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HKDSE Chemistry Paper 2 Practice Paper (Elective Part) with Answers | 30 questions, 60 minutes, 100 marks
Practice paper · HKDSE Chemistry Paper 2
HKDSE Chemistry Paper 2 elective-part practice paper: 25 multiple-choice and 5 structured questions on industrial and analytical chemistry, 60 minutes, 100 marks, full worked answers and PDFs.
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Paper details Level: S6 (HKDSE) | Subject: Chemistry (Paper 2, Elective Part) | Time: 60 minutes | Total: 100 marks | Questions: 30
Section A: Industrial Chemistry (30 marks)
1. Two marks each. Choose the best answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | In the Haber process 4.0 mol of nitrogen reacts completely with excess hydrogen (N₂ + 3H₂ ⇌ 2NH₃, Mr(NH₃) = 17). What is the maximum mass of ammonia formed? | Mole ratio N₂ : NH₃ = 1 : 2, so n(NH₃) = 2 × 4.0 = 8.0 mol; m = n × Mr = 8.0 × 17 = 136 g. (2 marks) |
| 2 | In the same batch only 102 g of ammonia is obtained. What is the percentage yield? | Percentage yield = actual ÷ theoretical × 100% = 102 ÷ 136 × 100% = 75%. (2 marks; 0 if the theoretical mass is not used as the denominator.) |
| 3 | What is the atom economy of the Haber process? | Every reactant atom ends up in the product, so atom economy = (2 × 17) ÷ (28 + 3 × 2) × 100% = 34 ÷ 34 × 100% = 100%. (2 marks) |
| 4 | In the Contact process, the catalyst and typical conditions for oxidising sulfur dioxide to sulfur trioxide are: | The catalyst is vanadium(V) oxide (V₂O₅), at about 450 ℃ and 1–2 atm. Iron is the Haber catalyst and must not be confused with it. (2 marks; 1 for the catalyst, 1 for conditions.) |
| 5 | In the Contact process 200 mol SO₂ is mixed with 120 mol O₂ (2SO₂ + O₂ ⇌ 2SO₃). Which is limiting, and what is the maximum mass of SO₃ in kg (Mr = 80)? | O₂ needed = 200 ÷ 2 = 100 mol < 120 mol, so SO₂ is limiting (O₂ left = 20 mol); n(SO₃) = 200 mol; m = 200 × 80 g = 16 kg. (2 marks) |
| 6 | Which change does NOT raise the equilibrium yield of SO₃ in 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (ΔH < 0)? | Raising the temperature shifts the equilibrium in the endothermic (reverse) direction, so the SO₃ yield falls. Higher pressure, excess O₂ and continuous removal of SO₃ all raise it. (2 marks) |
| 7 | Take the molar volume of a gas as 24.0 dm³ mol⁻¹ at room conditions. What volume is occupied by 0.50 mol of carbon dioxide? | V = n × molar volume = 0.50 × 24.0 = 12.0 dm³. (2 marks) |
| 8 | In the electrolysis of brine (chlor-alkali industry), how many grams of sodium hydroxide (Mr = 40) are formed per 1.0 mol of chlorine? | 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂; the ratio Cl₂ : NaOH = 1 : 2, so n(NaOH) = 2.0 mol and m = 2.0 × 40 = 80 g (chlorine formed = 71 g). (2 marks) |
| 9 | What minimum mass of sodium chloride (Mr = 58.5) is needed to produce 1.0 mol of chlorine by electrolysis? | n(NaCl) = 2 × 1.0 = 2.0 mol; m = 2.0 × 58.5 = 117 g. (2 marks) |
| 10 | In the Solvay process sodium hydrogencarbonate is heated: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. What mass of sodium carbonate (Mr = 106) is obtained from 2.0 mol of NaHCO₃? | Ratio 2 : 1, so n(Na₂CO₃) = 1.0 mol; m = 1.0 × 106 = 106 g. (2 marks) |
| 11 | In a reaction 100 g of reactants gives 88 g of the desired product, the rest being by-products. What is the atom economy? | Atom economy = mass of desired product ÷ total mass of reactants × 100% = 88 ÷ 100 × 100% = 88%. (2 marks; dividing actual by theoretical gives percentage yield, not atom economy.) |
| 12 | Which statement about the role of a catalyst in an industrial reaction is correct? | A catalyst provides a route of lower activation energy, speeding up the forward and reverse reactions equally so equilibrium is reached sooner; it does not change the enthalpy change, the position of equilibrium or the equilibrium constant. (2 marks; 0 for “raises the yield”.) |
| 13 | Flue-gas desulfurisation passes SO₂ through limestone slurry: CaCO₃ + SO₂ + ½O₂ → CaSO₄ + CO₂. What mass of CaCO₃ (Mr = 100.1) is needed to absorb 64 g of SO₂ completely? | n(SO₂) = 64 ÷ 64 = 1.0 mol, in 1 : 1 ratio with CaCO₃, so 1.0 mol is needed; m = 1.0 × 100.1 = 100.1 g. (2 marks) |
| 14 | Compared with a petrol engine, the most direct advantage of running a car on a hydrogen–oxygen fuel cell is that: | The only product is water, so the vehicle emits no carbon dioxide, carbon monoxide or nitrogen oxides; whether the whole chain is low-carbon still depends on how the hydrogen is made. (2 marks; 0 for “hydrogen is cheap”.) |
| 15 | In a hydrogen–oxygen fuel cell each mole of hydrogen releases about 2 mol of electrons (2H₂ + O₂ → 2H₂O). Taking the Faraday constant as 96500 C mol⁻¹, what charge does 1 mol of hydrogen deliver? | Q = n(e⁻) × F = 2 × 96500 = 193000 C (about 1.9×10⁵ C). (2 marks) |
Section B: Analytical Chemistry (20 marks)
2. Two marks each. Choose the best answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 16 | In a flame test, what colour does a potassium ion give? | Potassium gives a lilac flame. Sodium is golden yellow, calcium brick red and copper blue-green. (2 marks; 1 mark for “purple” alone.) |
| 17 | Dilute sodium hydroxide solution is added drop by drop to a solution containing zinc ions until it is in excess. What is observed? | A white precipitate of Zn(OH)₂ forms first and redissolves in excess NaOH to give a colourless solution, because zinc hydroxide is amphoteric. (2 marks; 1 mark for the white precipitate only.) |
| 18 | Which is the best test for the carbonate ion in a solid? | Add dilute hydrochloric acid; effervescence that turns limewater milky shows carbonate, because carbon dioxide is released. (2 marks; 1 mark if only litmus is used.) |
| 19 | Which reagents and observation correctly test for the sulfate ion in solution? | Add dilute hydrochloric acid first to remove carbonate and sulfite, then barium chloride solution: a white precipitate that does not dissolve in dilute acid confirms sulfate. (2 marks; 1 mark if the acid step is omitted.) |
| 20 | 25.0 cm³ of 0.100 M sodium hydroxide exactly neutralises 25.0 cm³ of hydrochloric acid. What is the concentration of the acid? | n(NaOH) = 0.100 × 25.0 ÷ 1000 = 2.50e-03 mol; the ratio is 1 : 1, so n(HCl) is the same; c = 2.50e-03 ÷ 25.0 × 1000 = 0.100 M. (2 marks) |
| 21 | In an iodometric titration I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. How many moles of thiosulfate are needed to titrate 0.0050 mol of I₂? | From the equation the ratio I₂ : S₂O₃²⁻ = 1 : 2, so 2 × 0.0050 = 0.0100 mol is needed. (2 marks) |
| 22 | Three concordant titres are 21.40, 21.50 and 21.60 cm³. Which mean titre should be used in the calculation? | The three titres agree within 0.20 cm³, so take the arithmetic mean: (21.40 + 21.50 + 21.60) ÷ 3 = 21.50 cm³. (2 marks; 0 for discarding any of them.) |
