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Chemistry 9701/43 May/June 2020: question paper and mark scheme

Cambridge AS & A Level Chemistry (9701), Paper 4 A Level Structured Questions, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.

9701_s20_qp_439701_s20_ms_439701/43/M/J/20

Paper
Paper 4 A Level Structured Questions
Total marks
100
Questions
10
Code on the paper
9701/43/M/J/20
9701_s20_ms_43.pdf

9701/43 mark scheme, question by question

Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.

Question 113 marks

Question

(a) An aqueous solution of cobalt(II) contains the [Co(H2O)6]2+ complex ion. (i) Define the term complex ion. . …… . …… [1] (ii) Samples of [Co(H2O)6]2+ are reacted separately with aqueous sodium hydroxide and with an excess of aqueous ammonia. Give the following information about these reactions. ●● the reaction of [Co(H2O)6]2+ with aqueous sodium hydroxide colour and state of the cobalt-containing species . …… ionic equation …… type of reaction …… ●● the reaction of [Co(H2O)6]2+ with an excess of aqueous ammonia colour and state of the cobalt-containing species . …… ionic equation …… type of reaction …… [6] (b) When concentrated hydrochloric acid is added to a solution containing [Co(H2O)6]2+, a blue solution of [CoCl 4]2– is formed and the following equilibrium is established. [Co(H2O)6]2+ + 4Cl – [CoCl 4]2– + 6H2O Use Le Chatelier’s principle to suggest the expected observations when silver nitrate solution is added dropwise to the blue solution of [CoCl 4]2–. Explain your answer. . …… . …… . …… [2] (c) The [Co(NH3)3Cl 3] complex shows stereoisomerism. Complete the three-dimensional diagrams to show the two isomers of [Co(NH3)3Cl 3]. Suggest the type of stereoisomerism. Co Co type of stereoisomerism …… [2] (d) Compound X, C6H18N4, is a tetradentate ligand. X H H 2N N N NH2 H (i) Suggest why one molecule of X can form four dative bonds. . …… . …… . …… [1] (ii) C6H18N4 reacts with aqueous cobalt(II) ions, [Co(H2O)6]2+, in a 1 : 1 ratio to form a new complex ion. Construct an equation for this reaction. . …… [1] [Total: 13]

Mark scheme

1(a)(i) (a molecule or ion) formed by a (central) metal atom / ion surrounded by / bonded to (one or more) ligands 1 1(a)(ii) M1: blue ppt/solid 6 M2: [Co(H2O)6]2+ + 2OH- Co(OH)2 + 6H2O OR [Co(H2O)6]2+ + 2OH- [Co(H2O)4(OH)2] + 2H2O M3: precipitation/ acid-base M4: blue solution M5: [Co(H2O)6]2+ + 6NH3 [Co(NH3)6]2+ + 6H2O M6: ligand exchange/displacement/substitution/replacement 1(b) • solution turns blue → pink 2 • a white ppt. of AgCl forms • equilibrium shifts to the left / [Cl-] decreases Two correct responses = 1 mark Three correct responses = 2 marks 1(c) NH3 NH3 2 Cl NH3 H3N Cl Co Co Cl NH3 Cl NH3 Cl Cl geometric ALLOW cis-trans Two correct responses = 1 mark Three correct responses = 2 marks 1(d)(i) each nitrogen / the four nitrogen’s has a lone pair of electrons (to the metal ion) 1 Two correct responses = 1 mark 1(d)(ii) [Co(H2O)6]2+ + C6H18N4 [Co(C6H18N4)]2+ + 6H2O 1 OR [Co(H2O)6]2+ + C6H18N4 [Co(C6H18N4)(H2O)2]2+ + 4H2O

Question 26 marks

Question

(a) (i) Describe and explain the trend in the solubility of the Group 2 hydroxides down the group. . …… . …… . …… . …… . …… . …… [4] Group 2 hydroxides decompose on heating to give the corresponding metal oxide and water vapour. (ii) Suggest which of Mg(OH)2 and Sr(OH)2 will decompose at a lower temperature. Explain your answer. . …… . …… . …… . …… [2] [Total: 6]

Mark scheme

2(a)(i) M1 solubility increases down the group 4 M2 ΔHlatt and ΔHhyd both become less exothermic / less negative M3 ΔHlatt changes more (than ΔHhyd as OH– being smaller than M2+) M4 ΔHsol becomes more exothermic / more negative 2(a)(ii) M1 Mg(OH)2 AND Mg2+ has a smaller ionic radii/ Mg2+ has a higher charge density 2 M2 OH- ion is polarised/distorted more

Question 37 marks

Question

The overall reaction for photosynthesis is shown. 6CO2 + 6H2O → C6H12O6 + 6O2 Water is oxidised in this process according to the following half-equation. 2H2O → O2 + 4H+ + 4e– half-equation 1 (a) (i) Use these equations to deduce the half-equation for the reduction of carbon dioxide in this process. [2] (ii) Draw a fully labelled diagram of the apparatus that should be used to measure the standard electrode potential, E o, of O2(g) in half-equation 1 under standard conditions. Include all necessary chemicals. [4] (iii) For the cell drawn in (a)(ii), use the Data Booklet to calculate the and deduce which electrode is positive. = …… V identity of the positive electrode = …… [1] [Total: 7]

Mark scheme

3(a)(i) 6CO2 + 24H+ + 24e- C6H12O6 + 6H2O 2 ALLOW 6CO2 + 12H+ + 12e- C6H12O6 + 3O2 for both marks ALLOW one mark for an unbalanced equation showing the correct species of either equation 3(a)(ii) salt bridge (indicated) voltmeter / V labelled 4 O2 good delivery system H2 good delivery system Pt electrode H+ / HCl / H2SO4 solution labelled (at least once) 1 atm 1 mol dm–3 quoted Every two correct responses = 1 mark 3(a)(iii) Eocell = (+) 1.23 V AND positive electrode = O2 half-cell identified 1

Question 47 marks

Question

(a) The molecular formulae of three nitrogen‑containing compounds are given. S C6H5CONH2 T C6H5NH2 U C6H5CH2NH2 Describe and explain the relative basicities of S, T and U. …… > …… > …… most basic least basic . …… . …… . …… . …… . …… . …… [3] (b) Compound U can be prepared by two different methods as shown. O NH2 Br reaction 1 U reaction 2 NH2 (i) Suggest reagents and conditions for reaction 1 and for reaction 2. reaction 1 . …… reaction 2 . …… [2] (ii) State the type of reaction in reaction 1 and name the mechanism in reaction 2. type of reaction in reaction 1 …… mechanism of reaction 2 …… [2] [Total: 7]

Mark scheme

4(a) M1 phenylmethanamine / U > phenylamine / T > benzamide / S [1] 3 any two from: • alkyl group is electron donating so lone pair more able to accept a proton • lone pair on N overlaps with delocalised system so less able to accept a proton • presence of electron-withdrawing oxygen / carbonyl group means lone pair is not available to accept a proton OR amides are neutral 4(b)(i) reaction 1 LiAlH4 2 reaction 2 heat NH3 under pressure/ heat NH3 in a sealed tube 4(b)(ii) reaction 1 reduction 2 reaction 2 nucleophilic substitution

