Chemistry 9701/42 October/November 2022: question paper and mark scheme

Question

Calcium chloride, CaCl 2, is an ionic solid. The values of some energy changes are shown in Table 1.1. Table 1.1 energy change value / kJ mol–1 lattice energy, , CaCl 2(s) –2237 standard enthalpy change of atomisation of calcium +193 first ionisation energy of calcium +590 second ionisation energy of calcium +1150 standard enthalpy change of atomisation of chlorine +121 first electron affinity of chlorine –364 (a) Define lattice energy. . …… . …… [1] (b) Use the data in Table 1.1 to calculate the standard enthalpy change of formation, , of calcium chloride. It may be helpful to draw an energy cycle. Show all your working. (CaCl 2(s)) = …… kJ mol–1 [2] (c) Three possible values for the first electron affinity of bromine are shown in Table 1.2. One of them is correct. Place a tick by the correct value. Explain your choice. Table 1.2 possible values place one tick (✓) in this column –342 kJ mol–1 –364 kJ mol–1 –386 kJ mol–1 explanation …… . …… [1] (d) The enthalpy change of hydration of the chloride ion can be calculated using the lattice energy of calcium chloride and the data shown in Table 1.3. Table 1.3 energy change value / kJ mol–1 standard enthalpy change of solution of CaCl 2(s) –83 standard enthalpy change of hydration of Ca2+(g) –1650 (i) Define the following terms. enthalpy change of solution . …… . …… enthalpy change of hydration …… . …… [2] (ii) Calculate the standard enthalpy change of hydration of the chloride ion, Cl –(g). It may be helpful to draw an energy cycle. Show all your working. (Cl –(g)) = …… kJ mol–1 [2] (e) Calcium fluoride, CaF2(s), can be synthesised directly from its elements. The value of (CaF2(s)) is –1214 kJ mol–1. (i) Predict the sign of the entropy change, ∆S o, for this synthesis. Explain your answer. The sign of the entropy change is …… . explanation …… . …… [1] (ii) Use the value of (CaF2(s)) given in (e) and your answer to (e)(i) to predict how the feasibility for this synthesis will change with increasing temperature. . …… . …… . …… . …… [2] [Total: 11]

Question

(a) Nitrogen monoxide, NO, reacts with ozone, O3. NO(g) + O3(g) → NO2(g) + O2(g) This reaction is first order with respect to both NO and O3. At 298 K, the rate constant k = 11 500 mol–1 dm3 s–1. (i) Complete the rate equation for this reaction. rate = …… [1] (ii) A reaction is carried out in which the initial concentrations of NO and O3 are both 1.20 × 10–6 mol dm–3. Calculate the initial rate of the reaction. State its units. rate of reaction = …… units = …… [2] (iii) The reaction described in (a)(ii) is monitored over a period of time. Predict whether or not the graph of [NO] against time, under these conditions, shows that the reaction has a constant half-life. Explain your answer. prediction …… explanation …… . …… [1] (b) Nitrous oxide, N2O, decomposes into its elements. 2N2O(g) → 2N2(g) + O2(g) At a high temperature, a small amount of platinum wire is added to a large amount of nitrous oxide. The reaction follows zero order kinetics. The platinum wire behaves as a catalyst. (i) Sketch a graph, on the axes below, of reaction rate against time for the catalysed decomposition of N2O under these conditions. reaction rate time [1] (ii) Sketch a graph, on the axes below, of [N2O] against time for this reaction. [N2O] time [1] (iii) Platinum behaves as a heterogeneous catalyst in this reaction. Describe the mode of action of a heterogeneous catalyst. . …… . …… . …… . …… [3] (iv) Suggest a reason why this reaction has zero order kinetics when the amount of nitrous oxide is large and the amount of platinum is small. . …… . …… [1] [Total: 10] Question 3 starts on the next page.

Question

Data should be selected from Table 3.1 in order to answer some parts of this question. Table 3.1 electrode reaction Eo/V Mg2+ + 2e– Mg –2.38 Mn2+ + 2e– Mn –1.18 Mn3+ + e– Mn2+ +1.49 MnO2 + 4H+ + 2e– Mn2+ + 2H2O +1.23 MnO4– + e– MnO42– +0.56 MnO4– + 4H+ + 3e– MnO2 + 2H2O +1.67 MnO4– + 8H+ + 5e– Mn2+ + 4H2O +1.52 (a) An electrochemical cell can be constructed from a Mg2+ / Mg half-cell and a MnO4– / Mn2+ half‑cell. The standard cell potential of this cell can be calculated using the standard electrode potentials of the two half-cells. (i) Define standard electrode potential. Include details of the standard conditions used. . …… . …… . …… . …… [2] (ii) Complete the diagram below to show an electrochemical cell constructed from a Mg2+ / Mg half‑cell and a MnO4– / Mn2+ half-cell. Label your diagram. [3] (iii) Use a positive (+) sign and a negative (–) sign to identify the polarity of each of the two electrodes in your diagram. Use an arrow and the symbol ‘e’ to show the direction of electron flow in the external circuit.[1] (iv) Calculate the standard cell potential, , of this cell. = …… V [1] (v) Construct an equation for the cell reaction. . …… [1] (vi) Predict how the cell reaction will change, if at all, when the solution in the Mg2+ / Mg half‑cell is diluted by the addition of a large volume of water. Explain your answer. . …… . …… . …… [1] (b) A molten magnesium salt is electrolysed for 15.0 minutes by a constant current. 4.75 × 1022 magnesium atoms are produced at the cathode. Calculate the value of the current used. current = …… A [2] [Total: 11]

Question

(a) The value of the solubility product, Ksp, of iron(III) hydroxide, Fe(OH)3, is given by the following expression. Ksp = [Fe3+][OH–]3 = 2.0 × 10–39 mol4 dm–12 (i) Calculate the solubility of Fe(OH)3 in water. solubility = …… mol dm–3 [1] (ii) Calculate the solubility of Fe(OH)3 in 0.010 mol dm–3 barium hydroxide, Ba(OH)2(aq). solubility = …… mol dm–3 [2] (iii) Fe(OH)3 is less soluble in Ba(OH)2(aq) than it is in pure water. Name this effect. . …… [1] (b) The numerical value of the Ka of HBrO is 2.00 × 10–9. X is a solution of HBrO which contains 4.00 × 10–3 mol of HBrO in 100 cm3 of solution. In this solution the following equilibrium is established in which there are two conjugate acid-base pairs. HBrO + H2O BrO– + H3O+ (i) Define conjugate acid-base pair. . …… . …… [1] (ii) Identify the two conjugate acid-base pairs shown in the equation above. pair one …… …… acid base pair two acid base [1] (iii) Calculate the pH of solution X. Show all your working. pH = …… [2] (iv) A solution containing 2.00 × 10–3 mol of NaOH is added to solution X. A buffer solution is formed. Calculate the pH of this buffer solution. pH = …… [1] [Total: 9]

