Cambridge A Level Chemistry 9701 — 2013 May/June Paper 2 · Variant 1

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. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fl uid. DO NOT WRITE IN ANY BARCODES. Answer all questions. Electronic calculators may be used. You may lose marks if you do not show your working or if you do not use appropriate units. 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. CHEMISTRY 9701/21 Paper 2 Structured Questions AS Core May/June 2013 1 hour 15 minutes Candidates answer on the Question Paper. Additional Materials: Data Booklet UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certifi cate of Education Advanced Subsidiary Level and Advanced Level This document consists of 10 printed pages and 2 blank pages. [Turn over IB13 06_9701_21/2RP © UCLES 2013 *7635733974* For Examiner’s Use 1 2 3 4 5 Total

2 9701/21/M/J/13 © UCLES 2013 For Examiner’s Use Answer all the questions in the spaces provided. 1 A sample of a fertiliser was known to contain ammonium sulfate, (NH4)2SO4, and sand only. A 2.96 g sample of the solid fertiliser was heated with 40.0 cm3 of NaOH(aq), an excess, and all of the ammonia produced was boiled away. After cooling, the remaining NaOH(aq) was exactly neutralised by 29.5 cm3 of 2.00 mol dm–3 HCl. In a separate experiment, 40.0 cm3 of the original NaOH(aq) was exactly neutralised by 39.2 cm3 of the 2.00 mol dm–3 HCl. (a) (i) Write balanced equations for the following reactions. NaOH with HCl … (NH4)2SO4 with NaOH … (ii) Calculate the amount, in moles, of NaOH present in the 40.0 cm3 of the original NaOH(aq) that was neutralised by 39.2 cm3 of 2.00 mol dm–3 HCl. (iii) Calculate the amount, in moles, of NaOH present in the 40.0 cm3 of NaOH(aq) that remained after boiling the (NH4)2SO4. (iv) Use your answers to (ii) and (iii) to calculate the amount, in moles, of NaOH that reacted with the (NH4)2SO4.

3 9701/21/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (v) Use your answers to (i) and (iv) to calculate the amount, in moles, of (NH4)2SO4 that reacted with the NaOH. (vi) Hence calculate the mass of (NH4)2SO4 that reacted. (vii) Use your answer to (vi) to calculate the percentage, by mass, of (NH4)2SO4 present in the fertiliser. Write your answer to a suitable number of signifi cant fi gures. [9] (b) The uncontrolled use of nitrogenous fertilisers can cause environmental damage to lakes and streams. This is known as eutrophication. What are the processes that occur when excessive amounts of nitrogenous fertilisers get into lakes and streams? … … … [2] (c) Large quantities of ammonia are manufactured by the Haber process. Not all of this ammonia is used to make fertilisers. State one large-scale use for ammonia, other than in the production of nitrogenous fertilisers. … [1] [Total: 12]

4 9701/21/M/J/13 © UCLES 2013 For Examiner’s Use 2 Ammonium nitrate fertiliser is manufactured from ammonia. The fi rst reaction in the manufacture of the fertiliser is the catalytic oxidation of ammonia to form nitrogen monoxide, NO. This is carried out at about 1 × 103 kPa (10 atmospheres) pressure and a temperature of 700 to 850 °C. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ∆H = –906 kJ mol–1 (a) Write the expression for the equilibrium constant, Kp, stating the units. Kp = units … [2] (b) What will be the effect on the yield of NO of each of the following? In each case, explain your answer. (i) increasing the temperature … … … (ii) decreasing the applied pressure … … … [4]

5 9701/21/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) The standard enthalpy changes of formation of NH3(g) and H2O(g) are as follows. NH3(g), ∆Hf = –46.0 kJ mol–1 H2O(g), ∆Hf = –242 kJ mol–1 Use these data and the value of given below to calculate the standard enthalpy change of formation of NO(g). Include a sign in your answer. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ∆H = –906 kJ mol–1 [4] [Total: 10]

