rate laws with non-integer or unexpected orders. Thus, observed reaction orders encode mechanistic information.

11.6.8 Catalysis: Changing the Pathway (Not the Thermodynamics)

A catalyst is a substance that increases the reaction rate without being consumed in the overall reaction. Catalysts work by providing an alternative mechanism with a lower effective activation energy.

Catalyst vs. intermediate

A catalyst differs from an intermediate in one crucial way:

An intermediate is produced then consumed, and does not appear in the overall reaction.

A catalyst is consumed in an early step and regenerated in a later step, so it does not appear in the overall reaction either, but it is present at the start and end of the mechanism.

Thus, catalysts participate in the mechanism but are not used up.

Energetic viewpoint: lowering Ea

From the Arrhenius equation,

a smaller Ea increases k at the same temperature, often dramatically. Catalysts lower the effective barrier by changing the pathway, commonly through:

forming a more reactive intermediate,

stabilizing the transition state,

providing a surface that brings reactants together in a favorable orientation,

enabling multi-step routes with smaller barriers for each step.

Catalysts do not change ΔH or equilibrium. Catalysts change kinetics, not thermodynamics. They do not alter the energies of reactants and products and therefore do not change:

the overall enthalpy change ΔH,

the equilibrium constant K,

the equilibrium position.

A catalyst increases the rates of both forward and reverse reactions by providing lower-barrier pathways in both directions, so equilibrium is reached faster but is not shifted.

11.6.9 Types of Catalysis

Homogeneous catalysis

In homogeneous catalysis, catalyst and reactants are in the same phase (often all in solution). The mechanism typically involves formation of a catalyst–substrate complex and subsequent conversion to products with catalyst regeneration.

General pattern.

A + Cat ↔ A ⋅ Cat |  |
A ⋅ Cat + B → Products + Cat |  |

Such mechanisms can produce rate laws that include catalyst concentration explicitly; increasing [Cat] often increases the reaction rate until other factors become limiting.

Heterogeneous catalysis

In heterogeneous catalysis, catalyst and reactants are in different phases, most commonly a solid catalyst with gas-phase or solution-phase reactants. Many industrial processes (e.g., hydrogenation on metal surfaces) rely on this type.

Key features include:

adsorption of reactants onto the surface,

surface diffusion and reaction at active sites,

desorption of products.

Surface saturation can lead to zero-order behavior with respect to a reactant at high concentration/pressure, because once all active sites are occupied, adding more reactant does not increase the rate.

Enzymatic catalysis

Enzymes are biological catalysts (typically proteins) that exhibit high specificity and enormous rate enhancements. A simplified kinetic scheme is

E + S ↔ ES |  |
ES → E + P |  |

where E is the enzyme and S the substrate. Enzymatic rates can show saturation behavior: at high [S], nearly all enzyme exists as ES, and the rate becomes limited by the turnover step ES → E + P. This again parallels zero-order kinetics in substrate at high concentration, but with strong mechanistic grounding.

11.6.10 Mechanistic Thinking as a Unifying Tool

By this point in the chapter, three layers of kinetic understanding connect:

Rate laws and orders describe how rate responds to concentration (empirical).

Arrhenius behavior and activation energy describe how k responds to temperature (energetic barrier).

Mechanisms explain why those dependencies arise by specifying elementary steps, intermediates, and bottlenecks.

Catalysis fits naturally into this framework: a catalyst is best understood as a mechanistic modification that lowers activation barriers and changes which steps control the overall rate.

Looking ahead. The problems at the end of this chapter will require you to move between these layers: extracting rate laws from data, integrating to predict concentrations, using temperature dependence to infer Ea, and testing whether proposed mechanisms are consistent with observed kinetics.

11.7 Problems

11.7.1 Rates from Concentration–Time Data (Average and Instantaneous)

The reaction A → B is monitored spectroscopically, and the concentration of A is obtained at several times:

t (s) | [A] (m)
0 | 0.800
20 | 0.664
40 | 0.552
60 | 0.459
80 | 0.382
100 | 0.318

Calculate the average rate of disappearance of A over the intervals 0–20 s, 40–60 s, and 80–100 s.

Estimate the instantaneous rate of disappearance of A at t = 40 s using a symmetric secant approximation (use the data at 20 s and 60 s).

Estimate the instantaneous rate at t = 0 s using the best available finite-difference approximation from the table.

For the reaction 2A → B, explain how the numerical value of the reaction rate differs from −d[A]∕dt and compute the reaction rate over 0–20 s

under that stoichiometry.

11.7.2 Method of Initial Rates: Determine a Differential Rate Law

For the reaction

initial-rate experiments at a fixed temperature give the following data:

Experiment | [A]₀ (m) | [B]₀ (m) | Initial rate (m s⁻¹)
1 | 0.100 | 0.100 | 1.20 × 10⁻³
2 | 0.200 | 0.100 | 2.40 × 10⁻³
3 | 0.100 | 0.200 | 4.80 × 10⁻³
4 | 0.150 | 0.300 | 1.62 × 10⁻²

Determine the orders m and n in the rate law rate = k[A]m[B]n.

Determine k (with correct units) using each experiment and comment on consistency.

Using your rate law, predict the initial rate for an experiment with [A]₀ = 0.250 M and [B]₀ = 0.0500 M.

If [A]₀ is tripled while [B]₀ is halved, by what factor does the initial rate change?

11.7.3 Integrated Rate Law Application (First Order Pollutant Decay)

A pollutant X in a lake decays by a first-order process:

At 25 ∘C, k = 3.20 × 10⁻⁶ s⁻¹.

If [X]₀ = 2.50 × 10⁻⁵ M, what is [X] after 7.00 days? (Be careful with units of time.)

How long does it take for [X] to decrease to 1.00 × 10⁻⁶ M?

What fraction of X remains after 30.0 days?

If the temperature increases and k doubles, how do your answers to parts (1)–(3) change qualitatively? Identify which quantities scale directly with k and which scale inversely with k.

11.7.4 Graphical Analysis: Identify Reaction Order from Time-Series Data

The decomposition of a species Y is studied at constant temperature. The following concentration–time data are obtained:

t (s) | [Y ] (m)
0 | 0.500
50 | 0.400
100 | 0.333
150 | 0.286
200 | 0.250
250 | 0.222
300 | 0.200

Determine whether the reaction is zero, first, or second order in Y by testing linearity of [Y ] vs. t, ln[Y ] vs. t, and 1∕[Y ] vs. t. Show your transformed data in a table.

Use the best linear plot to determine k (with units).

Using your fitted k, predict the time at which [Y ] = 0.125 M.

Briefly discuss how you would detect if the reaction order changes at later times.

11.7.5 Half-Life of a Second-Order Dimerization Reaction

A solute dimerizes in solution:

The reaction is second order in A with k = 1.80mol⁻¹ dm⁻³ min⁻¹ at a given temperature.

Derive (from the second-order integrated rate law) an expression for the half-life t1∕2 in terms of k and [A]₀.

Calculate t1∕2 for [A]₀ = 0.250 M.

Calculate t1∕2 for [A]₀ = 0.0500 M.

Explain, in words, why the half-life increases as [A]₀ decreases for second-order kinetics.

11.7.6 Arrhenius Two-Point Calculation: Determine Activation Energy

For a reaction following the same rate law over the temperature range studied, the rate constants are measured as:

Use the two-point Arrhenius equation to calculate the activation energy Ea (in kJmol⁻¹).

Using your Ea, estimate the pre-exponential factor A (with units) using either data point.

Estimate k at T = 308.0K using the Arrhenius equation.

Briefly describe one experimental reason why an Arrhenius plot might deviate from linearity.

11.7.7 Mechanism Validity: Consistency with an Observed Rate Law

The overall reaction is

Experimentally, the rate law is

A proposed mechanism is:

NO + O2 ↔ NO3 | (fast equilibrium) |  |
NO3 + NO →2NO2 | (slow) |  |

Show that the mechanism sums to the overall reaction.

Derive the rate law predicted by this mechanism using the fast pre-equilibrium assumption and the slow-step rate law.

Determine whether the predicted rate law matches the experimental rate law. If not, identify the discrepancy.

Propose a minimal change (e.g., change which step is slow, add a step, or modify the equilibrium) that could plausibly produce the observed dependence on [NO] and [O2], and justify your proposal qualitatively.

11.7.8 Radioactive Decay and Carbon-14 Dating (First-Order Kinetics)

Carbon-14 decays by first-order kinetics with a half-life of 5730 years.

Derive the relationship between the first-order rate constant k and the half-life t1∕2 and compute k for 14C (in year⁻¹).

A sample of charcoal contains 18.0% of the 14C activity of a living sample (assume activity is proportional to 14C concentration). Estimate the age of the charcoal.

How many years would it take for the activity to drop to 1.00% of its original value?

Discuss briefly one chemical or physical factor (not mathematical) that could cause a radiocarbon age estimate to be inaccurate.

11.7.9 Catalyst Effect via a Change in Activation Energy

A catalyst changes the activation energy of a reaction but does not significantly change the pre-exponential factor A. For the uncatalyzed reaction, Ea,u = 78.0kJmol⁻¹. With catalyst, Ea,c = 58.0kJmol⁻¹. Assume both follow Arrhenius behavior with the same A.

Derive an expression for the ratio kc∕ku at temperature T in terms of Ea,u, Ea,c, R, and T.

Calculate kc∕ku at T = 298.0K.

Calculate kc∕ku at T = 350.0K.

Explain why the catalytic advantage (ratio kc∕ku) typically decreases as temperature increases, even though both rate constants increase with temperature.

11.7.10 Advanced Challenge: Steady-State Approximation with a Reversible First Step

Consider the mechanism:

A + B ↔ I | k₁, k₋₁ |  |
I + B → P | k₂ |  |

where I is an intermediate and P is product. Assume that P does not significantly react back to form I on the time scale considered.

Write the differential equation for d[I]∕dt in terms of [A], [B], and [I].

Apply the steady-state approximation (d[I]∕dt ≈ 0) to solve for [I] as a function of [A] and [B].

Using rate = d[P]∕dt = k₂[I][B], derive the predicted overall rate law in terms of [A] and [B] only.

Analyze limiting behavior: simplify your rate law for (i) low [B] and (ii) high [B], and interpret the apparent reaction orders in each limit.

Chapter 12
Chemical Equilibrium

Contrary to the static implication of a finished reaction, chemical systems often settle into a state of dynamic balance where forward and reverse processes occur at equal rates. This chapter formalizes this state through the Law of Mass Action, bridging the gap between reaction kinetics and thermodynamic stability. We will rigorously define the equilibrium constant, explore its dependency on temperature, and develop systematic methods for calculating the composition of reaction mixtures, ultimately providing the tools to control chemical outcomes in both industrial and biological contexts.

12.1 Dynamic Equilibrium

In earlier chapters, we treated many reactions as proceeding essentially “to completion,” meaning that (given the initial amounts) the limiting reactant is consumed and products accumulate until the reaction stops for practical purposes. That model is appropriate when the reverse reaction is negligibly slow under the conditions of interest, or when products are continuously removed. However, a large class of reactions are reversible: as products form, they can react to regenerate reactants. In a closed system, such reactions do not simply stop; instead, they approach a state called chemical equilibrium.

12.1.1 Reversible reactions and the meaning of equilibrium

A reversible reaction is written with a double arrow to emphasize that both directions are possible:

Here, A and B are reactants in the forward direction, while C and D are products in the forward direction (and reactants in the reverse direction). The coefficients a,b,c,d are stoichiometric coefficients.

A common misconception is that equilibrium means “the reaction has stopped.” In fact, equilibrium is not defined by the absence of reaction, but by a balance of reaction rates. At equilibrium:

Because the two rates are equal, there is no net change in composition, so concentrations (or partial pressures) become constant in time. But at the molecular level, chemical events continue: bonds break and form in both directions at the same rate. This is why equilibrium is described as dynamic rather than static.

12.1.2 A kinetic definition: equilibrium as synchronized forward and reverse rates

To make the kinetic picture concrete, consider the reversible isomerization

In a simple kinetic model, each direction has its own rate law:

rateforward | = kf[A], |  |
ratereverse | = kr[B], |  |

where kf and kr are rate constants (with units of s⁻¹ for this first-order example), and square brackets denote molar concentration.

Suppose a closed container initially contains only A. At t = 0:

[A] is large and [B] = 0,

rateforward = kf[A] is significant,

ratereverse = kr[B] = 0.

As time passes, A is converted to B. Two things happen simultaneously:

[A] decreases, so the forward rate decreases.

[B] increases, so the reverse rate increases.

Eventually, the system reaches a point where the two rates become equal. Beyond this point, A continues converting to B and B continues converting back to A, but the rates match. Therefore, [A] and [B] stop changing: the system has reached equilibrium.

This kinetic definition is general. For a reaction with multiple reactants and products, the functional forms of the rate laws may be more complicated, but the equilibrium condition is always the equality of forward and reverse rates under fixed conditions (especially temperature).

12.1.3 Static vs. dynamic equilibrium (and why chemistry uses the dynamic idea)

It is helpful to contrast chemical equilibrium with a truly static situation:

Static equilibrium: no motion and no ongoing processes. For example, a book resting on a table is “in equilibrium” because forces balance, but nothing is happening microscopically that changes the book into something else.

Dynamic equilibrium: opposing processes occur simultaneously at equal rates, producing no net change. For example, in a sealed container of liquid water, evaporation and condensation continue; the liquid level stays constant once the two rates match.

Chemical equilibrium is dynamic because molecules continue to collide and react. The macroscopic constancy (no change in concentration with time) is the observable consequence of balanced microscopic processes.

12.1.4 Concentrations at equilibrium are constant, but not necessarily equal

Another common misconception is that equilibrium implies “equal amounts of reactants and products.” Equilibrium does not mean equal concentrations; it means unchanging concentrations.

In the isomerization A ↔ B, equilibrium might favor A (so [A]eq ≫ [B]eq), or favor B, or yield comparable amounts. Which outcome occurs depends on the relative tendencies of the forward and reverse processes under the given conditions. Later in this chapter, we will quantify this tendency using the equilibrium constant K and connect it to thermodynamic stability.

12.1.5 Approach to equilibrium: what you would observe in the laboratory

Imagine monitoring a reversible reaction in a closed flask at constant temperature. Two kinds of experimental observations are typical:

Concentration vs. time: Reactant concentrations typically decrease from their initial values and then level off. Product concentrations typically increase and then level off. The leveling off reflects the disappearance of net change, not the stopping of reaction.

Rate vs. time: The forward rate usually begins large (if reactants are initially abundant) and decreases as reactants are consumed. The reverse rate typically begins small (often zero if no products are initially present) and increases as products accumulate. At equilibrium, the two curves meet

and coincide: both rates are equal and constant.

These ideas are summarized in Figure, which illustrates both concentration-time behavior and the corresponding rate-time behavior.

Figure 12.1: Concentration–time (top) and rate–time (bottom) behavior for a generic reversible reaction A ↔ B approaching dynamic equilibrium. Concentrations plateau when the forward and reverse rates become equal. The plateau values need not be equal.

12.1.6 What changes when conditions change?

Equilibrium is defined for a particular set of conditions. If temperature is held constant, a closed reacting system will approach a definite equilibrium state. If the temperature is changed, the rate constants kf and kr change (often substantially), and the equilibrium state generally changes as well. If concentrations are disturbed (for example, by adding more reactant), the forward and reverse rates will no longer match immediately, and the system will respond by evolving toward a new equilibrium state.

These responses are the subject of later sections: the equilibrium constant provides the quantitative description of the equilibrium state, and Le Châtelier’s principle provides a qualitative prediction of how equilibrium shifts under imposed stresses. For now, the essential point is:

Chemical equilibrium is a dynamic state characterized by equal forward and reverse reaction rates, producing constant (not necessarily equal) concentrations.

12.2 The Equilibrium Constant

The previous section defined dynamic equilibrium in kinetic terms: at equilibrium the forward and reverse reaction rates are equal, so concentrations (or partial pressures) no longer change with time even though molecular reactions continue. This raises a quantitative question: What is special about the equilibrium composition? For a given reaction at a fixed temperature, different starting mixtures can lead to the same equilibrium relationship among concentrations. That relationship is summarized by the equilibrium constant.

12.2.1 The Law of Mass Action and the equilibrium expression

Consider the general reaction

Experiment shows that, at a fixed temperature, the equilibrium concentrations satisfy a specific ratio:

This statement is called the Law of Mass Action. The subscript c in Kc indicates that the expression uses molar concentrations.

Several features of Eq. (??) are essential:

Only equilibrium values belong in the definition of Kc. The ratio formed from nonequilibrium concentrations is not Kc; it is the reaction quotient Qc (introduced in the next section). At equilibrium, Qc = Kc.

Exponents come from stoichiometric coefficients. If the balanced equation contains c moles of C per reaction event, then [C] appears raised to the c power.

K is constant for a given reaction at a given temperature. Changing initial concentrations changes the equilibrium composition, but not the value of K (provided the temperature is unchanged).

The magnitude of K indicates the extent of reaction. If Kc ≫ 1, equilibrium favors products (the numerator tends to be large relative to the denominator); if Kc ≪ 1, equilibrium favors reactants. This interpretation is qualitative; it must be used with care because the expression includes powers and may involve multiple species.

12.2.2 A kinetic motivation for the equilibrium form (a simple case)

While the Law of Mass Action is best grounded in thermodynamics (via activities and minimization of Gibbs free energy), it is instructive to see why an equilibrium ratio naturally arises from the kinetic picture introduced previously.

For the elementary reversible reaction

a simple rate model is

rf | = kf[A], |  |
rr | = kr[B]. |  |

At equilibrium, rf = rr, so

Thus the equilibrium constant emerges as a ratio of rate constants for an elementary step. For complex reactions (multi-step mechanisms), the experimentally observed K still exists and is still constant at fixed T, but the connection to individual rate constants is more subtle. The important takeaway is that a fixed equilibrium ratio is a natural consequence of the equilibrium condition “forward rate equals reverse rate.”

12.2.3 Units and the meaning of “constant”

In introductory chemistry, Kc is often computed from concentration ratios as in Eq. (??). Because concentrations carry units (typically molL⁻¹), a naive dimensional analysis might suggest that Kc should have units such as (molL⁻¹)Δν, where

In rigorous thermodynamics, equilibrium constants are defined in terms of activities (dimensionless effective concentrations), which makes K dimensionless. In general chemistry calculations, we proceed with concentration forms, treating Kc as a numerical constant appropriate to the stated standard-state conventions. What matters operationally is consistency: use the same concentration units throughout the expression.

12.2.4 Writing correct equilibrium expressions

Use the balanced chemical equation exactly as written

The equilibrium constant depends on how the reaction is written. For example, for

the equilibrium constant is

If the equation is written in reverse,

then the constant is the reciprocal:

If all coefficients are multiplied by a factor n, the new equilibrium constant is raised to the nth power. These transformations follow directly from exponent rules.

Sum of reactions corresponds to multiplication of equilibrium constants

When reactions are added to form an overall reaction, the corresponding equilibrium constants multiply. If

A ↔ B | K₁, |  |
B ↔ C | K₂, |  |

then for the net reaction A ↔ C,

This property foreshadows the thermodynamic interpretation: adding reactions adds their ΔG∘ values, and multiplying K values corresponds to adding lnK values.

12.2.5 From concentration to pressure: defining Kp

For gas-phase equilibria, it is often more convenient to work with partial pressures rather than molar concentrations. For a gaseous reaction

the pressure-based equilibrium constant is defined as

where Pi denotes the partial pressure of species i.

12.2.6 Deriving the relationship Kp = Kc(RT)Δn

To relate Kp and Kc, assume the gas mixture behaves ideally. For an ideal gas,

Divide by V and recognize that ni∕V is the molar concentration [i]:

Substitute Eq. (??) into Eq. (??):

Kp | = |  |
| = ⋅ |  |
| = Kc (RT)(c+d)−(a+b). |  |

Define

to obtain the standard relationship:

Interpreting Δn. Only gaseous species contribute to Δn because Kp is built from gas partial pressures. For example:

For N2O4(g) ↔2NO2(g), Δn = 2 − 1 = 1, so Kp = Kc(RT)¹.

For H2(g) + I2(g) ↔2HI(g), Δn = 2 − 2 = 0, so Kp = Kc.

Temperature matters in two distinct ways. Equation (??) contains T explicitly, but do not confuse this algebraic relationship with the deeper fact that K itself changes with temperature. At a new temperature, Kc and Kp generally have different numerical values than before. Equation (??) tells you how to convert between Kc and Kp at the same temperature for an ideal-gas system.

12.2.7 Heterogeneous equilibria and the exclusion of pure solids and liquids

Many important equilibria involve more than one phase. For example, the thermal decomposition of calcium carbonate is

If one were to follow Eq. (??) mechanically, one might write

But this is not the correct practical form, because the “concentrations” of pure solids (CaCO3(s) and CaO(s)) do not change in the way gas or solute concentrations do. In a heterogeneous equilibrium, pure solids and pure liquids are assigned an activity of 1 (constant under the usual assumptions of constant temperature and pressure). As a result, they do not appear in the equilibrium expression.

For reaction (??), the equilibrium constant is therefore

Physical interpretation. At fixed T, equilibrium is reached when the partial pressure of CO2 in the closed container reaches a specific value. Adding more CaCO3(s) does not change that equilibrium pressure; it only changes how much solid remains once equilibrium is established (provided both solids are present).

Important caution. Solids and liquids are omitted only when they are pure phases. If a substance is dissolved (aqueous), present as a gas, or part of a mixture (e.g., a solution of liquids), it generally does appear in the expression (via its concentration or activity).

12.2.8 Summary of key ideas

The Law of Mass Action states that, at equilibrium and fixed temperature, concentrations satisfy a characteristic ratio defining Kc.

For gas-phase systems, an analogous constant Kp is defined using partial pressures.

Assuming ideal-gas behavior, the two forms are related by

where Δn is the change in moles of gaseous species (products minus reactants).

In heterogeneous equilibria, pure solids and pure liquids are omitted from K expressions because their activities are effectively constant (taken as 1).

12.3 The Reaction Quotient

The previous section introduced the equilibrium constant K as the numerical summary of the equilibrium composition for a given reaction at a fixed temperature. In practice, however, we often encounter reacting systems that are not at equilibrium: a mixture may be freshly prepared, a product may have been removed, a reactant may have been added, or conditions may have been changed. To predict how such a system will evolve, we need a quantity that mirrors the form of K but is computed from the current (instantaneous) composition. This quantity is the reaction quotient, Q.

12.3.1 Definition of the reaction quotient

For the general reaction

the reaction quotient in concentration form is defined as

where the bracketed quantities are the current concentrations, not necessarily equilibrium values.

For gas-phase systems, the analogous pressure form is

computed from the current partial pressures.

Key point. Q is computed from where the system is now; K describes where the system must end up at equilibrium (at that T).

12.3.2 How to use Q: predicting the direction of net change

At a fixed temperature, compare the current value of Q to the equilibrium constant K.

If Q < K: the ratio of products to reactants is too small compared with equilibrium. The system must produce more products to reach equilibrium. The net reaction proceeds in the forward direction (shift right).

If Q > K: the ratio of products to reactants is too large. The system must consume products (form reactants) to reach equilibrium. The net reaction proceeds in the reverse direction (shift left).

If Q = K: the system is already at equilibrium. There is no net change in composition (forward and reverse processes still occur at equal rates).

This comparison is one of the most powerful qualitative tools in equilibrium chemistry because it converts a potentially complex time-dependent problem into a simple inequality.

12.3.3 Why the Q vs. K comparison works (kinetic viewpoint)

The logic of the Q vs. K test is consistent with the dynamic, rate-based definition of equilibrium from the first section.

Consider again a reversible reaction. At any moment, the forward and reverse rates depend on the current composition. For many reactions (and for all elementary steps), the rates increase with increasing concentrations of the reactants for that direction. Therefore:

When products are scarce relative to reactants, the reverse reaction lacks “fuel” and is slow, while the forward reaction is comparatively faster. The net change is toward products. This corresponds to Q being smaller than the equilibrium ratio K.

When products are abundant relative to reactants, the reverse reaction becomes comparatively faster. The net change is toward reactants. This corresponds to Q being larger than K.

At equilibrium, the rates are equal; the product-to-reactant ratio has adjusted itself to the particular value enforced by the energetics at that temperature, summarized by K.

