Question:
For the reaction $A_{(g)} \rightarrow(B)_{(g)}$, the value of the equilibrium constant at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$ is equal to $100.0$. The value of $\Delta_{\mathrm{r}} \mathrm{G}$ for the reaction at $300 \mathrm{~K}$ and $1 \mathrm{~atm}$ in $\mathrm{J} \mathrm{mol}^{-1}$ is $-\mathrm{xR}$, where $\mathrm{x}$ is__________ (Rounded of to the nearest integer) $\left(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right.$ and $\left.\ln 10=2.3\right)$
Solution:
$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{Kp}$
$=-R(300)(2) \ln (10)$
$=-R(300 \times 2 \times 2.3)$
$\Delta \mathrm{G}^{\circ}=-1380 \mathrm{R}$