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] Alternative method: Total no. of moles of LiOH supplied = 5.2 × 1000 (6.9+16+1) = 54393 mol Total no. of moles of CO2 can be absorbed = 54393/2 = 27196.5 mol Total mass of CO2 that can be absorbed = 27196.2 × (12+16+16) = 1196600 g = 1196.6 kg Max no. of days = 1196.6/(1.07 × 8) = 139.8 = 139 days

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme/Paper 2 3 (v) For the same number of moles, the mass of LiOH is smallest hence it is easier to transport in a space craft. [1] (c) (i) Li2CO3(s) → Li2O(s) + CO2(g) [1] Li+ is a cation with small ionic radius and high charge density [1], it polarises the electron cloud of carbonate to a large extent, hence weakening the C -O bonds [1]. Hence less energy is needed to break the C-O bonds and it decomposes easily when heated. (ii) ion Li+ Na+ Mg2+ Ca2+ ionic radius / nm 0.06 0.095 0.065 0.099 q+ r+ 16.7 10.5 30.7 20.2 Marking points [2] • quote ionic radius for all 4 ions • suggest a temperature between 635 and 851 (excluding the two numbers) • correct explanation in terms of charges and ionic radii: calculate q+ r+ and conclude that charge density of Li+ is between those of Na+ and Ca2+ 3 (a) (i) 50%. The optically inactive mixture of P contains equal proportions of two enantiomers [0.5]. Enzyme amidase has specific activity and only one enantiomer can bind to the active sites of the enzyme [0.5]. (ii) Amidase is a biological catalyst which is regenerated during the reaction / remains chemically unchanged / not used up / lower E a [0.5]. Hence, its concentration stays constant [0.5]. H2O is a solvent and is present in large excess [0.5]. Hence, its concentration can be treated as a constant (or differs very slightly or pseudo-zeroth order) [0.5]. (iii) At lower concentrations of P From experiments 1 and 2 (accept if compare expt 2 and 3), when [P] doubles from 0.50 to 1.00 mmol dm⁻³, the initial rate doubles from 1.25 to 2.50 × 10⁻⁴ mol dm⁻³ s⁻¹.[1] Reaction is first order with respect to [P]. [0.5] At higher concentrations of P From experiments 5 to 6, when [P] doubles from 8.00 to 16.00 mmol dm ⁻³, the initial rate remains constant at 8.00 × 10 ⁻⁴ mol dm⁻³ s⁻¹.[1] Reaction is zeroth order with respect to [P]. [0.5] or From experiments 4 to 5, when [ P] doubles from 4.00 to 8.00 mmol dm ⁻³, the initial rate remains almost constant at 7.50 × 10 ⁻⁴ mol dm⁻³ s⁻¹ (accept increases slightly). Reaction is zeroth order with respect to [P]. (iii) At low [P], rate of reaction increases proportionally with increasing [P] (or follow first order kinetics) [0.5] as [amidase] is greater than [ P] / [P] is the limiting factor ( accept there are many active sites available on the enzyme amidase / binds ->reaction->released / form enzyme-substrate complex). [0.5]

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme/Paper 2 4 At high [ P], rate of reaction remains constant with increasing [ P] (or follow zero th order kinetics) [0.5] as the active sites on the enzyme amidase are saturated (accept all active sites are filled) with P molecules. [0.5] (b) [1] pH 3 [1] pH 14 (c) [1] (d) 4 (a) (i) disproportionation [1] (ii) 3 Cl2 + 6 NaOH → 5 NaCl + 1 NaClO3 + 3 H2O [1] (b) (i) Eo (Cl2/Cl –) = +1.36 V, Eo (O2/H2O) = +1.23 V [1] Since Eo (O2/H2O) is less positive than Eo (Cl2/Cl –) [1], H2O will be selectively oxidised. (ii) The high concentration of Cl – (5.0 mol dm–3) [0.5] causes the position of equilibrium Cl2 + 2e– ⇌ 2Cl– to shift to the left . [0.5] This causes E(Cl2/Cl –) to become less positive than +1.23 V, resulting in the selective oxidation of Cl –over H2O. [1] (c) total charge passed = 5000 × 60 × 60 = 18 000 000 C [0.5] actual charge to produce Cl2 = 18 000 000 × 0.88 = 15 840 000 C [0.5] amount of e – = 15 840 000 = 164.145 mol [0.5] 96500 amount of Cl2(g) = 164.145 = 82.07 mol [0.5] 2 volume = (82.07)(8.31)(70 + 273) = 2.34 m3 105

2025 HCI C2 H2 Chemistry Prelim Exam Mark Scheme/Paper 2 5 (d) (i) NaHg + H2O → NaOH + ½H2 + Hg [1] (ii) It can be used in fuel cells / to hydrogenate alkenes / in the Haber Process. [1] (e) (i) The electronegativity of an atom is a measure of its ability to attract the electrons in a covalent bond to itself. [1] (ii) chemical bond explanation ionic There is a transfer of electrons from the less electronegative Na to more electronegative Hg to form Na + and Hg-rich polyanions like Hg44– and Hg22–. metallic Electrons are lost from low electronegativity Na atoms to form a sea of delocalised electrons. [1] each correct explanation 5 (a) The order of thermal stability is: H−Cl > H−Br > H−I [0.5]. Down the group, atomic radius (or size of halogen atom) increases, hence orbital overlap (with H atom