2019 ACJC Prelim H2 Chem P3 ANS

[Turn over ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY H2 9729/03 Paper 3 Free Response 2 September 2019 2 hours Candidates answer on separate paper. Additional Materials: Cover Page Answer Paper Insert READ THESE INSTRUCTIONS FIRST Write your index number and name, form class and tutorial class on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. Start each question on a new page of writing paper. Fasten the insert in front of all writing paper at the end of the examination. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. 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. This document consists of 21 printed pages and 1 blank page. 9729/03/Prelim/19 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2019 Department of Chemistry

2 © ACJC2019 9729/03/Prelim Exam/2019 [Turn over Section A Answer all the questions from this section. 1 (a) Use of the insert is necessary for this question. Fasten the insert in front of all writing paper at the end of the examination. An iodoalkane, R I, is hydrolysed by aqueous sodium hydroxide. The results obtained from two experiments a re plotted on the insert. In each experiment, the overall [NaOH(aq)] remained virtually constant at the value given beside each graph. (i) Use the graphs on the insert to determine the following. Show all workings clearly. I Use the half -life method to deduce the order of reaction with respect to the iodoalkane. II Use the initial rates method to deduce the order of reaction with respect to sodium hydroxide. III Construct a rate equation for the reaction and use it to calculate a value for the rate constant. Include its units. [7] From working on graph, when [OH–] = 0.10 M, half-life = 58 min OR From working on graph, when [OH–] = 0.15 M, half-life = 39 min [1] ∴ reaction is first order w.r.t. RI. [1] From working on graph, when [OH–] = 0.10 M, initial rate = 0.0100/96 = 0.000104 mol dm–3 min–1 & From working on graph, when [OH–] = 0.15 M, initial rate = 0.0100/64 = 0.000156 mol dm–3 min–1 [1] When [OH–] x1.5, rate x1.5, ∴ reaction is first order w.r.t. OH–. [1] rate = k[RI][OH–] [1] (if SN1, do not award this mark) k = 0.000104 [0.010][0.10] = 0.104 mol–1 dm3 min–1 OR k = 0.000156 [0.010][0.15] = 0.104 mol–1 dm3 min–1 [1+1] Allow ecf if rate equation is rate = k[RI], then k = 0.000104 [0.010] = 0.0104 mol–1 dm3 min–1 OR k = 0.000156 [0.010] = 0.0156 mol–1 dm3 min–1 [1+1] (D.N.A. if rate = k[OH–])

3 © ACJC2019 9729/03/Prelim Exam/2019 [Turn over (ii) Hence draw a fully labelled energy profile diagram of the reaction. [3] Shape and axes [1] Reactants, Products [1] (if SN1, mark for intermediate) Ea and ∆H [1] Allow ecf if deduced SN1 in part (i) (b) Silver(I) iodide and iodine are two solids which have low solubility in water. However when both are mixed together with water and left to stand , they dissolve completely to give a coloured solution of silver(I) triiodide. (i) State the expected colour of the silver(I) triiodide solution. [1] Yellow or Brown (answer is concentration dependent) (ii) Write three relevant equilibria equations and use them to explain how the two solids can completely dissolve when mixed with water. [3] AgI(s) ⇌ Ag+(aq) + I–(aq)--------- (1) [✓] I2(s) + aq ⇌ I2(aq)------------------ (2) [✓] I–(aq) + I2(aq) ⇌ I3–(aq)----------- (3) [✓] Equilibria (1) and (2) exist for each of the poorly soluble compounds. However, equilibrium (3) is set up in the presence of I–(aq) and I2(aq) in solution and the position of equilibrium lies strongly to the right hand side [✓]. With [I–] and [ I2] constantly falling in (1) and (2), their positions of equilibria will constantly shift to the right[✓][✓] as well, causing AgI(s) and I2(s) to dissolve. OWTTE [✓] for each eqn and [✓] for accompanying POE shift. 6[✓]-3m, 4[✓]-2m, 2[✓]-1m (c) Another triiodide compound, nitrogen triiodide, NI3 is a simple covalent molecule that is very sensitive to shock and will decompose rapidly. A touch of a feather or even alpha particles from radioactive decay can trigger an explosion.

4 © ACJC2019 9729/03/Prelim Exam/2019 [Turn over 2NI3(s) ⟶ N2(g) + 3I2(g) ∆H = –290 kJ mol–1 76.5 kJ mol–1 of energy is needed to convert solid NI3 to gaseous NI3. Use the above information and relevant information from the Data Booklet to calculate the N–I bond energy. [2] 2NI3(s) ⟶ N2(g) + 3I2(g) 2NI3(g) ⟶ 2N(g) + 6I(g) –290 = 2(+76.5) + 6BDE(N–I) + (–944) + 3(–151) BDE(N–I) = 159 kJ mol–1 [1] using –290, +76.5, –944 & –151, no need coefficients and sign. [1] answer 159, no ECF (d) Alpha particles, He2+ 2 4 are produced from the radio active decay of certain isotopes . An example of one is from the decay of Americium-241. Am95 241 → He2+ 2 4 + M (i) Use the Data Booklet to identify species M that is formed. [1] Np2− 93 237 , 237Np2– minimally, no need for 93. (ii) Calculate the angle of deflection for M in a uniform electric field if the angle of deflection for He2+ 2 4 is +7.11o. [1] Angle of deflection, AOD = k Q m 𝐴𝑂𝐷1𝑚1 𝑄1 = 𝐴𝑂𝐷2𝑚2 𝑄2 AODNp2– = +7.11×4×(−2) (+2)237 = –0.12o (2 or 3 sf) Americium is often used in smoke detectors as the compound AmO 2 and is made from the thermal decomposition of americium(III) ethanedioate, Am2(C2O4)3. Am2(C2O4)3 ⟶ 2AmO2 + 4CO + 2CO2 (iii) Explain why the thermal decomposition is a redox reaction in terms of oxidation state changes. [2] Am in Am2(C2O4)3 is oxidised from +3 oxidation state to +4 in AmO2. [1] C in Am2(C2O4)3 is reduced from +3 oxidation state to +2 in CO and oxidised +4 in CO2. [1] [Total: 20] –290 2(+76.5) 3(–151) 2(–944) 6BDE(N–I)

5 © ACJC2019 9729/03/Prelim Exam/2019 [Turn over 2 (a) On 12 June 2019, Hong Kong police fired tear gas into the crowds as many took to the streets to protest against a proposed extradition Bill. The active compound in tear gas is 2-chlorobenzalmalononitrile. 2-chlorobenzalmalononitrile 2-chlorobenzalmalononitrile is a solid at room temperature and is dispersed as an aerosol dissolved in a suitable organic solvent, typically dichloromethane, CH2Cl2. Its effects are felt when the solvent evaporates in air and the dry powder touches the eyes, nose and mouth. (i) Suggest, with reasoning, two physical properties that make dichloromethane a suitable solvent. [2] 1. Polar to favour pd-pd interactions. 2. High volatility/low BP to dry/evaporate in the air. 2-chlorobenzalmalononitrile can be synthesised via the 2 -stage Knoevenagel condensation.

6 © ACJC2019 9729/03/Prelim Exam/2019 [Turn over (ii) State the type of the reaction in stage 2. [1] Elimination (iii) Instead of purchasing malononitrile as a starting reagent, it can be made from dichloromethane. Suggest the reagent and conditions needed to convert dichloromethane to malononitrile. [1] KCN (alc), heat under reflux The following 3-step mechanism illustrates stage 1.  Malononitrile is deprotonated by a weak base, R 2NH to form its conjugate base, (NC)2CH–.  The conjugate base of malononitrile undergoes nucleophilic addition with 2-chlorobenzaldehyde to give as an intermediate.  The conjugate acid of R2NH is depro