Use the following data : $ \begin{array}{|c|c|c|} \hline \text{Substance} & \frac{\Delta_f H^{\ominus}(500…

Use the following data : $ \begin{array}{|c|c|c|} \hline \text{Substance} & \frac{\Delta_f H^{\ominus}(500 K)}{kJmol^{-1}} & \frac{S^{\ominus}(500 K)}{JK^{-1} mol^{-1}} \\ \hline AB(g) & 32 & 222 \\ \hline A_{2}(g) & 6 & 146 \\ \hline B_{2}(g) & x & 280 \\ \hline \end{array} $ 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. $A_{2}(g)+B_{2}(g) \rightleftharpoons 2 AB(g)$ The value of $x(in kJ mol^{-1})$ is $\_\_\_\_$. (Nearest integer) (Given : $\log K=2.2 \quad R=8.3 J K^{-1} mol^{-1}$)

Solution

Reaction: $A_2(g) + B_2(g) \rightleftharpoons 2AB(g)$
$\Delta G° = -2.303RT \log K = -2.303 \times 8.3 \times 500 \times 2.2 = -21026$ J/mol = $-21.03$ kJ/mol
$\Delta S° = 2(222) - [146 + 280] = 444 - 426 = 18$ J K⁻¹ mol⁻¹
$\Delta H° = 2(32) - [6 + x] = (58 - x)$ kJ/mol
Using $\Delta G° = \Delta H° - T\Delta S°$:
$-21.03 = (58 - x) - (500 \times 0.018)$
$-21.03 = 58 - x - 9 = 49 - x$
$x = 49 + 21.03 = 70.03 \approx 70$ kJ/mol

Asked in: JEE Main 2026 (21 Jan Shift 1)

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