Question 514 marks

Question

(a) Benzene reacts with bromine in the presence of an aluminium bromide catalyst, Al Br3, to form bromobenzene. This is a substitution reaction. No addition reaction takes place. (i) Explain why no addition reaction takes place. . …… . …… [1] Al Br3 reacts with bromine to generate an electrophile, Br+. (ii) Draw the mechanism of the reaction between benzene and Br+ ions. Include all relevant arrows and charges. [3] (iii) Write an equation to show how the Al Br3 catalyst is reformed. . …… [1] (b) Suggest why bromination of phenol occurs more readily than bromination of benzene. . …… . …… . …… . …… . …… [2] (c) (i) There are four different carbocations with the same formula, C4H9+. One structure is given in the table. Suggest the structural formulae of the three other carbocations. structure 1 structure 2 structure 3 structure 4 CH3CH2CH2CH2+ [3] (ii) Benzene reacts with each of these carbocations in separate Friedel-Crafts alkylation reactions. In each reaction an organic compound with formula C10H14 is formed. The number of peaks observed in the carbon‑13 NMR spectrum of each compound is given. Suggest the structures for the three other compounds. number of peaks in carbon-13 NMR = 8 number of peaks in carbon-13 NMR = 6 number of peaks in carbon-13 NMR = 7 number of peaks in carbon-13 NMR = 8 [4] [Total: 14]

Mark scheme

5(a)(i) The substitution product is stabilised by delocalisation of (6)π-electrons 1 OR The addition product is not stabilised by delocalisation of (6)π-electrons [1] 5(a)(ii) Br 3 + Br + H+ Br+ H [1] intermediate [1] curly arrow + H+ lost [1] • first curly arrow • intermediate • 2nd curly arrow, product and H+ formed / lost 5(a)(iii) AlBr4- + H+ AlBr3 + HBr 1 5(b) lone pair of oxygen is delocalised into the ring 2 any one from: • phenol has a higher electron density in the ring • phenol can polarise/induce a dipole in Br2 5(c)(i) CH3CH2CH+CH3 (CH3)2CHCH2+ (CH3)3C+ Each correct structure = 1 mark 3 5(c)(ii) 4 number of peaks in carbon-13 NMR = 8 number of peaks in carbon-13 NMR = 6 number of peaks in carbon-13 NMR = 7 number of peaks in carbon-13 NMR = 8 Two correct organic products = 1 mark three correct organic products = 2 marks all products linked correctly to NMR = 2 marks

Question 623 marks

Question

(a) Compare and explain the relative acidities of 2‑chloropropanoic acid, 3‑chloropropanoic acid, and propanoic acid. Explain your answer. …… > …… > …… most acidic least acidic explanation …… . …… . …… . …… . …… . …… [3] (b) (i) The numerical values of Ka for methanoic acid, HCO2H, and pyruvic acid, CH3COCO2H, are given. acid Ka HCO2H 1.78 × 10–4 CH3COCO2H 4.07 × 10–3 An equilibrium mixture containing the two acid-base pairs is formed. HCO2– + CH3COCO2H HCO2H + CH3COCO2– Use the Ka values to calculate the equilibrium constant, Keq, for this equilibrium. Keq = …… [1] (ii) Use your value of Keq to predict the position of this equilibrium. Indicate this by placing a tick () in the appropriate box in the table. Explain your answer. equilibrium lies equilibrium lies equilibrium lies to the left in the middle to the right . …… . …… [1] (iii) Ethanedioic acid, HO2CCO2H, has two dissociation constants, Ka1 and Ka2, whose pKa values are 1.23 and 4.19. Suggest equations to show the two dissociations that give rise to these pKa values. pKa1 1.23 …… pKa2 4.19 …… [2] (iv) State the mathematical relationship between pKa and the acid dissociation constant Ka. . …… [1] (c) Three tests were carried out on separate samples of the organic acids shown in the table. The following results were obtained. = observed change = no observed reaction reagent(s) test HCO2H CH3COCO2H HO2CCO2H observed change and conditions 1 2 3 Complete the table with the reagent(s) and conditions and the observed change for each test. Assume these organic acids all have a similar acid strength. [5] (d) A sample of pyruvic acid, CH3COCO2H, is analysed by carbon‑13 NMR spectroscopy. Three peaks are observed. Complete the table by: ●● circling the carbon atom responsible for the chemical shift ●● stating the hybridisation of the circled carbon atom. chemical carbon atom responsible hybridisation of the shift (δ) for chemical shift circled carbon atom H O O 27 H C C C H O H H O O 163 H C C C H O H H O O 192 H C C C H O H [2] (e) An ester of pyruvic acid, F, is dissolved in CDCl 3 and analysed by proton NMR spectroscopy. F O C O CH3 H 3C C CH2 O The proton NMR spectrum of F is shown. 5 4 3 2 1 0 δ / ppm Use the proton NMR spectrum of F to complete the table. chemical group responsible number of 1H atoms splitting pattern shift (δ) for the peak responsible for the peak 1.3 2.2 4.0 [3] (f) Deuterium oxide, D2O, where D is 2H, can be used as a solvent in proton NMR spectroscopy. The proton NMR spectrum of alanine in CDCl 3 has 4 peaks. The proton NMR spectrum of alanine in D2O has 2 peaks. alanine NH2 H3C C COOH H On the diagram of alanine, circle the protons that show peaks in both NMR spectra. Explain your answer. . …… . …… . …… [2] (g) The ionic product, Kw, for D2O has a value of 1.35 × 10–15 mol2 dm–6 at 298 K. (i) Write the expression for the Kw of D2O. Kw = [1] (ii) Calculate the pH of pure, neutral D2O at 298 K. Assume [D+] is equivalent to [H+] for pH calculations. pH = …… [2] [Total: 23]

Mark scheme

6(a) M1 2-chloropropanoic acid > 3-chloropropanoic acid > propanoic acid [1] 3 M2 CH3CHClCO2H / ClCH2CH2CO2H (are more acidic) as they contain an electronegative Cl atom so weaken O-H bond / stabilise carboxylate anion [1] M3 CH3CHClCO2H (is more acidic than ClCH2CH2CO2H) as the Cl atom is closer to CO2H so weaken O-H bond more / stabilise carboxylate anion more [1] 6(b)(i) Keq = 4.07x10-3/1.78x10-4 = 22.9 1 6(b)(ii) 3rd box ticked [to the right] AND as the Keq is greater than one ecf on Keq 1 6(b)(iii) pKa 1.23 HO2CCO2H + H2O ⇌ HO2CCO2- + H3O+ 2 OR HO2CCO2H ⇌ HO2CCO2- + H+ pKa 4.19 HO2CCO2- + H2O ⇌ -O CCO - 2 2 + H3O+ OR HO2CCO2- ⇌ -O CCO - 2 2 + H+ 6(b)(iv) pKa = – logKa 1 6(c) 5 reagents and conditions observed change M1 Tollen’s reagent, warm silver mirror test 1 OR Fehling’s solution, warm (brick)-red ppt. M2 aqueous alkaline iodine yellow ppt. test 2 OR 2,4-DNPH orange ppt. test 3 M3 acidified MnO4- , warm decolourises (and bubbles) Two correct observations = 1 mark Three correct observations = 2 marks 6(d) 2 environment of hybridisation of chemical shift (δ) the carbon atom the carbon atom 27 CH3 circled sp3 163 COOH circled sp2 192 C=O(COOH) circled sp2 Award one mark for each correct column 6(e) 3 chemical group responsible splitting number of 1H atoms shift (δ) for the peak pattern responsible for the peak 1.3 alkane / CH / CH3 triplet 3 CH3CO or 2.2 singlet 3 alkyl / CH next to C=O CH2O or quartet / 4.0 alkyl / CH next to 2 quadruplet electronegative atom / C=O Award one mark for every three correct responses. 6(f) CH AND CH3 circled 2 these protons do not exchange with D2O OR OH and NH protons exchange with D2O 6(g)(i) Kw = [D+][DO-] 1 6(g)(ii) M1 [D+] = √1.35 x 10-15 = 3.67 x10-8 2 M2 pH = -log [D+] = 7.4(3) min 2sf

