Cambridge A Level Chemistry 9701 — 2005 May/June Paper 4 · Variant 1

This document consists of 11 printed pages and 1 blank page. SPA (MML 8078 3/04) S77899/2 © UCLES 2005 [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 2005 1 hour 15 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, Centre number and candidate number in the spaces at the top of this page. 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. The number of marks is given in brackets [ ] at the end of each question or part question. 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. Centre Number Candidate Number Name If you have been given a label, look at the details. If any details are incorrect or missing, please fill in your correct details in the space given at the top of this page. Stick your personal label here, if provided. For Examiner’s Use 5 4 3 2 1 6 TOTAL

Answer all the questions in the spaces provided. 1 A student decided to determine the value of the Faraday constant by an electrolysis experiment. The following incomplete diagram shows the apparatus that was used. (a) (i) Apart from connecting wires, what two additional pieces of equipment are needed for this experiment? … … (ii) Complete the diagram, showing additional equipment connected in the circuit, and showing the powerpack connected to the correct electrodes. (iii) List the measurements the student would need to make in order to use the results to calculate a value for the Faraday constant. … … … … [7] (b) (i) Using an equation, state the relationship between the Faraday constant, F, the Avogadro constant, L, and the charge on the electron, e. … (ii) The value the student obtained was: 1 Faraday = 9.63 × 104 Coulombs Use this value and your equation in (b)(i) to calculate the Avogadro constant (take the charge on the electron to be 1.60 × 10–19 Coulombs) … … [2] [Total: 9] hydrogen being collected power pack inert electrodes – + dilute H2SO4 2 For Examiner's Use 9701/04/M/J/05 © UCLES 2005

2 (a) What do you understand by the term order of reaction? … [1] (b) Cyanohydrins can be made by reacting ketones with an acidified solution of sodium cyanide. (CH3)2C=O + H+ + CN– →(CH3)2C(OH)CN In a series of experiments, the reaction was carried out with different concentrations of the three reagents, and the following relative initial rates were obtained. (i) Use the data in the table to deduce the order of the reaction with respect to propanone … hydrogen ions … cyanide ions … (ii) Hence write a rate equation for this reaction. … Two different mechanisms have been suggested for this reaction Mechanism A: (CH3)2C=O + H+ → (CH3)2COH+ (CH3)2COH+ + CN– → (CH3)2C(OH)CN Mechanism B: (CH3)2C=O + CN– → (CH3)2C(O–)CN (CH3)2C(O–)CN + H+ → (CH3)2C(OH)CN (iii) Which mechanism is consistent with the rate equation you deduced in (ii), and which step in this mechanism is the slower (rate determining) step? Explain your answer. … … … … [7] [Total: 8] 3 [Turn over For Examiner's Use 9701/04/M/J/05 © UCLES 2005 experiment number [(CH3)2CO] / mol dm–3 [H+] / mol dm–3 [CN–] / mol dm–3 1 0.020 0.060 0.060 2 0.020 0.050 0.050 3 0.020 0.050 0.060 4 0.025 0.050 0.050 relative initial rate / mol dm–3sec–1 1.00 0.833 1.00 1.042

3 Limestone is an important raw material, used in building, steel making and agriculture. The first stage in using limestone is often to heat it in a kiln. CaCO3(s) →CaO(s) + CO2(g)
H = +178 kJ mol–1 reaction 1 Water is then added to the ‘quicklime’ produced in the kiln, to make ‘slaked lime’. CaO(s) + H2O(l) →Ca(OH)2(s)
H = –82 kJ mol–1 reaction 2 (a) (i) Suggest two reasons why reaction 1 needs heating to a high temperature. … … (ii) Explain whether MgCO3 would require a higher or a lower temperature than CaCO3 for its decomposition. … … … [5] Before the widespread use of cement, bricks and stones used for buildings were bonded together with a mixture of slaked lime, sand and water, known as lime mortar. On exposure to the air, the lime mortar gradually set hard due to the following reaction. Ca(OH)2(s) + CO2(g) →CaCO3(s) + H2O(l) (b) Use the data given above to calculate the enthalpy change for this reaction. … … … [1] 4 For Examiner's Use 9701/04/M/J/05 © UCLES 2005

(c) One of the major ores of magnesium is the mixed carbonate called dolomite, CaMg(CO3)2. Calculate the percentage loss in mass that would be observed when a sample of dolomite is heated at a high temperature until the reaction had finished. … … … [2] [Total: 8] 5 [Turn over For Examiner's Use 9701/04/M/J/05 © UCLES 2005

4 (a) (i) State the electronic configuration of the iron atom. … (ii) Apart from its electronic structure, state two properties of iron or its compounds that are characteristic of a transition element. … … [3] (b) Acidified solutions of iron(II) salts can be titrated using a dilute solution of potassium manganate(VII), KMnO4. (i) Use the Data Booklet to calculate the standard cell potential and to write a balanced ionic equation for the reaction that takes place during the titration. … … … (ii) Explain why no indicator is required for this titration. What colour change would you see at the end point? … … … [4] (c) Use the reaction between Fe3+ ions and water molecules to explain the meanings of the terms ligand and complex formation. … … … … [2] 6 For Examiner's Use 9701/04/M/J/05 © UCLES 2005

(d) An important biological molecule containing iron is haemoglobin. (i) What is the role of haemoglobin in the body? … (ii) Use your answer to (i) to explain why carbon monoxide is poisonous. … … … [2] (e) In a possible industrial synthesis of ethanol, the complex Fe(CO)5 catalyses the reaction between carbon monoxide, hydrogen and methanol according to the following equation. 200 °C CH3OH + 2CO + H2 –––––→CH3CH2OH + CO2 300 atm Describe a test (reagents and observations) that would distinguish ethanol from methanol. reagents … observation with methanol … observation with ethanol… [2] [Total: 13] 7 [Turn over For Examiner's Use 9701/04/M/J/05 © UCLES 2005

5 (a) Give an expression for Ka as applied to the weak acid RCO2H. … … [1] (b) The Ka values for three carboxylic acids are listed in the table below. (i) Describe and explain the trend in acid strength illustrated by these values. … … … … (ii) Calculate the pH of a 0.100 mol dm–3 solution of ClCH2CO2H. … … … (iii) Calculate the pKa value for Cl 2CHCO2H. … … [5] 8 For Examiner's Use 9701/04/M/J/05 © UCLES 2005 acid Ka/ mol dm–3 CH3CO2H 1.8 × 10–5 ClCH2CO2H 1.4 × 10–3 Cl2CHCO2H 5.5 × 10–2

(c) The acid ClCH2CO2H features in the industrial synthesis of the important weedkiller 2,4-D. (i) Suggest a possible reagent for reaction I. … (ii) What type of reaction is reaction I, … reaction II? … (iii) Describe a test (reagents and observations) that would distinguish phenol from compound A. reagents … observation with phenol … observation with compound A … [5] [Total: 11] CH2OH compound A OH OH Cl Cl OCH2CO2H Cl Cl phenol + NaOH + ClCH2CO2H I II 2,4-D 9 [Turn over For Examiner's Use 9701/04/M/J/05 © UCLES 2005

6 The antipyretic (fever-reducing) drug antifebrin can be made from benzene and ethanoic acid by the following route. (a) (i) What type of reaction is reaction I? … (ii) Suggest the reagents and conditions for reaction I. … (iii) Complete the following scheme showing the mechanism of reaction I, by drawing appropriate formulae in the three boxes. [6] (b) (i) What type of reaction is reaction II? … (ii) Suggest the reagents and conditions for reaction II. … [2] NO2 + + NO2 NH2 NHCOCH3 I CH3CO2H CH3COCl antifebrin III IV II 10 For Examiner's Use 9701/04/M/J/05 © UCLES 2005

(c) Suggest the reagents and conditions for reaction III. … [1] (d) (i) Apart from the benzene ring, name the functional group in antifebrin. … (ii) What reagents and conditions are needed to hydrolyse antifebrin? … [2] [Total: 11] 11 For Examiner's Use 9701/04/M/J/05 © UCLES 2005

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the June 2005 question paper 9701 CHEMISTRY 9701/04 Paper 4 (Structured Questions A2 Core), 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. • CIE will not enter into discussion or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the June 2005 question papers for most IGCSE and GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Grade thresholds for Syllabus 9701 (Chemistry) in the June 2005 examination. minimum mark required for grade: maximum mark available A B E Component 4 60 45 40 22 The thresholds (minimum marks) for Grades C and D are normally set by dividing the mark range between the B and the E thresholds into three. For example, if the difference between the B and the E threshold is 24 marks, the C threshold is set 8 marks below the B threshold and the D threshold is set another 8 marks down. If dividing the interval by three results in a fraction of a mark, then the threshold is normally rounded down.

