Answer:
The process will be spontaneous above 702 K.
Explanation:
Step 1: Given data
- Standard enthalpy of the reaction (ΔH°): 308 kJ/mol
- Standard entropy of the reaction (ΔS°): 439 J/mol.K
Step 2: Calculate the temperature range in which the process will be spontaneous
The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
When ΔG° < 0,
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (308,000 J/mol)/(439 J/mol.K)
T > 702 K
The process will be spontaneous above 702 K.
