Gibbs Free Energy Change & Gibbs Equation (CIE A Level Chemistry)

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21 October 2024

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^-1mol^-1

Calculate the free energy change for the following reaction:

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

Answer:

Step 1: Convert the entropy value in kilojoules
- ΔS^θ= +344 J K^-1mol^-1÷ 1000 = +0.344 kJ K^-1mol^-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^θthe value for ΔS_system^θ must be divided by 1000

J K^-1mol^-1 $\overset{\div 1000}{\to}$ kJ K^-1mol^-1

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 = $\frac{∆ {H_{r e a c t i o n}}^{θ} - ∆ G^{θ}}{∆ {S_{s y s t e m}}^{θ}}$

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

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^θinto 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

the (exam) results speak for themselves:

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