At temperature T , compound AB 2( g ) dissociates as AB 2( g ) rightleftharpoons AB ( g ) +1/2 B 2( g ) having degree of dissociation x (small compared to unity). The correct expression for x in terms of K p and p is

Chemistry · Chemical Equilibrium

JEE Main 2025 — 29 January, Morning Shift — Question 3

At temperature $T$ , compound $\mathrm{AB}_{2(\mathrm{~g})}$ dissociates as $\mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}$ having degree of dissociation x (small compared to unity). The correct expression for x in terms of $\mathrm{K}_{\mathrm{p}}$ and p is

Option A:
$\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}}{\mathrm{p}}}$
Option B:
$\sqrt[4]{\frac{2 \mathrm{~K}_{\mathrm{p}}}{\mathrm{p}}}$
Option C:
$\sqrt[3]{\frac{2 \mathrm{~K}_{\mathrm{p}}^{2}}{\mathrm{p}}}$
Correct
Option D:
$\sqrt{\mathrm{K}_{\mathrm{p}}}$

Answer: C

Step-by-step solution

$\quad \mathrm{AB}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{AB}_{(\mathrm{g})}+\frac{1}{2} \mathrm{~B}_{2(\mathrm{~g})}$ $\mathrm{t}_{\text {eq. }} \frac{(1-\mathrm{x})}{1+\frac{x}{2}} P \frac{x P}{1+\frac{x}{2}} \frac{\left(\frac{x}{2}\right) P}{1+\frac{x}{2}}$ $\Rightarrow \mathrm{x} \ll 1 \Rightarrow 1+\frac{\mathrm{x}}{2} \simeq 1$ and $1-\mathrm{x} \simeq 1$ $\Rightarrow k_{P}=\frac{(x p) \cdot\left(\frac{x p}{2}\right)^{\frac{1}{2}}}{P}$ $\Rightarrow \mathrm{k}_{\mathrm{P}}^{2}=\mathrm{x}^{2} \cdot \frac{\mathrm{xP}}{2}$ $\mathrm{x}=\sqrt[3]{\frac{2 \mathrm{k}_{\mathrm{P}}^{2}}{\mathrm{P}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient