Edexcel A-Level Chemistry AS Paper 2, June 2024: Question 4

11 marks · Medium difficulty · Calculations

Calculate the enthalpy change of dimerisation of ethene to cyclobutane using mean bond enthalpies, evaluate accuracy, complete a Hess's law cycle, and determine the enthalpy of formation.

Practise this question

Question

Question 4 presents the reaction where two ethene molecules react to form cyclobutane. Part (a)(i) gives a table of mean bond enthalpies: C-H (413 kJ/mol), C-C (347 kJ/mol), and C=C (612 kJ/mol), asking to calculate the enthalpy change of dimerisation. Part (a)(ii) asks to explain why using bond enthalpies rather than mean bond enthalpies gives a more accurate value. Part (b) provides enthalpy data: enthalpy of combustion of cyclobutane (-2721 kJ/mol), enthalpy of formation of carbon dioxide (-394 kJ/mol), and enthalpy of formation of water (-286 kJ/mol). Part (b)(i) provides an incomplete Hess's law cycle diagram with boxes and arrows for the formation of C4H8(g) to be completed with reactants, products, and state symbols. Part (b)(ii) asks to calculate the enthalpy change of formation of cyclobutane.
Question text

4 This question is about the formation of cyclobutane, a gas at 298K.

(a) Cyclobutane can be made by the dimerisation of ethene.

H H

H H H H H H

C C C C

+ →

C C C C

H H

H H H H

H H

(i) Some mean bond enthalpy values are given in the table.

Bond Mean bond enthalpy / kJ mol–1

C H 413

C C 347

C C 612

Calculate the enthalpy change of this dimerisation by selecting appropriate

data from the table.

(3)

(ii) A different value for the enthalpy change of the reaction can be calculated

using bond enthalpies instead of mean bond enthalpies.

This value is more accurate.

Explain why the use of bond enthalpies gives a more accurate enthalpy

change value.

(2)

(b) The enthalpy change of formation of cyclobutane can be calculated using the

enthalpy change data in the table.

Enthalpy change Value / kJ mol–1

Enthalpy change of combustion of cyclobutane –2721

12 Enthalpy change of formation of carbon dioxide –394

Enthalpy change of formation of water*P76894A01232*–286

(i) Complete the enthalpy cycle using Hess’s Law.

Include reactants, products, state symbols and arrows in your cycle.

(4)

ΔfH

C4H8(g)

… + …

… + …

(ii) Calculate the enthalpy change of formation of cyclobutane. 13

*P76894A01332* (2)

Mark scheme

Show the mark scheme Mark scheme for Question 4. For (a)(i), breaks down bonds broken (2 × C=C = 1224) and made (4 × C-C = 1388) to give delta H = -164 kJ/mol (3 marks). For (a)(ii), explains that mean bond enthalpies are averages over many different molecules, while specific bond enthalpies apply to the particular bonds in these molecules (2 marks). For (b)(i), shows completed Hess's cycle with 4C(s) + 4H2(g) in the reactant box, downward arrows labeled with + 6O2, and the bottom box containing 4CO2(g) + 4H2O(l) (4 marks). For (b)(ii), calculation gives (4 × -394) + (4 × -286) - (-2721) = -2720 - (-2721) = +1 kJ/mol (2 marks).

Question

Answer Additional Guidance Mark

Number

4(a)(i) Example of calculation (3)

Either

• calculation of energy required to break two C=C (1) 2 × 612 = 1224 (kJ / kJ mol‒1)

(1) 4 × 347 = 1388 (kJ / kJ mol‒1)

• calculation of energy required to make 4 C‒C

(1) 1224 ‒ 1388 = ‒164 (kJ mol‒1)

• calculation of enthalpy change for the reaction including

the minus sign

Correct answer with some correct working scores (3)

Allow making and breaking all bonds

OR

(1) (8 × 413) + (2 × 612) = 4692 (kJ / kJ mol‒1)

• calculation of energy required to break all bonds

(1) (8 × 413) + (4 × 347) = 4528 (kJ / kJ mol‒1)

• calculation of energy required to make all bonds

(1) 4528 – 4692 = –164 (kJ mol-1)

• calculation of enthalpy change for the reaction

including the minus sign -1

Allow (2) for +164 (kJ mol ) with some working

Allow TE throughout

Correct answer with some working scores (3)

Question

Answer Additional Guidance Mark

Number

4(a)(ii) An explanation that makes reference to the following points: (2)

Either

• (because) mean bond enthalpies are the average value Allow mean bond enthalpies are averages

for all bonds of that type (1)

• bond enthalpy values are for a particular bond (in that (1) Allow bond enthalpies are specifically for that reaction

molecule) (and so are more accurate)

