OxfordAQA A-Level Chemistry: Alkanes (9620) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set tests section 3.3.2 Alkanes from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, across points 3.3.2.1 (distilling crude oil), 3.3.2.2 (cracking), 3.3.2.3 (combustion and pollution) and 3.3.2.4 (methane and chlorine). The whole section is International AS, so no question here is International A-level only. It is assessed in Unit 2 (Organic 1 and Physical 1), and Unit 5 can draw on any content.

Learn the content first from the alkanes study guide and the alkanes revision notes. See also the course hub, the printable checklist and a free 10-minute diagnostic. Flue-gas masses are practised in the Group 2 practice questions.

Data where needed: gas constant R = 8.31 J K⁻¹ mol⁻¹; relative atomic masses H 1.0, C 12.0, O 16.0, S 32.1.

Questions

1. State two ways in which the conditions for catalytic cracking differ from those for thermal cracking. [2]

2. Alkane P has six carbon atoms in an unbranched chain. Alkane Q has eighteen. Both are present in a crude oil sample that is fractionally distilled. State which alkane is collected higher up the column, and explain why. [3]

3. Heptadecane, C₁₇H₃₆, is cracked.

(a) One molecule of heptadecane forms one molecule of an alkane with eleven carbon atoms, one molecule of ethene and one molecule of another alkene only. Give the equation. [1] (b) State which type of cracking a company would choose if it wanted mainly alkenes. [1] (c) Name the catalyst used in catalytic cracking and state the two main types of product it is used to make. [2]

4. A refinery produces far more heavy gas oil (long-chain alkanes) than its customers buy, and it cannot make enough fuel for cars. Explain why the refinery cracks some of its heavy gas oil. [4]

5. Write balanced equations for:

(a) the complete combustion of hexadecane, C₁₆H₃₄ [1] (b) the incomplete combustion of propane in which carbon monoxide and carbon form in a 1 : 2 mole ratio, together with water [1] (c) the complete combustion of butan-1-ol, C₄H₉OH. [1]

6. 15 cm³ of a gaseous alkane burns completely in 120 cm³ of oxygen (an excess). After cooling to room temperature, so that water is liquid, 82.5 cm³ of gas is left. Sodium hydroxide solution absorbs the carbon dioxide only, leaving 22.5 cm³ of oxygen. All volumes are measured at the same temperature and pressure. Deduce the molecular formula of the alkane, showing your working. [4]

7. A gas burner uses ethane, C₂H₆.

(a) Write an equation for the complete combustion of ethane. [1] (b) What volume of oxygen, in dm³, does complete combustion of 4.50 g of ethane use at 100 kPa and 290 K? Give your answer to 3 significant figures. [3] (c) Assume that air is 20% oxygen by volume. Calculate the volume of air needed, in dm³. [1]

8. Exhaust gases from a petrol engine contain nitrogen monoxide, carbon monoxide, carbon particles and unburned hydrocarbons.

(a) Explain how nitrogen monoxide forms in the engine. Include an equation. [2] (b) Over the converter’s catalyst, hexane (C₆H₁₄) that escaped burning is turned by nitrogen monoxide into CO₂, H₂O and N₂ and nothing else. Give the balanced equation. [1] (c) State the condition in the cylinder that leads to carbon monoxide and carbon particles. [1] (d) Give the equation for the converter reaction that removes CO and NO together. [1]

9. A fuel sample is contaminated with a sulfur compound, methanethiol, CH₃SH.

(a) Write an equation for the complete combustion of methanethiol to carbon dioxide, sulfur dioxide and water. [1] (b) A power station sprays powdered calcium carbonate into its waste gases to cut SO₂ emissions. Explain why this works, and include an equation. [3]

10. Methane and chlorine react in ultraviolet light.

(a) Name the type of mechanism and explain the role of the ultraviolet light. [2] (b) Write an equation for the initiation step. [1] (c) Give the pair of propagation steps leading to chloromethane. [2] (d) A trace of 1,2-dichloroethane, CH₂ClCH₂Cl, is found in the product. Write an equation for the step that forms it. [1] (e) Suggest how the proportions of the reactants should be chosen to obtain mainly chloromethane. Explain your answer. [2]

11. A company cracks icosane, C₂₀H₄₂, which it obtains from a low-value heavy fraction.

(a) Each molecule of icosane forms one molecule of dodecane and some ethene only. Write an equation for this reaction. [1] (b) The company uses thermal cracking. State the conditions. [2] (c) Calculate the maximum mass of ethene, in tonnes, that can be made from 1.00 tonne of icosane. Give your answer to 3 significant figures. [3] (d) Write an equation for the complete combustion of dodecane, C₁₂H₂₆. [1] (e) Give two economic reasons why the company cracks icosane rather than selling it as it is. [2]

Answers

1. Catalytic cracking uses a slight pressure rather than a high pressure [1]. It uses a zeolite catalyst; thermal cracking uses no catalyst [1]. [2] Examiner insight: Temperature is high in both types, so “a higher temperature” is not a difference and is unlikely to gain credit.

2. P is collected higher up [1]. P has a shorter chain, so fewer electrons and less surface contact, so weaker van der Waals forces between molecules [1]. Less energy is needed to overcome them, so P has a lower boiling point and condenses in the cooler upper part of the column [1]. [3] Examiner insight: The force must be named as acting between molecules; an answer that refers to breaking C–C or C–H bonds contradicts the explanation.

3. (a) C₁₇H₃₆ → C₁₁H₂₄ + C₂H₄ + C₄H₈ [1] (b) Thermal cracking [1] (c) Zeolite [1]; motor fuels and aromatic hydrocarbons [1] [4] Examiner insight: In (a), check the leftover fragment: C₄H₈ fits CₙH₂ₙ, so it is an alkene; writing C₄H₁₀ leaves hydrogen unbalanced, so the equation is unlikely to gain credit.

4. Distillation gives a fixed mix of fractions that does not match demand [1]. There is a surplus of heavy gas oil and a shortage of lighter fractions such as motor fuel [1]. The lighter fractions are in higher demand, so they are more valuable; cracking converts the surplus into higher-value products [1]. Cracking also produces alkenes, which are valuable feedstocks for polymers and other chemicals [1]. [4] Examiner insight: Link each point to the refinery described; a bare list of products does not explain why cracking pays.

5. (a) C₁₆H₃₄ + 24½O₂ → 16CO₂ + 17H₂O [1] (b) C₃H₈ + 2½O₂ → CO + 2C + 4H₂O [1] (c) C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O [1] [3] Examiner insight: In (c), the products hold 13 oxygen atoms but one is already in the alcohol, so 6 O₂ is needed, not 6½; whole-number multiples of a correct equation are equally acceptable.

6. Volume of CO₂ = 82.5 − 22.5 = 60.0 cm³ [1]. 15 cm³ of alkane gives 60 cm³ of CO₂, so each molecule contains 4 carbon atoms [1]. Volume of O₂ used = 120 − 22.5 = 97.5 cm³, so 6.5 mol O₂ react per mole of alkane [1]. 13 O atoms: 8 go into 4CO₂ and 5 into H₂O, so there are 10 H atoms: C₄H₁₀ [1]. [4] Examiner insight: Taking 120 cm³ as the oxygen used, without subtracting the 22.5 cm³ left, gives the wrong hydrogen count.

7. (a) C₂H₆ + 3½O₂ → 2CO₂ + 3H₂O [1] (b) n(C₂H₆) = 4.50 ÷ 30.0 = 0.150 mol [1]. n(O₂) = 3.5 × 0.150 = 0.525 mol [1]. V = nRT ÷ p = 0.525 × 8.31 × 290 ÷ 100 000 = 1.265 × 10⁻² m³ = 12.7 dm³ [1]. (c) 12.65 ÷ 0.20 = 63.3 dm³ (63.5 dm³ from the rounded 12.7 is also fine) [1]. [5] Examiner insight: Carry the unrounded volume into (c), and convert m³ to dm³ before giving the answer.

