Cambridge A Level Chemistry 9701 — 2019 Oct/Nov Paper 4 · Variant 2

1 An electrochemical cell is constructed using two half-cells. ●● a Br2/Br – half-cell ●● an Mn3+/Mn2+ half-cell (a) State the material used for the electrode in each half-cell. Br2/Br – half-cell ........................................................................................................................... Mn3+/Mn2+ half-cell ...................................................................................................................... [1] (b) The cell is operated at 298 K. The Br2/Br – half-cell has standard concentrations. The Mn3+/Mn2+ half-cell has [Mn3+] = 0.500 mol dm–3 and [Mn2+] = 0.100 mol dm–3. (i) Use the Nernst equation to calculate the electrode potential, E, of the Mn3+/Mn2+ half‑cell under these conditions. E = .............................. V [2] (ii) Calculate the Ecell under these conditions. Ecell = .............................. V [1] (iii) Write an equation for the overall cell reaction that occurs. ....................................................................................................................................... [2] (c) An aqueous solution of copper(II) sulfate is electrolysed using copper electrodes. A current of 1.50 A is passed for 3.00 hours. 5.09 g of copper is deposited on the cathode. The charge on one electron is –1.60 × 10–19 C. The relative atomic mass of copper is 63.5. Use these data to calculate an experimentally determined value for the Avogadro constant, L. Give your answer to three significant figures. L = .............................. mol–1 [5] (d) Explain why magnesium metal cannot be obtained by the electrolysis of dilute aqueous magnesium sulfate. Your answer should include data from the Data Booklet. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] [Total: 13]

Mark scheme: 1(a) Platinum and platinum 1 1(b)(i) M1: Nernst quoted correctly E = Eo + 0.0590 / zlog [ox] / [red] or E = 1.49 + 0.0590log5 M2: (+)1.53 V minimum 2 sig. fig. Correct answer scores 2 marks 2 1(b)(ii) + / – 0.46 minimum 2 sig. fig. 1 1(b)(iii) M1: Mn3+ + 2Br- → Mn2+ + Br2 M2: 2Mn3+ + 2Br- → Mn2+ + Br2 2 1(c) M1: 16200 C M2: 1.0125 × 1023 electrons (use of 1.60 × 10–19) M3: 0.0802 moles of copper (use of 5.09 and 63.5) M4: 0.1603 moles electrons M5: L = 6.32 × 1023 (correct answer [5] other approaches acceptable including: M1: 16200 C M2: 1.0125 × 1023 electrons (use of 1.60 × 10–19) M3: 5.0625 × 1022 copper atoms M4: 0.0802 moles of copper (use of 5.09 and 63.5) M5: L = 6.32 × 1023 (correct answer [5]) 5 1(d) M1: Mg2+ + 2e– ⇌ Mg Eo = –2.38 and 2H+ + 2e– ⇌ H2 Eo = 0.00 M2: hydrogen produced instead / hydrogen easier to reduce / hydrogen preferentially reduced / hydrogen has more positive Eo 2

More questions on Electrolysis

2 (a) Explain what is meant by the following terms. half-life of a reaction ................................................................................................................... .................................................................................................................................................... rate-determining step ................................................................................................................. .................................................................................................................................................... [2] (b) The reaction between hydroxide ions and bromomethane is first order with respect to [OH–] and first order with respect to [CH3Br]. CH3Br + OH– CH3OH + Br – Suggest a practical method that would confirm that the reaction is first order with respect to [OH–]. ●● Your method should include details of measurements that would be taken in order to calculate the rate of the reaction. ●● You should include a method of presenting the results to show that the reaction is first order with respect to [OH–]. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [4] (c) The hydrolysis of methyl ethanoate, CH3CO2CH3, by hydroxide ions, OH–, is first order with respect to [CH3CO2CH3] and also first order with respect to [OH–]. In a particular experiment, ●● [CH3CO2CH3] = 0.100 mol dm–3 ●● [OH–] = 0.100 mol dm–3 ●● rate of reaction = 2.06 × 10–3 mol dm–3 s–1. Write a rate equation for this reaction and calculate the value of the rate constant, k, under these conditions. State the units of k. rate = .......................................................................................................................................... k = .............................. units = .............................. [3] [Total: 9]

