The absolute standard entropies of $X_{2(g)}, Y_{2(g)}$ and $X Y_{3(g)}$ are 50, 40, and $60 ext{

Question

The absolute standard entropies of $X_{2(g)}, Y_{2(g)}$ and $X Y_{3(g)}$ are 50, 40, and $60 	ext{ J K}^{-1} 	ext{ mol}^{-1}$ respectively. For the synthesis reaction $\frac{1}{2}X_{2(g)} + \frac{3}{2}Y_{2(g)} ightleftharpoons X Y_{3(g)}$, the standard enthalpy change is $\Delta H^\circ = -20 	ext{ kJ/mol}$. To establish thermodynamic equilibrium, the system must be maintained at what specific temperature?

Professor PC Thomas Classes · Answer

(d) : First, calculate the standard entropy change of the reaction: $\Delta S^\circ = S^\circ(	ext{products}) - \sum S^\circ(	ext{reactants}) = 60 - [\frac{1}{2}(50) + \frac{3}{2}(40)] = 60 - [25 + 60] = 60 - 85 = -25 	ext{ J K}^{-1} 	ext{ mol}^{-1}$. At thermodynamic equilibrium, the standard Gibbs free energy change is zero: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0$. Therefore, $T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{-20000 	ext{ J mol}^{-1}}{-25 	ext{ J K}^{-1} 	ext{ mol}^{-1}} = 800 	ext{ K}$.