Mark Scheme

Section A — Structured Questions

1. (a) (up to 2) Electrolysis = the breaking down/decomposition of an ionic compound (molten or in solution) using an electric current. [2] (b) Cathode: lead (Pb) (1); Anode: bromine (Br₂) (1). [2] (c) (up to 3) In the solid, the ions are held in fixed positions and cannot move; when molten, the ions are free to move and carry charge; charged particles must be free to move for electrolysis/conduction to occur. [3] (d) Pb²⁺ + 2e⁻ → Pb (1 for species/balancing, 1 for electrons on the correct side). [2]

2. (a) General formula of alkanes = CₙH₂ₙ₊₂ (1). [1] (b) CH₄ + 2O₂ → CO₂ + 2H₂O (1 for reactants/products, 1 for balancing O₂, 1 for balancing H₂O). [3] (c) (up to 4) Alkanes are saturated (only single C–C bonds); alkenes are unsaturated (contain a C=C double bond) (2). Test: add bromine water — it stays orange/brown with an alkane but is decolourised (turns colourless) by an alkene (2). [4] (d) (up to 2) Carbon monoxide (CO) (1); it is dangerous because it is a toxic gas — it is colourless and odourless and reduces the blood's ability to carry oxygen (1). [2]

3. (a) (up to 3) Increasing concentration means there are more reactant particles in the same volume; so the particles collide more frequently (more collisions per second); more frequent successful collisions means a faster rate. [3] (b) (up to 3) A catalyst lowers the activation energy by providing an alternative reaction pathway, so more collisions are successful and the rate increases (2); unlike a reactant, a catalyst is not used up in the reaction (it can be reused) (1). [3] (c) average rate = 50 ÷ 20 = 2.5 cm³/s (M1 A1); the rate decreased because the reactants were being used up, so their concentration fell and collisions became less frequent (A1). [3]

4. (a) Mr(NaCl) = 23 + 35.5 = 58.5 (M1); moles = mass ÷ Mr = 11.7 ÷ 58.5 = 0.20 mol (M1 A1). [3] (b) 500 cm³ = 0.5 dm³ (M1); concentration = moles ÷ volume = 0.20 ÷ 0.5 = 0.40 mol/dm³ (M1 A1). [3] (c) percentage yield = (actual ÷ theoretical) × 100 = (6.0 ÷ 8.0) × 100 = 75% (M1 A1). [2]

5. (a) (up to 2) Dip a clean wire in the sample and hold it in a (blue/roaring) Bunsen flame and observe the colour (1); sodium gives a yellow/orange flame (1). [2] (b) (up to 2) Bubble the gas through limewater (1); carbon dioxide turns limewater cloudy/milky (1). [2] (c) (up to 4) Add dilute nitric acid, then silver nitrate solution (2); a white precipitate (of silver chloride) forms if chloride ions are present (2). [4]

Section B — Extended Response

6. (a) Nitrogen from the air (1); hydrogen from natural gas / methane (or reacting methane with steam) (1). [2] (b) (up to 6) The forward reaction is exothermic, so a lower temperature would give a higher yield of ammonia at equilibrium, but a lower temperature also makes the reaction too slow; 450 °C is a compromise between a reasonable yield and a fast enough rate. Increasing pressure shifts the equilibrium towards the side with fewer gas molecules (2 NH₃ vs 4 reactant molecules), giving a higher yield, but very high pressures are expensive and dangerous (equipment/energy costs); 200 atm is a compromise between yield and cost/safety. Award up to 3 for temperature compromise, up to 3 for pressure compromise. [6]

7. (up to 8) Fractional distillation of crude oil: the crude oil is heated until most of it evaporates; the vapours enter a fractionating column which is hot at the bottom and cooler at the top; as the vapours rise they cool, and each fraction condenses at the level where the temperature matches its boiling point; fractions with larger molecules have higher boiling points and condense lower down (near the bottom), while fractions with smaller molecules have lower boiling points and condense higher up (near the top); the different fractions are collected at different levels. Levels: 7–8 = clear, detailed and accurate, explaining the link between molecule size, boiling point and position; 4–6 = several points, some explanation; 1–3 = basic. [8]

Sample Answers with Examiner Commentary

Question 2(c) — Sample Answers

Grade A response. "An alkane is a saturated hydrocarbon, which means it contains only single bonds between the carbon atoms (C–C). An alkene is an unsaturated hydrocarbon, which means it contains at least one carbon–carbon double bond (C=C). To distinguish between them, you can add bromine water to each. With an alkane, the bromine water stays orange/brown because there is no reaction. With an alkene, the bromine water is decolourised — it turns from orange to colourless — because the bromine adds across the C=C double bond. So a positive test (decolourising the bromine water) shows the presence of an alkene."

Mark: 4/4. Examiner commentary: A full-mark answer. The candidate defines both terms precisely using the key idea of saturation (single bonds) versus unsaturation (a C=C double bond), then gives the correct chemical test with both the reagent (bromine water) and — crucially — the different results for each: unchanged with an alkane, decolourised with an alkene. Stating both the 'before' and 'after' colours makes the test unambiguous, which is exactly what the question requires.

Grade C response. "An alkane has single bonds and an alkene has a double bond. You can test them with bromine water and the alkene reacts."

Mark: 2/4. Examiner commentary: The candidate correctly distinguishes the two using single versus double bonds, earning credit, and knows that bromine water is the test and that the alkene reacts. However, the answer does not state the result of the test — that the bromine water is decolourised (goes from orange to colourless) with an alkene and stays orange with an alkane. A chemical test is only complete when the observation is given, so describing the colour change would gain the remaining marks. Using the terms 'saturated' and 'unsaturated' would also strengthen it.

Question 6(b) — Sample Answers

Grade A response (extract). "The choice of temperature is a compromise. The forward reaction that makes ammonia is exothermic, so a lower temperature would move the equilibrium to the right and give a higher yield of ammonia. However, at a low temperature the reaction is very slow, so it would take too long to make useful amounts. A temperature of about 450 °C is chosen as a compromise: the yield is lower than it would be at a very low temperature, but the reaction is fast enough to be economic.

The pressure is also a compromise. There are four molecules of gas on the left (N₂ + 3H₂) and only two on the right (2NH₃), so increasing the pressure shifts the equilibrium towards the side with fewer molecules — the ammonia — giving a higher yield. But building equipment that can withstand very high pressures is expensive and can be dangerous, so about 200 atmospheres is used as a compromise between a good yield and reasonable cost and safety."

Mark: 6/6. Examiner commentary: A full-mark answer that correctly explains both compromises. For temperature, the candidate links the exothermic forward reaction to a higher yield at low temperature, but balances this against the reaction being too slow. For pressure, they use the number of gas molecules on each side to explain why high pressure increases yield, then balance this against cost and safety. Recognising that each condition is a trade-off between yield and another factor (rate; cost/safety) is exactly the reasoning the question is testing.

Grade C response (extract). "A high pressure is used because it gives more ammonia. A temperature of 450 °C is used to make the reaction fast. An iron catalyst speeds it up."

Mark: 3/6. Examiner commentary: The candidate correctly states that high pressure increases the yield of ammonia and that the temperature helps the rate, and mentions the catalyst, earning three marks. However, the answer does not explain the idea of a compromise, which is the focus of the question. For temperature, it needs to say that a lower temperature would give a higher yield (exothermic reaction) but too slow a rate, so 450 °C is a balance. For pressure, it should explain that high pressure favours the side with fewer gas molecules but is expensive/dangerous, so 200 atm is a balance. Adding the trade-offs would gain the remaining marks.