Review AP Chem Unit 5 to understand how chemists measure and control reaction speeds, from writing rate laws using experimental data to tracing the step-by-step path of a multistep mechanism. This unit connects molecular collisions, energy profiles, and catalysis into one coherent picture of why some reactions are fast and others are slow.
Use the topic guides, key terms, and practice questions available for this unit to work through rate law calculations, integrated rate law graphs, and mechanism analysis before your exam.
Kinetics asks a simple question: how fast does a reaction go, and why? Unit 5 answers that question at three levels. At the macroscopic level, you measure concentration changes over time and express them as rate laws. At the molecular level, you use collision theory and energy profiles to explain why only some collisions produce products. At the mechanistic level, you break an overall reaction into elementary steps and identify which step controls the overall rate.
You cannot read reaction orders from the coefficients of a balanced equation. Orders must be determined from experimental data, either by comparing initial rates or by identifying which concentration-time graph is linear. The rate constant k has units that depend on the overall reaction order.
Each reaction order has a characteristic linear plot: [A] vs. t for zeroth order, ln[A] vs. t for first order, and 1/[A] vs. t for second order. The slope of the linear plot gives k. First-order reactions also have a constant half-life equal to 0.693/k, which is independent of initial concentration.
A proposed mechanism is valid only if its elementary steps add up to the overall balanced equation and the rate law derived from the slowest step matches the experimentally determined rate law. Intermediates appear in elementary steps but cancel out of the overall equation.
Every rate law reflects what is happening at the molecular level. The rate constant k increases with temperature because more collisions exceed the activation energy threshold, as described by the Arrhenius equation. A catalyst speeds up a reaction by providing a new pathway with lower activation energy, which changes the mechanism without being consumed overall. Understanding this connection lets you move fluently between experimental data, energy diagrams, and mechanistic explanations on the AP exam.
Define reaction rate as concentration change per unit time, use stoichiometric coefficients to relate rates of different species, and identify factors that increase or decrease rate.
Write rate = k[A]^m[B]^n, determine reaction orders using the method of initial rates, calculate k, and assign correct units based on overall reaction order.
Apply integrated rate laws to find concentration at any time, identify reaction order from linear concentration-time plots, and calculate first-order half-life using t1/2 = 0.693/k.
Write rate laws directly from the stoichiometry of elementary steps, identify molecularity as unimolecular or bimolecular, and explain why termolecular steps are rare.
Explain that effective collisions require both sufficient energy and correct orientation, use the Maxwell-Boltzmann distribution to show how temperature affects the fraction of reactive collisions.
Draw and interpret single-step energy profiles, label activation energy and transition state, connect Ea to the Arrhenius equation, and distinguish exothermic from endothermic profiles.
Identify reactants, products, intermediates, and catalysts in a mechanism, confirm that elementary steps sum to the overall balanced equation, and explain how detecting an intermediate supports a proposed mechanism.
Derive the overall rate law from the molecularity of the rate-determining first step, and verify that the derived rate law is consistent with experimental data.
Apply the pre-equilibrium approximation when the first step is fast and reversible, substitute the intermediate's concentration into the slow step's rate law, and express the final rate law in terms of reactants only.
Construct energy profiles for multistep mechanisms, label each transition state and intermediate, identify the rate-determining step as the highest barrier, and read the overall energy change from the diagram.
Explain how catalysts lower activation energy and change the reaction mechanism, distinguish homogeneous, heterogeneous, and enzyme catalysis, and show catalyst regeneration in a mechanism.
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Review Pre-Equilibrium Approximation with attention to how the concept appears in AP-style source and evidence questions.
Review Multistep Reaction Energy Profile with attention to how the concept appears in AP-style source and evidence questions.
Review Reaction Mechanism and Rate Law with attention to how the concept appears in AP-style source and evidence questions.
Review Concentration Changes Over Time with attention to how the concept appears in AP-style source and evidence questions.