12.3.4 Thermodynamic connection: the sign of ΔG

A deeper reason for the predictive power of Q is its connection to Gibbs free energy, which determines spontaneity at constant temperature and pressure. For a reaction mixture at some composition (not necessarily equilibrium), the Gibbs free energy change for the reaction is

where R is the gas constant and T is the absolute temperature.

At equilibrium, by definition, the system has no driving force for net change, so ΔG = 0. Setting ΔG = 0 in Eq. (??) gives

Substitute this expression back into Eq. (??):

Now the Q vs. K criterion becomes a statement about the sign of ΔG:

If Q < K, then Q∕K < 1, so ln(Q∕K) < 0 and ΔG < 0: the forward direction is spontaneous.

If Q > K, then Q∕K > 1, so ln(Q∕K) > 0 and ΔG > 0: the reverse direction is spontaneous (the forward direction is nonspontaneous).

If Q = K, then ΔG = 0: equilibrium.

This interpretation is valuable because it emphasizes that the “shift” toward equilibrium is not arbitrary; it is the system moving downhill toward a minimum in Gibbs free energy.

12.3.5 Computing Q correctly: practical rules

Use the balanced equation as written

Q must be constructed with exponents matching the stoichiometric coefficients in the balanced chemical equation. If the equation is reversed, the correct Q is the reciprocal.

Omit pure solids and pure liquids

Just as for equilibrium constants, pure solids and pure liquids do not appear in Q expressions for heterogeneous systems because their activities are effectively constant. For example, for

If PCO2 is below the equilibrium value Kp, the net reaction produces CO2. If it is above, CO2 is consumed (forming more solid carbonate), provided the relevant solids are present.

Choose Qc or Qp to match Kc or Kp

Compare like with like:

If you have Kc, compute Qc from concentrations.

If you have Kp, compute Qp from partial pressures.

For ideal gases, you can convert between concentration and pressure forms using Pi = [i]RT and the relationship between Kp and Kc derived previously, but the simplest approach is usually to stay within one framework.

12.3.6 Worked direction predictions (no equilibrium solving)

The following examples illustrate the logic of the Q vs. K test without yet calculating equilibrium concentrations (that is the task of the next section).

Example 1: a gas-phase equilibrium

For the reaction

Suppose at some moment PNO2 = 0.40atm and PN2O4 = 0.80atm. Then

If (at this temperature) Kp = 0.50, then Qp < Kp, so the system has too little product NO2 relative to equilibrium and will undergo net reaction to the right, producing more NO2 until Qp increases to Kp.

Example 2: a solution equilibrium

For

If a mixture is prepared with a large excess of SCN– , then the denominator is large and Qc is small compared with Kc. The prediction is a net shift right (formation of the complex ion) until the ratio reaches the equilibrium value.

12.3.7 Common misconceptions and how to avoid them

Mistaking Q for K. If the system is not known to be at equilibrium, do not set Q = K. Compute Q from current conditions and compare.

Equating “shift right” with “products dominate.” A rightward net change only means more products form than are consumed on the way to equilibrium. Depending on the magnitude of K and the starting amounts, reactants may still remain substantial at equilibrium.

Using incorrect exponents. Exponents come from the balanced equation. If the equation is not balanced, the Q expression is meaningless.

Including pure solids/liquids in Q. Their omission is not a trick; it reflects their constant activity. Including them would incorrectly predict that changing the mass of a solid changes the direction of net change under

fixed T.

The reaction quotient Q is the equilibrium-expression-shaped ratio computed from instantaneous concentrations or partial pressures.

Comparing Q to K predicts the direction of net change:

Thermodynamically,

so the sign of ΔG is determined by whether Q is less than, greater than, or equal to K.

12.4 Calculating Equilibrium Concentrations

The previous sections established three pillars of equilibrium reasoning:

Equilibrium is dynamic: forward and reverse rates are equal.

The equilibrium state at fixed temperature is summarized by an equilibrium constant K (in concentration form Kc or pressure form Kp).

The reaction quotient Q predicts the direction of net change by comparison with K.

We now turn to the central quantitative task in equilibrium chemistry: given initial conditions and a value of K at a stated temperature, calculate the equilibrium composition.

12.4.1 The ICE-table framework

An equilibrium calculation is, at its core, an accounting problem constrained by stoichiometry and the equilibrium constant expression. The most reliable way to organize that accounting is the ICE table:

General setup for a single reaction

For

let the initial concentrations be [A]₀, [B]₀, [C]₀, [D]₀. Introduce a single unknown x representing the extent of reaction in concentration units (typically molL⁻¹) that occurs as the system moves toward equilibrium.

| A | B | C | D
Initial (I) | [A]₀ | [B]₀ | [C]₀ | [D]₀
Change (C) | −ax | −bx | +cx | +dx
Equilibrium (E) | [A]₀ − ax | [B]₀ − bx | [C]₀ + cx | [D]₀ + dx

Sign convention for x. The table above assumes the net change proceeds forward (toward products), so reactants decrease and products increase. If the initial mixture has “too much product” (i.e., Q > K), the net change will be reverse; you can still use the same table by allowing x to be negative. In practice, many students find it clearer to choose the direction of net change using Q and then assign x ≥ 0 in that direction.

Physical constraints. Equilibrium concentrations must be nonnegative. During algebra, discard any root that would make an equilibrium concentration negative, because it is unphysical.

The algorithmic core

Once the ICE table is written, the workflow is always the same:

Use the ICE table to express equilibrium concentrations (or partial pressures) in terms of x.

Substitute those expressions into the equilibrium constant expression.

Solve for x.

Use x to compute equilibrium concentrations (or partial pressures).

Check that Q = K using the computed equilibrium values (and confirm that approximations, if any, were valid).

12.4.2 Decision tree for solving the algebra (overview)

Some equilibrium problems produce simple algebra; others lead to quadratics (or higher-order polynomials). Figure provides a practical decision tree.

Figure 12.2: Decision tree for selecting a mathematical approach in equilibrium calculations after setting up an ICE table.

12.4.3 Choosing the direction of change using Q

Before filling in the “Change” row, determine whether the reaction must proceed forward or reverse to reach equilibrium.

Compute Q from the initial mixture.

Compare with K (at the stated temperature).

Then:

This step prevents sign errors and helps you estimate whether the change should be small or large.

12.4.4 Worked setup: a typical Kc problem leading to a quadratic

Consider

with Kc = 50.0 at a given temperature. Suppose initially

First, Qc = 0 < Kc, so the net change is forward. Set up an ICE table with x ≥ 0:

| H2 | I2 | HI
I | 0.200 | 0.200 | 0.000
C | −x | −x | +2x
E | 0.200 − x | 0.200 − x | 2x

Write the equilibrium expression:

Taking square roots is efficient here because the denominator is a perfect square:

From this point, solve for x, then compute equilibrium concentrations. This example illustrates a general lesson: look for algebraic structure (perfect squares, common factors) before defaulting to the quadratic formula.

12.4.5 The small-x approximation and the 5% rule

Many equilibria in general chemistry have very small K (reactant-favored) or very large K (product-favored). In such cases, the change needed to reach equilibrium may be small relative to the initial amount of a major component. This motivates an approximation that simplifies the algebra.

When is “x is small” plausible?

A common scenario is a reaction with Kc ≪ 1 starting from reactants only. For a simple example,

starting with [A]₀ and [B]₀ = 0 gives:

If Kc is very small, equilibrium should contain mostly A, suggesting x ≪ [A]₀ and thus [A]₀ − x ≈ [A]₀.

Implementing the approximation

In the expression for K, replace terms like [A]₀ − x by [A]₀ when justified. This usually converts a quadratic into a linear equation.

Example pattern (reactant-favored). For

so x ≈ Kc[A]₀.

The 5% rule (validating the approximation)

Approximations must be checked. After solving for x using the simplification, test whether the neglected x is small compared with the quantity it was subtracted from.

If you replaced [A]₀ − x by [A]₀, compute

If this percentage is ≤ 5%, the approximation is typically considered acceptable in general chemistry (your final concentrations will usually be within typical significant-figure expectations). If it exceeds 5%, you should solve the exact equation (often a quadratic) instead.

Why 5%? It is a practical threshold, not a fundamental law. The true error in the final answer depends on the structure of the K expression and how the approximation propagates. The 5% rule provides a simple, conservative check that prevents obviously inaccurate results.

12.4.6 Quadratic equations: when you must solve exactly

When the 5% rule fails, or when K is not extremely small/large, the algebra typically leads to a quadratic equation in x.

General approach

After substituting ICE expressions into K, rearrange into standard form:

Then apply the quadratic formula:

Choose the physically meaningful root by enforcing nonnegative equilibrium concentrations.

Practical advice.

Keep at least 2–3 guard digits during intermediate algebra to avoid rounding error.

Reject roots that make any equilibrium concentration negative.

After selecting a root, recompute Q from the resulting equilibrium concentrations and confirm that Q = K (within rounding).

12.4.7 Method of successive approximations (iterative solving)

Some equilibrium expressions (especially when multiple terms like (a − x) and (b + x) appear in powers) can be cumbersome to solve analytically, even if the underlying equation is only quadratic-like. An alternative is successive approximation (iteration), which is also the conceptual basis for many numerical solvers.

Basic idea

Rewrite the equilibrium equation in the form

choose a reasonable starting guess x₀, and generate improved estimates:

Stop when x changes negligibly between iterations (for example, when |xn+1 −xn| is below a chosen tolerance).

Choosing a starting guess. A good first guess often comes from:

the small-x approximation (even if it is not ultimately valid), or

bounding logic: x cannot exceed the smallest initial reactant amount divided by its stoichiometric coefficient.

Caution. Not every algebraic rearrangement produces stable convergence. In general chemistry, iteration is best used as a computational aid (or as a check) rather than the primary method unless guided carefully.

12.4.8 Special notes on Kp and gas mixtures

If you are given Kp and initial partial pressures, construct the ICE table in partial pressures:

| PA | PB | PC
I | ⋯ | ⋯ | ⋯
C | −νAx | −νBx | +νCx
E | ⋯ | ⋯ | ⋯

The mathematics is identical to concentration problems; only the symbols and units differ.

If you are given Kc but partial pressures (or vice versa), convert consistently using the relationship

and, for ideal gases, Pi = [i]RT. Maintain a single representation throughout the calculation to avoid mixing incompatible quantities.

12.4.9 Checklist for equilibrium concentration problems

Write the balanced chemical equation and the correct K expression.

Determine direction using Q vs. K when needed.

Set up an ICE table with a single variable x.

Substitute equilibrium expressions into K to obtain an equation in x.

Decide: approximation (with 5% check), perfect-square simplification, quadratic formula, or iteration.

Compute equilibrium concentrations/pressures and verify Q = K.

This structured approach turns equilibrium calculations into a repeatable procedure rather than a collection of special tricks. In the next section, we will use this quantitative foundation to predict how equilibria respond qualitatively to external stresses (concentration, pressure/volume, and temperature changes).

12.5 Le Châtelier’s Principle

The previous sections developed the quantitative framework of equilibrium: a system at equilibrium has constant composition because forward and reverse reaction rates are equal; the equilibrium constant K specifies the equilibrium state at a given temperature; and the reaction quotient Q indicates the direction of net change when a system is displaced from equilibrium. In many laboratory and industrial situations, we do not want to compute the full new equilibrium composition. Instead, we often need a rapid, reliable qualitative prediction of how the system responds to a disturbance.

Le Châtelier’s principle provides that prediction.

12.5.1 Statement and interpretation

Le Châtelier’s principle: If a system at equilibrium is subjected to a stress (a change in concentration, pressure/volume, or temperature), the system will shift in the direction that partially counteracts the imposed stress and establishes a new equilibrium.

Two points prevent common misunderstandings:

The principle describes the direction of the net shift, not the magnitude. The amount of shift depends on K and the starting composition.

The system does not “try” to oppose the change; rather, the shift is a consequence of how the disturbance changes Q relative to K. The system then proceeds in the direction that drives Q back toward K.

12.5.2 A unifying viewpoint: disturbances change Q

For a reaction

the reaction quotient is

A stress typically changes one or more concentrations or partial pressures, changing Q instantly. If the temperature is unchanged, K is unchanged. Therefore:

This Q-based logic makes Le Châtelier’s principle precise and prevents memorization errors.

12.5.3 Changes in concentration

Adding reactants or products

If a reactant is added, its concentration increases immediately. Because reactants appear in the denominator of Qc, Qc decreases. Thus Q < K and the net reaction proceeds forward (toward products) until a new equilibrium is reached.

Similarly, adding a product increases the numerator, so Qc increases; typically Q > K and the net reaction proceeds in reverse (toward reactants).

Example. For

Adding SCN– increases the denominator, making Qc smaller. The system shifts right, forming more FeSCN2+ (often observed as a deeper red color).

Removing reactants or products

Removing a product decreases the numerator of Qc, making Qc smaller; the system shifts forward to replace the removed product. Removing a reactant decreases the denominator, making Qc larger; the system shifts in reverse to regenerate reactants.

Important distinction: changing concentration vs. changing amount. In solution, adding water changes concentrations of all dissolved species simultaneously (dilution). A consistent approach is to compute the new concentrations immediately after dilution (before reaction occurs), calculate Q, and compare with K to determine the shift direction.

12.5.4 Pressure and volume changes (gas-phase equilibria)

Pressure/volume changes affect equilibria only when gases are present, because Qp is built from partial pressures. For an ideal gas mixture at fixed T:

and if the volume decreases, the total pressure increases. Equivalently, at fixed T, decreasing V increases all partial pressures proportionally immediately.

The direction of shift is governed by the change in moles of gas:

Compression favors fewer moles of gas

If volume decreases (compression), all partial pressures increase. In the expression

multiplying each Pi by a common factor α > 1 multiplies Qp by αΔngas.

If Δngas > 0, then Qp increases upon compression, so typically Qp > Kp and the system shifts left (toward fewer gas moles).

If Δngas < 0, then Qp decreases upon compression, so typically Qp < Kp and the system shifts right (again toward fewer gas moles, because products have fewer moles).

If Δngas = 0, then Qp is unchanged by uniform compression/expansion; there is no shift due solely to pressure/volume.

Thus, a concise rule emerges (valid at constant T): Increasing pressure (decreasing volume) shifts equilibrium toward the side with fewer moles of gas; decreasing pressure shifts toward more moles of gas.

Pressure section visualization: N2O4/NO2 under compression

A classic demonstration uses the equilibrium

N2O4 is colorless, while NO2 is brown. Here, Δngas = 2 − 1 = +1. Compression increases Qp (because Qp ∝ α⁺¹), so immediately after compression Qp > Kp and the system shifts left, consuming NO2 and forming more N2O4. The observable result is a transient darkening (instant concentration increase) followed by a lightening as the new equilibrium composition contains a higher fraction of colorless N2O4.

12.5.5 Temperature changes: treating heat as a reagent

Temperature is the most subtle “stress” because it changes the value of K itself. Concentration and pressure changes primarily change Q while leaving K unchanged (if T is constant). Temperature changes both the rates and the equilibrium constant.

A practical way to apply Le Châtelier’s principle to temperature is to include heat in the reaction as if it were a reagent:

Exothermic forward reaction (ΔH < 0): heat is a product.

Raising T is like adding a product, shifting equilibrium left. Lowering T shifts right.

Endothermic forward reaction (ΔH > 0): heat is a reactant.

Raising T is like adding a reactant, shifting equilibrium right. Lowering T shifts left.

Why catalysts do not change equilibrium. A catalyst lowers activation energies for both forward and reverse directions, increasing the rates of both. This helps the system reach equilibrium faster but does not change the equilibrium state, because K (a thermodynamic quantity) is unchanged by a catalyst.

12.5.6 Inert gases and common misconceptions

Adding an inert gas at constant volume

In a rigid container (constant V and T), adding an inert gas increases the total pressure but does not change the partial pressures of the reacting gases:

Since ni, T, and V for each reacting species are unchanged, all Pi are unchanged, so Qp is unchanged. Therefore, no shift occurs.

Adding an inert gas at constant pressure

If the container can expand to keep total pressure constant, adding an inert gas increases volume, which decreases the partial pressures of the reacting gases. This is effectively a volume increase. The system will then shift toward the side with more

moles of gas (the same rule as for decreasing pressure).

Changing the amount of a pure solid or liquid

Adding more of a pure solid (or pure liquid) present in a heterogeneous equilibrium does not change Q and therefore does not shift equilibrium. It can, however, affect how much total product can form before a limiting phase is consumed.

12.5.7 A structured procedure for Le Châtelier predictions

For reliable predictions, use the following steps:

Write the balanced equilibrium reaction and (mentally) identify gaseous species and whether the reaction is endo/exothermic (if temperature changes are involved).

Identify the stress: concentration change, pressure/volume change, or temperature change.

Translate the stress into how it changes Q (or how it changes K, for temperature).

Predict the direction of net change needed to restore Q = K.

Le Châtelier’s principle predicts the direction of the net shift when an equilibrium system is disturbed.

Concentration changes alter Q directly; the system shifts to restore Q = K.

Pressure/volume changes affect gas-phase equilibria: compression favors fewer moles of gas; expansion favors more moles of gas; if Δngas = 0, there is no shift.

Temperature changes shift equilibrium in the direction that consumes the added “heat” (treat heat as a reactant for endothermic reactions and a product for exothermic reactions). Temperature changes also change K.

Catalysts affect rates but not the position of equilibrium.

12.6 Problems

12.6.1 Problem 1: Determining Kc from equilibrium concentrations

For the reaction

a sealed 2.00 L vessel at a constant temperature is charged with SO2 and O2. At equilibrium, the measured concentrations are

Calculate Kc for this reaction at this temperature.

If the reaction is written instead as SO3(g) ↔ SO2(g)+O2(g), determine the corresponding equilibrium constant Kc′.

Briefly explain (in one or two sentences) why the numerical value changes when the equation is rewritten.

12.6.2 Problem 2: Converting between Kc and Kp

For the gas-phase equilibrium

the equilibrium constant at T = 350.0K is Kp = 0.145.

Calculate Kc at 350.0 K.

At the same temperature, compute Kp for the reverse reaction 2NO2(g) ↔ N2O4(g).

Indicate the value of Δngas and explain how it affects the relationship between Kp and Kc.

12.6.3 Problem 3: Using Q to predict direction of net change (mixture not at equilibrium)

At T (constant), the reaction

has Kp = 50.0.

A rigid container (constant volume) contains a mixture with

Calculate Qp.

Predict the direction of net change required to reach equilibrium.

Without solving for equilibrium composition, state which partial pressure(s) must increase and which must decrease as equilibrium is approached.

12.6.4 Problem 4: ICE table with quadratic solution (no approximation)

For the reaction

Kc = 0.160 at a given temperature.

A flask is charged initially with [NOBr]₀ = 0.500molL⁻¹ and no products.

Set up an ICE table and derive an equation in x.

Solve for the equilibrium concentrations [NOBr]eq, [NO]eq, and [Br2]eq.

Verify your results by computing Qc from your equilibrium concentrations.

12.6.5 Problem 5: Small-x approximation and the 5% rule

For the dissociation equilibrium

Kp = 1.10 × 10⁻³ at a specified temperature.

A container is prepared with PPCl5 = 1.00atm and PPCl3 = PCl2 = 0 initially.

Use an ICE table in partial pressures to compute the equilibrium partial pressures. Use the small-x approximation if appropriate.

Apply the 5% rule to check whether your approximation is valid.

If the approximation fails, solve the exact equation and compare the results.

12.6.6 Problem 6: Heterogeneous equilibrium (omitting solids) and equilibrium pressure

The decomposition

has Kp = 0.120 at a given temperature.

A rigid, evacuated container is charged with excess CaCO3(s) and a small amount of CaO(s) (both solids present). The container is sealed and brought to the stated temperature.

Determine the equilibrium partial pressure of CO2.

Suppose that, at equilibrium, additional CaCO3(s) is added (without changing T or V ). Predict whether the equilibrium CO2 pressure changes and justify your answer in terms of Qp and Kp.

Now suppose instead that some CO2(g) is removed from the container at equilibrium (solids remain). Predict the direction of net change and the final equilibrium PCO2.

12.6.7 Problem 7: Equilibrium after a sudden concentration change (compute new equilibrium)

For the aqueous equilibrium

assume Kc = 1.8 × 10⁻⁵ at 25 ∘C.

A solution is prepared with [CH3COOH]₀ = 0.200molL⁻¹ and no added H+ or CH3COO– .

Compute the equilibrium concentrations of all species using an ICE table. Clearly state and check any approximation used.

After equilibrium is established, 0.0100 mol of solid sodium acetate (CH3COONa) is added to 1.00 L of the solution (assume complete dissolution and negligible volume change). Immediately after addition (before any shift), compute Qc.

Predict the direction of net change and then compute the new equilibrium concentrations.

12.6.8 Problem 8: Competing initial products (nonzero product initially)

For the reaction

Kc = 1.56 at a particular temperature.

A 1.00 L vessel initially contains

Compute Qc and predict the direction of net change.

Set up an ICE table consistent with your direction prediction and compute the equilibrium concentrations.

Report the equilibrium value of [H2] to three significant figures.

12.6.9 Problem 9: Coupling Q with Le Châtelier: compression and re-equilibration

Consider again

at a fixed temperature where Kp = 0.145.

A mixture is initially at equilibrium in a cylinder with a movable piston:

The piston is suddenly pushed in so that the volume is halved at constant temperature.

Immediately after compression (before any reaction shift), compute the new partial pressures and the new Qp.

Predict the direction of net change.

Let x be the change (in atm) in PN2O4 as the system relaxes to the new equilibrium. Write expressions for the final equilibrium partial pressures in terms of x and solve for the new equilibrium state.

Compare the final PNO2 to the value immediately after compression and explain the trend qualitatively.

12.6.10 Problem 10: Industrial equilibrium scenario (Haber–Bosch loop: K, Q, and removal of product)

Ammonia synthesis occurs via

At a certain temperature, Kp = 6.0 × 10⁻².

A reactor feed contains the following partial pressures at the reactor inlet:

Assume ideal-gas behavior and constant temperature.

Compute Qp at the inlet and predict the direction of net change.

The mixture reaches equilibrium in the reactor. Using an ICE table in partial pressures, compute the equilibrium partial pressures.

The reactor effluent is cooled so that ammonia condenses and is removed, leaving the gas-phase partial pressures of N2 and H2 unchanged during the brief removal step (assume T is restored afterward to the original reactor temperature before re-equilibration). Immediately after removing half of the NH3(g) present at equilibrium, compute the new Qp and predict the direction of net change upon returning to reactor conditions.

Briefly explain why removing NH3 can increase overall conversion in a recycle process, using the language of Q vs. K (one paragraph maximum).

Chapter 13
Acids and Bases

From the enzymatic regulation of biological fluids to the industrial optimization of catalysts, acid-base chemistry is the lexicon of proton transfer. This chapter transcends elementary litmus definitions to establish a rigorous thermodynamic framework for acidity. By analyzing the interplay between molecular structure, solvent leveling effects, and equilibrium constants, we develop the predictive power necessary to model complex aqueous systems, calculate pH with high precision, and understand the fundamental reactivity of ions in solution.

13.1 Acid-Base Definitions and Molecular Structure

Earlier chapters developed three ideas that will now converge: (1) molecular structure controls electron distribution (polarity and bond strength), (2) chemical reactions can often be described as the movement of electrons or nuclei to reach lower free energy, and (3) equilibrium concepts quantify the extent of reaction. Acid–base chemistry is the arena in which these ideas become especially visible. This section builds the conceptual foundation by developing three definitions of acids and bases (Arrhenius, Brønsted–Lowry, and Lewis) and then connecting those definitions to molecular structure-why some species donate protons readily while others resist, and why some species accept protons (or donate electron pairs) strongly.

13.1.1 Why multiple acid–base definitions exist

No single definition captures every context in which chemists use the words acid and base. Instead, acid–base theory evolved in layers:

Arrhenius theory (historically first) is specialized for aqueous solutions and focuses on the formation of H3O+ and OH– .

Brønsted–Lowry theory generalizes acid–base reactions to proton transfer and applies in many solvents (and even gas-phase chemistry).

Lewis theory is the broadest; it defines acid–base reactions in terms of electron-pair transfer and unifies proton transfer with coordination chemistry and many electrophile–nucleophile reactions.

As the definitions become more general, they become less tied to a particular solvent and more directly tied to electronic structure.

13.1.2 Arrhenius acids and bases (aqueous ion production)

Arrhenius acid: a substance that increases the concentration of H3O+ (or H+ as a shorthand) in water.

Arrhenius base: a substance that increases the concentration of OH– in water.

Examples and scope

Common Arrhenius acids include molecular acids that react with water to form hydronium:

HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq) |  |
HNO3(aq) + H2O(l) → H3O+(aq) + NO 3−(aq) |  |

Common Arrhenius bases include ionic hydroxides that dissociate to release OH– :

NaOH(s) → Na+(aq) + OH−(aq) |  |
Ba(OH)2(s) → Ba2+(aq) + 2OH−(aq) |  |

Limitations.

It is water-centered: it does not directly describe acid–base reactions in other solvents or in the gas phase.

It treats bases primarily as hydroxide sources, which excludes important bases such as NH3.

It does not provide a direct structural explanation for acid or base strength beyond “how much H3O+ or OH– is produced.”

These limitations motivate the proton-transfer model.

13.1.3 Brønsted–Lowry acids and bases (proton transfer)

Brønsted–Lowry acid: a proton donor.

Brønsted–Lowry base: a proton acceptor.

In this view, an acid–base reaction is a proton-transfer reaction. Importantly, a species can be a base without containing OH– . For example:

Here, NH3 is a Brønsted–Lowry base because it accepts a proton from water, even though it contains no hydroxide ion initially.

Conjugate acid–base pairs

A central structural insight of the Brønsted–Lowry model is that acids and bases come in conjugate pairs. When an acid donates a proton, it becomes its conjugate base. When a base accepts a proton, it becomes its conjugate acid.

For a general acid HA in water:

HA / A– is a conjugate acid–base pair.

H2O / H3O+ is a conjugate base–acid pair.

Key principle: The stronger an acid, the weaker its conjugate base (and vice versa).

Amphoteric and amphiprotic substances

Some species can act as either an acid or a base depending on the reaction partner.

Amphiprotic species can both donate and accept H+. Water is the classic example:

H2O + HCl → H3O+ + Cl− (H 2O acts as a base) |  |
H2O + NH3 ↔ OH− + NH 4+ (H 2O acts as an acid) |  |

Many hydrogen-containing anions are amphiprotic, such as HCO3 – and HSO4 – , because they contain a transferable proton and also carry lone pairs/negative charge that can accept a proton.

The Brønsted–Lowry framework is the primary language for aqueous equilibria developed later in this chapter.

13.1.4 Lewis acids and bases (electron-pair transfer)

Lewis base: an electron-pair donor.

Lewis acid: an electron-pair acceptor.

Lewis theory removes the requirement of an explicit proton and focuses instead on the underlying electronic event: formation of a bond by donation of a lone pair into an empty orbital (or an electron-poor region) of another species.

Examples: coordination and beyond

A classic Lewis acid–base reaction is adduct formation between ammonia and boron trifluoride:

Here, NH3 donates a lone pair (Lewis base). BF3 is electron-deficient; boron has an incomplete octet and an empty orbital that can accept electron density (Lewis acid).

Proton transfer is also a Lewis acid–base reaction: H+ is a Lewis acid, and any Brønsted–Lowry base is necessarily a Lewis base.

Brønsted–Lowry acid–base chemistry is a subset of Lewis acid–base chemistry.

Mechanistic picture: what “accepting an electron pair” means

In molecular-orbital language, a Lewis base has a filled orbital (often a lone pair in a nonbonding orbital, the HOMO) and a Lewis acid has a low-energy empty orbital (often the LUMO). The interaction is stabilizing when:

the base has high electron density available, and

the acid has an accessible, low-energy acceptor orbital.

13.1.5 Connecting the definitions: one phenomenon, different lenses

In water, an Arrhenius acid produces H3O+. Mechanistically, that requires proton transfer, so every Arrhenius acid is also a Brønsted–Lowry acid.

Every Brønsted–Lowry acid–base reaction involves formation of a new bond to H+ using a lone pair, so it is also a Lewis acid–base reaction.

The reverse implications are not always true: many Lewis acid–base reactions do not involve H+ at all (e.g., BF3 + NH3).

13.1.6 Molecular structure and acid strength: what makes H+ easy to lose?

Within the Brønsted–Lowry framework, an acid HA is stronger when it more readily donates H+:

Structurally, HA will be a stronger acid when its conjugate base A– is especially stable. Several molecular features stabilize A– :

Bond polarity: strong polarization toward A makes the proton more electrophilic.

Bond strength: a weaker H–A bond is easier to break.

Stability of negative charge in A–: dispersion or delocalization of charge stabilizes the conjugate base.

13.1.7 Binary hydrides H–X: trends across a period and down a group

Binary hydrides (acids of the form HX or H2X) illustrate two distinct structural controls on acidity.

Across a period: electronegativity controls bond polarity

Consider the second-period hydrides CH4, NH3, H2O, and HF. As electronegativity increases, the H–X bond becomes more polarized:

Thus, acidity increases left-to-right:

The conjugate bases (CH3 – , NH2 – , OH– , F– ) become progressively more stabilized.

Down a group: bond strength and anion size dominate

For the hydrohalic acids HF, HCl, HBr, HI, acidity increases strongly down the group:

The H–X bond weakens as X becomes larger, and the larger conjugate base spreads negative charge over a greater volume.

Takeaway for binary hydrides.

Across a period: increasing electronegativity and bond polarity dominate.

Down a group: decreasing bond strength and increasing anion size dominate.

13.1.8 Oxyacids: the role of electronegative substituents and resonance

Many important acids are oxyacids, with the general connectivity H–O–Y. Their acidity depends strongly on the identity and oxidation state of the central atom Y and on the number of oxygen atoms.

A structural baseline: acidity comes from the O–H bond

In most oxyacids, the acidic proton is bonded to oxygen. The key question becomes: how stable is the conjugate base after deprotonation?

Inductive effects: pulling electron density away from O–H

Electronegative atoms or electron-withdrawing groups attached to the O–Y framework draw electron density away through sigma bonds (the inductive effect). This polarizes the O–H bond and stabilizes the conjugate base.

Resonance stabilization: delocalizing the conjugate base charge

Many oxyacid conjugate bases can delocalize the negative charge over multiple oxygen atoms via resonance. For example, nitrate NO3 – distributes the negative charge over three equivalent oxygens, making HNO3 a strong acid.

The “more oxygens, stronger acid” pattern (within a family)

Within oxyacids that share the same central atom, acidity typically increases as the number of terminal oxygens increases:

More oxygen atoms withdraw electron density inductively.

The conjugate base can delocalize negative charge over more oxygen atoms.

Same number of oxygens: electronegativity of the central atom matters

For oxyacids with the same number of oxygen atoms and similar structures, the acid is stronger when the central atom is more electronegative. Among HOCl, HOBr, and HOI:

13.1.9 Organic acids: inductive effects and local structure

Two structural themes are central.

Carboxylic acids: a resonance-stabilized conjugate base

Carboxylic acids (e.g., acetic acid CH3CO2H) are more acidic than alcohols (e.g., ethanol CH3CH2OH) because the carboxylate conjugate base is resonance-stabilized over two oxygen atoms, whereas an alkoxide localizes negative charge mainly on one oxygen.

RCO2H ↔ H+ + RCO 2− |  |
ROH ↔ H+ + RO− |  |

Inductive substituent effects in organic acids

Electron-withdrawing substituents (e.g., F, Cl, NO2) increase acidity when placed near the acidic functional group, because they stabilize the conjugate base by inductive withdrawal. The effect diminishes with distance:

13.1.10 Molecular structure and base strength: what makes an electron pair available?

Base strength generally increases when:

the lone pair is high in electron density and accessible,

the lone pair is not delocalized by resonance, and

any negative charge is not overly stabilized.

Electronegativity and basicity

Within a row, bases become weaker as the atom bearing the lone pair becomes more electronegative. For the conjugate bases of the second-period hydrides:

Resonance decreases basicity

If a lone pair is delocalized by resonance, it is less available to bond to H+. For example, a neutral amide is a weak base because the nitrogen lone pair is delocalized into the carbonyl group, whereas an amine lone pair is localized and more basic.

13.1.11 A unifying summary: predicting acid strength from structure

Identify the acidic proton or electron-rich site.

Draw the conjugate base and ask: Is it stabilized?

Apply, in order: resonance stabilization; atom effects (electronegativity and size); inductive effects; bond strength.

These qualitative ideas will later be made quantitative through equilibrium constants and the pH scale.

13.2 The pH Scale and Water Autoionization

The previous section established acids and bases as proton donors and acceptors (Brønsted–Lowry) and, more fundamentally, as electron-pair acceptors and donors (Lewis). To use these ideas quantitatively in aqueous solution, we need a way to measure “how acidic” or “how basic” a solution is. This section introduces the central quantitative anchor for aqueous acid–base chemistry: the autoionization of water and the ion-product constant Kw. From Kw we build the pH and pOH scales, derive the relationship between them, and set careful rules for logarithms and significant figures. Finally, we distinguish the ideal definition of pH in terms of activity from the practical concentration-based calculations used in most general chemistry problems.

13.2.1 Autoionization of water: the origin of H3O+ and OH– in pure water

Even in the absence of any added acid or base, liquid water contains small but nonzero concentrations of hydronium and hydroxide ions. This occurs by autoionization (also called autoprotolysis), a proton-transfer reaction between two water molecules:

One water molecule acts as a Brønsted–Lowry acid (proton donor) and the other as a Brønsted–Lowry base (proton acceptor). This is a concrete example of water’s amphiprotic behavior.

Writing the equilibrium constant

For Eq. (??), the equilibrium constant in terms of activities is

Because liquid water is the solvent, its activity is very close to 1 in dilute aqueous solutions (aH2O ≈ 1). We therefore define the ion-product constant of water:

In typical general chemistry calculations, we approximate activities by molar concentrations in sufficiently dilute solutions:

Standard value at 25 ∘C

At 25 ∘C,

This value is so widely used that it is often treated as a constant; however, it depends on temperature (Section ??).

Neutral water at 25 ∘C

In pure water, the stoichiometry of Eq. (??) produces equal amounts of H3O+ and OH−. Let

Then Eq. (??) gives

At 25 ∘C, using Eq. (??),

This is the quantitative basis of the familiar statement that “neutral water has pH 7”—but as emphasized later, this is only exactly true at 25 ∘C and only under the concentration approximation.

13.2.2 The pH and pOH scales: logarithms as chemical bookkeeping

Hydronium concentrations in aqueous chemistry can range over many orders of magnitude (from near 10 M in very concentrated strong acids to below 1 × 10⁻¹³ M in strongly basic solutions). A logarithmic scale compresses this enormous range into manageable numbers.

Definition of pH (activity form and concentration form)

The rigorous thermodynamic definition of pH uses hydronium activity:

In dilute solutions, we often approximate aH3O+ ≈ [H3O+]∕c∘, where c∘ = 1M is the standard-state concentration. Because c∘ is 1 M, it is usually omitted explicitly, and we write the practical working definition:

Here [H3O+] must be expressed in molarity so that the argument of the logarithm is dimensionless (implicitly [H3O+]∕1M).

Definition of pOH

Analogously,

What pH numbers mean physically

Because pH is a negative base-10 logarithm, each change of 1 pH unit corresponds to a factor of 10 change in hydronium activity (or concentration, approximately):

Thus, lower pH means higher hydronium concentration (more acidic), and higher pH means lower hydronium concentration (more basic).

13.2.3 The relationship between pH, pOH, and Kw

Starting from Eq. (??),

take −log ₁₀ of both sides:

− log ₁₀Kw | ≈−log ₁₀([H3O+][OH−]) |  |
| = −log ₁₀[H3O+] − log 10[OH−]. |  |

Define

and recognize pH ≈−log ₁₀[H3O+] and pOH ≈−log 10[OH−]. Then

At 25 ∘C

Using Kw = 1.0 × 10⁻¹⁴,

so at 25 ∘C,

A common pitfall

Equation (??) is not universally valid. It is only valid when Kw = 1.0 × 10⁻¹⁴, i.e., near 25 ∘C under typical conditions. The more general identity is Eq. (??).

13.2.4 Neutrality, acidity, and basicity: what “neutral pH” really means

Neutrality is defined by equality of [H3O+] and [OH−]

A solution is neutral when

or more rigorously aH3O+ = aOH−. Using Eq. (??),

Therefore, the neutral pH is

At 25 ∘C, pKw = 14.00, so pHneutral = 7.00. At other temperatures, the neutral pH is not exactly 7.

Acidic vs. basic (definition in water)

In aqueous solution:

Acidic: [H3O+] > [OH−], equivalently pH < pH neutral.

Basic: [H3O+] < [OH−], equivalently pH > pH neutral.

Notice that “acidic” does not mean pH < 7 unless the temperature is 25 ∘C (or unless pKw happens to be 14.00).

13.2.5 Temperature dependence of Kw and the shifting meaning of “pH 7”

Autoionization (Eq. (??)) is an equilibrium process whose position depends on temperature. As temperature changes, Kw changes, which shifts the neutral point through Eq. (??). The practical consequence is that the statement “neutral water has pH 7” is a temperature-specific statement.

Conceptual direction (qualitative)

As temperature increases, autoionization becomes more favorable (water forms more ions), so Kw increases and pKw decreases. Then Eq. (??) implies that pH neutral = pKw decreases. Thus, at higher temperatures, pure water is still neutral, but its pH is less than 7.

Practical guidance for calculations

When a problem states a temperature other than 25 ∘C, do not assume Kw = 1.0 × 10⁻¹⁴. Use the provided Kw (or pKw) at that temperature and apply Eq. (??).

13.2.6 Significant figures and logarithms: reporting pH correctly

Because pH is a logarithm, the usual significant-figure rules require special care.

Rule 1: pH has digits after the decimal that reflect significant figures in [H3O+]

If [H3O+] is reported with n significant figures, then pH should be reported with n digits after the decimal point.

Example

If [H3O+] = 1.0 × 10⁻³ M (2 significant figures),

Two significant figures in concentration → two digits after the decimal in pH.

If [H3O+] = 1.00 × 10⁻³ M (3 significant figures),

Rule 2: interpreting the digits in a pH value

In a pH value, only the digits to the right of the decimal point are significant. The digits to the left of the decimal indicate order of magnitude.

Example

pH = 2.43 means [H3O+] ≈ 3.7 × 10⁻³ M, and the two digits after the decimal imply [H3O+] is known to roughly 2 significant figures.

13.2.7 From pH to concentration and back: essential conversions

The definition pH = −log ₁₀[H3O+] can be inverted:

Similarly,

And when Kw is known,

These relationships will be used repeatedly in later sections to compute pH for strong acids/bases (stoichiometry-dominated) and weak acids/bases (equilibrium-dominated).

13.2.8 Activities versus concentrations: what pH meters actually measure

Why concentration is an approximation

The thermodynamic driving forces in equilibria depend on activity, not concentration. In real solutions, ions interact electrostatically with each other and with solvent molecules. These interactions mean that an ion’s “effective concentration” in equilibrium expressions is

where γi is the activity coefficient. At low ionic strength, γi ≈ 1, and concentration-based calculations are accurate to within typical general-chemistry expectations.

When the distinction matters

Activity effects become important in:

relatively concentrated electrolyte solutions (ionic strength not small),

solutions with multivalent ions (stronger electrostatic interactions),

high-precision work (analytical chemistry, physical chemistry).

In this textbook, unless explicitly stated otherwise, we will assume γ ≈ 1 and use concentrations. This keeps the focus on developing equilibrium reasoning without introducing additional layers of modeling.

13.2.9 A concise summary for later problem-solving

Water autoionizes: 2H2O ↔ H3O+ + OH−.

Kw ≡ aH3O+aOH− ≈ [H3O+][OH−]; at 25 ∘C, K w = 1.0 × 10⁻¹⁴.

pH = −log ₁₀(aH3O+) ≈−log ₁₀[H3O+], and pOH ≈−log 10[OH−].

pH + pOH = pKw (general). At 25 ∘C, pH + pOH = 14.00.

Neutrality means [H3O+] = [OH−], so pH neutral = pKw (not always 7).

With these tools, we can now translate the qualitative language of acids and bases into numerical predictions of solution composition and acidity.

13.3 Strong Acids and Bases

The previous sections established (1) acids and bases as proton-transfer partners

(Brønsted–Lowry) and electron-pair acceptors/donors (Lewis), and (2) pH as a quantitative measure of hydronium activity (approximated by concentration in dilute solutions), anchored by the water autoionization equilibrium and Kw. We now apply these tools to the simplest and most important limiting case: strong acids and strong bases, which in water react essentially to completion. In this regime, pH is determined primarily by stoichiometry rather than by equilibrium solving.

13.3.1 What “strong” means in aqueous acid–base chemistry

Strong acid: essentially complete proton transfer to water

A strong acid is an acid that transfers its proton(s) to water essentially completely:

The equilibrium lies so far to the right that, for most general chemistry calculations, we treat the reaction as going to completion. As a consequence, the concentration of undissociated HA at equilibrium is negligible relative to its initial concentration (in dilute solutions).

Strong base: essentially complete generation of OH–

A strong base in water is a substance that produces hydroxide ion essentially completely. The most common strong bases are soluble ionic hydroxides:

for alkali metals (LiOH, NaOH, KOH, …), and

for the more soluble alkaline earth hydroxides (notably Ca(OH)2, Sr(OH)2, Ba(OH)2).

Important distinction. Calling NaOH a “strong base” refers to its extent of dissociation in water, not to its solubility. Some hydroxides (e.g., Mg(OH)2) are only sparingly soluble; the dissolved portion dissociates completely, but the solution may still have limited [OH−] because little dissolves. Solubility equilibria are treated separately.

13.3.2 Common strong acids and strong bases in water

For most introductory work, the following are treated as strong acids in water:

Strong acid | Notes
HCl, HBr, HI | Hydrohalic acids (except HF, which is weak).
HNO3 | Nitric acid.
HClO4 | Perchloric acid (strong; also a strong oxidizer when concentrated).
H2SO4 | First proton dissociates essentially completely; second is not complete.

Common strong bases include:

Strong base | Notes
LiOH, NaOH, KOH, RbOH, CsOH | Alkali metal hydroxides; dissociate completely when dissolved.
Ca(OH)2, Sr(OH)2, Ba(OH)2 | Alkaline earth hydroxides; Ba(OH)2 most soluble among these.

Why lists are not the main idea. The deeper concept is that in water, strong acids and bases react so completely that the dominant species can be determined from stoichiometry alone. This is a consequence of large equilibrium constants for proton transfer in water, which is tied to the solvent leveling effect (Section ??).

13.3.3 Stoichiometry-first calculations for strong acids

Monoprotic strong acids

For a monoprotic strong acid HA (e.g., HCl, HNO3) in dilute water:

where Ca is the formal (analytical) concentration of the acid after mixing/dilution.

Example: dilution. If 25.0mL of 0.100moldm⁻³ HCl is diluted to 250.0mL,

so

Strong acids with multiple ionizable protons: “formal” versus “effective” H+ yield

If an acid can donate more than one proton, the maximum hydronium concentration

depends on how many protons dissociate essentially completely.

A key example: sulfuric acid. Sulfuric acid is diprotic, but in water:

the first dissociation is essentially complete (strong acid behavior),

the second dissociation is incomplete (weak acid behavior) and must be treated by equilibrium methods introduced later.

The first step:

Thus, immediately after the first step (before considering the second), a solution prepared at formal concentration C has approximately:

Only in a subsequent equilibrium treatment would HSO4 – partially donate a second proton:

Practical implication. For many problems, you will be explicitly instructed whether to treat only the first proton as strong (common in general chemistry) or to include the second dissociation with Ka2 (a more advanced calculation).

13.3.4 Stoichiometry-first calculations for strong bases

Monohydroxide bases: MOH

For strong bases like NaOH and KOH,

where Cb is the formal base concentration after any mixing/dilution. Then

At 25∘C, pKw = 14.00, so pH ≈ 14.00 − pOH.

Dihydroxide bases: M(OH)2

For Ba(OH)2 (and similar soluble dihydroxides),

because each formula unit yields two hydroxide ions. This is a stoichiometric factor, not an equilibrium subtlety.

Example. A 0.0250moldm⁻³ Ba(OH)2 solution has

so

13.3.5 Strong acid–strong base reactions: neutralization as limiting-reagent stoichiometry

When a strong acid and strong base react in water, proton transfer is essentially complete. The net ionic equation is:

Spectator ions (e.g., Na+, Cl– ) remain in solution and typically do not affect pH significantly in dilute solutions.

General procedure (before buffers and weak-acid equilibria)

Convert volumes and molarities to moles of H3O+ produced by strong acid and moles of OH− produced by strong base.

Use Eq. (??) as a limiting-reagent problem.

Determine the excess H3O+ or OH−.

Divide excess moles by total volume to get the concentration of the excess species.

Convert to pH (or pOH), using pKw appropriate to the temperature.

Example (conceptual). If OH− is in excess, you find [OH−], compute pOH, then compute pH via pH = pKw − pOH.

13.3.6 Very dilute strong acid or base: when water autoionization matters

The stoichiometric approach assumes that the added strong acid or base overwhelmingly determines [H3O+] or [OH−]. This can fail at extremely low concentrations, where the ∼ 1.0 × 10⁻⁷ moldm⁻³ ions from water autoionization are comparable to the amount added.

A conceptual threshold

At 25∘C, pure water has [H3O+] = [OH−] = 1.0 × 10⁻⁷ moldm⁻³. Therefore:

If a strong acid solution has Ca ≫ 1.0 × 10⁻⁷ moldm⁻³ (e.g., 1 × 10⁻⁴ moldm⁻³ or larger), then [H3O+] ≈ C a is usually an excellent approximation.

If Ca ≲ 1.0 × 10⁻⁷ moldm⁻³, you must include water autoionization to obtain a physically consistent answer (and the pH will not be simply −log Ca).

A similar statement holds for very dilute strong bases.

A minimal quantitative model (idea, not full equilibrium solving)

For a very dilute strong acid of formal concentration Ca, one can reason that total hydronium arises from two sources: added acid and water. The correct treatment uses mass balance and Kw simultaneously; it leads to a quadratic relationship. We defer the systematic ICE-table approach to later sections, but the key lesson is immediate:

At extreme dilution, even a “strong” electrolyte does not guarantee that [H3O+] is set solely by stoichiometry, because water itself contributes ions.

13.3.7 The solvent leveling effect in water

Statement of the effect

In a given solvent, there is a limit to how strong an acid can appear: all acids stronger than the solvent’s conjugate acid are “leveled” to the same apparent strength because they react completely with the solvent.

For water, the strongest acid that can exist in appreciable amount is H3O+. Any acid stronger than H3O+ transfers its proton to water essentially completely:

Thus, in water, many different strong acids produce the same strongest acidic species: H3O+. Their strengths are leveled by the solvent.

Why superacids are not “super” in aqueous solution

So-called superacids (acids stronger than 100% sulfuric acid) can exist and be studied in nonaqueous media. However, in water, their additional proton-donating power cannot be expressed because water itself is a sufficiently strong base to capture the proton, forming H3O+. The “extra” acidity is therefore not observable as a higher intrinsic acid strength in water; it is converted into complete protonation of water.

A parallel statement for bases

The strongest base that can persist in water is OH−, the conjugate base of water. Any base stronger than OH− is leveled by reaction with water to form OH−. For example, the oxide ion O2− cannot exist in water; it is protonated immediately:

Similarly, very strong bases such as NH2 – (amide) are instantly protonated by water:

Structural connection (link to Lewis theory). From the Lewis perspective, water is both a Lewis base (lone pairs on oxygen) and a Brønsted–Lowry acid/base. A species that is an extremely strong Lewis base will abstract a proton from water; a species that is an extremely strong Brønsted–Lowry acid will donate a proton to water. In either case, the solvent funnels the chemistry into the same limiting species (H3O+ or OH−).

13.3.8 Summary: when strong electrolytes simplify pH calculations

In aqueous solutions of strong acids and bases (at non-extreme dilutions), pH is usually determined by stoichiometry:

Strong acids: [H3O+] ≈ formal acid concentration (accounting for how

many protons dissociate completely).

Strong bases: [OH−] ≈ hydroxide stoichiometric yield × formal base concentration.

Neutralization: treat H3O+ and OH− as reacting to completion; compute the excess and convert to pH using Kw.

The solvent leveling effect explains why many different strong acids appear similarly “strong” in water: they all produce H3O+ as the strongest acid species available in this solvent.

