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Cambridge (CIE) A Level
Chemistry
Gibbs Free Energy Change, ΔG
Contents
Gibbs Free Energy Change & Gibbs Equation
Reaction Feasibility
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Gibbs Free Energy Change & Gibbs Equation
The Gibbs Equation
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:
ΔGθ = ΔHreactionθ - TΔSsystemθ
The units of ΔGθ are in kJ mol-1
The units of ΔHreactionθ are in kJ mol-1
The units of T are in K
The units of ΔSsystemθ are in J K-1 mol-1
Worked Example
Calculate the free energy change for the following reaction:
2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)
ΔHθ = +135 kJ mol-1
ΔSθ = +344 J K-1 mol-1
Answer:
Step 1: Convert the entropy value in kilojoules
ΔSθ = +344 J K-1 mol-1 ÷ 1000 = +0.344 kJ K-1 mol-1
Step 2: Substitute the terms into the Gibbs Equation
ΔGθ = ΔHreaction – TΔSsystem
The temperature is 298 K since standard values are quoted in the question
Page 2 of 8
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ΔGθ = +135 – (298 x 0.344)
ΔGθ = +32.49 kJ mol-1
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Examiner Tips and Tricks
Careful: When calculating ΔGθ the value for ΔSsystemθ must be divided by 1000
÷ 1000
J K-1 mol-1 ⎯⎯⎯⎯⎯⎯⎯ kJ K-1 mol-1
The Gibbs Equation: Calculations
The Gibbs equation can be used to calculate the Gibbs free energy change of a reaction
ΔGθ = ΔHreactionθ - TΔSsystemθ
The equation can also be rearranged to find values of ΔHreaction , ΔSsystem or the temperature, T
For example, if for a given reaction, the values of ΔG , ΔHreaction and ΔSsystem are given, the
temperature can be found by rearranging the Gibbs equation as follows:
∆ H re a ction θ − ∆ G θ
T=
∆ S s ystem 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) ΔHrθ = -47 kJ mol-1
ΔSθ [CH3OH (l)] = +240 J K-1 mol-1
ΔSθ [HBr (g)] = +99.0 J K-1 mol-1
ΔSθ [H2O (l)] = +70.0 J K-1 mol-1
ΔSθ [CH3Br (g)] = +246 J K-1 mol-1
Answer:
Step 1: Calculate ΔSsystemθ
ΔSsystemθ = ΣΔSproductsθ - ΣΔSreactantsθ
ΔSsystemθ = (ΔS [CH3Br (g)] + ΔSθ [H2O (l)]) - (ΔSθ [CH3OH (l)] + ΔSθ [HBr (g)])
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ΔSsystemθ = (246 + 70.0) - (240 + 99.0)
ΔSsystemθ = -23.0 J K-1 mol-1
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Step 2: Convert ΔSθ into kJ K-1 mol-1
ΔSsystemθ =
−23 . 0
= 0.023 kJ K-1 mol-1
1000
Step 3: Calculate ΔG
ΔGθ = ΔHreactionθ - TΔSsystemθ
ΔGθ = -47 - (298 x -0.023)
ΔGθ = -40.146 kJ mol-1
ΔGθ = -40.1 kJ mol-1
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Reaction Feasibility
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Reaction Feasibility
The Gibbs equation can be used to calculate whether a reaction is feasible or not
ΔGθ = ΔHreactionθ - TΔSsystemθ
When ΔGθ is negative, the reaction is feasible and likely to occur
When ΔGθ is positive, the reaction is not feasible and unlikely to occur
Worked Example
Calculate the Gibbs free energy for the following reaction at 298 K and determine whether the
reaction is feasible.
2Ca (s) + O2 (g) → 2CaO (s) ΔHθ = -635.5 kJ mol-1
Sθ[Ca (s)] = 41.00 J K-1 mol-1
Sθ[O2 (s)] = 205.0 J K-1 mol-1
Sθ[CaO (s)] = 40.00 J K-1 mol-1
Answer:
Step 1: Calculate ΔSsystemθ
ΔSsystemθ = ΣΔSproductsθ - ΣΔSreactantsθ
ΔSsystemθ = (2 x ΔSθ [CaO(s)]) - (2 x ΔSθ [Ca(s)] + ΔSθ [O2(g)])
ΔSsystemθ = (2 x 40.00) - (2 x 41.00 + 205.0)
ΔSsystemθ = -207.0 J K-1 mol-1
Step 2: Convert ΔSθ to kJ K-1 mol-1
Δ Sθ =
−207 . 0
= -0.207 kJ K-1 mol-1
1000
Step 3: Calculate ΔGθ
ΔGθ = ΔHreactionθ - TΔSsystemθ
ΔGθ = -635.5 - (298 x -0.207)
ΔGθ = -573.8 kJ mol-1
Step 4: Determine whether the reaction is feasible
Since the ΔGθ is negative the reaction is feasible and likely to occur
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Reaction Feasibility & Temperature Changes
The feasibility of a reaction can be affected by the temperature
The Gibbs equation will be used to explain what will affect the feasibility of a reaction for exothermic
and endothermic reactions
Exothermic reactions
In exothermic reactions, ΔHreactionθ is negative
If the ΔSsystemθ is positive:
Both the first and second term will be negative
Resulting in a negative ΔGθ so the reaction is feasible
Therefore, regardless of the temperature, an exothermic reaction with a positive ΔSsystemθ will
always be feasible
If the ΔSsystemθ is negative:
The first term is negative and the second term is positive
At high temperatures, the -TΔSsystemθ will be very large and positive and will overcome ΔHreactionθ
Therefore, at high temperatures ΔGθ is positive and the reaction is not feasible
The reaction is more feasible at low temperatures, as the second term will not be large enough to
overcome ΔHreactionθ resulting in a negative ΔGθ
This corresponds to Le Chatelier’s principle which states that for exothermic reactions an increase in
temperature will cause the equilibrium to shift position in favour of the reactants, i.e. in the
endothermic direction
In other words, for exothermic reactions, the products will not be formed at high temperatures
The reaction is not feasible at high temperatures
Summary of factors affecting Gibbs free energy for exothermic
reactions
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If ΔH ....
And if ΔS ....
Then ΔG is
Spontaneous?
Because
is negative
is positive
always negative
Always
<0
>0
<0
Forward reaction spontaneous at
any T
exothermic
more
disorder
is negative
is negative
Spontaneous only at low T
<0
<0
Dependent
on T
exothermic
more order
negative at
low T
positive high T
T ΔS < H
Endothermic reactions
In endothermic reactions, ΔHreactionθ is positive
If the ΔSsystemθ is negative:
Both the first and second term will be positive
Resulting in a positive ΔGθ so the reaction is not feasible
Therefore, regardless of the temperature, endothermic with a negative ΔSsystemθ will never be
feasible
If the ΔSsystem is positive:
The first term is positive and the second term is negative
At low temperatures, the -TΔSsystemθ will be small and negative and will not overcome the larger
ΔHreactionθ
Therefore, at low temperatures ΔGθ is positive and the reaction is less feasible
The reaction is more feasible at high temperatures as the second term will become negative
enough to overcome the ΔHreactionθ resulting in a negative ΔGθ
This again corresponds to Le Chatelier’s principle which states that for endothermic reactions an
increase in temperature will cause the equilibrium to shift position in favour of the products
In other words, for endothermic reactions, the products will be formed at high temperatures
The reaction is therefore feasible
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Summary of factors affecting Gibbs free energy for endothermic
reactions
If ΔH ....
And if ΔS ....
Then ΔG is
Spontaneous?
Because
is positive
is negative
always positive
Never
>0
<0
>0
Reverse reaction spontaneous at
any T
endothermic
more order
is positive
is positive
Spontaneous only at high T
>0
>0
Dependent
on T
endothermic
more
disorder
negative at
high T
positive low T
T ΔS > H
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