FHSS Chem Prelim Ans

1 FUHUA SECONDARY SCHOOL Sec 4E Chemistry 6092 Preliminary Examinations 2025 – Mark Scheme PAPER 1 1 2 3 4 5 6 7 8 9 10 C A B A D A A B A A 11 12 13 14 15 16 17 18 19 20 D C C B B A D D C B 21 22 23 24 25 26 27 28 29 30 B A D B A A C C B D 31 32 33 34 35 36 37 38 39 40 A B D B C C B C D C

1 FUHUA SECONDARY SCHOOL Sec 4E Chemistry 6092 Preliminary Examinations 2025 – Mark Scheme PAPER 2 Section A [70 marks] Q Answer Mark Remarks 1a B and E 1 b A, D and F 1 c A 1 d Two B3+ ions Three D2- ions 1 1 Total 5 marks 2a No. I disagree with the student. A catalyst provides an alternative pathway with lower activation energy for the reaction to proceed but the enthalpy change remains unchanged as the energy levels of reactant and product remain the same. Accept both not accepted be catalyst. 1 1 b Stage 1: S is oxidised as the oxidation state of sulfur increases from 0 in S to +4 in SO2 . O2 is reduced as the oxidation state of oxygen decreases from 0 in O2 to -2 in SO2 . Stage 2: SO2 is oxidised as the oxidation state of sulfur increases from +4 in SO2 to +6 in SO3 and O2 is reduced as its oxidation state decreases from 0 in O2 to -2 in SO3. ; ; ; ; 4; [3] c Stage 2 is a reversible reaction AND some of the sulfur trioxide formed decompose to form sulfur dioxide and oxygen. 1 di Shape of graph Label axes with units and include correct reactant and product Enthalpy change and activation energy, arrow must be correct ; ; ; 3; [2] ii Moles of SO3 = 500000 (32+3×16) = 625 Moles of SO2 = SO3 = 625 Total amount of heat released = 625/2 × 196 = 61 300 kJ 1 1 Total 10 marks H = +196 kJ/mol Activation energy

2 3ai 2H+(aq) + 2e →H2 (g) 1 ii Solution turned from colourless to reddish-brown. Chlorine is produced at electrode Q as chloride ions are selectively oxidised. Chlorine is more reactive than bromine [;] and displaces bromine from its halide solution.[;] Aqueous bromine formed is reddish-brown. 1 1 1 2; [1] iii No gas collected in the gas syringe / reading on syringe is 0 cm3 [;] as Cu2+ ions are selectively reduced at electrode P to form Cu. [;] No visible change / solution remained colourless [;] Oxygen gas, instead of chlorine gas is produced at electrode Q.[;] as OH⎯ ions are selected oxidised at electrode Q to form oxygen gas. [;] 3 5; [3] 3-4;[2] 1-2; [1] bi Electrons flow from Mg to Pb. 1 ii Electrode 3 [;] as Ag+ ions are selectively reduced by gaining electrons at electrode 3 to form Ag deposited on toy car. [;] Ag+(aq) + e → Ag(s) [;] 3 ; [2] iii pH decreases to below 7. As Ag+ ions and OH- ions are selectively reduced and oxidised at electrode 3 and 4 respectively [;], H+ and NO3 – ions are left in the electrolyte / concentration of H+ ions in electrolyte increases / electrolyte becomes increasingly acidic.[;] 1 1 2; [1] Total 12 marks 4a 38 and 71 1 b F2 (g) +2I−(aq) → 2F−(aq) + I2(aq) 1 c Rate of diffusion of gases is directly proportional to 1/time taken AND for Rate to be directly proportional to 1/t and 1/t has to be directly proportional to Mr of the gas, the product of Mr × time taken must be equal. Data Processing For F2, Mr × t = 38 × 16 = 608 s For Cl2, Mr × t = 71 × 29 = 2059 s Since the products are different, the measurements do not support the data. OR Rate of diffusion of gases is indirectly proportional to time taken AND for Mr of the gas to be indirectly proportional to time taken, the product of Mr × time taken must be equal. 1 1 Total 4 marks