June 2005 GCE A LEVEL MARK SCHEME MAXIMUM MARK: 60 SYLLABUS/COMPONENT: 9701/04 CHEMISTRY Paper 4 (Structured Questions A2 Core)

Page 1 Mark Scheme Syllabus Paper A LEVEL – JUNE 2005 9701 4 © University of Cambridge International Examinations 2005 1 (a) (i) Ammeter/galvanometer [1] Clock/watch/timer (or rheostat) [1] (For items above 2 in number, e.g. voltmeter, penalise [1]) (ii) Diagram to show ammeter (allow symbol) in circuit, and [1] complete circuit with ⊖ terminal of power pack connected to LH electrode [1] (iii) Volume/amount of hydrogen/gas [1] Time [1] Current/amps/ammeter reading (ignore extra measurements) [1] Part (a): [7] (b) (i) F = L x e [1] (ii) L = 9.63 x 104/1.6 x 10-19 = 6.02 x 1023 (must show working) [1] Allow 6.0 but not 6 or 6.01 Part (b): [2] Total: [9] 2 (a) The power/index/exponent to which a concentration term is raised in a rate equation or a in rate = k[A]a (k is needed – or can use rateα [A]a) [1] Part (a): [1] (b) (i) 1st order w.r.t. propanone [1] Zero order w.r.t. H+ ions [1] 1st order w.r.t. CN- ions [1] (ii) Rate = k [propanone][CN-] (e.c.f. from (i)) [1] (iii) Mechanism B (or A – see grid below), with the first (or second – see grid below) step being the slow step, [1] (since H+ does not appear in rate equation) it must be involved after the slow step or [H+] is not involved in slow step [1] Grid for e.c.f. in first mark of (iii) Deductions in (i) or (ii) E.C.F. deductions in (iii) [Propanone] [CN-] [H+] Mechanism Slow step 1 1 0 B 1st 1 0 1 A 1st 1 1 1 A or B 2nd Any other No e.c.f. mark can be awarded Part (b): [6] Total: [7]

Page 2 Mark Scheme Syllabus Paper A LEVEL – JUNE 2005 9701 4 © University of Cambridge International Examinations 2005 3 (a) (i) It is an endothermic reaction, or taking in heat [1] It has a high activation energy/Ea [1] (ii) MgCO3 will decompose at a lower temperature/needs less energy [1] Mg2+ is a smaller (ion) than Ca2+ or Mg2+ has high charge density [1] So polarises/distorts the anion CO3 2- ion more easily [or LE(MgO) > LE(CaO)] [1] Part (a): [5] (b) ∆H = 82 – 178 = -96 (kJ mol-1) [1] Part (b): [1] (c) [CaMg(CO3)2 → CaO + MgO + 2CO2] Mr(CaMg(CO3)2) = 40.1 + 24.3 + 24 + 96 = 184.4 [1] Mr(2CO2) = 2 x 44 = 88 ∴% loss in mass = 100 x 4 . 184 88 = 47.7% (e.c.f. in 184.4) [1] Allow 48%. Also allow 48.8% if Mr = 184 Part (c): [2] Total: [8]

