NYJC Prelim CHEM P2 ANS

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 Answers CANDIDATE NAME CLASS TUTOR’S NAME CHEMISTRY 9746/02 Paper 2 Structured 23 September 2009 1 hour 30 minutes Candidates answer on the Question Paper Additional Materials: Answer Paper 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 soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. 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 9 2 16 3 8 4 10 5 13 6 4 Total 60 This document consists of 12 printed pages and 0 blank page. [Turn over

2 H2 Chemistry 9746/02/NYJC J2/09 PX [Turn Over For Examiner's Use Answer all questions. 1 (a) The use of the Data Booklet is relevant to thi s question. Nitrogen dioxide disproportionates in ac idic solution to nitrous acid and nitrate ions. (i) Write a balanced equation for the disproportionation reaction. [1] 2NO2 + H2O → HNO2 + NO3 ─ + H+ (ii) Use the data below and any other rele vant data in the Data Booklet to determine if this disproportionation r eaction would actually occur under standard conditions NO2 + H+ + e HNO2 Eθ = +1.10 V [2] NO2 + H+ + e HNO2 Eθ = +1.10 V NO3 ─ + 2H+ + e NO2 + H2O Eθ = +0.81 V 2NO2 + H2O → HNO2 + NO3 ─ + H+ Eθ cell = (+1.10) - (+0.81) = +2.9 V Hence disproportionation is favourable (b) A 0.360 g sample of gaseous aluminium chloride takes up a volume of 52 cm3 at a temperature of 200oC and a pressure of 1.02 x 105 Pa. (i) Under what conditions of temperat ure and pressure would you expect the behaviour of gaseous aluminium chloride to be most like that of an ideal gas? [1] High temperature and low pressure. (ii) Calculate the Mr of the vapour at this temperature. [2] Mr = 56 0.360 8.31 (273 200) 2671.02 10 52 10 RT massPV      (c) Compounds of aluminium have many important uses industrially. For

3 For Examiner's Use example in the H aber process, Al 2O3 are mixed with ir on catalyst to enhance efficiency of the iron catalyst. Ammonia is manufactured in the Haber process according to the equation: N2 (g) + 3H2 (g) 2NH3 (g) ΔH = -92 kJmol-1 (i) Write an expression for Kp for this reaction [1] 3 22 2 3 () () NH p NH PK PP  (ii) When a 1:3 mixture of N 2 and H 2 was allowed to reach equilibrium at 200 atm and 500oC, the partial pressure of NH3 was found to be 40 atm. Using the expression in c) (i) to calculate the value for Kp. [2] 3 2 2 3 22 2 2 52 33 200 , 40 1 (200 40) 404 3 (200 40) 1204 () 40 2.31 10( ) 40 120 total NH N H NH p NH Pa t m P a t m Pa tm Pa tm PKa tmPP           [Total:9] 2 (a) CH2=CHCO2H (l) + H2(g) CH3CH2CO2H (l)  H = -380 kJ mol-1 compound A (i) Given that the absolute value of S for the above reaction is 68 J mol-1K-1. Predict the sign of S, stating your reasons. [2] S expected to be –ve. (1 mol of gas + 1 mol of liquid) in reactant give 1 mol of liquid product, system becomes more ordered. (ii) Hence determine the temperature for the reaction to be non- spontaneous. [2] G = H - TS For reaction to be non-spontaneous G >0 -380 – (T) (–0.068) > 0 T(0.068) > 380 T > 5588 K (iii) With the aid of an energy cycle, calculate the enthalpy change of combustion of compound A by using the data provided. [3] H2 Chemistry 9746/02/NYJC J2/09 PX [Turn Over

