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Equilbrium Constant Calculations (Edexcel A Level Chemistry): Revision Note

Stewart Hird

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Equilibrium Constant 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

 

Equilibria Equation for concentration, downloadable AS & A Level Chemistry revision notes

Equation to calculate concentration from number of moles and volume

Worked Example

Calculating Kc of ethanoic acid

Ethanoic acid and ethanol react according to the following equation:

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

At equilibrium, 500 cm3 of the reaction mixture 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.

Calculate a value of Kc for this reaction

Answer

  • Step 1: Calculate the concentrations of the reactants and products

    • [CH3COOH] equals space fraction numerator 0.235 over denominator 0.500 end fraction space equals space0.470 mol dm-3 

    • [C2H5OH] equals space fraction numerator 0.035 over denominator 0.500 end fraction space equals space0.070 mol dm-3 

    • [CH3COOC2H5] equals space fraction numerator 0.182 over denominator 0.500 end fraction space equals space0.364 mol dm-3 

    • [H2O] equals space fraction numerator 0.182 over denominator 0.500 end fraction space equals space0.364 mol dm-3 

  • Step 2: Write out the balanced chemical equation with the concentrations of beneath each substance

Equilibrium Constant Calculations WE Step 1 equation 2

  • Step 3: 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 4: Substitute the equilibrium concentrations into the expression

Kc = equals fraction numerator left square bracket 0.364 right square bracket cross times left square bracket 0.364 right square bracket over denominator left square bracket 0.070 right square bracket cross times left square bracket 0.0470 right square bracket end fraction equals4.03

  • Step 5: Deduce the correct units for Kc 

Kc = fraction numerator left square bracket mol space dm to the power of negative 3 end exponent right square bracket cross times left square bracket mol space dm to the power of negative 3 end exponent right square bracket over denominator left square bracket mol space dm to the power of negative 3 end exponent right square bracket cross times left square bracket mol space dm to the power of negative 3 end exponent right square bracket end fraction

All units cancel out

Therefore, Kc = 4.03

  • 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 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)

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 1dm3. 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

Equilibrium Constant Calculations WE Step 2 equation

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

Equilibrium Constant Calculations WE Step 3 equation

  • Step 4: Substitute the equilibrium concentrations into the expression

K subscript c space equals space fraction numerator left parenthesis 0.0346 right parenthesis space cross times space left parenthesis 0.0346 right parenthesis over denominator left parenthesis 0.0654 right parenthesis space cross times space left parenthesis 0.0654 right parenthesis end fraction

Kc = 0.28

  • Step 5: Deduce the correct units for Kc

K subscript c space equals space fraction numerator left parenthesis mol space dm to the power of negative 3 end exponent right parenthesis cross times left parenthesis mol space dm to the power of negative 3 end exponent right parenthesis over denominator left parenthesis mol space dm to the power of negative 3 end exponent right parenthesis cross times left parenthesis mol space dm to the power of negative 3 end exponent right parenthesis end fraction

All units cancel out

Therefore, Kc = 0.288

Calculations involving Kp

  • In the equilibrium expression the p represent the partial pressure of the reactants and products in Pa

  • The units of Kp therefore depend on the form of the equilibrium expression

Worked Example

Calculating Kp of a gaseous reaction:

In the reaction:

2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

the equilibrium partial pressures at constant temperature are

SO2 = 1.0 × 106 Pa, O2 = 7.0 × 106 Pa, SO3 = 8.0 × 106 Pa

Calculate the value for Kp for this reaction.

Answer

  • Step 1: Write the equilibrium constant for the reaction in terms of partial pressures

K subscript p space equals space fraction numerator p squared space SO subscript 3 over denominator p squared space SO subscript 2 space cross times p straight O subscript 2 end fraction

  • Step 2: Substitute the equilibrium concentrations into the expression

K subscript p space equals space fraction numerator left parenthesis 8.0 space cross times 10 to the power of 6 right parenthesis squared over denominator left parenthesis 1.0 space cross times 10 to the power of 6 right parenthesis squared space cross times left parenthesis 7.0 space cross times 10 to the power of 6 right parenthesis end fraction

Kp = 9.1 x 10-6

  • Step 3: Deduce the correct units of Kp

K subscript p space equals space fraction numerator P a squared over denominator P a squared space cross times P a end fraction

The units of Kp are Pa-1

Therefore, Kp = 9.1 x 10-6 Pa-1

  • Some questions only give the number of moles of gases present and the total pressure

  • The number of moles of each gas should be used to first calculate the mole fractions

  • The mole fractions are then used to calculate the partial pressures

  • The values of the partial pressures are then substituted in the equilibrium expression

Worked Example

Calculating Kp of a hydrogen iodide equilibrium reaction:

The equilibrium between hydrogen, iodine and hydrogen iodide at 600 K is as follows:

H2 (g) + I2 (g) ⇌ 2HI (g)

At equilibrium the number of moles present are:

H2 = 1.71 × 10-3

I2 = 2.91 × 10-3

HI = 1.65 × 10-2 

The total pressure is 100 kPa.

Calculate the value of Kp for this reaction.

Answer

  • Step 1: Calculate the total number of moles

Total number of moles = 1.71 x 10-3 + 2.91 x 10-3 + 1.65 x 10-2

= 2.112 x 10-2

  • Step 2: Calculate the mole fraction of each gas

straight H subscript 2 equals fraction numerator 1.71 cross times 10 to the power of negative 3 end exponent over denominator 2.112 cross times 10 to the power of negative 2 end exponent end fraction equals 0.0810 straight I subscript 2 space equals space fraction numerator 2.91 space cross times 10 to the power of negative 3 end exponent over denominator 2.112 space cross times 10 to the power of negative 2 end exponent end fraction equals space 0.1378 HI space equals space fraction numerator 1.65 space cross times 10 to the power of negative 2 end exponent over denominator 2.112 space cross times 10 to the power of negative 2 end exponent end fraction equals space 0.7813

  • Step 3: Calculate the partial pressure of each gas

H2 = 0.0810 x 100 = 8.10 kPa

I2 = 0.1378 x 100 = 13.78 kPa

HI = 0.7813 x 100 = 78.13 kPa

  • Step 4: Write the equilibrium constant in terms of partial pressure

K subscript p space equals fraction numerator p squared HI over denominator p straight H subscript 2 space cross times p straight I subscript 2 end fraction

  • Step 5: Substitute the values into the equilibrium expression

K subscript p space equals space fraction numerator 78.13 squared over denominator 8.10 space cross times 13.78 end fraction

Kp = 54.7

  • Step 6: Deduce the correct units for Kp

K subscript p space equals space fraction numerator P a squared over denominator P a space cross times P a end fraction

All units cancel out

Therefore, Kp = 54.7

  • Other questions related to equilibrium expressions may involve calculating quantities present at equilibrium given appropriate data

Examiner Tips and Tricks

  • If a reaction has reached equilibrium:

    • The moles of reactants and products will remain unchanged

    • This means that the concentrations / mole fractions / partial pressures of each chemical will remain unchanged 

    • Consequently, the value of the equilibrium constant (Kc or Kp) will also remain unchanged

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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.