# Question #658ea

Jan 22, 2015

$K = 3 \times {10}^{5}$

$\Delta G = - R T \ln K$

$\ln K = - \frac{\Delta G}{R T}$

$\ln K = - \left(\frac{- 32.5 \times {10}^{3}}{8.31 \times 310}\right)$

$\ln K = 12.61$

$K = 3 \times {10}^{5}$

Strictly speaking $\Delta {G}^{0}$ should be under standard conditions. This is at 310K.