AQA A-Level Chemistry Paper 2, 2017: Question 5

9 marks · Medium difficulty · State/Explain/Numerical

Draw the structural formula of the diester formed from ethanoic acid and ethane-1,2-diol, complete an equilibrium moles table, write the Kc expression, explain why volume cancels, and calculate the equilibrium moles of ethanoic acid.

Practise this question

Question

Question 05 shows the reaction 2CH3COOH(l) + HOCH2CH2OH(l) ⇌ C6H10O4(l) + 2H2O(l). Part 05.1 asks to draw a structural formula for the diester C6H10O4 (1 mark). Part 05.2 provides Table 1 with initial amounts: CH3COOH = 0.470 mol, HOCH2CH2OH = 0.205 mol, C6H10O4 = 0 mol, H2O = 0 mol, and equilibrium amount of CH3COOH = 0.180 mol, asking to complete the remaining equilibrium amounts (3 marks). Part 05.3 asks to write an expression for Kc and justify why the total volume does not need to be measured (2 marks). Part 05.4 provides Table 2 at a different temperature where Kc = 6.45, with equilibrium amounts: HOCH2CH2OH = 0.264 mol, C6H10O4 = 0.802 mol, and H2O = 1.15 mol, asking to calculate the amount in mol of ethanoic acid to the appropriate number of significant figures (3 marks).
Question text

05 Ethanoic acid and ethane-1,2-diol react together to form the diester (C6H10O4)

as shown.

2CH3COOH(l) + HOCH2CH2OH(l) ⇌ C6H10O4(l) + 2H2O(l)

05.1 Draw a structural formula for the diester C6H10O4

[1 mark]

05.2 A small amount of catalyst was added to a mixture of 0.470 mol of

ethanoic acid and 0.205 mol of ethane-1,2-diol.

The mixture was left to reach equilibrium at a constant temperature.

Complete Table 1.

Table 1

Amount in the mixture / mol

CH3COOH HOCH2CH2OH C6H10O4 H2O

At the start 0.470 0.205 0 0

At equilibrium 0.180

[3 marks]

Do

Space for working 11 ou

05.3 Write an expression for the equilibrium constant, Kc, for the reaction.

The total volume of the mixture does not need to be measured to allow a

correct value for Kc to be calculated.

Justify this statement.

[2 marks]

Expression

Justification

05.4 A different mixture of ethanoic acid, ethane-1,2-diol and water was prepared

and left to reach equilibrium at a different temperature from the experiment in

Question 5.2

The amounts present in the new equilibrium mixture are shown in Table 2.

Table 2

Amount in the mixture / mol

CH3COOH HOCH2CH2OH C6H10O4 H2O

At new To be

0.264 0.802 1.15

equilibrium calculated

The value of Kc was 6.45 at this different temperature.

Use this value and the data in Table 2 to calculate the amount, in mol, of

ethanoic acid present in the new equilibrium mixture.

Give your answer to the appropriate number of significant figures.

[3 marks]

Amount of ethanoic acid mol

Mark scheme

Show the mark scheme Mark scheme for Question 05. 05.1: Structural formula showing H3C-C(=O)-O-CH2-CH2-O-C(=O)-CH3 (1 mark). 05.2: Mol HOCH2CH2OH = 0.0600, Mol C6H10O4 = 0.145, Mol H2O = 0.290 (1 mark each, total 3 marks). 05.3: Kc = [C6H10O4][H2O]^2 / ([CH3COOH]^2[HOCH2CH2OH]) (1 mark); Justification: volume cancels out OR equal number of moles/molecules on each side of the equation (1 mark). 05.4: Rearranging for (Mol CH3COOH)^2 = (0.802 * 1.15^2) / (6.45 * 0.264) = 0.623 (M1), taking the square root gives Mol CH3COOH = 0.789 mol to 3 significant figures (M2, M3, total 3 marks).

Question Answers Mark Additional Comments/Guidance

O Allow CH3COOCH2CH2OOCCH3

OR CH3COOCH2CH2OCOCH3

C CH2 O CH3 O

05.1 H3C O CH2 C 1

O

O O

OR O

-2 – – –

Mol HOCH2CH2OH = 6.00 × 10 OR 0.06(00) 1

05.2 Mol C H O = 1.45 × 10-1 OR 0.145 1

6 10 4

Mol H O = 2.90 × 10-1 OR 0.29(0) 1

[ester] [H O]2 Allow words for acid and alcohol

(K =) 2

c 2 1

[CH3COOH] [HOCH 2CH2OH]

The volume cancels out (Penalise a contradictory justification

05.3 from expression if the volumes do not cancel out)

