AQA A-Level Chemistry Paper 2, 2018: Question 4
13 marks · Medium difficulty · State/Explain/Numerical
Calculate equilibrium amounts of B, C, and D, state the Kc expression and units, calculate the equilibrium concentration of A, and justify the effect of dilution on the amount of A.
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Question text
04 Compounds A and B react together to form an equilibrium mixture containing
compounds C and D according to the equation
2A + B ⇌ 3C + D
A beaker contained 40 cm3 of a 0.16 mol dm–3 aqueous solution of A.
04.1
9.5 × 10–3 mol of B and 2.8 × 10–2 mol of C were added to the beaker and the mixture
was left to reach equilibrium.
The equilibrium mixture formed contained 3.9 × 10–3 mol of A.
Calculate the amounts, in moles, of B, C and D in the equilibrium mixture.
[5 marks]
Amount of B mol
Amount of C mol
Amount of D mol
04.2 Give the expression for the equilibrium constant (Kc) for this equilibrium and its units.
[2 marks]
Kc
11 Units
04.3 A different equilibrium mixture of these four compounds, at a different temperature,
contained 0.21 mol of B, 1.05 mol of C and 0.076 mol of D in a total volume of
5.00 × 102 cm3 of solution.
*10* At this temperature the numerical value of Kc was 116
Calculate the concentration of A, in mol dm–3, in this equilibrium mixture.
Give your answer to the appropriate number of significant figures.
[3 marks]
Concentration of A mol dm–3
04.4 Justify the statement that adding more water to the equilibrium mixture in
Question 04.3 will lower the amount of A in the mixture.
[3 marks]
Mark scheme
Show the mark scheme
![Mark scheme for Question 04 showing a 13-mark breakdown. 04.1 awards marks for initial moles of A (6.4 × 10⁻³ mol), finding change x = 1.25 × 10⁻³ mol, and equilibrium amounts: B = 8.25 × 10⁻³ mol, C = 0.0318 mol, D = 1.25 × 10⁻³ mol. 04.2 awards 1 mark for Kc = [C]³[D] / ([A]²[B]) and 1 mark for units mol dm⁻³. 04.3 awards marks for rearranging for [A], calculating concentrations using 0.5 dm³, and final answer [A] = 0.17 mol dm⁻³ (2 sig figs). 04.4 awards 3 marks for noting that all concentrations decrease, equilibrium shifts to the side with more moles (RHS), to oppose the decrease in concentration / restore Kc.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2FmarkSchemes%252FQ18.L2.04-1790711447695.png%3Falt%3Dmedia&w=1000)
Question Answers Mark Additional Comments/Guidance
Initial amount of A = 6.4 × 10–3 If M1 wrong can score max 3
M1
Equ A = 6.4 × 10–3 2x x = 1.25 × 10–3 M2 If incorrect x can score max 3
04.1 B = 9.5 × 10–3 x = 8.25 × 10–3 M3 Allow 2 or more sig figs
C = 2.8 × 10–2 + 3x = 0.0318 M4
D = x = 1.25 × 10–3 M5
[C]3[D] 1
Kc = Penalise ( ) but mark on in 4.2 & 4.3
04.2 [A]2[B]
Units = mol dm–3 If K wrong no mark for units
1 c
[C]3[D] [C]3[D]
M1 for correct rearrangement [A]2= or [A] = If Kc wrong in 4.2 can score 1 for dividing by correct
Kc [B] Kc [B] M1 volume
04.3
M2 for division of mol of B, C and D by correct volume If Kc correct but incorrect rearrangement can score
Can see M2 1 for dividing by correct volume
4.2 [A]2= or 0.0289 or 0.0290
M3
M3 for final answer: [A] = 0.17 (must be 2 sfs)
(All) conc fall: (ignore dilution) 1 OR Kc = mole ratio × 1/V
Equm moves to side with more moles 1 If vol increases, mole ratio must increase
04.4 To oppose the decrease in conc 1 To keep Kc constant
If only conc of A falls CE=0
If pressure falls CE=0
Total 13
How to answer it
Equilibrium Quantities, Kc Calculations & Dilution Effects
📋 What This Question Tests
This multi-step question assesses core Year 1 Physical Chemistry equilibrium concepts:
- Constructing an initial-change-equilibrium (ICE) table using reaction stoichiometry ( 2A + B ⇌ 3C + D ).
- Writing homogenous equilibrium constant expressions ( Kc ) with correct square bracket convention and determining units.
- Rearranging equilibrium expressions to solve for an unknown concentration, converting volumes correctly ( cm³ → dm³ ), and applying appropriate significant figures.
- Explaining the effect of dilution (adding water) on an aqueous equilibrium position using Le Chatelier's Principle or the reaction quotient expression.
