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The Equilibrium Constant & Gibbs Energy (HL) (HL IB Chemistry)

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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
(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

Examiner Tip

  • 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

bold ln bold italic K bold space bold equals bold space bold minus fraction numerator bold capital delta bold G over denominator bold R bold T end fraction

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)

  1. Calculate the value of K 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
    • 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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    Caroline

    Author: Caroline

    Expertise: Physics Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.