Or

• (because) the bond enthalpy of a C-H / C-C / C=C is

different in different molecules

• so the mean is less accurate than using the actual bond

enthalpies of each bond (in that molecule)

Question

Answer Additional Guidance Mark

Number

4(b)(i) (4)

All 8 points scores (4) ΔfH

4C(s) + 4H2(g) C4H8(g)

6 or 7 points scores (3)

4 or 5 points scores (2)

2 or 3 poinst scores (1)

(6)O2 (6)O2

• one or two arrows going down from LHS (6O2(g))

• C and H2 in reactants box 4CO (g) + 4H O(l)

• on the left-hand dotted line any one of

(6)O2 / –2720 / Do not award incorrect balancing numbers for O2 except 1

(4 ×) –394 and (4 ×) –286 / (4 ×) –680 / Ignore multipliers other than 4 even if incorrect

(4 ×) fH CO2 and (4 ×) fH H2O / Ignore multipliers other than 4 even if incorrect

(4 ×) cH C and (4 ×) cH H2 Ignore multipliers other than 4 even if incorrect

fH / cH Ignore just fH / cH

• CO2 and H2O in products box

Dependent on substances in both boxes or near miss, e.g. H

• balancing

instead of H2

Accept C(graphite) / C(s, graphite) for C(s)

• correct state symbols in both boxes

No TE

NOTE: Ignore an additional arrow going up as working (see

• one arrow going down from RHS

practice)

• on the right-hand dotted line any one of Do not award incorrect balancing numbers for O except 1

(6)O2 / –2721 /

Allow just cH

cH C4H8

Question

Answer Additional Guidance Mark

Number

4(b)(ii) Example of calculation (2)

• calculation of enthalpy of formation of 4 ((4 × ‒394) + (4 × ‒286)) = ‒2720 (kJ / kJ mol‒1)

moles of ‒1576 + ‒1144 = ‒2720 (kJ / kJ mol‒1)

carbon dioxide and water (1) May be seen on the cycle

‒2720 ‒ ‒2721 = (+)1 (kJ mol‒1)

• calculation of enthalpy change of formation

of cyclobutane (1)

Allow TE on the arrow in the cycle

Allow TE for M2 on adding 2721 to (subtracting -2721 from)

the calculated value in M1 or a value from the cycle

Allow (1) for ‒2721 ‒ ‒2720 = ‒1 (kJmol‒1)

Ignore SF

Final answer with some working scores 2

(Total for Question 4 = 11 marks)

How to answer it

Thermochemistry: Formation & Dimerisation of Cyclobutane

🎯 What this question tests

This question assesses core energetics topics from AS Topic 8 (Energetics):

  • Mean bond enthalpy calculations: Using the net change in covalent bonds broken and formed to find ΔrH.
  • Theoretical concepts: Explaining the limitation of mean bond enthalpies compared to compound-specific bond enthalpies.
  • Hess's Law cycles: Constructing an accurate, balanced thermochemical cycle connecting enthalpy of formation (ΔfH) and combustion data (ΔcH), including complete state symbols.
  • Indirect enthalpy calculations: Applying Hess's Law with correct signs, multipliers, and stoichiometry.
Part (a)(i) — 3 Marks

Enthalpy Change of Dimerisation from Bond Enthalpies

Reaction: 2 C₂H₄(g) → C₄H₈(g)

📐 Step-by-Step Calculation

Method 1: Shortcut (Bonds actually changing)

  1. Bonds broken: 2 × (C=C) = 2 × 612 = +1224 kJ mol⁻¹ [1 mark]
  2. Bonds formed: 4 × (C—C) = 4 × 347 = 1388 kJ mol⁻¹ [1 mark]
  3. ΔH = Σ(Bonds Broken) − Σ(Bonds Formed):
    ΔH = +1224 − 1388 = -164 kJ mol⁻¹ [1 mark]

Method 2: Breaking all bonds

  • Bonds broken: (8 × 413) + (2 × 612) = 4692 kJ mol⁻¹
  • Bonds formed: (8 × 413) + (4 × 347) = 4528 kJ mol⁻¹
  • ΔH = 4692 − 4528 = -164 kJ mol⁻¹

❌ Common Errors & Pitfalls

  • Missing the negative sign: Quoting +164 kJ mol⁻¹ will lose the final mark (award 2/3 only). Dimerisation forms more stable single bonds overall, releasing energy.
  • Miscounting bonds: Counting only 1 C=C broken instead of 2 (remember there are two ethene molecules), or miscounting the ring as 2 C—C instead of 4 C—C bonds.
  • Inverting the formula: Calculating formed − broken instead of broken − formed.
Mark Scheme Breakdown:
• M1: Calculation of energy required to break bonds (+1224 kJ mol⁻¹ or +4692 kJ mol⁻¹)
• M2: Calculation of energy required to make bonds (1388 kJ mol⁻¹ or 4528 kJ mol⁻¹)
• M3: Final value including the negative sign ( -164 kJ mol⁻¹ ). Full marks awarded for correct answer with working.
Part (a)(ii) — 2 Marks

Mean vs. Actual Bond Enthalpy

Explaining why compound-specific bond enthalpies provide higher accuracy

✅ Model Answer

  • Mean bond enthalpies are averages taken over a wide range of different chemical environments / molecules [1 mark].
  • Bond enthalpies for specific molecules take into account the actual, unique chemical environment of the bonds in that particular molecule (ethene and cyclobutane) [1 mark].

