Gibbs Free Energy (HL) (DP IB Chemistry)

ΔG^ꝋ from ΔH^ꝋ and ΔS^ꝋ values

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

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

• ΔH^ꝋ = +135 kJ mol^-1       

• ΔS^ꝋ = +344 J K^-1 mol^-1

Answer:

• Step 1: Convert the entropy value in kilojoules

  • ΔS^ꝋ =    = +0.344 kJ K^-1 mol^-1 

• Step 2: Substitute the terms into the Gibbs Equation

  • ΔG^ꝋ = ΔH_reaction^ꝋ – TΔS_system^ꝋ

  • ΔG^ꝋ = +135 – (298 x 0.344)

  • ΔG^ꝋ = +32.49 kJ mol^-1 

ΔG^ꝋ from other ΔG^ꝋ values

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

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

Answer:

• This can be calculated in the same way as you complete enthalpy calculations

  • 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 

• This can also be done by drawing a Hess cycle - find the way that is best for you

• 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!