OR

there are equal no of moles on each side of the equation

OR

there are equal no of molecules on each side of the equation

(8.02 10 1 /V )(1.15 /V )2 0.789 scores 3

(Mol CH COOH/V)2 =

31 (8.02 10 1)(1.15)2

6.45 (2.64 10 /V ) Allow without V : (nCH COOH)2 =

M1 3 1

6.45 (2.64 10 )

If (nCH COOH)2 =0.623 then award M1 and M2

3 19 of 32

05.4 If Kc is correct in 05.3 but incorrect rearrangement, then

(8.02 10 1) (1.15)2

CE=0 except if upside down rearrangement then M3 only

Mol CH COOH = √ = 0.623 M2

31 awarded for 1.27

6.45 (2.64 10 )

If Kc is incorrect in 05.3 then only M1 can be awarded for

Mol CH3COOH = 0.789 (must be 3 sfs) Allow 0.788 – 0.790 M3 correct rearrangement.

Total 9

How to answer it

AQA A-Level Chemistry • Physical & Organic Chemistry

Esterification, Diesters & Equilibrium Constant (Kc) Calculations

What this question tests

This multi-step question assesses core Year 2 equilibrium concepts coupled with carbonyl/ester chemistry:

  • Structural formulas: Deducing the structure of a diester formed from a dicarboxylic alcohol and a monocarboxylic acid.
  • Equilibrium quantities (ICE): Applying stoichiometry (2 : 1 : 1 : 2 ratio) to calculate equilibrium moles of reactants and products.
  • Kc expressions & unit cancellation: Writing homogenous equilibrium constant expressions and explaining why total volume cancels out.
  • Algebraic rearrangement of Kc: Solving for an unknown squared reactant term, handling square roots, and quoting answers to correct significant figures.

Question 05.1: Structural Formula of the Diester

1 Mark • Target: Organic Structure Representation

✅ Correct Answer

Any unambiguous structural or displayed formula of ethane-1,2-diyl diethanoate:

CH₃COOCH₂CH₂OOCCH₃

or displayed / skeletal formula:

CH₃–C(=O)–O–CH₂–CH₂–O–C(=O)–CH₃

[1 Mark] for correct structure showing both ester linkages correctly orientated.

💡 Key Knowledge

  • Ethane-1,2-diol (HOCH₂CH₂OH) has two –OH groups.
  • Each –OH reacts with one molecule of ethanoic acid (CH₃COOH) via condensation (loss of H₂O).
  • Overall reaction produces a diester and 2H₂O.

❌ Common Errors

  • Reversed ester oxygen connectivity: Writing –COOCH₂CH₂COOCH₃ (which would imply a dicarboxylic acid reacted with methanol, completely wrong stoichiometry and formula).
  • Omitting carbons or writing an ether linkage instead of an ester.

🧠 Exam Technique

Always count your atoms against the given molecular formula: C₆H₁₀O₄ .

  • Carbons: 2 (from diol) + 2×2 (from ethanoic acid) = 6 C.
  • Hydrogens: 4 (from –CH₂CH₂–) + 2×3 (from –CH₃) = 10 H.
  • Oxygens: 4 O.

Question 05.2: Calculating Equilibrium Moles (ICE Method)

3 Marks • Target: Reaction Stoichiometry & Amounts

📐 Step-by-Step Calculation

Equation: 2CH₃COOH + HOCH₂CH₂OH ⇌ C₆H₁₀O₄ + 2H₂O

Stage CH₃COOH HOCH₂CH₂OH C₆H₁₀O₄ H₂O
Initial (I) 0.470 mol 0.205 mol 0 mol 0 mol
Change (C) − 0.290 mol − 0.145 mol + 0.145 mol + 0.290 mol
Equilibrium (E) 0.180 mol (given) 0.060 mol 0.145 mol 0.290 mol
  1. Step 1: Determine moles of ethanoic acid reacted
    Δn(CH₃COOH) = 0.470 − 0.180 = 0.290 mol reacted.
  2. Step 2: Ethane-1,2-diol at equilibrium
    Ratio of acid to diol is 2 : 1.
    Moles of diol reacted = 0.290 / 2 = 0.145 mol.
    Equilibrium moles = 0.205 − 0.145 = 0.060 mol (or 6.00 × 10⁻² mol). [1 Mark]
  3. Step 3: Diester at equilibrium
    Ratio of acid to diester is 2 : 1.
    Moles of diester formed = 0.290 / 2 = 0.145 mol (or 1.45 × 10⁻¹ mol). [1 Mark]
  4. Step 4: Water at equilibrium
    Ratio of acid to water is 2 : 2 (1 : 1).
    Moles of water formed = 0.290 mol (or 2.90 × 10⁻¹ mol). [1 Mark]

❌ Common Errors

  • Ignoring the 2 : 1 stoichiometry: Assuming 1 mol of acid reacts with 1 mol of diol, yielding 0.290 mol of diol reacted (which would give an impossible negative equilibrium amount!).
  • Forgetting water is a product: Forgetting that 2 moles of H₂O are produced alongside the diester.

