Answer:
-203 kJ/mol
Explanation:
Step 1: Given data
- Standard enthalpy of the reaction (ΔH°): -314 kJ/mol
- Standard entropy of the reaction (ΔS°): -0.372 kJ/K.mol
- Absolute temperature (T): 298 K
Step 2: Calculate the standard Gibbs free energy of the reaction (ΔG°)
We will use the following expression.
ΔG° = ΔH° - T × ΔS°
ΔG° = (-314 kJ/mol) - 298 K × (-0.372 kJ/K.mol) = -203 kJ/mol
By convention, when ΔG° < 0, the reaction is spontaneous.
