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

Calculate the free energy change for the following reaction:

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

• ΔH^θ = +135 kJ mol^-1       

• ΔS^θ = +344 J K^-1 mol^-1

Answer:

• Step 1: Convert the entropy value in kilojoules

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

• Step 2: Substitute the terms into the Gibbs Equation

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

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

  • ΔG^θ = +135 – (298 x 0.344)

  • ΔG^θ = +32.49 kJ mol^-1 

Careful: When calculating ΔG^θ the value for ΔS_system^θ must be divided by 1000 

J K^-1 mol^-1  kJ K^-1 mol^-1

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)     ΔH_r^θ = -47 kJ mol^-1

• ΔS^θ [CH₃OH (l)] = +240 J K^-1 mol^-1

• ΔS^θ [HBr (g)] = +99.0 J K^-1 mol^-1

• ΔS^θ [H₂O (l)] = +70.0 J K^-1 mol^-1

•  ΔS^θ [CH₃Br (g)] = +246 J K^-1 mol^-1

Answer:

• Step 1: Calculate ΔS_system^θ 

  • ΔS_system^θ = ΣΔS_products^θ - ΣΔS_reactants^θ

  • ΔS_system^θ = (ΔS^ꝋ [CH₃Br (g)] + ΔS^θ [H₂O (l)]) -  (ΔS^θ [CH₃OH (l)] + ΔS^θ [HBr (g)])

  • ΔS_system^θ = (246 + 70.0) - (240 + 99.0)

  • ΔS_system^θ = -23.0 J K^-1 mol^-1

• Step 2: Convert ΔS^θ into kJ K^-1 mol^-1

  • ΔS_system^θ = = 0.023 kJ K^-1 mol^-1

• Step 3: Calculate ΔG^ꝋ

  • ΔG^θ = ΔH_reaction^θ - TΔS_system^θ

  • ΔG^θ = -47 - (298 x -0.023)

  • ΔG^θ = -40.146 kJ mol^-1

  • ΔG^θ = -40.1 kJ mol^-1