JEE Main 2026 — Jan 21, Shift 1 — Chemistry solution

JEE Main2026Jan 21, Shift 1Chemistry

Use the following data:

Substance | $Δ_{f} H^{⊖} (500 K)$ (kJ mol $^{- 1}$ ) | $S^{⊖} (500 K)$ (J K $^{- 1}$ mol $^{- 1}$ )

$A B (g)$ | 32 | 222

$A_{2} (g)$ | 6 | 146

$B_{2} (g)$ | X | 280

One mole each of $A_{2} (g)$ and $B_{2} (g)$ are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K.

$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$

The value of x (in kJ mol $^{- 1}$ ) is ________. (Nearest integer)

(Given: $lo g K = 2.2$ , $R = 8.3$ J K $^{- 1}$ mol $^{- 1}$ )

Your answer

Solution

$Δ H^{\circ} = 2 (32) - (6 + x) = (58 - x)$ kJ.

$Δ S^{\circ} = 2 (222) - (146 + 280) = 18$ J/K.

$Δ G^{\circ} = - R T ln K = - 8.314 \times 500 \times 2.2 \times 2.303/1000 = - 21.06$ kJ.

$Δ H^{\circ} - T Δ S^{\circ} = - 21.06 \Rightarrow (58 - x) - 500 (18/1000) = - 21.06 \Rightarrow x = 70.06 \approx 70$ .

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