Gibbs Free Energy, ΔG (OCR A Level Chemistry A): Revision Note
Exam code: H432
Gibbs free energy, ΔG
As we have seen in the previous sections, the feasibility of a reaction is determined by two factors, the enthalpy change and the entropy change
The two factors come together in a fundamental thermodynamic concept called the Gibbs free energy (G)
The Gibbs equation is:
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
The units are:
ΔGꝋ = kJ mol-1
ΔHreactionꝋ = kJ mol-1
T = K
ΔSsystemꝋ = J K-1 mol-1
So, the units of ΔSsystemꝋ must be converted to kJ K-1 mol-1 by dividing by 1000
Calculating ΔGꝋ
There are two ways you can calculate the value of ΔGꝋ
From ΔHꝋand ΔSꝋ values
From ΔGꝋ values of all the substances present
Worked Example
Calculate the free energy change for the following reaction:
2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)
ΔHꝋ= +135 kJ mol-1 ΔS = +344 J K-1 mol-1
Answer
Convert the entropy value in kilojoules:
ΔSꝋ = +344 J K-1 mol-1 ÷ 1000 = +0.344 kJ K-1 mol-1
Substitute the terms into the Gibbs equation and evaluate:
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
ΔGꝋ = +135 - (298 x 0.344)
ΔGꝋ = +32.49 kJ mol-1
The temperature is 298 K since standard values are quoted in the question
Worked Example
What is the standard free energy change, ΔGꝋ, for the following reaction?
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
Substance | ΔGꝋ, kJ mol-1 |
|---|---|
C2H5OH (l) | -175 |
O2 (g) | 0 |
CO2 (g) | -394 |
H2O (g) | -229 |
Answer:
Use the appropriate ΔGꝋ equation:
ΔGꝋ = ΣΔGproductsꝋ - ΣΔGreactantsꝋ
Substitute the terms into the equation:
ΔGꝋ = [(2 x CO2 ) + (3 x H2O )] - [(C2H5OH) + (3 x O2)]
Substitute the values into the equation and evaluate:
ΔGꝋ = [(2 x -394 ) + (3 x -229 )] - [-175 + 0]
ΔGꝋ = -1300 kJ mol-1
Free energy & equilibrium
There is a direct link between free energy change and chemical equilibrium:
ΔG = –RT lnK
Where:
R is the gas constant (8.314 J mol⁻¹ K⁻¹)
T is the temperature in kelvin (K)
K is the equilibrium constant
A negative ΔG corresponds to K > 1, favouring the products
A positive ΔG corresponds to K < 1, favouring the reactants
Free energy and reaction direction
As a reaction progresses, the Gibbs free energy of the system typically decreases
At equilibrium, ΔG = 0
This is the minimum energy point
A spontaneous reaction proceeds in the direction that reduces ΔG
This explains why reactions always move towards equilibrium, where ΔG stops changing
Examiner Tips and Tricks
Although not explicitly required by the OCR specification, graphs involving ΔG may appear in exams
You might be asked to interpret a ΔG vs. reaction progress graph
This shows how ΔG falls as a reaction proceeds and reaches a minimum at equilibrium (ΔG = 0)
Alternatively, a ΔG vs. temperature graph may be shown
You are expected to estimate the minimum temperature at which a reaction becomes feasible (ΔG = 0)
Limitations of using ΔG
The Gibbs free energy change, ΔGꝋ, predicts whether a reaction is thermodynamically feasible
But, only under standard conditions
A negative ΔG means a reaction could happen
But, feasibility does not guarantee an observable rate
This is because ΔG tells us about thermodynamics, not kinetics
Some reactions are thermodynamically feasible (ΔG < 0)
But, they proceed very slowly if the activation energy (Ea) is high
The decomposition of hydrogen peroxide
One example is the decomposition of hydrogen peroxide at 25 oC:
H2O2 (l) → H2O (l) + ½O2 (g) ∆G = -117 kJ mol-1
The reaction is:
Thermodynamically feasible (ΔG < 0)
Very slow without a catalyst
The reaction has a high activation energy, Ea
This means that it commonly catalysed with manganese dioxide (MnO2)
If the reaction was left for long enough, the hydrogen peroxide would eventually decompose
However the addition of the MnO2 catalyst provides an alternative pathway with a lower activation energy
This allows the reaction to proceed at a measurable rate
Examiner Tips and Tricks
ΔG only indicates whether a reaction is thermodynamically possible.
It does not tell you how fast the reaction happens
You must consider thermodynamics and kinetics when assessing reaction feasibility
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