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Questions on alkanes: fractional distillation, cracking and combustion
Alkanes as saturated hydrocarbons with the general formula CₙH₂ₙ₊₂, boiling point against chain length and branching, fractional distillation of crude oil, thermal and catalytic cracking, reforming, complete and incomplete combustion, the pollutants CO, NOₓ, SO₂, soot and unburnt hydrocarbons, photochemical smog and PAN, NO₂ as a catalyst in acid rain, catalytic converters, flue-gas desulfurisation and the greenhouse effect.
20 original questions · 64 marks · the alkanes: fractional distillation, cracking and combustion notes · Alkanes and halogenoalkanes · Flashcards for this unit
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Explain the two trends in the boiling points of the alkanes given in the table: along the unbranched series, and among the three isomers of C₈H₁₈.
Alkane Formula Boiling point / °C Propane C₃H₈ −42 Pentane C₅H₁₂ 36 Octane C₈H₁₈ 126 Decane C₁₀H₂₂ 174 2-methylheptane C₈H₁₈ 118 2,2,4-trimethylpentane C₈H₁₈ 99 Mark scheme
Indicative content
- along the unbranched series the boiling point rises steadily as the chain gets longer, from −42 °C for propane to 174 °C for decane
- the only forces between alkane molecules are van der Waals forces between induced dipoles, and no covalent bonds break when an alkane boils
- a longer chain has more electrons, so larger induced dipoles form, and it has a larger surface in contact with neighbouring molecules
- the van der Waals forces between longer molecules are therefore stronger, and more energy is needed to overcome them, so the boiling point is higher
- the three C₈H₁₈ isomers have the same number of electrons, yet the boiling point falls from 126 °C to 118 °C to 99 °C as the branching increases
- a more branched molecule is more compact, cannot lie as closely alongside its neighbours and has less surface contact, so its van der Waals forces are weaker
Levels
- Level 3 (5–6 marks) both trends are described with values from the table and both are explained: the rise along the series through more electrons and more surface contact giving stronger van der Waals forces, and the fall with branching through compact molecules touching less; the answer says that only intermolecular forces, not covalent bonds, are overcome
- Level 2 (3–4 marks) the rise along the series is put down to stronger van der Waals forces from more electrons and branching is said to lower the boiling point; why branching does so is not explained through surface contact, or the table's values go unused
- Level 1 (1–2 marks) the boiling point is said to rise with chain length or fall with branching, typically because longer chains have more covalent bonds to break
Explain, using the general formula of the alkanes, what is meant by describing an alkane as saturated, and why cracking a long-chain alkane always produces at least one alkene.
Mark scheme
- B1 the general formula of the alkanes is CₙH₂ₙ₊₂
- B1 saturated means the molecule has only single bonds between carbon atoms, so no more atoms can be added to it
- B1 two alkanes sharing n carbon atoms would need 2n + 4 hydrogen atoms between them, but the alkane CₙH₂ₙ₊₂ has only 2n + 2
- B1 so at least one product must have fewer hydrogens, CₙH₂ₙ, and contain a C=C double bond; allow a correct example equation that shows this
Describe how crude oil is separated into fractions by fractional distillation.
Mark scheme
- B1 the crude oil is heated in a furnace so that most of it vaporises, and the vapour is fed into the bottom of a column
- B1 the column has a temperature gradient, hottest at the bottom and coolest at the top
- B1 vapour rises and each hydrocarbon condenses on the tray where the temperature falls below its boiling point
- B1 fractions with similar boiling points are drawn off at different heights; short chains near the top, long chains near the bottom, and the residue leaves at the base
Compare thermal cracking with catalytic cracking.
Mark scheme
- B1 thermal cracking uses 400 to 900 °C, whereas catalytic cracking runs at a lower temperature, about 450 °C; accept any single value in each range
- B1 thermal cracking uses a high pressure and no catalyst; catalytic cracking uses a slight pressure and a zeolite catalyst
- B1 thermal cracking goes through radicals, catalytic cracking through carbocations
- B1 thermal cracking gives a high proportion of alkenes, catalytic cracking mainly branched alkanes, cycloalkanes and aromatic hydrocarbons for petrol
Give an equation for the reaction of carbon monoxide with nitrogen monoxide in a catalytic converter, and name a metal used as its catalyst. Give an equation for the removal of sulfur dioxide from flue gases by calcium carbonate, and name the useful product made from it.
Mark scheme
- B1 2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g); ignore state symbols
- B1 platinum, palladium or rhodium; reject 'a transition metal' alone
- B1 CaCO₃(s) + SO₂(g) → CaSO₃(s) + CO₂(g)
- B1 the calcium sulfite is oxidised to calcium sulfate, sold as gypsum for plasterboard
Evaluate the claim that a car fitted with a catalytic converter produces a harmless exhaust.
Mark scheme
- B1 the converter removes most CO, NOₓ and unburnt hydrocarbons, converting them to CO₂, N₂ and H₂O
- B1 it adds carbon dioxide, a greenhouse gas, and does not remove any
- B1 it does not work until it is hot, so pollutants pass through on short or cold journeys; it does not remove soot particles from a diesel engine
- B1 the claim is therefore not justified: the exhaust is much less toxic, but not harmless
Explain why 2,2-dimethylpropane boils at 10 °C while its isomer pentane boils at 36 °C.
Mark scheme
- B1 both have the formula C₅H₁₂ and so the same number of electrons, but 2,2-dimethylpropane is branched and more compact, closer to spherical
- B1 its molecules cannot pack as closely and have a smaller surface area in contact with each other
- B1 so the van der Waals forces between its molecules are weaker and less energy is needed to separate them; reject any answer in which covalent bonds break
Calculate the mass of carbon dioxide formed when 500 g of hexane, C₆H₁₄, burns completely. Ar: H 1.0, C 12.0, O 16.0.
