Chemistry · Chemical Equilibrium

JEE Main 2026 — 6 April, Evening Shift — Question 63

In a closed flask at 600 K , one mole of $\mathrm{X}_{2} \mathrm{Y}_{4}(\mathrm{~g})$ attains equilibrium as given below: $\mathrm{X}_{2} \mathrm{Y}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_{2}(\mathrm{~g})$ At equilibrium, $75 \% \mathrm{X}_{2} \mathrm{Y}_{4}(\mathrm{~g})$ was dissociated and the total pressure is 1 atm . The magnitude of $\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) at this temperature is $\_\_\_\_$ . (Nearest Integer) (Given: $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ln 10=2.3$ , $\log 2=0.3, \log 3=0.48, \log 5=0.69, \log 7=0.84)$

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{X}_{2} \mathrm{Y}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_{2}(\mathrm{~g})$

\begin{aligned} & \mathrm{t}=0 \quad \mathrm{P}^{\circ} & \mathrm{t}=\mathrm{t} \quad \mathrm{P}^{\circ}(1-\alpha) \quad 2 \mathrm{P}^{\circ} \alpha & \mathrm{P}^{\circ}(1+\alpha)=1 & \alpha=\frac{3}{4} ; \mathrm{P}^{\circ}=\frac{4}{7} & \mathrm{k}_{\mathrm{P}}=\frac{4 \mathrm{P}_{\mathrm{o}} \alpha^{2}}{1-\alpha^{2}}=\frac{36}{7} & \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{p}} & \left|\Delta \mathrm{G}^{\circ}\right|=8.314 \times 600 \ln \left(\frac{36}{7}\right) & =8169.5 \mathrm{~J} \mathrm{~mol}{ }^{-1} & =8.169 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient