For a certain reaction $\Delta H=-225 \mathrm{~kJ}$ and $\Delta S=-150 \mathrm{JK}^{-1}$. Find the…
- 1500 K
- 1450 K
- 1340 K
- 1300 K
Solution
The temperature where $\Delta G$ becomes zero corresponds to the condition $\Delta G = \Delta H - T\Delta S = 0$.
For $\Delta H = -225\ \mathrm{kJ}$ and $\Delta S = -150\ \mathrm{JK}^{-1}$, convert $\Delta H$ to consistent units: $\Delta H = -225,\!000\ \mathrm{J}$.
Solving $0 = \Delta H - T\Delta S$ yields $T = \frac{\Delta H}{\Delta S} = \frac{-225,\!000\ \mathrm{J}}{-150\ \mathrm{JK}^{-1}} = 1,\!500\ \mathrm{K}$.
The temperature at which $\Delta G = 0$ is therefore $1,\!500\ \mathrm{K}$.
The correct option is A.
Asked in: MHT CET 2025 (05 May Shift 2)