Given reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$ Change in number of gaseous moles, $\Delta n_g = n_p - n_r = 2 - 0 = 2$ Given: $\Delta U = 2.1 \text{ kcal} = 2100 \text{ cal}$ $\Delta S = 20 \text{ cal K}^{-1}$ $T = 300 \text{ K}$ Universal gas constant, $R \approx 2 \text{ cal K}^{-1}\text{mol}^{-1}$ First, calculate the enthalpy change ($\Delta H$): $\Delta H = \Delta U + \Delta n_g RT$ $\Delta H = 2100 + (2 \times 2 \times 300) = 2100 + 1200 = 3300 \text{ cal}$ Next, calculate the Gibbs free energy change ($\Delta G$) using the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$ $\Delta G = 3300 - (300 \times 20) = 3300 - 6000 = -2700 \text{ cal}$ $\Delta G = -2.7 \text{ kcal}$