Cambridge A Level Chemistry 9701 — 2006 May/June Paper 4 · Variant 1

This document consists of 11 printed pages and 1 blank page. SPA SJF3704 T12878/2 © UCLES 2006 [Turn over UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level CHEMISTRY 9701/04 Paper 4 Structured Questions A2 Core May/June 2006 1 hour 15 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen in the spaces provided on the Question Paper. You may use a pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions. You may lose marks if you do not show your working or if you do not use appropriate units. A Data Booklet is provided. You may use a calculator. 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. Centre Number Candidate Number Name For Examiner’s Use 1 2 3 4 5 Total

2 9701/04/M/J/06 Answer all the questions in the spaces provided. 1 The oxidation of nitrogen monoxide occurs readily according to the following equation. NO(g) +
 O2(g) ⎯→ NO2(g) The following table shows how the initial rate of this reaction depends on the concentrations of the two reactants. (a) (i) Use the data to determine the order of reaction with respect to each of the reagents. order with respect to NO ………………………… order with respect to O2 ………………………… (ii) Write the rate equation for the reaction, and use it to calculate a value for the rate constant, k, stating its units. rate equation … numerical value of k = ………………………… units of k ………………………… (iii) Use your rate equation in (ii) to calculate the rate of reaction when [NO] = [O2] = 0.0025 mol dm–3. rate of reaction =………………………… [6] For Examiner’s Use © UCLES 2006 [NO] / mol dm–3 0.0050 0.0050 0.010 [O2] / mol dm–3 0.0050 0.0075 0.0075 initial rate / mol dm–3s–1 0.02 0.03 0.12

3 9701/04/M/J/06 [Turn over (b) Nitrogen monoxide plays an important catalytic role in the oxidation of atmospheric SO2 in the formation of acid rain. (i) State the type of catalysis shown in this process. … (ii) Explain the steps involved in this process by writing equations for the reactions that occur. … … … [3] [Total: 9] For Examiner’s Use © UCLES 2006

4 9701/04/M/J/06 2 Monuments made of marble or limestone, such as the Taj Mahal in India and the Mayan temples in Mexico, are suffering erosion by acid rain. The carbonate stone is converted by the acid rain into the relatively more soluble sulphate. CaCO3(s) + H2SO4(aq) →CaSO4(s) + H2O(l) + CO2(g) acid rain (a) (i) Write an expression for the solubility product, Ksp, of CaSO4, stating its units. … (ii) The Ksp of CaSO4 has a numerical value of 3 x 10–5. Use your expression in (i) to calculate [CaSO4] in a saturated solution. … (iii) Hence calculate the maximum loss in mass of a small statue if 100 dm3 of acid rain falls on it. Assume the statue is made of pure calcium carbonate, and that the acid rain becomes saturated with CaSO4. … … … [5] (b) The life of such monuments is now being extended by treating them with a mixture of urea and barium hydroxide solutions. After soaking into the pores of the carbonate rock, the urea gradually decomposes to ammonia and carbon dioxide. The carbon dioxide then reacts with the barium hydroxide to form barium carbonate. (NH2)2CO(aq) + H2O(l) ⎯→ 2NH3(g) + CO2(g) Ba(OH)2(aq) + CO2(g) ⎯→ BaCO3(s) + H2O(l) Acid rain then converts the barium carbonate to its sulphate. BaCO3(s) + H2SO4(aq) ⎯→ BaSO4(s) + H2O(l) + CO2(g) Barium sulphate is much less soluble than calcium sulphate. A saturated solution contains [Ba2+] = 9.0 x 10–6mol dm–3. (i) Explain why barium sulphate is less soluble than calcium sulphate. … … … … For Examiner’s Use © UCLES 2006

5 9701/04/M/J/06 [Turn over (ii) Write an expression for the Ksp of barium sulphate and use the data to calculate its value. … … [4] (c) (i) Explain what is meant by the term lattice energy. … … (ii) Predict, with a reason, how the lattice energy of BaSO4 might compare with that of MgSO4. … … … [3] [Total: 12] For Examiner’s Use © UCLES 2006

6 9701/04/M/J/06 3 (a) A transition element X has the electronic configuration [Ar] 4s2 3d3. (i) Predict its likely oxidation states. … (ii) State the electronic configuration of the ion X3+. … [2] (b) Potassium manganate(VII), KMnO4, is a useful oxidising agent in titrimetric analysis. (i) Describe how you could use a 0.0200 mol dm–3 solution of KMnO4 to determine accurately the [Fe2+] in a solution. Include in your description how you would recognise the end-point in the titration, and write an equation for the titration reaction. … … … … … … (ii) A 2.00 g sample of iron ore was dissolved in dilute H2SO4 and all the iron in the salts produced was reduced to Fe2+(aq). The solution was made up to a total volume of 100 cm3. A 25.0 cm3 portion of the solution required 14.0 cm3 of 0.0200 mol dm–3 KMnO4 to reach the end-point. Calculate the percentage of iron in the ore. … [8] For Examiner’s Use © UCLES 2006

