Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 22 January, Morning Shift — Question 70

Dissociation of a gas $\mathrm{A}_{2}$ takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300 K . $\mathrm{A}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})$ The standard Gibbs energy of formation of the involved substances has been provided below:

Substance	$\Delta \mathrm{G}_{\mathrm{t}}^{\circ} / \mathrm{kJ} \mathrm{mol}^{-1}$
$\mathrm{~A}_{2}$	-100.00
$\mathrm{~A}$	-50.832

The degree of dissociation of $\mathrm{A}_{2}(\mathrm{~g})$ is given by $\left(\mathrm{x} \times 10^{-2}\right)^{1 / 2}$ where $\mathrm{x}=$ $\_\_\_\_$ . (Nearest integer). [Given : $\mathrm{R}=8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \log 2$ = 0.3010, $\log 3=0.48]$

Answer: 33

Numerical answer — enter this value.

Step-by-step solution

$-1.664 \times 10^{3}=-8.3 \times 300 \ln \mathrm{~K}_{\mathrm{p}}$ $\operatorname{lnK}_{\mathrm{P}}=0.693$ $\mathrm{K}_{\mathrm{p}}=2$ $2=\frac{4 \alpha^{2} \mathrm{P}_{0}}{1-\alpha^{2}}$ $\alpha=\frac{1}{\sqrt{3}}$ $\alpha=\left(\frac{100}{3} \times 10^{-2}\right)^{1 / 2}$ $=\left(33.33 \times 10^{-2}\right)^{1 / 2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry