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

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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 [2] (iv) Given that the equilibrium in (a) was established using hydrogen and bromine only, determine the mass of bromine (in kg) that was used. Since only hydrogen and bromine were present in the vessel in the beginning, all the hydrogen bromide formed must have come from Br2 initially. Initial no. of moles of Br2 = No. of moles of Br2 at equilibrium + ½ x no. of moles of HBr = 2 x 103 + ½ x (1.00 x 104) = 7.00 x 103 mol Mass of Br2 = 7.00 x 103 x 79.9 x 2 = 1118 kg  1120 kg [2]

4 © ACJC 2021 JC2 H2 Prelim 2021 [Turn over (v) Predict, with explanation, the effect on the equilibrium position if the volume was decreased at constant temperature. In this reaction, there is no change in the number of gaseous molecules when the reactants form the products. Hence, changes of pressure have no effect on the position of the equilibrium. [1] (b) The preparation of the acid, HBr , in the laboratory can be carried out by the reaction between bromine, sulfur dioxide and water only. The only by-product formed is also a strong acid. (i) Write the equation for the above laboratory preparation of HBr. Br2 + SO2 + 2H2O  2HBr + H2SO4 [1] (ii) When concentrated solutions of the two products formed in the above preparation react with each other, Br2 is regenerated. State and explain the type of reaction that HBr undergoes. Oxidation Oxidation state of Br changes from -1 in HBr to 0 in Br2. [2] (c) Inter and intramolecular bondings respectively can explain the trends in the volatility and thermal stability of the hydrogen halides. (i) State and explain the trend in the volatility of hydrogen halides from HF to HI. HCl to HI molecules: Decreasing volatility from HCl to HBr to HI HF molecules: HF boiling point is exceptionally high. Volatility lowest. HCl to HI molecules: As the electron cloud size of halogen atom increases from Cl to Br to I, there is greater ease of distortion/polarisation of the electron cloud of HX leading to stronger instantaneous dipole -induced dipole interactions between the molecules. As the amount of energy required to overcome the forces increases, volatility decreases down the group. HF mol ecules: Due to hydrogen bonding between HF molecules, HF has an exceptionally low volatility. [3]

5 © ACJC 2021 JC2 H2 Prelim 2021 [Turn over (ii) State and explain the trend in the relative thermal stabilities of hydrogen halides from HF to HI. Thermal stability of the halides decreases down the group as HX bond energy /bond strength decreases. As the size of halogen atom increases, its p orbitals become more diffused and overlap less effectively with the s orbital of the hydrogen atom resulting in longer and thus weaker H-X covalent bonds. [2] [Total: 15] 2 This question is about nitrogen and its oxides. (a) (i) N3 is isoelectronic with F and Na+. Define the term isoelectronic. Same number of electrons [1] (ii) Sketch a graph of the successive ionisation energies of all the electrons of a nitrogen atom. [2] (iii) State and explain how the ionic radius of N3 compares to that of P3. Ionic radius of N3 is smaller than P3. P3: valence electrons are further away from the nucleus due to increase in number of principal quantum shell. OR The valence electrons experiences weaker nuclear attraction due to additional screening effect. Therefore, P3 has a larger ionic radius. [2]

6 © ACJC 2021 JC2 H2 Prelim 2021 [Turn over (b) N2O3 is a pale blue solid at very low temperatures. As the temperature is raised, N2O3 dissociates to form colourless nitrogen monoxide gas, NO and brown nitrogen dioxide gas, NO 2. The interesting feature of this reaction is that both products are molecular radicals, each with an unpaired electron on the nitrogen atom. N2O3(s) ⇌ NO(g) + NO2(g) Draw the dot -and-cross diagram for N 2O3, stating the shapes