The Equilibrium Constant & Gibbs Energy (HL) (DP IB Chemistry)
Revision Note
The Equilibrium Constant & Gibbs Energy
How is Gibbs energy related to the equilibrium constant?
The equilibrium constant, K, gives no information about the individual rates of reaction
It is independent of the kinetics of the reaction
The equilibrium constant, K, is directly related to the Gibbs energy change, ΔGꝊ, according to the following Gibbs energy equation:
ΔGꝊ = -RT lnK
ΔGꝊ= Gibbs energy change (kJ mol–1)
R = gas constant (8.31 J K-1 mol-1)
T = temperature (Kelvin, K)
K = equilibrium constant
This equation is provided in section 1 of the data booklet
This relationship between the equilibrium constant, K, and Gibbs energy change, ΔGꝊ, can be used to determine whether the forward or backward reaction is favoured
The relationship between the equilibrium constant, K, and Gibbs energy change, ΔG
Equilibirum constant, K | Description | Gibbs energy change, ΔG |
---|---|---|
K > 1 | Products favoured | ΔG < 0 |
K = 1 | Reaction at equilibirum | ΔG = 0 |
K < 1 | Reactants favoured | ΔG > 0 |
At a given temperature, a negative ΔG value for a reaction indicates that:
The reaction is feasible / spontaneous
The equilibrium concentration of the products is greater than the equilibrium concentration of the reactants
The value of the equilibrium constant is greater than 1
As ΔG becomes more negative:
The forward reaction is favoured more
The value of the equilibrium constant increases
Examiner Tips and Tricks
When completing calculations using the ΔGꝊ = –RT lnK equation, you have to be aware that:
ΔGꝊ is measured in kJ mol–1
R is measured in J K-1 mol–1
This means that one of these values will need adjusting by a factor of 1000
Gibbs energy is also referred to as 'Gibbs free energy', or just 'free energy'
Free Energy & Equilibrium Calculations
The relationship between Gibbs energy change, ΔGꝊ, temperature and the equilibrium constant, K, is described by the equation:
ΔGꝊ = -RT lnK
The rearrangement of this equation makes it possible to:
Calculate the equilibrium constant
Deduce the position of equilibrium for the reaction
Worked Example
Calculating K
Ethanoic acid and ethanol react to form the ester ethyl ethanoate and water as follows:
CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)
At 25 oC, the free energy change, ΔGꝊ, for the reaction is -4.38 kJ mol-1. (R = 8.31 J K-1 mol-1)
Calculate the value of K for this reaction
Using your answer to part (1), predict and explain the position of the equilibrium
Answer 1:
Step 1: Convert any necessary values
ΔGꝊ into J mol-1:
-4.38 x 1000 = -4380 J mol-1
T into Kelvin
25 + 273.15 = 298.15 K
Step 2: Write the equation:
ΔGꝊ = -RT lnK
Step 3: Substitute the values:
-4380 = -8.31 x 298.15 x lnK
Step 4: Rearrange and solve the equation for K:
ln K = -4380 ÷ (-8.31 x 298.15)
ln K = 1.7678
K = e1.7678
K = 5.86
Answer 2:
From part (1), the value of Kc is 5.86
Therefore, the equilibrium lies to the right / products side because the value of K is positive
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