2025 Prelim P2 MS (for exchange)_H2 Chem_HCI

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme/Paper 2 1 HWA CHONG INSTITUTION 2025 C2 H2 CHEMISTRY PRELIMINARY EXAMINATION MARK SCHEME Paper 2 1 (a) Na⁺, Mg²⁺, Al³⁺ and Si⁴⁺ are isoelectronic (or have the same number of electrons) and P³⁻, S²⁻ and Cl⁻ are isoelectronic, hence shielding effect is the same. Across each series, number of protons and thus nuclear charge increases. [1] Thus, effective nuclear charge increases, attraction of the nucleus on the valence/outermost electrons become stronger (or outermost electrons pulled closer to the nucleus), and ionic radius decreases. There is a sharp increase in radius from Si⁴ ⁺ to P³ ⁻, as anions have one more quantum/electron shell than cations. [1] (b) P4O10(s) + 6H2O(l) → 4H3PO4(aq) [0.5] pH of resulting mixture is 2 (accept 1 – 3). [0.5] (c) Al2O3 is amphoteric. [0.5] Al2O3 (s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2O(l) [0.5] Al2O3 (s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq) [1] (d) (i) A Lewis acid is a species that accepts an electron pair. [1] (ii) x

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme/Paper 2 2 (iii) [1] 2 (a) (i) The heat evolved when 1 mole of water is formed from an acid-base reaction at 298 K and 1 bar. [1] (ii) q = mcT = [(25+45) × 1.00](4.18)(30.5 – 27.8) = 733.6 J [1] no. of moles of H2O = no. of moles of LiOH (limiting reagent) = 25/1000 × 0.5 = 0.0125 mol ∆𝐻 𝑛𝑒𝑢𝑡 ꝋ = – q nH2O = – 733.6 0.0125 = –58688 J mol–1 = –58.7 kJ mol–1 (3 s.f.) [1] (b) (i) ∆𝐻 2 ꝋ 2HCl(aq) + 2LiOH(s) + CO2(g) Li2CO3(s) + H2O(l) + 2HCl(aq) 2(–21) –29.0 2HCl(aq) + 2Li+(aq) + 2OH–(aq) + CO2(g) 2LiCl(aq) + CO₂(g) + 2H₂O(l) [2] 2(–58.7) ∆𝐻 2 ꝋ = 2(–21) + 2(–58.7) + 29.0 = –130.4 kJ mol–1 = –130 kJ mol–1 (3 s.f.) [1] *If used suggested value in question (∆𝐻 𝑛𝑒𝑢𝑡 ꝋ = −48.2 kJ mol−1)* ∆𝐻 2 ꝋ = 2(–21) + 2(–48.2) + 29.0 = –109.4 = –109 kJ mol–1 [1] (ii) Since the no. of moles of gas decreases from 1 to 0 in this reaction, the entropy change of reaction is negative. [1] G = H – TS When temperature increases, –TS becomes more positive . Since H is negative, G becomes less negative, and reaction is less spontaneous. [1] (iii) Volume of CO2 exhaled per crewmember per day = 25 × 15 × 60 × 24 = 540000 cm3/CM-d = 540 dm3/CM-d [1] Mass of CO2 exhaled = 1.98 × 540 = 1069.2 g/CM-d = 1.07 kg/CM-d [1] (iv) No. of moles of CO2 /CM-d = 1069.2 / (12+16+16) = 24.3 mol No. of moles of LiOH/ CM-d = 24.3 × 2 = 48.6 mol [0.5] Mass of LiOH /CM-d = 48.6 × (6.9+16+1) = 1161.54 g = 1.16 kg [0.5] Mass of LiOH brought on trip = 250 × 5.2 = 1300 kg [0.5] No. of days = 1300 / (8 × 1.16) =139.9 = 139 days [0.5] Alte