Gibbs Free Energy Calculations (College Board AP® Chemistry) : Study Guide
Gibbs Free Energy, Enthalpy & Entropy
Enthalpy change (ΔH°) and entropy change (ΔS°) come together in a fundamental thermodynamic concept called the Gibbs free energy (G)
The Gibbs free-energy equation is:
ΔG° = ΔH° - TΔS°
The units of ΔG° are in kJ mol-1
The units of ΔH° are in kJ mol-1
The units of T are in K
The units of ΔS° are in J K-1 mol-1(and must therefore be converted to kJ K-1 mol-1 by dividing by 1000
When ΔG° is negative, the reaction is thermodynamically favoured and likely to occur
When ΔG° is positive, the reaction is not thermodynamically favoured and unlikely to occur
Worked Example
Calculate ΔG° for the following reaction at 298K.
2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)
ΔH° = +135 kJ mol-1
ΔS° = +344 J K-1 mol-1
Answer:
Step 1: Convert the entropy value to kilojoules
ΔS°= +344 J K-1 mol-1 ÷ 1000 = +0.344 kJ K-1 mol-1
Step 2: Substitute the terms into the Gibbs Equation
ΔG° = ΔH°- TΔS°
ΔG° = +135 - (298 x 0.344)
ΔG° = +32.49 kJ mol-1
So, the reaction is thermodynamically unfavourable at 298K
Worked Example
Determine whether the following reaction is thermodynamically favored at 298 K.
2Ca (s) + O2 (g) → 2CaO (s) ΔH = –635.5 kJ mol-1
So [Ca (s)] = 41.00 J K-1 mol-1
So [O2 (g)] = 205.0 J K-1 mol-1
So [CaO (s)] = 40.00 J K-1 mol-1
Answer:
Step1: Calculate ΔSo
ΔSo = ΣΔSo products – ΣΔSo reactants
ΔSo = (2 x ΔSo [CaO (s)]) – (2 x ΔSo [Ca (s)] + ΔSo [O2 (g)])
ΔSo = (2 x 40.00) – (2 x 41.00 + 205.0)
ΔSo = -207.0 J K-1 mol-1
Step 2:Convert ΔSo to kJ K-1 mol-1
ΔSo = –207.0 J K-1 mol-1 ÷ 1000 = –0.207 kJ mol-1
Step 3: Calculate ΔGo
ΔGo= ΔHo – TΔSo
ΔGo = –635.5 – (298 x -0.207)
ΔGo= –573.8 kJ mol-1
Since ΔGo is negative, the reaction is thermodynamically favoured
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