ACJC H2 Chem 2021 Paper 2 (Suggested Solutions)

Name Index Number Form Class Tutorial Class Subject Tutor ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY Preliminary Examination CHEMISTRY 9729/02 Higher 2 Paper 2 Structured Questions 25 August 2021 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet READ THESE INSTRUCTIONS CAREFULLY Write your name, index number, form class, tutorial class and subject tutor’s name 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. Answer all questions in the spaces provided on the Question Paper. 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 Question no. Marks 1 / 15 2 / 15 3 / 12 4 / 5 5 / 19 6 / 9 Presentation of answers TOTAL / 75 This document consists of 25 printed pages and 1 blank page. 9729/02/Prelim/2021 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2021 Department of Chemistry [Turn over

2 © ACJC 2021 JC2 H2 Prelim 2021 [Turn over BLANK PAGE

3 © ACJC 2021 JC2 H2 Prelim 2021 [Turn over 1 Hydrobromic acid, HBr, is one of the strongest mineral acids known and is even stronger than hydrochloric acid. Both the industrial and laboratory syntheses of HBr are well documented due to the usefulness of HBr in many chemical reactions. (a) The primary industrial preparation of HBr involves the platinum -catalysed reaction between hydrogen and bromine at temperatures between 450 K to 700 K. H2(g) + Br2(g) ⇌ 2HBr(g) The reaction was carried out in a 250 m3 vessel at 334 C. (i) Calculate the amount of HBr present at equilibrium, given that the equilibrium partial pressure of HBr is 2 atm. PV = nRT n = RT PV n = (2 x 101325)(250) 8.31 x 607 n = 1.00 x 104 mol [1] (ii) At 334 C, the equilibrium constant, Kp, for the above reaction is 2.50. Write the expression for the equilibrium constant, Kp, for the above reaction. Kp = 22 BrH 2 HBr( PP P ) [1] (iii) At equilibrium, the number of hydrogen molecules is ten times that of bromine molecules. Calculate the amount of bromine present at equilibrium. Let x be the number of moles of Br2. Number of moles of H2= 10x Kp = 22 BrH 2 HBr( PP P ) =   (10x)(x) 1x10 24 =2.50 x = 2.00 x 103 mol