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Equilibrium Constant, Kc (Edexcel International A Level Chemistry): Revision Note

Stewart Hird

Last updated

Kc Expressions - Deduction

Equilibrium expression & constant

  • The equilibrium constant expression is an expression that links the equilibrium constant, K, 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 corresponding equilibrium constant expression is written as:

bold italic K bold space bold equals bold space fraction numerator stretchy left square bracket C stretchy right square bracket to the power of bold c stretchy left square bracket D stretchy right square bracket to the power of bold d over denominator stretchy left square bracket A stretchy right square bracket to the power of bold a stretchy left square bracket B stretchy right square bracket to the power of bold b end fraction

  • Where:

    • [A] and [B] = equilibrium reactant concentrations (mol dm-3)

    • [C] and [D] = equilibrium product concentrations (mol dm-3)

    • a, b, c and d = number of moles of corresponding reactants and products

  • Solids are ignored in equilibrium constant expressions

  • The Kc of a reaction is specific to a given reaction and only changes if the temperature of the reaction changes

Worked Example

Deducing equilibrium expressions

Deduce the equilibrium constant expression for the following reactions

  1. Ag+ (aq) + Fe2+ (aq) ⇌ Ag (s) + Fe3+ (aq)

  2. N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

  3. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

 

Answer 1:

   K space equals fraction numerator open square brackets Fe to the power of 3 plus end exponent open parentheses aq close parentheses close square brackets over denominator open square brackets Fe to the power of 2 plus end exponent space open parentheses aq close parentheses close square brackets space open square brackets Ag to the power of plus space open parentheses aq close parentheses close square brackets end fraction

  • [Ag (s)] is not included in the equilibrium constant expression as it is a solid

Answer 2:

   K space equals fraction numerator open square brackets NH subscript 3 space open parentheses straight g close parentheses close square brackets over denominator stretchy left square bracket straight N subscript 2 space open parentheses straight g close parentheses stretchy right square bracket space stretchy left square bracket straight H subscript 2 space open parentheses straight g close parentheses stretchy right square bracket cubed end fraction

Answer 3: 

   K space equals fraction numerator stretchy left square bracket SO subscript 3 space open parentheses straight g close parentheses stretchy right square bracket squared over denominator stretchy left square bracket SO subscript 2 space stretchy left parenthesis straight g stretchy right parenthesis stretchy right square bracket squared space stretchy left square bracket straight O subscript 2 space stretchy left parenthesis straight g stretchy right parenthesis stretchy right square bracket end fraction

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

  • Kc = fraction 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

  • Kc = fraction 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 

  • Kc = fraction 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

Examiner Tips and Tricks

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

  • Kc = fraction 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

  • Kc = fraction 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 

  • Kc = fraction 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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    Stewart Hird

    Author: Stewart Hird

    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 Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.