Mark scheme
- M1 Mr(C₆H₁₄) = 86.0, so the amount of hexane is 500 ÷ 86.0 = 5.814 mol
- M1 each mole of hexane gives 6 mol of CO₂, so 6 × 5.814 = 34.88 mol of CO₂
- A1 mass = 34.88 × 44.0 = 1535 g = 1.53 × 10³ g, or 1.53 kg, to three significant figures
Calculate the volume of oxygen needed for the complete combustion of 2.40 dm³ of propane, and the volume of carbon dioxide formed, both at room temperature and pressure.
Mark scheme
- M1 C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
- M1 equal volumes of gas hold equal amounts, so the volumes are in the mole ratio 1 : 5 : 3
- A1 12.0 dm³ of oxygen and 7.20 dm³ of carbon dioxide, to three significant figures
Explain how nitrogen monoxide is formed in a petrol engine, and give one environmental problem it causes.
Mark scheme
- B1 nitrogen and oxygen from the air drawn into the engine react; the nitrogen does not come from the fuel
- B1 the reaction needs the high temperature of the spark and the burning fuel: N₂(g) + O₂(g) → 2NO(g)
- B1 NO is oxidised in air to NO₂, which causes acid rain or contributes to photochemical smog
Suggest why the catalyst in a catalytic converter is spread as a thin layer over a ceramic honeycomb, and why a converter removes little pollution during the first few minutes of a journey.
Mark scheme
- B1 the honeycomb gives a very large surface area, so more gas is adsorbed onto the catalyst at once
- B1 a thin layer means only a small mass of an expensive metal is needed
- B1 the converter is cold at first and the reactions on the surface are too slow until it heats up
Suggest why calcium oxide removes sulfur dioxide from the flue gases of a coal-fired power station, give an equation for the reaction, and suggest why the sulfur is removed after burning rather than from the coal.
Mark scheme
- B1 sulfur dioxide is an acidic oxide and calcium oxide is a base, so they react together
- B1 CaO(s) + SO₂(g) → CaSO₃(s)
- B1 sulfur bound into a solid fuel cannot easily be removed before it burns, whereas petrol and diesel are liquids that are desulfurised at the refinery
Explain how carbon dioxide in the atmosphere causes the greenhouse effect.
Mark scheme
- B1 the Earth's surface, warmed by the Sun, emits infrared radiation
- B1 carbon dioxide molecules absorb some of this infrared, because it makes their bonds vibrate
- B1 they re-emit it in all directions, some back towards the surface, so the lower atmosphere is warmer than it would otherwise be
Explain why refineries reform straight-chain alkanes such as hexane into branched and cyclic hydrocarbons for use in petrol.
Mark scheme
- B1 straight-chain alkanes can ignite from compression alone, before the spark, causing knocking in a petrol engine
- B1 branched alkanes, cycloalkanes and arenes resist this early ignition and burn more smoothly
- B1 so the engine can run at a higher compression and the fuel is used more efficiently
Explain, with equations, how nitrogen dioxide in the air acts as a catalyst in the formation of acid rain from sulfur dioxide.
Mark scheme
- B1 SO₂(g) + NO₂(g) → SO₃(g) + NO(g)
- B1 NO(g) + ½O₂(g) → NO₂(g), so the nitrogen dioxide is regenerated and is not used up
- B1 the sulfur trioxide dissolves in rain to give sulfuric acid: SO₃(g) + H₂O(l) → H₂SO₄(aq)
Show that the complete combustion of 1.00 kg of octane, C₈H₁₈, releases about 48 MJ, given that the enthalpy of combustion of octane is −5470 kJ mol⁻¹. Ar: H 1.0, C 12.0.
Mark scheme
- M1 Mr of C₈H₁₈ = 8 × 12.0 + 18 × 1.0 = 114.0, so the amount of octane is 1000 ÷ 114.0 = 8.77 mol
- M1 energy released = 8.77 mol × 5470 kJ mol⁻¹, the negative sign of ΔH showing that the energy leaves the fuel
- A1 energy = 4.80 × 10⁴ kJ = 48.0 MJ, to three significant figures, which is about 48 MJ; error carried forward from the amount of octane
Explain why the residue from the fractionating column is distilled again at reduced pressure.
Mark scheme
- B1 reducing the pressure lowers the boiling points of the long-chain hydrocarbons in the residue
- B1 so they can be vaporised and separated at a temperature below that at which they would decompose
Complete the equation for the cracking of tetradecane into octane and propene only: C₁₄H₃₀(g) → C₈H₁₈(g) + ...
Mark scheme
- B1 the other product is propene, C₃H₆(g)
- B1 two molecules of it are needed: C₁₄H₃₀(g) → C₈H₁₈(g) + 2C₃H₆(g), with 14 carbon and 30 hydrogen atoms on each side
Suggest an equation for the reforming of heptane, C₇H₁₆, into methylbenzene, C₇H₈, and suggest a use for the other product.
Mark scheme
- B1 C₇H₁₆(g) → C₇H₈(g) + 4H₂(g), with 7 carbon and 16 hydrogen atoms on each side
- B1 the hydrogen can be used at the refinery, for example to remove sulfur from fuels
Suggest why nitrogen and oxygen in the air do not react at room temperature, although they do inside a car engine.
Mark scheme
- B1 the N≡N triple bond is very strong, 945 kJ mol⁻¹, and non-polar, so the activation energy is very high
- B1 only at the high temperature in the cylinder do enough collisions have energy equal to or above the activation energy
Practise alkanes: fractional distillation, cracking and combustion one question at a time