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AP Chemistry · Unit 5 of 9
Kinetics
7–9% of the exam6 lessons · 82 min19 terms
What this unit covers
The topics below follow the published Chemistry course framework for Unit 5. This unit is worth 7–9% of the exam, so budget your time against that rather than against how long the unit takes to teach.
Reaction ratesRate lawsReaction mechanismsCatalysis
Lessons in this unit
- Reaction Rates & Collision Theory12 min · 3 objectivesDefine reaction rate and express it in terms of a reactant or product · Use collision theory and activation energy to explain why reactions occur · Predict how concentration, temperature, surface area, and a catalyst change the rate
- Rate Laws & Reaction Order14 min · 3 objectivesWrite a rate law and identify the rate constant k and reaction orders · Determine reaction order from initial-rate data by comparing trials · Calculate k and give its units from a determined rate law
- Integrated Rate Laws & Half-Life14 min · 3 objectivesMatch zero-, first-, and second-order reactions to their integrated rate laws and linear plots · Use the first-order half-life relationship t½ = 0.693/k · Calculate concentration or time remaining after a whole number of half-lives
- Reaction Mechanisms & Catalysis13 min · 3 objectivesDescribe a reaction as a sequence of elementary steps and identify the rate-determining step · Distinguish an intermediate from a catalyst in a mechanism · Explain how a catalyst speeds a reaction by lowering the activation energy
- Integrated Rate Laws & Graphical Analysis15 min · 3 objectivesMatch zero-, first-, and second-order reactions to the plot that comes out linear ([A], ln[A], or 1/[A] vs t) · Determine reaction order from data and extract k from the slope with the correct sign · Select the half-life expression for each order and read its dependence on [A]₀
- Arrhenius, Reaction Coordinate Diagrams & Catalysis14 min · 3 objectivesUse k = A·e^(−Eₐ/RT) to reason about how temperature and activation energy control the rate constant · Read a reaction-coordinate diagram for Eₐ (forward and reverse), ΔH, the transition state, and any intermediate · Explain how a catalyst lowers Eₐ without changing ΔH or the position of equilibrium
Formulas in Unit 5
Average reaction rate
rate = −Δ[reactant] / Δt = +Δ[product] / Δt
For a A → b B, divide each term by its coefficient so all species give the same rate: −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt.
Factors that speed a reaction
↑ concentration, ↑ temperature, ↑ surface area, add catalyst → ↑ rate
Concentration, temperature, and surface area raise the collision rate; temperature and a catalyst also change the fraction of collisions that succeed.
General rate law
rate = k[A]ᵐ[B]ⁿ
m is the order in A, n the order in B, and (m + n) the overall order. Orders are usually 0, 1, or 2 and come only from data.
Integrated rate laws & linear plots
Zero: [A] = [A]₀ − kt (plot [A] vs t) | First: ln[A] = ln[A]₀ − kt (plot ln[A] vs t) | Second: 1/[A] = 1/[A]₀ + kt (plot 1/[A] vs t)
Whichever plot is a straight line reveals the order; the slope gives k (−k for zero/first order, +k for second order).
First-order half-life
t½ = 0.693 / k
0.693 is ln 2. Because [A]₀ cancels out, first-order half-life is independent of starting concentration. (Zero order: t½ = [A]₀/2k; second order: t½ = 1/(k[A]₀) — these do depend on [A]₀.)
Spotting each species in a mechanism
Intermediate: appears first as a product, then a reactant. Catalyst: appears first as a reactant, then a product.
Both cancel out of the overall equation — the order in which they appear (made-then-used vs. used-then-remade) is what tells them apart.
Integrated rate laws as straight lines
Zero: [A] = −kt + [A]₀ (plot [A] vs t) | First: ln[A] = −kt + ln[A]₀ (plot ln[A] vs t) | Second: 1/[A] = kt + 1/[A]₀ (plot 1/[A] vs t)
In each case y = mx + b: the y-intercept is the starting value and the slope is ±k. Zero and first order give slope = −k; second order gives slope = +k.
Half-life by order
Zero: t½ = [A]₀ / (2k) First: t½ = 0.693 / k Second: t½ = 1 / (k[A]₀)
Only first-order t½ is independent of [A]₀ (0.693 = ln 2). If successive half-lives are equal → first order; if they get shorter → zero order; if they get longer → second order.
Arrhenius equation
k = A·e^(−Eₐ/RT)
R = 8.314 J·mol⁻¹·K⁻¹ (use Eₐ in J/mol). Two-temperature form: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). A plot of ln k vs 1/T is linear with slope = −Eₐ/R.
Energy relationships on the diagram
Eₐ(forward) = E(transition state) − E(reactants) Eₐ(reverse) = E(transition state) − E(products) ΔH = Eₐ(forward) − Eₐ(reverse)
ΔH also equals E(products) − E(reactants). If Eₐ(forward) < Eₐ(reverse) the reaction is exothermic (ΔH < 0); if Eₐ(forward) > Eₐ(reverse) it is endothermic (ΔH > 0).
