Equilbrium Constant Calculations | Edexcel A Level Chemistry Revision Notes 2017

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

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

Equation to calculate concentration from number of moles and volume

Worked example

Calculating K_c 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 K_c for this reaction

Answer

Step 1: Calculate the concentrations of the reactants and products
- [CH₃COOH] $= \frac{0.235}{0.500} =$ 0.470 mol dm^-3
- [C₂H₅OH] $= \frac{0.035}{0.500} =$ 0.070 mol dm^-3
- [CH₃COOC₂H₅] $= \frac{0.182}{0.500} =$ 0.364 mol dm^-3
- [H₂O] $= \frac{0.182}{0.500} =$ 0.364 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

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 = $= \frac{[0.364] \times [0.364]}{[0.070] \times [0.0470]} =$ 4.03

Step 5: Deduce the correct units for K_c

K_c = $\frac{[mol {dm}^{- 3}] \times [mol {dm}^{- 3}]}{[mol {dm}^{- 3}] \times [mol {dm}^{- 3}]}$

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

Calculating K_c 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 K_cfor 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

$K_{c} = \frac{(0.0346) \times (0.0346)}{(0.0654) \times (0.0654)}$

K_c= 0.28

- Step 5: Deduce the correct units for K_c

$K_{c} = \frac{(mol {dm}^{- 3}) \times (mol {dm}^{- 3})}{(mol {dm}^{- 3}) \times (mol {dm}^{- 3})}$

All units cancel out

Therefore, K_c = 0.288

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

Calculating K_pof 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 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

_{$K_{p} = \frac{P a^{2}}{P a^{2} \times P a}$}

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

Calculating K_pof 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 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

= 2.112 x 10^-2

- Step 2: Calculate the mole fraction of each gas

$H_{2} = \frac{1.71 \times 10^{- 3}}{2.112 \times 10^{- 2}} = 0.0810 I_{2} = \frac{2.91 \times 10^{- 3}}{2.112 \times 10^{- 2}} = 0.1378 HI = \frac{1.65 \times 10^{- 2}}{2.112 \times 10^{- 2}} = 0.7813$

- 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

$K_{p} = \frac{P a^{2}}{P a \times P a}$

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

Examiner Tip

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 (K_c or K_p) will also remain unchanged

Equilbrium Constant Calculations (Edexcel A Level Chemistry)

Revision Note

Equilibrium Constant Calculations

Calculations involving K_c

Worked example

Calculating K_c of ethanoic acid

Worked example

Calculations involving K_p

Worked example

Worked example

Examiner Tip

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Equilbrium Constant Calculations (Edexcel A Level Chemistry)

Revision Note

Calculations involving Kc

Calculating Kc of ethanoic acid

Calculations involving Kp

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Author: Sonny

Calculations involving K_c

Calculating K_c of ethanoic acid

Calculations involving K_p