Consider the following elementary mechanism for the reaction $A+B\to P$: 1. $A+Bxrightarrow{k}_{1}I$ (slow) 2. $Ixrightarrow{k}_{2}P$ (fast) Use the steady‑state approximation for the intermediate $I$ to derive the rate law for the overall reaction, and identify the rate‑determining step.

Step-by-step solution

1. Write the steady‑state condition: $d[I]/dt=0=k_1[A][B]-k_2[I]$.
2. Solve for the intermediate concentration: $[I]=\frac{k_1[A][B]}{k_2}$.
3. Write the rate of product formation: $r=k_2[I]$.
4. Substitute $[I]$ into $r$: $r=k_{2}cdot\frac{k_1[A][B]}{k_2}=k_1[A][B]$.
5. Thus the overall rate law is $r=k_1[A][B]$.
6. Identify the slow step as the rate‑determining step because it limits the overall rate.
7. Therefore step 1 is the RDS.

Concept tested: Steady‑state approximation in mechanism analysis

Published 6 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked