For the reaction taking place at certain temperature $\mathrm{NH}_{2} \mathrm{COONH}_{4}(\mathrm{~s})…
if equilibrium pressure is $3 \mathrm{X}$ bar then $\Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ would be
- $-\mathrm{RT} \ln 9-3 \mathrm{RT} \ln \mathrm{X}$
- $\mathrm{RT} \ln 4-3 \mathrm{RT} \ln \mathrm{X}$
- $-3 \mathrm{RT} \ln \mathrm{X}$
- None of these
Solution
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \left(4 \mathrm{X}^{3}ight)$
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln 4-3 \mathrm{RT} \ln \mathrm{X}$
Asked in: JEE-TOPICTESTS-CHEMISTRY