Chemistry 2024 May/June | Past Paper

Q1(a)

Short Answer 1 mark

Describe the trend in the solubility of the sulfates of magnesium, calcium and strontium. Explain your answer . most soluble least soluble [4] latt.

Q1(c)

Short Answer 1 mark

State and explain the main factors that affect the magnitude of lattice energies.

Q1(d)

Short Answer 1 mark

Table 1.1 shows some energy changes. Table 1.1 energy change value / kJ mol standard enthalpy change of atomisation of potassium +89 first ionisation energy of potassium +419 second ionisation energy of potassium +3070 standard enthalpy change of atomisation of sulfur +279 S±S bond energy +265 first ionisation energy of sulfur +1000 second ionisation energy of sulfur +2260 first electron affinity of sulfur ±200 second electron affinity of sulfur +640 standard enthalpy change of formation of potassium sulfide, K S

Q1(d)(s)

Short Answer 1 mark

±381 * 0019655314902 *

Q1(i)

Short Answer 1 mark

Born±Haber cycles can be used to determine the lattice energies of ionic compounds. Complete the Born±Haber cycle in Fig. 1.1 for potassium sulfide, K S

Q1(i)(s)

Short Answer 1 mark

. Include state symbols for all of the species.

Q1(g)

Short Answer 1 mark

+ S

Q1(g)(g)

Short Answer 1 mark

+ 2e 2K

Q1(g)(s)

Short Answer 1 mark

[3] latt, of K S

Q1(g)(s)

Short Answer 1 mark

using relevant data from Table 1.1. Show your working. latt of K S

Q1(g)(s)

Short Answer 1 mark

= kJ mol [2] [Total: 13] * 0019655314903 *

Q2(a)

Short Answer 1 mark

(i) Lithium nitrate, LiNO , decomposes on heating in a similar way to Group 2 nitrates to give the metal oxide, a brown gas and oxygen. Write an equation for the decomposition of LiNO

Q2(ii)

Short Answer 1 mark

The other Group 1 nitrates, MNO , decompose on heating to form the metal nitrite,MNO , and oxygen. The thermal stability of these nitrates increases down the group. Suggest why the thermal stability of MNO increases down the group.

Q2(b)

Short Answer 1 mark

Acidified manganate(VII) ions, MnO , can be used to analyse solutions containing nitrite ions, NO , by titration. X is a solution of NaNO 250.0 cm of X is added to 50.0 cm of 0.125 mol dm acidified MnO

Q2(aq)

Short Answer 1 mark

The MnO

Q2(aq)

Short Answer 1 mark

ions are in excess; all the NO ions are oxidised in the reaction. The unreacted MnO

Q2(aq)

Short Answer 1 mark

required 22.50 cm of 0.0400 mol dm

Q2(aq)

Short Answer 1 mark

to reach the + H + 2H + 2eMnO + 8H + 5e + 4H + e Calculate the concentration, in mol dm , of NaNO in X. concentration of NaNO in X = m ol dm [3] * 0019655314904 *

Q2(c)

Short Answer 1 mark

Table 2.1 shows electrode potentials for some electrode reactions involving m anganese compounds. Table 2.1 / V + 2e Mn ±1.18MnO + 4H + 2e + 2H O +1.23MnO + e MnO +0.56MnO + 4H + 3e MnO + 2H O +1.67MnO + 8H + 5e + 4H O +1.52MnO + 2H O + 3e MnO + 4OH +0.59MnO + 2H O + 2e MnO + 4OH +0.60MnO + 4H + 2e MnO + 2H O +1.70

Q2(i)

Short Answer 1 mark

Aqueous manganate(VI) ions, MnO , are unstable in acidic conditions and undergo a disproportionation reaction. cellfor this reaction is +1.14 V. Construct an overall ionic equation for this disproportionation reaction.

