The questions are on the rate law test. Take it under real conditions before reading the mark scheme.
Quick answer: A rate law, rate = k[A]^m[B]^n, shows how reaction rate depends on reactant concentrations, with orders found by experiment. This mark scheme gives the model answers and the marks for the 8 questions of the AP Chemistry test on rate law, from rate law from data to order from half-lives, with the equivalent answers to accept.
Each model answer is followed by a marking guide. Award method marks even if a final number is slightly off through rounding, and do not penalise a correct alternative route such as using half-lives instead of the logarithm.
Level: AP Chemistry · Curriculum: AP Chem Unit 5
Question 1 Rate law from data (3 marks)
(a) (1)→(2): [ICl] ×2, rate ×2: order 1. (2)→(3): [H₂] ×4, rate ×4: order 1.
(b) k = 1.5 × 10⁻³/(0.10 × 0.10) = 0.15 M⁻¹ s⁻¹.
Marking: 1 mark for each order with reasoning; 1 mark for the value and units of k.
Question 2 Units (2 marks)
(a) Overall order 2; k in M⁻¹ s⁻¹.
(b) s⁻¹.
Marking: 1 mark each.
Question 3 First-order calculation (3 marks)
(a) [A] = 0.250 × e^(−5.0 × 10⁻⁴ × 1200) = 0.250 × e^(−0.60) = 0.250 × 0.549 = 0.137 M.
(b) t½ = 0.693/5.0 × 10⁻⁴ = 1.39 × 10³ s.
Marking: 1 for the correct equation and substitution, 1 for the answer, 1 for the half-life.
Question 4 Graph reading (2 marks)
(a) 1.20 → 0.60 M takes 30 s, and 0.60 → 0.30 M takes another 30 s: constant half-life of 30 s, so first order.
(b) k = 0.693/30 = 0.023 s⁻¹.
Marking: 1 for half-life and order, 1 for k.
Question 5 Second-order time (3 marks)
1/[A] = 1/[A]₀ + kt → t = (1/0.10 − 1/0.30)/0.40 = (10.0 − 3.33)/0.40 = 16.7 s.
Marking: 1 for the second-order equation, 1 for 10.0 − 3.33, 1 for the answer with unit.
Question 6 Explaining the effect of concentration (2 marks)
More particles in the same volume means collisions between reactant particles are more frequent, so more successful collisions per second occur and the rate rises. The fraction of collisions with enough energy does not change.
Marking: 1 for greater collision frequency, 1 for linking it to more successful collisions per unit time.
Question 7 Plots (2 marks)
(a) First order.
(b) The slope is −k: negative, with magnitude equal to the rate constant (units s⁻¹).
Marking: 1 each.
Question 8 Order from half-lives (3 marks)
(a) The half-life doubles each time [A]₀ is halved, so t½ ∝ 1/[A]₀: second order.
(b) k = 1/(t½[A]₀) = 1/(40 × 0.80) = 0.031 M⁻¹ s⁻¹.
Marking: 1 for order with reasoning; 1 for the formula; 1 for the value and units.
