2022 HCI Prelim P2 Mark Scheme

2022 HCI C2 H2 Chemistry Prelims / Paper 2 HWA CHONG INSTITUTION 2022 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS (PAPER 2) 1 (a) (i) Hydrocarbons / Alkanes / Alkenes / Arenes [1] 1 (a) (ii) N2 + O2 → 2NO [0.5] 2NO + O2 → 2NO2 [0.5] 1 (a) (iii) Forms photochemical smog / leads to acid rain [1] 1 (b) (i) pV = nRT n(gas) = pV ÷ RT = (10 × 105)(63.2 ÷ 1000) ÷ (8.31)(900+273) = 6.48 mol [1] correct substitution and conversions [1] answer (ecf) 1 (b) (ii) Amount of SO2 = 0.0183 ÷ 64.1 = 2.85 × 10–4 mol Number of SO2 particles = 2.85 ×10–4 × L (where L is Avogadro constant) Number of particles in fuel exhaust = 6.48 × L Amount of SO2 in ppm = [(2.85 × 10–4 × L) ÷ (6.48 × L)] × 1 × 106 = 44.1 ppm The sample meets the sulfur emission standards [1] Finding the number of SO2 particles [1] Calculating the concentration of SO2 in ppm and correct conclusion (ecf given from part (b)(i)). ([–0.5] if comparison is correct but answer in ppm is absent)

1 (c) • SO3 dissolves in water in a violent / very exothermic reaction • forming a strongly acidic solution around pH 1 • SO3(l) + H2O(l) → H2SO4(aq) • Na2O dissolves completely in water in a vigorous / exothermic reaction • forming a strongly alkaline colourless solution around pH 13 • Na2O(s) + H2O(l) → 2Na+(aq) + 2OH−(aq) • Al2O3 is insoluble / has no reaction in water • The resulting mixture is neutral / pH 7 8 points – [4] 5 to 7 points – [3] 4 points – [2] 2 to 3 points – [1] 1 point – [0.5] [4] total 2 (a) (i) Intermediate A [1], step 1: C l2(aq) [1], step 2: excess conc H 2SO4, heat OR concentrated H3PO4, heat OR Al2O3, heat [1] WRONG INTERMEDIATE A LOSES MARKS FOR STEP 1 AND STEP 2. Intermediate B [1]

2022 HCI C2 H2 Chemistry Prelims / Paper 2 2 (a) (ii) The –CyO2H is nearer the electron withdrawing amine group, that better disperses the negative charge on the conjugate carboxylate (–CyO2–), stabilising th is conjugate carboxylate more, this makes the –CyO2H group more acidic and hence is deprotonated first. All 4 points below = [2]. 2 out of 4 points = [1] - amine group (NH2) is nearer –CyO2H than to –CxO2H (compares distance) - amine group is electron withdrawing - negative charge on the conjugate base (–CyO2–) is better dispersed - conjugate base of –CyO2H is more stable (than that of –CxO2H) 2 (a) (iii) Naturally occurring MSG consists of only 1 enantiomer. MSG synthesised in the laboratory will be a racemic mixture, the enantiomer produced in lab synthesis may interact with the taste bud differently. [1] for realising the lab synthesis results in a racemic mixture, or at least acknowledge different ratios of enantiomers may be produced in lab synthesis vs the natural process. 2 (b) Relative mass of MSG = 183 Amount of Na+ in 1 g of MSG = 1 ÷ 183 = 5.46 × 10–3 mol [1] Relative mass of NaCl = 58.5 Amount of Na+ in 1 g of NaCl = 1 ÷ 58.5 = 1.71 × 10–2 mol [1] There are less sodium ions in 1 g of MSG than in 1 g of table salt. OR correct calculations of percentage by mass OR actual mass of Na+ ions 3 (a) C [1] 3 (b) At pH 1:

