Standard entropies of , , and are 60, 40, and 50 J/K-mol. For the reaction , the enthalpy change is — Thermodynamics and Thermochemistry Chemistry Question
Question
Standard entropies of $X_2$, $Y_2$, and $XY_3$ are 60, 40, and 50 J/K-mol. For the reaction $0.5 X_2 + 1.5 Y_2 \rightleftharpoons XY_3$, the enthalpy change $\Delta H$ is -30 kJ. What must the exact temperature be for the reaction to be at thermodynamic equilibrium?
Answer: A
💡 Solution & Explanation
The standard entropy change is $\Delta S^\circ = 50 - [0.5(60) + 1.5(40)] = 50 - [30 + 60] = -40$ J/K-mol. At thermodynamic equilibrium, $\Delta G = 0$, meaning $\Delta H = T\Delta S$. Solving for temperature: $T = \Delta H / \Delta S = -30,000 / -40 = 750$ K.
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