To solve this problem, we need to understand the relationship between Gibbs free energy change$\\$and the equilibrium constant for a reversible process at equilibrium.$\\$The Gibbs free energy change $\Delta G$ for a reaction is given by:$\\ \Delta G = \Delta G^{\circ} + RT \ln Q \\$where $\Delta G^{\circ}$ is the standard Gibbs free energy change, $R$ is the universal gas constant,$\\ T$ is the temperature in Kelvin, and $Q$ is the reaction quotient.$\\$At equilibrium, the reaction quotient $Q$ is equal to the equilibrium constant $K. \\$Thus, at equilibrium:$\\ \Delta G = 0 \\$Therefore, the equation becomes:$\\ 0 = \Delta G^{\circ} + RT \ln K \\$Rearranging for $\Delta G^{\circ},$ we get:$\\ \Delta G^{\circ} = -RT \ln K \\$To convert the natural logarithm to base 10, we use the conversion:$\\ \ln K = 2.303 \log K \\$Substituting this into the equation, we have:$\\ \Delta G^{\circ} = -2.303 RT \log K \\$This corresponds to Option 2: $\Delta G^{\circ} = -2.30 RT \log K. \\$Therefore, the correct statement for a reversible process in a state of equilibrium is Option 2.