Consider the following mechanism for the reaction between A, B and C: 1. $A+B\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}I$ (fast equilibrium) 2. $I+C\stackrel{k_2}{\to }P$ (slow) Using the pre‑equilibrium approximation, (a) derive the rate law for the overall reaction, (b) state the overall order with respect to each reactant, and (c) identify the rate‑determining step.

Step-by-step solution

1. Write the rate of the slow step: $r=k_2[I][C]$.
2. Because step 1 is a fast equilibrium, apply the equilibrium expression: $K=\frac{[I]}{[A][B]}=\frac{k_1}{k_{-1}}$, so $[I]=\frac{k_1}{k_{-1}}[A][B]$.
3. Substitute $[I]$ into the rate expression: $r=k_2\left(\frac{k_1}{k_{-1}}[A][B]\right)[C]$.
4. Combine constants: $r=\frac{k_1{k}_2}{k_{-1}}[A][B][C]$.
5. Thus the reaction is first order in each of A, B and C, giving an overall third‑order reaction.
6. Since step 2 is the slow step and controls the overall rate, it is the rate‑determining step.

Concept tested: Pre‑equilibrium approximation and overall rate law

Published 7 September 2026 · How this solution was prepared and checked