Gibbs Free Energy, ΔG (OCR A Level Chemistry)

Revision Note

Test yourself
Philippa

Author

Philippa

Last updated

Gibbs Free Energy, ΔG

Gibbs free energy

  • As we have seen in the previous sections, 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 ΔGare in kJ mol-1
  • The units of ΔHreactionare in kJ mol-1
  • The units of T are in K
  • The units of ΔSsystemare in J K-1 mol-1(and must therefore be converted to kJ K-1 mol-1 by dividing by 1000)

Calculating ΔG
  • There are two ways you can calculate the value of ΔG
    • From ΔHand ΔSvalues
    • From ΔGvalues of all the substances present

Worked example

Calculate the free energy change for the following reaction:

2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)

ΔH= +135 kJ mol-1        ΔS = +344 J K-1 mol-1

Answer

Step 1: Convert the entropy value in kilojoules

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

Step 2: Substitute the terms into the Gibbs Equation

   ΔG = ΔHreaction - TΔSsystem

   = +135 - (298 x 0.344)

   = +32.49 kJ mol-1 

The temperature is 298 K since standard values are quoted in the question

Worked example

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

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

Worked Example 2- Free energy calculation, downloadable AS & A Level Chemistry revision notes

Answer

   ΔG= ΣΔGproducts - ΣΔGreactants

   ΔG= [(2 x CO2 ) + (3 x H2O )] - [(C2H5OH) + (3 x O2)]

   ΔG= [(2 x -394 ) + (3 x -229 )] - [-175 + 0]

   ΔG= -1300 kJ mol-1 

Free Energy & Equilibrium

Limitations of using ΔG

  • Go can only be used to predict the feasibility of a reaction under standard conditions
    • Under non-standard conditions, ∆G must be calculated
  • It is important to note that just because a reaction is feasible does not mean that it will occur at an observable rate
  • While ∆G can be used to determine the feasibility of a reaction, it does not take into account the kinetics of the reaction i.e. rate of reaction
  • There might be a large energy barrier (Ea) which the reacting species have to overcome before a reaction can occur
  • Some reactions are feasible since ∆G is negative, but kinetically not feasible since it just occurs too slowly
  • Such reactions are feasible but very slow

  • An example is the decomposition of hydrogen peroxide at 25 oC

H2O2 (l) → H2O (l) + ½O2 (g)          ∆= -117 kJ mol-1 

  • This reaction has a very high Ea so must be catalysed using manganese dioxide, MnO2
  • If the reaction was left for long enough, the hydrogen peroxide would eventually decompose, however the addition of the MnO2 allows the reaction to take place via an alternative route with a lower Ea
  • Although the value for G indicates the reaction is feasible, it does not take into account the kinetics of the reaction

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Philippa

Author: Philippa

Expertise: Chemistry

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.