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HKDSE Chemistry Paper 2 Elective Units Practice Paper | 9 questions, 60 marks, 1 hour 30 minutes
HKDSE Chemistry Paper 2
HKDSE Chemistry Paper 2 elective units practice paper: 9 questions, 60 marks, 1 hour 30 minutes, covering industrial, materials and analytical chemistry with full worked solutions and marks.
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Paper details Level: S6 (HKDSE) | Subject: Chemistry | Time: 1 hour 30 minutes | Marks: 60 | Questions: 9
Section A (20 marks)
Elective questions: show your working and give units. Total 60 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) Ammonia is made by the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), with a negative ΔH. Explain why a low temperature favours the equilibrium yield, yet about 450 °C is used in industry. (2 marks) (b) At equilibrium in a 2.0 dm³ vessel at a certain temperature, there are 0.40 mol N₂, 0.60 mol H₂ and 0.80 mol NH₃. Calculate the equilibrium constant Kc at this temperature and state its unit. (3 marks) (c) In industry the ammonia formed is continuously condensed and removed. Explain the effect of this on the yield and on the rate of reaction. (2 marks) |
(a) The forward reaction is exothermic, so a low temperature shifts the equilibrium to the right and raises the yield [1]; however the rate would be far too slow, so about 450 °C is used with a catalyst to reach equilibrium sooner [1] (b) [N₂] = 0.20, [H₂] = 0.30, [NH₃] = 0.40 mol dm⁻³ [1]; Kc = 0.40² ÷ (0.20 × 0.30³) [1]; = 29.6 mol⁻² dm⁶ [1] (c) Removing the product shifts the equilibrium right and raises the yield [1]; but the concentrations fall so both forward and backward rates decrease [1] |
| 2 | (a) Ethanol can be made by fermenting glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. Calculate the atom economy of this reaction for ethanol as a percentage, to 1 decimal place. (3 marks) (b) 500 kg of glucose is fermented with a 90% yield of ethanol. The density of ethanol is 0.789 g cm⁻³. Calculate the volume of ethanol obtained in dm³, to 1 decimal place. (3 marks) (c) State one advantage of fermentation over the catalytic hydration of ethene. (1 mark) |
(a) Total relative molecular mass of ethanol = 2 × 46 = 92 [1]; of the reactant = 180 [1]; atom economy = 92 ÷ 180 × 100% = 51.1% [1] (b) Glucose reacted = 500 000 × 90% = 450 000 g, i.e. 2500 mol [1]; ethanol = 5000 mol, mass = 230000 g [1]; volume = 291.5 dm³ [1] (c) The raw material is renewable (crops) and the conditions are mild, so less energy is used [1] |
| 3 | (a) The chlor-alkali industry makes chlorine, hydrogen and sodium hydroxide by electrolysing concentrated brine. A current of 20.0 A is passed for 2.0 hours. Calculate the masses of chlorine and sodium hydroxide formed (1 decimal place; 1 mol of electrons = 96 500 C). (4 marks) (b) Give two reasons why modern plants use the ion-exchange membrane cell instead of the old mercury cell. (2 marks) |
(a) Charge = 20.0 × 7 200 = 144 000 C [1]; moles of electrons = 144 000 ÷ 96 500 = 1.49 mol [1]; chlorine = 1.49 ÷ 2 × 71 = 53.0 g [1]; sodium hydroxide = 1.49 × 40 = 59.7 g [1] (b) Mercury is toxic, so the membrane cell avoids mercury pollution [1]; it uses less energy (lower cell voltage) and gives purer products without a mercury-removal step [1] |
Section B (20 marks)
Elective questions: show your working and give units. Total 60 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | (a) Nylon-6,6 is made by condensing adipic acid with hexamethylenediamine. Name the linkage formed and state the class of polymer it belongs to. (2 marks) (b) Polystyrene is made by the addition polymerisation of styrene (C₈H₈, relative molecular mass 104). A sample has a relative molecular mass of 104 000. Calculate its degree of polymerisation. (3 marks) (c) Explain why addition polymers such as poly(ethene) do not degrade readily in the environment. (2 marks) |
(a) Amide linkage (-CO-NH-) [1]; it is a condensation polymer (a polyamide) [1] (b) Degree of polymerisation = relative molecular mass of polymer ÷ that of the monomer [1]; = 104 000 ÷ 104 [1]; = 1000 [1] (c) The backbone is made of strong C-C single bonds with no polar groups that water or microbes can attack [1]; the chains pack closely and are highly crystalline, so water and enzymes cannot penetrate [1] |
| 5 | (a) The standard electrode potential of Zn²⁺/Zn is −0.76 V and that of Fe²⁺/Fe is −0.44 V. Explain why galvanising an iron gate with zinc still protects the iron even when the zinc layer is scratched. (2 marks) (b) A galvanised steel sheet measures 2.0 m × 1.0 m and carries 80 g m⁻² of zinc. Given the density of zinc is 7.14 g cm⁻³, calculate the average thickness of the zinc layer in µm, to 1 decimal place. (3 marks) (c) Aluminium is more reactive than iron, yet aluminium articles resist corrosion well. Explain. (2 marks) |
