A reaction proceeds via the following mechanism: 1. $displaystyleA+Bxrightleftharpoons[k_{-1}]k_{1}I$ (fast equilibrium) 2. $displaystyleI+Cxrightarrow{k}_{2}P$ (slow) Derive the rate law in terms of $[A]$, $[B]$, $[C]$, the equilibrium constant $K$ and $k_2$. State the overall order of the reaction with respect to each reactant and identify the rate‑determining step.

Step-by-step solution

1. Write the equilibrium constant for step 1: $K=\frac{[I]}{[A][B]}$.
2. Express the intermediate concentration: $[I]=K[A][B]$.
3. The slow step (step 2) determines the rate: $r=k_2[I][C]$.
4. Substitute $[I]$ into the rate expression: $r=k_{2}K[A][B][C]$.
5. Thus the rate law is $r=k_{2}K[A][B][C]$.
6. The reaction is third‑order overall, with each reactant appearing to the first power.
7. Step 2 is the rate‑determining step because it is slow and controls the overall rate.

Concept tested: Pre‑equilibrium rate law derivation

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked