2019 HCI Prelim H2 Chem P2 ANS

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 Paper 2 1 (a) (i) HCN with trace amount of KCN (or NaOH(aq)), cold [1] (ii) [1] for each correct structure A B C (b) (i) Mechanism: Bimolecular Nucleophilic Substitution (SN2) [1] [1] for:  OH– is the nucleophile that attacks C+ of C–I (attack from the side that’s opposite to the I)  partial charges: C+–Cl–  curly arrow going from lone pair of OH– to C+ of C–I  curly arrow going from C–I bond, to I [1] for:  transition state negatively charged (need not label “transition state”)  I– is produced at the end of the reaction  no slow or fast step −½m for every mistake HWA CHONG INSTITUTION 2019 C2 H2 CHEMISTRY PRELIMINARY EXAM SUGGESTED SOLUTIONS

(ii) [1] shape of graph [1] labelling graph  label Ea  label ∆H  label “reactants” and “products”  label axes −½m for every mistake (c) The halogens Cl2, Br2 and I2 exist as simple non-polar covalent molecules. From Cl2 to Br2 to I2, the size of the electron cloud and hence, the polarisability of the halogen molecule increases [1]. More energy is needed to overcome the stronger dispersion forces between the molecules. Hence, the volatility of the halogens decreases [1] from chlorine to iodine. (d) (i) 2HX → H2 + X2 [1] (ii) Down the group, as atomic radius increases from Cl to I, the bond length of the H–X bond increases / bond strength decreases [1]. Hence, less energy is needed to break the H –X bond . Thus, the thermal stability of the hydrogen halides decreases down the group [1]. Remarks: can also justify by quoting the bond energy data, H–Cl = +431, H–Br = +366, H–I = +299 kJ mol –1. Weaker H–X bond needs less energy to break during thermal decomposition (e) From Data Booklet, Eo/V Br2 + 2e 2Br– +1.07 I2 + 2e− 2I− +0.54 Bromine reacts with S2O32– to form S4O62–, which will be further oxidised to SO2 and subsequently SO42–. Eocell = +1.07 − (+0.09) = +0.98 V > 0 (spontaneous) Eocell = +1.07 − (+0.51) = +0.56 V > 0 (spontaneous) Eocell = +1.07 − (+0.17) = +0.90 V > 0 (spontaneous) Iodine reacts with S2O32– to form S4O62– but there is no further oxidation of S 4O62– to SO2. Eocell = +0.54 − (+0.09) = +0.45 V > 0 (spontaneous) Eocell = +0.54 − (+0.51) = +0.03 V > 0 (spontaneous but extent of reaction is too small) Energy CH3CHICH2CH3 + NaOH CH3CH(OH)CH2CH3 + NaI H < 0 Ea Progress of reaction

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 [1] Eocell for reaction with bromine or showing that the Eo(Br2/Br-) > all the 3 Eo [1] Eocell for reaction with iodine or showing that Eo(I2/I-) > +0.09 V but only slightly larger than +0.51 V [½] spontaneous since Eocell > 0 [½] spontaneous but extent of reaction too small 2 (a) (i) Ka = [H+][A–] / [HA] Ka = (10–5.5)(1) / 4 Ka = 7.91 × 10–7 mol dm–3 pKa = – lg(7.91 × 10–7) = 6.10 [1] OR pH = pKa + lg[A–]/[HA] 5.5 = pKa + lg(1/4) pKa = 6.10 [1] (ii) 1:1 [1] (iii) Let x be the volume of NaOH required to achieve maximum buffer capacity For visualization: 0 cm3 x 20.00 cm3 Since the ratio of the [A–]:[HA] in the buffer is 1:4, a titre value of 20.00 cm3 must correspond to 4 units. Hence, maximum buffer capacity must occur when there are 2.5 units of [A–] and 2.5 units of [HA]. [1] Therefore, the volume of NaOH required, x, corresponds to 1.5 units. Volume of NaOH required = 20.00/4 × 1.5 = 7.5 cm3 [1] OR Let x be the volume of NaOH required to achieve maximum buffer capacity Since the ratio of the [A–]/[HA] in the buffer is 1:4, (20 – 2x) / (20) = 1/4 [1] 2(10 – x) = 20/4 x = 7.5 cm3 [1] (iv) [1]

