CIE AS Chemistry (9701) exams from 2022

Revision Notes

1.7.4 Equilibrium Constant Calculations

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

Equilibria Worked example - Calculating Kc of ethanoic acid, downloadable AS & A Level Chemistry revision notes

Answer

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

Equilibrium Constant Calculations WE Step 1 equation 1, downloadable AS & A Level Chemistry revision notes

  • 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

Equilibrium Constant Calculations WE Step 1 equation 3

  • Step 4: Substitute the equilibrium concentrations into the expression

Equilibrium Constant Calculations WE Step 1 equation 4

  • Step 5: Deduce the correct units for Kc

Equilibrium Constant Calculations WE Step 1 equation 5

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

Equilibria Worked example - Calculating Kc of ethyl ethanoate, downloadable AS & A Level Chemistry revision notes

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

Equilibrium Constant Calculations WE Step 4 equation

= 0.28

  • Step 5: Deduce the correct units for Kc

Equilibrium Constant Calculations WE Step 5 equation

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

Equilibria Worked example - Calculating Kp of a gaseous reaction, downloadable AS & A Level Chemistry revision notes

Answer

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

Equilibrium Constant Calculations WE 3 Step 1 equation

  • Step 2: Substitute the equilibrium concentrations into the expression

Equilibrium Constant Calculations WE 3 Step 2 equation

= 9.1 x 10-6

  • Step 3: Deduce the correct units of Kp

Equilibrium Constant Calculations WE 3 Step 3 equation

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 hydrogen iodide equilibrium reaction

Equilibria Worked example - Calculating Kp of hydrogen iodide equilibrium reaction, downloadable AS & A Level Chemistry revision notes

  • 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

Equilibrium Constant Calculations WE 4 Step 2 equation

  • 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

Equilibrium Constant Calculations WE 4 Step 4 equation

  • Step 5: Substitute the values into the equilibrium expression

Equilibrium Constant Calculations WE 4 Step 5 equation

= 54.7

  • Step 6: Deduce the correct units for Kp

Equilibrium Constant Calculations WE 4 Step 6 equation

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: Calculating partial pressures

Equilibria Worked example - Calculating partial pressures, downloadable AS & A Level Chemistry revision notes

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

Total pressure = sum of equilibrium (Σ) 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

Author: Francesca

Fran has taught A level Chemistry in the UK for over 10 years. As head of science, she used her passion for education to drive improvement for staff and students, supporting them to achieve their full potential. Fran has also co-written science textbooks and worked as an examiner for UK exam boards.
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