JEE Main 2026 Apr 6 Shift 2, Chemistry Q24: Gibbs Energy Change and Equilibrium

JEE Main2026Apr 6, Shift 2Chemistry

In a closed flask at 600 K, one mole of X $_{2}$ Y $_{4}$ (g) attains equilibrium as given below:

X $_{2}$ Y $_{4}$ (g) $⇌$ 2 XY $_{2}$ (g)

At equilibrium, 75% X $_{2}$ Y $_{4}$ (g) was dissociated and the total pressure is 1 atm. The magnitude of $Δ_{r} G^{\circ}$ (in kJ mol $^{- 1}$ ) at this temperature is _____. (Nearest integer)

(Given: R = 8.3 J mol $^{- 1}$ K $^{- 1}$ ; ln 10 = 2.3, log 2 = 0.3, log 3 = 0.48, log 5 = 0.69, log 7 = 0.84)

Your answer

Solution

Let initial pressure of X $_{2}$ Y $_{4}$ be $P_{0}$ and dissociation $α$ . At equilibrium: $P (X_{2} Y_{4}) = P_{0} (1 - α)$ , $P (XY_{2}) = 2 P_{0} α$ , total $= P_{0} (1 + α) = 1$ atm.

With $α = 0.75$ : $P_{0} = \frac{1}{1.75} = \frac{4}{7}$ atm.

$K_{P} = \frac{( 2 P _{0} α ) ^{2}}{P _{0} ( 1 - α )} = \frac{4 P _{0} α ^{2}}{1 - α} = \frac{4 \times \frac{4}{7} \times ( 0.75 ) ^{2}}{0.25} = \frac{36}{7}$ .

$Δ G^{\circ} = - R T ln K_{P} = - R T \times 2.303 lo g (\frac{36}{7})$ .

$lo g (36/7) = lo g 36 - lo g 7 = 2 lo g 6 - lo g 7 = 2 (0.78) - 0.84 = 0.72$ . So $ln (36/7) = 2.303 \times 0.72 \approx 1.66$ .

$∣Δ G^{\circ} ∣ = 8.3 \times 600 \times 1.66 \approx 8265$ J $\approx 8.3$ kJ.

Nearest integer (using the values rounded in the published key) = 8 kJ mol $^{- 1}$ .

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q24: Gibbs Energy Change and Equilibrium

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