AP Chemistry Unit 5 review: Kinetics
Unit 5 explains reaction rates through concentration data, collisions and reaction mechanisms. It is 7–9% of multiple choice. Expect to infer a rate law from measurements, choose the right integrated-rate plot and test whether a proposed mechanism fits the evidence.
Unit 5 on the May 2027 exam
| 2027 exam | |
|---|---|
| Weight | 7–9% of the multiple-choice section |
| Exam date | Thursday, May 6, 2027, afternoon session |
| Multiple choice | 60 questions in 90 minutes, worth 50% of the exam score |
| Free response | 7 questions in 105 minutes, worth 50%: three long questions (10 points each) and four short questions (4 points each). |
| Calculator | Both sections allow scientific or approved graphing calculators. Four-function calculators are allowed but not recommended. Bluebook includes both Desmos scientific and graphing calculators. |
| Class time | About 13–14 class periods in College Board's pacing guide |
| 2026 results | 15% of students scored a 5 on the whole exam, and about 76% scored 3 or higher |
Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.
What Unit 5 covers
College Board's course description splits the unit into 11 topics. The second column is what you should be able to do for each one.
| Topic | What you should be able to do |
|---|---|
| 5.1 Reaction Rates | Calculate average or instantaneous rates and relate species rates through coefficients. |
| 5.2 Introduction to Rate Law | Determine rate-law exponents and the rate constant from experimental data. |
| 5.3 Concentration Changes Over Time | Use integrated rate laws and half-lives to analyze concentration over time. |
| 5.4 Elementary Reactions | Write the rate law for an elementary step from its molecularity. |
| 5.5 Collision Model | Explain temperature and orientation effects using collisions and activation energy. |
| 5.6 Reaction Energy Profile | Read activation energy and reaction enthalpy from a single-step energy diagram. |
| 5.7 Introduction to Reaction Mechanisms | Identify intermediates and catalysts and add steps to recover the overall reaction. |
| 5.8 Reaction Mechanism and Rate Law | Compare a mechanism’s predicted rate law with the observed law. |
| 5.9 Pre-Equilibrium Approximation | Eliminate an intermediate using a fast pre-equilibrium before a slow step. |
| 5.10 Multistep Reaction Energy Profile | Identify intermediates and transition states on a multistep profile. |
| 5.11 Catalysis | Explain how a catalyst changes the pathway and rate without changing equilibrium. |
Choose the order from the straight-line plot
These rows apply to a rate law in one reactant, rate = k[A]ⁿ. Check the plotted variable before reading a slope.
| Order | Linear plot and slope | Half-life and k units |
|---|---|---|
| Zero | [A] versus t; slope −k | t½ = [A]₀/(2k); k in M s⁻¹. |
| First | ln[A] versus t; slope −k | t½ = ln 2/k; k in s⁻¹. |
| Second | 1/[A] versus t; slope +k | t½ = 1/(k[A]₀); k in M⁻¹ s⁻¹. |
| Initial-rate comparison | Rate ratio = concentration ratio raised to the order | Hold other reactant concentrations constant. |
| Temperature increase | A larger fraction of collisions exceeds the barrier | k changes; rate-law exponents are not simply temperature multipliers. |
| Catalyst | Alternative pathway with a lower effective activation barrier | Changes approach to equilibrium, not K or ΔH. |
Common mistakes
Each of these loses points on questions you otherwise understand.
- Reading the slope before checking the graph axes. A straight ln[A] plot and a straight 1/[A] plot imply different orders and k units. CED printed p. 91 flags graph and unit errors.
- Using overall reaction coefficients as rate-law exponents. That shortcut is justified for an elementary step, not a general multistep reaction. CED topics 5.2 and 5.4.
- Leaving an intermediate in the final observable rate law when it can be eliminated. Use the stated pre-equilibrium relationship. CED topic 5.9.