| 23 | 25.0 cm³ of white vinegar (density taken as 1.00 g cm⁻³) is titrated with a mean 21.50 cm³ of 0.100 M sodium hydroxide (1 : 1 reaction, Mr(CH₃COOH) = 60). What is the percentage by mass of ethanoic acid? | n(NaOH) = 0.100 × 21.50 ÷ 1000 = 2.150e-03 mol = n(acid); m(acid) = 2.150e-03 × 60 = 0.129 g; sample mass = 25.0 g (density 1.00), so 0.129 ÷ 25.0 × 100% ≈ 0.52% (w/w). (2 marks; 1 mark if the density conversion is missed.) |
| 24 | In paper chromatography a coloured spot moves 3.2 cm from the start line while the solvent front moves 8.0 cm. What is its Rf value? | Rf = distance moved by the spot ÷ distance moved by the solvent front = 3.2 ÷ 8.0 = 0.40 (no unit). (2 marks) |
| 25 | An organic compound shows a strong infrared absorption near 1700 cm⁻¹. Which functional group does this most likely indicate? | A strong band near 1700 cm⁻¹ is the C=O stretch of a carbonyl group. A broad 2500–3300 cm⁻¹ band indicates the O—H of a carboxylic acid, and a sharp 3300 cm⁻¹ band is usually an alcohol O—H or an N—H. (2 marks) |
Section C: Structured questions (50 marks)
3. Answer all questions. Show your working, units and significant figures.
| # | Question | Answer and marking notes |
|---|---|---|
| 26 | Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹. (a) Write the equilibrium equation and name the industrial catalyst. (2 marks) (b) Explain why the operating temperature is about 450 ℃ rather than lower or higher. (3 marks) (c) 200 mol N₂ reacts with 700 mol H₂: identify the limiting reagent and find the theoretical mass of ammonia in kg; if 5.4 kg is obtained, find the percentage yield. (3 marks) (d) Explain why continuously liquefying and removing ammonia raises the yield. (2 marks) |
(a) N₂(g) + 3H₂(g) ⇌ 2NH₃(g); the catalyst is iron (with K₂O and Al₂O₃ promoters). (2 marks) (b) The reaction is exothermic, so a low temperature favours the equilibrium but is far too slow; a high temperature is fast but shifts the equilibrium left and lowers the yield; 450 ℃ is the compromise, made workable by the catalyst. (3 marks) (c) H₂ needed = 3 × 200 = 600 mol < 700 mol, so N₂ is limiting; n(NH₃) = 2 × 200 = 400 mol; m = 400 × 17 ÷ 1000 = 6.8 kg; percentage yield = 5.4 ÷ 6.8 × 100% = 80%. (3 marks) (d) Removing the product lowers its concentration so Q < K and the equilibrium shifts right, increasing the amount of ammonia formed. (2 marks) |
| 27 | White vinegar is titrated with standard sodium hydroxide to find its ethanoic acid content. Each sample is 25.0 cm³; three titres are 21.40, 21.50 and 21.60 cm³ of 0.100 M NaOH. The reaction is 1 : 1 and Mr(CH₃COOH) = 60. (a) Explain why the mean of three titres is used and calculate it. (3 marks) (b) Find the concentration of ethanoic acid in the vinegar in M. (3 marks) (c) Find its mass concentration in g dm⁻³. (2 marks) (d) What effect would an uncalibrated burette have on the result? (2 marks) |
(a) The three titres agree within 0.20 cm³, so they are concordant; the mean reduces random error: (21.40 + 21.50 + 21.60) ÷ 3 = 21.50 cm³. (3 marks) (b) n(NaOH) = 0.100 × 21.50 ÷ 1000 = 2.15e-03 mol; the ratio is 1 : 1 so n(acid) is the same; c = 2.15e-03 ÷ 25.0 × 1000 = 0.0860 M. (3 marks) (c) Mass concentration = 0.0860 × 60 = 5.16 g dm⁻³. (2 marks) (d) An uncalibrated burette gives a systematic error: every reading is displaced the same way, the mean cannot remove it, and the result is consistently biased; the burette should be calibrated against a standard solution. (2 marks) |