Reaction rate measures how quickly reactant concentrations decrease or product concentrations increase over time, expressed in units of mol/L/s. Because stoichiometric coefficients relate the rates of different species, you must divide each species' rate of change by its coefficient to get a single reaction rate. For example, if 2 mol of NO2 are consumed for every 1 mol of O2 produced, the rate of disappearance of NO2 is twice the rate of appearance of O2. Factors that increase rate include higher reactant concentration, higher temperature, greater surface area, and the presence of a catalyst.
| Factor | Effect on rate | Reason |
|---|---|---|
| Increase concentration | Increases | More frequent collisions |
| Increase temperature | Increases | More collisions exceed activation energy |
| Increase surface area | Increases | More reactant particles exposed |
| Add catalyst | Increases | Lower activation energy pathway |
A rate law has the form rate = k[A]^m[B]^n, where m and n are the orders with respect to each reactant and must be determined experimentally using the method of initial rates. To find an order, compare two experiments where only one reactant concentration changes and calculate how the rate changes. The rate constant k is temperature-dependent and has units that vary with overall order: M/s for zeroth order, 1/s for first order, and 1/(M·s) for second order.
| Order | Rate law form | Units of k |
|---|---|---|
| Zeroth | rate = k | M/s |
| First | rate = k[A] | 1/s |
| Second | rate = k[A]^2 or k[A][B] | 1/(M·s) |
Integrated rate laws relate concentration directly to time, allowing you to predict concentration at any point or determine how long a reaction takes. The key skill is identifying which plot is linear: [A] vs. t for zeroth order, ln[A] vs. t for first order, and 1/[A] vs. t for second order. The slope of the linear plot equals -k for zeroth and first order, and +k for second order. First-order reactions have a constant half-life t1/2 = 0.693/k, meaning each successive half-life reduces concentration by half regardless of starting concentration. Radioactive decay is a classic first-order process.
| Order | Linear plot | Slope | Half-life |
|---|---|---|---|
| Zeroth | [A] vs. t | -k | t1/2 = [A]0 / 2k |
| First | ln[A] vs. t | -k | t1/2 = 0.693/k (constant) |
| Second | 1/[A] vs. t | +k | t1/2 = 1/(k[A]0) |
An elementary reaction is a single-step process, and its rate law can be written directly from its stoichiometry: a bimolecular step A + B produces rate = k[A][B]. This is the only situation where you can read orders from coefficients. For a collision to produce a reaction, it must meet two requirements: sufficient energy to overcome the activation energy barrier and correct geometric orientation of the reacting molecules. The Maxwell-Boltzmann distribution shows that only a fraction of molecules at a given temperature have enough energy to react, and that fraction increases as temperature rises.
A reaction energy profile plots potential energy along the reaction coordinate. For a single-step reaction, the diagram shows reactants on the left, a single peak at the transition state, and products on the right. The activation energy Ea is the energy difference between the reactants and the transition state peak. The overall energy change is the difference between reactant and product energy levels. A catalyst lowers the peak but does not change the energy of reactants or products. The Arrhenius equation k = Ae^(-Ea/RT) quantifies how Ea and temperature together determine the rate constant.
| Feature | Exothermic profile | Endothermic profile |
|---|---|---|
| Products vs. reactants | Products lower in energy | Products higher in energy |
| Overall energy change | Negative (energy released) | Positive (energy absorbed) |
| Ea forward vs. reverse | Ea forward < Ea reverse | Ea forward > Ea reverse |
A reaction mechanism is a sequence of elementary steps that together produce the overall balanced equation. Intermediates are species produced in one step and consumed in a later step; they do not appear in the overall equation. When the first step of a mechanism is the slowest (rate-determining step), the rate law for the overall reaction equals the rate law for that slow step, written from its molecularity. You validate a proposed mechanism by checking that the steps sum to the overall equation and that the derived rate law matches the experimental rate law.
| Species | Appears in overall equation? | Consumed and regenerated? |
|---|---|---|
| Reactant | Yes, as starting material | No |
| Product | Yes, as final product | No |
| Intermediate | No | Produced then consumed |
| Catalyst | No (net) | Consumed then regenerated |
When the first step of a mechanism is fast and reversible rather than rate-limiting, you cannot directly use its molecularity to write the overall rate law. Instead, apply the pre-equilibrium approximation: set the forward and reverse rates of the fast first step equal to each other, solve for the concentration of the intermediate, and substitute that expression into the rate law for the slow second step. The result is a rate law expressed only in terms of actual reactants, which can then be compared to experimental data.
