Temasek 2024 Prelim O Chem P2 MS

2024 4E Prelim Pure Chemistry 6092 Section A Mark Scheme S/N Mark Scheme Common Mistakes A1 (a) FeSO4 (b) Al2O3 (c) Ag2CO3 (d) (NH4)2SO4 (e) SiO2 or Al2O3 A2(a) The metal electrodes have a giant metallic lattice structure. [1] It consists of ‘sea of delocalised electrons’ that are mobile to act as charged carriers. [1] A2(b) Q, S, P, R A2(c) X: 1.6 V Y: 0.8 V Accept values ± 0.2 V of answer. A2(d) Oxidation A2(e) Yes [stop marking if wrong] Metal P is more reactive than Metal R, it is able to reduce the oxide of metal R / displace R from the oxide of R. [1]

A3(a) D A3(b) C [1] C is not soluble in solvent 2. [1] A3(c) 1 m for each correct plot of A and D. A3(d) Spray locating agent on the chromatogram. A4(a) It is used as a feedstock for chemicals. B D A C

A4(b) 0.5 m for pentane and propane, 1 m for correct drawing of ethene A4(c) Product 2. [0.5] [stop marking if wrong.] It has the same molecular formula [0.5] of C6H14 [0.5] as hexane, but different structural formula. [0.5] A5(a) Graph X [1] Ethanoic acid ionised partially in water to give a low concentration of hydrogen ions, hence the starting pH should be higher than HCl. [1] Or This is because ethanoic acid is a weak acid hence the starting pH should be higher than that of hydrochloric acid. [1] A5(b) Both graphs show that both acids require the same volume (50 cm3) of NaOH to neutralise.

A5(c) Phenolphthalein is a more suitable indicator. [1] The end-point falls within the colour change interval of phenolphthalein. [1] A6(a) Amide linkage A6(b) and 1m each A6(c) Mr of 1 repeat unit = 14x 2 +2 + 12 x 2 + 16 x 2 + 10 x (12 +2) = 226 [1] No. of repeat units in polymer C = 20800 / 226 = 92 [1]

A7(a) A7(b) 2N2H4(g) → 2NH3(g) + N2(g) + H2(g) Total energy absorbed for bond breaking = 2 x 160 + 8 x 390 = 3440 kJ [1] Total energy released for bond forming = 2 x 3 x 390 + 944 + 436 = 3720 kJ [1] Enthalpy change = 3440 – 3720 [1] = -280 kJ [shown]

A7(c) 1 m for correct named products lower than reactants 1 m for correct direction of arrow of enthalpy change -0.5 m for adding in activation profile A7(d) Dinitrogen monoxide can cause global warming which leads to: Any of the following: • Melting of ice cap, hence flooding of low-lying lands • Droughts A8(a) (i) Hydrogen gas that is produced escaped to the environment.

(ii) No. of moles of magneisum = 1.2/ 24 = 0.05 [0.5] No. of moles of nitric acid = 0.25 x 0.5 = 0.125 [0.5] Moles ratio of Mg : HNO3 1 : 2 0.05 : 0.100 [1] 0.05 moles of Mg requires 0.100 moles of HNO3, but 0.125 moles are given. Hence HNO3 is in excess and Mg is the limiting reagent. [1] A8(b) Either of the following: - Lower temperature - Larger lumps of Mg - Lower concentration of acid (make sure acid still in excess) A8(c) I agree. [1] stop marking if wrong. Mg is more reactive than copper and can displace copper from copper(II) sulfate./ (mention of displacement reaction occuring) [1] Hence, less Mg to react with acid, hence less hydrogen gas will be produced, and less loss of mass will occur. [1] A9(a) Presence of oxygen and water/ moisture A9(b) To act as a barrier to prevent the iron metal from having contact with moisture and oxygen. (prevent rusting)

A9(c) When iron is in contact with copper, it acts as a sacrificial metal and corrodes in place of copper. [1] A9(d) Copper(II) carbonate A9(f) (i) Cathode: Cu2+ + 2e → Cu Anode: Cu → Cu2+ + 2e (ii) The concentration of the solution remains unchanged. [1] For every one mole of Cu 2+ discharged at the cathode, one mole of Cu2+ is formed at the anode. [1] Or The amount of Cu2+ ions formed at the anode is the same as the amount of Cu2+ ions that is discharged at the cathode. A10(a) (i) 2H+ + 2e → H2 (iii) Graphite is cheaper / more readily available. A10 (b)(i) substances oxidation state of chlorine ClO - +1 ClO3 - +5 Cl - -1 All 3 correct – 2m

2 correct – 1m 1 correct - 0 A10(b) (ii) Oxidation state of CI increases from 0 in CI2 to +1 in CIO- / +5 in CIOI3-. Therefore, CI2 is oxidised. [1] Oxidation state of CI decreases from 0 in CI2 to -1 in CI-. Therefore, CI2 is reduced. [1] A10(c) Number of moles of NaCI in 1 tonne of brine = 1000000 x 0.25 / (23 + 35.5) = 4273.5 moles [0.5] Mole ratio: NaCl: H2 is 2: 1 Number of moles of H2 = 4273.5/2 = 2136.8 moles [0.5] Volume of H2 = 2136.8x 24 = 51282 = 51300 dm3 [3 s.f.] [1 for correct 3 s.f.] A10(d) Membrane cell operates at a lower voltage of 3.3V as compared to diaphragm cell of 3.8V, hence cheaper to operate. A10(e) Agree. Chloride ions are preferentially discharged to hydroxide ions due to higher concentration. [1] Hydrogen ions are preferentially discharged to sodium ions as hydrogen is lower than sodium in the reactivity series. [1] Hence, sodium ions and hydroxide ions remain behind.

Section B Mark Scheme S/N Mark Scheme Common Mistakes B11(a) H2O2(aq) + 2H+(aq) + 2 I ‾(aq) → 2H2O(l) + I2(aq) (b) Reducing agent: iodide ions/ I- [1] Stop marking if wrong Explanation: Iodide ions reduces oxygen in H2O2 to H2O causing oxygen to gain electrons. [1] (c) Colourless solution turns yellow/ yellow brown/ brown. [1] (d)(i) Comparing experiment 1 and 4, increasing concentration of KI from 0.1 to 0.3 mol/dm3, increases speed of reaction from 0.00017 to 0.00051 mol/dm3 per second. 1m – trend 1m – quoting values Accept any other comparison of expt. (d)(ii) Comparing experiment 1 and 3, increasing concentration of sulfuric acid from 0.1 to 0.2 mol/dm3 has no effect of the speed of reaction as it remains constant at 0.00017 mol/dm3 per second. 1m – trend 1m – quoting values Accept any other comparison of expt.