Q22 / 150 · MHT CET 2026 (15 April Shift 2) · Chemistry
For a certain reaction, $\Delta H^0$ is $-345$ kJ and $\Delta S^0$ is $-123\ \text{JK}^{-1}$. At what temperature the change over from spontaneous to nonspontaneous will occur?
- A$1052$ K
- B$1956$ K
- C$2568$ K
- Correct answerD$2805$ K
Solution
The changeover from spontaneous to nonspontaneous occurs when $\Delta G^0 = 0$.
Using the Gibbs free energy equation:
$\Delta G^0 = \Delta H^0 - T\Delta S^0$
Setting $\Delta G^0 = 0$ gives:
$T = \dfrac{\Delta H^0}{\Delta S^0}$
Substituting the given values (converting $\Delta H^0$ to Joules):
$T = \dfrac{-345 \times 10^3 \text{ J}}{-123 \text{ JK}^{-1}}$
$T = 2804.87 \text{ K}$
$T \approx 2805 \text{ K}$
Answer: $2805$ K
Using the Gibbs free energy equation:
$\Delta G^0 = \Delta H^0 - T\Delta S^0$
Setting $\Delta G^0 = 0$ gives:
$T = \dfrac{\Delta H^0}{\Delta S^0}$
Substituting the given values (converting $\Delta H^0$ to Joules):
$T = \dfrac{-345 \times 10^3 \text{ J}}{-123 \text{ JK}^{-1}}$
$T = 2804.87 \text{ K}$
$T \approx 2805 \text{ K}$
Answer: $2805$ K
Hence, the correct answer is option: 'D'.