| Scenario | Rate law source | Approximation needed? |
|---|---|---|
| First step is slow (rate-limiting) | Molecularity of slow step directly | No |
| First step is fast and reversible | Slow step after substituting intermediate | Yes, pre-equilibrium |
A multistep reaction energy profile has one peak per elementary step. Each peak is a transition state and each valley between peaks is an intermediate. The tallest peak corresponds to the rate-determining step and has the largest activation energy. The overall energy change is still the difference between the final products and the initial reactants. A catalyst provides an alternative mechanism with a lower activation energy, which appears on the diagram as a new, shorter peak. The catalyst is consumed in one step and regenerated in a later step, so its net concentration does not change. Homogeneous catalysts are in the same phase as reactants; heterogeneous catalysts are in a different phase and often work by adsorbing reactants onto a surface.
| Catalyst type | Phase relationship | How it lowers Ea |
|---|---|---|
| Homogeneous | Same phase as reactants | Provides alternative reaction pathway in solution |
| Heterogeneous | Different phase from reactants | Adsorbs reactants onto surface, improving orientation |
| Enzyme | Same phase (aqueous) | Binds substrate at active site, stabilizing transition state |
Try stimulus-based AP practice questions and written prompts after you review the notes.
| Term | Definition |
|---|---|
| Activation Energy | The minimum energy required for a collision to result in a reaction; the energy difference between the reactants and the transition state on an energy profile. |
| Rate Laws | Mathematical expressions of the form rate = k[A]^m[B]^n that relate reaction rate to reactant concentrations; orders must be determined experimentally. |
| Rate Constant | The proportionality factor k in the rate law; its value depends on temperature and its units depend on the overall reaction order. |
| Integrated Rate Law | An equation relating reactant concentration to time; the form differs by order: [A] vs. t (zeroth), ln[A] vs. t (first), 1/[A] vs. t (second). |
| Half-Life | The time required for reactant concentration to fall to half its initial value; for first-order reactions, t1/2 = 0.693/k and is constant regardless of initial concentration. |
| Collision Theory | The model stating that reactions occur only when molecules collide with both sufficient energy to exceed the activation energy and the correct geometric orientation. |
| Effective Collisions | Collisions that result in a reaction because they meet both the energy requirement (Ea) and the orientation requirement for bond rearrangement. |
| Transition State | The highest-energy arrangement of atoms along the reaction coordinate; represents the activated complex that must form for reactants to become products. |
| Elementary Step | A single-step reaction in a mechanism whose rate law can be written directly from the stoichiometry of the colliding particles. |
| Molecularity | The number of particles that collide in an elementary step: unimolecular (1), bimolecular (2), or termolecular (3, rare). |
| Reaction Intermediate | A species produced in one elementary step and consumed in a later step; it does not appear in the overall balanced equation. |
| Rate-Determining Step | The slowest elementary step in a mechanism; when it is the first step, its molecularity directly sets the rate law for the overall reaction. |
| Catalyst | A substance that increases reaction rate by providing a pathway with lower activation energy; it is consumed in one step and regenerated in a later step, so its net concentration is unchanged. |
| Reaction Order | The exponent of a reactant's concentration in the rate law; the sum of all exponents gives the overall order of the reaction. |
Reaction orders must come from experimental data. The coefficients in the overall balanced equation do not equal the exponents in the rate law unless the reaction is a single elementary step and that is explicitly stated.
An intermediate is produced in one step and consumed in a later step; it is not present at the start. A catalyst is present at the start, consumed in one step, and regenerated later. Both are absent from the overall equation, but for different reasons.
Only first-order reactions have a constant half-life (t1/2 = 0.693/k). For zeroth-order reactions, half-life decreases as concentration decreases. For second-order reactions, half-life increases as concentration decreases.
If the first step is fast and reversible, you cannot directly use its molecularity. You must apply the pre-equilibrium approximation to express the intermediate's concentration in terms of reactants before writing the final rate law.
