The Equilibrium Constant & Gibbs Energy (HL) | HL IB Chemistry Revision Notes 2025

The Equilibrium Constant & Gibbs Energy

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, ΔG^Ꝋ, according to the following Gibbs energy equation:

ΔG^Ꝋ = -RT lnK

- ΔG^Ꝋ= 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, ΔG^Ꝋ, can be used to determine whether the forward or backward reaction is favoured

Equilibirum constant, K	Description	Gibbs energy change, ΔG
K > 1	Products favoured	ΔG < 0 (negative)
K = 1	Reaction at equilibirum Neither reactants nor products are favoured	ΔG = 0
K < 1	Reactants favoured	ΔG > 0 (positive)

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

When completing calculations using the ΔG^Ꝋ = –RT lnK equation, you have to be aware that:
- ΔG^Ꝋ 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'

The relationship between Gibbs energy change, ΔG^Ꝋ, temperature and the equilibrium constant, K, is described by the equation:

ΔG^Ꝋ = -RT lnK

The rearrangement of this equation makes it possible to:
- Calculate the equilibrium constant
- Deduce the position of equilibrium for the reaction

$\ln K = - \frac{Δ G}{R T}$

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, ΔG^Ꝋ, 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
- ΔG^Ꝋ into J mol^-1:
  - -4.38 x 1000 = -4380 J mol^-1

Step 4: Rearrange and solve the equation for K:
- ln K = -4380 ÷ (-8.31 x 298.15)
- ln K = 1.7678
- K = e^1.7678
- K = 5.86

Answer 2:

From part (1), the value of K_c is 5.86
Therefore, the equilibrium lies to the right / products side because the value of K is positive