2022 JPJC JC2 H2 Chem Prelim P3 Suggested Answers

© Jurong Pioneer Junior College [Turn Over NAME CLASS 21S JURONG PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2022 CHEMISTRY 9729/03 Higher 2 Paper 3 Free Response Questions 19 September 2022 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. If additional space is required, you should use the pages at end of this booklet. The question number must be clearly shown. Answer all questions in Section A. Answer 1 question in Section B. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 21 2 20 3 19 4 or 5 20 Penalty (delete accordingly) Lack 3sf in final answer –1 / NA Missing/wrong units in final ans –1 / NA Bond linkages –1 / NA Total 80 This document consists of 31 printed pages inclusive of 1 blank page.

2 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2022 Section A Answer all the questions in this section. 1 (a) Bromoethane reacts with “acetylide” anion, CH3C≡C:− to form new carbon−carbon bonds. This reaction takes place in two steps. In step 1, an acid-base reaction occurs. CH3C≡C:− is formed from the reaction of propyne and a strong base, sodium amide, NaNH2. step 1: CH3C≡CH + :NH2− → CH3C≡C:− + NH3 In step 2, the intermediate anion reacts with bromoethane to form the product. step 2: CH3CH2Br + CH3C≡C:− → CH3CH2C≡CCH3 + Br− For Examiner’s use Name and suggest the mechanism for step 2. Show relevant dipoles, using curly arrows to indicate the movement of electron pairs. Type of mechanism: nucleophilic substitution [3] (b) 4-bromopentanol can be used to synthesise Compound C by the three -step route shown in Fig. 1.1. Br OH O O A B C step 1 step 2 step 3 4-bromopentanol Fig. 1.1 State the structures for compounds A and B, and reagents and conditions for steps 1, 2 and 3 in this route. [4] step 1 : NaCN (or KCN), ethanol, heat (✓) step 2 : H2SO4(aq) (or HCl(aq)), heat (✓)

3 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2022 [Turn Over step 3 : concentrated sulfuric acid, heat(✓) (c) Describe and explain the relative ease of hydrolysis of the following three chlorine-containing compounds. . [3] Ease of hydrolysis of C6H5Cl < C6H5CH2Cl < C6H5COCl C6H5Cl is inert or unreactive to hydrolysis as the p–p orbital overlap results in the delocalisation of lone pair of electrons on Cl atom of C6H5Cl into the –electrons system of benzene ring, leading to the formation of partial double bond character of C–Cl bond and hence, strengthening C–Cl bond. Thus, C6H5Cl is resistant towards nucleophilic attack. C6H5COCl is most readily hydrolysed because the C of –COCl is highly electron–deficient or has the highest partial positive charge as it is bonded to two electronegative atoms, O and C l. Hence, C6H5COCl is more susceptible towards nucleophilic attack than C6H5CH2Cl. (d) Ethanol is formed when bromoethane is heated with NaOH(aq). The standard enthalpy change of combustion of ethanol is –1367 kJ mol−1. In an experiment, 0.23 g of ethanol was burned under a container, using a spirit lamp. An unknown mass of water was heated from 30 oC to its boiling point. The process was found to be 70 % efficient. Calculate the mass of water that could be brought to the boiling point by burning this amount of ethanol. [Given specific heat capacity of water is 4.18 J g-1 K-1] [2] Amount of ethanol burnt = 0.23 =46.0 5.00  10−3 mol Heat evolved by combustion of ethanol = 5.00  10−3  1367 =6.835 kJ Heat absorbed by x g of water = 70/100 x 5.00  10−3  1367 = 4.78 kJ mass of water, x = 4.78  103 / (4.18  70) = 16.3 g (e) (i) Nitrosyl bromide, Br-N=O is an inorganic halogen-containing compound. It decomposes to NO and Br2 as shown below. 2NOBr(g) → 2NO(g) + Br2(g), Hr Given that the bond energy of N−Br is +120 kJ mol−1, use appropriate bond energy data from the Data Booklet to calculate the enthalpy change of decomposition of nitrosyl bromide. [2] (ii) Enthalpy changes of formation of NOBr(g) and NO(g) and the enthalpy change of vaporisation of Br2(l) are given below. Hf (NOBr(g)) y kJ mol−1 Hf (NO(g)) +90 kJ mol−1 Hvap(Br2(l)) +31 kJ mol−1 With the aid of an energy cycle, use your answer in (i) and the given data to calculate the enthalpy change of formation of NOBr(g). [3]

4 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2022 (e) (i) 2 O=N–Br(g) → 2 N=O(g) + Br–Br(g), Hr = 2  Hdecomposition(NOBr) Hr = +2E(N=O) + 2E(N–Br) – 2E(N=O) – E(Br−Br) = 2E(N–Br) – E(Br−Br) = 2(+120) – (+193) | = +47 kJ mol–1 Hdecomposition (NOBr) = ½ (+47) = +23.5 kJ mol–1 (ii) N2(g) + O2(g) + Br2(l) 2NOBr(g) 2NO(g) + Br2(l) 2NO(g) + Br2(g) By Hess’ Law, 2y = 2(+90) + (+31) – (+47) y = Hf(NOBr) = +82.0 kJ mol–1 (f) (i) State how the reactivity of the halogens as oxidising agents varies down the group, and relate this variation to relevant Eo values. [2] (ii) Describe a reaction that illustrates the relative oxidising abilities of two halogens of your choice. [1] (iii) Iodine and chlorine react together to form solid iodine trichloride, ICl3. Given the following enthalpy changes, calculate the standard enthalpy change of formation of ICl3(s). I2(g) + 3Cl2(g) → 2ICl3(s) HO = −214 kJ mol−1 I2(s) → I2(g) HO = +38 kJ mol−1 [1] (f) (i) Down the group, E o (X2/X−) becomes less positive, implying that the tendency of X2 to be reduced decreases. Hence, the oxidising power of X2 decreases down the group. [2] (ii) Eg. Br2 + 2I- → 2Br- + I2 (Halogen displacement reaction where the stronger oxidising halogen can oxidise the halides of the weaker oxidising halogen. or Eg. Cl2 + 2Fe2+ → 2Cl - + 2Fe3+ Reaction of X2 with Fe2+ (iii) Let x = HfOICl3(s) HO =HOf products ─ HOf reactants -214= 2x -3(+38) x = HfOICl3(s) = -88.0 kJ mol− [Total: 21] Hvap =+31 2y 2(+90) +47

5 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2022 [Turn Over 2 (a) Explain why transition element complexes are usually coloured. For Examiner’s Use

6 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2022 The presence of ligands causes the d orbitals to split into 2 different energy levels with a small energy gap. Visible light is absorbed when an electron transits from a lower energy d orbital to a higher energy d orbital which is partially filled. Hence, transition element complexes are coloured and the colour observed is the complement of the colours absorbed. (b) In a given electroplating experiment, a solution of CrCl3(aq) is electrolysed using a current of 3.50 A. Calculate the time, in min, required to produce 4.60 g of chromium by this electrolysis. [2] Amount of Cr = 4.60 / 52.0 = 0.0885 mol Cathode: Cr3+(aq) + 3e− → Cr(s) Q = neF = It Hence Q = 3  0.0885  96500