Chemistry 2023 May/June | Past Paper

Q1(a)

Short Answer 1 mark

Group 2 carbonates decompose when heated to form the metal oxide and carbon dioxide.

Q1(i)

Short Answer 1 mark

Suggest a mechanism for the decomposition of the carbonate ion by adding two curly arrows in Fig. 1.1. C CO

Q1(ii)

Short Answer 1 mark

Describe the variation in the thermal stability of Group 2 carbonates. Explain your answer.

Q1(b)

Short Answer 1 mark

(i) Define lattice energy. latt(MCO ), becomes less exothermic down the group. latt(MO), also becomes less exothermic down the group. latt(MCO latt(MO) change by different amounts going down the group. Suggest how the standard enthalpy change of the decomposition reaction for Group 2 carbonates changes down the group. Explain your reasoning in terms of the relative sizes of the anions and the relative changes in lattice energy down the group.

Q1(c)

Short Answer 1 mark

Potassium sulfite, K , is used as a food additive. The concentration of sulfite ions, SO , can be determined by titration using aqueous acidified manganate(VII) ions, MnO solution contains 3.40 g of impure K of this solution requires 22.40 cm of 0.0250 mol dm acidified MnO reach the end-point. All the SO ions are oxidised. None of the other species in the impure K are oxidised. The reaction occurs as shown by the two half-equations. O + SO + 2H + 2eMnO + 8H + 5e + 4H

Q1(i)

Short Answer 1 mark

Give the ionic equation for the reaction between SO and acidified MnO

Q1(ii)

Short Answer 3 marks

Calculate the percentage purity of the sample of K Show your working. percentage purity of K =

Q1(d)

Short Answer 1 mark

Potassium disulfite, K , is another food additive. The disulfite ion, S , has the displayed formula shown in Fig. 1.2. Deduce the geometry (shape) around the S(α) atom in S geometry around S(α) [1] [Total: 13]

Q2(a)

Short Answer 1 mark

State two typical chemical properties of a transition element. 1 2

Q2(b)

Short Answer 1 mark

Aqueous solutions of cobalt(II) salts contain the complex ion [Co(H

Q2(i)

Short Answer 1 mark

Define complex ion.

Q2(ii)

Short Answer 1 mark

Samples of [Co(H are reacted separately with an excess of aqueous ammonia, with an excess of concentrated HCl and with an excess of aqueous sodium hydroxide, excess

Q2(aq)

Short Answer 1 mark

excess NaOH(aq) excess concentrated HCl [Co(H

Q2(aq)

Short Answer 1 mark

A C Complete Table 2.1 about the reactions shown by [Co(H2+. Table 2.1 reagent added to [Co(H

Q2(aq)

Short Answer 1 mark

formula of cobalt species formed colour and state of cobalt species formed type of reaction an excess of NH

Q2(aq)

Structured 4 marks

A = an excess of concentrated HClB = an excess of NaOH(aq)C =

Q2(c)

Short Answer 1 mark

The ethanedioate ion, C , can act as a bidentate ligand.

Q2(i)

Short Answer 1 mark

Explain what is meant by a bidentate ligand.

Q2(ii)

Short Answer 3 marks

The complex [Co(H exists as stereoisomers. Complete the three-dimensional diagrams in Fig. 2.2 to show four stereoisomers of [Co(H The C ligand is represented using O O.

Q2(iii)

Short Answer 1 mark

State the oxidation state of cobalt in this complex and a type of stereoisomerism shown. oxidation state of cobalt type of stereoisomerism [1] [Total: 12]

Q3(a)

Short Answer 1 mark

Complete Table 3.1 by placing one tick (3) in each row to indicate the sign of each type of energy change under standard conditions. Table 3.1 energy change always positive always negative can be either negative or positive bond energy enthalpy change of atomisation enthalpy change of formation

Q3(b)

Short Answer 1 mark

Define standard enthalpy change of atomisation.

Q3(c)

Short Answer 1 mark

Table 3.2 shows some energy changes. Table 3.2 energy change value / kJ mol standard enthalpy change of atomisation of silver +285 first ionisation energy of silver +731 second ionisation energy of silver +2074 bond energy of O=O +496 bond energy of O-O +150 first electron affinity of oxygen -141 second electron affinity of oxygen +798 first ionisation energy of oxygen +1314 standard enthalpy change of formation of silver oxide, Ag O

Q3(c)(s)

Short Answer 1 mark

-31 latt, of Ag O

Q3(c)(s)

Short Answer 1 mark

using relevant data from Table 3.2. It may be helpful to draw a labelled energy cycle. Show your working. latt of Ag O

Q3(c)(s)

Short Answer 1 mark

= kJ mol

Q3(d)

Short Answer 2 marks

Suggest the trend in the magnitude of the lattice energies of the silver compounds Ag O and AgSe. Explain your answer. least exothermic most exothermic

Q3(e)

Short Answer 1 mark

Silver sulfite, Ag

Q3(s)

Short Answer 1 mark

, is sparingly soluble in water.

Q3(i)

Short Answer 1 mark

Give an expression for the solubility product, K , of Ag

Q3(ii)

Short Answer 1 mark

Calculate the equilibrium concentration of Ag in a saturated solution of Ag at 298 K. : Ag , 1.50 × 10-14mol at 298 K] [Ag ] = mol dm [1] sol, of AgNO

Q3(s)

Short Answer 1 mark

in water is +22.6 kJ mol Suggest how the feasibility of dissolving AgNO

Q3(s)

Short Answer 1 mark

in water changes with temperature. Explain your answer. [Total: 12]

Q4(a)

Short Answer 1 mark

In aqueous solution, iron(III) ions react with iodide ions, as shown.2Fe + 2I 2Fe+ I A series of experiments is carried out using different concentrations of Fe and I , as shown in Table 4.1. Table 4.1 experiment[Fe ] / mol dm ] / mol dm initial rate / mol dm 1 0.0400 0.0200 2.64 × 10 2 0.1200 0.0200 7.92 × 10 3 0.0800 0.0400 2.11 × 10

Q4(i)

Short Answer 1 mark

Explain what is meant by overall order of reaction.

Q4(ii)

Short Answer 1 mark

Use the data in Table 4.1 to deduce the order of reaction with respect to Fe and with respect to I Explain your reasoning.

Q4(iii)

Short Answer 1 mark

Use your answer to (a)(ii) to construct the rate equation for this reaction. rate =

Q4(iv)

Short Answer 2 marks

Use your answer to (a)(iii) and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = units

Q4(v)

Short Answer 1 mark

Describe qualitatively the effect of an increase in temperature on the rate constant and on the rate of this reaction.

Q4(b)

Short Answer 1 mark

In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown.+ H + 2H + 2H The initial rate of reaction is found to be first order with respect to I , first order with respect to and zero order with respect to H Fig. 4.1 shows a possible four-step mechanism for this reaction. step 1 H+ I + H step 2 H + IO HIO step 3 HIO + I + OH step 4 OH+ H

Q4(i)

Short Answer 1 mark

Suggest which of the steps, 1, 2, 3 or 4, in this mechanism is the rate-determining step. Explain your answer.

Q4(ii)

Short Answer 1 mark

Identify a step in Fig. 4.1 that involves a redox reaction. Explain your answer in terms of oxidation numbers.

Q4(iii)

Short Answer 1 mark

Suggest the role of HIO in this mechanism. Explain your reasoning. [Total: 10]

Q5(a)

Short Answer 1 mark

Methylbenzene can undergo different reactions, as shown in Fig. 5.1. reaction 1 / Pt Br / UV light / FeBr reaction 2 reaction 3

Q5(i)

Short Answer 3 marks

Draw structures in Fig. 5.1 for the possible organic products of the three reactions shown.

Q5(ii)

Short Answer 2 marks

Complete Table 5.1. Table 5.1 type of reaction mechanism reaction 1 reaction 2

Q5(b)

Short Answer 1 mark

When methylbenzene reacts with an electrophile, a substitution reaction occurs. No addition reaction takes place under these conditions. Explain why no addition reaction takes place.

Q5(c)

Short Answer 1 mark

The reaction of methylbenzene with thionyl bromide, SOBr , in the presence of an iron(III) bromide catalyst, FeBr , is shown in Fig. 5.2. FeBr +SOBr +HBr The mechanism of this reaction is similar to that of the bromination of benzene. The first step of the mechanism generates the SOBr electrophile, as shown. SOBr + FeBr SOBr + FeBr

Q5(i)

Short Answer 3 marks

The reaction of methylbenzene with SOBr ions is shown in Fig. 5.3. Complete the mechanism in Fig. 5.3. Include all relevant curly arrows and charges. Draw the structure of the organic intermediate. intermediate +

Q5(ii)

Short Answer 1 mark

The reaction shown in Fig. 5.2 produces a small amount of a by-product, P, with the molecular formula COS. Suggest a structure for by-product P.

