Free Energy and Reaction Spontaneity Example Problem

This example problem demonstrates how to calculate and use changes in free energy to determine a reaction's spontaneity.

Problem:

Using the following values for ΔH, ΔS, and T, determine the change in free energy and if the reaction is spontaneous or nonspontaneous.

I) ΔH = 40 kJ, ΔS = 300 J/K, T = 130 K
II) ΔH = 40 kJ, ΔS = 300 J/K, T = 150 K
III) ΔH = 40 kJ, ΔS = -300 J/K, T = 150 K

Solution

The free energy of a system can be used to determine if a reaction is spontaneous or nonspontaneous. Free energy is calculated with the formula

ΔG = ΔH - TΔS

where

ΔG is the change in free energy
ΔH is the change in enthalpy
ΔS is the change in entropy
T is the absolute temperature

A reaction will be spontaneous if the change in free energy is negative. It will not be spontaneous if the total entropy change is positive.

**Watch your units! ΔH and ΔS must share the same energy units.**

System I

ΔG = ΔH - TΔS
ΔG = 40 kJ - 130 K x (300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 130 K x 0.300 kJ/K
ΔG = 40 kJ - 39 kJ
ΔG = +1 kJ

ΔG is positive, therefore the reaction will not be spontaneous.

System II

ΔG = ΔH - TΔS
ΔG = 40 kJ - 150 K x (300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 150 K x 0.300 kJ/K
ΔG = 40 kJ - 45 kJ
ΔG = -5 kJ

ΔG is negative, therefore the reaction will be spontaneous.

System III

ΔG = ΔH - TΔS
ΔG = 40 kJ - 150 K x (-300 J/K x 1 kJ/1000 J)
ΔG = 40 kJ - 150 K x -0.300 kJ/K
ΔG = 40 kJ + 45 kJ
ΔG = +85 kJ

ΔG is positive, therefore the reaction will not be spontaneous.

Answer:

A reaction in system I would be nonspontaneous.
A reaction in system II would be spontaneous.
A reaction in system III would be nonspontaneous.