Question

For the reaction \[ \text{H}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)} \rightarrow \text{H}_2 \text{O}(\ell), \Delta H = -285.8 \, \text{kJ/mol}^{-1} \] \text{The value of free energy change at 27°C for the reaction is:}

• A. \( -236.9 \, \text{kJ/mol}^{-1} \)
• B. \( -9 \, \text{kJ/mol}^{-1} \)
• C. \( -281 \, \text{kJ/mol}^{-1} \)
• D. \( +334.7 \, \text{kJ/mol}^{-1} \)

Hint:

The change in free energy can be found using the relation \( \Delta G = \Delta H - T \Delta S \).

Answer 1

The Correct Option is C

The change in free energy \( \Delta G \) is related to the enthalpy change \( \Delta H \) and the entropy change \( \Delta S \) by the equation: \[ \Delta G = \Delta H - T \Delta S \] Using the given \( \Delta H \) and the temperature of 27°C (300K), we calculate the free energy change to be \( -281 \, \text{kJ/mol}^{-1} \).
Final Answer: \[ \boxed{-281 \, \text{kJ/mol}^{-1}} \]