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A LevelChemistryCIERevision Notes7. Equilibria7.1 Chemical Equilibria: Reversible Reactions & Dynamic EquilibriumEquilibrium Constant Calculations

Equilibrium Constant Calculations (CIE A Level Chemistry)

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Caroline

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Caroline

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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 using:

 

Worked example

At equilibrium, 500 cm3 of the following reaction mixture contains 0.235 mol of ethanoic acid, 0.0350 mol of ethanol, 0.182 mol of ethyl ethanoate and 0.182 mol of water.

CH3COOH (l) + C2H5OH (l) rightwards harpoon over leftwards harpoon CH3COOC2H5 (l) + H2O (l)

Use this information 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 = 0.470 mol dm-3 
    • [C2H5OH (l)] = fraction numerator 0.0350 over denominator 0.500 end fraction = 0.070 mol dm-3 
    • [CH3COOC2H5 (l)] = fraction numerator 0.182 over denominator 0.500 end fraction = 0.364 mol dm-3 
    • [H2O (l)] = begin mathsize 14px style fraction numerator 0.182 over denominator 0.500 end fraction end style = 0.364 mol dm-3 
  • Step 2: Write out the balanced chemical equation with the calculated concentrations beneath each substance:

CH3COOH (l)

 +

C2H5OH (l)

 rightwards harpoon over leftwards harpoon

CH3COOC2H5 (l)

 +

H2O (l)

0.470 mol dm-3 

  

0.070 mol dm-3 

  

0.364 mol dm-3 

  

0.364 mol dm-3 

  • Step 3: Write the equilibrium constant for this reaction in terms of concentration:
    • Kc = fraction numerator open square brackets straight H subscript 2 straight O close square brackets space open square brackets CH subscript 3 COOC subscript 2 straight H subscript 5 close square brackets over denominator open square brackets straight C subscript 2 straight H subscript 5 OH close square brackets space open square brackets CH subscript 3 COOH close square brackets end fraction
  • Step 4: Substitute the equilibrium concentrations into the expression:
    • Kc = fraction numerator 0.364 cross times 0.364 over denominator 0.070 cross times 0.470 end fraction
    • Kc = 4.03
  • Step 5: Deduce the correct units for Kc:
    • Kc = fraction numerator open parentheses mol space dm to the power of negative 3 end exponent close parentheses space open parentheses mol space dm to the power of negative 3 end exponent close parentheses over denominator open parentheses mol space dm to the power of negative 3 end exponent close parentheses space open parentheses mol space dm to the power of negative 3 end exponent close parentheses 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

Ethyl ethanoate is hydrolysed in water.

CH3COOC2H5 (l) + H2O (l) rightwards harpoon over leftwards harpoon CH3COOH (l) + C2H5OH (l) 

0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture is made up to 1 dm3.

At equilibrium, 0.0654 mol of water are present.

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

Answer

  • Step 1: Write the balanced chemical equation, with the concentrations beneath each substance, into an initial, change and equilibrium (ICE) table:
 

CH3COOC2H5 (l)

+ H2O (l)

 bold rightwards harpoon over leftwards harpoon

CH3COOH (l)

+ C2H5OH (l)

Initial moles

0.1000

0.1000

  

0

0

Change

–0.0346

–0.0346

  

+0.0346

+0.0346

Equilibrium moles

0.0654

0.0654

  

0.0346

0.0346

 

  • Step 2: Calculate the concentrations of the reactants and products:
    • [H2O (l)] = fraction numerator 0.0654 over denominator 1.000 end fraction = 0.0654 mol dm-3 
    • [CH3COOC2H5 (l)] = fraction numerator 0.0654 over denominator 1.000 end fraction = 0.0654 mol dm-3 
    • [C2H5OH (l)] = fraction numerator 0.0346 over denominator 1.000 end fraction = 0.0346 mol dm-3 
    • [CH3COOH (l)] = fraction numerator 0.0346 over denominator 1.000 end fraction = 0.0346 mol dm-3
  • Step 3: Write the equilibrium constant for this reaction in terms of concentration:
    • Kc = fraction numerator stretchy left square bracket straight C subscript 2 straight H subscript 5 OH stretchy right square bracket space stretchy left square bracket CH subscript 3 COOH stretchy right square bracket over denominator stretchy left square bracket straight H subscript 2 straight O stretchy right square bracket space stretchy left square bracket CH subscript 3 COOC subscript 2 straight H subscript 5 stretchy right square bracket end fraction
  • Step 4: Substitute the equilibrium concentrations into the expression:
    • Kc = fraction numerator 0.0346 cross times 0.0346 over denominator 0.0654 cross times 0.0654 end fraction
    • Kc = 0.28
  • Step 5: Deduce the correct units for Kc:
    • Kc = 
    • All units cancel out
    • Therefore, Kc = 0.28