Mark scheme: 2(a) M1: the time taken for the amount/concentration or a reactant to halve M2: the slowest step 2 2(b) • Use an excess of CH3Br • (Several experiments with) different initial [OH] • control / equilibrate temperatures • measure time • find [OH–] by sample and titrate or use of pH probe or find [Br –] by sample and reference to use of Ag+. • processing of results – plot graph of [OH–] vs rate or evaluate rate is proportional to [OH–] numerically Alternative approach: • Use an excess of CH3Br • One experiment with known initial [OH–] • control / equilibrate temperatures • measure time • find [OH–] by sample and titrate or use of pH probe or find [Br –] by sample and reference to use of Ag+ and describes how to calculate [OH–] . • processing of results – plot graph of [OH–] vs time and look for constant half-life Award 1 mark for each correctly identified point. 4 Question Answer Marks 2(c) M1: rate = k[ester][OH–] M2: value of k = 0.206 M3: units mol–1dm3s–1 3

More questions on Simple rate equations, orders of reaction and rate constants

3 Ammonium chloride, NH4Cl (s), dissolves in water to form an acidic solution. This is due to the dissociation of the ammonium ions. NH4+(aq) H+(aq) + NH3(aq) (a) The ammonium ion is a weak acid. The pH of a 0.300 mol dm–3 solution of ammonium chloride is 4.89 under standard conditions. (i) Calculate the [H+] in a 0.300 mol dm–3 solution of ammonium chloride. [H+] = .............................. mol dm–3 [1] (ii) Calculate the value of pKa of the ammonium ion. pKa = .............................. [2] (b) A buffer solution can be made by mixing ammonium chloride with ammonia solution. (i) Explain, with the aid of an equation, how this solution can behave as a buffer when a small amount of a strong acid is added. ............................................................................................................................................. ....................................................................................................................................... [1] (ii) Explain, with the aid of an equation, how this solution can behave as a buffer when a small amount of a strong base is added. ............................................................................................................................................. ....................................................................................................................................... [1] (c) (i) Use the value of Kw to calculate [H+] in pure water under standard conditions. Show your working. [H+] = .............................. mol dm–3 [1] (ii) The pH of pure water at 50 °C is 6.64. Calculate the numerical value of Kw at 50 °C. Kw = .............................. [2] [Total: 8]

Mark scheme: 3(a)(i) 1 3(a)(ii) M1: Ka expression used correctly and Ka = 5.5(3) × 10–10 M2: pKa = 9.26 Award 2 marks for correct answer 2 3(b)(i) NH3 + H+ → NH4+ 1 3(b)(ii) NH4+ + OH– → NH3 + H2O or H+ + OH– → H2O and reference to expression in Q shifting R 1 3(c)(i) quotes Kw = 1 × 10–14 or 1 × 10–1 [H+][OH–] [H+] = 1 × 10–7 1 3(c)(ii) M1: [H+] = 2.3 × 10–7 (calculator value 2.290867 × 10–7) and Kw = [2.3 × 10–7]2 M2: Kw = 5.2 × 10–14 calculator 5.248074 × 10–14 Award 2 marks for correct answer 2

More questions on Acids and bases

4 The table shows some standard entropy data. standard entropy, S o substance / J K–1 mol–1 PbO2(s) 77 PbO(s) 69 O2(g) 205 Lead(IV) oxide, PbO2, decomposes to lead(II) oxide, PbO, and oxygen when heated. 2PbO2(s) 2PbO(s) + O2(g) ΔH o = +118 kJ mol–1 (a) Use the data to calculate the value of ΔS o for this reaction. ΔS o = .............................. J K–1 mol–1 [2] (b) Use the value of ΔH o and your answer to (a) to calculate the temperature at which this reaction becomes feasible. T = .............................. K [3] (c) Solid lead(II) oxide can be made by heating lead metal in air. Predict the sign of the standard entropy change of this reaction. Explain your answer. .................................................................................................................................................... .............................................................................................................................................. [1] [Total: 6]

Mark scheme: 4(a) M1: correct use of stoichiometry M2: answer + 189 2 4(b) M1: States or uses correct form of Gibbs equation ∆G = ∆H – T∆S M2: appreciates / includes ∆G = 0 at temperature required M3: uses 1000 correctly and answer +624(.339) Award 3 marks for correct answer 3 4(c) negative and decrease in number / amount of gas molecules 1