6 9701/21/M/J/13 © UCLES 2013 For Examiner’s Use 3 This question refers to the elements in the section of the Periodic Table shown below. H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca … transition elements … Ga Ge As Se Br Kr (a) From this list of elements, identify in each case one element that has the property described. Give the symbol of the element. (i) An element that fl oats on cold water and reacts readily with it. … (ii) An element that forms an oxide that is a reducing agent. … (iii) The element that has the smallest fi rst ionisation energy. … (iv) The element which has a giant molecular structure and forms an oxide which has a simple molecular structure. … (v) The element in Period 3 (Na to Ar) that has the smallest anion. … (vi) The element in Period 3 (Na to Ar) which forms a chloride with a low melting point and an oxide with a very high melting point. … [6]

7 9701/21/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (b) Use the elements in Period 3 (Na to Ar) in the section of the Periodic Table opposite to identify the oxide(s) referred to below. In each case, give the formula of the oxide(s). (i) An oxide which when placed in water for a long time has no reaction with it. … (ii) An oxide which dissolves readily in water to give a strongly alkaline solution. … (iii) Two acidic oxides formed by the same element. … and … (iv) An oxide which is amphoteric. … [5] (c) Fluorine reacts with other elements in Group VII to form a number of different compounds. Two such compounds and their boiling points are given in the table. compound Cl F3 BrF3 boiling point / °C 12 127 (i) The two molecules have similar electronic confi gurations. Showing outer electrons only, draw a 'dot-and-cross' diagram of the bonding in Cl F3. (ii) The two molecules have the same shape. Suggest why the boiling points are signifi cantly different. … … … … [4] [Total: 15]

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9 9701/21/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use 4 Organic chemistry is the chemistry of carbon compounds. The types of organic reactions that you have studied are listed below. addition elimination hydrolysis oxidation reduction substitution Addition and substitution reactions are further described as follows. electrophilic nucleophilic free radical Complete the table below. Fill in the central column by using only the types of reaction given in the lists above. Use both lists when appropriate. In the right hand column give the formula(e) of the reagent(s) you would use to carry out the reaction given. organic reaction type of reaction reagent(s) CH3CH2CH2CH2Br → CH3CH2CH2CH2NH2 CH3CH2CH2CH2OH → BrCH2CH2CH2CH2OH CH3COCH3 → CH3C(OH)(CN)CH3 CH3CH(OH)CH2CH3 → CH3CH=CHCH3 [Total: 11]

10 9701/21/M/J/13 © UCLES 2013 For Examiner’s Use 5 Crotonaldehyde, CH3CH=CHCHO, occurs in soybean oils. (a) In the boxes below, write the structural formula of the organic compound formed when crotonaldehyde is reacted separately with each reagent under suitable conditions. If you think no reaction occurs, write 'NO REACTION' in the box. reaction reagent product A Br2 in an inert organic solvent B PCl 3 C H2 and Ni catalyst D NaBH4 E K2Cr2O7 / H+ [5] (b) Crotonaldehyde exists in more than one stereoisomeric form. Draw the displayed formulae of the stereoisomers of crotonaldehyde. Label each isomer. [3]

11 9701/21/M/J/13 © UCLES 2013 [Turn over For Examiner’s Use (c) Draw the skeletal formula of crotonaldehyde. [1] (d) The product of reaction E in the table opposite will react with a solution containing acidifi ed manganate(VII) ions. Draw the structural formulae of the organic products when the reagent is (i) cold, dilute; (ii) hot, concentrated. [3] [Total: 12]

12 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 Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. 9701/21/M/J/13 © UCLES 2013 BLANK PAGE

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9701 CHEMISTRY 9701/21 Paper 2 (AS Structured Questions), maximum raw mark 60 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 1 (a) (i) NaOH + HCl → NaCl + H2O (1) [1] (NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4 + 2H2O (1) [1] allow ionic equations in each case (ii) n(NaOH) = n(HCl) = 1000 00 . 2 × 2 . 39 = 0.0784 (1) [1] (iii) n(NaOH) = n(HCl) = 1000 00 . 2 × 5 . 29 = 0.059 (1) [1] (iv) n(NaOH) = 0.0784 – 0.059 = 0.0194 (1) [1] (v) n[(NH4)2SO4] = 2 0194 . 0 = 9.7 × 10-3 (1) [1] (vi) mass of (NH4)2SO4 = 9.7 × 10-3 × 132.1 = 1.2814 g (1) (vii) % of (NH4)2SO4 = 1.2814 × 100 = 43.30405405 = 43.3 2.96 give one mark for the correct expression (1) [1] give one mark for answer given as 43.3 – i.e. to 3 sig. fig. (1) [1] allow ecf where appropriate [9] (b) fertiliser in the river causes excessive growth of aquatic plants/algae or algal bloom (1) [1] when plants and algae die O2 is used up or fish or aquatic life die (1) [2] (c) manufacture of HNO3 or explosives or nylon or as a cleaning agent or as a refrigerant not detergent (1) [1] [Total:12]

Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 2 (a) ( ) ( ) ( ) ( ) 5 2 4 3 6 2 4 P O NH O H NO p p p p = K (1) [1] atmospheres or Pa or kPa (1) allow ecf on incorrect powers [2] (b) (i) increasing temperature yield of NO is decreased or reaction moves to LHS (1) [1] forward reaction is exothermic (1) [1] (ii) decreasing the pressure yield of NO is increased or reaction moves to RHS (1) [1] more moles/molecules of gas on RHS or fewer moles/molecules of gas on LHS (1) [4] (c) let ∆Hf o for NO be y kJ mol–1 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ∆Hf o 4 × (–46.0) 4y 6 × (–242) (1) [1] ∆Ho reaction = 4y + [6 × (–242)] – [4 × (–46.0)] (1) [1] = 4y – 1452 + 184 ∆Ho reaction is –906 kJmol–1 so 4y = –906 + 1452 – 184 = 362 (1) [1] whence y = ∆Hf o for NO = +90.5 kJ mol–1 + sign is required (1) [4] [Total: 10]

Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 3 (a) penalise (–1) for names of elements (i) Na or K or Li (1) [1] (ii) S or C or N or P (1) [1] (iii) K (1) [1] (iv) C (1) [1] (v) Cl (1) [1] (vi) Al or Si (1) [6] (b) (i) Al2O3 or SiO2 (1) [1] (ii) Na2O (1) [1] (iii) P2O3 or P4O6 and P2O5 or P4O10 or SO2 and SO3 (1+1) [1] (iv) Al2O3 (1) [5] (c) (i) 3 bonding pairs and 2 lone pairs around Cl atom (1) [1] 3 lone pairs on each of the F atoms (1) [1] (ii) either referring to van der Waals’ forces in BrF3 van der Waals’ or intermolecular forces are greater/stronger (1)) [1] because there are more electrons in BrF3 than in ClF3 (1) [1] OR referring to permanent dipoles permanent dipole or intermolecular forces are stronger/greater in BrF3 (1) [1] because BrF3 has a larger permanent dipole than ClF3 OR because difference in electronegativity is larger between Br and F than between Cl and F (1) [1] part (ii) has a maximum of 2 marks (max 2) [4] [Total: 15] Cl F F F

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 4 Types of reaction used must come from the list in the question. organic reaction type of reaction reagent(s) CH3CH2CH2CH2Br → CH3CH2CH2CH2NH2 nucleophilic (1) substitution (1) NH3 (1) CH3CH2CH2CH2OH→ BrCH2CH2CH2CH2OH free radical (1) substitution (1) Br2 or Br2 in an organic solvent (1) not Br2(aq) CH3COCH3 → CH3C(OH)(CN)CH3 nucleophilic (1) addition (1) HCN or HCN and CN– or NaCN/KCN + H+ (1) CH3CH(OH)CH2CH3 → CH3CH=CHCH3 elimination (1) not dehydration conc. H2SO4 or P4O10 or Al2O3 or H3PO4 (1) [Total: 11]

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 5 (a) reaction reagent product A Br2 in an inert organic solvent CH3CHBrCHBrCHO B PCl3 NO REACTION C H2 and Ni catalyst CH3CH2CH2CH2OH D NaBH4 CH3CH=CHCH2OH E K2Cr2O7/H+ CH3CH=CHCO2H one mark for each correct answer [5] (b) trans or E cis or Z two correct structures (1) [1] both correctly labelled (1) [1] correctly displayed -CHO group (1) [3]

Page 7 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9701 21 © Cambridge International Examinations 2013 (c) (1) [1] (d) (i) CH3CH(OH)CH(OH)CO2H (1) [1] (ii) CH3CO2H (1) [1] HO2CCO2H (1) [3] allow ecf on candidate’s answer to E in (a) [Total: 12]