| 28 | A colourless solution contains one cation and one anion, possibly among K⁺, Na⁺, Ca²⁺ and Cl⁻, CO₃²⁻, SO₄²⁻. Tests give: ① flame test: lilac flame; ② dilute nitric acid then silver nitrate solution: white precipitate that dissolves in dilute ammonia; ③ no effervescence with dilute nitric acid. (a) Deduce the cation and anion and name the compound. (4 marks) (b) Explain why dilute nitric acid, not dilute hydrochloric acid, is added before silver nitrate. (3 marks) (c) Describe one further test that confirms the anion, giving the reagent and the expected observation. (3 marks) |
(a) ① lilac flame → K⁺; ② white precipitate soluble in dilute ammonia → Cl⁻ (AgCl); ③ no effervescence → no CO₃²⁻, and no acid-insoluble white precipitate → no SO₄²⁻. The compound is potassium chloride, KCl. (4 marks) (b) Dilute hydrochloric acid itself contains Cl⁻, which would give a false positive white AgCl precipitate with silver nitrate; dilute nitric acid removes interfering ions such as carbonate without adding chloride. (3 marks) (c) Add dilute nitric acid then barium nitrate solution: no white precipitate means no sulfate, so with the cation test the salt is a chloride. Alternatively heat the solid with manganese(IV) oxide and concentrated sulfuric acid; the gas evolved turns damp blue litmus red then bleaches it, confirming chloride. (3 marks) |
| 29 | The Contact process starts with burning sulfur: S + O₂ → SO₂; 2SO₂ + O₂ ⇌ 2SO₃; SO₃ + H₂O → H₂SO₄ (concentrated sulfuric acid is used in practice). (a) Write the three equations and state the catalyst and temperature for the second step. (4 marks) (b) 320 kg of sulfur is burned and every step goes to completion: find the theoretical masses of SO₂, SO₃ and H₂SO₄ in kg. (3 marks) (c) If the overall yield is 95%, what mass of sulfuric acid is obtained? (1 mark) (d) Suggest two measures to reduce SO₂ emissions. (2 marks) |
(a) S + O₂ → SO₂; 2SO₂ + O₂ ⇌ 2SO₃ (catalyst V₂O₅, about 450 ℃, 1–2 atm); SO₃ + H₂O → H₂SO₄. (4 marks) (b) n(S) = 320 × 1000 ÷ 32.0 = 10000 mol; sulfur atoms are conserved, so n(SO₂) = n(SO₃) = n(H₂SO₄) = 10000 mol; m(SO₂) = 640 kg, m(SO₃) = 800 kg, m(H₂SO₄) = 980 kg. (3 marks) (c) Actual mass = 980 × 95% = 931 kg. (1 mark) (d) Flue-gas desulfurisation with limestone slurry, low-sulfur fuel, scrubbers, or recovering the SO₂ as sulfuric acid — any two. (2 marks) |
| 30 | Brine is electrolysed (chlor-alkali industry) at 10.0 A for 2.00 hours: 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂ (Faraday constant 96500 C mol⁻¹, molar gas volume at room conditions 24.0 dm³ mol⁻¹). (a) Find the charge passed in C. (2 marks) (b) Find the moles of chlorine formed. (3 marks) (c) Find the volume of chlorine at room conditions in dm³. (3 marks) (d) What mass of sodium hydroxide is formed at the same time? (2 marks) |
(a) Q = It = 10.0 × 2.00 × 3600 = 72000 C. (2 marks; 0 if × 3600 is forgotten.) (b) n(e⁻) = Q/F = 72000 ÷ 96500 = 0.7461 mol; each Cl₂ needs 2 electrons, so n(Cl₂) = 0.7461 ÷ 2 = 0.3731 mol. (3 marks) (c) V = n × 24.0 = 0.3731 × 24.0 = 8.95 dm³. (3 marks) (d) The ratio Cl₂ : NaOH = 1 : 2, so n(NaOH) = 0.7461 mol and m = 0.7461 × 40 = 29.8 g. (2 marks) |
A TutorZone practice paper; original questions and passages, not publisher material.
