consider the elementary reaction A ( g )+ B ( g ) rightarrow C ( g )+ D ( g ) If the volume of reaction mixture is suddenly reduced to 1/3 of its initial volume, the reaction rate will become ' x ' times of the original reaction rate. The value of x is :

Chemistry · Chemical Kinetics

JEE Main 2025 — 28 January, Evening Shift — Question 26

consider the elementary reaction

$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{D}(\mathrm{g})$

If the volume of reaction mixture is suddenly reduced to $\frac{1}{3}$ of its initial volume, the reaction rate

will become ' $x$ ' times of the original reaction rate. The value of $x$ is :

Option A:
$\frac{1}{9}$
Option B:
9
Correct
Option C:
$\frac{1}{3}$
Option D:
3

Answer: B

Step-by-step solution

$\mathrm{R}_{1}=\mathrm{K}[\mathrm{A}]^{1}[\mathrm{~B}]^{1}$

$\mathrm{R}_{1}=\mathrm{K}\left[\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\right]^{1}\left[\frac{\mathrm{n}_{\mathrm{B}}}{\mathrm{V}}\right]^{\mathrm{I}}$

$\mathrm{R}_{2}=\mathrm{K}\left[\frac{3 \mathrm{n}_{\mathrm{A}}}{\mathrm{V}}\right]^{1}\left[\frac{3 \mathrm{n}_{\mathrm{B}}}{\mathrm{V}}\right]^{1}$

$\mathrm{R}_{2}=9 \mathrm{R}_{1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant