The reaction $\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$ follows the mechanism $\mathrm{A}_{2} \underset{\mathrm{k}_{-1}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}} \mathrm{~A}+\mathrm{A}$ (fast) $\mathrm{A}+\mathrm{B}_{2} \xrightarrow{\mathrm{k}_{2}} \mathrm{AB}+\mathrm{B}$ (slow) $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}$ (fast) The overall order of the reaction is:

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Chemical Kinetics

The reaction

A_{2} + B_{2} \to 2 AB

follows the mechanism

A_{2} k_{- 1} ⇌ k_{1} A + A

(fast)

A + B_{2} k_{2} AB + B

(slow)

A + B \to AB

(fast)
The overall order of the reaction is:

Answer: (A)

The rate law is fixed by the slow step:

rate = k_{2} [A] [B_{2}]

. Since

[A]

is a reactive intermediate, it must be eliminated using the fast pre-equilibrium

A_{2} ⇌ 2 A

, for which

\frac{k _{1}}{k _{- 1}} = \frac{[ A ] ^{2}}{[ A _{2} ]}

, giving

[A] = \frac{k _{1}}{k _{- 1}} [A_{2}]

. Substituting,

rate = k_{2} \frac{k _{1}}{k _{- 1}} [A_{2}]^{1/2} [B_{2}]^{1}

The overall order is the sum of the exponents on the observable species:

\frac{1}{2} + 1 = 1.5

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option