💡 Key Knowledge

The exact strength of a C—C or C—H bond depends on the surrounding atoms and geometry. For example, cyclobutane has considerable ring strain (bond angles ~90° instead of 109.5°), meaning its C—C bonds are weaker than an average C—C bond in an open-chain alkane. Mean values overlook this strain.

Examiner Insight: To secure both marks, you must clearly contrast both ideas: (1) state that mean bond enthalpies are averaged over many different molecules, and (2) state that specific bond enthalpies apply specifically to that exact bond/molecule.
Part (b)(i) — 4 Marks

Constructing the Hess's Law Enthalpy Cycle

Connecting Formation of Cyclobutane with Combustion Products

Hess Cycle Architecture
Reactants Box:
4C(s) + 4H₂(g)
─── ΔfH ───►
Target Product:
C₄H₈(g)
│
▼ Arrow DOWN from LHS
Label: + 6O₂(g) / -2720 kJ mol⁻¹
or 4ΔcH[C] + 4ΔcH[H₂]
│
▼ Arrow DOWN from RHS
Label: + 6O₂(g) / -2721 kJ mol⁻¹
or ΔcH[C₄H₈]
Bottom Box (Combustion Products):
4CO₂(g) + 4H₂O(l)

🧠 8 Marking Checkpoints (Sliding Scale)

  • 1. C and H₂ present in reactants box
  • 2. Correct balancing: 4C, 4H₂
  • 3. Correct state symbols: 4C(s) [or C(s, graphite)] and 4H₂(g)
  • 4. CO₂ and H₂O present in bottom box
  • 5. Bottom box balanced: 4CO₂ + 4H₂O
  • 6. Correct state symbols: 4CO₂(g) + 4H₂O(l)
  • 7. Arrow pointing down from LHS with correct value/expression (-2720 or 6O₂)
  • 8. Arrow pointing down from RHS with correct value/expression (-2721 or 6O₂)

❌ Common Errors That Cost Marks

  • State of water: Writing H₂O(g) instead of H₂O(l). Standard enthalpy of combustion/formation definitions require products in standard states at 298 K (water is a liquid!).
  • Arrow directions: Pointing combustion arrows up towards reactants/products instead of down towards the combustion products (CO₂ and H₂O).
  • State of carbon: Forgetting (s) on C.
Score Conversion: 8 points = 4 marks | 6–7 points = 3 marks | 4–5 points = 2 marks | 2–3 points = 1 mark.
Part (b)(ii) — 2 Marks

Calculation of ΔfH of Cyclobutane

Using the Hess cycle and combustion data

📐 Step-by-Step Solution

  1. Calculate combustion enthalpy of LHS elements:
    Note that ΔcH(C) = ΔfH(CO₂) = -394 kJ mol⁻¹
    and ΔcH(H₂) = ΔfH(H₂O) = -286 kJ mol⁻¹.
    LHS = 4(-394) + 4(-286) = -1576 + (-1144) = -2720 kJ mol⁻¹ [1 mark]
  2. Apply Hess's Law:
    Following the alternative route clockwise:
    ΔfH + ΔcH(C₄H₈) = LHS combustion products
    ΔfH + (-2721) = -2720
    ΔfH = -2720 − (-2721) = +1 kJ mol⁻¹ [1 mark]

🧠 Top Tip: Formula vs. Cycle

For cycles where all arrows go down to common combustion products:

ΔfH = ΣΔcH(reactants) − ΣΔcH(products)

ΔfH = [4(-394) + 4(-286)] − [-2721]
ΔfH = -2720 − (-2721) = +1 kJ mol⁻¹

Allow 1 mark for -1 kJ mol⁻¹ (sign error). Significant figures are not penalised here.

Mark Scheme Breakdown:
• M1: Summing combustion of 4 mol C and 4 mol H₂ to give -2720 kJ mol⁻¹ (can be credited if seen on cycle in b(i)).
• M2: Subtraction of -2721 to give +1 kJ mol⁻¹ (or 1 kJ mol⁻¹).

Topics

Physical Chemistry · Topic 8: Energetics I

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