Gibbs Free Energy Change & Gibbs Equation (Cambridge International A-Level Chemistry)

The Gibbs Equation

Spec point — 23.4.1

Gibbs free energy

• The feasibility of a reaction does not only depend on the entropy change of the reaction but can also be affected by the enthalpy change
• Therefore, using the entropy change of a reaction only to determine the feasibility of a reaction is inaccurate
• The Gibbs free energy (G) is the energy change that takes into account both the entropy change of a reaction and the enthalpy change
• The Gibbs equation is:

ΔG ^θ = ΔH_reaction^θ - TΔS_system^θ

• The units of ΔG^θ are in kJ mol^-1

  • The units of ΔH_reaction^θ are in kJ mol^-1
  • The units of T are in K
  • The units of ΔS_system^θ are in J K^-1 mol-1

Worked Example Calculate the free energy change for the following reaction at a temperature of 298 K:

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

ΔG^θ = +135 – (298 x 0.344)

ΔG^θ = +32.49 kJ mol^-1

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

J K^-1 mol^-1 $\overset{\div1000}{\rightarrow}$ kJ K^-1 mol^-1

Gibbs Equation Calculations

Spec point — 23.4.1

The Gibbs Equation: Calculations

• The Gibbs equation can be used to calculate the Gibbs free energy change of a reaction

ΔG ^θ = ΔH_reaction^θ - TΔS_system^θ

• The equation can also be rearranged to find values of ΔH_reaction^ꝋ, ΔS_system^ꝋ or the temperature, T
• For example, if for a given reaction, the values of ΔG^ꝋ, ΔH_reaction^ꝋ and ΔS_system^ꝋ are given, the temperature can be found by rearranging the Gibbs equation as follows:

T = $\frac{\Delta H_{reaction}^{θ}-\Delta G^{θ}}{\Delta S_{system}^{θ}}$

Worked Example 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 **

• S^θ [CH₃OH (l)] = +240 J K^-1 mol^-1
• S^θ [HBr (g)] = +99.0 J K^-1 mol^-1
• S^θ [H₂O (l)] = +70.0 J K^-1 mol^-1
• S^θ [CH₃Br (g)] = +246 J K^-1 mol^-1

Answer:

• Step 1: Calculate ΔS**_system^{**θ **}

ΔS_system^θ = ΣS_products^θ - ΣS_reactants^θ

ΔS_system^θ = (S^ꝋ [CH₃Br (g)] + S^θ [H₂O (l)]) - (S^θ [CH₃OH (l)] + S^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 =** $\frac{-23.0}{1000}$= -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