To solve this problem, we need to calculate the standard Gibbs free energy change $\Delta G^{\circ}$ for the reaction.$\\ \\$The reaction given is:$\\$A + B $\rightleftharpoons$ C + D$\\ \\$The equilibrium concentrations are:$\\ [A] = 2 \,$mol L$^{-1}, \\ [B] = 3 \,$mol L$^{-1}, \\ [C] = 10 \,$mol L$^{-1}, \\ [D] = 6 \,$mol L$^{-1} \\ \\$The equilibrium constant $K_c$ for the reaction is given by:$\\ K_c = \frac{[C][D]}{[A][B]} \\ \\$Substitute the equilibrium concentrations:$\\ K_c = \frac{10 \times 6}{2 \times 3} \\ K_c = \frac{60}{6} \\ K_c = 10 \\ \\$The relationship between $\Delta G^{\circ}$ and $K_c$ is given by:$\\ \Delta G^{\circ} = -RT \ln K_c \\ \\$Where:$\\ R = 2 \,$cal/mol K$\\ T = 300 \,$K$\\ \\$Substitute the values into the equation:$\\ \Delta G^{\circ} = -2 \times 300 \times \ln 10 \\ \\$We know that $\ln 10 \approx 2.302 \\ \\$Substitute $\ln 10$ into the equation:$\\ \Delta G^{\circ} = -2 \times 300 \times 2.302 \\ \Delta G^{\circ} = -1381.2 \,$cal$\\ \\$Therefore, the standard Gibbs free energy change $\Delta G^{\circ}$ for the reaction is approximately $-1381.8 \,$cal.$\\ \\$This corresponds to Option 3.