8. (a) At the very high temperature in the cylinder, atmospheric N₂ and O₂ combine [1]: N₂ + O₂ → 2NO [1]. (b) C₆H₁₄ + 19NO → 6CO₂ + 7H₂O + 9½N₂ [1] (c) A limited supply of oxygen (incomplete combustion) [1] (d) 2CO + 2NO → 2CO₂ + N₂ [1] [5] Examiner insight: For (a), an answer saying the nitrogen comes from the fuel is unlikely to gain credit, because alkanes contain no nitrogen.

9. (a) CH₃SH + 3O₂ → CO₂ + SO₂ + 2H₂O [1] (b) SO₂ is an acidic oxide [1]. Calcium carbonate is a base that reacts with acids, so it neutralises the SO₂, trapping it as calcium sulfite [1]: CaCO₃ + SO₂ → CaSO₃ + CO₂ [1] [4] Examiner insight: “It absorbs the gas” names no chemistry; say SO₂ is acidic and the calcium compound is basic.

10. (a) Free-radical substitution [1]. UV light supplies the energy to break the Cl–Cl bond homolytically, forming radicals [1]. (b) Cl₂ → 2Cl• [1] (c) CH₄ + Cl• → HCl + •CH₃ [1]; •CH₃ + Cl₂ → CH₃Cl + Cl• [1] (d) •CH₂Cl + •CH₂Cl → CH₂ClCH₂Cl [1] (e) Use an excess of methane [1]. Then a chlorine radical is far more likely to collide with a methane molecule than with chloromethane, so further substitution is reduced [1]. [8] Examiner insight: A propagation step that forms H• and CH₃Cl, or a dot drawn on the wrong atom, does not match the accepted mechanism even if the atoms balance.

11. (a) C₂₀H₄₂ → C₁₂H₂₆ + 4C₂H₄ [1] (b) High temperature [1] and high pressure [1]. (c) Mr(C₂₀H₄₂) = 282.0, so n = 1.00 × 10⁶ ÷ 282.0 = 3546 mol [1]. n(C₂H₄) = 4 × 3546 = 1.418 × 10⁴ mol [1]. Mass = 1.418 × 10⁴ × 28.0 = 3.97 × 10⁵ g = 0.397 tonnes [1]. (d) C₁₂H₂₆ + 18½O₂ → 12CO₂ + 13H₂O [1] (e) Heavy fractions such as icosane are in surplus and have low value, while the smaller products are in higher demand and worth more [1]. Ethene is an alkene, a valuable feedstock for polymers and other chemicals [1]. (Also accept: dodecane is a shorter alkane better suited to use as a fuel.) [9] Examiner insight: In (c), convert tonnes to grams before dividing by Mr, and back to tonnes at the end.

Where marks are usually lost

• Naming temperature as a difference between thermal and catalytic cracking, when both are high.
• Referring to bonds breaking when explaining boiling points in distillation.
• Writing an alkene fragment as CₙH₂ₙ₊₂ in a cracking equation.
• Explaining cracking with a list of products but no link to surplus, demand and value.
• Not subtracting the oxygen already present in an alcohol.
• Forgetting leftover oxygen in gas-volume data, or leaving pressure in kPa.
• Tracing the N in NOₓ to the fuel instead of to air.
• “Absorbs SO₂” with no acid–base reasoning.
• Radical steps with dots in the wrong place, or termination steps that produce radicals.

Next steps

• Alkanes revision notes
• Alkanes study guide
• Amount of substance
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Section 3.3.2 Alkanes, points 3.3.2.1 to 3.3.2.4, headed: Fractional distillation of crude oil; Modification of alkanes by cracking; Combustion of alkanes; Chlorination of alkanes.