AP Chemistry / Topic

Kinetics

Unit 5 covers reaction rates: rate laws from initial-rates data, integrated rate laws and half-life, collision theory, energy profiles, catalysts, and reaction mechanisms with rate-determining steps.

32 practice questions1 FRQ14 skill areas

What the exam asks

  • Determine reaction orders and the rate constant from experimental data tables
  • Use integrated rate laws and identify order from linear plots
  • Apply half-life reasoning for first-order processes
  • Explain how concentration, temperature, surface area, and catalysts change rates via collision theory
  • Evaluate proposed mechanisms: elementary steps, intermediates, catalysts, and consistency with the rate law

Key formulas and rules

Generic rate law

rate = k[A]^m[B]^n - orders come from experiment, not coefficients

Zeroth order

[A] vs t is linear; rate independent of [A]

First order

ln[A] vs t linear; ln[A]t = −kt + ln[A]0; t½ = 0.693/k (constant)

Second order

1/[A] vs t is linear; 1/[A]t = kt + 1/[A]0

Mechanism rule

rate law of the slow (rate-determining) step must match the observed rate law

Catalyst

lowers activation energy, appears unchanged; intermediate is made then consumed

Question bank breakdown

Multiple choice

32

Free response

1

Difficulty mix

11 easy · 13 medium · 8 hard

Skills covered

Rate laws and method of initial ratesRate lawsIntegrated rate laws and half-lifeIntegrated rate lawsCatalysis and activation energyReaction mechanismsReaction Rates and StoichiometryIntegrated Rate Laws and Half-LifeCollision TheoryEnergy ProfilesCatalysisRate LawsMethod of Initial RatesReaction Mechanisms

Every question in the bank comes with a worked explanation, and most add a per-choice breakdown so you learn why each wrong answer is wrong - not just the key.

Sample question

From the free tier of the ScoreMint AP Chemistry bank - try it, then check the answer.

For the reaction A + B → C, the following initial rate data were collected: Experiment 1: [A] = 0.10 M, [B] = 0.10 M, Rate = 2.0 × 10⁻³ M/s Experiment 2: [A] = 0.20 M, [B] = 0.10 M, Rate = 8.0 × 10⁻³ M/s Experiment 3: [A] = 0.10 M, [B] = 0.20 M, Rate = 4.0 × 10⁻³ M/s What is the rate law for this reaction?

  • A. Rate = k[A][B]
  • B. Rate = k[A]²[B]
  • C. Rate = k[A][B]²
  • D. Rate = k[A]²[B]²
Show answer and explanation

Answer: B. Correct. Compare Exp 1 and 2: [A] doubles, [B] constant, rate quadruples (8.0/2.0 = 4 = 2²). So the reaction is second order in A. Compare Exp 1 and 3: [B] doubles, [A] constant, rate doubles (4.0/2.0 = 2 = 2¹). So the reaction is first order in B. Rate law: Rate = k[A]²[B]. Check with Exp 1: k = 2.0 × 10⁻³ / [(0.10)²(0.10)] = 2.0 × 10⁻³ / 1.0 × 10⁻³ = 2.0 M⁻²s⁻¹. Check with Exp 2: Rate = 2.0 × (0.20)²(0.10) = 2.0 × 4.0 × 10⁻³ = 8.0 × 10⁻³ ✓

Study tip

Memorize the three linear plots ([A], ln[A], 1/[A] versus time). The exam's favorite kinetics question is handing you a straight-line graph and asking for the order or the meaning of the slope.