13.4 Weak Acids and Ionization Constants

Strong acids simplify pH calculations because proton transfer to water is essentially complete, so stoichiometry dominates. Many important acids, however, ionize only partially in water. In these cases, the equilibrium position must be calculated, and the pH depends on both the acid’s intrinsic tendency to donate a proton and the solution conditions (notably the initial concentration). This section develops the quantitative framework for weak acids using the acid ionization constant Ka, systematic ICE tables, approximation strategies, and the quadratic equation when needed. We also introduce percent ionization and explain why weak acids become more ionized upon dilution.

13.4.1 Weak acid ionization and the definition of Ka

Ionization equilibrium in water

For a monoprotic weak acid HA, the Brønsted–Lowry reaction with water is

Because H2O is the solvent, its activity is approximately 1. The equilibrium constant (in activity form) is therefore written as the acid ionization constant:

In dilute solution, we approximate activities with concentrations:

What Ka means. A larger Ka means the equilibrium in Eq. (??) lies further to the right: more H3O+ and A– form, and the acid is stronger. A smaller Ka means the acid remains mostly as HA and generates relatively little H3O+.

The pKa scale

Because Ka values span many orders of magnitude, it is common to define

A smaller pKa corresponds to a larger Ka and therefore a stronger acid.

13.4.2 Setting up weak-acid equilibrium calculations: ICE tables

For most weak-acid pH problems, the core task is to determine [H3O+] at equilibrium. The ICE table is a structured way to apply mass action (Eq. (??)) while keeping track of stoichiometry.

Generic ICE table for a monoprotic weak acid

Suppose a solution is prepared with formal (analytical) concentration C of HA. Neglecting the small contribution from water autoionization (appropriate unless C is extremely small), the initial concentrations are:

Let x be the amount (in molarity) of HA that ionizes. Then:

| HA | H3O+ | A–
Initial (M) | C | 0 | 0
Change (M) | −x | +x | +x
Equilibrium (M) | C − x | x | x

Substitute into Eq. (??):

Once x is determined, [H3O+] = x and

Conceptual checkpoint. Equation (??) encodes two competing tendencies:

the acid’s intrinsic tendency to ionize (captured by Ka),

the “penalty” for building up products [H3O+] and [A−], which drives the equilibrium back toward HA.

The dependence on C arises because starting with more HA changes how large x can become relative to the remaining HA.

13.4.3 The x-is-small approximation and when it is valid

Solving Eq. (??) exactly requires a quadratic equation. In many weak-acid problems, however, the acid ionizes only a small fraction, so x ≪ C and C − x ≈ C. Then

This immediately yields

Checking the approximation (the 5% rule)

A common practical criterion is the 5% rule:

If x∕C > 5%, then replacing C −x by C can introduce non-negligible error, and the quadratic solution should be used.

Important nuance. The 5% rule is a guideline, not a theorem. Whether the approximation is “good enough” depends on the expected precision of the problem. In many laboratory contexts, even smaller errors matter; in many textbook contexts, 5% is a reasonable cutoff.

13.4.4 Exact solution: the quadratic equation

Starting from Eq. (??),

rearrange to standard quadratic form:

Ka(C − x) | = x² |  |
KaC − Kax | = x² |  |
x² + K ax − KaC | = 0. |  |

Using the quadratic formula,

Only the positive root is physically meaningful (concentrations cannot be negative), so

Then [H3O+] = x and pH = −log 10x.

Interpretation. When Ka ≪ C, the square root term is dominated by 4KaC, and Eq. (??) reduces to x ≈, matching the approximation in Eq. (??). The quadratic formula therefore provides a consistent “always correct” method, while the approximation provides speed and insight.

13.4.5 Percent ionization and its dependence on concentration

Definition

For a monoprotic weak acid HA of formal concentration C, the percent ionization (also called percent dissociation) is

Because x = [H3O+]eq in the simple model (neglecting water autoionization), percent ionization is also

Why percent ionization increases upon dilution

Using the x-is-small approximation (often valid for weak acids at moderate concentration),

Then the fraction ionized is

This scaling relation contains the key qualitative result:

For a given weak acid (fixed Ka), the fraction (and percent) ionized increases as the initial concentration C decreases.

This is an expression of Le Châtelier’s principle: dilution lowers the concentrations of products and reactants; the equilibrium shifts toward producing more ions to partially counteract that change.

Visualizing the trend

The inverse relationship in Eq. (??) is easiest to see on a logarithmic concentration axis. The curve approaches 0% ionization at high concentration and rises toward 100% at very low concentration (though in real solutions the simple model eventually breaks down due to activity effects and water autoionization).

Reading the plot. At high C, the ratio x∕C is small: the weak acid remains mostly unionized. At low C, the equilibrium shifts toward ion formation, and the percent ionization rises sharply. The curve does not imply that the solution becomes more acidic upon dilution; rather, [H3O+] = x still typically decreases as C decreases, even though the fraction ionized increases.

13.4.6 Polyprotic weak acids: sequential ionization and why the first step dominates

Many acids can donate more than one proton. A diprotic acid H2A ionizes in two steps:

H2A + H2O ↔ H3O+ + HA− K a1 |  |
HA− + H 2O ↔ H3O+ + A2− K a2 |  |

Similarly, triprotic acids ionize in three steps with Ka1, Ka2, Ka3.

Typical magnitude pattern

For most polyprotic acids,

Each successive proton is harder to remove because the species becomes progressively more negatively charged, and it is increasingly unfavorable to create additional negative charge and additional H3O+.

Implication for pH calculations

In many polyprotic-acid pH calculations at moderate concentration, [H3O+] is dominated by the first ionization step, and later steps contribute only a small additional amount. Thus a common (often accurate) strategy is:

compute pH using only the first equilibrium (treat H2A as if it were a monoprotic weak acid with Ka = Ka1),

then, if needed, check whether the second step significantly changes [H3O+].

Later in the chapter, we will refine this reasoning for amphiprotic species and salts.

13.4.7 Where weak-acid methods fit in the broader acid–base toolkit

Weak-acid problems illustrate a general pattern used throughout equilibrium

chemistry:

Write the chemical equilibrium (identify the dominant acid–base reaction).

Write the equilibrium constant expression (Ka for acids).

Use an ICE table to express equilibrium concentrations in terms of one variable.

Solve (approximation if justified; otherwise, quadratic).

Convert [H3O+] to pH and apply significant-figure rules.

This framework will be mirrored for weak bases (Kb) and will later be combined with mass balance and charge balance ideas when we analyze salts and more complex systems.

13.5 Weak Bases and Salt Properties

Weak acids required equilibrium methods because ionization was incomplete. Weak bases behave analogously: they react with water to produce OH– only partially, and their basicity is quantified by an equilibrium constant Kb. Beyond individual acids and bases, aqueous solutions frequently contain salts—ionic compounds that may be neutral, acidic, or basic depending on whether their ions react with water. This section develops (1) equilibrium calculations for weak bases, (2) the relationship between Ka and Kb for conjugate pairs, and (3) a systematic method for predicting and calculating salt-solution pH via hydrolysis reactions.

13.5.1 Weak bases and the base ionization constant Kb

Base reaction with water

A Brønsted–Lowry base accepts a proton from water, producing its conjugate acid and hydroxide ion. For a generic weak base B:

The base ionization constant is

Interpreting Kb. A larger Kb corresponds to a stronger base (more OH– produced at equilibrium). A smaller Kb corresponds to a weaker base.

ICE-table framework for a weak base

If a solution is prepared with formal concentration C of B, and if we neglect water autoionization initially (valid unless the base is extremely dilute), we can write:

| B | BH+ | OH–
Initial (M) | C | 0 | 0
Change (M) | −x | +x | +x
Equilibrium (M) | C − x | x | x

Substituting into Eq. (??) gives

Solve for x = [OH−], then compute

The small-x approximation and the 5% check

If the base is weak enough that x ≪ C, then C − x ≈ C and

Check validity using the 5% rule:

If the criterion fails, solve the quadratic from Eq. (??):

keeping the physically meaningful (positive) root.

13.5.2 Conjugate acid–base pairs and the relationship KaKb = Kw

Derivation from two equilibria

Consider a conjugate pair HA/A– . The acid ionization is

The base ionization of A– is

Multiply the expressions:

Thus, for any conjugate acid–base pair in water,

Consequence: strength trade-off. If HA is a strong acid (large Ka), then its conjugate base A– has a very small Kb and is a very weak base. Conversely, if B is a weak base, its conjugate acid BH+ has a comparatively larger Ka.

13.5.3 Salt hydrolysis: why some salts change pH

A salt is an ionic compound that dissociates into cations and anions in water. Whether a salt solution is acidic, basic, or neutral depends on whether the ions undergo acid–base reactions with water (hydrolysis).

The key idea: ions can be conjugate acids or bases

Many anions are conjugate bases of acids. If the parent acid is weak, the anion may be basic enough to react with water and produce OH– .

Many cations are conjugate acids of bases. If the parent base is weak, the cation may donate a proton to water and produce H3O+.

Ions that are conjugates of strong acids or bases are typically so weak that they do not measurably hydrolyze; they act as spectators in pH.

13.5.4 Predicting whether a salt solution is acidic, basic, or neutral

Spectator ions from strong acids and strong bases

Anions of strong acids. Anions such as Cl– , Br– , I– , NO3 – , and ClO4 – are conjugate bases of strong acids. They have negligible basicity in water, so they do not raise pH by hydrolysis.

Cations of strong bases. Cations such as Li+, Na+, K+ (and many other monatomic metal cations with very low charge density) are conjugate acids of strong bases; they have negligible acidity in water.

Example: neutral salt. NaCl contains Na+ (from strong base NaOH) and Cl– (from strong acid HCl). Neither ion hydrolyzes appreciably, so the solution is approximately neutral:

Salts that produce acidic solutions: cations that are weak acids

A common source of acidic salt solutions is a cation that is the conjugate acid of a weak base. The classic example is ammonium:

Hydrolysis reaction:

Thus salts like NH4Cl yield acidic solutions because NH4+ produces H3O+.

Salts that produce basic solutions: anions that are weak bases

Anions that are conjugate bases of weak acids hydrolyze to form OH– . Example: acetate CH3CO2 – is the conjugate base of acetic acid CH3CO2H (weak acid). Hydrolysis:

Therefore salts like NaCH3CO2 yield basic solutions because CH3CO2 – produces OH– .

Salts where both ions hydrolyze: competing acidity and basicity

If the cation is acidic and the anion is basic (each derived from a weak parent), then the pH depends on the relative magnitudes of Ka and Kb.

Example: ammonium acetate NH4CH3CO2 contains NH4+ (weak acid) and CH3CO2 – (weak base). The net pH depends on whether NH4+ donates protons more strongly than CH3CO2 – accepts them.

13.5.5 Quantitative treatment of salt hydrolysis

Computing the hydrolysis constant from Ka or Kb

If an anion A– is the conjugate base of a weak acid HA, its base hydrolysis is Eq. (??)-type:

The equilibrium constant for this reaction is Kb(A−), and it is related to K a(HA) by Eq. (??):

Similarly, if a cation BH+ is the conjugate acid of a weak base B, its acid hydrolysis is Eq. (??)-type:

with

Example framework: pH of a salt of a weak acid (basic salt)

Consider NaA, where A– is the conjugate base of weak acid HA. Since Na+ is a spectator, pH is set by A– hydrolysis:

Let the formal salt concentration be C, so initially [A−] 0 = C. Set up an ICE table:

| A– | HA | OH–
Initial (M) | C | 0 | 0
Change (M) | −x | +x | +x
Equilibrium (M) | C − x | x | x

Then

Solve for x = [OH−], then compute pOH and pH (using the appropriate pK w if temperature differs from 25 ∘C).

Example framework: pH of a salt of a weak base (acidic salt)

Consider BHCl, where BH+ is the conjugate acid of weak base B, and Cl– is a spectator. Hydrolysis:

With formal concentration C, use:

| BH+ | B | H3O+
Initial (M) | C | 0 | 0
Change (M) | −x | +x | +x
Equilibrium (M) | C − x | x | x

Then

Solve for x = [H3O+], then compute pH.

Salts with both weak ions: a useful shortcut

For salts formed from a weak acid HA and a weak base B (so the ions are BH+ and A– ), the pH often depends primarily on the relative strengths of the conjugates. A common qualitative rule is:

Because Kb(A−) = K w∕Ka(HA) and Ka(BH+) = K w∕Kb(B), this comparison can be made from tabulated Ka and Kb values.

13.5.6 Structural basis of amine basicity: inductive effects and resonance

Weak bases in aqueous solution are often organic, especially amines. Their basicity is best understood through the Lewis picture: an amine is basic because the nitrogen lone pair can donate electron density to H+ (or to H3O+) to form R3NH+.

Inductive effects

Electron-donating alkyl groups push electron density toward nitrogen through sigma bonds, generally increasing amine basicity relative to ammonia:

Electron-withdrawing groups (e.g., CF3 – , NO2 – , carbonyl-adjacent substituents) pull electron density away, lowering basicity.

Resonance effects

If the nitrogen lone pair is delocalized by resonance, it is less available to bind H+, making the base weaker. This is a major reason that amides are far less basic than amines: the amide nitrogen lone pair is delocalized into the carbonyl group, stabilizing the unprotonated form and reducing proton affinity.

13.5.7 Decision tree for salt solutions (qualitative prediction)

In most problems, the first task is not calculation but classification: will the salt solution be acidic, basic, or neutral? The following decision tree summarizes the logic.

13.5.8 Summary: a unified view of weak bases and salt pH

Weak bases are treated by equilibrium: B+H2O ↔ BH+ +OH− with K b.

Conjugate pairs satisfy KaKb = Kw, linking acid and base strength in water.

Salt solutions can be acidic, basic, or neutral depending on whether the ions hydrolyze.

ions from strong parents are usually spectators,

conjugates of weak parents hydrolyze measurably,

if both ions are weak, pH depends on comparing Ka and Kb.

These principles complete the core toolkit needed to predict and calculate pH for most simple aqueous acid–base systems: strong electrolytes (stoichiometry), weak acids/bases (equilibrium), and salts (hydrolysis driven by conjugate relationships).

13.6 Problems

13.6.1 Conjugate pairs in a non-aqueous proton-transfer reaction

Liquid ammonia can act as both a Brønsted–Lowry acid and base, analogous to water. Consider the reaction in liquid ammonia:

Write the net ionic reaction in liquid ammonia by identifying the acid and base species actually involved in proton transfer.

Identify all conjugate acid–base pairs present in the net ionic reaction.

Recast the same reaction in Lewis terms: identify the electron-pair donor and acceptor in the proton-transfer step.

13.6.2 pH of pure water at elevated temperature from a nonstandard Kw

At 60.0 ∘C, the ion-product constant of water is Kw = 9.6 × 10⁻¹⁴.

Calculate [H3O+] and [OH−] in pure water at 60.0 ∘C.

Determine the neutral pH at 60.0 ∘C.

A student claims that a solution at 60.0 ∘C with pH = 7.00 must be neutral. State whether this is correct and justify using quantitative reasoning.

13.6.3 Dilute strong acid where water autoionization is significant

A solution is prepared by dissolving sufficient HCl to make a formal concentration of Ca = 1.0 × 10⁻⁸ molL⁻¹ at 25 ∘C.

Explain qualitatively why pH≠8.00 even though −log ₁₀(Ca) = 8.00.

Let [H3O+] = x at equilibrium. Using K w = 1.0×10⁻¹⁴, write an equation relating x to Ca that accounts for water autoionization.

Solve for x (you may use the quadratic formula) and determine the pH.

Compare your result to pure water at 25 ∘C and interpret the change.

13.6.4 Determining Ka from pH and formal concentration

A monoprotic weak acid HA is prepared at formal concentration C = 0.125molL⁻¹. The measured pH at 25 ∘C is 2.62.

Calculate [H3O+] from the pH.

Assuming water autoionization is negligible relative to the acid, determine [A−] and [HA] at equilibrium.

Calculate Ka and pKa.

State whether the “x is small” approximation would have been justified a priori for this system by checking the percent ionization.

13.6.5 Weak-acid equilibrium requiring the quadratic formula

A solution is made by dissolving HF to obtain a formal concentration of C = 0.0100molL⁻¹ at 25 ∘C. Use Ka(HF) = 6.8 × 10⁻⁴.

Set up an ICE table for HF ionization and derive the quadratic equation for [H3O+].

Solve the quadratic equation for [H3O+] and calculate the pH.

Evaluate whether the 5% rule supports or rejects the small-x approximation.

Compute the pH that would be obtained using the approximation and report the percent error in [H3O+].

13.6.6 Percent ionization at two concentrations

A weak monoprotic acid HA has Ka = 1.3 × 10⁻⁶ at 25 ∘C.

Calculate the percent ionization of HA in a 0.500molL⁻¹ solution.

Calculate the percent ionization of HA in a 5.00 × 10⁻⁴ molL⁻¹ solution.

Explain, using equilibrium reasoning, why the percent ionization increases as the solution is diluted even though the solution becomes less acidic overall.

13.6.7 Polyprotic acid speciation: the H2CO3 system

Carbonic acid is diprotic:

A solution is prepared with formal concentration C = 0.0500molL⁻¹ H2CO3 at 25 ∘C.

Estimate the pH using only the first dissociation step, clearly stating approximations.

Using your pH estimate, determine whether the second dissociation step contributes significantly to [H3O+] by estimating [CO 32−]∕[HCO 3−] at equilibrium.

Determine the dominant carbonate-containing species (H2CO3, HCO3−, or CO32−) under these conditions and justify quantitatively.

13.6.8 pH of an acidic salt solution: NH4Cl

A 0.150molL⁻¹ aqueous solution of NH4Cl is prepared at 25 ∘C. The base ionization constant of ammonia is Kb(NH3) = 1.8 × 10⁻⁵.

Identify the ion responsible for hydrolysis and write the hydrolysis equilibrium reaction.

Determine Ka for NH4+ using K aKb = Kw.

Calculate [H3O+] and the pH of the salt solution using an ICE table.

Verify whether the small-x approximation is justified.

13.6.9 Amphiprotic salt: predicting whether NaHSO3 is acidic or basic

The hydrogen sulfite ion HSO3− is amphiprotic, acting as either an acid or a base in water. For sulfurous acid:

A 0.100molL⁻¹ solution of NaHSO3 is prepared at 25 ∘C.

Write the two competing equilibria for HSO3− acting as (i) an acid and (ii) a base.

Using Ka2 and Ka1, determine whether the solution should be acidic or basic without solving a full system of equations (justify by comparing the relevant Ka and Kb).

Estimate the pH using an appropriate simplifying approximation for amphiprotic species (state the approximation clearly).

13.6.10 Ranking acidity from molecular structure: inductive effects, bond strength, and oxyacid patterns

Rank the following sets from strongest acid to weakest acid, and for each set provide a brief structural justification based on concepts from this chapter (electronegativity, size/bond strength, resonance, and inductive effects). No numerical Ka values are required.

HF, HCl, HBr, HI

HClO, HClO2, HClO3, HClO4

CH3CO2H (acetic acid), ClCH2CO2H (chloroacetic acid), Cl2CHCO2H (dichloroacetic acid), CCl3CO2H (trichloroacetic acid)

H2O, H2S, H2Se, H2Te

Chapter 14
Aqueous Ionic Equilibria

While simple acid-base reactions provide a framework for proton transfer, the chemical reality of biological and geological systems rarely involves isolated species. This chapter explores the sophisticated interplay of ions in solution, where the presence of one species profoundly dictates the behavior of another-controlling pH in the human body through buffers, determining the fate of pollutants via solubility, and driving the precise analysis of unknown substances through titration.

14.1 The Common Ion Effect

In earlier chapters you learned to describe chemical equilibrium quantitatively with equilibrium constants (K) and qualitatively with Le Châtelier’s principle. You also applied these ideas to acid–base equilibria (using Ka and Kb) and to solubility equilibria (using Ksp). The common ion effect is a unifying application of those same equilibrium tools to aqueous ionic systems: adding an ion that already participates in an equilibrium shifts the equilibrium to consume part of the added ion. This shift suppresses ionization of weak electrolytes and decreases solubility of slightly soluble salts.

14.1.1 Definition and qualitative picture

A common ion is an ion that appears in an equilibrium system and is introduced (typically as a strong electrolyte that dissociates essentially completely). For an equilibrium written as

adding a product species drives the equilibrium left; adding a reactant drives the equilibrium right. Because many aqueous equilibria have ions as products, adding a salt that contributes one of those ions commonly drives the equilibrium back toward the undissociated (or solid) form.

Two central cases dominate general chemistry:

Weak acid or weak base ionization suppressed by a strong electrolyte containing the conjugate ion.

Solubility of a sparingly soluble salt suppressed by a soluble salt containing a lattice ion.

Although the chemical contexts differ, the mathematics is the same: the equilibrium constant does not change when you add a common ion (at constant temperature), but the equilibrium composition must change to keep the constant ratio of activities (or concentrations, in the typical dilute approximation).

14.1.2 Common ion effect for weak acids

Consider a weak monoprotic acid HA in water:

with

(Using bracket notation for concentrations; in rigorous thermodynamic form, Ka is defined in terms of activities. In this section we use concentrations as a practical approximation for moderately dilute solutions.)

Why adding A– suppresses ionization

If a soluble salt such as NaA is added, it dissociates essentially completely:

The added A− is a product of the acid ionization equilibrium. Le Châtelier’s principle predicts a shift to the left to consume A−, forming more HA and lowering [H3O+]. The acid becomes less ionized (lower percent ionization) in the presence of its conjugate base.

Quantitative treatment (ICE table approach)

Suppose a solution is prepared with a known analytical concentration of weak acid CHA and a known analytical concentration of its conjugate base (from a salt) CA⁻,0. Before the acid reacts appreciably, the dominant sources are:

Then let x be the amount of HA that ionizes:

[H3O+] eq | = x |  |
[A−] eq | = CA⁻,0 + x |  |
[HA]eq | = CHA − x |  |

Substitute into Ka:

Common ion approximation. When the added common ion concentration is much larger than the amount produced by ionization, CA⁻,0 ≫ x, then CA⁻,0 + x ≈ CA⁻,0 and often CHA − x ≈ CHA. This yields the simple estimate:

The key dependence is the inverse proportionality to the added conjugate base concentration: the more common ion you add, the smaller [H3O+] becomes.

Percent ionization. For a weak acid solution, the percent ionization is

With a common ion, x becomes much smaller than it would be in pure water; thus the percent ionization decreases sharply.

Example: acetic acid with and without sodium acetate

Acetic acid ionizes as

Let Ka = 1.8 × 10⁻⁵ (at 25∘C).

Case 1: 0.10M acetic acid in pure water. With no added acetate, an ICE setup gives (initially [CH3CO2−] ≈ 0):

Thus [H3O+] ≈ 1.34 × 10⁻³ M and

Percent ionization:

Case 2: 0.10M acetic acid plus 0.10M sodium acetate. Now CHA = 0.10M and CA⁻,0 = 0.10M. Using Eq. (??):

So [H3O+] ≈ 1.8 × 10⁻⁵ M and

Percent ionization:

Interpretation. Adding acetate (the common ion) decreased percent ionization from about 1.34% to about 0.018%—a reduction by nearly two orders of magnitude. Simultaneously, the pH rose substantially because the equilibrium produced far less H3O+ once acetate was already abundant.

14.1.3 Common ion effect for weak bases

A weak base B reacts with water:

Adding a soluble salt containing BH+ (for example, BHCl) introduces the conjugate acid BH+, which is a product of the base equilibrium. The equilibrium shifts left, decreasing [OH−] and lowering the solution’s basicity.

Example setup

If CB is the analytical concentration of base and CBH⁺,0 is the added conjugate acid concentration, and if x = [OH−] eq produced by the base reaction, then approximately

When CBH⁺,0 ≫ x and CB ≫ x,

mirroring the weak acid result.

14.1.4 A unifying viewpoint: equilibrium expressions enforce the shift

It is useful to connect the qualitative and quantitative ideas explicitly. For the weak acid equilibrium,

If [A−] is increased by adding NaA, then the ratio temporarily becomes too large (the reaction quotient Q exceeds Ka). The only way for the system to return to Ka is to decrease the numerator and/or increase the denominator: [H3O+] and [A−] must decrease (by reaction to form HA), and [HA] must increase. This is Le Châtelier’s principle expressed in the language of Q and K.

14.1.5 Common ion effect in solubility equilibria

The common ion effect is not limited to weak acids/bases. It strongly influences solubility when a sparingly soluble ionic solid is in equilibrium with its ions.