Every term in Unit 5
All 19 terms we publish for Kinetics, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.
- Reaction rate
- Change in concentration per unit time, always expressed as a positive quantity and divided by the stoichiometric coefficient.
- Rate law
- Rate = k[A]^m[B]^n. The orders m and n are determined experimentally and cannot be read off the balanced equation.
- Method of initial rates
- Compare trials in which one concentration changes. Doubling a reactant and doubling the rate means first order; quadrupling means second order.
- Zero-order kinetics
- [A] vs t is linear with slope −k; rate is independent of concentration. Half-life shortens as the reaction proceeds.
- First-order kinetics
- ln[A] vs t is linear with slope −k; half-life t½ = 0.693/k is constant, independent of starting concentration.
- Second-order kinetics
- 1/[A] vs t is linear with slope +k; half-life lengthens as concentration falls.
- Determining order graphically
- Plot [A], ln[A] and 1/[A] against time; whichever is linear identifies the order as zero, first or second.
- Collision theory
- Reaction requires collisions with sufficient energy and correct orientation. Most collisions produce nothing.
- Activation energy
- The minimum energy for a successful collision. Raising temperature increases the fraction of molecules above it, which is why rate is so temperature-sensitive.
- Arrhenius equation
- k = Ae^(−Ea/RT). A plot of ln k against 1/T is linear with slope −Ea/R.
- Reaction mechanism
- The sequence of elementary steps. It must sum to the overall equation and be consistent with the experimental rate law.
- Rate-determining step
- The slowest elementary step. The rate law follows from its molecularity, which is why the rate law reveals mechanism.
- Intermediate vs catalyst
- An intermediate is produced then consumed; a catalyst is consumed then regenerated. Neither appears in the overall equation.
- Catalysis
- Provides an alternative pathway with lower activation energy. It does not change ΔH, the equilibrium position, or the products.
- Why order cannot be read from coefficients
- The balanced equation describes the overall change; the rate law reflects only the rate-determining step, which may involve different species entirely.
- Pseudo-first-order conditions
- Flooding the system with one reactant keeps its concentration effectively constant, so the observed order reflects the other reactant alone.
- Steady-state intermediates
- An intermediate formed in a fast pre-equilibrium can be substituted out of the rate law using the equilibrium expression of the earlier step.
- Energy profile diagram
- Peaks are transition states and wells are intermediates. A two-step mechanism has two peaks, and the taller one is the rate-determining step.
- Why a catalyst does not shift equilibrium
- It lowers the activation energy of forward and reverse steps by the same amount, so both rates rise equally and K is unchanged.
What examiners penalize here
- When asked to *justify* a rate change, name the mechanism: more frequent collisions, a greater fraction of collisions exceeding Eₐ, or a lower Eₐ (catalyst only). Naming the factor without the mechanism rarely earns the point.
- The orders in a rate law come from **experimental data only** — never from the stoichiometric coefficients of the overall balanced equation. Reading exponents off the coefficients is the single most common kinetics mistake on the exam.
- A **constant** half-life is the signature of first order; a linear ln[A]-vs-t plot confirms it. If half-life instead grows as the reaction proceeds, suspect second order; if a plain [A]-vs-t plot is linear, it is zero order.
- To recover the RDS from data: if the experimental rate law is rate = k[A]²[B], the slow step must consume 2 A and 1 B (or an equivalent set once an intermediate is expressed via a fast equilibrium). Mechanism, RDS, and rate law must all agree.
- On free-response, justify the order by naming **which plot is linear**, then read k off the slope with the right sign (−k for zero/first order, +k for second order) and give **units by overall order** (M·s⁻¹, s⁻¹, M⁻¹·s⁻¹). A bare "second order" without pointing to the linear 1/[A] plot usually loses the justification point.
- To annotate a diagram fast: mark the **peak as the transition state**, arrow **up to it from reactants (Eₐ forward)** and **up to it from products (Eₐ reverse)**, and read **ΔH from reactant level to product level**. Check with ΔH = Eₐ(forward) − Eₐ(reverse). Count humps: two humps means a two-step mechanism with an intermediate in the valley.
Practice Chemistry
Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.
Questions about this unit
How much of the AP Chemistry exam is Unit 5?
Unit 5, Kinetics, is worth 7–9% of the Chemistry multiple-choice section according to the published course framework. Across all 9 units that makes it a middling share, roughly what an even split across units would give.
What topics are covered in Chemistry Unit 5?
Kinetics covers Reaction rates, Rate laws, Reaction mechanisms and Catalysis. We publish 19 terms with definitions for this unit, all of them on this page.
How should I study Chemistry Unit 5?
Read the 6 lessons below first — about 80 minutes — then drill the 19 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.
All 9 units of AP Chemistry
Unit names, topics and exam weights follow the published College Board course framework for AP Chemistry. AP® is a trademark registered by the College Board, which does not endorse this site.