JEE Main 2026 — Chemical Equilibrium Question with Solution

The given equilibrium reaction is:
${\text{X}}_2{\text{Y}}_4(g)\rightleftharpoons 2{\text{XY}}_2(g)$

Initial moles: $1$
Degree of dissociation, $\alpha =0.75$
Moles of ${\text{X}}_2{\text{Y}}_4$ at equilibrium $=1-\alpha =1-0.75=0.25$
Moles of ${\text{XY}}_2$ at equilibrium $=2\alpha =2\times 0.75=1.5$
Total moles at equilibrium $=0.25+1.5=1.75$

Given total pressure, $P=1$ atm.
Partial pressure of ${\text{X}}_2{\text{Y}}_4$, $P_{{\text{X}}_2{\text{Y}}_4}=\frac{0.25}{1.75}\times 1=\frac{1}{7}$ atm
Partial pressure of ${\text{XY}}_2$, $P_{{\text{XY}}_2}=\frac{1.5}{1.75}\times 1=\frac{6}{7}$ atm

The equilibrium constant $K_p$ is given by:
$K_p=\frac{(P_{{\text{XY}}_2}{)}^2}{P_{{\text{X}}_2{\text{Y}}_4}}=\frac{(\frac{6}{7}{)}^2}{\frac{1}{7}}=\frac{36}{7}$

The standard Gibbs free energy change is:
${\Delta }_r{G}^{\ominus }=-RT\ln K_p=-2.3RT\log K_p$

Calculating $\log K_p$:
$\log \left(\frac{36}{7}\right)=\log 36-\log 7=\log (2^2\times 3^2)-\log 7$
$\log \left(\frac{36}{7}\right)=2\log 2+2\log 3-\log 7$
$\log \left(\frac{36}{7}\right)=2(0.3)+2(0.48)-0.84=0.6+0.96-0.84=0.72$

Substituting the values into the ${\Delta }_r{G}^{\ominus }$ equation:
${\Delta }_r{G}^{\ominus }=-2.3\times 8.3\times 600\times 0.72$
${\Delta }_r{G}^{\ominus }=-8246.88{\text{ J mol}}^{-1}=-8.24688{\text{ kJ mol}}^{-1}$

The magnitude of ${\Delta }_r{G}^{\ominus }$ is $8.24688{\text{ kJ mol}}^{-1}$.
Rounding to the nearest integer, we get $8$.

Answer: $8$