AQA A-Level Chemistry Paper 2, 2024: Question 2

8 marks · Medium difficulty · State/Explain/Numerical

Equilibrium Calculations: Concentrations, Kc and Effect of Dilution

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AQA A-Level Chemistry Paper 2, 2024: Question 2
Question text

02 This question is about an equilibrium.

2P(aq) + Q(aq) ⇌ R(aq) + 3S(aq)

A 25.0 cm3 sample of a solution of P is added to a 20.0 cm3 sample of a solution of Q.

The mixture is allowed to reach equilibrium.

The amounts in the equilibrium mixture are

P = 0.0145 mol Q = 0.0275 mol R = 0.0115 mol S = 0.0345 mol

02.1 Calculate the amount, in moles, of P before the reaction with Q.

Use your answer to calculate the concentration, in mol dm–3, of P in the

initial 25.0 cm3 sample.

[2 marks]

Amount of P mol

Concentration7 mol dm–3

02.2 Give the expression for the equilibrium constant, Kc

Calculate the value of Kc and deduce its units.

[4 marks]

Kc

Value of Kc Units

02.3 Explain why the amount of S increases when water is added to the

equilibrium mixture.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2024: Question 2

Question Answers Additional Comments/Guidelines Mark

M1 mol P = 0.0145 + (2 × 0.0115) = 0.0375

02.1 M1

M2 [P] = = 1.50 mol dm–3 ECF from incorrect M1 (2 x AO2)

0.025

[R][S]3 M1 Must be square brackets in expression

M1 Kc = [P]2 [Q]

0.0115 0.0345 3 3

( )( ) (0.256)(0.767)3 M2 Inserts values and divides by volume in dm

M2 K = 0.045 0.045 or =

c 2 2

0.0145 0.0275 (0.322) (0.611)

( ) ( )

0.045 0.045

02.2

M3 Evaluates expression (4 x AO2)

M3 = 1.81 to 1.82

If no use of volume lose M2 but can score M3 for

0.0817

M4 units mol dm–3 M4 Allow consequential to their expression

– A-LEVEL CHEMISTRY – – JUNE 2024

M1 equilibrium shifts to side with most moles

M2 to oppose decrease in concentration of all reactants and products Allow

/ dilution of everything M2 oppose the decrease in concentration of S

OR

02.3

(2 x AO3)

M1 K is expressed as a function of concentrations and concentration K = RS3/P2Q × 1/V (where R,S etc are amounts)

c c

equals amount over volume. So, if V increases R and S must increase relative to

P and Q to keep Kc constant

M2 If Volume increases the amount of R and S must increase in order

to keep Kc constant.

How to answer it

Equilibrium from Amounts: finding initial moles, Kc & effect of dilution

What this question tests

Converting amounts at equilibrium into concentrations, back-calculating the initial amount of a reactant using stoichiometry for 2 P(aq) + Q(aq) ⇌ R(aq) + 3 S(aq), forming the correct Kc expression with concentrations (not moles), evaluating Kc and its units, and explaining how adding water shifts the equilibrium position using Le Chatelier’s principle (or the Kc expression).

Part (a)

Initial amount and concentration of P

📐 Calculations

  1. Extent of reaction from products: n(R) = 0.0115 mol → 2 mol P used per 1 mol R.
    P consumed = 2 × 0.0115 = 0.0230 mol .
  2. Equilibrium P = 0.0145 mol → Initial P = 0.0145 + 0.0230 = 0.0375 mol .
  3. Initial volume of P sample = 25.0 cm³ = 0.0250 dm³ → [P] = 0.0375 / 0.0250 = 1.50 mol dm⁻³ .

✅ Answers

  • Amount of P initially = 0.0375 mol
  • Concentration of P in the 25.0 cm³ sample = 1.50 mol dm⁻³
2 marks

🧠 Exam technique

Use the stoichiometric link via R: every 1 mol of R formed uses 2 mol of P.

Part (b)

Expression and value of Kc

💡 Expression

Kc = [R][S]³ / ( [P]² [Q] )

Square brackets mean concentrations in mol dm⁻³.

📐 Numbers (total volume = 25.0 + 20.0 = 45.0 cm³ = 0.0450 dm³)

  • [R] = 0.0115 / 0.0450 = 0.256
  • [S] = 0.0345 / 0.0450 = 0.767
  • [P] = 0.0145 / 0.0450 = 0.322
  • [Q] = 0.0275 / 0.0450 = 0.611

So Kc = (0.256 × 0.767³) / (0.322² × 0.611) = 1.81 (to 3 s.f.).

Units: overall power = 4 (top) − 3 (bottom) → mol dm⁻³ .

✅ Final values

Kc = 1.81   with units   mol dm⁻³

4 marks

❌ Common errors

  • Using amounts instead of concentrations (not dividing by 0.0450 dm³).
  • Missing the power of 3 on [S].
  • Forgetting units for Kc.
Part (c)

Why adding water increases the amount of S

💡 Le Chatelier explanation

Adding water increases the total volume → all concentrations decrease. The equilibrium shifts to the side with more moles of aqueous species to oppose this change. Products have 4 mol (R + 3 S) vs 3 mol (2 P + Q), so it shifts right, producing more S.

🧠 Alternative using Kc

Since Kc = [R][S]³/( [P]²[Q] ) is constant at fixed T, dilution lowers every concentration; to restore the same Kc, the numerator must increase relatively, so R and especially S increase.

2 marks

Final takeaways

💡 Key knowledge

  • Always convert amount → concentration before using Kc.
  • Count powers to get the units of Kc.
  • For dilution questions, compare total moles on each side or reason directly with the Kc expression.

❌ Common errors

  • Ignoring the combined volume (25.0 + 20.0 cm³).
  • Using equilibrium amounts to find initial P without the 2:1 stoichiometry via R.

Topics

Physical Chemistry · 3.1.2 Amount of Substance · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.