For the given reaction: $X_2O_4(l) \rightarrow 2XO_2(g)$ The change in the number of moles of gaseous species, $\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - 0 = 2$. 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 change in enthalpy ($\Delta H$) : $\Delta H = \Delta U + \Delta n_g RT$ $\Delta H = 2100 \text{ cal} + (2 \times 2 \text{ cal K}^{-1} \text{mol}^{-1} \times 300 \text{ K}) = 2100 \text{ cal} + 1200 \text{ cal} = 3300 \text{ cal} = 3.3 \text{ kcal}$

Now, calculate the change in Gibbs free energy ($\Delta G$) : $\Delta G = \Delta H - T\Delta S$ $\Delta G = 3300 \text{ cal} - (300 \text{ K} \times 20 \text{ cal K}^{-1}) = 3300 \text{ cal} - 6000 \text{ cal} = -2700 \text{ cal} = -2.7 \text{ kcal}$.