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 © ACJC20