Solubility equilibrium and Ksp

For a slightly soluble salt

(Again, activities are the rigorous form; concentrations are commonly used for dilute solutions.)

If the salt dissolves to a molar solubility s in pure water, then

so s = for a 1 : 1 salt in pure water.

Adding a common ion suppresses solubility

Suppose a soluble salt contributes X− at concentration C X⁻,0 (for example, adding NaX). Let s be the new molar solubility of MX in the presence of the common ion. Then

and

If CX⁻,0 ≫ s, then CX⁻,0 + s ≈ CX⁻,0 and

Thus solubility is reduced approximately in inverse proportion to the added common-ion concentration.

Example: solubility of AgCl in chloride solutions

For

with a representative value Ksp = 1.8 × 10⁻¹⁰ (at 25∘C).

In pure water.

In 0.10M NaCl. Here CCl⁻,0 = 0.10M, so using Eq. (??),

The solubility is decreased by about four orders of magnitude relative to pure water.

Interpretation. Because chloride is already abundant, dissolving additional AgCl would increase [Cl−] only slightly, but it must increase [Ag+] by the same amount as the dissolution. The Ksp constraint forces [Ag+] to remain extremely small when [Cl−] is large; therefore only a tiny amount of solid can dissolve.

14.1.6 When approximations are valid (and when they fail)

The “common ion approximation” (treating Ccommon + x ≈ Ccommon) is powerful, but it must be justified. A simple, practical check is:

Solve using the approximation to get x (or s).

Verify that x ≤ 0.05Ccommon (and often also x ≤ 0.05Cweak for weak acids/bases).

If the check fails, you should retain x (or s) in the equilibrium expression and solve the resulting quadratic (or higher-order) equation.

14.1.7 Conceptual connections: why the common ion effect matters

The common ion effect is not a niche curiosity; it is a central organizing idea for aqueous equilibria.

Buffers are common-ion systems. A buffer typically contains a weak acid and its conjugate base (or a weak base and its conjugate acid). The resistance of pH to added acid/base is rooted in the same suppression of ionization described here.

Selective precipitation relies on common ions. In mixtures, controlling the concentration of a common ion can lower the solubility of one salt enough to precipitate it while others remain dissolved, a key idea in qualitative analysis and separation schemes.

Equilibrium constants stay fixed; compositions shift. This section reinforces a frequent equilibrium misconception: adding a reactant or product does not change K (at fixed T); it changes the equilibrium concentrations so that Q returns to K.

The common ion effect is the shift in an ionic equilibrium caused by adding an ion already present in the equilibrium expression.

For weak acids, adding the conjugate base A− suppresses ionization, lowering [H3O+] and decreasing percent ionization.

For weak bases, adding the conjugate acid BH+ suppresses base ionization, lowering [OH−].

For sparingly soluble salts, adding a lattice ion decreases solubility because Ksp constrains the product of ion concentrations.

Quantitatively, in the strong common-ion limit, [H3O+] ≈ for a weak acid system and s ≈ for a 1 : 1 salt.

14.2 Buffered Solutions

The common ion effect provides the central idea behind buffered solutions: adding a substantial concentration of a conjugate partner (a “common ion”) suppresses the extent to which a weak acid or weak base must react to establish equilibrium. A buffer is an aqueous solution that resists changes in pH when moderate amounts of strong acid or strong base are added. Buffer action is not mysterious; it is simply an application of equilibrium shifting and stoichiometry.

14.2.1 What a buffer is (and is not)

A practical buffer contains both members of a conjugate acid–base pair in appreciable amounts:

a weak acid HA and its conjugate base A− (often supplied as a soluble salt, e.g., NaA), or

a weak base B and its conjugate acid BH+ (often supplied as a soluble salt, e.g., BHCl).

Key requirement: comparable concentrations. Buffers work best when the concentrations of the conjugate pair are of the same order of magnitude (commonly within a factor of about 10). This condition keeps the solution away from extremes where one component is nearly exhausted and pH changes rapidly.

What is not a buffer.

A solution of only a weak acid (no conjugate base added) is not a good buffer; its pH can shift substantially upon addition of base because there is no reservoir of conjugate base to consume added H3O+ and no large common-ion suppression.

A solution of a strong acid and its salt (e.g., HCl + NaCl) is not a buffer; the conjugate base of a strong acid (Cl– ) is too weak to meaningfully react with H3O+.

14.2.2 How buffers respond to added strong acid or base

Consider an acid buffer composed of HA and A−. The weak acid equilibrium is

Because both HA and A− are present, the equilibrium position is strongly influenced by the common ion A− (Section 14.1). The buffer’s resistance arises primarily from fast stoichiometric reactions with added strong acid/base, followed by a small equilibrium readjustment.

Addition of strong base

Let a small amount of strong base (source of OH– ) be added. The dominant reaction

is

This reaction consumes OH– nearly completely (to the extent that HA is available), converting HA to A– . Therefore, the pH does not jump as it would in pure water; instead, it changes modestly because the ratio [A−]∕[HA] changes.

Addition of strong acid

If strong acid (source of H3O+) is added, the dominant reaction is

This consumes added H3O+ by converting A– to HA. Again, pH changes modestly

because the equilibrium composition shifts but does not collapse.

The idea of a “reservoir”

A buffer contains a reservoir of weak acid to neutralize added base (Eq. ??) and a reservoir of conjugate base to neutralize added acid (Eq. ??). Once either reservoir is largely depleted, the buffer fails and pH changes sharply.

14.2.3 Derivation of the Henderson–Hasselbalch equation

For an acid buffer, start with the acid dissociation constant expression:

Solve for [H3O+]:

Take the negative base-10 logarithm of both sides:

− log[H3O+] | = −log K a − log |  |
pH | = pKa − log |  |
pH = pKa + log |  |

Equation (??) is the Henderson–Hasselbalch equation. It makes two ideas immediate:

When [A−] = [HA], pH = pK a.

pH depends on the ratio of conjugate base to acid, not their absolute magnitudes (though absolute magnitudes control buffer capacity).

Important note on concentrations vs. activities. The exact equilibrium expression uses activities; replacing activities by concentrations is most accurate at low ionic strength. In many general-chemistry buffer calculations (moderate dilutions, typical laboratory practice) the Henderson–Hasselbalch equation provides excellent estimates. At higher ionic strength, activity coefficients shift the effective pKa and the measured pH; more advanced treatments incorporate activity corrections.

Base-buffer form. For a buffer made from a weak base B and its conjugate acid BH+, it is often simplest to use the conjugate acid equilibrium for BH+ with K a (where KaKb = Kw):

Then

14.2.4 Visual comparison: buffered vs. unbuffered response to added base

Immediately after Eq. (??), it is useful to visualize what “resists pH change” means operationally: compare adding the same amount of strong base to pure water versus to a buffer. The unbuffered pH rises sharply because nearly all added OH– remains as OH– . In a buffer, added OH– is consumed by Eq. ?? until the weak acid is largely exhausted; only then does pH rise quickly.

14.2.5 Buffer range and why pH ≈ pKa matters

Equation (??) shows that pH is controlled by the ratio [A−]∕[HA]. If that ratio becomes very large or very small, the logarithm changes rapidly with small composition changes, and buffering becomes weak.

A common operational definition of the buffer range is:

which corresponds to

Within this window both components are present in appreciable amounts, so the buffer can respond to added acid or base without exhausting one component too quickly.

14.2.6 Buffer capacity and exhaustion

Definition and conceptual meaning

Buffer capacity is a measure of how much strong acid or strong base a buffer can neutralize before the pH changes by an unacceptably large amount. Qualitatively:

Buffer capacity increases as the total concentration of buffer components increases (more moles of HA and A− available to react).

Buffer capacity is largest near pH = pKa (where [HA] ≈ [A−]), because the buffer can neutralize both added acid and added base effectively.

Stoichiometric exhaustion

Buffer failure is ultimately a stoichiometric limitation. For an acid buffer, added OH– consumes HA (Eq. ??). If the moles of added base approach the initial moles of HA, the buffer approaches exhaustion:

At that point the Henderson–Hasselbalch equation no longer applies because the ratio [A−]∕[HA] becomes ill-defined and the chemistry is no longer governed by a weak acid equilibrium; excess strong base dominates pH.

14.2.7 How to calculate buffer pH in practice

Nearly all buffer problems involve two linked steps:

Stoichiometry (reaction accounting) for the strong acid/base addition.

Equilibrium (Henderson–Hasselbalch) using the updated amounts of conjugate pair.

General recipe for an acid buffer

For a buffer with HA∕A−:

Write the neutralization reaction with the added strong reagent: if adding strong base, use Eq. ??; if adding strong acid, use Eq. ??.

Use mole accounting to compute new moles n(HA) and n(A−) after neutralization.

Compute pH with

If volume changes are small or the same factor applies to numerator and denominator, you may use mole ratios directly:

because [A−]∕[HA] = n(A−)∕V ∕n(HA)∕V .

Worked example: pH change upon adding strong base

Problem. A buffer is prepared by mixing 50.0 mL of 0.200 molL⁻¹ acetic acid with 50.0 mL of 0.200 molL⁻¹ sodium acetate. Calculate the pH before and after adding 5.00 mL of 1.00 molL⁻¹ NaOH. Use Ka = 1.8 × 10⁻⁵.

Step 1: initial pH. Initial moles:

Since n₀(A−) = n 0(HA), Eq. (??) gives

Step 2: stoichiometry of added base. Moles of OH– added:

Neutralization consumes HA and produces A– (Eq. ??):

n(HA) | = 0.0100 − 0.00500 = 0.00500mol |  |
n(A−) | = 0.0100 + 0.00500 = 0.0150mol. |  |

Step 3: apply Henderson–Hasselbalch.

Interpretation. Adding 5.00 mmol OH– increased pH by only about 0.48 pH units. In contrast, adding the same base amount to 105 mL of pure water would produce [OH−] ≈ 0.0476molL⁻¹ and pH ≈ 12.68—a dramatic difference in response.

14.2.8 Designing buffers: choosing a conjugate pair and proportions

Buffer design begins with a target pH and proceeds by selecting a weak acid (or base) whose pKa is close to that pH.

Selection principle

To design an acid buffer at a desired pH:

If the target pH differs from pKa, the needed ratio follows from Eq. (??):

Example: set the ratio for a target pH

Problem. You want a buffer at pH = 5.00 using acetic acid/acetate (pKa = 4.74). What ratio [A−]∕[HA] is required?

Calculation.

Thus you need about 1.8 times as much acetate as acetic acid (on a concentration basis, or equivalently on a mole basis for a fixed volume).

14.2.9 Common pitfalls and how to avoid them

Using Henderson–Hasselbalch without stoichiometry. If strong acid/base is added, always account for neutralization first (moles change), then apply Eq. (??).

Forgetting that buffering is finite. No buffer can absorb unlimited acid/base. Check whether the added amount is large enough to substantially deplete one component.

Assuming dilution never matters. If volume changes are large (e.g., adding significant volumes of titrant), compute concentrations explicitly rather than relying only on mole ratios.

Using an inappropriate conjugate pair. If the target pH is far from pKa, the required ratio becomes extreme and buffering weak.

A buffer contains a weak acid/base and its conjugate partner in comparable amounts; buffer action is a direct consequence of the common ion effect and rapid neutralization reactions.

The Henderson–Hasselbalch equation,

relates buffer pH to the conjugate pair ratio.

Buffers are most effective near pH ≈ pKa and within about ±1 pH unit of pKa.

Buffer capacity increases with total buffer concentration and is limited by stoichiometric exhaustion of one component.

14.3 Acid-Base Titrations

Acid–base titration is an experimental method for determining the concentration of an acid or base by reacting it with a solution of known concentration (the titrant) until stoichiometric completion is reached. The central output of an acid–base titration is the titration curve: a plot of pH versus volume of titrant added. Interpreting titration curves requires a coordinated use of (1) stoichiometry, (2) equilibrium (weak-acid/weak-base chemistry), and (3) the buffer concepts developed in the previous section.

14.3.1 Key definitions: equivalence point, end point, and indicators

Equivalence point

The equivalence point is the point in the titration at which the moles of titrant added are in exact stoichiometric proportion to the analyte being titrated. For a monoprotic acid titrated with a strong base:

the equivalence condition is

At equivalence, the initial acid has been converted entirely (in the stoichiometric sense) into its conjugate base A– .

End point

The end point is the experimentally observed point (often indicated by a color change) used to stop the titration. A well-chosen indicator makes the end point coincide closely with the equivalence point; otherwise, a systematic titration error results.

Indicators

An acid–base indicator is typically a weak acid (or weak base) whose conjugate forms have different colors. For a common indicator treated as a weak acid HIn:

The color depends on the ratio [In−]∕[HIn], so the visible transition occurs over a pH interval where both forms are present in comparable amounts.

14.3.2 General strategy for titration calculations

Regardless of titration type, the pH at any point is obtained by identifying which chemistry dominates:

Stoichiometric reaction accounting. Strong acid/base neutralization is essentially complete. Compute moles remaining and moles produced after reaction.

Select the appropriate equilibrium model. Depending on what species remain: a strong acid/base controls pH directly (via excess H3O+ or OH– ); a buffer mixture controls pH via Henderson–Hasselbalch; a conjugate base/acid of a weak species controls pH at equivalence via hydrolysis.

Account for dilution. Concentrations change as volume increases: C = n∕V total.

14.3.3 Classification of titration types and qualitative curve shapes

Titration curves differ because the identity and strength of the acid/base present changes over the course of the titration.

Titration type | Equivalence-point pH | Key curve feature
Strong acid–strong base | ≈ 7 (at 25∘C) | Very steep pH jump near equivalence; no buffer region.
Weak acid–strong base | > 7 | Buffer region before equivalence; half-equivalence at pH = pKa.
Strong acid–weak base | < 7 | No true buffer region; equivalence governed by conjugate acid hydrolysis.
Weak acid–weak base | Depends on Ka and Kb | Small pH jump; indicators often poor; pH may require pH meter.

The most instructive comparison is between (i) strong acid–strong base and (ii) weak acid–strong base. Both have a sharp rise near equivalence, but the weak-acid case has a distinctive buffer region and an equivalence point above pH 7.

14.3.4 Strong acid–strong base titration: pH at the four stages

Consider titrating V ₀ L of a strong monoprotic acid (e.g., HCl) of concentration Ca with a strong base (e.g., NaOH) of concentration Cb. The reaction

goes essentially to completion.

Define the moles initially present:

At a volume V b of base added:

Stage 1: initial pH (V b = 0)

Because the acid is strong:

Stage 2: pre-equivalence (0 < V b < V eq)

Excess H3O+ remains after neutralization:

Stage 3: equivalence (V b = V eq)

At equivalence:

No excess strong acid/base remains; in an ideal dilute solution at 25 ∘C,

(acknowledging that the exact pH can deviate slightly due to activity effects).

Stage 4: post-equivalence (V b > V eq)

Excess OH– controls pH:

14.3.5 Weak acid–strong base titration: buffer region and basic equivalence point

Now consider titrating V ₀ L of a weak monoprotic acid HA (concentration Ca) with a strong base of concentration Cb:

In addition to stoichiometry, the weak acid equilibrium (and its conjugate base hydrolysis) matters.

Let nHA,0 = CaV ₀ and nOH− = CbV b.

Stage 1: initial pH (V b = 0)

The initial pH is set by weak acid dissociation:

often approximated by [H3O+] ≈ when ionization is small.

Stage 2: pre-equivalence buffer region (0 < V b < V eq)

Added OH– converts some HA to A– . After stoichiometric neutralization:

Both species are present, forming a buffer. Using Henderson–Hasselbalch:

Half-equivalence point. At half-equivalence, nOH− = nHA,0 so n(A−) = n(HA) and the Henderson–Hasselbalch equation gives

This provides an experimental route to pKa from the titration curve.

Stage 3: equivalence (V b = V eq)

At equivalence, all HA has been converted to A– :

The pH is now determined by base hydrolysis of A– :

Its equilibrium constant is

Assuming [A−] 0 at equivalence is known, solve

to estimate x = [OH−], then compute pH. Because K b > 0 for the conjugate base of a weak acid, the equivalence point is basic:

Stage 4: post-equivalence (V b > V eq)

Excess strong base dominates; treat as in the strong–strong case using excess OH– .

14.3.6 Comparing strong–strong vs. weak–strong titration curves

The contrast between these two titrations is best captured by plotting both curves on the same axes:

The strong acid begins at a lower pH (fully dissociated), then shows a very steep rise centered around pH 7 at equivalence.

The weak acid begins at a higher pH (partially dissociated), exhibits a buffer region (gradual slope) before equivalence, passes through pH = pKa at half-equivalence, and reaches an equivalence point at pH > 7.

14.3.7 Strong acid–weak base titration: acidic equivalence point

When a strong acid titrates a weak base B, the stoichiometric reaction forms the conjugate acid BH+:

At equivalence, the solution contains primarily BH+, a weak acid:

Thus, the equivalence point is acidic:

The pH at equivalence is found by treating BH+ as a weak acid with

and then solving for [H3O+] using the formal concentration of BH+ at the total volume.

14.3.8 Weak acid–weak base titration: limited pH jump

When both reactants are weak, the pH change near equivalence is typically much smaller than in strong–strong titrations. This has practical consequences:

The equivalence point pH depends on both Ka and Kb.

Visual indicators often perform poorly because there may be no steep pH region that fully spans an indicator’s transition range.

A pH meter (potentiometric titration) is commonly used for accurate endpoint determination.

14.3.9 Indicator theory and selecting an indicator

Indicator equilibrium and transition range

For an indicator HIn:

so

Human vision typically detects a distinct color change when the ratio of the two forms changes from about 1∕10 to 10∕1. Therefore, the indicator transition range is approximately

Indicator selection criterion

To minimize systematic error, choose an indicator whose transition range lies within the steep portion of the titration curve near the equivalence point.

Strong acid–strong base: the pH jump is large and centered near pH 7, so many indicators work (e.g., bromothymol blue is a common choice).

Weak acid–strong base: equivalence is basic (pH > 7), so indicators with higher transition ranges are preferred (e.g., phenolphthalein often performs well).

Strong acid–weak base: equivalence is acidic (pH < 7), so indicators with lower transition ranges are preferred (e.g., methyl orange can be appropriate).

Connecting the curve to indicator error

The magnitude of indicator error depends on how much titrant volume corresponds to the indicator’s color-change interval. In a steep curve, a large pH change occurs with a tiny volume change, so the error is small. In a shallow curve (especially weak–weak titrations), the same pH interval may correspond to several milliliters of titrant, leading to significant concentration error.

14.3.10 Practical notes: reading titration curves

Inflection point and equivalence. For strong–strong titrations, the equivalence point coincides closely with the inflection point (maximum slope). For weak–strong titrations, the inflection point remains near equivalence, but the equivalence pH shifts away from 7 due to conjugate base/acid hydrolysis.

Half-equivalence as a pKa measurement. In weak acid–strong base titrations, pH = pKa at half-equivalence provides a convenient experimental determination of pKa.

Buffer region interpretation. A shallow slope region before equivalence indicates significant amounts of both HA and A– (a buffer), consistent with the common ion effect and the Henderson–Hasselbalch equation.

Titration curves reflect changing solution composition; pH calculations combine stoichiometry with equilibrium.

Strong acid–strong base titrations have equivalence near pH 7 and a large pH jump.

Weak acid–strong base titrations exhibit a buffer region, half-equivalence at pH = pKa, and a basic equivalence point due to conjugate base hydrolysis.

Strong acid–weak base titrations have acidic equivalence points due to conjugate acid hydrolysis.

Indicator choice is guided by matching the indicator transition range to the steep portion of the titration curve near equivalence.

14.4 Solubility Equilibria (Ksp)

Acid–base equilibria involve the competition between molecular and ionic forms in water. A closely related class of equilibria arises when an ionic solid is in contact with water. Many ionic compounds are described as “insoluble,” but in equilibrium terms this usually means slightly soluble: a small, but nonzero, concentration of ions is present in a saturated solution. The quantitative tool that governs such systems is the solubility product constant, Ksp.

This section develops Ksp systematically and shows how to compute molar solubility, ion concentrations in saturated solutions, and the effect of changing conditions (excluding the common ion effect and complex-ion formation, which are treated separately).

14.4.1 Dissolution as an equilibrium

Consider a pure ionic solid MX(s) placed in water. Two opposing processes occur:

ions detach from the lattice and become hydrated (dissolution),

hydrated ions collide with the solid surface and re-form the lattice (precipitation).

When the rates become equal, a dynamic equilibrium is established and the solution is saturated. The equilibrium can be represented as

14.4.2 The solubility product constant

Definition

For the equilibrium in Eq. (??), the equilibrium constant expression is

because the activity of the pure solid MX(s) is 1 and does not appear in the expression. The special name solubility product constant is used:

In typical general-chemistry calculations for moderately dilute solutions, activities are approximated by molar concentrations:

Units and interpretation. Strictly, equilibrium constants are dimensionless when written in terms of activities. When concentrations are used, Ksp is commonly reported with implicit standard-state conventions. Conceptually, a smaller Ksp indicates a stronger tendency for ions to remain in the solid lattice (lower solubility).

General form for arbitrary stoichiometry

For a solid with formula MaXb that dissolves as

the solubility product is

The exponents match the stoichiometric coefficients of the ions in the dissolution reaction.

14.4.3 Molar solubility s vs. Ksp

Definitions

Molar solubility, s, is the number of moles of solid that dissolve per liter of solution to form a saturated solution (units: molL⁻¹).

Ksp is an equilibrium constant that constrains the ion activities (or concentrations) in any saturated solution at a fixed temperature.

These are related but not identical. Ksp is a constant (at fixed T), while s depends on the dissolution stoichiometry and on solution conditions such as pH and the presence of other solutes (which can change ion activities). In this section we focus on the baseline relationship between Ksp and s in pure water and in simple cases without common ions or complex formation.

Deriving s from Ksp: common stoichiometries

The core technique is to express equilibrium ion concentrations in terms of s, then substitute into the Ksp expression.

Case 1: MX (1:1 salt). If MX(s) dissolves to molar solubility s in pure water:

so

Case 2: MX2 (1:2 salt).

If molar solubility is s:

so

Case 3: M2X3 (2:3 salt).

If molar solubility is s:

so

Worked example: computing solubility and ion concentrations

Example. A salt BaF2(s) dissolves as

If Ksp = 1.7 × 10⁻⁶ at a given temperature, find the molar solubility s in pure water and the equilibrium concentrations of Ba2+ and F– .

Solution outline. Let s be molar solubility:

Therefore,

14.4.4 The reaction quotient Qsp and the meaning of “saturated”

The reaction quotient for solubility, Qsp, has the same form as Ksp but uses current (not equilibrium) ion concentrations:

Comparing Qsp to Ksp predicts whether a precipitate will form:

Qsp < Ksp: solution is unsaturated (more solid can dissolve).

Qsp = Ksp: solution is saturated (equilibrium with solid).

Qsp > Ksp: solution is supersaturated (precipitation is thermodynamically favored).

Although the detailed use of Qsp for predicting precipitation in mixtures is emphasized later, it is useful here to connect Ksp to the physical states of a solution.

14.4.5 Approximations and solving Ksp problems rigorously

When the s-only assumption is valid

In pure water for many slightly soluble salts, it is valid to assume that the only significant source of ions is dissolution of the solid itself. Then the stoichiometric relationships (e.g., [X−] = 2s for MX 2) are exact in the concentration model, and Ksp leads directly to s.

When quadratics appear

Quadratic (or higher-order) equations arise when an ion concentration is not simply proportional to s. Two common situations:

An ion is present initially from another source (treated in depth in the common-ion section and later in precipitation problems).

A dissolved ion participates in a secondary equilibrium that changes its free concentration (complexation, hydrolysis, or acid–base reactions). Hydrolysis and complexation are treated later, but one important case appears already: sparingly soluble hydroxides in water can couple to water autoionization and pH.

In such cases, the systematic approach is:

define unknown concentrations,

write mass-balance and charge-balance relationships as needed,

apply equilibrium expressions (Ksp and any other K values),

solve algebraically (often approximately, with checks).

14.4.6 Factors affecting solubility (excluding common ions and complex ions)

At fixed temperature, Ksp is constant, but “solubility” as measured experimentally can depend on conditions that alter ion activities or shift coupled equilibria. Two broad factors are important here.