A catalyst lowers the activation energy and changes the mechanism, but it does not change the energies of the reactants or products. The overall enthalpy change for the reaction is the same with or without a catalyst.
AP Chemistry frequently presents a table of initial rate data and asks you to determine reaction orders, write the rate law, and calculate k. A related task is identifying reaction order from a set of concentration-time graphs by recognizing which plot is linear. Be prepared to justify each order with a ratio calculation rather than just stating the answer.
A common task presents a proposed two- or three-step mechanism and asks you to identify intermediates, derive the rate law from the rate-determining step, and evaluate whether the mechanism is consistent with an experimentally determined rate law. You may also need to apply the pre-equilibrium approximation when the first step is fast and reversible.
Questions often ask you to draw or annotate a reaction energy profile, labeling activation energies, transition states, intermediates, and overall energy change. A follow-up task typically asks you to explain how a catalyst changes the diagram and the mechanism, or to compare homogeneous and heterogeneous catalysis in terms of how each lowers the activation energy.
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open calculatorAP Chem Unit 5 covers 11 topics in kinetics: Reaction Rates, Introduction to Rate Law, Concentration Changes Over Time, Elementary Reactions, the Collision Model, Reaction Energy Profile, Introduction to Reaction Mechanisms, Reaction Mechanism and Rate Law, Pre-Equilibrium Approximation, Multistep Reaction Energy Profile, and Catalysis. Together these topics explain why reactions happen at different speeds, how rate laws are written and used, and how catalysis speeds up reactions without being consumed. See AP Chem Unit 5 for matched practice on each topic.
AP Chem Unit 5 makes up 7-9% of the AP exam. That weight covers all of kinetics, including rate law, reaction mechanisms, the collision model, and catalysis. It's a focused unit, but the concepts show up in FRQs regularly, so understanding how reaction rates are determined and manipulated pays off on exam day.
The AP Chem Unit 5 progress check includes both MCQ and FRQ parts drawn from all 11 kinetics topics. MCQ questions test rate law interpretation, concentration changes over time, the collision model, and reaction energy profiles. FRQ questions typically ask you to determine rate laws from experimental data, analyze reaction mechanisms, or explain how catalysis affects activation energy. The progress check is College Board's built-in check on whether you can apply these concepts, not just recall them. Practicing with questions matched to each topic on AP Chem Unit 5 is a solid way to prepare for both parts.
AP Chem Unit 5 FRQs most often ask you to determine a rate law from experimental data, identify the rate-determining step in a reaction mechanism, or explain how catalysis lowers activation energy. These questions reward clear, step-by-step reasoning, so writing out your logic matters as much as the final answer. Good practice steps: - Work through rate law problems where you solve for reaction orders from initial rate data. - Practice drawing and interpreting reaction energy profiles for both catalyzed and uncatalyzed reactions. - Write out mechanism steps and connect them to the overall rate law. You can find FRQ-style practice questions organized by topic at AP Chem Unit 5.
The best place to find AP Chem Unit 5 practice questions, including MCQ and full practice test sets, is AP Chem Unit 5. Questions there are organized by topic, so you can target rate law, the collision model, reaction mechanisms, or catalysis specifically rather than hunting through a mixed set. For the most useful practice, start with MCQ on the topics you find hardest, then move to FRQ-style questions that ask you to interpret data or explain mechanisms in writing. That combination covers both question formats you'll see on the real exam.
Start AP Chem Unit 5 by building a solid understanding of rate law before moving to mechanisms. If you can't write a rate law from experimental data, the later topics on reaction mechanisms and the pre-equilibrium approximation won't click. A practical study sequence: 1. Learn how to determine reaction orders from initial rate tables (5.2). 2. Practice integrated rate laws and half-life calculations for concentration changes over time (5.3). 3. Study the collision model and reaction energy profiles together, since activation energy connects both (5.5, 5.6). 4. Work through multistep mechanisms and practice identifying the rate-determining step (5.7, 5.8). 5. Finish with catalysis, focusing on how it changes the energy profile without shifting equilibrium (5.11). Kinetics is one of those units where doing problems beats re-reading notes. Use AP Chem Unit 5 to get topic-specific practice at each step.