Equilibrium Constant Calculations (Cambridge (CIE) AS 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)  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)] = = 0.470 mol dm^-3

[C₂H₅OH (l)] = = 0.070 mol dm^-3

[CH₃COOC₂H₅ (l)] = = 0.364 mol dm^-3

[H₂O (l)] = = 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 = 

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

K_c = 

K_c = 4.03

• Step 5: Deduce the correct units for K_c:

K_c = 

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 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)] = = 0.0654 mol dm^-3

[C₂H₅OH (l)] = = 0.0654 mol dm^-3

[CH₃COOC₂H₅ (l)] = = 0.0346 mol dm^-3

[H₂O (l)] = = 0.0346 mol dm^-3

• 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 K_c:

All units cancel out

Therefore, K_c = 0.28

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 = 

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

K_p = 

K_p = 9.1 x 10^–6

• Step 3: Deduce the correct units of K_p:

K_p = 

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 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

1. Step 2: Calculate the mole fraction of each gas:

H₂ =  = 0.0810

I₂ =  = 0.1378

HI =  = 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 = 

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

K_p = 

K_p = 54.7

• Step 6: Deduce the correct units for K_p:

K_p = 

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)  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