Q23 / 75 · JEE Main 2026 (21 January Shift 1) · Chemistry
Use the following data :
\(\begin{array}{|c|c|c|} \hline \text {Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJmol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{g}) & x & 280 \\ \hline \end{array}\)
One mole each of $\mathrm{A}_{2}(\mathrm{~g})$ and $\mathrm{B}_{2}(\mathrm{~g})$ are taken in a 1 L 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 $x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$. (Nearest integer)
(Given : $\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
\(\begin{array}{|c|c|c|} \hline \text {Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJmol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{g}) & x & 280 \\ \hline \end{array}\)
One mole each of $\mathrm{A}_{2}(\mathrm{~g})$ and $\mathrm{B}_{2}(\mathrm{~g})$ are taken in a 1 L 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 $x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$. (Nearest integer)
(Given : $\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
Correct answer: 70
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
$\Delta G° = -2.303RT \log K = -2.303 \times 8.3 \times 500 \times 2.2 = -21026$ 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
Hence, the correct answer is 70.