Syllabus Edition

First teaching 2023

First exams 2025

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Gibbs Free Energy (HL) (HL IB Chemistry)

Revision Note

Philippa

Author

Philippa

Expertise

Chemistry

Gibbs Free Energy

Gibbs free energy

  • The feasibility of a reaction is determined by two factors, the enthalpy change and the entropy change
  • The two factors come together in a fundamental thermodynamic concept called the Gibbs free energy (G)
  • The Gibbs equation is:

ΔG = ΔHreaction – TΔSsystem

    • The units of ΔG are in kJ mol1
    • The units of ΔHreaction are in kJ mol1
    • The units of T are in K
    • The units of ΔSsystem are in J K-1 mol1 (and must therefore be converted to kJ Kmol1 by dividing by 1000)

Calculating ΔG

  • There are two ways you can calculate the value of ΔG
    1. From the Gibbs equation, using enthalpy change, ΔH, and entropy change, ΔS, values
    2. From ΔG values of all the substances present

Worked example

ΔGꝋ from ΔHꝋ and ΔSꝋ values

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

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

  • ΔHꝋ = +135 kJ mol-1       
  • ΔSꝋ = +344 J K-1 mol-1

 

Answer:

  • Step 1: Convert the entropy value in kilojoules
    • ΔSꝋ = begin mathsize 14px style fraction numerator plus 344 space straight J space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent over denominator 1000 end fraction end style   = +0.344 kJ K-1 mol-1 
  • Step 2: Substitute the terms into the Gibbs Equation
    • ΔG = ΔHreaction – TΔSsystem
    • ΔG = +135 – (298 x 0.344)
    • ΔG = +32.49 kJ mol-1 

Worked example

ΔGꝋ from other ΔGꝋ values

What is the standard free energy change, ΔG, for the following reaction?

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (g)

Substance ΔGꝋ kJ mol-1
C2H5OH (l) -175
O2 (g) 0
CO2 (g) -394
H2O (g) -229

 

Answer:

  • This can be calculated in the same way as you complete enthalpy calculations
    • ΔGꝋ = ΣΔGproducts – ΣΔGreactants
    • ΔGꝋ = [(2 x CO) + (3 x H2O )] – [(C2H5OH) + (3 x O2)]
    • ΔGꝋ = [(2 x -394 ) + (3 x -229 )] – [-175 + 0]
    • ΔGꝋ -1300 kJ mol-1 
  • This can also be done by drawing a Hess cycle - find the way that is best for you

Gibbs energy cycle for ethanol

Exam Tip

  • The idea of free energy is what’s ‘leftover’ to do useful work when you’ve carried out the reaction
  • The enthalpy change is the difference between the energy you put in to break the chemical bonds and the energy out when making new bonds
  • The entropy change is the ‘cost’ of carrying out the reaction, so free energy is what you are left with!

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Philippa

Author: Philippa

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.