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:
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- 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 equation used is:
- The second is by considering the following factors:
- Enthalpy change
- Entropy change
- These come together to give the following equation:
- ΔG° = ΔH° - TΔS°
- The first is using standard Gibbs free energy of formation of the reactants and products
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- 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.
Exam Tip
Both equations are given to you in the exam!
