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IB DP Chemistry: HL復習筆記17.1.2 Gibbs Free Energy & the Equilibrium Constant

Category: IB課程, 教材筆記, 福利干貨 Date: 2022年7月29日 下午12:15

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 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

Exam Tip

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

  • ΔG??is measure 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

  • 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

17.1.2-The-relationship-between-the-equilibrium-constant-Kc-and-Gibbs-free-energy-change-%CE%94G%EA%9D%8B

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

lnK = - frac{Delta G}{RT}

Worked Example

Calculating?KcEthanoic 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

Answers

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 positive

 

 

 

 

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