(a) Zinc has the more negative electrode potential and is oxidised preferentially as a sacrificial anode [1]; after scratching, iron, zinc and the electrolyte form a cell, electrons flow from zinc to iron and the iron becomes the protected cathode [1] (b) Volume of zinc per square metre = 80 ÷ 7.14 = 11.2 cm³ [1]; 1 m² = 10 000 cm² [1]; thickness = 11.2 ÷ 10 000 = 1.12 × 10⁻³ cm = 11.2 µm [1] (c) Aluminium oxidises rapidly to give a dense, firmly adherent oxide layer [1]; the layer is impermeable to water and oxygen and stops further corrosion [1] |
| 6 | (a) A concrete mix uses cement : sand : crushed stone = 1 : 2 : 4 by mass, with water equal to 0.5 times the mass of cement. To make 3 500 kg of concrete, calculate the masses of cement, sand, crushed stone and water (to the nearest kilogram). (4 marks) (b) Explain why steel bars are embedded in structural concrete. (2 marks) |
(a) Total parts = 1 + 2 + 4 + 0.5 = 7.5 [1]; one part = 3 500 ÷ 7.5 = 466.67 kg [1]; cement 467 kg, sand 933 kg, crushed stone 1867 kg [1]; water 233 kg [1] (b) Concrete is strong in compression but weak in tension [1]; the steel bars carry the tensile force and the composite can resist bending loads [1] |
Section C (20 marks)
Elective questions: show your working and give units. Total 60 marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | (a) 25.0 cm³ of white vinegar is diluted to 250.0 cm³. A 25.0 cm³ portion of the diluted solution needs 22.4 cm³ of 0.100 mol dm⁻³ sodium hydroxide to reach the end point. Calculate the concentration of ethanoic acid in the original vinegar in g/100 cm³ (relative molecular mass of CH₃COOH = 60, to 2 decimal places). (3 marks) (b) Phenolphthalein, not methyl orange, should be used for this titration. Explain. (2 marks) (c) State one experimental error that would make the result too high, and explain why. (2 marks) |
(a) Moles of NaOH = 0.100 × 0.0224 = 0.00224 mol [1]; the 25.0 cm³ of original vinegar contains 0.00224 × 10 = 0.02240 mol, i.e. 1.344 g [1]; concentration = 1.344 ÷ 25.0 × 100 = 5.38 g/100 cm³ [1] (b) The end point gives sodium ethanoate, a weakly alkaline solution [1]; the phenolphthalein range 8.3-10.0 matches it, while methyl orange changes in the acidic range and would turn too early [1] (c) The burette was not rinsed with the sodium hydroxide solution, so the solution is diluted [1]; a larger volume is delivered and the calculated acid concentration is too high [1] |
| 8 | (a) In a colorimetric determination, a standard solution of concentration 2.0 × 10⁻⁴ mol dm⁻³ gives an absorbance of 0.45, while an unknown solution in the same cell gives 0.72. Calculate the concentration of the unknown solution. (3 marks) (b) In thin-layer chromatography the solvent front moves 6.0 cm and two spots move 3.6 cm and 2.4 cm. Calculate the Rf values of the two spots. (2 marks) (c) Explain why the spot must not be placed below the surface of the developing solvent. (2 marks) |
(a) In the same cell the absorbance is proportional to concentration [1]; c = 2.0 × 10⁻⁴ × 0.72 ÷ 0.45 [1]; = 3.2 × 10⁻⁴ mol dm⁻³ [1] (b) Rf = distance moved by the spot ÷ distance moved by the solvent front [1]; they are 0.60 and 0.40 [1] (c) The sample would simply dissolve into the developing solvent [1]; the spot would be washed away near the origin, so no separation occurs and Rf cannot be found [1] |
| 9 | (a) A compound gives a mass spectrum with a molecular ion peak at m/z = 58 and a strong fragment peak at m/z = 43. Its infrared spectrum has a strong absorption at 1 710 cm⁻¹ but no broad absorption at 2 500-3 300 cm⁻¹. Identify the compound as propane, propanal or propanoic acid, and explain. (2 marks) (b) Another compound absorbs at both 1 710 cm⁻¹ and 2 500-3 300 cm⁻¹ and has a relative molecular mass of 88. Write its structural formula. (2 marks) (c) 0.1000 g of a monoprotic carboxylic acid (relative molecular mass 88) is dissolved in water and needs 11.36 cm³ of 0.100 mol dm⁻³ sodium hydroxide for neutralisation. Use these data to calculate the relative molecular mass of the acid. (2 marks) |
(a) Propanal [1]; m/z 58 matches C₃H₆O, the 1 710 cm⁻¹ band is the C=O stretch and the absence of a broad band shows there is no O-H, so it is not propanoic acid; propane has no C=O absorption [1] (b) Butanoic acid, CH₃CH₂CH₂COOH [1]; 1 710 cm⁻¹ is C=O, 2 500-3 300 cm⁻¹ is the carboxylic O-H, and M = 88 fits [1] (c) Moles of NaOH = 0.100 × 0.01136 = 0.00114 mol [1]; the acid is monoprotic so it has the same number of moles, M = 0.1000 ÷ 0.00114 = 88.0 [1] |
A TutorZone practice paper; original questions and passages, not publisher material.