More questions on Entropy change, ΔS

5 (a) Describe fully what would be seen when magnesium and strontium are heated separately in oxygen. magnesium ................................................................................................................................. strontium .................................................................................................................................... [2] (b) (i) Write an equation for each of the following processes. Include state symbols. calcium is burned in air ............................................................................................................................................. calcium carbonate is heated strongly ............................................................................................................................................. [2] (ii) Calcium hydroxide is formed when water is added to calcium oxide. Calcium hydroxide and calcium carbonate are both used in agriculture. Describe the main benefit of adding calcium hydroxide or calcium carbonate to soil. ............................................................................................................................................. ....................................................................................................................................... [1] (iii) Explain why the Group 2 hydroxides become more soluble down the group. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [3] (c) Describe the observations, if any, that you would make when: ●● a few drops of NaOH(aq) are added to BaCl 2(aq) .................................................................................................................................................... ●● a few drops of H2SO4(aq) are added to BaCl 2(aq). .................................................................................................................................................... [2] (d) Describe and explain how the thermal stability of the Group 2 carbonates varies down the group. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [3] [Total: 13]

Mark scheme: 5(a) M1: Mg – white flame and Sr – red flame M2: white solid product once 2 5(b)(i) M1: 2Ca(s) + O2(g) → 2CaO(s) CaCO3(s) → CaO(s) + CO2(g) all substances, balanced M2: all state symbols 2 5(b)(ii) neutralises acid / raises pH 1 5(b)(iii) M1: ∆Hlat and ∆Hhyd decrease down group M2: ∆Hlat decreases / changes more M3: ∆Hsol becomes more exo / more –ve / less endo / less +ve 3 Question Answer Marks 5(c) • no change (for hydroxide) / colourless solution • white (for sulfate) • precipitate (for sulfate) Award 1 mark for two points, award 2 marks for all three points 2 5(d) M1: stability increases / higher T needed for decompose M2: larger ionic radius M3: harder to distort / polarise anion / carbonate ion or harder to polarise / weaken C–O or C=O bond. 3

More questions on Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds

6 Many copper compounds, such as CuSO4 and Cu(NO3)2 contain Cu2+ ions. Aqueous solutions of this ion contain the [Cu(H2O)6]2+ complex ion, in which water behaves as a monodentate ligand. (a) Explain what is meant by the term monodentate ligand. .................................................................................................................................................... .............................................................................................................................................. [1] (b) Copper(II) sulfate solution, CuSO4(aq), undergoes visible changes with certain reagents. Complete the table below. formula of the copper(II) reagent added to a observations compound or complex ion solution of CuSO4(aq) that is formed a few drops of NH3(aq) an excess of NH3(aq) an excess of NaOH(aq) an excess of concentrated HCl (aq) [4] (c) When water is added to concentrated aqueous cobalt(II) chloride the colour of the solution changes from blue to pink. Explain this observation. No equation is needed, but you should include reference to electron movement between orbitals in your answer. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [3] (d) When chlorine gas is bubbled into FeSO4(aq) the colour of the solution changes from pale green to yellow. Use data from the Data Booklet to explain this observation. Include an equation in your answer. Reference to electron movement between orbitals is not needed. .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [2] (e) If a solution of chloride ions is added to a solution containing [Fe(H2O)6]3+ an equilibrium is established. [Fe(H2O)6]3+ + 4Cl – [FeCl 4]– + 6H2O (i) Write an expression for the stability constant of [FeCl 4]–, Kstab. Kstab = [1] (ii) For the above equilibrium the numerical value of Kstab = 0.080. Calculate the concentration of [Fe(H2O)6]3+ in a solution in which the concentration of Cl – is 2.0 mol dm–3 and the concentration of [FeCl 4]– is 0.10 mol dm–3. concentration of [Fe(H2O)6]3+ = .............................. mol dm–3 [1] [Total: 12]

Mark scheme: 6(a) donates one pair of electrons / forms one coordinate bond 1 6(b) Reagent added to a solution of CuSO4(aq) Observations Formula of the copper(II) compound or complex ion that is formed a few drops of dilute ammonia blue ppt Cu(OH)2 or Cu(OH)2(H2O)4 an excess of dilute ammonia deep blue solution [Cu(NH3)4]2+ or [Cu(NH3)4(H2O)2]2+ an excess of aqueous sodium hydroxide blue ppt Cu(OH)2 or Cu(OH)2(H2O)4 an excess of conc HCl. green-yellow / yellow-green / yellow [CuCl4]2– Award 1 mark for each correct observation and formula in a row of the table. 4 Question Answer Marks 6(c) • ligand exchange • magnitude of d-orbital splitting changes / ∆E for d-orbitals changes / energy gap between d-orbitals changes • change in colour / frequency / wavelength of light absorbed • electrons are promoted/excited to higher d Award 1 mark for two points, award 2 marks for three points, award 3 marks for all four points 3 6(d) M1: Eo values 1.36 and 0.77 quoted M2: 2FeSO4 + Cl2 → Fe2(SO4)2Cl2 or 2Fe2+ + Cl2 → 2Fe3+ + 2Cl – 2 6(e)(i) [[FeCl4]–] / [ [Fe(H2O)6]3+][Cl –]4 1 6(e)(ii) 0.078(125) 1

