Use the following data: | Substance | $\frac{\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJ} \mathrm{mol}^{-1}}$ | $\frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}}$ | |-----------|---------|----------| | $\mathrm{AB}(\mathrm{g})$ | 32 | 222 | | $\mathrm{~A}_{2}(\mathrm{~g})$ | 6 | 146 | | $\mathrm{~B}_{2}(\mathrm{~g})$ | X | 280 | One mole each of $\mathrm{A}_{2}(\mathrm{~g})$ and $\mathrm{B}_{2}(\mathrm{~g})$ are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K. $\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g})$ The value of $\mathrm{x}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$) is _____ (Nearest integer) (Given: $\log \mathrm{K}=2.2$, $\mathrm{R}=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$)

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Chemical Thermodynamics

Use the following data:

One mole each of

A_{2} (g)

and

B_{2} (g)

are taken in a 1L closed flask and allowed to establish the equilibrium at 500 K.

A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)

The value of

x

(in

kJ mol^{- 1}

) is _____ (Nearest integer)

(Given:

lo g K = 2.2

R = 8.3 JK^{- 1} mol^{- 1}

)

Answer: 70

For A

_{2}

+

_{2}

⇌

2AB the enthalpy change is

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

and the entropy change is

Δ S^{\circ} = 2 (222) - (146 + 280) = 18 J K^{- 1}

. From

Δ G^{\circ} = - 2.303 R T lo g K = - 2.303 \times 8.3 \times 500 \times 2.2 = - 21.0 kJ

and

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

- 21.0 = (58 - x) - 500 (0.018) = 49 - x

. Solving gives

x = 70 kJ mo l^{- 1}

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer