Edexcel A-Level Chemistry Paper 1, November 2020: Question 6

12 marks · Hard difficulty · Calculations

Calculate mean bond enthalpies, enthalpy of combustion, entropy changes, and reaction feasibility for prop-2-en-1-ol and a carbon dioxide reduction process.

Practise this question

Question

An exam question about prop-2-en-1-ol containing a chemical structure, a table of bond enthalpies, a combustion equation, an entropy change calculation based on a reaction equation with carbon dioxide and water, and several sub-questions involving bond enthalpy, entropy, and reaction feasibility.
Question text

6 Prop-2-en-1-ol is an unsaturated alcohol with the structure shown.

H

H

H C OH

C C

H H

(a) A student planned to use bond enthalpy data to calculate a value for the

enthalpy change of combustion of prop-2-en-1-ol.

(i) When researching the bond enthalpy data, the student claimed that it was not

necessary to find the value for the C=C bond as they could use the value for a

C–C bond and multiply it by two.

Explain why the student is incorrect.

(2)

(ii) Calculate a value for the enthalpy of combustion of prop-2-en-1-ol using the

data shown.

C3H6O(g) + 4O2(g) → 3CO2(g) + 3H2O(g)

Bond C−C C=C C−O C=O O−H C−H O=O

Bond enthalpy /

−1 347 612 358 805 464 413 498

kJ mol

(3)

(iii) Explain, in terms of entropy, why the combustion of prop-2-en-1-ol is always

feasible in the gaseous state.

(2)

(b) Chemists are researching a process to make ethanol and ethene directly from

16 carbon dioxide and water.

4CO (g) + 5H O(l) →*P62668A01624*CHCHOH(l)+CH(g)+6O(g)∆HO=+2778 kJ mol−1

22 3 2 2 4 2

CO2(g) H2O(l) CH3CH2OH(l) C2H4(g) O2(g)

SO / J K−1 mol−1 213.6 69.9 160.7 219.5 205.0

Calculate ∆SO for the reaction and hence determine whether the reaction is

total

feasible under standard conditions.

(5)

(Total for Question 6 = 12 marks)

Mark scheme

Show the mark scheme The mark scheme providing step-by-step guidance for calculating bond enthalpies, explaining why a double bond differs from two single bonds in terms of sigma and pi bonds, entropy calculations for systems and surroundings, and determining reaction feasibility.

How to answer it

Thermochemistry & Entropy Study Guide: Prop-2-en-1-ol Combustion

📌 What this question tests

This multi-part Edexcel A-Level Chemistry question evaluates core physical chemistry concepts including mean bond enthalpies, orbital bonding theory (sigma vs pi bonds), entropy changes in systems and surroundings ( ΔS_system and ΔS_surroundings ), and reaction feasibility ( ΔS_total / ΔG ).

Part (a)(i): Bond Enthalpy Theory

Explain why the student's assumption about C=C and C-C bonds is incorrect. (2 marks)

✅ Correct Answer

  • The C=C double bond is weaker than two individual C-C single bonds.
  • A C=C double bond consists of one sigma (σ) bond and one pi (π) bond, whereas two C-C bonds would be two sigma bonds.

💡 Key Knowledge

Pi bonds involve sideways overlap of p-orbitals, resulting in lower electron density between the nuclei and less effective orbital overlap compared to the head-on overlap in sigma bonds. Hence, π bonds are weaker.

Mark Scheme Breakdown: 1 mark for stating C=C is weaker than 2 × C-C. 1 mark for explaining it contains a pi and a sigma bond rather than two sigma bonds. Examiner note: Ignore references to pi bonds formed by sideways/less effective orbital overlap without the direct comparison of bond strength and bond type.

