Equilibrium Constant Calculations (Cambridge International A-Level Chemistry)

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:

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

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) $\rightleftharpoons$ 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 = $\frac{[ \mathrm{H}_{2} \mathrm{O} ] [ CH_{3} COOC_{2} \mathrm{H}_{5} ]}{[ \mathrm{C}_{2} \mathrm{H}_{5} \text{OH} ] [ CH_{3} \text{COOH} ]}$

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

K_c = $\frac{0.364\times0.364}{0.070\times0.470}$

K_c = 4.03

• Step 5: Deduce the correct units for K_c:

K_c = $\frac{( \text{mol} dm^{- 3} ) ( \text{mol} dm^{- 3} )}{( \text{mol} dm^{- 3} ) ( \text{mol} dm^{- 3} )}$

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 not 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 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 CH₃COOC₂H_5****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

	CH3COOC2H5** (I) +**	H2O (I) ⇌	CH3COOH (I) +	C2H5OH (I)
Initial moles	0.100	0.100	0.000	0.000
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:

[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* *= $\frac{[ \mathrm{C}_{2} \mathrm{H}_{5} \text{OH} ] [ CH_{3} \text{COOH} ]}{[ \mathrm{H}_{2} O ] [ CH_{3} COOC_{2} \mathrm{H}_{5} ]}$

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

K_c* *= $\frac{0.0346\times0.0346}{0.0654\times0.0654}$ = 0.280

• Step 5: Deduce the correct units for K_c:

K_c* *= $\frac{[ \mathrm{C}_{2} \mathrm{H}_{5} \text{OH} ] [ CH_{3} \text{COOH} ]}{[ \mathrm{H}_{2} O ] [ CH_{3} COOC_{2} \mathrm{H}_{5} ]}$

All units cancel out

Therefore, K_c* *= 0.280 (no units)

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) $\rightleftharpoons$ 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 = $\frac{p^{2}\mathrm{SO}_{3}}{p^{2}\mathrm{SO}_{2}\times pO_{2}}$

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

So, the units of K_p* *are Pa^-1

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

• 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 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 = $\frac{1.71\times10^{-3}}{2.112\times10^{-2}}$ = 0.0810

I₂ = $\frac{2.91\times10^{-3}}{2.112\times10^{-2}}$ = 0.1378

HI = $\frac{1.65\times10^{-2}}{2.112\times10^{-2}}$ = 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 = $\frac{p^{2}\mathrm{HI}}{pH_{2}\times pI_{2}}$

• Step 5: Substitute the values into the equilibrium expression:

K_p = $\frac{78.13^{2}}{8.10\times13.78}$

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

• 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) $\rightleftharpoons$ 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