More questions on General characteristic chemical properties of the first set of transition elements, titanium to copper

7 (a) A polymer, P, can be made from the monomers benzene‑1,4‑dioyl chloride and benzene‑1,3‑diamine. benzene-1,4-dioyl chloride benzene-1,3-diamine O O H2N NH2 C C Cl Cl (i) Draw a section of the polymer chain of P. Your structure should include two repeat units. P [2] (ii) Place one tick in the table to describe polymer P and its method of polymerisation. type of polymerisation type of polymer addition condensation polyalkene polyamide polyester [1] (iii) State whether or not polymer P is biodegradable. Give a reason for your answer. ............................................................................................................................................. ....................................................................................................................................... [1] (iv) State a suitable reagent to produce benzene‑1,4‑dioyl chloride from benzene‑1,4‑dicarboxylic acid. ....................................................................................................................................... [1] (v) Benzene‑1,3‑diamine can be made from benzene using the reaction sequence shown. O2N O2N NO2 H2N NH2 step 1 step 2 step 3 State the reagents used for steps 1 and 3. step 1 .................................................................................................................................. step 3 .................................................................................................................................. [2] (b) Proteins are natural polymers. A protein is said to have a primary, secondary and tertiary structure. (i) Describe what is meant by each of these terms. primary structure ................................................................................................................. ............................................................................................................................................. secondary structure ............................................................................................................. ............................................................................................................................................. tertiary structure .................................................................................................................. ............................................................................................................................................. [3] (ii) Name the forces or bonds responsible for holding together the primary structure of a protein molecule. ....................................................................................................................................... [1] (iii) Name the forces or bonds responsible for the stabilisation of the secondary structure of a protein molecule, and identify the groups of atoms within the protein molecule that are held together by these forces or bonds. ............................................................................................................................................. ....................................................................................................................................... [2] [Total: 13]

Mark scheme: 7(a)(i) two or more repeat units • correct orientation of groups on all four rings and rings correct • trailing bonds shown • amide links all correct Award 1 mark for two points, award 2 marks for all four points 2 7(a)(ii) polyamide and condensation 1 7(a)(iii) yes and can be hydrolysed 1 7(a)(iv) PCl3 or PCl5 or SOCl2 1 7(a)(v) M1: conc nitric acid + conc sulfuric acid M2: Sn + HCl 2 Question Answer Marks 7(b)(i) M1: sequence / order of amino acids M2: α-helix or β-sheet M3: folding of chain or 3-D shape 3 7(b)(ii) covalent bonds / peptide bonds / amide bonds 1 7(b)(iii) M1: hydrogen bonds M2: between C=O and N–H 2

More questions on Condensation polymerisation

8 Butanedioic acid can be made in a three-step synthesis using ethane as the starting material. ethane A H H H H step 1 step 2 H C C H Br C C Br B H H H H step 3 butanedioic acid H H O O C C C C HO OH H H (a) Describe the reagents and conditions needed for step 1. .............................................................................................................................................. [1] (b) Name the other product also formed in step 1 which is an isomer of A. .............................................................................................................................................. [1] (c) Give the structural formula of B. [1] (d) State the reagents and conditions needed for step 2. .............................................................................................................................................. [2] (e) Butanedioic acid cannot be oxidised by a warm, aqueous solution of any commonly used oxidising agents but ethanedioic acid can. (i) Identify the oxidising agent that could be used to oxidise ethanedioic acid. ....................................................................................................................................... [1] (ii) State the product(s) of the reaction in (e)(i). ....................................................................................................................................... [1] (f) Compare and explain the relative acidities of hexanoic acid, hexan‑1‑ol and phenol. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .............................................................................................................................................. [3] (g) Methylpropanedioic acid is an isomer of butanedioic acid. methylpropanedioic acid CH3 O O C C C HO OH H The proton NMR spectrum of methylpropanedioic acid in CCl 4 is shown. δ 1.1 δ 12.7 δ 3.3 14 12 10 8 6 4 2 0 δ / ppm (i) Identify the protons in the methylpropanedioic acid molecule that are responsible for each area of the proton NMR spectrum. δ 12.7 .................................................................................................................................. δ 3.3 .................................................................................................................................... δ 1.1 .................................................................................................................................... [2] (ii) Name the splitting pattern shown at δ 3.3 and explain how it arises. ............................................................................................................................................. ....................................................................................................................................... [1] The carbon‑13 NMR spectra of butanedioic acid, HO2CCH2CH2CO2H, and methylpropanedioic acid, HO2CCH(CH3)CO2H are different. (iii) State the number of peaks ● in the carbon‑13 NMR spectrum of butanedioic acid ............................................................................................................................................. ● in the carbon‑13 NMR spectrum of methylpropanedioic acid. ............................................................................................................................................. [1] [Total: 14]

