Use the following data : Substance frac Δ f H ominus (500 K ) kJ mol -1 frac S ominus (500 K ) JK -1 mol -1 --- --- --- AB ( g ) 32 222 A 2 ( g ) 6 146 B 2 ( g ) X 280 One mole each of A 2 ( g ) and B 2 ( g ) are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K . A 2 ( g )+ B 2 ( g ) rightleftharpoons 2 AB ( g ) The value of x ( in kJ mol -1 ) is \ \ \ \ (Nearest integer) (Given: log K =2.2 R =8.3 JK -1 mol -1 )

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2026 — 21 January, Morning Shift — Question 65

Use the following data :

Substance	$\frac{\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJ} \mathrm{mol}^{-1}}$	$\frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}}$
$\mathrm{AB}(\mathrm{g})$	32	222
$\mathrm{~A}_{2}(\mathrm{~g})$	6	146
$\mathrm{~B}_{2}(\mathrm{~g})$	X	280

One mole each of $\mathrm{A}_{2}(\mathrm{~g})$ and $\mathrm{B}_{2}(\mathrm{~g})$ are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K . $\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{g})$ The value of $\mathrm{x}\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$ (Nearest integer) (Given: $\log \mathrm{K}=2.2 \mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

Answer: 70

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A}_{2}+\mathrm{B}_{2} \xrightarrow{500 \mathrm{~K}} 2 \mathrm{AB} \log \mathrm{K}=2.2$ $\Delta \mathrm{H}^{\circ}=(2 \times 32)-(6+\mathrm{x})=(58-\mathrm{x}) \mathrm{kJ}$ $\Delta \mathrm{S}^{\circ}=(2 \times 222)-(146+280)=18$ Joule $\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$ $\Delta \mathrm{G}^{\circ}=-\frac{8.314 \times 500 \times 2.2 \times 2.303}{1000}$ $\Delta \mathrm{G}^{\circ}=-21.06$ $\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}=-21.06$ $58-\mathrm{x}-500\left(\frac{18}{1000}\right)=-21.06$ $\mathrm{x}=70.06 \mathrm{KJ} / \mathrm{mol}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry