The problems are on the rate law worksheet. Try them on your own first, then compare with the answers below.
Quick answer: A rate law, rate = k[A]^m[B]^n, shows how reaction rate depends on reactant concentrations, with orders found by experiment. These worked answers correct the 20 problems of the AP Chemistry worksheet on rate law, from reading a rate law to two first-order reactions racing: each solution shows the formula, the substitution and the final answer with its unit.
Full methods are given: formula, substitution and result in bold. Compare your reasoning with the model, not only the final number, because AP graders award marks for method. Small differences in the last digit come from rounding, so keep an extra figure until the end.
Level: AP Chemistry · Curriculum: AP Chem Unit 5
Problem 1 Reading a rate law
1. Order in A = 2, order in B = 1, overall order = 3.2. rate (M s⁻¹) = k M³, so k has units M⁻² s⁻¹.
3. The rate is proportional to [B], so doubling [B] doubles the rate.
Problem 2 Units of the rate constant
1. (X) M s⁻¹; (Y) s⁻¹; (Z) M⁻¹ s⁻¹.2. Reaction X (zero order) has a rate that does not depend on concentration.
Problem 3 What the change in rate tells you
1. Factor 2 gives factor 2 = 2¹: order 1.2. Factor 3 gives factor 9 = 3²: order 2.
3. Factor 2 gives factor 1 = 2⁰: order 0.
Problem 4 Half-life of a first-order reaction
1. t½ = 0.693/0.0462 = 15.0 min.2. 45.0 min = 3 half-lives, so the fraction left is (½)³ = 1/8 = 0.125 (12.5 %).
Problem 5 Reading a concentration–time graph
1. The graph is a straight line falling steadily: zero order.2. Slope = (0.30 − 0.90)/(60 − 0) = −0.010 M/s.
3. For zero order slope = −k, so k = 0.010 M s⁻¹.
Problem 6 Choosing the straight-line plot
1. A straight ln[A] plot means first order.2. k = −slope = 0.025 s⁻¹.
3. t½ = 0.693/0.025 = 28 s (27.7 s).
Problem 7 Second-order half-life
1. t½ = 1/(0.25 × 0.40) = 10 s.2. t½ = 1/(0.25 × 0.20) = 20 s.
3. In a second-order reaction t½ ∝ 1/[A]₀. With less reactant, collisions between A molecules are rarer, so the halving takes longer.
Problem 8 Initial rates with two reactants
1. Experiments 1 and 2: [P] ×2, rate ×2, so the order in P is 1.2. Experiments 2 and 3: [Q] ×3, rate ×(5.4 × 10⁻³/6.0 × 10⁻⁴ = 9), so the order in Q is 2.
3. rate = k[P][Q]². k = 3.0 × 10⁻⁴/(0.20 × 0.10²) = 0.15 M⁻² s⁻¹.
Problem 9 Time for a first-order change
1. t = ln(0.60/0.15)/k = ln 4/(2.5 × 10⁻³) = 1.386/(2.5 × 10⁻³) = 5.5 × 10² s (554 s).2. t½ = 0.693/(2.5 × 10⁻³) = 277 s. 0.60 → 0.30 → 0.15 M is two half-lives = 554 s. ✓
Problem 10 Finding k from two measurements
1. k = ln(0.80/0.50)/120 = 0.470/120 = 3.9 × 10⁻³ s⁻¹.2. t½ = 0.693/3.92 × 10⁻³ = 1.8 × 10² s.
3. [A] = 0.80 × e^(−3.92 × 10⁻³ × 300) = 0.80 × e^(−1.176) = 0.80 × 0.3085 = 0.25 M.
Problem 11 Pesticide breakdown on a farm
1. k = 0.693/8.0 = 0.0866 day⁻¹.2. t = ln(5.0/1.0)/0.0866 = 1.609/0.0866 = 19 days (18.6).
Problem 12 Second-order concentration
1. 1/[A]₀ = 1/0.50 = 2.0 M⁻¹.2. 1/[A] = 2.0 + 0.80 × 25 = 22.0 M⁻¹, so [A] = 0.045 M.
Problem 13 Zero-order surface reaction
1. t = [A]₀/k = 0.0600/(1.5 × 10⁻⁴) = 4.0 × 10² s.2. t½ = [A]₀/(2k) = 2.0 × 10² s.
3. The surface sites of the metal are saturated with ammonia; they work at full capacity, so adding more gas does not speed the reaction.
Problem 14 Reading half-lives from a graph
1. 0.80 → 0.40 M takes 20 s and 0.40 → 0.20 M takes another 20 s: 20 s, 20 s.2. The half-life is constant, so the reaction is first order.
3. k = 0.693/20 = 0.035 s⁻¹ (0.0347).
4. 60 s is three half-lives: 0.80/8 = 0.10 M.
Problem 15 Converting the rate constant
1. k = 3.6 × 10⁻²/60 = 6.0 × 10⁻⁴ s⁻¹.2. t½ = 0.693/0.036 = 19 min (19.25), which is 0.693/6.0 × 10⁻⁴ = 1.2 × 10³ s.
Problem 16 Deciding the order from data
1. 1/[A] = 1.00, 2.0, 3.0, 4.0 M⁻¹ (to two figures: 1/0.33 = 3.0 and 1/0.25 = 4.0).2. 1/[A] increases by 1.0 M⁻¹ every 50 s: a straight line, so second order.
3. k = slope = 1.0/50 = 0.020 M⁻¹ s⁻¹.
4. t½ = 1/(0.020 × 1.00) = 50 s, and the data show 0.50 M at 50 s. ✓
Problem 17 Three reactants
1. (1)→(2): [X] ×2, rate ×2, order 1. (1)→(3): [Y] ×2, rate ×4, order 2. (1)→(4): [Z] ×3, rate unchanged, order 0.2. rate = k[X][Y]². k = 4.0 × 10⁻³/(0.10 × 0.10²) = 4.0 M⁻² s⁻¹.
3. rate = 4.0 × 0.30 × (0.05)² = 4.0 × 0.30 × 0.0025 = 3.0 × 10⁻³ M/s.
Problem 18 Gas pressure and first order
1. t = ln(80.0/25.0)/1.2 × 10⁻³ = 1.163/1.2 × 10⁻³ = 9.7 × 10² s.2. 10.0 min = 600 s; fraction left = e^(−1.2 × 10⁻³ × 600) = e^(−0.72) = 0.487, so 51 % reacted.
3. At constant temperature and volume, the pressure of a gas is proportional to its concentration (p = cRT), so the ratio is the same.
Problem 19 Pseudo-first-order kinetics
1. B is in 500-fold excess, so even when all A reacts, [B] changes by only 0.2 %.2. Slope magnitude = k[B]: k = 3.0 × 10⁻³/0.50 = 6.0 × 10⁻³ M⁻¹ s⁻¹.
3. k[B] = 6.0 × 10⁻³ × 0.25 = 1.5 × 10⁻³ s⁻¹.
4. t½ = 0.693/1.5 × 10⁻³ = 4.6 × 10² s.
Problem 20 Two first-order reactions racing
1. Reaction 1: 4 half-lives, fraction (½)⁴ = 1/16. Reaction 2: 1 half-life, fraction 1/2.2. Ratio = 2^(t/10) ÷ 2^(t/40) = 2^(3t/40) = 4 = 2², so 3t/40 = 2 and t = 27 min (26.7).