Question 710 marks

Question

(a) Silver carbonate, Ag2CO3, is sparingly soluble in water. The numerical value of the solubility product, Ksp, for silver carbonate is 6.3 × 10–12 at 25 °C. (i) Write an expression for the solubility product, Ksp, of Ag2CO3, and state its units. Ksp = units = …… [2] (ii) Calculate the equilibrium concentration of Ag+ in a saturated solution of Ag2CO3 at 25 °C. [Ag+] = …… mol dm–3 [1] (iii) Solid Ag2CO3 is stirred at 25 °C with 0.050 mol dm–3 AgNO3 until no more Ag2CO3 dissolves. Calculate the concentration of carbonate ions, [CO32–], in this solution. [CO32–] = …… mol dm–3 [1] (iv) An electrochemical cell is set up to measure the electrode potential, E, for the Ag+ / Ag half‑cell using the saturated Ag2CO3(aq) with a standard hydrogen electrode. Use the Data Booklet, your answer to (a)(ii), and the Nernst equation to calculate the electrode potential, E, for this Ag+ / Ag half-cell. E for Ag+ / Ag half-cell = …… V [2] (b) Silver chloride, AgCl, is sparingly soluble in water. The equation for the enthalpy change of solution is shown. AgCl (s) → Ag+(aq) + Cl –(aq) = +65.5 kJ mol–1 Standard entropies are shown in the table. species AgCl (s) Ag+(aq) Cl –(aq) S o / J K–1 mol–1 +96.2 +72.7 +56.5 (i) Calculate the standard entropy change of solution, ∆S o. ∆S o = …… J K–1 mol–1 [1] (ii) Explain, with the aid of a calculation, why AgCl is insoluble in water at 25 °C. You should use data from this question and your answer to (b)(i). . …… . …… [3] [Total: 10]

Mark scheme

7(a)(i) M1 Ksp = [Ag+]2[CO32-] 2 M2 units = mol3 dm-9 7(a)(ii) 1 x = 3√6.3 x 10-12/4 = 1.16 x 10-4 (mol dm-3) [Ag+] = 1.16 x 10-4 x 2 = 2.33 x 10-4 (mol dm-3) min 2sf 7(a)(iii) 6.3 x 10-12 = [0.05]2[CO32-] 1 [CO32-] = 2.52 x 10-9 (mol dm-3) min 2sf 7(a)(iv) M1 E=Eo + 0.059log[Ag+] 2 M2 E=0.80 + 0.059log(1.2 x 10-4) = 0.57 V ecf from (a)(ii) min 2sf 7(b)(i) ΔSo = 72.7 + 56.5 -96.2 = +33.0 J K-1 mol-1 1 7(b)(ii) M1 ΔG = ΔHo - TΔSo 3 M2 ΔG = (65.5)- (298x0.033) = +55.7 kJ mol-1 min 3sf M3 ΔG = positive so not feasible/spontaneous

Question 88 marks

Question

(a) Explain what is meant by the term buffer solution. . …… . …… . …… [2] (b) (i) Write an expression for the acid dissociation constant, Ka, for ammonium ions, NH4+(aq). Ka = [1] (ii) Write two equations to describe how a solution containing ammonium ions, NH4+(aq), and ammonia, NH3(aq), can act as a buffer. . …… . …… [2] (iii) The numerical value of Ka for NH4+(aq) is 5.6 × 10–10 at 298 K. A buffer solution was prepared by adding 0.80 dm3 of 0.25 mol dm–3 ammonia, an excess, to 0.20 dm3 of 0.20 mol dm–3 hydrochloric acid. Calculate the pH of the buffer solution formed at 298 K. pH = …… [3] [Total: 8]

Mark scheme

8(a) M1 a solution that resists changes in pH 2 M2 when small amounts of acid and alkali are added to it 8(b)(i) Ka = [NH3][H+] 1 [NH4+] 8(b)(ii) M1 NH4+ + OH- NH3 + H2O 2 M2 NH3 + H3O+ NH4+ + H2O 8(b)(iii) M1 moles NH3(initial) = 0.25 x 0.80 = 0.200 AND 3 moles HCl = 0.20 x 0.20 = 0.040 (= moles NH4+eqm) M2 moles NH3(eqm) = 0.20 – 0.04 = 0.160 [H+] = (5.6 x10-10 x 0.04)/(0.16) = 1.4 x 10-10 (mol dm-3) ecf on M1 M3 pH = -log(1.4 x 10-10) = 9.85 ecf on M2 min 2sf

Question 97 marks

Question

(a) Manganese(IV) oxide, MnO2, catalyses the decomposition of hydrogen peroxide, H2O2, as shown. MnO2 2H2O2(aq) 2H2O(l) + O2(g) The mechanism involves the formation of the intermediate species, Mn2+, in the first step which is subsequently used up in the second step. State and use relevant electrode potentials, E o, to construct two equations to show how MnO2 can catalyse this reaction. . …… . …… . …… . …… . …… equation 1 . …… equation 2 . …… [3] (b) The equation for the decomposition of hydrogen peroxide without a catalyst is shown. 2H2O2(aq) → 2H2O(l) + O2(g) Under certain conditions this reaction is found to be first order with respect to hydrogen peroxide, with a rate constant, k, of 2.0 × 10–6 s–1 at 298 K. Calculate the initial rate of decomposition of a 0.75 mol dm–3 hydrogen peroxide solution at 298 K. initial rate = …… mol dm–3 s–1 [1] (c) A four-step mechanism is suggested for the reaction between hydrogen peroxide and iodide ions in an acidic solution. step 1 H2O2 + I– → IO– + H2O step 2 H+ + IO– → HIO step 3 HIO + I– → I2 + OH– step 4 OH– + H+ → H2O Step 1 is the rate-determining step. (i) State what is meant by the term rate-determining step. . …… . …… [1] (ii) Use this mechanism to construct a balanced equation for this reaction. . …… [1] (iii) Deduce the order of reaction with respect to each of the following. H2O2 = …… I– = …… H+ = …… [1] [Total: 7]

Mark scheme

9(a) M1 data seen H2O2/H2O +1.77V and MnO2/Mn2+ +1.23 V and O2/H2O2 +0.68 V 3 OR Ecell= 0.55 V (first step) and 0.54 V (second step) M2 MnO2 + H2O2 + 2H+ Mn2+ + O2 + 2H2O M3 Mn2+ + H2O2 MnO2 + 2H+ 9(b) rate = 2.0 x 10-6 x 0.75 = 1.5 x 10-6 1 9(c)(i) slowest step in overall reaction 1 9(c)(ii) H2O2 + 2H+ + 2I- I2 + 2H2O 1 OR H2O2 + 2HI I2 + 2H2O 9(c)(iii) H2O2 = 1 AND I- = 1 AND H+ = 0 1

Question 105 marks

Question

(a) The electronic configuration of transition element Q is [Ar] 3d2 4s2. Predict the likely oxidation states of element Q in compounds. . …… [1] (b) Suggest why transition elements often show variable oxidation states in their compounds, but typical s‑block elements such as calcium do not. . …… . …… [1] (c) Many enzymes contain transition element complexes. Describe, with the aid of a suitably labelled diagram, how an enzyme catalyses the breakdown of a substrate molecule. . …… . …… . …… . …… [3] [Total: 5]

Mark scheme

10(a) +4 and any of +1, +2, +3 1 10(b) close similarity of energy of the 4s and 3d sub-shells 1 10(c) diagrams 3 M1 (can be in words or diagram) substrate shape is complementary to active site M2 (can be in words or diagram) the substrate bind / bonds / fits into the active site M3 (can be in words or diagram) products are released M4 (words) lower EA / bonds weakened (in substrate) Any three points

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Chemistry 9701/43 for May/June 2020 is out of 100 marks, across 10 questions.

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