AQA A-Level Chemistry Paper 2, 2024: Question 3
10 marks · Medium difficulty · State/Explain/Numerical
Hydrocarbons: Cracking, NMR, Combustion Gas Volumes and Environmental Issues
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Question text
03 This question is about hydrocarbons.
03.1 Eicosane (C20H42) can be cracked by heating to 700 K in the presence of a catalyst.
The products are
• an aromatic hydrocarbon C8H10
• an alkane C6H14
• another alkane.
C20H42 ⟶ C8H10 + C6H14 +
Complete the equation for this reaction.
Give a suitable catalyst for this reaction.
[2 marks]
Catalyst
03.2 Figure 2 shows the 13C NMR spectrum for the aromatic hydrocarbon C H
8 10
Figure 2
Which of these is the structure of C8H10?
[1 mark]
Tick ( ) one box.
A B C D
03.3 Cracking can also be done without a catalyst, using a temperature of 1200 K and a
pressure of 7000 kPa
State the type of product that is formed in high percentage in this type of cracking.
[1 mark]
03.4 A sample of butane has a volume of 20 cm3 at room temperature and pressure.
The sample is burned completely in 1350 cm3 of air.
The final mixture is cooled to room temperature and pressure.
C4H10 + 6 O2 ⟶ 4CO2 + 5H2O
Calculate the total volume of gas in the final mixture.
Assume that air contains 21% by volume of oxygen.
[4 marks]
Total volume of gas remaining cm3
03.5 Natural gas is used in power stations to produce electricity.
Natural gas contains sulfur impurities. Sulfur dioxide forms when these impurities are
burned.
State an environmental problem caused by sulfur dioxide.
Give the formula of a compound that is used to help remove sulfur dioxide from the
combustion products.
[2 marks]
Environmental problem
Formula of compound
Mark scheme
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Question Answers Additional Comments/Guidelines Mark
M1 + 3C2H6
M2 Zeolite / Aluminosilicate / Aluminium oxide
03.1
(2 x AO1)
Option B 1
03.2
(1 x AO3)
Alkenes 1
03.3
(1 x AO1)
– A-LEVEL CHEMISTRY – – JUNE 2024
Alternative route:
M1 Initial volume O = 0.21 × 1350 = 283.5 (cm3) M1 Vol Air decreases by 6.5 × 20 = 130 cm3
33 17
M2 Volume of O2 remaining = M1 – (6.5 × 20) = 153.5 cm M2 = 1220 cm
33 4
03.4 M3 Volume of CO2 formed = 20 × 4 = 80 cm M3 Vol CO2 produced = 4 × 20 = 80 cm
(4 x AO2)
M4 Total volume of gas left = M2 + M3 + (0.79 × 1350) = 1300 cm3 M4 Total Vol Air + CO = 1220 + 80 = 1300 cm3
M1 Acid rain M1 Allow damages (limestone) buildings or statues
/ death of aquatic organisms / air pollution 2
03.5
M2 CaO or CaCO3 (2 x AO1)
How to answer it
Hydrocarbons: Cracking, ¹³C NMR, Combustion Gas Volumes & Desulfurisation
What this question tests
Using formulae to complete cracking equations and naming a suitable catalyst; interpreting a ¹³C NMR spectrum by counting carbon environments; recalling products of thermal cracking; applying gas-volume ratios for complete combustion in air (with water condensed at room temperature); and stating an environmental problem of SO₂ and the solid used to remove it from flue gases.
Catalyst and missing product
✅ Correct answers
- Equation: C₂₀H₄₂ → C₈H₁₀ + C₆H₁₄ + 3 C₂H₆
- Catalyst: zeolite / aluminosilicate / aluminium oxide.
❌ Common errors
- Writing “C₆H₁₈” as one molecule instead of 3 C₂H₆ (must be a viable alkane).
🧠 Exam technique
Balance carbon and hydrogen then check the product is a real compound with the correct general formula CnH2n+2.
Identify the aromatic C₈H₁₀ from its spectrum
✅ Choice
Option B (the only structure with five different carbon environments).
💡 Key knowledge
Five peaks (four in the 120–140 ppm aromatic region plus one aliphatic near ~20 ppm) mean five distinct ¹³C environments.
❌ Common misreads
Not linking number of peaks ↔ number of unique carbons (symmetry reduces the count).
Main type of product at 1200 K, 7000 kPa (no catalyst)
✅ Answer
Alkenes
💡 Why
High temperature and short residence time favour C–C scission to smaller, often unsaturated molecules.
Burning 20 cm³ butane completely in 1350 cm³ air, then cooling
📐 Calculations (volume ratios ≡ mole ratios at same T,P)
- Reaction: C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O
- O₂ in the air sample: 0.21 × 1350 = 283.5 cm³
- O₂ used: 6.5 × 20 = 130 cm³ → O₂ remaining = 153.5 cm³
- CO₂ formed: 4 × 20 = 80 cm³
- When cooled to room temperature, H₂O becomes liquid → not counted in gas volume.
- Other air components (79%) unchanged: 0.79 × 1350 = 1066.5 cm³
✅ Final total gas volume
1300 cm³ (= 153.5 O₂ left + 1066.5 “air other gases” + 80 CO₂)
❌ Common errors
- Trying to use the ideal gas equation unnecessarily.
- Including water vapour in the final gas volume after cooling.
Environmental issue and removal
✅ Answers
- Problem: acid rain (damages buildings/statues; harms aquatic life).
- Compound for removal: CaO or CaCO₃ (used in flue gas desulfurisation).
💡 Key knowledge
SO₂ + base → sulfite/sulfate; wet scrubbing with powdered limestone or lime removes SO₂ efficiently.
Final takeaways
💡 Key knowledge
- Catalytic cracking uses zeolites and can form aromatics plus alkanes/alkenes.
- ¹³C NMR: number of peaks = number of distinct carbon environments (consider symmetry).
- At fixed T,P: gas volume ratios = mole ratios in balanced equations.
🧠 Exam technique
- For gas-volume questions, keep water as liquid after cooling unless told otherwise.
- Always sanity-check unknown products: must obey general formulae and be realistic compounds.
❌ Common errors
- Treating “C₆H₁₈” as a product instead of “3 C₂H₆”.
- Picking an aromatic isomer with the wrong number of ¹³C signals.
Topics
Physical Chemistry · Organic Chemistry · 3.1.2 Amount of Substance · 3.3.2 Alkanes · 3.3.10 Aromatic Chemistry · 3.3.15 Nuclear Magnetic Resonance Spectroscopy
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.