5.4.8 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^ꝋ = ΔH_reaction^ꝋ - TΔS_system^ꝋ

• The units of ΔG^ꝋare in kJ mol^-1
• The units of ΔH_reaction^ꝋare in kJ mol^-1
• The units of T are in K
• The units of ΔS_system^ꝋare in J K^-1 mol^-1(and must therefore be converted to kJ K^-1 mol^-1 by dividing by 1000)

Calculating ΔG^ꝋ

• There are two ways you can calculate the value of ΔG^ꝋ
  • From ΔH^ꝋand ΔS^ꝋ values
  • From ΔG^ꝋ values of all the substances present

Answer

Step 1: Convert the entropy value in kilojoules

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

Step 2: Substitute the terms into the Gibbs Equation

   ΔG^ꝋ = ΔH_reaction^ꝋ - TΔS_system^ꝋ

   = +135 - (298 x 0.344)

   = +32.49 kJ mol^-1 

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

Answer

   ΔG^ꝋ = ΣΔG_products^ꝋ - ΣΔG_reactants^ꝋ

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

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

   ΔG^ꝋ = -1300 kJ mol^-1 

Limitations of using ΔG

• ∆G^o 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 (E_a) 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)          ∆G = -117 kJ mol^-1 

• This reaction has a very high E_a 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 E_a
• Although the value for ∆G indicates the reaction is feasible, it does not take into account the kinetics of the reaction