7 9701/04/M/J/06 [Turn over (c) High-strength low-alloy (HSLA) steels are used to fabricate TV masts and long span bridges. They contain very low amounts of phosphorus and sulphur, but about 1% copper, to improve resistance to atmospheric corrosion. When dissolved in nitric acid, a sample of this steel gives a pale blue solution. (i) What species is responsible for the pale blue colour? … (ii) Describe and explain what you would see when dilute aqueous ammonia is added to this solution. … … … … [4] [Total: 14] For Examiner’s Use © UCLES 2006

8 9701/04/M/J/06 4 The amino acids tyrosine, lysine and glycine are constituents of many proteins. (a) State the reagents and conditions you could use to break proteins down into amino acids. … [2] (b) Draw a ring around each chiral centre in the above molecules. [1] (c) In aqueous solution amino acids exist as zwitterions. Draw the zwitterionic structure of glycine. … [1] (d) For each of the following reactions, draw the structure of the organic compound formed. (i) glycine + excess NaOH(aq) … (ii) tyrosine + excess NaOH(aq) … For Examiner’s Use © UCLES 2006 CH2 C O CH OH tyrosine H2N OH (CH2)4 NH2 C O CH lysine H2N OH H C O CH glycine H2N OH

9 9701/04/M/J/06 [Turn over (iii) lysine + excess HCl(aq) … (iv) tyrosine + excess Br2(aq) … [5] (e) Draw the structural formula of a tripeptide formed from all three of these amino acids, showing clearly the peptide bonds. … [2] (f) The formula of part of the chain of a synthetic polyamide is shown below. (i) Identify the repeat unit of the polymer by drawing square brackets around it on the above formula. (ii) Draw the structures of the two monomers from which the polymer could be made. … [3] [Total: 14] For Examiner’s Use © UCLES 2006 CH2 CH2 CH2 NH NH NH CO CO CO CH2 CH2 NH NH CO

10 9701/04/M/J/06 5 Benzocaine is an important local anaesthetic used in skin creams for sprains and other muscular pains. It can be made by the following route. (a) Suggest reagents and conditions for each of the above four reactions. I … II … III … IV … [6] (b) Draw steps to show the mechanism of reaction I. [2] (c) Another local anaesthetic is amylocaine, which can be made from compound X. (i) Apart from the benzene ring, name two functional groups in the molecule of compound X. … … For Examiner’s Use © UCLES 2006 CH3 I CH3 NO2 II CO2H NO2 III CO2H NH2 IV CO2CH2CH3 NH2 benzocaine amylocaine X CH2 CH3 CH2CH3 NH2 C C O O CH2 CH3 CH3 CH3 CH2CH3 N C C O O

11 9701/04/M/J/06 (ii) Explain whether compound X would be more or less basic than benzocaine. … … [3] [Total: 11] For Examiner’s Use © UCLES 2006

12 9701/04/M/J/06 BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level and GCE Advanced Subsidiary Level MARK SCHEME for the May/June 2006 question paper 9701 CHEMISTRY 9701/04 Paper 4 Maximum raw mark 60 This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which Examiners were initially instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began. Any substantial changes to the mark scheme that arose from these discussions will be recorded in the published Report on the Examination. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes must be read in conjunction with the question papers and the Report on the Examination. The minimum marks in these components needed for various grades were previously published with these mark schemes, but are now instead included in the Report on the Examination for this session. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2006 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 1 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 04 © University of Cambridge International Examinations 2006 1 (a) (i) order w.r.t. NO = 2 [1] order w.r.t.O2 = 1 [1] (ii) rate = k[NO]2[O2] [1] taking the first row: k = rate/([NO]2[O2]) = 0.020/(0.0052 x 0.005) = 1.6 x 105 ecf [1] units = mol-2dm6sec-1 ecf [1] (iii) rate = k[NO]2[O2] = 1.6 x 105 x 0.00252 x 0.0025 = 2.5 x 10-3 (mol dm-3 s-1) ecf [1] [6] (b) (i) homogeneous [1] (ii) NO + ½ O2 → NO2 SO2 + NO2 → SO3 + NO (SO3 + H2O → H2SO4) [2] [3] [Total: 9] 2 (a) (i) Ksp = [Ca2+][SO4 2-] [1] units are: mol2dm-6 ecf [1] (ii) [CaSO4] = √Ksp = 5.5 x 10-3 (5.477 x 10-3)(mol dm-3) ecf [1] (iii) n(CaSO4) in 100 dm3 = 5.5 x 10-3 x 100 = 0.55 moles ecf from (ii) [1] Mr(CaSO4) = 136.1 Thus mass(CaSO4) = 0.55 x 136.1 = 74.8g (0.55 x Mr) [1] (if the accurate [CaSO4] is held throughout the calculation, ans = 74.5g) [5] (b) (i) down the group: the ∆Hsolution becomes more endothermic; both lattice energy and ∆Hhydration become less (exothermic); due to ionic radius (of M2+) increasing; but ∆Hhydration changes more than lattice energy any three points [3] (ii) Ksp = [Ba2+][SO4 2-] = (9 x 10-6)2 = 8.1 x 10-11 NO ecf [1] [4] (c) (i) LE is the energy change when 1 mole of (ionic) solid [1] is formed from its gaseous ions [1] (ii) LE(BaSO4) < LE(MgSO4), due to larger radius of Ba2+ both points [1] [3] [Total: 12]

