Question:
The equilibrium constant for a reaction is $10 .$ What will be the value of $\Delta G^{\theta} ? \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, T=300 \mathrm{~K}$.
Solution:
From the expression,
$\Delta G^{\theta}=-2.303 \mathrm{RT} \log K_{e q}$
$\Delta G^{\theta}$ for the reaction,
$=(2.303)\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K}) \log 10$
$=-5744.14 \mathrm{Jmol}^{-1}$
$=-5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}$