Mark scheme: 8(a) bromine / Br2 and uv / light / heat 1 8(b) 1,1-dibromoethane 1 8(c) NCCH2CH2CN / CH2CNCH2CN 1 8(d) M1: KCN / NaCN / CN– M2: boil/heat/reflux and ethanol as solvent 2 8(e)(i) acidified manganate(VII) or dichromate(VI) 1 8(e)(ii) carbon dioxide and water 1 Question Answer Marks 8(f) M1: most acidic: hexanoic acid > phenol > hexan-1-ol :least acidic • the other O atom in CO2H group of hexanoic acid either − withdraws charge from OH group or is electronegative and weakens O–H bond or − stabilises resultant anion/negative ion / –CO2– group/carboxylate ion • benzene / aromatic / C6H5 ring in phenol delocalises either − lone pair on O atom and weakens O–H bond or − lone pair on resultant anion/negative ion / phenoxide ion this stabilises resultant anion negative ion / –CO2– group/carboxylate ion • the alkyl group in hexan-1-ol donates electrons this strengthens O–H bond Award 1 mark for each bullet point identified. 3 8(g)(i) M1: δ12.7 is COOH M2: δ3.3 is CH and δ1.1 is CH3 2 8(g)(ii) quadruplet / quartet 3 H / protons on neighbouring / adjacent carbon / carbons / C 1 8(g)(iii) 2 (butanedioic acid) and 3 (methylpropanedioic acid) 1

More questions on Carboxylic acids

9 The structure of nicotinamide is shown. nicotinamide O NH2 N (a) The nitrogen atom in the six‑membered ring has one electron in an unhybridised p‑orbital. This electron becomes delocalised, becoming part of a single delocalised system of electrons. This delocalised system also includes: ●● electrons from the carbon atoms in the six-membered ring ●● the two electrons in the π bond of the C=O group ●● the two electrons in the lone pair on the nitrogen atom of the amide group. (i) State the number of delocalised electrons in one nicotinamide molecule. ....................................................................................................................................... [1] (ii) Predict the H–N–H bond angle in the NH2 group in nicotinamide. ....................................................................................................................................... [1] (b) Nicotinamide can be synthesised from nicotinic acid. The synthesis involves two steps. nicotinic acid nicotinamide E O O step 1 step 2 OH NH2 N N (i) Draw the structural formula of E in the box. [1] (ii) Give the name or formula of a suitable reagent for step 2. ....................................................................................................................................... [1] (c) Nicotinamide reacts separately with aqueous acid and aqueous alkali. The six‑membered ring remains unchanged in these reactions. Complete the reaction scheme below to give the structural formula of the organic product of each reaction. O HCl (aq) NaOH(aq) NH2 reflux reflux N [2] (d) Nicotinamide can be reduced to compound X. X NH2 N (i) Identify a suitable reducing agent for this reaction. ....................................................................................................................................... [1] (ii) Predict and explain the relative basicities of the NH2 groups in phenylamine, C6H5NH2, nicotinamide and compound X. ............................................................................................................................................. ............................................................................................................................................. ............................................................................................................................................. ....................................................................................................................................... [3] (e) The height of the M peak in a mass spectrum of nicotinamide is 80. Calculate the expected height of the M+1 peak. [2] [Total: 12]

Mark scheme: 9(a)(i) 10 1 9(a)(ii) 120 1 9(b)(i) correct acid chloride 1 9(b)(ii) NH3 or ammonia 1 Question Answer Marks 9(c) M1: (C5NH4)COOH or (C5NH5)+COOH M2: (C5NH4)COO–(Na+) or (C5NH4)COONa 2 9(d)(i) LiAl H4 1 9(d)(ii) M1: most basic: X > phenylamine > nicotinamide :least basic M2: LP in X cannot be delocalised M3: LP in phenylamine delocalised over the benzene ring or LP in amide delocalised (more effectively) by C=O 3 9(e) M1: M + 1 / M = (1.1 × ?) / 100 M2: Ans 5.28 Award 2 marks for correct answer 2