(v) Given that the region of rapid pH change will be at about pH 9.4, I would use an indicator with a working range that coincides with that pH range [1], which would be cresolphthalein. [1] (b) (i) Initially, the graph is a s traight line/increases steadily. This is because the reaction is first order wrt to [sugar]/ rate is directly proportional to [sugar]/ more enzyme-substrate complex can be formed by increasing [sugar]. [1] Thus, the rate of reaction increase. The graph p lateaus/becomes horizontal/becomes zero order wrt [sugar] because the enzyme is now saturated /all active sites are occupied [1], thus, the rate of reaction cannot be increased by increasing [sugar]. (ii) [1] Key points  Initial rate increases more quickly  Plateau occurs at a higher concentration of sugar  Plateau occurs at a higher rate (iii) Order wrt to [sugar]: Comparing experiment 1 and 2, When [sugar] is doubled from 0.20 mol dm–3 to 0.40 mol dm–3, the relative rate of reaction doubled. Thus, [sugar] is directly proportional to the rate of reaction, and is first order. [1] for explanation Order wrt to [zymase]: Comparing experiment 1 and 3, When [sugar] is halved from 0.20 mol dm–3 to 0.10 mol dm –3, the relative rate of reaction is expected to halve. When [zymase] is doubled from 0.010 mol dm–3 to 0.020 mol dm–3, the relative rate is doubled from ½ to 1 Thus, [zymase] is directly proportional to the rate of reaction, and is first order. [1] for explanation [½] x 2 for each correctly identified order

2019 HCI C2 H2 Chemistry Preliminary Exam / Paper 2 3 (a) (i) Anion: additional electrons of opposite symbol were accepted additional electrons of third symbol were accepted dative bonds were rejected [½] for each ion (ii) CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) [1] [1] balanced equation with correct state symbols (b) (i) [1] for correct bond angle (only one needs to be labelled on the diagram itself), and correct Lewis structure which must show linear shape. Type of hybridisation: sp [1] (ii) Two characteristics of ethyne, and explanation: 1. Ethyne has a small molecular size , which means the volume of the particles of ethyne is small compared to the volume of the container, and so can be considered negligible just as for ideal gases; and 2. Ethyne is a non-polar molecule, so it has relatively weak intermolecular dispersion forces, which could therefore be considered negligible, just as for ideal gases. [1] x 2 for each characteristic (c) (i) Total initial pressure = x + y = 760 mm Hg [1] (ii) Since p mm Hg is the change in partial pressure of ethyne during the combustion: C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) Initial partial pressure / mm Hg x y 0 - Change / mm Hg -p -5/2p +2p - Final partial pressure / mm Hg x - p y – 5/2p 2p - Total pressure after combustion = (x – p) + (y – 5 2p) + 2p = [(x + y) – 3 2p] mm Hg

[1] Correct expression of total pressure after combustion in terms of x, y and p (iii) Total initial pressure = (x + y) mm Hg Total pressure after combustion = (x + y) – 3 2p mm Hg Since a reaction occurs, p must be > 0 so (x + y) – 3 2p < (x + y), i.e. (final pressure) < (initial pressure) and ∴ the term “– 3 2p” represents a fall in pressure from the original (x + y) mm Hg (shown) [1] Convincing argument in which there is a co mparison of initial and final pressures in terms of x, y and p. (iv) From (ii) and (iii), we know that the fall in pressure inside the flask corresponds to difference in height (atmospheric pressure remains unchanged): 3 2p = 65 p = 65 ÷ (3/2) = 43.3 (to 1 d.p.) (in mm Hg) [1] Answer must be to 1 d.p. to get the full credit. (v) (I) Since final partial pressure of CO2 is 2p ∴ PCO2 = 2p = 86.7 mm Hg [1] (to 3 s.f.) (II) If all oxygen was used up, then y – 5/2p = 0 ∴ y = 5/2p y = 5/2 (65 ÷ 3/2) = 108.3 mm Hg (1 d.p.) = 108 mm Hg (3 s.f.) [½] And x = 760 – 108.3 = 651.7 mm Hg (1 d.p.) = 652 mm Hg (3 s.f.) [½]