3 5ai displayed formula for organic compounds 1 1 1 ii butanol and methanoic acid 1 iii Add aqueous bromine to a sample of each solution. If reddish brown aqueous bromine decolourise rapidly, the solution is A. Add magnesium/sodium carbonate to remaining three solutions, if effervescence is observed, the solution is D. Add acidified potassium manganate (VII) to a sample of remaining two solutions and warm them. If purple potassium manganate (VII) turns colourless, the solution is B. The remaining unidentified solution is C. ; ; ; ; Sequence to be correct to be awarded full credit. 4; [3] bi correct formula for all the structures [1] balanced chemical equation [1] + 3 CH3OH → 3 2 ii A glycerol molecule contains 3 hydroxyl/-OH functional groups. 1 iii 1 Total 11 marks 6a Ethylamine acts as a base as it accepts a proton from water. 1 bi Moles of sulfuric acid = 20/1000 ×0.25 = 0.005 Moles of sodium hydroxide = 0.01 Concentration of sodium hydroxide = 0.01 / (25/1000) = 0.400 mol/dm3 1 1 + H2O + Cl2 + HCl

4 ii ethylamine is a weak base that ionises partially to give a lower concentration of OH- ions and hence a lower initial pH. Hydrochloric acid is a monobasic acid and hence one acid unit gives half the no. of moles of H+ ions and hence double the volume of acid required to neutralise the ethylamine. 1 1 1 Total 6 marks 7a Mass of CO2 given off = 0.25 g Moles of PbCO3 = Moles of CO2 = 0.25 / 44 = 0.0056818 Mass of PbCO3= 0.0056818 × (207 + 12 + 3 × 16) = 1.5170 g % purity of PbCO3 sample = 1.517 / 2.00 × 100% = 75.9% 1 1 Ignore if no ‘g’ Ignore if > 3sg but reject if no ‘%’ bi steeper gradient but yield lower, final mass above 125.75 1 ii Lead (II) carbonate reacts with sulfuric acid to form insoluble lead(II) sulfate which coats around lead(II) carbonate prevents it from further reaction with acid, giving a low yield of carbon dioxide gas Sulfuric acid is a dibasic acid which will give twice the concentration of H+ ions leading to a faster rate of reaction. 1 1 Starting pH 8 – 10 volume at 40 cm3 final pH at 0 - 2 experiment 2

5 ci gentler gradient and yield same at 125.75. 1 ii Ethanoic acid dissociate partially in aqueous solution to produce H+ ions [;] Ethanoic acid produces a lower initial concentration of H+ ions [;] in aqueous solution than hydrochloric acid. Hence produces a slower initial rate of reaction. However, same mass/moles of the limiting reactant PbCO3 is used and hence gives the same yield of gas. 1 1 2;[1] d Add aqueous lead(II) nitrate to a fixed volume of aqueous sodium carbonate till no precipitate is formed. Filter the mixture to remove lead(II) carbonate as residue. AND Wash the residue with distilled water and leave the precipitate to dry. 1 1 Total 10 marks 8ai PET: HDPE: 1 1 bi PET: 0.03 kg × 1.9 = 57 g CO₂ HDPE: 0.04 kg × 1.8 = 72 g CO₂ Tritan: 0.15 kg × 4.5 = 675 g CO₂ ; ; ; 3; [2] 1-2; [1] ii copolyester per use = 675 ÷ 50 = 13.5 g CO₂/use 1 experiment 3

6 c PET HDPE copolyester Factor 1: atom economy PET = (238- 36)/238 × 100% = 84.2% HDPE = highest (100%), no by- products copolyester = (310-36)/310 × 100% = 88.4% HDPE is most sustainable as it has 100% economy, no wastage of raw material and is widely recycled. [1] Factor 2: carbon footprint PET = 57 g CO₂ per use HDPE = 36 g CO₂ per use copolyester = lowest per use, 13.5 g Copolyester yields the lowest carbon footprint per use. [1] Factor 3 recyclability widely recycled widely recycled not recycled 1 1 1 [1] for calculated data on atom economy [1] for comparison d Mr of repeat unit = [C8H6O4- 2OH] + [C8H16O2- 2H] = 132 + 142 = 274 When Mr = 42 000, number of repeating units = 42 000/274 = 153.28 [;] = 154 [round up] When Mr = 50 400, number of