JEE Main 2026 Apr 5 Shift 2, Chemistry Q4: Gibbs Energy Change and Equilibrium

JEE Main2026Apr 5, Shift 2Chemistry

Consider the following data for the reaction $X_{2} (g) + Y_{2} (g) ⇌ 2 X Y (g)$ at 600 K. The $Δ_{r} G^{\circ}$ (in kJ mol $^{- 1}$ ) for the reaction is:

Compound	$Δ_{f} H_{600 K}^{\circ}$ (kJ mol $^{- 1}$ )	$S_{600 K}^{\circ}$ (J mol $^{- 1}$ K $^{- 1}$ )
$X Y (g)$	42	200
$X_{2} (g)$	8	140
$Y_{2} (g)$	80	250

A
$- 21000$
B
$- 10$
C
$- 1000$
D
$- 9.012$

Solution

$Δ_{r} H^{\circ} = 2 (42) - 8 - 80 = - 4$ kJ/mol.

$Δ_{r} S^{\circ} = 2 (200) - 140 - 250 = + 10$ J/(K·mol).

$Δ_{r} G^{\circ} = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ} = - 4000 - 600 \times 10 = - 10000$ J/mol $= - 10$ kJ/mol.

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JEE Main 2026 Apr 5 Shift 2, Chemistry Q4: Gibbs Energy Change and Equilibrium

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