Page 2 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 04 © University of Cambridge International Examinations 2006 3 (a) (i) +2, +3, +4, +5 (ignore 0 and +1) all four [1] (ii) [Ar]3d2 [1] [2] (b) (i) take a fixed amount/aliquot/pipette-full of the Fe2+ solution [1] titrate with KMnO4 in the burette [1] until the first permanent pink colour (or change from colourless to pink) [1] repeat until two titres are within 0.1 cm3 [1] MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ [1] (or molecular equn.) (ii) n(MnO4 -) = 0.02 x 14/1000 = 2.8 x 10-4 moles [1] n(Fe2+) in 25 cm3 = 2.8 x 10-4 x 5 (x 5) [1] (= 1.4 x 10-3 moles) n(Fe2+) in 100 cm3 = 1.4 x 10-3 x 4 (x 4) [1] (= 5.6 x 10-3 moles) mass of Fe in 2.0 g ore = 5.6 x 10-3 x 55.8 = 0.31 g percentage = 100 x 0.31/2 = 15.6% (use of 55.8 or 56 and %) [1] [9] (c) (i) Cu2+(aq) or [Cu(H2O)6]2+ [1] (ii) pale blue ppt. (of Cu(OH)2(s)) [ignore any refs. to iron hydroxides] [1] (which dissolves to give…) a deep blue solution [1] which contains [Cu(NH3)4]2+ ions (can be read into equn, below) [1] formed by ligand displacement [1] or an equation such as Cu(OH)2 + 4NH3 → [Cu(H2O)4]2+ + 2OH- or [Cu(H2O)6]2+ + 4NH3 → [Cu(H2O)4]2+ + 6H2O [5] [Total: 16 max 14]

Page 3 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 04 © University of Cambridge International Examinations 2006 4 (a) HCl or H2SO4 or H+ or acid [1] conc(if HCl only)/dilute/aqueous + heat [1] [2] (b) two rings only (1 ring around the α-C of tyrosine & 1 around the α-C of lysine) [1] [1] (c) +NH3CH2CO2 - (or displayed formula) [1] [1] (d) (i) NH2CH2CO2 - (Na+) (either -CO2 -Na+ or -CO2Na but NOT –CO-O-Na) [1] (ii) (Na+) –O-C6H4-CH2CH(NH2)CO2 - (Na+) [1] + [1] (iii) (Cl-)+NH3(CH2)4CH(NH3 +)CO2H (Cl-) [1] + [1] (iv) HO-C6H2Br2-CH2CH(NH2)CO2H (if shown, Br at 2,6 to OH group) [1] [6] (e) H2N CH2 C NH CH C NH CH CO2H O O CH2 OH CH2 CH2 CH2 NH2 CH2 structure [1] at least one peptide group identified [1] [2] (f) (i) e.g. CH2 NH CO CO NH CH2 CH2 NH CO CO NH CH2 CH2 NH [1] (ii) HO2C CO2H H2N CH2 CH2 NH2 or ClCO---------------COCl [1] [1] [3] [Total: 15 max 14]

Page 4 Mark Scheme Syllabus Paper GCE A/AS Level – May/June 2006 9701 04 © University of Cambridge International Examinations 2006 5 (a) I: HNO3 + H2SO4 (or names) [1] (both) conc. and at 50oC < T < 60oC ✓ [1] II: KMnO4 (+OH-) + heat [1] III: Sn + (conc) HCl [1] IV: CH3CH2OH (or name) [1] + c. H2SO4 + heat [1] [6] (b) CH3 NO2 + CH3 H NO2 CH3 NO2 + H+ intermediate, including ⊕ [1] NO2 + at start and H+ at finish [1] (no marks for curly arrows, but if present, they must be in correct direction) [2] (c) (i) ester and (primary) amine [2] (ii) more basic: amine group is not adjacent to benzene ring both points [1] (or lone pair (on N) is not delocalised) [3] [Total: 11]