The reaction A 2 + B 2 rightarrow 2 AB follows the mechanism A 2 underset k -1 stackrel k 1 rightleftharpoons A + A (fast) A + B 2 xrightarrow k 2 AB + B (slow) A + B rightarrow AB (fast) The overall order of the reaction is :

Chemistry · Chemical Kinetics

JEE Main 2025 — 29 January, Morning Shift — Question 11

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} \text { (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 :

Option A:
1.5
Correct
Option B:
3
Option C:
1.5
Option D:
2

Answer: A

Step-by-step solution

$\quad$ rate $=\mathrm{k}_{2}[\mathrm{~A}]\left[\mathrm{B}_{2}\right] \ldots \ldots . .(1)$

\begin{aligned} & \left(\frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}\right)=\left(\frac{[\mathrm{A}]^{2}}{\left[\mathrm{~A}_{2}\right]}\right) & \Rightarrow[\mathrm{A}]=\sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}} \cdot \sqrt{\left[\mathrm{~A}_{2}\right]} \end{aligned}

Substituting in (1); we get Rate $=\mathrm{k}_{2} \sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}} \cdot\left[\mathrm{~A}_{2}\right]^{\frac{1}{2}} \cdot\left[\mathrm{~B}_{2}\right]$ $\therefore$ order $=\left(\frac{3}{2}\right)=1.5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Molecularity and Practical Methods to Determine Order of Reaction