Question

Copper is a transition element. It forms a wide variety of compounds. (a) Define transition element. . …… . …… [1] (b) An aqueous solution of copper(II) sulfate, CuSO4, contains [Cu(H2O)6]2+ complex ions. If an excess of concentrated hydrochloric acid is added to this solution a ligand exchange reaction occurs and [CuCl 4]2– complex ions are formed. (i) Complete Table 5.1 to state the geometry, the coordination number of copper, and one bond angle in each of the two complex ions. Table 5.1 complex ion geometry coordination number bond angle [Cu(H2O)6]2+ [CuCl 4]2– [3] (ii) In an isolated Cu2+ ion the d-orbitals are all degenerate. In a complex ion such as [Cu(H2O)6]2+ the d-orbitals are non-degenerate. Define degenerate and non-degenerate in this context. degenerate …… non-degenerate …… [1] (iii) Explain why the solutions of the two complex ions in Table 5.1 are different colours. . …… . …… [1] (c) Cu2+ forms a complex ion containing water molecules and ethanedioate ions, C2O42–, as ligands. The formula of the complex is [Cu(C2O4)2(H2O)2]2–. The ethanedioate ion is a bidentate ligand. (i) Explain what is meant by bidentate. . …… . …… [1] (ii) There are three stereoisomers with the formula [Cu(C2O4)2(H2O)2]2–. Complete the three-dimensional diagrams to show these three stereoisomers. stereoisomer 1 stereoisomer 2 Cu Cu stereoisomer 3 Cu [2] (iii) Use your answer to (c)(ii) to answer this question. Stereoisomers 1, 2 and 3 show two different types of isomerism. Name these two types of isomerism. For each type of isomerism identify the pair of stereoisomers that demonstrate this isomerism. type of isomerism pair of stereoisomers and and [2] (d) A solution contains 3.70 g of Na2[Cu(C2O4)2(H2O)2] dissolved in 100 cm3 of solution. A 25.0 cm3 sample of this solution is warmed and then oxidised by 0.0100 mol dm–3 acidified potassium manganate(VII). The equation for the redox reaction is shown. 5C2O42– + 2MnO4– + 16H+ → 10CO2 + 2Mn2+ + 8H2O Calculate the minimum volume of 0.0100 mol dm–3 acidified potassium manganate(VII) needed to oxidise all of the ethanedioate ions, C2O42–, in the 25.0 cm3 sample. Show all your working. [Mr: Na2[Cu(C2O4)2(H2O)2], 321.5] minimum volume = …… cm3 [3] (e) Copper(II) nitrate, Cu(NO3)2, and barium nitrate, Ba(NO3)2, both decompose when heated. Copper(II) nitrate decomposes at a lower temperature than barium nitrate. Suggest a reason for this difference. Explain your answer. . …… . …… . …… . …… [2] [Total: 16]

Question

An aqueous solution of cobalt(II) chloride is a pink colour due to the presence of [Co(H2O)6]2+ complex ions. (a) (i) Complete Table 6.1 to describe what is observed when the named reagent is added to an aqueous solution of cobalt(II) chloride. Table 6.1 colour of state of reagent cobalt-containing product cobalt-containing product NaOH(aq) an excess of conc. HCl [2] (ii) Write an equation for the reaction between [Co(H2O)6]2+ ions and NaOH(aq). . …… [1] (iii) Write an equation for the reaction between [Co(H2O)6]2+ ions and an excess of conc. HCl. . …… [1] (b) (i) Define stability constant. . …… . …… [1] (ii) Write an expression for the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion. Kstab = [1] (iii) Give the units of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion. units = …… [1] (iv) The numerical value of the stability constant, Kstab, of the [Co(NH3)6]2+ complex ion is 7.7 × 104. In an aqueous solution the concentration of the [Co(NH3)6]2+ complex ion is 0.0740 mol dm–3 and the concentration of NH3 is 0.480 mol dm–3 at equilibrium. Calculate the concentration of [Co(H2O)6]2+ in this solution. concentration = …… mol dm–3 [1] [Total: 8]