4 For Examiner's Use Data Hf (H2O) = -286 kJ mol-1 Hc (CH3CH2COOH) = -1450 kJ mol-1 CH2 CH CO 2H 3CO2 + 2H2O CH3 CH2 CO2H + H2 3O2 + 1/2 O2 H3= -286 kJ mol-1 H2 = -1450 kJ mol-1 H1 = -380 kJ mol-1 +7/2 O2 Hc + H2 3CO2 + 3H2O + (l) (aq) (g) (g)(l) (l) (g) (l) –380 = Hc(A) – 286 + 1450  Hc(A) = –1544 kJ mol ⎯1 (b) Copper is an important metal used extensively in pipes and electrical wires. It can be made extremely pure and corrodes very slowly. (i) Copper corrodes in moist air to first gi ve a thin layer of copper(II) oxide and this process happens much more sl owly than the rusting of iron to form the iron(III) oxides. With th e use of relevant data from the Data Booklet, explain this difference. [2]  Cu 2+ + 2e Cu E θ = +0.34 V  Fe 3+ + 3e Fe E θ = – 0.04 V  O 2 + 2H2O + 4e 4OH - Eθ = +0.40 V (MOIST AIR)  O2 + 2H2O + 2Cu  4OH- + 2Cu2+ Eθ cell = +0.40 – (+0.34) = + 0.06V (close to 0) (feasible)  3O2 + 6H2O + 4Fe  12OH- + 4 Fe3+ Eθ cell = +0.40 – (-0.04) = +0.44V > 0 (feasible)  Eθ cell for the oxidation of Fe is much more positive than that for the oxidation of Cu, hence more spontaneous. (ii) Another possible oxidation state that copper can hav e is +1, such as in copper(I) sulphate, Cu2SO4. When this sulphate is added to water, blue solution of copper(II) sulphate and pi nk deposit of copper metal forms. Write a balanced equation for this pr ocess and using relevant data from the Data Booklet, show that this reaction is feasible. [2] 2 Cu+  Cu + Cu2+ H2 Chemistry 9746/02/NYJC J2/09 PX [Turn Over

5 For Examiner's Use Cu+ + e Cu + 0.52 V Cu2+ + e Cu+ + 0.15 V Eθ = + 0.52 – (+ 0.15) = + 0.37 V > 0 (feasible) (iii) A copper metal plate was dipped into an aqueous solution of 1.0 mol dm –3 copper(II) sulphate solu tion and this half-cell was connected via a salt bridge, to the following half-cell, 2CO2 + 2H+ + 2e H2C2O4 The overall cell e.m.f was found to be +0.83 V and the size of the copper plate increased after some time. Give the cell notation and hence, calculate the reduction potential, E θ (CO2 / H2C2O4) [2] The size of Cu increased shows that Cu 2+ ions is reduced to Cu. Hence oxidation must be occurring at the other CO2 half-cell. Pt (s) | H2C2O4 (aq) , CO2 (g) || Cu2+ (aq) | Cu (s) E θ = Eθ R – Eθ o +0.83 = +0.34 - Eθ o Eθ o = - 0.49 V = Eθ (CO2 / H2C2O4) (c) To obtain a pure metal, electrolysis can be conducted using an aqueous copper(II) sulphate solution and copper electrodes. The results of the experiment are shown below in the graph. (i) Write balanced half-equations, with state symbols, for the reactions at the anode and the cathode. Mass of c opper deposited /g 20 40 60 time /min 0.05 0.04 0.03 0.02 0.01 [1] H2 Chemistry 9746/02/NYJC J2/09 PX [Turn Over

6 For Examiner's Use Cathode: Cu2+(aq) + 2e  Cu(s) Anode: Cu(s)  Cu2+(aq) + 2e (ii) Calculate the current used during electrolysis. [1] Q = nzF = (0.04/63.5)(2)(96500) = 121.6 C I = Q / t = 121.6 / (20 x 60) = 0.101 A (iii) It is not always possible to accurately predict the electrode reactions that occur during electrolysis. Suggest a reason for this. [1] Electrolysis may not occur at standard conditions. OR Rate of reaction may be too slow. OR The reaction has a high activation energy. [Total:16] 3 (a) Aluminium