2024 H2 Chem P1 Q10 — Kinetics Solution

The answer is C (38.1).
• Order in P: with Q in large excess, [P] against time has a constant half-life, so the reaction is (pseudo) first order with respect to P.
• Order in Q: [P] is constant in both experiments while [Q] increases by \(\frac{0.0600}{0.0250} = 2.4\) times. The rate increases by \(\frac{2.74 \times 10^{-3}}{4.76 \times 10^{-4}} = 5.76\) times, and \(2.4^2 = 5.76\), so the reaction is second order with respect to Q.
• Rate equation: rate \(= k[\text{P}][\text{Q}]^2\)
• Using experiment 1: \(k = \frac{4.76 \times 10^{-4}}{0.0200 \times 0.0250^2} = \frac{4.76 \times 10^{-4}}{1.25 \times 10^{-5}} = 38.1\)
• A is wrong: 0.762 comes from rate \(= k[\text{Q}]^2\), leaving out [P].
• B is wrong: 0.952 comes from rate \(= k[\text{P}][\text{Q}]\), taking Q as first order.
• D is wrong: 47.6 comes from rate \(= k[\text{P}]^2[\text{Q}]\), swapping the orders of P and Q.