Gibbs Free Energy Change & Gibbs Equation (Cambridge (CIE) A Level Chemistry): Revision Note

Exam code: 9701

Philippa Platt

Last updated

The Gibbs Equation

Gibbs free energy

  • The feasibility of a reaction does not only depend on the entropy change of the reaction but can also be affected by the enthalpy change

  • Therefore, using the entropy change of a reaction only to determine the feasibility of a reaction is inaccurate

  • The Gibbs free energy (G) is the energy change that takes into account both the entropy change of a reaction and the enthalpy change

  • The Gibbs equation is:

ΔGθ = ΔHreactionθ - TΔSsystemθ

  • The units of ΔGθ are in kJ mol-1

    • The units of ΔHreactionθ are in kJ mol-1

    • The units of T are in K

    • The units of ΔSsystemθ are in J K-1 mol-1

Worked Example

Calculate the free energy change for the following reaction at a temperature of 298 K:

2NaHCO(s) → Na2CO3 (s) + H2O (l) + CO2 (g)

  • ΔHθ = +135 kJ mol-1       

  • ΔSθ = +344 J K-1 mol-1

Answer:

  • Step 1: Convert the entropy value in kilojoules

ΔSθ = +344 J K-1 mol-1  ÷ 1000 = +0.344 kJ K-1 mol-1 

  • Step 2: Substitute the terms into the Gibbs Equation

ΔGθ = ΔHreaction – TΔSsystem

ΔGθ = +135 – (298 x 0.344)

ΔGθ +32.49 kJ mol-1 

Examiner Tips and Tricks

Careful: When calculating ΔGθ the value for ΔSsystemθ must be divided by 1000 

J K-1 mol-1 rightwards arrow with divided by space 1000 on top kJ K-1 mol-1

The Gibbs Equation: Calculations

  • The Gibbs equation can be used to calculate the Gibbs free energy change of a reaction

ΔGθ = ΔHreactionθ - TΔSsystemθ

  • The equation can also be rearranged to find values of ΔHreaction, ΔSsystem or the temperature, T

  • For example, if for a given reaction, the values of ΔG, ΔHreaction and ΔSsystem are given, the temperature can be found by rearranging the Gibbs equation as follows:

Tfraction numerator bold increment bold H subscript bold r bold e bold a bold c bold t bold i bold o bold n end subscript to the power of bold theta bold minus bold space bold increment bold G to the power of bold theta over denominator bold increment bold S subscript bold s bold y bold s bold t bold e bold m end subscript to the power of bold theta end fraction

Worked Example

Calculate the Gibbs free energy for the reaction of methanol, CH3OH, with hydrogen bromide, HBr, at 298 K.

CH3OH (l) + HBr (g) → CH3Br (g) + H2O (l)     ΔHrθ = -47 kJ mol-1

  • ΔSθ [CH3OH (l)] = +240 J K-1 mol-1

  • ΔSθ [HBr (g)] = +99.0 J K-1 mol-1

  • ΔSθ [H2O (l)] = +70.0 J K-1 mol-1

  •  ΔSθ [CH3Br (g)] = +246 J K-1 mol-1

Answer:

  • Step 1: Calculate ΔSsystemθ 

ΔSsystemθ = ΣΔSproductsθ - ΣΔSreactantsθ

ΔSsystemθ = (ΔS [CH3Br (g)] + ΔSθ [H2O (l)]) -  (ΔSθ [CH3OH (l)] + ΔSθ [HBr (g)])

ΔSsystemθ = (246 + 70.0) - (240 + 99.0)

ΔSsystemθ = -23.0 J K-1 mol-1

  • Step 2: Convert ΔSθ into kJ K-1 mol-1

ΔSsystemθ = begin mathsize 14px style fraction numerator negative 23.0 over denominator 1000 end fraction end style= 0.023 kJ K-1 mol-1

  • Step 3: Calculate ΔG

ΔGθ = ΔHreactionθ - TΔSsystemθ

ΔGθ = -47 - (298 x -0.023)

ΔGθ = -40.146 kJ mol-1

ΔGθ = -40.1 kJ mol-1

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Philippa Platt

Author: Philippa Platt

Expertise: Chemistry Content Creator

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener