Edexcel International A Level Chemistry

Revision Notes

4.4.1 Equilibrium Constant, Kc

Kc Expressions - Deduction

Equilibrium expression & constant

  • The equilibrium expression links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account
  • So, for a given reaction:

aA + bB ⇌ cC + dD

          the Kc is defined as follows:

Equilibria Equilibrium Expression, downloadable AS & A Level Chemistry revision notes

Equilibrium expression linking the equilibrium concentration of reactants and products at equilibrium

  • Solids are ignored in equilibrium expressions
  • The Kc of a reaction is specific and only changes if the temperature of the reaction changes

Worked example

Deducing equilibrium expressions

Deduce the equilibrium expression for the following reactions:

  1. Ag+ (aq) + Fe2+ (aq) ⇌ Ag (s) + Fe3+ (aq)
  2. N2 (g) + 3H2 (g) ⇌ 2NH(g)
  3. 2SO(g) + O(g) ⇌ 2SO(g)

Equilibrium Constant Concentrations Worked Example equations, downloadable AS & A Level Chemistry revision notes

Kc Expressions - Calculations

Calculations involving Kc

  • In the equilibrium expression each figure within a square bracket represents the concentration in mol dm-3
  • The units of Kc therefore depend on the form of the equilibrium expression
  • Some questions give the number of moles of each of the reactants and products at equilibrium together with the volume of the reaction mixture
  • The concentrations of the reactants and products can then be calculated from the number of moles and total volume

Equation to calculate concentration from number of moles and volume

Worked example

Calculating Kc of ethanoic acid

In the reaction:

CH3COOH (I) + C2H5OH (I) ⇌ CH3COOC2H5 (I) + H2O (I)

ethanoic acid     ethanol          ethyl ethanoate       water

500 cm3 of the reaction mixture at equilibrium contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water. Use this data to calculate a value of Kc for this reaction.

Answer

Step 1: Calculate the concentrations of the reactants and products

    • [CH3COOH (l)] = fraction numerator 0.235 over denominator 0.500 end fraction space equals space 0.470 space mol space dm to the power of negative 3 end exponent
    • [C2H5OH (l)] =fraction numerator 0.035 over denominator 0.500 end fraction space equals space 0.070 space mol space dm to the power of negative 3 end exponent
    • [CH3COOC2H5 (l)] = fraction numerator 0.182 over denominator 0.500 end fraction space equals space 0.364 space mol space dm to the power of negative 3 end exponent
    • [H2O (l)] = fraction numerator 0.182 over denominator 0.500 end fraction space equals space 0.364 space mol space dm to the power of negative 3 end exponent

Step 2: Write the equilibrium constant for this reaction in terms of concentration

    • Kcfraction numerator left square bracket straight H subscript 2 straight O right square bracket space left square bracket CH subscript 3 COOC subscript 2 straight H subscript 5 right square bracket over denominator left square bracket straight C subscript 2 straight H subscript 5 OH right square bracket space left square bracket CH subscript 3 COOH right square bracket end fraction

Step 3: Substitute the equilibrium concentrations into the expression

    • Kcfraction numerator left square bracket 0.364 right square bracket space cross times space left square bracket 0.364 right square bracket over denominator left square bracket 0.070 right square bracket space cross times space left square bracket 0.470 right square bracket end fraction space equals space 4.03

Step 4: Deduce the correct units for Kc 

    • Kcfraction numerator left square bracket m o l space d m to the power of negative 3 end exponent right square bracket space cross times space left square bracket m o l space d m to the power of negative 3 end exponent right square bracket over denominator left square bracket m o l space d m to the power of negative 3 end exponent right square bracket space cross times space left square bracket m o l space d m to the power of negative 3 end exponent right square bracket end fraction
    • All units cancel out

Therefore, Kc = 4.03

Exam Tip

Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures

  • Some questions give the initial and equilibrium concentrations of the reactants but not the products
  • An initial, change and equilibrium table should be used to determine the equilibrium concentration of the products using the molar ratio of reactants and products in the stoichiometric equation

Worked example

Calculating Kc of ethyl ethanoate

Ethyl ethanoate is hydrolysed by water:

CH3COOC2H5 (I) + H2O (I) ⇌ CH3COOH (I) + C2H5OH (I)

ethyl ethanoate       water        ethanoic acid     ethanol

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1.00 dm3. At equilibrium, 0.0654 mol of water are present.

Use this data to calculate a value of Kc for this reaction.

Answer

Step 1: Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Equilibria Calculating Kc of ethyl ethanoate table

Step 2: Calculate the concentrations of the reactants and products

    • [CH3COOC2H5 (l)] = fraction numerator 0.0654 over denominator 1.00 end fraction space equals space 0.0654 space mol space dm to the power of negative 3 end exponent
    • [H2O (l)] = fraction numerator 0.0654 over denominator 1.00 end fraction space equals space 0.0654 space mol space dm to the power of negative 3 end exponent
    • [CH3COOH (l)] = fraction numerator 0.0346 over denominator 1.00 end fraction space equals space 0.0346 space mol space dm to the power of negative 3 end exponent
    • [C2H5OH (l)] =fraction numerator 0.0346 over denominator 1.00 end fraction space equals space 0.0346 space mol space dm to the power of negative 3 end exponent

Step 3: Write the equilibrium constant for this reaction in terms of concentration

    • Kcfraction numerator left square bracket straight C subscript 2 straight H subscript 5 OH right square bracket space left square bracket CH subscript 3 COOH right square bracket space over denominator left square bracket straight H subscript 2 straight O right square bracket space left square bracket CH subscript 3 COOC subscript 2 straight H subscript 5 right square bracket end fraction

Step 4: Substitute the equilibrium concentrations into the expression

    • Kcfraction numerator left square bracket 0.346 right square bracket space cross times space left square bracket 0.346 right square bracket over denominator left square bracket 0.654 right square bracket space cross times space left square bracket 0.654 right square bracket end fraction space equals space 0.28

Step 4: Deduce the correct units for Kc 

    • Kcfraction numerator left square bracket m o l space d m to the power of negative 3 end exponent right square bracket space cross times space left square bracket m o l space d m to the power of negative 3 end exponent right square bracket over denominator left square bracket m o l space d m to the power of negative 3 end exponent right square bracket space cross times space left square bracket m o l space d m to the power of negative 3 end exponent right square bracket end fraction
    • All units cancel out

Therefore, Kc = 0.28

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