Q5(d)

Short Answer 1 mark

Acyl bromides, RCOBr, can be synthesised by the reaction of a carboxylic acid and SOBr This is a similar reaction to the synthesis of acyl chlorides using SOCl

Q5(i)

Short Answer 1 mark

Give an equation for the reaction between ethanoic acid and SOBr

Q5(ii)

Short Answer 1 mark

Suggest the relative ease of hydrolysis of acyl bromides, RCOBr, acyl chlorides, RCOCl, and alkyl chlorides, RCl. Explain your answer. easiest to hydrolyse hardest to hydrolyse [3] [Total: 14]

Q6(a)

Short Answer 1 mark

Perindopril is a drug used to treat heart disease. perindopril

Q6(i)

Short Answer 1 mark

State the number of chiral carbon atoms present in one molecule of perindopril.

Q6(ii)

Short Answer 2 marks

Suggest one benefit and one disadvantage of producing a drug such as perindopril as a single pure optical isomer. benefit disadvantage

Q6(b)

Short Answer 1 mark

(i) Name all the functional groups in perindopril.

Q6(ii)

Short Answer 1 mark

A sample of perindopril is hydrolysed with hot aqueous acid. Draw the structures of the three organic products of the complete acid hydrolysis of perindopril. [3] [Total: 8]

Q7(a)

Short Answer 1 mark

Explain why phenol is brominated much more easily than benzene is brominated. catalyst. Predict whether the organic product will be bromobenzene or iodobenzene. Explain your answer.

Q7(c)

Short Answer 1 mark

Fig. 7.1 shows some reactions of phenol. OH OH OHN NOQ R reaction 1 reaction 2 excess Br

Q7(aq)

Short Answer 1 mark

Na(s)

Q7(i)

Short Answer 1 mark

Give an equation for the reaction of phenol with Na(s).

Q7(ii)

Short Answer 1 mark

Draw the structure of the organic product, R, formed when phenol reacts with an excess of Br

Q7(iii)

Short Answer 1 mark

State the reagents and conditions for reaction 1 and reaction 2 in Fig. 7.1. reaction 1 reaction 2 [2] [Total: 8]

Q8(a)

Short Answer 1 mark

Describe the difference in reactivity between chloroethane and chlorobenzene with OH

Q8(aq)

Short Answer 1 mark

Explain your answer.

Q8(b)

Short Answer 1 mark

Compound T, C Cl, is a useful synthetic intermediate. Fig. 8.1 shows some reactions of T. Z, C step 1 step 2 excess NaOH(aq) heat excess HCl (aq) heat

Q8(i)

Short Answer 1 mark

Give the systematic name for T.

Q8(ii)

Structured 4 marks

Draw the structures of W, X, Y and Z in Fig. 8.1.

Q8(iii)

Short Answer 2 marks

State the reagents and conditions for steps 1 and 2 in Fig. 8.1. step 1 step 2

Q8(c)

Short Answer 1 mark

The proton ( H) NMR spectrum of compound T, C Cl, in CDCl is shown in Fig. 8.2. 5 4 3 chemical shift δ / ppm δ 3.9 δ 3.7 δ 2.8 δ 1.2 2 1 0 Table 8.1 environment of proton example chemical shift range δ / ppm alkane -CH , -CH -, >CH- 0.9-1.7 alkyl next to C=O CH -C=O, -CH -C=O, >CH-C=O2.2-3.0 alkyl next to aromatic ring CH -Ar, -CH -Ar, >CH-Ar 2.3-3.0 alkyl next to electronegative atomCH -O, -CH -O, -CH -Cl 3.2-4.0 attached to alkene =CHR 4.5-6.0 attached to aromatic ring H-Ar 6.0-9.0 aldehyde HCOR 9.3-10.5 alcohol ROH 0.5-6.0 phenol Ar-OH 4.5-7.0 carboxylic acid RCOOH 9.0-13.0 alkyl amine R-NH- 1.0-5.0 aryl amine Ar-NH 3.0-6.0 amide RCONHR 5.0-12.0

Q8(i)

Short Answer 1 mark

Suggest why CDCl is used as a solvent instead of CHCl for the proton ( H) NMR spectrum.

Q8(ii)

Structured 4 marks

Complete Table 8.2 for the proton ( H) NMR spectrum of T. Table 8.2 chemical shift δ / ppm environment of proton splitting pattern number of H atoms responsible for the peak 1.2 2.8 3.7 3.9

Q8(iii)

Short Answer 3 marks

Explain the splitting pattern of the peak at δ 3.9 ppm. [Total: 15] cell. cell of a cell consisting of an Fe / Fe and a Cl half-cell. Draw a labelled diagram of this electrochemical cell. Include all necessary substances. It is not necessary to state conditions used.

