Q41 / 180 · COMEDK 2025 (10 May Shift 1) · Chemistry
For a reaction $\mathrm{X}_2(\mathrm{l})+\mathrm{Y}_2(\mathrm{~g}) \leftrightharpoons 2 \mathrm{XY}(\mathrm{g})$, the $\Delta \mathrm{H}^0$ and $\Delta \mathrm{S}^0$ are $+29.3 \mathrm{~kJ} / \mathrm{mol}$ and $104.1 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively at 298 K . Find the free energy change in $\mathrm{kJ} / \mathrm{mol}$.
- A$2.04$
- B$5.2$
- C$0.6$
- Correct answerD$1.72$
Solution
The Gibbs free energy change $\Delta G^0$ is given by the relation $\Delta G^0 = \Delta H^0 - T\Delta S^0$.
Given values are $\Delta H^0 = 29.3 \text{ kJ/mol}$ and $\Delta S^0 = 104.1 \text{ J K}^{-1} \text{ mol}^{-1} = 0.1041 \text{ kJ K}^{-1} \text{ mol}^{-1}$ at $T = 298 \text{ K}$.
Substituting these values into the equation:
$\Delta G^0 = 29.3 \text{ kJ/mol} - (298 \text{ K} \times 0.1041 \text{ kJ K}^{-1} \text{ mol}^{-1})$
$\Delta G^0 = 29.3 - 31.0218$
$\Delta G^0 = -1.7218 \text{ kJ/mol}$.
The magnitude of the free energy change is approximately $1.72 \text{ kJ/mol}$.
Answer: $1.72$
Given values are $\Delta H^0 = 29.3 \text{ kJ/mol}$ and $\Delta S^0 = 104.1 \text{ J K}^{-1} \text{ mol}^{-1} = 0.1041 \text{ kJ K}^{-1} \text{ mol}^{-1}$ at $T = 298 \text{ K}$.
Substituting these values into the equation:
$\Delta G^0 = 29.3 \text{ kJ/mol} - (298 \text{ K} \times 0.1041 \text{ kJ K}^{-1} \text{ mol}^{-1})$
$\Delta G^0 = 29.3 - 31.0218$
$\Delta G^0 = -1.7218 \text{ kJ/mol}$.
The magnitude of the free energy change is approximately $1.72 \text{ kJ/mol}$.
Answer: $1.72$
Hence, the correct answer is option: 'D'.