Part (a)(ii): Enthalpy of Combustion Calculation

Calculate a value for the enthalpy of combustion of prop-2-en-1-ol. (3 marks)

C₃H₆O(g) + 4O₂(g) → 3CO₂(g) + 3H₂O(g)

📐 Step-by-Step Calculation

  1. Reactant bonds broken: 5(C-H) + 1(C=C) + 1(C-C) + 1(C-O) + 1(O-H) + 4(O=O)
    = 5(413) + 612 + 347 + 358 + 464 + 4(498) = 5838 kJ mol⁻¹ (1 mark)
  2. Product bonds formed: 6(C=O) + 6(O-H)
    = 6(805) + 6(464) = 7614 kJ mol⁻¹ (1 mark)
  3. Enthalpy change (ΔH): Bonds broken − Bonds formed = 5838 − 7614 = -1776 kJ mol⁻¹ (1 mark)

❌ Common Errors & Exam Technique

  • Counting mistakes: Forgetting to count bonds inside functional groups (like the O-H bond in the alcohol or C-O bonds). Always draw out or check the full displayed formula.
  • Sign convention trap: Remember that energy input (breaking bonds) is positive and energy release (forming bonds) is negative. Mixing up the subtraction order yields +1776 , which loses the final mark.
Mark Scheme Breakdown: 1 mark for reactant bonds total, 1 mark for product bonds total, 1 mark for final subtraction with correct sign and units ( kJ mol⁻¹ ). Correct answer with no working scores full marks.

Part (a)(iii): Feasibility & Entropy

Explain, in terms of entropy, why combustion of prop-2-en-1-ol is always feasible in the gaseous state. (2 marks)

✅ Correct Answer

  • ΔS_total is always positive; OR both ΔS_surroundings and ΔS_system are positive.
  • Alternatively: ΔG is always negative because ΔH is negative and ΔS_system is positive.

💡 Key Knowledge

Look at the stoichiometry of the equation: 1 mol (g) + 4 mol (g) → 3 mol (g) + 3 mol (g) . Total gaseous moles increase from 5 to 6 moles, creating an increase in disorder ( ΔS_system > 0 ). Combustion is also strongly exothermic ( ΔH < 0 ), meaning ΔS_surroundings is also positive.

Mark Scheme Breakdown: 1 mark for stating ΔS_total is positive (or equivalent ΔG argument), and 1 mark for explaining why both contributing entropy terms support this.

Part (b): Full Feasibility Calculation

Calculate ΔS_total for the reaction and determine whether it is feasible under standard conditions. (5 marks)

4CO₂(g) + 5H₂O(l) → CH₃CH₂OH(l) + C₂H₄(g) + 6O₂(g)     ΔHº = +2778 kJ mol⁻¹

📐 Step-by-Step Calculation

  1. Calculate ΔS_system (S_products − S_reactants):
    ΔS_sys = [(1 × 160.7) + (1 × 219.5) + (6 × 205.0)] − [(4 × 213.6) + (5 × 69.9)]
    = [160.7 + 219.5 + 1230.0] − [854.4 + 349.5]
    = 1610.2 − 1203.9 = +406.3 J K⁻¹ mol⁻¹ (2 marks: expression + calculation)
  2. Calculate ΔS_surroundings (−ΔH / T):
    Convert ΔH to Joules: +2778 kJ mol⁻¹ = +2778000 J mol⁻¹
    ΔS_surr = −(2778000) / 298 = −9322.15 J K⁻¹ mol⁻¹ (2 marks: expression + calculation)
  3. Calculate ΔS_total and conclude:
    ΔS_total = ΔS_sys + ΔS_surr = 406.3 + (−9322.15) = −8915.85 J K⁻¹ mol⁻¹
    Conclusion: Since ΔS_total is negative, the reaction is not feasible under standard conditions. (1 mark)

❌ Common Calculation Traps

  • Unit mismatch: ΔH is given in kJ mol⁻¹ , whereas entropy values are in J K⁻¹ mol⁻¹ . You must multiply ΔH by 1000 before dividing by temperature ( 298 K ).
  • State symbols matter: Ensure you use the entropy value for liquid water H₂O(l) = 69.9 , not steam.
Mark Scheme Breakdown: 5 marks total — (1) expression for ΔS_system , (2) correct calculation of ΔS_system , (3) expression for ΔS_surroundings (including negative sign and conversion to Joules), (4) correct calculation of ΔS_surroundings , (5) correct ΔS_total value with units and valid feasibility statement based on sign.

Topics

Physical Chemistry · Organic Chemistry · Topic 8: Energetics I · Topic 13: Energetics II · Topic 6: Organic Chemistry I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.