Free energy change of a reaction is calculated by,

$∆\mathrm{G}=∆{\mathrm{G}}^{\circ }+\mathrm{RT} \mathrm{lnQ}$

At equilibrium, the free energy change of a reaction is zero $(∆\mathrm{G}=0)$, and the reaction quotient $\left(\mathrm{Q}\right)$ is equal to the equilibrium constant $\left(\mathrm{K}\right)$. So, the above relation takes the form,

$∆{\mathrm{G}}^{\circ }=-\mathrm{RT} \mathrm{lnK}$

$∆{\mathrm{G}}^{\circ }=-2.303\mathrm{RT} {\log }_{10}\mathrm{K}$

where $∆{\mathrm{G}}^{\circ }=$standard free energy change

Therefore,

$∆{\mathrm{G}}^{\circ }=-2.303\times 8.314\times 298\times \log \left(3\times {10}^{-29}\right) \mathrm{J} {\mathrm{mol}}^{-1}$

$∆{\mathrm{G}}^{\circ }=-5705.8\times \left[\log \left(3\right)+\log \left({10}^{-29}\right)\right] \mathrm{J} {\mathrm{mol}}^{-1}$

$∆{\mathrm{G}}^{\circ }=-5705.8\times \left[0.47-29\right] \mathrm{J} {\mathrm{mol}}^{-1}$

$∆{\mathrm{G}}^{\circ }=162787\mathrm{J} {\mathrm{mol}}^{-1}$

$∆{\mathrm{G}}^{\circ }\approx 163 \mathrm{kJ} {\mathrm{mol}}^{-1}$