- Saying a catalyst raises equilibrium yield. It changes the pathway and rates in both directions; equilibrium composition at fixed temperature does not change. CED topic 5.11.
6 practice questions with answers
Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.
Question 1 · 5.2
At fixed [B], doubling [A] makes the initial rate four times larger. What is the order in A?
- AZero
- BFirst
- CSecond
- DFourth
Show the answer
C. 2ⁿ = 4 gives n = 2. The controlled [B] allows the rate change to be attributed to [A]. Overall reaction coefficients are not needed for this inference.
Question 2 · 5.2
A reaction follows rate = k[A]². The rate is 0.0180 M s⁻¹ when [A] = 0.300 M. What is k?
- A0.200 M⁻¹ s⁻¹
- B0.0600 s⁻¹
- C0.00540 M s⁻¹
- D0.600 M⁻¹ s⁻¹
Show the answer
A. k = 0.0180/(0.300)² = 0.200 M⁻¹ s⁻¹. Dividing M s⁻¹ by M² gives M⁻¹ s⁻¹.
Question 3 · 5.3
A first-order reactant has a half-life of 12.0 min. Starting at 0.640 M, what concentration remains after 36.0 min?
- A0.320 M
- B0.160 M
- C0.0200 M
- D0.0800 M
Show the answer
D. 36.0 min is three half-lives. [A] = 0.640(1/2)³ = 0.0800 M. Equal half-lives give equal fractions lost, not equal concentration decreases.
Question 4 · 5.1
For 2A → B, [A] falls from 0.800 M to 0.500 M in 30.0 s at constant volume. What is the average formation rate of B?
- A0.0100 M s⁻¹
- B0.00500 M s⁻¹
- C0.0200 M s⁻¹
- D0.300 M s⁻¹
Show the answer
B. A disappears at 0.300/30.0 = 0.0100 M s⁻¹. The 2:1 ratio means B forms at half that rate: 0.00500 M s⁻¹.
Question 5 · 5.6
Reactants lie at 25 kJ mol⁻¹, a transition state at 90 kJ mol⁻¹ and products at 10 kJ mol⁻¹. What is the forward activation energy?
- A15 kJ mol⁻¹
- B80 kJ mol⁻¹
- C65 kJ mol⁻¹
- D90 kJ mol⁻¹
Show the answer
C. The forward barrier is transition state minus reactants: 90 − 25 = 65 kJ mol⁻¹. Product minus reactant energy, −15 kJ mol⁻¹, is the reaction enthalpy, not the activation barrier.
Question 6 · 5.9
A fast equilibrium A + B ⇌ I has K = [I]/([A][B]). A slow step I + B → P has rate k[I][B]. Which overall rate law follows?
- Arate = kK[A][B]²
- Brate = kK[A]²[B]
- Crate = k[A][B]/K
- Drate = kK[I][B]²
Show the answer
A. The fast equilibrium gives [I] = K[A][B]. Substitution into the slow-step law gives rate = kK[A][B]². The final expression no longer contains the intermediate I.
How to study Unit 5
Build a small initial-rate table and change one concentration at a time. Calculate rate ratios before using logarithms. For concentration–time data, compare [A], ln[A] and 1/[A] plots and derive the units of k from the chosen law.
For a mechanism, add all elementary steps, cancel intermediates and check that the overall equation matches. Then derive the predicted rate law under the stated slow-step or pre-equilibrium assumptions. Agreement is necessary for a plausible mechanism, but it does not uniquely prove that mechanism.
Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.
- AP Chemistry: all 9 units, exam format and scores
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- AP Chemistry Unit 2: Compound Structure and Properties
- AP Chemistry Unit 3: Properties of Substances and Mixtures
- AP Chemistry Unit 4: Chemical Reactions
- AP Chemistry Unit 6: Thermochemistry
- AP Chemistry Unit 7: Equilibrium
- AP Chemistry Unit 8: Acids and Bases
- AP Chemistry Unit 9: Thermodynamics and Electrochemistry
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