Question

If the standard enthalpy change and entropy change of a reaction at 298 K are 11.1 kJ mol$^{-1}$ and 42 J K$^{-1}$ mol$^{-1}$, which one of the following options is correct?

• A. \( \Delta G^\circ = -1.416 \, \text{kJ mol}^{-1}, \) \text{the reaction is spontaneous}
• B. \( \Delta G^\circ = 1.416 \, \text{kJ mol}^{-1}, \) \text{the reaction is non-spontaneous}
• C. \( \Delta G^\circ = 1.416 \, \text{kJ mol}^{-1}, \) \text{the reaction is spontaneous}
• D. \( \Delta G^\circ = -1.416 \, \text{kJ mol}^{-1}, \) \text{the reaction is non-spontaneous}

Hint:

For a reaction to be spontaneous, the Gibbs free energy change (\( \Delta G^\circ \)) should be negative.

Answer 1

The Correct Option is A

The Gibbs free energy change \( \Delta G^\circ \) is given by: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] Substituting the values, we get: \[ \Delta G^\circ = 11.1 \, \text{kJ mol}^{-1} - 298 \times (42 \, \text{J K}^{-1} \, \text{mol}^{-1}) \] \[ \Delta G^\circ = 11.1 - 1.416 = -1.416 \, \text{kJ mol}^{-1} \] Since \( \Delta G^\circ \) is negative, the reaction is spontaneous.