Temperature

Most dissolution processes have nonzero enthalpy changes, so Ksp generally depends on temperature. If dissolution is endothermic, Ksp typically increases with temperature; if exothermic, it typically decreases. Quantitative temperature dependence requires thermodynamic data or measurements at multiple temperatures (often via a van’t Hoff analysis), which is beyond the scope of this section but consistent with general equilibrium principles:

Solvent composition and ionic strength (activities)

In real solutions, ions interact electrostatically; as ionic strength increases, activity coefficients deviate from 1. Because Ksp is defined by activities,

the concentration-based “Ksp” can appear to change unless activity coefficients are accounted for. In introductory calculations, solutions are treated as ideal enough that γ ≈ 1; in more accurate treatments, activity corrections become essential, especially for multivalent ions.

14.4.7 Common Ksp expressions and s relationships

The following summary is a useful reference for common dissolution stoichiometries in pure water.

Salt | Dissolution reaction | Relation between Ksp and s (pure water)
MX | MX(s) ↔ M+ + X− | K sp = s², s =
MX2 | MX2(s) ↔ M2+ + 2 X− | K sp = 4s³, s = 1∕3
M2X3 | M2X3(s) ↔2M3+ + 3 X2− | K sp = 108s⁵, s = 1∕5

Slightly soluble ionic solids establish a dynamic equilibrium between solid and hydrated ions in water; saturated solutions satisfy a fixed solubility product constant, Ksp, at a given temperature.

Ksp is written as the product of ion activities raised to stoichiometric powers; pure solids do not appear in the equilibrium expression.

Molar solubility s is linked to Ksp through dissolution stoichiometry (e.g., Ksp = s² for a 1:1 salt in pure water).

The reaction quotient Qsp provides a saturation test: Qsp < Ksp (unsaturated), Qsp = Ksp (saturated), Qsp > Ksp (supersaturated, precipitation favored).

Temperature and non-ideality (activity coefficients) influence observed solubility; common-ion and complex-ion effects are treated in later sections.

14.5 Precipitation and Complex Ions

The solubility product constant, Ksp, quantifies the equilibrium between a slightly soluble ionic solid and its ions in solution. In practice, we often mix ionic solutions and ask two closely related questions:

Will a precipitate form? (precipitation prediction)

If a precipitate forms, can we dissolve it by adding a ligand? (complex-ion effects on solubility)

Both questions are answered by extending the equilibrium tools developed earlier: (i) compare the reaction quotient Q to K to predict the direction of change, and (ii) treat complex formation as a coupled equilibrium that changes the concentration of free metal ions, thereby shifting solubility equilibria.

14.5.1 Predicting precipitation using Qsp versus Ksp

The central criterion

For a salt

the solubility product is

and the corresponding ionic reaction quotient is

Then:

If Qsp < Ksp, the solution is unsaturated with respect to that solid: no precipitation (and additional solid could dissolve, if present).

If Qsp = Ksp, the solution is exactly saturated: equilibrium.

If Qsp > Ksp, the ion product is too large: precipitation occurs until Qsp is reduced to Ksp.

Why free-ion concentrations matter

The expressions for Ksp and Qsp involve the concentrations (more precisely, activities) of free ions in solution. If an ion is tied up in another form—for example, metal ions bound to ligands as complex ions—then the free-ion concentration may be far smaller than the analytical (total) concentration. This distinction becomes critical later in this section.

14.5.2 Computing Qsp after mixing solutions

To evaluate Qsp, you must compute ion concentrations after mixing (dilution is often decisive).

Procedure

Suppose V ₁ of solution 1 containing Mn+ at C M,1 is mixed with V ₂ of solution 2 containing Xm− at C X,2. Assume no reaction occurs initially except mixing. Then:

Insert these into Qsp and compare with Ksp.

Example: precipitation prediction for AgCl

Problem. Mix 25.0 mL of 1.0 × 10⁻³ molL⁻¹ AgNO3 with 25.0 mL of 1.0 × 10⁻³ molL⁻¹ NaCl. Will AgCl(s) precipitate? Use Ksp(AgCl) = 1.8 × 10⁻¹⁰.

Mixing step. Total volume = 50.0mL = 0.0500L.

Then

Since Qsp ≫ Ksp, precipitation is favored; AgCl(s) will form until the ion product is reduced to Ksp.

14.5.3 Selective (fractional) precipitation: separating ions by Ksp

When a solution contains multiple cations (or anions) that can form precipitates with a common counterion, the least soluble compound (smallest Ksp under the relevant conditions) tends to precipitate first as the counterion concentration is increased. This underlies fractional precipitation as a separation technique.

Onset of precipitation condition

Consider a mixture of metal ions Mn+ and Nn+ in the presence of an anion Xm− that can precipitate both:

Precipitation of MaXb begins when

Solving for the critical Xm− concentration at the onset of precipitation gives

The salt with the smaller required [Xm−]crit precipitates first (for the same initial metal concentration).

Example framework: halide separations (qualitative analysis)

In classical qualitative analysis, Ag+, Pb2+, and Hg22+ are precipitated as chlorides by adding Cl– , whereas many other cations remain in solution because their chlorides are much more soluble. Although modern laboratories often use instrumental methods, the underlying equilibrium reasoning remains valuable: the order of precipitation is controlled by Ksp, concentrations, and (as we will see) complexation.

14.5.4 From precipitation to dissolution: complex ions as a solubility lever

The common ion effect decreases solubility by increasing a lattice ion concentration. In contrast, complex-ion formation often increases solubility by decreasing the concentration of a free metal ion through binding to a ligand.

Lewis acid–base perspective

A metal cation acts as a Lewis acid (electron-pair acceptor), while many ligands (such as NH3, CN– , OH– , F– , or Cl– ) act as Lewis bases (electron-pair donors). Complexation can be represented as

or, more generally,

Formation constants

The stability of a complex is quantified by a formation constant (Kf):

A large Kf means that, at equilibrium, most of the metal exists as the complex when ligand is available. This can drastically reduce the free metal-ion concentration [Mn+], even if the total metal concentration remains substantial.

14.5.5 How complexation increases solubility: coupling Ksp and Kf

Conceptual mechanism

For a sparingly soluble salt such as AgCl(s):

If a ligand L binds Ag+:

then complexation removes free Ag+ from solution. The solubility equilibrium responds (Le Châtelier) by dissolving more AgCl to replace the depleted free Ag+. Thus a precipitate can dissolve upon addition of ligand.

Overall (net) equilibrium constant

It is often useful to combine equilibria by adding reactions and multiplying constants.

For example, suppose Ag+ forms [Ag(NH3)2]+:

Add this to the dissolution of AgCl:

to obtain the net process:

The equilibrium constant for the net reaction is

When Kf is large, Knet can be large enough that dissolution is strongly favored in the presence of sufficient ligand.

Interpretation. Knet quantifies the idea that solubility is not an intrinsic property of a solid alone; it depends on what other equilibria in solution remove or supply its ions.

14.5.6 A practical calculation pattern: conditional solubility in a ligand solution

In many real problems, a ligand is present in large excess, so its concentration can be treated as approximately constant. This leads to a conditional solubility that is greater than the solubility in pure water.

Example template: 1:1 salt with a dominant complex

Let a salt dissolve as

and suppose the metal forms a dominant complex

Define the total dissolved metal (the solubility) as

From Kf,

So

Meanwhile, dissolution of the salt provides

The Ksp expression involves the free metal ion:

Use Eq. (??) to eliminate [M+]:

Then

Therefore, the solubility is increased to

What Eq. (??) means. If Kf[L]p ≫ 1, then s ≈ [L]p∕2 (for this simplified 1:1 case). The stronger the complex and the higher the ligand concentration, the larger the solubility.

Caution. Equation (??) assumes one dominant complex and neglects activity effects. Real systems may require accounting for multiple complexes (stepwise formation constants), ligand protonation equilibria (acid–base chemistry), and

competition from other ligands.

14.5.7 Complex ions and selective precipitation: complexation can reverse expectations

Because complexation reduces free metal-ion concentration, it can:

prevent precipitation that would otherwise occur (by lowering Qsp),

dissolve an existing precipitate (by shifting the solubility equilibrium toward dissolution),

change the order of fractional precipitation in mixtures (by stabilizing one metal ion more than another).

Thus, selectivity is not determined by Ksp alone; it depends on the coupled network of equilibria in solution.

14.5.8 Laboratory perspective: precipitation tests and complexation

Two operational principles guide laboratory use:

Precipitation as identification. A characteristic precipitate forms under controlled conditions (known concentrations and pH), indicating the possible presence of a given ion.

Complexation as confirmation or separation. Adding a ligand that selectively dissolves one precipitate but not another can provide confirmatory evidence and enables separations.

A classic example is the dissolution of AgCl(s) in aqueous ammonia due to formation of [Ag(NH3)2]+, while other chlorides may not dissolve as readily. The chemical logic is the coupling of Ksp with a large Kf.

Precipitation is predicted by comparing Qsp to Ksp using post-mixing ion concentrations: Qsp > Ksp implies precipitation until Qsp = Ksp.

Fractional precipitation exploits differences in Ksp: the solid requiring the smallest counterion concentration to satisfy Qsp = Ksp precipitates first.

Complex ions form by Lewis acid–base interactions between metal ions (Lewis acids) and ligands (Lewis bases). Their stability is quantified by formation constants Kf.

Complexation can increase solubility by lowering the free metal-ion concentration, pulling the dissolution equilibrium toward products. Net equilibria can be obtained by adding reactions and multiplying constants (e.g., Knet = KspKf for a dissolution-plus-complexation process).

In real aqueous systems, precipitation behavior depends on coupled equilibria (Ksp, Kf, and often acid–base equilibria), not on Ksp alone.

14.6 Problems

14.6.1 Mixed weak acids: full equilibrium pH (no single-acid shortcut)

A solution is prepared by dissolving 0.0500 mol acetic acid (CH3CO2H, Ka = 1.8 × 10⁻⁵) and 0.0200 mol formic acid (HCO2H, Ka = 1.8 × 10⁻⁴) in water to a final volume of 1.000 L at 25∘C. Neglect activity effects.

Write the charge-balance and mass-balance relationships for all relevant species.

Compute the pH by solving the coupled equilibria (do not assume one acid dominates without justification).

Report the fractional compositions α of each acid system (fraction present as HA vs. A– ) at the computed pH.

14.6.2 Buffer capacity of a blood-plasma model (quantitative and conceptual)

A simplified blood plasma buffer is modeled as a H2CO3/HCO3 – system with pKa = 6.35 at 25∘C. A 1.00 L sample contains [HCO3−] = 0.0240 M and dissolved CO2 in equilibrium such that [H2CO3] = 0.00120 M.

Calculate the initial pH using the Henderson–Hasselbalch equation.

Define buffer capacity operationally as the moles of strong base required to raise pH by exactly 0.20 pH units. Assuming only stoichiometric conversion between H2CO3 and HCO3 – (ignore gas exchange), calculate the required moles of NaOH.

Repeat for addition of strong acid to lower pH by 0.20 pH units (moles of HCl).

Explain briefly why the capacities for acid and base additions may differ for the same initial buffer composition.

14.6.3 Diprotic acid titration: pH at the first equivalence point

A 25.00 mL sample of 0.1000 M diprotic acid H2A is titrated with 0.1000 M NaOH. The acid has Ka1 = 1.0 × 10⁻³ and Ka2 = 1.0 × 10⁻⁸ at 25∘C.

Determine the volume of base required to reach the first equivalence point.

At the first equivalence point, the solution contains primarily HA– . Derive an expression for pH at this point in terms of Ka1, Ka2, and Kw (show key steps).

Evaluate the pH numerically, clearly stating any justified approximations.

14.6.4 Estimating Ksp from electrochemical data (cross-reference)

At 25∘C, the standard reduction potential for the half-reaction

is E∘ = +0.7996 V. Consider a concentration cell built from:

Left half-cell: Ag(s) in 1.00 M AgNO3.

Right half-cell: Ag(s) in a saturated AgCl solution prepared by equilibrating pure AgCl(s) with pure water (no added electrolytes).

The measured cell potential is Ecell = 0.577 V.

Use the Nernst equation to determine [Ag+] in the saturated AgCl solution.

Use the solubility equilibrium to compute Ksp(AgCl).

Briefly discuss how activity coefficients could bias the result and in what direction you would expect the concentration-based estimate of Ksp to shift relative to the activity-based value.

14.6.5 Selective precipitation order for a mixture of halides

A 1.000 L solution contains Ag+, Pb2+, and Hg22+ each at 1.00 × 10⁻³ M. Chloride is added slowly as NaCl while maintaining constant volume (assume volume change negligible). Use:

For each cation, compute the critical [Cl−] at which precipitation begins.

Determine the order in which the chlorides precipitate as Cl– is increased.

At the moment AgCl just begins to precipitate, calculate the remaining free concentrations of the other two cations (assuming no complexation and that their solids have not yet formed).

14.6.6 Solubility of metal hydroxides in a pH-controlled buffer

A suspension of Mg(OH)2(s) is placed in a large volume of buffer solution that holds the pH constant at 9.50 at 25∘C. Assume the buffer has sufficient capacity to maintain constant pH and ignore complex formation. Use Ksp(Mg(OH)2) = 5.6 × 10⁻¹² and Kw = 1.0 × 10⁻¹⁴.

Compute [OH−] imposed by the buffer.

Determine the equilibrium free concentration [Mg2+] in the buffered solution.

Determine the molar solubility s of Mg(OH)2 under these conditions and compare it (qualitatively) to its solubility in pure water.

14.6.7 Coupled Kf and Ksp: dissolution of AgCl in ammonia

Solid AgCl is added to 1.00 L of aqueous ammonia such that the free ammonia concentration remains approximately constant at [NH3] = 0.200 M (assume ammonia is in large excess and neglect protonation of ammonia). Use:

Write the coupled equilibria and the net reaction for dissolution in the presence of ammonia.

Derive an expression for the total dissolved silver concentration stot = [Ag+] + [[Ag(NH 3)2]+] in terms of K sp, Kf, and [NH3] under the stated approximation.

Evaluate stot numerically.

Determine the fraction of dissolved silver present as free Ag+.

14.6.8 Buffer design for an industrial enzymatic reaction (constraints and optimization)

An enzyme-catalyzed reaction must be maintained at pH = 7.40 and ionic strength should be minimized. You are restricted to preparing a buffer from one of the following conjugate pairs:

Buffer system | Conjugate pair | pKa at 25∘C
Acetate | CH3CO2H/CH3CO2 – | 4.74
Phosphate (2nd dissociation) | H2PO4 – /HPO42– | 7.21
Ammonium | NH4+/NH3 | 9.25

You must prepare 2.000 L of buffer with total analytical concentration

Choose the best buffer system for pH 7.40 and justify your choice quantitatively.

Determine the required ratio of conjugate base to conjugate acid.

Determine the individual concentrations of the conjugate pair that satisfy both the pH requirement and CT.

Calculate the number of moles of each component required (you may assume ideal mixing and ignore activity effects).

Discuss briefly how the charge of the buffer components might influence ionic strength compared to a neutral buffer pair.

14.6.9 Indicator selection error in a weak base titration

A 25.00 mL sample of 0.1000 M weak base B (Kb = 1.0 × 10⁻⁵) is titrated with 0.1000 M HCl. Two indicators are available:

methyl orange: transition range pH 3.1–4.4,

bromothymol blue: transition range pH 6.0–7.6.

Determine the equivalence-point pH for this titration (treat the equivalence solution as a solution of BH+).

Based on the equivalence-point pH and the expected pH jump shape, decide which indicator is more appropriate and justify your choice.

Estimate the direction of the systematic error (too large or too small calculated analyte concentration) if the wrong indicator is used and the end point occurs significantly away from equivalence.

14.6.10 Multi-step fractional precipitation recovery (process design)

A 500.0 mL waste stream contains [Ag+] = 2.00×10⁻³ M and [Cu2+] = 1.00×10⁻² M. You wish to recover at least 95.0% of the silver as AgCl(s) by adding NaCl, while keeping the fraction of copper precipitated as CuCl(s) below 1.0%. Assume:

Ignore complex-ion formation and volume changes.

Determine the minimum [Cl−] required (in the final mixture) to precipitate 95.0% of Ag+.

At that [Cl−], calculate the remaining free [Cu+] required for CuCl saturation and determine whether CuCl precipitation is expected if copper were present as Cu+. Then discuss why the oxidation state of copper matters in applying Ksp data.

Propose a modified strategy (qualitative, no detailed kinetics) that uses complexation (choose either NH3 or CN– as a ligand) to improve selectivity for silver recovery, and explain the equilibrium reasoning in terms of Kf and Ksp.

Chapter 15
Thermodynamics and Electrochemistry

While the First Law of Thermodynamics guarantees energy conservation, it remains silent on the direction of time and change. This chapter unveils the driving forces of the universe-entropy and Gibbs free energy-defining the criteria for spontaneity. We then bridge these abstract thermodynamic potentials to the tangible world of electrochemistry, demonstrating how chemical free energy is harnessed to drive electrons, power civilizations, and manipulate matter through electrolysis.

15.1 Entropy and the Second Law

In earlier thermochemistry (First Law) we learned that energy is conserved: the internal energy U of a system changes by heat and work,

This statement powerfully constrains how much energy can be transferred, but it does not by itself determine whether a proposed change will occur spontaneously. For example, a hot object cools in contact with a cold one, and a gas expands into an evacuated bulb, but the reverse processes are never observed under the same conditions—even though they do not violate energy conservation.

The missing ingredient is a state function that tracks the “directionality” of change. That state function is entropy, S. In this section we develop entropy in two complementary ways:

Macroscopic (thermodynamic) entropy: defined through heat

transfer in a reversible path.

Microscopic (statistical) entropy: defined through the number of possible microscopic arrangements (microstates) consistent with the macroscopic state.

These perspectives converge in the Second Law of Thermodynamics, which provides a quantitative criterion for spontaneity.

15.1.1 The Second Law as a statement about direction

A process is called spontaneous if it proceeds in a given direction without continuous external intervention once initiated. Spontaneity is not the same as speed: some spontaneous changes are slow (rusting), while some nonspontaneous changes can be fast when driven (electrolysis).

Empirically, natural processes share a unifying feature:

Heat flows from hot to cold (not cold to hot without a device).

Gases spread out to fill available volume (not gather into a corner).

Solutes disperse in solvents (diffusion) until concentration differences diminish.

Each example is consistent with the idea that energy and matter tend to spread out among the accessible degrees of freedom. Entropy is the quantitative measure of that spreading (dispersal), and the Second Law declares that such dispersal has a preferred direction when the system and surroundings are considered together.

15.1.2 Thermodynamic definition: entropy change and reversible heat

Consider a system undergoing an infinitesimal change in which it absorbs a small

amount of heat δq. Not all heat transfers are equivalent: heating a gas reversibly at constant T differs fundamentally from a sudden, irreversible transfer. Clausius recognized that to define a state function connected to heat, one must reference a reversible path.

Definition (Entropy change) For a process that takes a system from state 1 to state 2, the entropy change of the system is defined as

where δqrev is the heat exchanged along a reversible path between the same endpoints and T is the absolute temperature.

Several crucial points follow immediately:

S is a state function. Although heat q depends on the path, the integral in Eq. (??) depends only on the initial and final states.

The unit of entropy is Jmol⁻¹ K⁻¹ when expressed per mole, reflecting “energy per temperature”.

The factor 1∕T makes entropy sensitive to where the energy is placed: transferring a given amount of heat at low T produces a larger entropy change than the same heat at high T. This is an early hint that entropy tracks the availability of microscopic energy distributions.

Entropy and reversibility (qualitative)

A reversible process is an idealized limiting case that proceeds through a continuum of equilibrium states. It can be reversed by an infinitesimal change in conditions. Real processes are irreversible: they involve finite gradients (temperature differences, pressure differences, concentration differences) and produce entropy. The reversible path is used in Eq. (??) not because nature is reversible, but because reversibility provides a well-defined thermodynamic benchmark.

15.1.3 Statistical definition: microstates and Boltzmann’s equation

The macroscopic state of a system (specified by T, P, V , and composition) can correspond to an enormous number of distinct microscopic arrangements. A microstate is a particular specification of the positions and energies of all particles consistent with the macroscopic constraints. The number of accessible microstates is denoted W.

Boltzmann connected macroscopic entropy to this microscopic multiplicity:

where kB = 1.380649 × 10⁻²³ JK⁻¹ is the Boltzmann constant.

Equation (??) turns the vague idea of “disorder” into a precise count:

If a change increases the number of accessible microstates (W₂ > W₁), then ΔS = kBln(W₂∕W₁) > 0.

The logarithm ensures additivity. For independent subsystems, Wtotal = WAWB, so

Entropy is therefore best interpreted as a measure of energy dispersal among microstates. A state with larger W corresponds to energy being distributed among more microscopic possibilities.

Why lnW (and why W grows so fast)

For a macroscopic sample, W is astronomically large (often on the order of exp(10²³)). A logarithm compresses that scale into experimentally meaningful values of S. More importantly, because multiplicities multiply when systems combine, ln is the unique simple function that converts multiplicative counting into additive entropy.

15.1.4 Microstate intuition: expansion and mixing

Two classic processes build immediate intuition:

Free expansion of a gas into a vacuum: the gas occupies a larger volume.

Mixing of gases: two different gases initially separated become uniformly mixed.

Both occur spontaneously because they increase the number of accessible microstates. When the gas has more volume available, each molecule has more possible positions. When gases mix, there are vastly more ways to arrange distinguishable particles throughout the container than to keep them segregated.

A simple counting model (conceptual)

Imagine a container divided into left and right halves with N molecules. If each molecule can be on the left or right, the number of position microstates is 2N when the molecule positions are unrestricted. If initially all molecules are confined to the left half, the positional multiplicity is effectively 1 (all on the left). Releasing the constraint multiplies the number of accessible positional microstates by 2N, giving

While this model is highly simplified (real space is not two boxes), it correctly captures the key idea: entropy increases because multiplicity increases exponentially with the number of particles.

15.1.5 Second Law of Thermodynamics (quantitative statement)

The Second Law is most useful when expressed for the universe, meaning the system plus its surroundings:

Criterion | Interpretation
ΔSuniv > 0 | Spontaneous in the forward direction (entropy of the universe increases).
ΔSuniv = 0 | Reversible (ideal limit); system is at equilibrium with surroundings for the process considered.
ΔSuniv < 0 | Nonspontaneous as written; spontaneous in the reverse direction.

This form makes the central message explicit: spontaneity is determined by the total entropy change, not by the system alone. A system’s entropy can decrease provided the surroundings’ entropy increases more, giving a net increase for the universe.

Surroundings entropy for heat exchange at constant T

In many chemical contexts (e.g., reactions in solution open to a large thermal reservoir), the surroundings can be approximated as a heat bath at constant temperature T. If the system absorbs heat qsys from the surroundings, then qsurr = −qsys, and for the surroundings (treated as effectively reversible because it is so large),

If the process occurs at constant pressure, qsys = ΔHsys, so

Equations (??) and (??) will later motivate the introduction of the Gibbs free energy, but for now they provide a practical route to assessing spontaneity by combining entropy and enthalpy effects.

15.1.6 Figure: macroscopic expansion vs. microscopic microstates

15.1.7 Connecting the two definitions: energy dispersal and microstate accessibility

The thermodynamic definition ΔS = ∫ δqrev∕T and the statistical definition S = kBlnW describe the same state function, so they must encode the same physics.

A useful conceptual bridge is:

Entropy increases when energy becomes spread out among more accessible microscopic possibilities.

This statement emphasizes energy dispersal rather than an ambiguous notion of “disorder.” In practice, W increases when:

Volume increases (more positional microstates).

Temperature increases (more energy-level populations become accessible).

Number of particles increases (more ways to distribute energy among more molecules).

Phases become more mobile (typically Sgas > Sliquid > Ssolid because translational degrees of freedom are more available in gases).

Mixing occurs (more compositional arrangements for distinguishable particles).

15.1.8 Standard molar entropies and calculating ΔSsys

In chemistry we often evaluate entropy changes without explicitly integrating Eq. (??). Instead we use tabulated standard molar entropies, S∘, defined for substances in their standard states (typically pure substances at 1 bar, with a specified temperature, commonly 298.15 K).

Like enthalpy, entropy is extensive. Therefore for a reaction,

where ν are stoichiometric coefficients.

