Standard entropies of , , and are , , and respectively. For the chemical reaction , the enthalpy cha — Thermodynamics and Thermochemistry Chemistry Question
Question
Standard entropies of $X_2$, $Y_2$, and $XY_3$ are $60$, $40$, and $50 \text{ J K}^{-1} \text{ mol}^{-1}$ respectively. For the chemical reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightleftharpoons XY_3$, the enthalpy change $\Delta H$ is $-30 \text{ kJ}$. What must the exact temperature be for the reaction to be at thermodynamic equilibrium?
💡 Solution & Explanation
The standard entropy change of the reaction is $\Delta S^\circ = S^\circ(XY_3) - [\frac{1}{2} S^\circ(X_2) + \frac{3}{2} S^\circ(Y_2)]$. Substituting the values: $\Delta S^\circ = 50 - [\frac{1}{2}(60) + \frac{3}{2}(40)] = 50 - [30 + 60] = 50 - 90 = -40 \text{ J K}^{-1} \text{ mol}^{-1}$. At thermodynamic equilibrium, $\Delta G = 0$, meaning $\Delta H = T\Delta S$. Solving for temperature: $T = \frac{\Delta H}{\Delta S} = \frac{-30 \times 10^3 \text{ J}}{-40 \text{ J K}^{-1}} = 750 \text{ K}$.