For the general reaction $\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}$, the equilibrium constant $K_c$ is calculated as the ratio of the product concentrations to the reactant concentrations: $K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]}$

Given the equilibrium concentrations: $[\text{A}] = 2\text{ mol L}^{-1}$ $[\text{B}] = 3\text{ mol L}^{-1}$ $[\text{C}] = 10\text{ mol L}^{-1}$ $[\text{D}] = 6\text{ mol L}^{-1}$

Substitute these values into the equilibrium expression: $K_c = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10$

The standard Gibbs free energy change ($\Delta G^\circ$) is related to the equilibrium constant by the equation: $\Delta G^\circ = -2.303 RT \log K_c$

Given values for the equation: $R = 2\text{ cal/mol K}$ $T = 300\text{ K}$ $K_c = 10$

Substitute the values to find $\Delta G^\circ$: $\Delta G^\circ = -2.303 \times 2 \times 300 \times \log(10)$ Since $\log(10) = 1$: $\Delta G^\circ = -2.303 \times 600 = -1381.8\text{ cal}$

Therefore, $\Delta G^\circ$ for the reaction is $-1381.80\text{ cal}$.