Page 3 Mark Scheme Syllabus Paper A LEVEL – JUNE 2005 9701 4 © University of Cambridge International Examinations 2005 4 (a) (i) 1s22s22p63s23p63d64s2 or [Ar] 3d64s2 [1] (ii) Coloured compounds/ions/solutions/ppts; paramagnetic; variable oxidation state/valency/more than one ion; dense metals; high melting point metals; are catalysts; form complexes (ANY 2) [1] + [1] Part (a): [3] (b) (i) MnO4 - + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ [1] Eo = 1.52 – 0.77 = 0.75V (allow e.c.f. 0.90V for MnO2 [1] (ii) MnO4 - is purple/highly coloured [1] End point is first (permanent) pink colour or colourless-to-pink (Allow yellow-to-pink but not purple-to-pink) [1] Part (b): [4] (c) Water molecules are ligands, in that they coordinate/form dative bonds (to the Fe ion) with their (lone) pairs of electrons or lone pairs are donated. [1] A complex ion is an ion/Fe3+ surrounded by/joined to ligands or [Fe(H2O)6]3+ [1] Part (c): [2] (d) (i) Haemoglobin transports oxygen in the blood or from lungs (to tissues) [1] (ii) CO forms stronger bonds to Hb/Fe2+ than does O2 or CO has higher affinity or bonds irreversibly or forms more stable complex [1] Part (d): [2] (e) Reagent: I2 + OH- [1] Observations - ethanol: yellow ppt./antiseptic smell; methanol: no change [1] Part (e): [2] Total: [13]

Page 4 Mark Scheme Syllabus Paper A LEVEL – JUNE 2005 9701 4 © University of Cambridge International Examinations 2005 5 (a) Ka = [RCO2 -][H+]/[RCO2H] [1] Part (a): [1] (b) (i) The more chlorine atoms in the molecule, the stronger the acid, [1] due to the electron-withdrawing (inductive) effect of Cl... [1] either...stabilising the anion, or spreading (-) charge more, or...weakening the O-H bond in the acid, or...increasing ionisation, or…facilitates H+ donation or…causing the equilibrium RCO2H ⇋ RCO2 - + H+ to lie further to the right. Mark is conditional on reference to the effect of presence of chlorine. [1] (ii) [H+] = √(0.1 x 1.4 x 10-3) = 0.0118 (mol dm-3) allow 0.012 [1] ∴ pH = -log10(0.0118) = 1.93 Allow 1.9 or 1.92 e.c.f. [1] (iii) pKa = -log10(5.5 x 10-2) = 1.26 Allow 1.3 [1] Part (b): [6] (c) (i) Cl2(aq) AlCl3 or UV negates [1] (ii) Electrophilic substitution or addition-elimination [1] Nucleophilic substitution or electrophilic substitution on OH group If neither mark is awarded, could give “salvage” mark for substitution x2 [1] (iii) Either: add Br2(aq) phenol decolourises it, or gives a white ppt. or: add FeCl3(aq) phenol give a purple colour or: add NaOH(aq) phenol dissolves or: add UI solution phenol goes yellow/orange (A stays green) or: add “diazonium”to solution in OH- phenol gives orange/red colour (in each case, A give no reaction) or: add Cr2O7 2-/H+/warm A changes colour from orange to green or: add MnO4 -/H+/warm A changes from purple to colourless or: add PCl5/POCl3/PCl3/SOCl2 A gives fumes or: add CH3CO2H + conc. H2SO4 A gives fruity smell (in each case, no change with phenol) Test + reagents [1] Both observations [1] Part (c): [5] Total: [12]

Page 5 Mark Scheme Syllabus Paper A LEVEL – JUNE 2005 9701 4 © University of Cambridge International Examinations 2005 6 (a) (i) Electrophilic substitution or nitration [1] (ii) HNO3 + H2SO4 [1] (both) conc., and at 50°C ≤ T ≤ 60°C [1] (iii) NO2 + [1] H NO2 H NO2 or etc. Any ⊕ on NO2 or H negates [1] H+ [1] Part (a): [6] (b) (i) Reduction [1] (ii) Sn/Fe/Zn/SnCl2 + HCl/H+/H2SO4 (but not conc. H2SO4) or H2 + Ni/Pt (not LiAlH4) [1] Part (b): [2] (c) PCl5/PCll3/SOCll2/POCl3 (+ heat) aq negates [1] Part (c): [1] (d) (i) An amide, not peptide [1] (ii) Heat with H3O+ or heat with OH-(aq) Or warm (not heat/reflux) with aqueous amidase/peptidase/protease not enzyme/trypsin/chymotrysin/pepsin/papain etc. [1] Part (d): [2] Total: [11]