Question

(a) The structural and displayed formulae of three aromatic compounds, A, B and C, are shown in Fig. 7.1. C6H5CH2CH2COOH CH3C6H4OH C6H5CH2CHCl CO2H HO O HO O OH Cl A B C Fig. 7.1 Compare the relative acidities of A, B and C. …… > …… > …… most acidic least acidic Explain your answer. . …… . …… . …… . …… [3] (b) Methylbenzene, C6H5CH3, can be made from benzene by an electrophilic substitution reaction. (i) Identify a compound that reacts with benzene to form methylbenzene. Identify the catalyst used. compound . …… catalyst …… [1] (ii) The first step in the reaction is the generation of the CH3+ electrophile. Write an equation for the reaction that generates this electrophile. . …… [1] (iii) Describe the mechanism for the reaction between benzene and the CH3+ electrophile. Include all relevant curly arrows and charges. CH3+ [3] (iv) Identify a suitable reagent to oxidise methylbenzene to benzoic acid, C6H5COOH. Write an equation for this reaction using [O] to represent one atom of oxygen from the oxidising agent. reagent …… equation . …… [2] (v) Methylbenzene and benzoic acid are both nitrated with a mixture of concentrated nitric acid and sulfuric acid to give mononitrated products. The structural formulae of these products are CH3C6H4NO2 and HOOCC6H4NO2 respectively. Draw the structures of these two products. [1] (c) A reaction scheme is shown in Fig. 7.2. reaction 1 CH3C6H4NO2 CH3C6H4NH2 reaction 2 reaction 3 CH3C6H4OH CH3C6H4N2+Cl – reaction 4 dye molecule Y Fig. 7.2 (i) Describe the reagents and conditions to produce CH3C6H4N2+Cl – from CH3C6H4NH2 in reaction 2. reagents . …… conditions …… [1] (ii) Describe how CH3C6H4OH can be produced from CH3C6H4N2+Cl – in reaction 3. . …… [1] (iii) Draw the structure of the dye molecule Y formed when CH3C6H4N2+Cl – and CH3C6H4OH react together in reaction 4. Describe the conditions for this reaction. structure conditions …… [2] [Total: 15]

Question

When answering this question it should be assumed that together all the hydrogen atoms in a benzene ring result in a single unsplit peak at δ = 7.2 in a proton (1H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C6H5COCH(CH3)2 S C6H5CH2CH2COCH3 Q C6H5COCH2CH2CH3 T C6H5CH(CH3)COCH3 R C6H5CH2COCH2CH3 (a) Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*). [1] (b) The proton (1H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig. 8.1. 11 10 9 8 7 6 5 4 3 2 1 0 / ppm Fig. 8.1 (i) Identify which of the compounds P, Q, R, S or T gives this spectrum. Draw the displayed formula of the compound you have identified. Identify the protons responsible for the peaks at δ = 3.7, δ = 2.5 and δ = 1.0 on the structure you have drawn. [2] (ii) Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern. . …… . …… [1] (c) Choose from the letters P, Q, R, S and T to identify: (i) the two compounds that each have a doublet peak in the proton (1H) NMR spectrum . …… [1] (ii) the compound with only three peaks in its proton (1H) NMR spectrum. . …… [1] (d) Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig. 8.1. . …… [1] (e) The proton (1H) NMR spectrum of compound T is compared in the presence of D2O and in the absence of D2O. Describe any difference between the two spectra. Explain your answer. . …… . …… [1] (f) Complete Table 8.1 below to give the number of peaks in the carbon-13 NMR spectrum of each compound. Table 8.1 compound number of peaks compound number of peaks [2] [Total: 10]

Question

Tyrosine and lysine, shown in Fig. 9.1, are naturally occurring amino acids found in proteins. O O H 2N OH OH NH2 NH2 HO tyrosine lysine Fig. 9.1 (a) The isoelectric point of lysine is 9.47. Draw the structure of lysine at the stated pH in the boxes below. lysine at pH 7.00 lysine at pH 9.47 lysine at pH 12.00 [2] (b) When ethanoic acid is treated with PCl 5 product D is formed. When D is added to tyrosine two different isomeric products, E and F, are formed. E has an ester linkage, F does not. Draw the structures of D, E and F in the boxes below. D E F [3] (c) Complete Table 9.1 by drawing the structure of the organic product formed when tyrosine reacts with each named reagent. Ignore the directing effect of the –CH2CHNH2COOH substituent. Table 9.1 reagent structure an excess of Br2(aq) an excess of NaOH(aq) an excess of HNO3(aq) [3] (d) A mixture of tyrosine and lysine can be separated by thin-layer chromatography. Under certain conditions the Rf value of lysine is 0.14 and the Rf value of tyrosine is 0.45. (i) Explain what is meant by Rf value. . …… . …… [1] (ii) Suggest an explanation for the difference in Rf values. . …… . …… [1] [Total: 10] Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 at www.cambridgeinternational.org after the live examination series. 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 24 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge