For the reaction: $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3$ The entropy change of the reaction ($\Delta S$) is: $\Delta S = S(XY_3) - \left[ \frac{1}{2}S(X_2) + \frac{3}{2}S(Y_2) \right]$ $\Delta S = 50 - \left[ \frac{1}{2}(60) + \frac{3}{2}(40) \right]$ $\Delta S = 50 - [30 + 60] = 50 - 90 = -40 \text{ J K}^{-1} \text{mol}^{-1}$ At equilibrium, the Gibbs free energy change $\Delta G = 0$. Since $\Delta G = \Delta H - T\Delta S$, we have: $\Delta H = T\Delta S \implies T = \frac{\Delta H}{\Delta S}$ Given $\Delta H = -30 \text{ kJ} = -30000 \text{ J}$ $T = \frac{-30000 \text{ J}}{-40 \text{ J K}^{-1}} = 750 \text{ K}$