Question

For a reaction \(X_2(l) + Y_2(g) \rightleftharpoons 2XY(g)\), the \(\Delta H^0\) and \(\Delta S^0\) are \(+29.3 \, \text{kJ/mol}\) and \(104.1 \, \text{J K}^{-1}\text{mol}^{-1}\) respectively at 298 K. Find the free energy change in kJ/mol.

• A. 0.6
• B. 2.04
• C. 5.2
• D. 1.72

Hint:

Always convert entropy into kJ before using \(\Delta G = \Delta H - T\Delta S\). A negative \(\Delta G\) indicates spontaneity of the reaction.

Answer 1

The Correct Option is D

Step 1: Formula for Gibbs free energy change.
The Gibbs free energy change is given by the relation:
\[ \Delta G = \Delta H - T\Delta S \] where \( \Delta H \) is enthalpy change, \( \Delta S \) is entropy change, and \( T \) is temperature in Kelvin.

Step 2: Converting entropy units.
Given entropy is in \( \text{J K}^{-1}\text{mol}^{-1} \), but enthalpy is in kJ/mol.
So, convert entropy into kJ units:
\[ \Delta S = 104.1 \, \text{J K}^{-1}\text{mol}^{-1} = \frac{104.1}{1000} = 0.1041 \, \text{kJ K}^{-1}\text{mol}^{-1} \]

Step 3: Substituting given values.
\[ \Delta H = 29.3 \, \text{kJ/mol}, \quad T = 298 \, K \] \[ \Delta G = 29.3 - (298 \times 0.1041) \]

Step 4: Calculating \(T\Delta S\).
\[ 298 \times 0.1041 = 31.0118 \, \text{kJ/mol} \]

Step 5: Calculating \(\Delta G\).
\[ \Delta G = 29.3 - 31.0118 = -1.7118 \, \text{kJ/mol} \]

Step 6: Interpreting the result.
The value of \(\Delta G\) is negative, indicating that the reaction is spontaneous under given conditions.
Taking magnitude as per options:
\[ |\Delta G| \approx 1.72 \, \text{kJ/mol} \]

Step 7: Final Answer.
Thus, the free energy change is:
\[ \boxed{1.72 \, \text{kJ/mol}} \] Hence, the correct answer is option (D).