Question

Standard entropies of \( X_2 \), \( Y_2 \), and \( X Y_3 \) are 60, 30, and 30 J mol\(^{-1}\) K\(^{-1}\) respectively. For the reaction \[ \frac{1}{2} X_2 + Y_2 \rightarrow X Y_3, \, \Delta H = -30 \, \text{kJ} \] at equilibrium, the temperature should be

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

Hint:

For reactions at equilibrium, the temperature can be determined using the relationship between enthalpy, entropy, and Gibbs free energy.

Answer 1

The Correct Option is B

Step 1: Use the Gibbs free energy equation.
The change in Gibbs free energy \( \Delta G \) is related to entropy change \( \Delta S \) and temperature by: \[ \Delta G = \Delta H - T \Delta S \] At equilibrium, \( \Delta G = 0 \). So, \[ 0 = \Delta H - T \Delta S \] \[ T = \frac{\Delta H}{\Delta S} \]
Step 2: Calculate the temperature.
The entropy change \( \Delta S = (S_{\text{products}} - S_{\text{reactants}}) \). After calculating \( \Delta S \), the temperature comes out to be 1000 K.
Final Answer: \[ \boxed{1000 \, \text{K}} \]