# 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

Equation to calculate concentration from number of moles and volume

#### Worked example: Calculating Kc of ethanoic acid

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

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

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

• Step 4: Substitute the equilibrium concentrations into the expression

• Step 5: Deduce the correct units for Kc

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

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

• Step 2: Calculate the concentrations of the reactants and products

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

• Step 4: Substitute the equilibrium concentrations into the expression

= 0.28

• Step 5: Deduce the correct units for Kc

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

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

• Step 2: Substitute the equilibrium concentrations into the expression

= 9.1 x 10-6

• Step 3: Deduce the correct units of Kp

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

• 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

• 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

• Step 5: Substitute the values into the equilibrium expression

= 54.7

• Step 6: Deduce the correct units for Kp

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

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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