Q31 / 150 · MHT CET 2026 (18 April Shift 1) · Chemistry
For a certain reaction, $\Delta\text{H}^\circ = 40\ \text{kJ}$ and $\Delta\text{S}^\circ = 80\ \text{JK}^{-1}$. Find the temperature so that $\Delta\text{G}^\circ = 0$
- Correct answerA$500\ \text{K}$
- B$400\ \text{K}$
- C$300\ \text{K}$
- D$600\ \text{K}$
Solution
Using the Gibbs free energy equation:
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
For $\Delta G^\circ = 0$, we get:
$T = \dfrac{\Delta H^\circ}{\Delta S^\circ}$
Given $\Delta H^\circ = 40\ \text{kJ} = 40000\ \text{J}$ and $\Delta S^\circ = 80\ \text{J K}^{-1}$.
Substituting the values:
$T = \dfrac{40000}{80} = 500\ \text{K}$
Answer: $500\ \text{K}$
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
For $\Delta G^\circ = 0$, we get:
$T = \dfrac{\Delta H^\circ}{\Delta S^\circ}$
Given $\Delta H^\circ = 40\ \text{kJ} = 40000\ \text{J}$ and $\Delta S^\circ = 80\ \text{J K}^{-1}$.
Substituting the values:
$T = \dfrac{40000}{80} = 500\ \text{K}$
Answer: $500\ \text{K}$
Hence, the correct answer is option: 'A'.