For the reaction between $A$ , $B$ and $C$ the following mechanism is proposed: 1. $A + B k_{1} k_{- 1} I$ (fast equilibrium, $K = \frac{k _{1}}{k _{- 1}}$ ) 2. $I + C k_{2} J$ (slow) 3. $J k_{3} P$ (fast)

hard 4 marks Class 12 · Chemistry · Chemical Kinetics

Answer

Rate law:

r = k_{2} K [A] [B] [C]

; order: first order in

A

, first order in

B

, first order in

C

(overall third order); rate‑determining step is step 2 because it is the slow step controlling the overall rate.

Step-by-step solution

Write the elementary rate for the slow step (step 2): $r = k_{2} [I] [C]$ .
Since step 1 is a fast pre‑equilibrium, express $[I]$ using the equilibrium constant: $K = \frac{[ I ]}{[ A ] [ B ]} \Rightarrow [I] = K [A] [B]$ .
Substitute $[I]$ into the rate expression: $r = k_{2} (K [A] [B]) [C]$ .
Combine constants: $r = (k_{2} K) [A] [B] [C]$ . This is the overall rate law.
The exponent of each concentration gives the order: $1$ in $A$ , $1$ in $B$ , $1$ in $C$ ; total order $= 3$ .
Step 2 is labelled as slow, meaning it has the highest activation barrier and limits the overall rate; steps 1 and 3 are fast and do not affect the rate. Hence step 2 is the rate‑determining step.

Common mistake: Using the steady‑state approximation for

I

instead of the pre‑equilibrium expression, leading to an incorrect dependence on reactants

Concept tested: Pre‑equilibrium & rate law derivation