JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Chemical Thermodynamics

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In a closed flask at 600 K, one mole of X₂Y₄(g) attains equilibrium as given below:

X₂Y₄(g) ⇌ 2XY₂(g)

At equilibrium, 75% X₂Y₄(g) was dissociated and the total pressure is 1 atm. The magnitude of ΔᵣG° (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)

Answer: 8

Solution

From one mole of

X_{2} Y_{4}

with 75% dissociation the equilibrium amounts are

0.25

mol

X_{2} Y_{4}

and

1.5

mol

X Y_{2}

, a total of

1.75

mol, so at 1 atm the partial pressures are

p_{X_{2} Y_{4}} = \frac{1}{7}

and

p_{X Y_{2}} = \frac{6}{7}

. Hence

K_{p} = \frac{( 6/7 ) ^{2}}{1/7} = \frac{36}{7}

, and

lo g \frac{36}{7} = 2 lo g 2 + 2 lo g 3 - lo g 7 = 0.72

, so

ln \frac{36}{7} = 2.3 \times 0.72 = 1.656

. Then

Δ_{r} G^{\circ} = - R T ln K_{p} = - 8.3 \times 600 \times 1.656 = - 8.25 \times 1 0^{3}

J mol

^{- 1}

, whose magnitude to the nearest integer is 8 kJ mol

^{- 1}

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Integer answer

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JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Chemical Thermodynamics

Solution

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