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Gibbs Free Energy & the Equilibrium Constant (DP IB Chemistry: HL)

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Stewart

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Gibbs Free Energy & the Equilibrium Constant

Gibbs Free Energy & the Equilibrium Constant

  • The equilibrium constant, Kc, gives no information about the individual rates of reaction
    • It is independent of the kinetics of the reaction

  • The equilibrium constant, Kc, is directly related to the Gibbs free energy change, ΔG, according to the following (van't Hoff's) equation:

ΔG = -RT lnK

    • ΔG= Gibbs free energy change (kJ mol1)
    • 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

Examiner Tip

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

  • ΔG is measure in kJ mol1
  • R is measured in J K-1 mol-1

This means that one of these values will need adjusting by a factor of 1000

  • This relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔG, can be used to determine whether the forward or backward reaction is favoured

The relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔGꝋ, downloadable IB Chemistry revision notes

The relationship between the equilibrium constant, Kc, and Gibbs free energy change, ΔG

  • 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

Free Energy & Equilibrium Calculations

  • The relationship between Gibbs free energy change, ΔG, temperature and the equilibrium constant, Kc, is described by the equation:

ΔG = -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 = negative fraction numerator capital delta G over denominator R T end fraction

Worked example

Calculating Kc

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, for the reaction is -4.38 kJ mol-1(R = 8.31 J K-1 mol-1)

  1. Calculate the value of Kc for this reaction
  2. Using your answer to part (1), predict and explain the position of the equilibrium

Answer 1:

  • Step 1: Convert any necessary values
    • ΔG into J mol-1:
      • -4.38 x 1000 = -4380 J mol-1
    • T into Kelvin
      • 25 + 273 = 298 K
  • Step 2: Write the equation:
    • ΔG = -RT lnKc
  • Step 3: Substitute the values:
    • -4380 = -8.31 x 298 x lnKc
  • Step 4: Rearrange and solve the equation for Kc:
    • lnKc = -4380 ÷ (-8.31 x 298)
    • lnKc = 1.77
    • Kc = e1.77
    • Kc = 5.87

Answer 2:

  • From part (1), the value of Kc is 5.87
  • Therefore, the equilibrium lies to the right / products side because the value of Kc is greater than 1

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Stewart

Author: Stewart

Expertise: Chemistry Lead

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Exam Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.