Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 9

9 marks · Medium difficulty · Calculations

Complete an enthalpy cycle and calculate the standard enthalpy change of formation of butane from combustion data, and state the standard conditions.

Practise this question

Question

Question 9 consists of three parts regarding the complete combustion of butane, C4H10(g) + 6.5 O2(g) -> 4 CO2(g) + 5 H2O(l). Part (a)(i) provides an incomplete enthalpy cycle with reactant and product boxes at the top and an empty box at the bottom, worth 4 marks. Part (a)(ii) gives standard enthalpy of combustion values: C(graphite/s) = -394 kJ mol^-1, H2(g) = -286 kJ mol^-1, and butane(g) = -2877 kJ mol^-1, asking to calculate the enthalpy of formation of butane (3 marks). Part (b) asks to state the relevant conditions indicated by the standard symbol in the standard enthalpy change of combustion of butane (2 marks).

Mark scheme

Show the mark scheme Mark scheme for Question 9. Part (a)(i) awards 4 marks: 1 for correct elements and states in the bottom box (4C(graphite/s) + 5H2(g) + 6.5 O2(g)), 1 for balancing, 1 for both arrows pointing upwards, and 1 for correctly labelling the arrows. Part (a)(ii) awards 3 marks: M1 for finding the RHS enthalpy (4 x -394 + 5 x -286 = -3006 kJ mol^-1), M2 for correct expression (-3006 + 2877), and M3 for evaluation giving -129 kJ mol^-1 with sign and units. Part (b) awards 2 marks for two conditions: pressure = 100 kPa / 1 atm, reactants and products in standard states, and temperature = 298 K.

How to answer it

Hess's Law: Enthalpy of Formation of Butane

📌 What this question tests

This question assesses your ability to construct an indirect Hess's Law cycle linking formation and combustion reactions, accurately balance equations with fractional coefficients and state symbols, calculate an unknown enthalpy change of formation (ΔfH) using enthalpy of combustion data, and state the exact definitions and conditions signified by the standard symbol (⦵).

Question 9(a)(i) • 4 Marks

Constructing the Enthalpy Cycle

Connecting elements, combustion reactants, and complete combustion products

Top Left: C₄H₁₀(g) + 6½O₂(g)   ───[ ΔcH [C₄H₁₀(g)] ]───►   Top Right: 4CO₂(g) + 5H₂O(l)

▲                      ▲

[ ΔfH [C₄H₁₀(g)] ]            [ 4 ΔcH [C(s)] + 5 ΔcH [H₂(g)] ]

Bottom Box: 4C(graphite/s) + 5H₂(g) + 6½O₂(g)

✅ Correct Answer Breakdown

  • Bottom box species & state symbols [1 mark]:
    4C(s) (or graphite) + 5H₂(g) + 6½O₂(g) .
  • Balancing [1 mark]: Exactly 4 carbons, 10 hydrogens (5H₂), and 13 oxygens (6½O₂) balancing both top boxes.
  • Arrow directions [1 mark]: Both vertical arrows point upwards from the elements to the compounds above.
  • Arrow labels [1 mark]:
    • Left: ΔfH [C₄H₁₀(g)]
    • Right: 4 ΔcH[C(s)] + 5 ΔcH[H₂(g)] (or 4(-394) + 5(-286) )

🧠 Exam Technique: Cycle Direction

Remember that enthalpy of formation is from elements to compound:

  • Because elements are at the bottom, formation arrows point UP.
  • Carbon plus oxygen burns to CO₂, and hydrogen plus oxygen burns to H₂O. Therefore, burning the elements directly yields the same products as the right box, so that arrow also points UP!
  • Alternatively, the right arrow represents the formation of 4 mol of CO₂(g) and 5 mol of H₂O(l).

❌ Common Errors in Cycle Questions

  • Missing oxygen in the bottom box: Candidates often write only 4C(s) + 5H₂(g) . If oxygen isn't balanced across all corners, mass is not conserved! (Note: The mark scheme allows + 6½O₂ to be written alongside the arrows, but it must be accounted for).
  • Missing or incorrect state symbols: Writing C instead of C(s) or C(graphite) , and H₂ without (g) .
  • Arrow pointing downwards: Drawing combustion arrows pointing down from top to bottom. Here, elements are at the bottom and combustion products are at the top-right, so the arrow must go upwards.
Question 9(a)(ii) • 3 Marks

Calculating the Enthalpy of Formation

Applying Hess's Law using combustion values

📐 Step-by-Step Calculation

According to Hess's Law, the energy change for the direct route equals the energy change for the indirect route:

Bottom Elements ➔ Products (Right):
ΔfH[butane] + ΔcH[butane] = 4 ΔcH[C] + 5 ΔcH[H₂]

  1. Calculate the right-hand arrow enthalpy (combustion of constituent elements):
    ΔHRHS = [4 × (-394)] + [5 × (-286)]
    ΔHRHS = -1576 + (-1430) = -3006 kJ mol⁻¹
    awarded Mark 1
  2. Rearrange Hess's Law to solve for ΔfH[butane]:
    ΔfH[butane] = ΔHRHS - ΔcH[butane]
    ΔfH[butane] = -3006 - (-2877) = -3006 + 2877
    awarded Mark 2 (correct use of data with cycle)
  3. Final Evaluation with sign and units:
    ΔfH[butane] = -129 kJ mol⁻¹
    awarded Mark 3 (correct value, negative sign, and kJ mol⁻¹)

💡 Alternative Formula Check

For combustion data:

ΔrH = Σ ΔcH(reactants) - Σ ΔcH(products)

For the formation of butane: 4C(s) + 5H₂(g) ➔ C₄H₁₀(g)

ΔfH = [4 × ΔcH(C) + 5 × ΔcH(H₂)] - [ΔcH(C₄H₁₀)]
ΔfH = -3006 - (-2877) = -129 kJ mol⁻¹

❌ Calculation Traps & Penalties

  • Sign inversion error: Calculating +129 kJ mol⁻¹ instead of -129 kJ mol⁻¹ . This typically scores 2 out of 3 marks. Be vigilant when subtracting a negative number: -(-2877) = +2877 .
  • Forgetting stoichiometric multipliers: Multiplying by incorrect coefficients (e.g. failing to multiply -394 by 4, or -286 by 5).
  • Missing units: Always write kJ mol⁻¹ explicitly.
Question 9(b) • 2 Marks

Standard Symbol (⦵) Conditions

What exact conditions are specified by the standard state symbol?

✅ Any Two Points Required [2 Marks]

  • Pressure: 100 kPa (or 10² kPa / 100 000 Pa / 101 kPa / 1 atm ) [1 mark]
  • Temperature: A specified temperature, standardly 298 K (or 25 °C ) [1 mark]
  • Standard states: Reactants and products in their normal/standard physical states (under standard conditions) [1 mark]

❌ Examiner Watch-outs

  • Do NOT mention concentration: References to solution concentration (e.g., 1.0 mol dm⁻³) are ignored here because all species are pure gases, liquids, or solids—there are no solutions.
  • Do NOT mention moles: Saying "1 mole of reactants" is part of the reaction definition, not a condition indicated by the ⦵ symbol.
  • Contradictory values: If you write more than one value for temperature or pressure (e.g., "100 kPa and 100 atm"), both must be correct or no mark is awarded.

Topics

Physical Chemistry · Topic 8: Energetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.