2024 H1 Prelim Answer HCI

2024 Prelim Exam H1 Chemistry For internal circulation only 2024 Preliminary Examination H1 Chemistry (8873) Suggested Solutions Paper 1 1 C 11 A 21 A 2 B 12 B 22 C 3 C 13 C 23 C 4 D 14 C 24 D 5 B 15 A 25 A 6 C 16 D 26 B 7 A 17 D 27 D 8 A 18 B 28 A 9 D 19 A 29 A 10 B 20 B 30 D Paper 2 Section A 1(a) Number of neutrons 78 [0.5] Number of electrongs 52 [0.5] (b) Te+ ion will be deflected towards the negatively charged plate while the electron will be deflected towards the positively charged plate. Te atom will not be deflected. [1] The angle of deflection of Te+ is much smaller than that for the electron. [1] (c)(i) Te (g) → Te+ (g) + e− [1] (c)(ii) The first ionisation energy increases across Period 3 [0.5] as the nuclear charge increases [0.5] while the shielding effect remains relatively constant or effective nuclear charge increases [0.5] and so more energy is required to remove the first outermost electron across the period. [0.5] (c)(iii) The paired electrons in the p orbital of Te experiences inter-electronic repulsion such that it takes less energy to remove the first outermost electron compared to the preceding element. [1] Note: 4d electrons are poor at shielding the outer electrons from the nuclear charge.

2 (d)(i) [0.5] general increase from 1 to 4 [0.5] general increase from 5 to 6 [0.5] jump from 4 to 5 [0.5] bigger jump from 6 to 7 (d)(ii) As electrons are successively removed from the Te atom, nuclear charge remains the same or number of protons remain the same [0.5] but the remaining electrons experiences stronger attraction by nucleus or shielding effect decreases or increase in effective nuclear charge.[0.5] (e) There are 6 bond pairs and 0 lone pairs of electrons around Te. [1] The 6 electron pairs are arranged as far apart as possible around Te to minimise repulsion and maximise stability. [1] (f)(i) TeF6 + 6H2O → Te(OH)6+ 6HF [1] (f)(ii) hydrolysis [1] 2(a)(i) NaCl [0.5] MgCl2 [0.5] [-0.5] for wrong entry (a)(ii) AlCl3 and SiCl4 and PCl5 [0.5] for any 2 correct ones [1] for all 3 (b)(i) –58.2 = 8 (264) + 4 (244) – 4 [ 264 + 2 (S–Cl) ] [1] quote of the correct values = 261 kJ mol–1 (3 s.f.) [1] multipliers [1] bond break – bond form, ecf (b)(ii) The enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states at 298 K and 1 bar. [1] (b)(iii) –58.2 S8(s) + 4Cl2(g) 4S2Cl2 (l) +4Cl2(g) 8 x ΔH 4(–40.6) +4Cl2(g) 8SCl2(l)

3 8 × Hf = –58.2 + 4(–40.6) [1] correct values at the arrows and = –27.6 kJ mol–1 [1] multipliers [1] apply the correct Hess’ Law, ecf (c)(i) S2Cl2 is oxidised to SCl2; oxidation number of sulfur increases from +1 to +2. [1] Cl2 is reduced to SCl2; oxidation number of chlorine decreases from 0 to –1. [1] (c)(ii) Cl2 + 2e– → 2Cl– Charge on 1 mole of electron = 6.02 × 1023 × (–1.60 × 10–19) = 96320 C [1] No. of moles of electrons = 2 × 1.5 = 3 mol [0.5] charge = 3 × 96320 = 288960 C ≈ 289000 C [0.5] (c)(iii) Position of equilibrium shifts to the right to decrease the number of moles of gases to offset the increase in pressure. [1] Equilibrium constant remains constant because it is only dependent on temperature [1] (c)(iv) S–Cl bonds are polar [1] and individual dipoles do not cancel off (or net dipole is not zero) so SCl2 is a polar molecule. [1] 3(a) The longer side chains in LDPE hinders the polymer chains from coming close together, resulting LDPE having weaker dispersion forces between the chains/molecules, so it is less strong / with less tensile strength and less useful. [1] (b) 4-methylhex-1-ene [1] Propene [1] [1] (c) Addition peak at 700-800 cm-1 [1] (d)(i) C15H10O2N2 [1] (d)(ii) [1] (d)(iii) Water molecules are able to form hydrogen bonds with the amide groups in the polyamide chains of Nylon-6, disrupting the existing hydrogen bonds between the chains and their arrangement. [1] When the water evaporates, new hydrogen bonds are formed between the chains, locking them in new positions, resulting in creases. [1] Polyesters like PET do not form extensive hydrogen bonds with water so the arrangement of the polymer chains are less disrupted hence they are less prone to creasing. [1]

