This example problem demonstrates how to use changes in entropy of a system and its surroundings to determine a reaction's spontaneity and whether or not the reaction will be exothermic or endothermic.
Problem:
Using the following values for entropy
a) determine if a reaction would be spontaneous.
b) determine if the reaction is exothermic or endothermic with respect to the system
I) ΔSsys = 30 J/K , ΔSsurr = 50 J/K
II) ΔSsys = 60 J/K, ΔSsurr = -85 J/K
III) ΔSsys = 140 J/K, ΔSsurr = -85 J/K
Solution
Part 1 - Determine if a reaction will be spontaneous.
A reaction will be spontaneous if the total entropy change is positive. It will not be spontaneous if the total entropy change is negative.
ΔStotal = ΔSsys + ΔSsurr
where
ΔStotal is the total entropy change
ΔSsys is the entropy change of the system
ΔSsurr is the entropy change of the surroundings
System I
ΔStotal = ΔSsys + ΔSsurr
ΔStotal = 30 J/K + 50 J/K
ΔStotal = 80 J/K
The total entropy change is positive, therefore the reaction will be spontaneous.
System II
ΔStotal = ΔSsys + ΔSsurr
ΔStotal = 60 J/K + -85 J/K
ΔStotal = -25 J/K
The total entropy change is negative, therefore the reaction will not be spontaneous.
System III
ΔStotal = ΔSsys + ΔSsurr
ΔStotal = 140 J/K + -85 J/K
ΔStotal = 55 J/K
The total entropy change is positive, therefore the reaction will be spontaneous.
Part 2 - Determine if the reaction is exothermic or endothermic
A reaction would be exothermic with respect to the system if it adds entropy to the surroundings and endothermic if it reduces entropy of the surroundings. This means
ΔSsurr > 0 - exothermic
ΔSsurr < 0 - endothermic
System I
ΔSsurr = 50 : Positive - exothermic
System II
ΔSsurr = -85 : negative - endothermic
System III
ΔSsurr = -85 : negative - endothermic
Answer:
A reaction in system I is spontaneous and exothermic.
A reaction in system II is not spontaneous and endothermic.
A reaction in system III is spontaneous and endothermic.