At pH 14: Any correct tripeptide drawn [1] At pH 1: Protonation of correct functional group i.e. amine group [1] At pH 14: Deprotonation of correct functional groups i.e. carboxylic acid and phenol groups [1] 3 (c) (i) The lone pairs on Nx and Ny can delocalise over the C=N group, making the lone pair on both N atoms much less available for donation to H+. [1] 3 (c) (ii) Concentration of arginine = 0.100  20 ÷ 10 = 0.200 mol dm−3 [1] 3 (c) (iii) Ka1 = x2 / (0.200 – x) = 10−2.17 Assume that x << 0.200, Ka1 = x2 / (0.200) = 10−2.17 x = [H+] = 0.03677 mol dm−3 pH = –lg 0.03677 = 1.43 Calculate [H+] correctly [1] allow ecf from (c)(ii) Calculate pH correctly [1] allow ecf from [H+] calculated 3 (c) (iv) 2.17 [1]

2022 HCI C2 H2 Chemistry Prelims / Paper 2 3 (c) (v) [1] As there is a large reservoir of the acid and conjugate base, [1] the addition of OH − will only lead to a small change in the ratio of the [acid] : [conjugate base]. Since [H +] = Ka  [acid]/[conjugate base] and Ka is constant, the pH change will be very gradual. [1] 3 (c) (vi) Put an X at the point when volume = 40 cm3 [1] 4 (a) (i) A ligand is an ion or molecule with one or more lone pairs of electrons available for donation to the vacant orbitals of a metal atom or ion. [1] 4 (a) (ii) (i) Nickel(II) contains partially filled d-orbitals. (ii) In the presence of ligands, the degenerate d -orbitals split into two distinct energy levels with an energy gap, E. (iii) When light shines on the solution, an electron from the lower energy level will absorb light of a specific wavelength and is promoted to a vacant orbital at a higher energy level. (iv) The wavelength of light absorbed corresponds to the wavelengths of visible light. (v) The complementary colours are seen, and so, the colour of each complex will be different. (vi) When the ligands change from H2O to NH3, the magnitude of the energy gap, E, also changes, so the solution changes colour. Explanation of the origin of colour (points (i) – (v) above) 5 points – [2] 3 to 4 points – [1] 1 to 2 points – [0] [1] Explain why a colour change occurs during ligand exchange (point (vi) above) 4 (b) (i) The entropy change is positive because there are more ways to distribute energy since there is an increase in the amount of aqueous species in the solution. The entropy change is large as the amount of products is significantly greater than the amount of reactants in this system. [1] For explaining why the change in entropy is positive [1] For explaining why the change in entropy is large

4 (b) (ii) Gor = Hor – TSor = –17.0 – 298(121/1000) = –53.1 kJ mol–1 [1] 4 (b) (iii) Gor is a negative number, this shows that the position of equilibrium lies to the right i.e. it is product favoured. Thus, ethane-1,2-diamine binds preferentially with Ni2+. [1] Correct identification of position of equilibrium based on (b)(ii) (ecf) [1] Correct identification of preferred ligand based on position of equilibrium (ecf) 4 (b) (iv) The coordination of the first –NH2 group is an intermolecular reaction, but the second coordinate bond is an intramolecular reaction , which occurs much more quickly than an intermolecular reaction. (or other words to that effect) OR The complexation of three bidentate ligands involve three sequential bimolecular reactions, which has a higher probability of occurring than the complexation of ammonia ligands, which will involves six sequential bimolecular reactions. Thus the rate of reaction involving en ligands is faster. (or other words to that effect) [1] for a suitable explanation using Collision Theory 4 (c) (i) [1] Correct drawing of [Ni(en)3]2+ [1] Correct drawing of enantiomer 4 (c) (ii) [Ni(en)3]2+ is chiral because its mirror image isomers are non-superimposable. OR [Ni(en)3]2+ is chiral because it has no internal plane of symmetry. [1]

2022 HCI C2 H2 Chemistry Prelims / Paper 2 4 (c) (iii) The enantiomers have the same boiling point / melting point / colour / density / electrical conductivity. The enantiomers rotate plane-polarised light in opposite directions. [1] Stating one similarity [1] Stating one difference 5 (a) (i) [1] diagram (–½ m per mistake in axes labels, starting from origin, single curve) A catalyst provides an alternative reaction pathway which requires a lower activation energy (Ea') than the uncatalysed reaction (Ea). [½] As represented by the shaded areas in the Boltzmann diagram, there is an increase in the proportion/ fraction/ number of reactant particles that have k