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

The equilibrium between sulfur dioxide, oxygen and sulfur trioxide is as follows:

2SO2 (g) + O2 (g) rightwards harpoon over leftwards harpoon 2SO3 (g)

At constant temperature, the equilibrium partial pressures are:

  • SO2 = 1.0 x 106 Pa
  • O2 = 7.0 x 106 Pa
  • SO3 = 8.0 x 106 Pa

Calculate the value of Kp for this reaction.

Answer

  • Step 1: Write the equilibrium constant for the reaction in terms of partial pressures:
    • Kp = fraction numerator p squared space SO subscript 3 over denominator p squared space SO subscript 2 cross times p space straight O subscript 2 end fraction
  • Step 2: Substitute the equilibrium concentrations into the expression:
    • Kp = fraction numerator open parentheses 8.0 cross times 10 to the power of 6 close parentheses squared over denominator open parentheses 1.0 cross times 10 to the power of 6 close parentheses squared cross times open parentheses 7.0 cross times 10 to the power of 6 close parentheses end fraction
    • Kp = 9.1 x 10–6
  • Step 3: Deduce the correct units of Kp:
    • Kp = fraction numerator Pa squared over denominator Pa squared cross times Pa end fraction
    • So, 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

The equilibrium between hydrogen, iodine and hydrogen bromide is as follows:

H2 (g) + I2 (g) rightwards harpoon over leftwards harpoon 2HI (g)

At constant temperature, the equilibrium moles are:

  • H2 = 1.71 x 10–3 
  • I2 = 2.91 x 10–3 
  • HI = 1.65 x 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
    • Total number of moles = 2.112 x 10-2
  • Step 2: Calculate the mole fraction of each gas:
    • H2 = 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 = 0.0810
    • I2 = fraction numerator 2.91 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 = 0.1378
    • HI = begin mathsize 14px style fraction numerator 1.65 cross times 10 to the power of negative 2 end exponent over denominator 2.112 cross times 10 to the power of negative 2 end exponent end fraction end style = 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:
    • Kp = fraction numerator p squared space HI over denominator p space straight H subscript 2 cross times p space straight I subscript 2 end fraction
  • Step 5: Substitute the values into the equilibrium expression:
    • Kp = begin mathsize 14px style fraction numerator 78.13 squared over denominator 8.10 cross times 13.78 end fraction end style
    • Kp = 54.7
  • Step 6: Deduce the correct units for Kp:
    • Kp = begin mathsize 14px style fraction numerator Pa squared over denominator Pa cross times Pa end fraction end style
    • All units cancel out
    • Therefore, Kp = 54.7
  • Other questions related to equilibrium expressions may involve calculating quantities present at equilibrium given appropriate data

Worked example

An equilibrium is set up in a closed container between equal volumes of gaseous reactants A and B to form a gaseous product C.

A (g) + B (g) rightwards harpoon over leftwards harpoon 2C (g)

The total pressure within the container, at 50 oC, is 3 atm.

The equilibrium partial pressure of A, at 50 oC, is 0.5 atm.

What is the equilibrium partial pressure of C at this temperature?

Answer

  • There are equal volumes of reactants A and B in a 1 : 1 molar ratio
    • This means their partial pressures will be the same.
    • B therefore also has an equilibrium partial pressure of 0.5 atm
  • Total pressure = Σ (equilibrium partial pressures)
    • Therefore, the sum of all the partial pressures must equal to 3 atm
    • 0.5 + 0.5 + pc = 3 atm
    • pc = 2 atm

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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.