For a reversible process, the Gibbs energy change ($\Delta G$) is related to the standard Gibbs energy change ($\Delta G^{\circ}$) and the reaction quotient ($Q$) by the equation: $\Delta G = \Delta G^{\circ} + RT \ln Q$. At equilibrium, the free energy change of the system is zero ($\Delta G = 0$), and the reaction quotient becomes equal to the equilibrium constant ($Q = K$). Substituting these values into the equation gives: $0 = \Delta G^{\circ} + RT \ln K$ $\Delta G^{\circ} = -RT \ln K$ Converting the natural logarithm to base 10 ($\ln K = 2.303 \log K$), we get: $\Delta G^{\circ} = -2.303 RT \log K \approx -2.30 RT \log K$.