Question

A process has ΔH = 200 Jmol^-1 and ΔS = 40 JK^-1 mol^-1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous 
1. 20 K
2. 4 K
3. 5 K
4. 12 K

Answer 1

Correct Answer - Option 3 : 5 K

Concept:

ΔG = ΔH – TΔS

The process will be spontaneous, when 

ΔG = - ve, i.e., |TΔS| > |ΔH|

Calculation:

Given:

ΔH = 200 Jmol^-1

ΔS = 40 JK^-1 mol^-1

\(\Rightarrow T > \frac{{\left| {\Delta H} \right|}}{{\left| {\Delta S} \right|}} = \frac{{200}}{{40}} = 5\;K\)

So, the minimum temperature for spontaneity of the process is 5 K.