Gibbs Free Energy Change (College Board AP® Chemistry)
Study Guide
Written by: Alexandra Brennan
Reviewed by: Stewart Hird
Gibbs Free Energy Change
Gibbs free-energy change is the maximum amount of energy available from any chemical reaction
It is represented by the symbol ΔG°
It is for chemical processes in which all the reactants and products are present in a standard state
This includes:
Pure substances
Solutions of 1.0 M
Gases at a pressure of 1.0 atm (or 1.0 bar)
When ΔG° is negative, the reaction is thermodynamically favored and likely to occur
When ΔG° is positive, the reaction is not thermodynamically favored and unlikely to occur
When a reaction is described as thermodynamically favorable it refers to whether the reaction takes place of its own accord
This is an outcome of the second law of thermodynamics which states that any physical or chemical change must result in an increase in the entropy of the universe
We can see examples of this all around us:
Cups fall off tables and spontaneously break into many pieces, never the other way around
Hot objects always cool and spread their heat into the surroundings, never the other way around
Earthquakes destroy buildings and create chaos and disorder
When living things die they decompose and change from complex ordered systems into disordered simple molecules
The standard Gibbs free energy change for a physical or chemical process can be determined in two ways:
The first is using standard Gibbs free energy of formation of the reactants and products
The equation used is:
ΔG° = ΣΔGf°products - ΣΔGf°reactants
The second is by considering the following factors:
Enthalpy change
Entropy change
These come together to give the following equation:
ΔG° = ΔH° - TΔS°
The units of ΔG° are in kJ mol-1
The units of ΔH° are in kJ mol-1
The units of T are in K
The units of ΔS° are in J K-1 mol-1(and must therefore be converted to kJ K-1 mol-1 by dividing by 1000
Worked Example
Calculate ΔG° for the combustion of propane.
C3H8 (l) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
| C3H8 | CO2 | H2O |
ΔGf° (kJ mol-1) | -23.5 | -394.4 | -228.6 |
Answer:
ΔG° = ΣΔGf°products - ΣΔGf°reactants
ΣΔGf°products = 3(-394.4) + 4(-228.6) = -2097.6 kJ mol-1
ΣΔGf°reactants= -23.5 kJ mol-1
ΔG° = -2097.6 - (-23.5)
ΔG° = 2074.1 kJ mol-1
Oxygen has a Gibbs free energy of formation of 0 due to being an element so is not included.
Examiner Tips and Tricks
Both equations are given to you in the exam!
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