Question

Standard entropy of X2, Y2 and X3Y is 60, 40 and 50 J K\(^{-1}\) mol\(^{-1}\), respectively. For the reaction, \[ 1/2 \, \text{X}_2 + 3/2 \, \text{Y}_2 \rightarrow \text{X}_3\text{Y}, \, \Delta H = -30 \, \text{kJ}, \, T = 1250 \, \text{K} \] the equilibrium temperature will be

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

Hint:

The equilibrium temperature can be found using the relationship between entropy change, enthalpy change, and temperature in the Gibbs free energy equation.

Answer 1

The Correct Option is C

Step 1: Entropy Change in Reactions.
The standard entropy change \( \Delta S \) for the reaction is: \[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \] Substitute the standard entropy values of X2, Y2, and X3Y to calculate \( \Delta S \). Using this and the enthalpy change, the equilibrium temperature can be calculated using the Gibbs free energy relation.
Step 2: Conclusion.
The correct answer is (C), 750 K.