Equilibrium Constant Calculations | CIE International A Level Chemistry Revision Notes 2025

Equilibrium Constant: Calculations

Calculations involving K_c

In the equilibrium expression, each figure within a square bracket represents the concentration in mol dm^-3
The units of K_c 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:

$concentration (mol {dm}^{- 3}) = \frac{number of moles}{volume ({dm}^{3})}$

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) $⇌$ CH₃COOC₂H₅ (l) + H₂O (l)

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

Answer

Step 1: Calculate the concentrations of the reactants and products:
- [CH₃COOH (l)] = $\frac{0.235}{0.500}$ = 0.470 mol dm^-3
- [C₂H₅OH (l)] = $\frac{0.0350}{0.500}$ = 0.070 mol dm^-3
- [CH₃COOC₂H₅ (l)] = $\frac{0.182}{0.500}$ = 0.364 mol dm^-3
- [H₂O (l)] = $\frac{0.182}{0.500}$ = 0.364 mol dm^-3
Step 2: Write out the balanced chemical equation with the calculated concentrations beneath each substance:

CH₃COOH (l)	+	C₂H₅OH (l)	$⇌$	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:
- K_c = $\frac{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}{[C_{2} H_{5} OH] [{CH}_{3} COOH]}$
Step 4: Substitute the equilibrium concentrations into the expression:
- K_c = $\frac{0.364 \times 0.364}{0.070 \times 0.470}$
- K_c = 4.03
Step 5: Deduce the correct units for K_c:
- K_c = $\frac{(mol {dm}^{- 3}) (mol {dm}^{- 3})}{(mol {dm}^{- 3}) (mol {dm}^{- 3})}$
- All units cancel out
- Therefore, K_c = 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) $⇌$ 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 K_c 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)	$⇌$	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)] = $\frac{0.0654}{1.000}$ = 0.0654 mol dm^-3
- [CH₃COOC₂H₅ (l)] = $\frac{0.0654}{1.000}$ = 0.0654 mol dm^-3
- [C₂H₅OH (l)] = $\frac{0.0346}{1.000}$ = 0.0346 mol dm^-3
- [CH₃COOH (l)] = $\frac{0.0346}{1.000}$ = 0.0346 mol dm^-3

Step 3: Write the equilibrium constant for this reaction in terms of concentration:
- K_c = $\frac{[C_{2} H_{5} OH] [{CH}_{3} COOH]}{[H_{2} O] [{CH}_{3} {COOC}_{2} H_{5}]}$

Step 4: Substitute the equilibrium concentrations into the expression:
- K_c = $\frac{0.0346 \times 0.0346}{0.0654 \times 0.0654}$
- K_c = 0.28
Step 5: Deduce the correct units for K_c:
- K_c =
- All units cancel out
- Therefore, K_c = 0.28

Calculations involving K_p

In the equilibrium expression the p represent the partial pressure of the reactants and products in Pa
The units of K_p 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) $⇌$ 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 K_p for this reaction.

Answer

Step 1: Write the equilibrium constant for the reaction in terms of partial pressures:
- K_p = $\frac{p^{2} {SO}_{3}}{p^{2} {SO}_{2} \times p O_{2}}$

Step 2: Substitute the equilibrium concentrations into the expression:
- K_p = $\frac{{(8.0 \times 10^{6})}^{2}}{{(1.0 \times 10^{6})}^{2} \times (7.0 \times 10^{6})}$
- K_p = 9.1 x 10^–6

Step 3: Deduce the correct units of K_p:
- K_p = $\frac{{Pa}^{2}}{{Pa}^{2} \times Pa}$
- So, the units of K_p are Pa^-1
- Therefore, K_p = 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) $⇌$ 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 K_p 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 = $\frac{1.71 \times 10^{- 3}}{2.112 \times 10^{- 2}}$ = 0.0810
- I₂ = $\frac{2.91 \times 10^{- 3}}{2.112 \times 10^{- 2}}$ = 0.1378
- HI = $\frac{1.65 \times 10^{- 2}}{2.112 \times 10^{- 2}}$ = 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_p = $\frac{p^{2} HI}{p H_{2} \times p I_{2}}$

Step 5: Substitute the values into the equilibrium expression:
- K_p = $\frac{78 . 13^{2}}{8.10 \times 13.78}$
- K_p = 54.7

Step 6: Deduce the correct units for K_p:
- K_p = $\frac{{Pa}^{2}}{Pa \times Pa}$
- All units cancel out
- Therefore, K_p = 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) $⇌$ 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 + p_c = 3 atm
- p_c = 2 atm

Equilibrium Constant Calculations (CIE A Level Chemistry)

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