Gibbs Free Energy Change (College Board AP® Chemistry)

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Alexandra Brennan

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Chemistry

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-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. 

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Alexandra Brennan

Author: Alexandra Brennan

Expertise: Chemistry

Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.