Question

Standard entropy of $X_2$, $Y_2$, and $XY_3$ are 60, 40, and 50 J K$^{-1}$ mol$^{-1}$, respectively. For the reaction, $$ \frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightarrow XY_3, $$ with $\Delta H = -30 \, \text{kJ}$, to be at equilibrium, the temperature will be:

• A. 1250 K
• B. 500 K
• C. 750 K
• D. 1000 K

Hint:

For reactions at equilibrium, use the relation \( \Delta G = \Delta H - T \Delta S \). When \( \Delta G = 0 \), the temperature can be calculated as \( T = \frac{\Delta H}{\Delta S} \).

Answer 1

The Correct Option is C

Entropy and Gibbs Free Energy Calculation
The entropy change (\(\Delta S\)) is given by: \[ \Delta S = S(XY_3) - \frac{1}{2} S(X_2) - \frac{3}{2} S(Y_2) \] Substituting values: \[ = 50 - 30 - 60 = -40 \, {J mol}^{-1} {K}^{-1} \] The enthalpy change (\(\Delta H\)): \[ \Delta H = -30 \, {kJ} = -30000 \, {J} \] The Gibbs free energy equation: \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \(\Delta G = 0\), so: \[ T = \frac{\Delta H}{\Delta S} = \frac{-30000}{-40} = 750 {K} \]