Equilibrium Mole Calculations (ICE Table)
Reaction: 2A + B ⇌ 3C + D
📐 Step-by-Step ICE Calculation
- Find initial moles of A:
Moles = concentration × volume (dm³) = 0.16 × (40 / 1000) = 6.4 × 10⁻³ mol - Determine mole change (2x):
Equilibrium moles of A = 3.9 × 10⁻³ mol
Moles of A reacted = 6.4 × 10⁻³ - 3.9 × 10⁻³ = 2.5 × 10⁻³ mol
Since 2 moles of A react per reaction unit ( 2x = 2.5 × 10⁻³ ):
x = 1.25 × 10⁻³ mol - Calculate equilibrium moles of B (1:1 ratio with x):
Initial B - x = 9.5 × 10⁻³ - 1.25 × 10⁻³ = 8.25 × 10⁻³ mol - Calculate equilibrium moles of C (initial moles present + 3x formed):
Initial C + 3x = (2.8 × 10⁻²) + 3(1.25 × 10⁻³) = 0.028 + 0.00375 = 0.0318 mol (or 3.18 × 10⁻² mol) - Calculate equilibrium moles of D (0 initial + x formed):
0 + x = 1.25 × 10⁻³ mol
| Species | 2A | B | 3C | D |
|---|---|---|---|---|
| Initial / mol | 6.4 × 10⁻³ | 9.5 × 10⁻³ | 2.8 × 10⁻² | 0 |
| Change / mol | -2x = -2.5 × 10⁻³ | -x = -1.25 × 10⁻³ | +3x = +3.75 × 10⁻³ | +x = +1.25 × 10⁻³ |
| Equilibrium / mol | 3.9 × 10⁻³ | 8.25 × 10⁻³ | 0.0318 | 1.25 × 10⁻³ |
✅ Final Answers
- Amount of B = 8.25 × 10⁻³ mol
- Amount of C = 0.0318 mol (or 3.18 × 10⁻² mol)
- Amount of D = 1.25 × 10⁻³ mol
❌ Common Traps & Errors
- Ignoring initial C: Students often assume initial product amounts are zero. The question explicitly states 2.8 × 10⁻² mol of C was added at the start!
- Power of 10 mismatch: C is given as 10⁻² , whereas A and B are 10⁻³ . Misreading the exponent leads to severe calculation loss.
- Forgetting stoichiometry: Forgetting that A has a coefficient of 2 (so change = 2x ) and C has a coefficient of 3 (change = +3x ).
Expression for Kc & Units
Reaction: 2A + B ⇌ 3C + D
✅ Correct Answer
Expression:
Kc = [C]³[D] / ([A]²[B])
Units:
mol dm⁻³
🧠 Exam Technique: Deriving Units
Substitute mol dm⁻³ into the expression and cancel terms:
Units = (mol dm⁻³)³ × (mol dm⁻³) / [(mol dm⁻³)² × (mol dm⁻³)]
= (mol dm⁻³)⁴ / (mol dm⁻³)³
= mol dm⁻³
Always show square brackets [ ] . Round brackets ( ) indicate partial pressures or non-concentrations and will lose the first mark!
Calculating Equilibrium Concentration of A
Given: Kc = 116 | Volume = 5.00 × 10² cm³
📐 Step-by-Step Calculation
- Convert volume to dm³:
Volume = 5.00 × 10² cm³ = 500 cm³ = 0.500 dm³ - Calculate equilibrium concentrations:
[B] = 0.21 / 0.500 = 0.42 mol dm⁻³
[C] = 1.05 / 0.500 = 2.10 mol dm⁻³
[D] = 0.076 / 0.500 = 0.152 mol dm⁻³ - Rearrange the Kc expression for [A]:
[A]² = ([C]³[D]) / (Kc × [B])
[A] = √(([C]³[D]) / (Kc × [B])) - Substitute and solve:
[A]² = ( (2.10)³ × 0.152 ) / ( 116 × 0.42 )
[A]² = ( 9.261 × 0.152 ) / 48.72 = 1.40767 / 48.72 ≈ 0.028893
[A] = √(0.028893) = 0.16998 mol dm⁻³ - Determine Significant Figures:
Given data: 0.21 mol (2 s.f.), 0.076 mol (2 s.f.), 1.05 mol (3 s.f.), 116 (3 s.f.).
The final answer must be given to 2 significant figures.
[A] = 0.17 mol dm⁻³
💡 Why Volume Does Not Cancel
The powers of concentration in the numerator equal 4 ( 3 + 1 ), while the denominator equals 3 ( 2 + 1 ). Because the total moles on each side differ, volume does not cancel out. You must divide every mole value by volume in dm³.
❌ Common Errors
- Forgetting the square root: Finding [A]² ≈ 0.029 and writing that as the final answer.
- Significant figures penalty: Leaving the answer as 0.170 or 0.1699 loses the final mark. The least precise input value (0.21 and 0.076) is 2 s.f.
Explaining the Effect of Adding Water (Dilution)
Target: Justify why adding water lowers the equilibrium amount (moles) of A
✅ Model Answer (Le Chatelier Approach)
- Adding water increases the total volume, so the concentrations of all species fall. (1 mark)
- The equilibrium shifts to the side with more moles (the right-hand side / product side: 4 moles vs 3 moles). (1 mark)
- The position of equilibrium shifts to oppose the decrease in concentration (consuming A, hence lowering the amount of A). (1 mark)
💡 Alternative Model Answer (Kc Quotient Approach)
- Expressing Kc in terms of moles and volume: Kc = (mole ratio) × (1/V) .
- When water is added, volume (V) increases.
- To keep Kc constant, the mole ratio must increase, so more products are formed and amount of A decreases.
❌ Critical Examiner Warnings (Contradictions & CE = 0)
- Chemical Error (CE = 0) for mentioning Pressure: This reaction occurs in aqueous solution. If a student mentions "pressure falls" or "moves to oppose decrease in pressure", a chemical error is penalised and scores 0/3.
- Chemical Error (CE = 0) for single species dilution: Writing that "only the concentration of A decreases" invalidates the whole question. Water dilutes all dissolved species simultaneously.
- Vague dilution statements: Saying simply "water dilutes the mixture" is not enough to secure Mark 1. You must state explicitly that concentrations fall.
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, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.