Free Energy

Chemistry: IB Diploma Higher Level, topic 4.3.3

Free energy is a simple way to determine if a reaction is feasible or thermodynamically favorable. Free energy includes both entropy and enthalpy.

For a reaction to happen, the total entropy of everything involved must increase.
This includes its molar entropies, but also the entropy change of the air when it is heated.
Instead of calculating the entropies of everything, we can define the Gibbs free energy change as ΔG = ΔH - TΔS.
There’s some complex maths, but it boils down to: If ΔG is negative, the overall entropy increases and a reaction will happen.
The reaction is said to be thermodynamically favorable.

The standard Gibbs free energy change is represented with the symbol ΔG°.
With ΔG°, all substances are pure, gases are at a pressure of 1 bar, and solutions are at a concentration of 1M.
The symbol "°" is used to represent reactions occuring under standard conditions.

If a reaction has a positive ΔH, it will only be feasible if the ΔS term is positive and larger than it. You can see a graph above of ΔG vs T for positive ΔH, and positive ΔS.

Entropy always increases. For a reaction to happen if the standard molar entropies of the reactants decreases, it must heat it's surroundings and cause the entropy of the surroundings to increase.
So the reaction must be exothermic.
Above is a graph of ΔG vs T for negative ΔH and ΔS.

Free Energy Calculations

You can calculate the temperature at which a reaction becomes feasible.

For a reaction to happen, ΔG° must be negative.
We can calculate the temperature at which a reaction switches from unfeasible to feasible by setting ΔG° equal to zero.
To find this temperature, we must rearrange the equation for Gibbs free energy:
ΔG° = ΔH° - TΔS° becomes 0 = ΔH° - TΔS°
So, T = $\frac{\Delta H^o}{\Delta S^o}$

A step in the extraction of bismuth from it's ore is the reduction of bismuth hydroxide (Bi(OH)₃) by hydrogen according to the equation:
2(Bi(OH)₃) + 3H₂ → 2Bi + 6H₂O
If the entropy change is: +400JK^-1mol^-1 and the enthalpy change is 50kJmol^-1, what temperature does it become feasible at?
See next slide for solution.

First, make sure your entropy change and enthalpy change both use Joules.
50kJmol^-1 = 50,000Jmol^-1
Insert this value into the rearranged Gibbs free energy equation to find the temperature at which this reaction is feasible:
T = 50,000 ÷ 400 = 125K
So, the reaction is feasible at temperatures above 125K.

By the free energy equation
ΔG° = ΔH° - TΔS°
Under what conditions will a reaction always be feasible?
Using T = $\frac{\Delta H^o}{\Delta S^o}$
A reaction has an enthalpy change of 2000Jmol^-1 and an entropy of +100JK^-1mol^-1
What is the temperature at which the reaction becomes feasible?
Which of the following is true when ΔG° = 0?