Q8(c)

Short Answer 1 mark

The cell reaction for the electrochemical cell in

Q8(c)(b)

Short Answer 1 mark

is shown. + 2Fe 2Fe + 2Cl cell = +0.59 V Calculate ΔG , in kJ mol , for this cell reaction.Δ G = kJ mol

Q8(d)

Short Answer 1 mark

Another experiment is set up using the same electrochemical cell. In this experiment the Fe concentration is 0.15 mol dm . All other concentrations remain at their standard values. The Nernst equation is shown. E = E + (0.059 / z) log [oxidised species] [reduced species]

Q8(i)

Short Answer 1 mark

Use the Nernst equation to calculate the electrode potential, E, for the Fe / Fe in this experiment. : Fe / Fe = +0.77 V] E = V

Q8(ii)

Short Answer 1 mark

Use your answer to (d)(i) to calculate E cell for this electrochemical cell. cell = V [1] [Total: 8] Important values, constants and standards molar gas constant R = 8.31 J Kmol Faraday constant F = 9.65 × 10 C mol Avogadro constant L = 6.022 × 10mol electronic charge e = -1.60 × 10-19 molar volume of gas V = 22.4 dmmol at s.t.p. (101 kPa and 273 K) = 24.0 dmmol at room conditions ionic product of water K = 1.00 × 10-14mol (at 298 K (25 °C)) specific heat capacity of waterc = 4.18 kJ kg (4.18 J g To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic ble of Elements hydroge 1.0 He helium 4.0 34567893Li lithium 6.9 beryllium 9.0 atomic number atomic symbolKey name relative atomic mas sodiu 23.0 magnesium 24.319K potassium 39.1 calcium 40.1 rubidiu 85.5 38Sr strontium 87.6 caesiu 132.9 barium 137.3 87Fr francium Ra radium boron 10.813A aluminiu 27.0 galliu 69.749I indium 1 14.8 81T thallium 204.4 carbon 12.0 14Si silicon 28.1 germanium 72.6tin 18.7 82 lead 207.2 22Ti titanium 47.9 40Zr zirconium 91.2 72Hf hafniu 178.5 104 Rf rutherfordiu23V vanadium 50.9 niobiu 92.9 73Ta tantalum 180.9 105 Db dubnium 24Cr chromium 52.0 molybdenum 95.974W tungsten 183.8 106 Sg seaborgium manganes 54.9 43Tc technetium rheniu 186.2 107 Bh bohriu 26Fe iron 55.8 ruthenium 101.1 osmium 190.2 108 Hs hassium Co cobalt 58.9 rhodiu 102.977I iridiu 192.2 109 Mt meitnerium 28Ni nickel 58.7 palladiu 106.4 78Pt platinum 195.1 darmstadtium copper 63.5 Ag silver 107.9 gold 197.0 111 Rg roentgeniu zinc 65.4 cadmiu 12.4 80 mercury 200.6 coperniciu fleroviu livermorium nitrogen 14.015P phosphorus 31.0 33As arseni 74.9 antimony 121.8 83Bi bismuth 209.0 oxygen 16.016S sulfur 32.1 seleniu 79.0 52Te tellurium 127.6 polonium fluorine 19.017C chlorin 35.5 35Br bromin 79.953I iodine 126.9 85At Ne neon 20.2 18Ar argon 39.9 36Kr krypton 83.8 Xe xenon 131.3 Rn radon nihonium moscoviu tennessine oganesso 21Sc scandium 45.039Y yttrium 88.9 57-71 lanthanoids 89-103 actinoids 57La lanthanum 138.9 89Ac lanthanoids actinoids actinium Ce cerium 140.1 90Th thorium 232.0 59Pr praseodymium 140.9 protactinium 231.0 neodymium 144.492U uraniu 238.0 promethium neptuniu samarium 150.4 plutonium europium 152.0 americiu gadoliniu 157.3 Cm curium 65Tb terbium 158.9 97Bk berkeliu dysprosiu 162.5 98Cf californium holmiu 164.9 99Es einsteinium 68Er erbium 167.3 100 Fm fermium Tm thuliu 168.9 101 Md mendeleviu ytterbium 173.1 102 No nobeliu 71Lu lutetium 175.0 103 Lr lawrencium

Board	CAIE
Level	A Level
Subject	Chemistry
Code	`9701`
Year	2023
Session	May/June
Q&A	79 questions

Chemistry 2023 · May/June

Questions & Model Answers 79 Qs

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