Question

If for a particular process, ΔH=308 kJmol and ΔS=439 Jmol K, in what temperature range will the process be spontaneous?

Answer 1

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.