🧠 Exam Technique

Always write down an explicit ICE table (Initial, Change, Equilibrium) in the rough working area. Note clearly that the Change row must strictly obey the balancing numbers in the chemical equation: −2x, −x, +x, +2x .

Question 05.3: Kc Expression & Volume Cancellation

2 Marks • Target: Equilibrium Constant & Justification

✅ Correct Answer

Expression:

Kc = [C₆H₁₀O₄][H₂O]² / ([CH₃COOH]²[HOCH₂CH₂OH])

[1 Mark] (Square brackets required; words like [ester] or [alcohol] accepted).

Justification:

The volume terms cancel out because there are an equal number of moles / molecules (3) on each side of the balanced chemical equation.

[1 Mark] for stating volume cancels out OR equal moles on both sides.

💡 Key Knowledge: Why Volume Cancels

Concentration = moles / volume (V):

Kc = [(nester/V) × (nH₂O/V)²] / [(nacid/V)² × (ndiol/V)]

Numerator volume term: (1/V)³

Denominator volume term: (1/V)³

Since both powers of V are equal (V³ in numerator and denominator), all V terms cancel completely!

❌ Common Errors

  • Forgetting powers: Missing the squared terms for [H₂O]² or [CH₃COOH]².
  • Round brackets instead of square: Using ( ) instead of [ ] for concentration expressions (strictly penalised).
  • Vague justification: Writing just "volume doesn't affect equilibrium" rather than specifically pointing out that the volume terms cancel or referencing the equal moles on both sides.

🧠 Exam Technique

Check the powers against the balanced equation: 2 moles reactants → 2 moles products? No! Here it's 2 + 1 = 3 moles on the left, and 1 + 2 = 3 moles on the right. Sum of powers = 3 on top, 3 on bottom.

Question 05.4: Calculating Equilibrium Amount of Ethanoic Acid

3 Marks • Target: Rearranging Kc & Significant Figures

📐 Step-by-Step Calculation

Given values at new equilibrium:

  • Kc = 6.45
  • n(HOCH₂CH₂OH) = 0.264 mol
  • n(C₆H₁₀O₄) = 0.802 mol
  • n(H₂O) = 1.15 mol
  1. Step 1: Substitute moles directly into Kc (since V cancels)

    6.45 = [ (0.802) × (1.15)² ] / [ (n(CH₃COOH))² × (0.264) ]

    [M1] for correct substitution into Kc expression (with or without V).

  2. Step 2: Rearrange to make (n(CH₃COOH))² the subject

    (n(CH₃COOH))² = [ 0.802 × (1.15)² ] / [ 6.45 × 0.264 ]

    (n(CH₃COOH))² = 1.060645 / 1.7028 = 0.62288...

    Take the square root of both sides:

    n(CH₃COOH) = √(0.62288...) = 0.7892... mol

    [M2] for correctly rearranging and taking the square root (evaluating √(0.623)).

  3. Step 3: Round to appropriate significant figures

    The input data are given to 3 significant figures (0.264, 0.802, 1.15, 6.45).

    Therefore, final answer must be quoted to 3 significant figures:

    Amount of ethanoic acid = 0.789 mol

    [M3] for 0.789 (range 0.788 – 0.790, strictly 3 sig figs).

❌ Common Errors & Pitfalls

  • Forgetting to take the square root: Leaving the answer as 0.623 mol.
  • Inverting the rearrangement: Getting 1.7028 / 1.0606 = 1.605, leading to √(1.605) = 1.27 mol. (Examiner rule: chemical error CE=0 if inverted, only allowed M3 if correctly rounded).
  • Significant figures penalty: Writing 0.79 mol (2 s.f.) or 0.7892 mol (4 s.f.). The question explicitly says "appropriate number of significant figures".
  • Premature rounding: Rounding intermediate numbers too early leading to values outside 0.788 – 0.790.

🧠 Top Examiner Tips

  • When an exam asks for "an appropriate number of significant figures", look at all given numerical values in the question table and text. If all are 3 s.f., your final answer must be 3 s.f.
  • Full marks (3/3) are awarded immediately for writing the correct answer 0.789 with no intermediate working, but showing each algebraic step protects your marks if you make a slip on the calculator!

Topics

Physical Chemistry · Organic Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.3.9 Carboxylic Acids and Derivatives

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.