Interpretation of tabulated values

Unlike standard enthalpies of formation, standard molar entropies are not defined relative to an arbitrary zero at 298 K. Instead, the Third Law of Thermodynamics (introduced later) establishes that a perfect crystal has S = 0 at 0 K, allowing absolute entropies to be tabulated. At ordinary temperatures, typical values reflect available degrees of freedom and molecular complexity:

Gases have large S∘ due to extensive translational freedom.

Larger, more complex molecules often have higher S∘ (more vibrational/rotational modes).

15.1.9 Calculating ΔSuniv to predict spontaneity

To apply the Second Law, we combine system and surroundings entropy changes (Eq. (??)). In many chemical processes at constant T and P (common laboratory conditions),

This expression highlights why neither ΔH nor ΔSsys alone can serve as a universal spontaneity criterion:

A process with ΔSsys > 0 tends to be favored (system entropy increases), but could be disfavored if it requires absorbing substantial heat from the surroundings (ΔHsys > 0 makes ΔSsurr < 0).

A process with ΔSsys < 0 can still be spontaneous if it releases enough heat to raise the surroundings’ entropy by a larger amount (large negative ΔHsys).

Worked conceptual example (no arithmetic)

Consider freezing liquid water at temperatures below 0 ∘C:

System: liquid → solid decreases molecular mobility, so ΔSsys < 0.

Enthalpy: freezing releases heat, so ΔHsys < 0 and thus ΔSsurr = −ΔHsys∕T > 0.

At sufficiently low T, the positive surroundings term outweighs the negative system term, and ΔSuniv > 0, so freezing is spontaneous. At temperatures above the melting point, the balance reverses. This illustrates how the Second Law naturally introduces temperature as a controlling parameter for spontaneity.

15.1.10 Entropy production and the arrow of time

For any real (irreversible) spontaneous process,

meaning that some entropy is produced by irreversibility (finite gradients, friction, uncontrolled mixing). In the reversible limit,

and the system can be returned to its initial state with no net change to the universe. Thus, entropy provides a quantitative “arrow of time” for macroscopic change: spontaneous processes are those that increase the universe’s entropy.

15.1.11 Summary of key ideas

The First Law (ΔU = q + w) enforces energy conservation but does not determine spontaneity.

Entropy is a state function defined thermodynamically by ΔS = ∫ δqrev∕T.

Entropy has a statistical meaning: S = kBlnW, where W counts microstates.

The Second Law provides the spontaneity criterion: ΔSuniv = ΔSsys + ΔSsurr; spontaneity requires ΔSuniv > 0.

Under many constant-T conditions, ΔSsurr ≈ −ΔHsys∕T, linking entropy to heat flow and anticipating the Gibbs free energy framework developed next.

15.2 Gibbs Free Energy

The Second Law provides the fundamental criterion for spontaneity:

This criterion is completely general, but it is often inconvenient because it requires tracking the entropy change of the surroundings. In chemistry, we usually want a spontaneity test written in terms of system properties that we can compute from tables or measure in the laboratory.

This section develops that test for the most common experimental conditions in chemistry: constant temperature and constant pressure. Under those conditions, spontaneity can be predicted using a single state function, the Gibbs free energy G.

15.2.1 Motivation: eliminating the surroundings from the Second Law

From the previous section, for processes occurring with the surroundings acting as a thermal reservoir at constant temperature T,

For many chemical changes carried out at constant pressure (open to the atmosphere

or in a constant-pressure calorimeter),

Combining Eqs. (??) and (??) gives

Substitute Eq. (??) into the Second Law:

ΔSuniv | = ΔSsys −. |  |

Multiply both sides by −T (with T > 0):

− TΔSuniv | = ΔHsys − TΔSsys. |  |

Now apply the spontaneity condition ΔSuniv > 0:

Equation (??) suggests defining a new state function for the system,

so that under constant T and p the sign of ΔG tells us whether ΔSuniv is positive.

15.2.2 Definition of Gibbs free energy and its physical meaning

Gibbs free energy

The Gibbs free energy is defined as

Because H, T, and S are state functions, G is also a state function. The usefulness of G in chemistry comes from two related interpretations:

Spontaneity criterion at constant T and p: ΔG predicts the direction of spontaneous change without explicitly considering the surroundings.

Maximum non-expansion work: at constant T and p, −ΔG is the maximum useful work obtainable from a process other than pV work (for example, electrical work in a galvanic cell). This interpretation will become central in electrochemistry.

Dimensional check and intuition

The units of G are energy units, typically kJmol⁻¹ for molar quantities, because TS has units

Thus G can be viewed as an “energy balance” that discounts enthalpy by an entropy term weighted by temperature.

A helpful qualitative picture is:

At fixed T and p, a process is favored when it either lowers enthalpy (ΔH < 0) or raises entropy (ΔS > 0), and it is opposed when it

does the reverse. The Gibbs free energy combines these competing tendencies into one quantity.

15.2.3 Derivation of the Gibbs spontaneity criterion

Starting from Eq. (??), consider a change at constant temperature:

ΔG | = ΔH − Δ(TS) | = ΔH − (TΔS + SΔT). |  |  |  |

At constant T, ΔT = 0, so

Now compare Eq. (??) with the right-hand side of Eq. (??). Under constant T and p, ΔHsys is the heat exchanged reversibly at constant pressure, and Eq. (??) applies. Therefore,

Since T > 0, the sign relationship follows immediately:

Condition (constant T and p) | Meaning
ΔG < 0 | Spontaneous (because ΔSuniv > 0).
ΔG = 0 | Equilibrium (because ΔSuniv = 0).
ΔG > 0 | Nonspontaneous as written (spontaneous in reverse).

Scope and limitations

The criterion above is exact under the assumptions used in its derivation:

Temperature is constant (system exchanges heat with a large reservoir).

Pressure is constant (the relevant heat is qp = ΔH).

If T and p are not constant, G still exists as a state function, but the simple

inequality “ΔG < 0 means spontaneous” may no longer apply without additional constraints. In general thermodynamics, spontaneity is always governed by ΔSuniv; G is the convenient reformulation for the most common chemical conditions.

15.2.4 Gibbs free energy as a balance of enthalpy and entropy

Equation (??) reveals the central competition:

The temperature factor means that entropy effects become more important as T increases.

To build intuition, analyze the sign of ΔG in terms of ΔH and ΔS:

ΔH < 0, ΔS > 0: always ΔG < 0 (both terms favor spontaneity).

ΔH > 0, ΔS < 0: never ΔG < 0 (both terms oppose spontaneity).

ΔH < 0, ΔS < 0: spontaneous only at low T (enthalpy dominates; −TΔS is positive and opposes).

ΔH > 0, ΔS > 0: spontaneous only at high T (entropy term −TΔS is negative and can overcome positive ΔH).

Temperature thresholds

When ΔH and ΔS have the same sign, there is a characteristic temperature at which ΔG changes sign. Solving Eq. (??) for ΔG = 0 gives

with the understanding that ΔH and ΔS must be expressed in consistent units (e.g., Jmol⁻¹ and Jmol⁻¹ K⁻¹).

This threshold is often used to interpret phase changes:

Melting: ΔHfus > 0 and ΔSfus > 0, so melting is favored at high T.

Freezing: reverse process; favored at low T.

At the normal melting point, the two phases are at equilibrium under 1 bar and ΔG = 0.

15.2.5 Standard Gibbs free energy change and tabulated data

For chemical reactions, we define a standard Gibbs free energy change, ΔG∘, corresponding to reactants and products in their standard states at a specified temperature (often 298.15 K). For a reaction written as

the standard free energy change is

In practice, chemists typically compute ΔGrxn∘ using tabulated standard Gibbs free energies of formation, ΔGf∘, analogous to the standard enthalpies of formation introduced earlier.

Standard Gibbs free energy of formation

Standard Gibbs free energy of formation: ΔGf∘ is the Gibbs free energy change for forming 1 mol of a compound from its constituent elements in their standard states at the specified temperature.

With formation values,

By convention, ΔGf∘ = 0 for elements in their standard states (e.g., O2(g), N2(g), C(graphite), H2(g)).

Relation to enthalpy and entropy under standard conditions

Because G = H − TS, the same combination holds for reaction quantities:

Equation (??) is widely used to interpret why a reaction has the spontaneity it does. However, a common pitfall is to treat ΔH∘ and ΔS∘ as temperature-independent over large ranges. Over modest temperature intervals this can be a useful approximation, but accurate temperature dependence generally requires heat capacity data and integration.

15.2.6 Interpreting ΔG: spontaneity vs. favorability

Spontaneity at specified conditions

At a given temperature and pressure (and given concentrations/partial pressures), the sign of ΔG refers to the instantaneous driving force for change:

ΔG < 0: the process proceeds forward spontaneously.

ΔG > 0: the process proceeds spontaneously in reverse.

ΔG = 0: the system is at equilibrium.

This statement concerns the direction of change from the current state.

What ΔG∘ does and does not tell you

ΔG∘ refers to the hypothetical change when all species are in their standard states (typically 1 bar for gases, unit activity for solutes). It is a measure of the intrinsic tendency of reactants to form products under standard conditions, but it does not guarantee that a reaction will go to completion.

For example, a reaction can have ΔG∘ < 0 and still reach equilibrium with substantial amounts of reactants remaining. The quantitative connection between ΔG∘ and the equilibrium constant K will be derived in the next section; for now, the key conceptual distinction is:

Quantity | What it answers
ΔG | “Is the system driven to change forward from its current composition?”
ΔG∘ | “Under standard conditions, is product formation favored relative to reactants?”

15.2.7 Gibbs free energy and maximum useful work (preview for electrochemistry)

At constant T and p, the Gibbs free energy change is linked to the maximum useful (non-pV ) work obtainable from a process:

This statement has two important implications:

If ΔG < 0, the process can, in principle, be harnessed to do useful work (for example, electrical work in a battery).

If ΔG > 0, useful work must be supplied to force the process forward (for example, using electricity to drive electrolysis).

Later, when we study galvanic cells, we will connect −ΔG directly to cell potential and electrical work.

Starting from the Second Law and ΔSsurr = −ΔHsys∕T (constant T, p), we define the Gibbs free energy G = H − TS.

Under constant temperature and pressure,

so ΔG < 0 indicates spontaneity, ΔG = 0 indicates equilibrium, and ΔG > 0 indicates nonspontaneity as written.

The relation ΔG = ΔH−TΔS captures the competition between enthalpy and entropy, with temperature controlling which effect dominates.

Standard free energies (ΔG∘, ΔGf∘) allow reaction free energies to be computed from tables and connect naturally to equilibrium and electrochemistry in subsequent sections.

15.3 Free Energy and Equilibrium

The Gibbs free energy criterion developed in the previous section gives a practical test for spontaneity at constant temperature and pressure:

However, chemical reactions rarely occur under standard-state conditions. As reactants are consumed and products accumulate, the system composition changes continuously, and so does the driving force. The central goal of this section is to make that driving force quantitative by connecting free energy to the reaction quotient Q, and by identifying equilibrium as the composition that minimizes the Gibbs free energy at fixed T and p.

15.3.1 Thermodynamic definition of equilibrium: a minimum in G

In thermodynamics, equilibrium is not defined by rates or by the absence of visible change; it is defined by a variational principle: at constant temperature and pressure, a closed reacting system is at equilibrium when its Gibbs free energy is at a minimum with respect to the extent of reaction.

The extent of reaction

Consider a general reaction

where νi are stoichiometric coefficients (positive for products, negative for

reactants). The extent of reaction ξ is a progress variable that tracks how far the reaction has proceeded. If ni is the amount (in moles) of species i, then

Thus, changing ξ changes all species amounts in the stoichiometric proportions of the balanced equation.

Equilibrium as a minimum

At constant T and p, the spontaneous direction of change is the one that lowers G. If we imagine the system “trying out” neighboring compositions by changing ξ slightly, then:

If increasing ξ lowers G, the reaction proceeds forward.

If decreasing ξ lowers G, the reaction proceeds in reverse.

At equilibrium, neither direction lowers G; G is locally minimal.

Mathematically, at equilibrium:

and stability requires that G curve upward at the minimum.

15.3.2 Reaction Gibbs free energy ΔrG and the reaction quotient Q

To connect the macroscopic function G to composition, we introduce the chemical potential μi, which is the molar Gibbs free energy of species i in the mixture:

For a reacting system at constant T and p,

dG | = ∑ iμi dni |  |
| = ∑ iμi νi dξ (using Eq. (??)). |  |

Therefore,

where ΔrG is the reaction Gibbs free energy (the instantaneous free energy change per infinitesimal advance of reaction at the current composition). The equilibrium condition in Eq. (??) is therefore equivalent to:

Chemical potential in terms of activity

For real mixtures, composition enters thermodynamics through activity ai, a dimensionless measure of “effective concentration” that accounts for nonideality. A central result is

where μi∘ is the standard chemical potential and R is the gas constant.

Substitute Eq. (??) into Eq. (??):

ΔrG | = ∑ iνi |  |
| = ∑ iνiμi∘ + RT ∑ iνi lnai |  |
| = ΔrG∘ + RT ln. |  |

Define the reaction quotient Q as

Then Eq. (??) becomes the fundamental link between free energy and composition:

Interpretation.

ΔrG∘ is the reaction free energy when all activities are at their standard values (so Q = 1).

RT lnQ is the composition-dependent correction; it tells how far the current mixture is from standard conditions.

15.3.3 Equilibrium condition and the relationship between ΔG∘ and K

At equilibrium, ΔrG = 0 by Eq. (??). Also, at equilibrium the reaction quotient equals the equilibrium constant: Q = K. Substitute into Eq. (??):

0 | = ΔrG∘ + RT lnK, |  |
ΔrG∘ | = −RT lnK. |  |

Equation (??) is one of the most important results in chemical thermodynamics. It connects a thermodynamic quantity (ΔrG∘) to an equilibrium property (K), allowing each to be determined from the other.

Why this result is so powerful

If ΔrG∘ < 0, then lnK > 0 and K > 1: products are favored at equilibrium.

If ΔrG∘ > 0, then K < 1: reactants are favored at equilibrium.

If ΔrG∘ = 0, then K = 1: neither side is favored under standard conditions.

Importantly, even a large K does not imply complete conversion to products; it implies the equilibrium mixture has a strong product bias.

15.3.4 Direction of spontaneous change: comparing Q to K

Equation (??) can be combined with Eq. (??) to express ΔrG directly in terms of Q and K:

ΔrG | = ΔrG∘ + RT lnQ |  |
| = −RT lnK + RT lnQ |  |
| = RT ln. |  |

Because RT > 0, the sign of ΔrG is controlled by the ratio Q∕K:

Comparison | Predicted direction (constant T, p)
Q < K | ΔrG < 0: reaction proceeds forward (toward products) to raise Q toward K.
Q = K | ΔrG = 0: equilibrium (no net driving force).
Q > K | ΔrG > 0: reaction proceeds in reverse (toward reactants) to lower Q toward K.

This is the thermodynamic basis of Le Châtelier’s principle: disturbances change Q, and the system shifts in the direction that drives Q back toward K.

15.3.5 Activities and common approximations for Q

The definition Q = ∏ aiνi is exact. To use it in introductory chemistry, we adopt standard activity conventions and approximations:

Gases (ideal-gas approximation)

For a gas i,

where P∘ is the standard pressure (often taken as 1bar). For an ideal gas mixture, this makes Q depend on partial pressures.

Solutes in dilute solution

For a solute i in an ideal dilute solution,

with c∘ = 1molL⁻¹. Under this approximation, Q is written using concentrations in the familiar way.

Pure solids and pure liquids

For a pure solid or pure liquid in its standard state, the activity is defined as a = 1, so these substances do not appear in Q or K. This is not because they are unimportant, but because their “effective concentration” is constant.

15.3.6 Visualizing equilibrium: G vs. extent of reaction

At fixed T and p, the system evolves toward the composition that minimizes G. The relationship between ΔrG and the slope of G(ξ) provides a geometric interpretation:

Thus:

ΔrG < 0 corresponds to a negative slope; increasing ξ lowers G (forward reaction).

ΔrG > 0 corresponds to a positive slope; decreasing ξ lowers G (reverse reaction).

ΔrG = 0 corresponds to zero slope; G is minimal (equilibrium).

Figure 15.1: At constant temperature and pressure, a reacting system moves spontaneously toward the composition that minimizes the Gibbs free energy G. The slope dG∕dξ equals the reaction Gibbs free energy ΔrG: negative slope (Q < K) drives the reaction forward, positive slope (Q > K) drives it in reverse, and zero slope at the minimum corresponds to equilibrium (Q = K).

15.3.7 Putting the relationships to work: a compact strategy

For many problems, the following workflow is effective:

Compute Q from the current composition (using activities or appropriate approximations).

Compare Q to K (or compute ΔrG directly).

Predict direction using ΔrG = RT ln(Q∕K):

At equilibrium (Q = K), compute K from ΔrG∘ using ΔrG∘ = −RT lnK, or compute ΔrG∘ from K if K is known.

At constant T and p, equilibrium corresponds to a minimum in the Gibbs free energy G as a function of reaction progress ξ.

The driving force for reaction at the current composition is the reaction Gibbs free energy:

Composition enters through activities: μi = μi∘ + RT lnai, leading to

At equilibrium, ΔrG = 0 and Q = K, giving

The sign of ΔrG is determined by Q∕K:

so the system spontaneously shifts in the direction that drives Q toward K.

15.4 Galvanic Cells and Standard Potentials

The previous sections connected thermodynamics to chemical change through the Gibbs free energy:

and showed that equilibrium corresponds to ΔG = 0 and Q = K. Electrochemistry provides a striking way to harness a spontaneous chemical reaction: if the reaction involves electron transfer (a redox process), the free energy decrease can appear as electrical work. A galvanic (voltaic) cell is the device that converts chemical energy into electrical energy.

This section focuses on the physical construction of galvanic cells and the practical use of standard reduction potentials to predict cell voltages and reaction directions.

15.4.1 What a galvanic cell does: separating half-reactions

A redox reaction involves two coupled processes:

Oxidation: loss of electrons.

Reduction: gain of electrons.

In a beaker, electrons transfer directly at the interface between reactants, usually releasing energy as heat. In a galvanic cell, we separate oxidation and reduction into two compartments so that electrons must travel through an external circuit. That forced electron flow is an electric current that can power a load.

Anode and cathode: definitions that never change

In any electrochemical cell (galvanic or electrolytic):

Electrode | Definition
Anode | Electrode at which oxidation occurs (electrons are produced).
Cathode | Electrode at which reduction occurs (electrons are consumed).

For a galvanic cell, the anode is negative (source of electrons) and the cathode is

positive. The oxidation/reduction definitions are the reliable anchors; the sign convention follows from whether the cell is producing or consuming electrical energy.

15.4.2 A prototype galvanic cell: the Daniell cell (Zn/Cu)

A classic example is the Daniell cell, which uses the redox reaction between zinc metal and copper(II) ions:

Here zinc is oxidized and copper(II) is reduced.

Required schematic: anatomy of the Daniell cell

15.4.3 Cell components and what each one accomplishes

Half-cells: where oxidation and reduction occur

Each half-cell contains:

An electrode (a conductor, often a metal) that allows electron transfer between the external circuit and the solution.

An electrolyte solution containing ions involved in the half-reaction.

In the Daniell cell:

External circuit: path for electrons

Oxidation at the anode generates electrons. Those electrons cannot travel through solution efficiently over macroscopic distances, but they move readily through a metal wire. The resulting electron flow constitutes an electric current that can perform work (power a device, charge a battery, etc.).

Salt bridge: maintaining electroneutrality

If the half-cells were not ionically connected, charge would rapidly build up:

At the anode, Zn2+ accumulates as zinc dissolves, making the anode compartment more positively charged.

At the cathode, Cu2+ is removed from solution as copper plates, leaving behind excess anion and making the cathode compartment relatively negative.

This charge separation would oppose further electron flow and quickly stop the reaction. The salt bridge contains an inert electrolyte (e.g., KNO3 or KCl) that provides ions to migrate and preserve approximate electroneutrality:

Anions migrate toward the anode compartment to offset the growing positive charge.

Cations migrate toward the cathode compartment to replace cations being consumed (or to offset the relative negative charge).

The salt bridge completes the circuit by providing an ionic pathway while keeping the solutions largely separate to prevent direct reaction.

15.4.4 Electromotive force and cell potential

The driving “push” that causes electrons to move through the wire is quantified by the cell potential (also called electromotive force, EMF), Ecell. It is measured in volts (V), where 1 V = 1 JC⁻¹.

Conceptually:

Ecell measures the tendency for the redox reaction to occur by driving electrons from anode to cathode.

A larger positive Ecell indicates a stronger tendency for spontaneous electron flow in the indicated direction.

The thermodynamic link between Ecell and ΔG will be derived in a later section; here we focus on predicting Ecell from tabulated data under standard conditions.

15.4.5 Standard reduction potentials and the Standard Hydrogen Electrode

Why potentials must be relative

Unlike temperature or mass, an absolute electric potential of a single half-cell cannot be measured directly. Any voltmeter measures a difference in potential between two electrodes. Therefore, electrochemistry adopts a reference half-cell assigned an exact potential, and all other half-cell potentials are measured relative to it.

The Standard Hydrogen Electrode (SHE)

The reference is the Standard Hydrogen Electrode (SHE), defined by the half-reaction

at T = 298.15K (unless otherwise specified), with a platinum electrode providing an inert conducting surface. By convention,

Standard reduction potential E∘

For any half-reaction written as a reduction, tabulations provide its standard reduction potential E∘, measured relative to the SHE under standard conditions (activities of solutes equal to 1, gases at 1 bar, pure solids and liquids with activity 1).

Two key points prevent common mistakes:

Potentials are intensive. Multiplying a half-reaction by a factor does not multiply E∘ by that factor.

Reversing a half-reaction changes the sign. If a reduction has potential E∘, the corresponding oxidation has potential −E∘.

15.4.6 Using standard potentials to predict spontaneous redox reactions

Assembling a galvanic cell from half-reactions

To construct a galvanic cell:

Choose two half-reactions from a standard reduction potential table.

The half-reaction with the more positive E∘ will tend to occur as a reduction at the cathode.

The other half-reaction is forced to run in reverse (as an oxidation) at the anode.

Balance electrons by scaling the half-reactions (without changing E∘ values).

Add the half-reactions to obtain the overall cell reaction.

Calculating Ecell∘

Standard cell potential is computed as

where both values are standard reduction potentials as tabulated. The subtraction automatically accounts for the fact that the anode process is an oxidation in the operating cell.

Example: Daniell cell. Using typical tabulated reduction potentials (at 298 K):

The cathode is copper (more positive E∘). Then

A positive Ecell∘ indicates that the reaction as written for the cell is spontaneous under standard conditions.

A compact “table-reading” rule

Standard reduction potentials are sometimes described as measuring “reduction

strength”:

More positive E∘: stronger tendency to be reduced (stronger oxidizing agent in oxidized form).

More negative E∘: stronger tendency to be oxidized (stronger reducing agent in reduced form).

Thus, in a galvanic cell, electrons flow spontaneously from the species with the more negative reduction potential (anode, oxidation) toward the species with the more positive reduction potential (cathode, reduction).

15.4.7 IUPAC cell notation

Electrochemical cells are concisely represented by a line notation:

Rules:

A single vertical line (|) denotes a phase boundary (e.g., metal/solution).

A double vertical line (||) denotes the salt bridge (or porous barrier).

Species in the same phase are separated by commas.

The anode (oxidation) is written on the left; the cathode (reduction) on the right.

For the Daniell cell:

15.4.8 Inert electrodes and nonmetal half-reactions

Not all half-reactions involve a solid metal that can serve as an electrode. If all reacting species are in solution or gas phase, an inert electrode (commonly Pt(s) or graphite) provides a surface for electron transfer without participating chemically.

Example (illustrative notation):

Here Pt(s) is included because the Fe3+/Fe2+ couple needs a conductor.

A galvanic cell converts the free energy decrease of a spontaneous redox reaction into electrical work by separating oxidation and reduction into two half-cells.

Anode is where oxidation occurs; cathode is where reduction occurs (true for all electrochemical cells).

Electrons flow through the external circuit from anode to cathode; ions migrate through the salt bridge to maintain electroneutrality.

Standard reduction potentials E∘ are measured relative to the Standard Hydrogen Electrode (SHE), defined as 0.000 V.

To predict cell voltage under standard conditions:

A positive Ecell∘ indicates a spontaneous galvanic reaction under standard conditions.

IUPAC cell notation compactly records the cell components and their arrangement: anode on the left, cathode on the right, with || representing the salt bridge.

