Equilibrium Constant Calculations (Cambridge (CIE) A Level Chemistry): Revision Note

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) $\rightleftharpoons$ 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)] = $\frac{0.235}{0.500{\mathrm{dm}}^3}$ = 0.470 mol dm^-3

[C₂H₅OH (l)] = $\frac{0.350}{0.500{\mathrm{dm}}^3}$ = 0.070 mol dm^-3

[CH₃COOC₂H₅ (l)] = $\frac{0.182}{0.500{\mathrm{dm}}^3}$ = 0.364 mol dm^-3

[H₂O (l)] = $\frac{0.182}{0.500{\mathrm{dm}}^3}$ = 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 = $\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

• 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 = $\frac{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}{\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)\left(\mathrm{mol}{\mathrm{dm}}^{-3}\right)}$

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

Ethyl ethanoate is hydrolysed by water:

CH₃COOC₂H₅ (I) + H₂O (I) ⇌ CH₃COOH (I) + C₂H₅OH (I)

0.1000 mol of ethyl ethanoate is 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 data to calculate a value of K_c for this reaction.

Answer:

• Step 1: Complete the ICE table for the reaction:

  • Write out the balanced chemical equation with the number of moles of each substance given in the question beneath using an initial, change and equilibrium table:

• Calculate the change in moles of water and add to the table (an increase is shown by + and a decrease is shown by -)

  • Equilibrium amount = Initial amount + Change in amount

  • 0.0654 = 0.100 + Change in amount

  • Change in amount = 0.0654 - 0.100 = –0.0346

• Use the stoichiometry of the equation to calculate the change in amounts of the remaining reactants/products and add to the table

  • There is a 1 : 1 reacting ratio between H₂O and all other reactants/products

  • As H₂O has decreased by 0.0346 mol, the other reactant CH₃COOC₂H₅ will decrease by 0.0346 mol 

  • Since CH₃COOH and C₂H₅OH are products, they will both increase by 0.0346 mol

• Calculate the number of moles at equilibrium of the remaining reactants / products to complete the table

  • Equilibrium amount = Initial amount + Change in amount

  • Equilibrium amount of CH₃COOC₂H₅  = 0.100 + (-0.0346) = 0.0654 mol

  • Equilibrium amount of CH₃COOH  = 0.000 + 0.0346 = 0.0346 mol

  • Equilibrium amount of C₂H₅OH = 0.000 + 0.0346 = 0.0346 mol

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

[CH₃COOH (l)] = $\frac{0.0346}{1.0}$ = 0.0346 mol dm^-3

[C₂H₅OH (l)] = $\frac{0.0346}{1.0}$ = 0.0346 mol dm^-3

[CH₃COOC₂H₅ (l)] = $\frac{0.0654}{1.0}$ = 0.0654 mol dm^-3

[H₂O (l)] = $\frac{0.0654}{1.0}$ = 0.0654 mol dm^-3

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

K_c = $\frac{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}{\left[{\mathrm{H}}_{2}O\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}$

• Step 4: Substitute the equilibrium concentrations into the expression:

K_c = $\frac{0.0346\times 0.0346}{0.0654\times 0.0654}$ = 0.280

• Step 5: Deduce the correct units for K_c:

K_c = $\frac{\left[{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right]\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}{\left[{\mathrm{H}}_{2}O\right]\left[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5\right]}$

All units cancel out

Therefore, K_c = 0.280 (no units)

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

2SO₂ (g) + O₂ (g) $\rightleftharpoons$ 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 = $\frac{p^2{\mathrm{SO}}_3}{p^2{\mathrm{SO}}_2\times p{\mathrm{O}}_2}$

• Step 2: Substitute the equilibrium concentrations into the expression:

K_p = $\frac{{\left(8.0\times {10}^6\right)}^2}{{\left(1.0\times {10}^6\right)}^2\times \left(7.0\times {10}^6\right)}$

K_p = 9.1 x 10^–6

• Step 3: Deduce the correct units of K_p:

K_p = $\frac{{\mathrm{Pa}}^2}{{\mathrm{Pa}}^2\times \mathrm{Pa}}$

So, the units of K_p are Pa^-1

Therefore, K_p = 9.1 x 10^-6 Pa^-1

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

H₂ (g) + I₂ (g) $\rightleftharpoons$ 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

1. 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 = $\frac{p^2\mathrm{HI}}{p{\mathrm{H}}_2\times p{\mathrm{I}}_2}$

• 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 = $\frac{{\mathrm{Pa}}^2}{\mathrm{Pa}\times \mathrm{Pa}}$

All units cancel out

Therefore, K_p = 54.7

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) $\rightleftharpoons$ 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