4 4(a) Sugarcane maybe considered to be ‘carbon neutral’ because during the process of photosynthesis, the plant removes carbon dioxide gas from the atmosphere. The fermentation of glucose to ethanol and the combustion of ethanol then release carbon dioxide gas back into the atmosphere. Overall, the process removes carbon dioxide gas from the atmosphere, and returns carbon dioxide gas to the atmosphere in the same quantities [1] explain how the amount of carbon remains the same from the source to the usage Ethanol produced by fermentation must be purified by distillation. Distillation requires high temperatures and hence requires energy which could come from the combustion of fossil fuels, a process which will release carbon dioxide gas into the atmosphere. OR Ethanol can also be prepared by an addition reaction between ethene and steam. This reaction does not remove any carbon dioxide gas from the atmosphere, but takes place at a high temperature (300 C) and pressure (70 atm.) which require a large amount of energy. This energy could be produced by burning fossil fuels which releases carbon dioxide gas into the atmosphere. [1] explain how the amount of carbon does not remain the same. (b)(i) For every 100 g of E80 fuel, there is 80 g of ethanol. Mass of oxygen in 100g of E80 fuel = 16 46 × 80 = 27.8 g Percentage mass of oxygen = 27.8% [1] (b)(ii) An increased percentage of oxygen in the fuel means that more of the fuel will undergo complete combustion/burns completely. This will release more energy, making the fuel more efficient. OR This will reduce the levels of harmful / toxic pollutants e.g. carbon monoxide. OR This will require less oxygen for complete combustion. [1] (c) Ethanol oxidises in air to form ethanoic acid. [0.5] Both ethanoic acid and ethanol undergoes esterification / condensation to form ethyl ethanoate. [0.5] (d) GGE of LPG = 34200 26500 = 1.3 [1] accept 1.29 To release the same amount of energy as 1 L of petrol, and hence in order to travel the same distance as it would be travelled on 1 L of petrol, the vehicle would have to store 1.3 L of LPG. The storage tank that is required to store LPG must be 1.3 times greater than that to store petrol for the same amount of energy output/same distance travelled. [1] must use 1.3 (e) Mass of octane in 1L = 0.75 𝑔 𝑐𝑚3 × 1000 𝑐𝑚3 = 750 g Amount of octane in 1L = 750 114.0 = 6.579 mol [0.5] Energy density per L of octane = 5470 × 6.579 = 36000 kJ/L [0.5] The energy density in Table 4.1 is for the fuel which contains a mixture of compounds / impurities while the approximate value assumes the fuel only contains octane as the only pure compounds. [1] (f)(i) (excess) concentrated sulfuric acid, heat [1]

5 (f)(ii) [1] (g)(i) Reactant molecules are adsorbed onto the active sites on the surface of the catalyst by forming weak interactions. [1] Adsorption brings reactant molecules closer together and weakens the covalent bonds within the molecules, hence lowering the activation energy. [1] After the reaction, the product molecules desorb from the surface and the active sites are available for further reaction. [1] (g)(ii) Nanomaterial is a material with at least one dimension on the nanoscale (1-100nm). Nanoparticle is a material with all 3 dimensions on the nanoscale (1-100nm). [1] (g)(iii) High surface area to volume ratio. [1] (h) [1] only 1 layer with at least 3 rings Graphene has a macromolecular/giant covalent structure, where each c