15.5 Cell Potential and Free Energy

Galvanic cells make a thermodynamic statement physically visible: when a redox reaction is spontaneous, it can drive electrons through an external circuit and deliver electrical work. In the previous sections we developed two parallel languages for spontaneity:

Thermodynamics: at constant T and p, spontaneity requires ΔG < 0, and at equilibrium ΔG = 0.

Electrochemistry: a galvanic cell produces a positive cell potential Ecell > 0 under the stated conditions.

This section unifies these ideas by deriving the quantitative link between cell potential and Gibbs free energy, and then connecting both to the equilibrium constant.

15.5.1 Electrical work and potential difference

Electric potential (voltage) is energy per unit charge:

If a charge q (in coulombs) moves through a potential difference E (in volts), the maximum electrical work obtainable (by the system) is

The sign convention is the thermodynamic one: w < 0 when the system does work on the surroundings. For a galvanic cell operating spontaneously, the system delivers electrical work to an external circuit, so welec is negative.

Charge moved when a redox reaction occurs

In a redox reaction, electrons are the charge carriers. One mole of electrons carries a charge equal to the Faraday constant:

If the balanced cell reaction transfers n moles of electrons per mole of reaction as written, then the charge transferred per mole of reaction is

For example, in the Daniell cell reaction

two electrons are transferred (n = 2) because zinc oxidation produces 2e− and copper(II) reduction consumes 2e−.

15.5.2 Deriving ΔG = −nFE

In the Gibbs free energy section, we established that at constant T and p, the maximum non-expansion work obtainable from a process is

For a galvanic cell, the useful work is electrical, so in the reversible limit,

Combine Eqs. (??), (??), and (??):

welec,max | = −qEcell = −(nF)Ecell, |  |
welec,max | = −ΔG. |  |

Therefore,

Meaning of the signs

Equation (??) is a compact translation between the two languages:

If Ecell > 0, then ΔG < 0 and the cell reaction is spontaneous as written.

If Ecell = 0, then ΔG = 0 and the system is at equilibrium.

If Ecell < 0, then ΔG > 0 and the reaction is nonspontaneous as written (it would proceed spontaneously in reverse).

This is why the sign of cell potential is a direct indicator of spontaneity under the stated conditions.

Standard conditions

Under standard conditions (standard activities, typically 1 bar for gases and unit activity for solutes),

Here Ecell∘ is computed from standard reduction potentials (previous section), and ΔG∘ is the standard Gibbs free energy change for the overall cell reaction.

15.5.3 Connecting cell potential to equilibrium

From the free energy–equilibrium relationship developed earlier,

where K is the thermodynamic equilibrium constant (defined in terms of activities). Combine Eq. (??) with Eq. (??):

− nFEcell∘ | = −RT lnK, |  |
nFEcell∘ | = RT lnK, |  |
lnK | = Ecell∘. |  |

Equivalently,

Interpretation

These equations explain why even modest-looking standard potentials correspond to

very large equilibrium constants. Because F is large and RT is comparatively modest at room temperature, the factor nF∕RT is typically on the order of 10². As a result:

A positive Ecell∘ implies K ≫ 1 (products strongly favored at equilibrium).

A negative Ecell∘ implies K ≪ 1 (reactants favored).

Electrochemical measurements (voltages) therefore provide a powerful experimental route to equilibrium constants.

15.5.4 Free energy, cell potential, and the reaction quotient

In nonstandard conditions, the Gibbs free energy depends on composition:

Combine with Eq. (??) and Eq. (??):

− nFEcell | = −nFEcell∘ + RT lnQ, |  |
nFEcell | = nFEcell∘− RT lnQ, |  |
Ecell | = Ecell∘−lnQ. |  |

Equation above is the Nernst equation (derived formally in the next section). Here it serves as an immediate demonstration that cell voltage is not a fixed property of the electrodes alone; it depends on the current composition through Q.

Equilibrium in electrochemical language

At equilibrium, ΔG = 0 and Q = K. Using Eq. (??):

Thus, a cell at equilibrium shows no net potential difference capable of driving current through an external load, even though microscopic oxidation and reduction events still occur (dynamic equilibrium). Substituting Q = K into the Nernst equation yields

consistent with Eq. (??).

15.5.5 Practical calculations: identifying n and avoiding common errors

Determining n

The integer n is the number of moles of electrons transferred per mole of overall reaction as written. To find n:

Write the oxidation and reduction half-reactions.

Balance electrons so the half-reactions cancel when added.

The number of electrons in the balanced half-reactions is n.

Example (Daniell cell).

Electrons cancel with n = 2.

Do not scale potentials when balancing

Balancing a redox reaction may require multiplying half-reactions to cancel electrons. This multiplies the amount of charge transferred per mole of overall reaction (changes n), but it does not multiply E∘ values. Potentials are intensive; the energy change scales through n in ΔG = −nFE.

Units and constants

Use consistent units:

F in Cmol⁻¹,

E in volts (JC⁻¹),

then nFE is in Jmol⁻¹.

If reporting ΔG in kJmol⁻¹, divide by 1000.

15.5.6 Conceptual summary: what cell potential really measures

A useful way to interpret the equations is:

Cell potential is the free-energy “steepness” per unit charge. A larger Ecell means that each coulomb of charge can, in principle, deliver more useful work, corresponding to a more negative ΔG for the cell reaction.

As the reaction proceeds, Q changes, the driving force ΔG changes, and the measured Ecell changes accordingly. The reaction stops (reaches equilibrium) when ΔG reaches zero and the cell voltage falls to zero.

Electrical work is related to potential difference by welec = −qE.

For a redox reaction transferring n moles of electrons per mole of reaction, q = nF.

At constant T and p, the maximum electrical work equals the maximum non-pV work, giving the central link:

Combining with ΔG∘ = −RT lnK yields

connecting standard cell potentials to equilibrium constants.

Combining ΔG = ΔG∘ + RT lnQ with ΔG = −nFE gives

showing explicitly how cell voltage depends on composition and reaching Ecell = 0 at equilibrium.

15.6 The Nernst Equation

In the previous section we established the central link between thermodynamics and electrochemistry:

and we connected chemical driving force to composition:

These two ideas together imply a powerful conclusion: cell potential is not a fixed number, even for the same electrodes. Instead, Ecell changes as concentrations (more precisely, activities) change during the reaction. The quantitative relationship between Ecell and composition is the Nernst equation.

This section develops the Nernst equation from thermodynamic principles and applies it to (i) nonstandard galvanic cells, (ii) equilibrium constants and equilibrium potentials, (iii) concentration cells, and (iv) potentiometric pH measurements.

15.6.1 Derivation of the Nernst equation from ΔG and Q

Start with the two expressions for ΔG:

ΔG | = −nFEcell, |  |
ΔG | = ΔG∘ + RT lnQ. |  |

Under standard conditions,

Substitute Eq. (??) into Eq. (??) and equate to Eq. (??):

− nFEcell | = −nFEcell∘ + RT lnQ, |  |
Ecell | = Ecell∘−lnQ. |  |

Equation (??) is the Nernst equation in natural-log form.

Base-10 form at 298.15 K

Because lnQ = 2.303log Q, Eq. (??) becomes

At T = 298.15K,

so

At other temperatures, use Eq. (??) or Eq. (??) with the actual T.

15.6.2 What exactly is Q for an electrochemical cell?

For an overall cell reaction written in balanced form,

the reaction quotient is defined thermodynamically as

where ai are activities (dimensionless) and νi are stoichiometric coefficients (positive for products, negative for reactants).

Typical activity approximations

In introductory problems we often approximate activities as:

Solutes in dilute solution: ai ≈ [i]∕1molL⁻¹.

Gases (ideal): ai ≈ Pi∕1bar.

Pure solids and pure liquids: ai = 1 by definition (they do not appear in Q).

These approximations convert Q into the familiar concentration/pressure expression, but it is important to remember that the true equilibrium constant and the true Nernst equation are written in terms of activities.

15.6.3 Interpreting the Nernst equation: how composition changes voltage

Equation (??) shows how Ecell responds to composition:

Key implications:

If Q < 1 (reactants “high” relative to products), then lnQ < 0 and Ecell > Ecell∘. The cell has a larger driving force than under standard conditions.

If Q > 1 (products “high”), then lnQ > 0 and Ecell < Ecell∘. The driving force is reduced.

As the cell reaction proceeds forward, Q generally increases, so Ecell typically decreases toward zero.

This aligns with the thermodynamic picture: as products accumulate, ΔG becomes less negative until equilibrium is reached.

15.6.4 Equilibrium and the equilibrium potential

At equilibrium, ΔG = 0, so Ecell = 0. Also, Q = K. Insert these into the Nernst equation:

0 | = Ecell∘−lnK, |  |
Ecell∘ | = lnK, |  |
lnK | = Ecell∘, |  |

which is the equilibrium link derived earlier. The Nernst equation is therefore a more general statement: it gives the voltage at any composition, and at equilibrium it reduces to the equilibrium-constant relationship.

15.6.5 Worked framework: the Daniell cell under nonstandard conditions

Consider the Daniell cell reaction:

Pure solids have activity 1, so

With n = 2,

This expression predicts:

Increasing [Cu2+] increases E cell (drives reduction more strongly).

Increasing [Zn2+] decreases E cell (product buildup suppresses further oxidation).

15.6.6 Concentration cells: voltage from a concentration difference alone

A concentration cell is a galvanic cell in which:

the electrodes are the same material,

the half-reactions are the same redox couple,

but the activities (concentrations/partial pressures) differ between compartments.

There is no intrinsic standard driving force because the standard potentials cancel; the voltage arises purely from the tendency to eliminate the concentration gradient.

Example: metal/metal-ion concentration cell

Consider two half-cells with the same redox couple:

Let the left half-cell have activity a₁ and the right have a₂. Because the half-reactions are identical, Ecell∘ = 0; the Nernst equation gives

To obtain Q, write the net reaction. The spontaneous direction is for ions to move (via oxidation/reduction at electrodes) such that the higher concentration is reduced and the lower concentration is increased, tending to equalize. The overall effect is:

A practical working form is

which is positive because ahigh > alow.

Physical interpretation

A concentration cell is a direct illustration that chemical potential differences can be converted into electrical work even when no “new” chemical species are formed overall. The cell consumes the free energy stored in the concentration gradient.

15.6.7 Measuring pH potentiometrically: Nernst behavior of an electrode

One of the most important applications of the Nernst equation is the measurement of hydrogen ion activity through an electrode potential. While practical pH electrodes use glass membranes and reference electrodes (details beyond the present scope), the underlying thermodynamics is captured by a hydrogen-ion-sensitive half-reaction.

Hydrogen electrode (conceptual model)

For the SHE half-reaction (written as reduction),

The Nernst equation for this half-cell (noting n = 2) is

Under standard hydrogen pressure a(H2) = 1 and E∘ = 0, so

E | = −ln |  |
| = lna(H+). |  |

Using pH ≡−log a(H+), rewrite Eq. (??) at 298.15 K:

E | = lna(H+) |  |
| = log a(H+) |  |
| = −pH |  |
| ≈−0.05916VpH (298.15K). |  |

Thus, at room temperature the potential changes by about 59 mV per pH unit (with sign depending on the chosen half-reaction direction and reference).

15.6.8 Common pitfalls and best practices

Use the balanced overall cell reaction to form Q

Always determine Q from the overall balanced cell reaction (or equivalently from activities and stoichiometric coefficients). Do not guess Q from half-cell expressions unless you are careful to combine them consistently.

Do not include solids and pure liquids in Q

Remember a = 1 for pure solids and pure liquids. Including them in Q introduces incorrect concentration dependence and wrong numerical results.

Temperature matters

The factor (RT∕nF) increases with temperature. If a problem specifies a non-room

temperature, do not use the shortcut unless you recompute it at that T.

Activities vs. concentrations

At moderate to high ionic strengths, ai≠[i]∕1molL⁻¹. Deviations are handled through activity coefficients (ai = γi[i]∕c∘). Introductory problems often assume γi ≈ 1, but in real electrochemical measurements this assumption is a major source of error if not controlled.

Combining ΔG = −nFEcell with ΔG = ΔG∘+RT lnQ yields the Nernst equation:

At 298.15 K,

Cell potential decreases as Q increases during the forward reaction, reaching Ecell = 0 at equilibrium where Q = K.

Concentration cells have Ecell∘ = 0 and generate voltage purely from a concentration gradient:

Electrode potentials can be used to infer activities, including hydrogen ion activity (pH), with an ideal slope of about 59 mV per pH unit at room temperature.

15.7 Electrolysis

Galvanic cells convert the free energy decrease of a spontaneous redox reaction into electrical work:

Electrolysis is the opposite mode of operation. In an electrolytic cell, an external power source supplies electrical energy to force a redox reaction that is nonspontaneous under the chosen conditions:

This section develops electrolysis from the thermodynamic viewpoint (how electrical energy changes ΔG), then addresses practical issues (overpotential, competing aqueous reactions), and finally presents quantitative predictions using Faraday’s laws.

15.7.1 Electrolytic cells: what changes relative to galvanic cells?

The definitions of anode and cathode do not change:

What changes is the polarity and the role of the external circuit.

Polarity in electrolytic cells

Because the power supply pulls electrons from the anode and pushes electrons to the cathode:

the anode is positive in an electrolytic cell,

the cathode is negative in an electrolytic cell.

This is the reverse of a galvanic cell, where the anode is negative and the cathode is positive.

Thermodynamic picture: driving an uphill process

For any cell reaction under specified conditions,

If the desired chemical change has ΔG > 0, then the corresponding Ecell (for that reaction direction) is negative. A power source must therefore impose an opposing potential difference of sufficient magnitude to make the overall process proceed.

15.7.2 Minimum applied voltage and the role of irreversibility

Reversible minimum (ideal limit)

Consider an electrolytic process operated reversibly at constant T and p. If the external supply provides an electrical potential Eapp that drives n moles of electrons per mole of reaction, the electrical work done on the system per mole of reaction is

In the reversible limit this work equals the free energy increase:

Thus the ideal minimum applied voltage is

Under standard conditions, using ΔG∘ = −nFEcell∘,

That is, the minimum voltage needed to drive the reverse of a spontaneous galvanic reaction is equal in magnitude and opposite in sign to the galvanic cell’s standard potential.

Real cells: overpotential and internal losses

Actual electrolysis requires a higher applied voltage than the reversible minimum because of irreversibilities:

Ohmic losses (iR drop): finite resistance of electrodes, wires, and solution.

Activation barriers at electrodes: electron transfer and bond-breaking/forming steps require an extra driving force.

Mass transport limitations: depletion of reactants near electrode surfaces and concentration gradients.

These effects are often summarized as overpotential (or overvoltage), η, the extra potential beyond the thermodynamic requirement:

Overpotential is not a fixed universal constant; it depends on electrode material, surface condition, current density, temperature, and solution composition.

15.7.3 Predicting what happens at each electrode

An electrolytic cell often presents multiple possible oxidation and reduction reactions, especially in aqueous solution. The reaction that actually occurs at each electrode is determined by a combination of:

thermodynamics (which half-reaction requires the least “uphill” potential),

kinetics (overpotentials and catalytic properties of electrode materials),

concentrations (Nernst shifts) and mass transport.

A practical approach is:

1. List plausible half-reactions at the cathode (reductions) and at the anode (oxidations).

2. Use standard reduction potentials and the Nernst equation to estimate potentials under the actual conditions.

3. Identify the reactions that require the smallest magnitude of applied voltage, while remembering that kinetics can override small thermodynamic differences.

15.7.4 Electrolysis of molten salts vs. aqueous solutions

Molten salts: usually straightforward

In a molten ionic compound, the only species present are the ions of the salt. For molten NaCl(l):

Overall:

Because water is absent, there is no competition from H2O reduction/oxidation.

Aqueous solutions: water competes

In water, both solute species and water itself can be oxidized or reduced. Two especially important water-involving half-reactions (in acidic form) are:

2H2O(l) + 2e− → H 2(g) + 2OH−(aq) | E∘≈−0.83V |  |
2H2O(l) → O2(g) + 4H+(aq) + 4e− | E∘≈ 1.23V |  |

(Equivalent basic forms can also be written.) These provide competing pathways:

At the cathode, if a solute cation is difficult to reduce (very negative E∘), water reduction to H2 may occur instead.

At the anode, if a solute anion is difficult to oxidize, water oxidation to O2 may occur instead.

Rule-of-thumb (with important caveats)

While real outcomes can depend strongly on overpotential and concentrations, a useful introductory guideline is:

Cathode (reduction): cations of less active metals (e.g., Ag+, Cu2+) tend to plate out; alkali and alkaline-earth cations (Na+, K+, Mg2+, Ca2+) tend not to reduce in water, so H2 is produced.

Anode (oxidation): halide ions (Cl– , Br– , I– ) can be oxidized to halogens, but water oxidation to O2 competes and can dominate depending on conditions and electrode material (notably because oxygen evolution often has large overpotential).

The phrase “tend to” matters: kinetics and overpotential can reverse simple thermodynamic expectations.

15.7.5 Faraday’s laws of electrolysis: quantitative stoichiometry

Electrolysis provides a direct and precise relationship between charge passed and amount of chemical change. This is the basis of Faraday’s laws.

Charge, current, and time

Current is charge per time:

Therefore,

with q in coulombs (C) when I is in amperes (A) and t is in seconds (s).

Moles of electrons transferred

Because F is the charge per mole of electrons,

Connecting electrons to product moles

If forming 1 mole of a product requires z moles of electrons (from the balanced half-reaction), then

If the product is a metal deposited at the cathode, the mass deposited is

where M is the molar mass.

Example framework: copper plating

For copper plating:

Given a current I applied for time t, the moles of copper deposited are

and the mass is m = nCuMCu.

15.7.6 Efficiency and side reactions

In real electrolytic processes, not all current goes into the desired product; competing reactions consume charge. The current efficiency (or faradaic efficiency) is defined as

When efficiency is less than 100%, Eq. (??) must be adjusted by multiplying by the efficiency (expressed as a fraction).

Side reactions commonly include:

H2 evolution at the cathode competing with metal plating,

O2 evolution at the anode competing with halide oxidation,

corrosion or dissolution of electrodes (if not inert),

recombination or secondary chemical reactions of products in solution.

Electrolysis uses an external power source to drive nonspontaneous redox reactions; in thermodynamic terms, it forces a process with ΔG > 0.

The ideal minimum applied voltage (reversible limit) satisfies nFEapp,rev = ΔG, so under standard conditions Eapp,rev∘ = −Ecell∘.

Real electrolysis requires extra voltage due to overpotential and resistance:

In aqueous electrolysis, water can be reduced to H2 or oxidized to O2, leading to competition with solute redox processes.

Faraday’s laws relate charge to chemical change:

These relationships enable quantitative predictions of mass deposited or gas evolved from current and time.

15.8 Problems

15.8.1 Entropy from microstates and macrostates

A container is divided into left and right halves by a removable partition. Initially, N = 20 distinguishable gas molecules are confined to the left half. The partition is removed and the gas expands to fill the entire container at constant temperature.

1. Using a two-compartment microstate model (each molecule can be in the left or right half), compute the ratio Wfinal∕Winitial and the corresponding entropy change ΔS = kB ln(Wfinal∕Winitial) for the sample (not per mole).

2. Convert your result to a molar entropy change for 1.00 mol of gas using NA and compare your expression to the ideal-gas result ΔS = nR ln(V ₂∕V ₁) for V ₂∕V ₁ = 2. Explain the origin of the agreement.

3. If the gas instead expands from V to 4V at constant T for 1.00 mol, write the expression for ΔS and compute its numerical value at T = 298.15K.

15.8.2 Entropy, enthalpy, and surroundings: predicting spontaneity via ΔSuniv

A reaction occurs in a large, well-stirred aqueous solution maintained at T = 298.15K and p = 1bar. For the reaction as written, the measured thermodynamic changes for the system are

1. Compute ΔSsurr and ΔSuniv under these conditions. State whether the reaction is spontaneous as written.

2. Determine the temperature at which the reaction changes from nonspontaneous to spontaneous (assume ΔH and ΔS are temperature-independent over the range considered).

3. At T = 350K, compute ΔG and interpret its sign in terms of spontaneity.

15.8.3 Computing ΔG∘ from formation data and finding K

At 298.15K, consider the gas-phase reaction

You are given the following standard molar Gibbs free energies of formation:

1. Compute ΔGrxn∘ at 298.15K.

2. Compute the equilibrium constant Kp at 298.15K.

3. If initially PNO2 = 0.400bar and PN2O4 = 0.200bar at 298.15K, compute Qp and determine the direction the system will shift to reach equilibrium.

15.8.4 Designing a galvanic cell and predicting Ecell∘

Use the following standard reduction potentials at 298.15K:

1. Construct a galvanic cell using the Zn2+/Zn and Ag+/Ag couples. Identify the anode and cathode, write the balanced overall reaction, and compute Ecell∘.

2. Write the IUPAC cell notation for the cell you designed.

3. Repeat parts (1) and (2) for a cell using the Cu2+/Cu and Ag+/Ag couples.

4. For each cell, state which one should deliver the larger standard electrical work per mole of overall reaction and justify your reasoning quantitatively.

15.8.5 Linking Ecell∘, ΔG∘, and K

A galvanic cell based on an overall reaction transfers n = 3 moles of electrons per mole of reaction. At 298.15K, the measured standard cell potential is Ecell∘ = 0.412V.

1. Compute ΔG∘ (in kJmol⁻¹).

2. Compute the equilibrium constant K at 298.15K.

3. If the cell is allowed to run until it is arbitrarily close to equilibrium at 298.15K, what is the limiting value of Ecell? Explain in terms of both ΔG and the Nernst equation.

15.8.6 Nonstandard cell potential via the Nernst equation

Consider the Daniell cell at 298.15K:

1. Write the reaction quotient Q for the cell reaction using activities, then write the common dilute-solution approximation in terms of molar concentrations.

2. Compute Ecell when [Zn2+] = 0.0100moldm⁻³ and [Cu2+] = 2.00moldm⁻³.

3. Compute Ecell when [Zn2+] = 1.00moldm⁻³ and [Cu2+] = 1.00 × 10⁻⁴ moldm⁻³.

4. For each case, state whether the cell reaction is spontaneous as written and justify your answer using both Ecell and the sign of ΔG.

15.8.7 Concentration cell and chemical potential

A concentration cell is constructed with identical electrodes based on the half-reaction

One half-cell contains [Ag+] = 1.0 × 10⁻² moldm⁻³ and the other contains [Ag+] = 1.0 × 10⁻⁶ moldm⁻³ at 298.15K. Assume ideal behavior.

1. Determine which side is the anode and which side is the cathode when the cell operates spontaneously.

2. Compute Ecell.

3. Write the net cell reaction in a form that emphasizes the overall effect on concentration.

4. Using ΔG = −nFE, compute ΔG per mole of electrons transferred and interpret its meaning in terms of the tendency to eliminate concentration gradients.

15.8.8 Electrolysis and Faraday’s laws: plating and gas evolution

An electrolytic cell is used to plate nickel metal from aqueous NiSO4 using the cathode half-reaction

A constant current of I = 3.50A is applied for t = 2.00h.

1. Compute the total charge passed.

2. Compute the moles of electrons transferred.

3. Compute the mass of Ni(s) deposited (use MNi = 58.69g mol⁻¹).

4. If the current efficiency for nickel deposition is only 85.0% because of competing H2 evolution, compute the corrected deposited mass and the moles of H2(g) produced. (Assume the only side reaction is 2H2O+2e− → H2 + 2OH−.)

15.8.9 Choosing products in aqueous electrolysis: thermodynamics and overpotential

An aqueous solution contains NaCl(aq) at high concentration. Electrolysis is carried out using inert electrodes. Consider the following possible electrode reactions (standard potentials given in reduction form at 298.15K):

1. Predict the most likely cathode reaction under typical aqueous conditions and justify your choice thermodynamically.

2. Predict the most likely anode reaction under typical aqueous conditions and justify your choice thermodynamically.

3. Write the overall electrolysis reaction corresponding to your predicted products.

4. Discuss qualitatively how overpotential (especially for oxygen evolution) and chloride concentration could shift the observed anode product between Cl2 and O2.

5. If the electrolysis is run at I = 5.00A for 30.0min and produces only H2 at the cathode and Cl2 at the anode with 100% current efficiency, compute the volumes of H2 and Cl2 produced at 298.15K and 1.00bar (assume ideal gas behavior).