Q46 / 180 · JIPMER 2015 · Chemistry
$\Delta H$ and $\Delta S$ for the reaction, $\mathrm{Ag}_2 \mathrm{O}(\mathrm{s}) \longrightarrow 2 \mathrm{Ag}(s)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$, are $30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $66.00 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature at which the free energy change for the reaction will be zero, is
- A3528 K
- Correct answerB463 K
- C73 K
- D144 K
Solution
From the Gibb's - Helmholtz equation,
$\Delta G=\Delta H-T \Delta S$
At equilibrium, $\Delta G=0$
Hence, $\quad \Delta H-T \Delta S=0$ or, $\Delta H=T \Delta S$
Or,$T=\frac{\Delta H}{\Delta S}$
Given that, $\Delta H=30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\begin{aligned} & =30560 \mathrm{~J} \mathrm{~mol}^{-1} \\ \therefore \quad \Delta S & =66.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ \therefore \quad T & =\frac{30560}{66.0}=463 \mathrm{~K}\end{aligned}$
Hence, the correct answer is option: 'B'.