Q2(ii)

Short Answer 1 mark

Suggest and explain how the E cell value of the disproportionation reaction changes with an increase in pH. [Total: 9] * 0019655314905 *

Q3(a)

Short Answer 1 mark

Carbon disulfide, CS , is flammable and reacts readily with oxygen, as shown in reaction 1. reaction 1 CS

Q3(g)

Short Answer 1 mark

+ 2SO

Q3(g)

Short Answer 1 mark

, for some substances. Table 3.1

Q3(g)

Short Answer 3 marks

f / kJ mol 116.7 0.0 ±393.5 ±296.8 / J Kmol 237.8 205.2 213.8 248.2 , in kJ mol , for reaction 1 at 25 °C. = kJ mol

Q3(b)

Short Answer 1 mark

Carbon disulfide reacts with chlorine to form tetrachloromethane, as shown in reaction 2. reaction 2 CS + 3Cl CCl+ S = ±261.6 kJ mol = ±365.5 J Kmol Calculate the maximum temperature, in K, for reaction 2 to be feasible. temperature = K [2] [Total: 5] * 0019655314906 *

Q4(a)

Short Answer 1 mark

(i) Explain why transition elements have variable oxidation states.

Q4(ii)

Short Answer 1 mark

Sketch the shape of a 3d 2 orbital in Fig. 4.1.

Q4(b)

Short Answer 1 mark

Samples of [Cu(H

Q4(aq)

Short Answer 1 mark

are reacted separately with an excess of solution A and with an excess of solution B. The reaction of [Cu(H

Q4(aq)

Short Answer 1 mark

with solution A is a precipitation reaction. The reaction of [Cu(H

Q4(aq)

Short Answer 1 mark

with solution B is a ligand substitution reaction. Suggest a possible identity for solution A and for solution B. Give relevant observations and solution A observations solution B observations [3] * 0019655314907 *

Q4(c)

Short Answer 1 mark

Solutions containing the [Ag(NH complex are colourless. Explain why this complex is colourless.

Q4(d)

Short Answer 1 mark

Two bidentate ligands are shown in Fig. 4.2. dpys N NNH Explain what is meant by a bidentate ligand., ,

Q4(e)

Short Answer 3 marks

Ruthenium(III) ions, Ru , form an octahedral complex, [Ru(dpys) , with the ligands dpys and chloride ions. This complex shows the same kind of stereoisomerism as [Ru(NH but also shows a different type of stereoisomerism. stereoisomers of [Ru(dpys) The dpys ligand can be represented using N N. isomer 1 isomer 2 isomer 3

Q4(ii)

Short Answer 1 mark

State the different types of stereoisomerism shown by [Ru(dpys)

Q4(iii)

Short Answer 1 mark

Deduce which stereoisomers in (e)(i) are non-polar. Explain your answer . [Total: 14] * 0019655314909 *

Q5(a)

Short Answer 1 mark

Nitrosyl chloride, NOCl, can be formed by the reaction between nitrogen m onoxide and chlorine, as shown. 2NO + Cl 2NOCl The initial rate of this reaction is investigated, starting with dif ferent concentrations of NO and . The results obtained are shown in Table 5.1. Table 5.1 experiment[NO] / mol dm [Cl ] / mol dm initial rate / mol dmmin 1 0.0250 0.0150 3.68 × 10 2 0.0750 0.0150 3.32 × 10 3 0.0500 0.0600 5.89 × 10

Q5(i)

Short Answer 1 mark

Use the data in Table 5.1 to deduce the rate equation for this reaction. Explain your reasoning.

Q5(ii)

Short Answer 1 mark

Use your rate equation from (a)(i) and the data from experiment 1 to calculate the rate constant, k, for this reaction. Include the units of k. k = units [2] * 0019655314910 *

Q5(b)

Short Answer 2 marks

NO Cl is another compound containing nitrogen, oxygen and chlorine. In sunlight, NO Cl can undergo homolytic fission to release chlorine radicals which can catalyse the conversion of ozone, O , into oxygen. Complete the mechanism for this process. initiation (homolytic fission) Cl + propagation step 1 + propagation step 2 + +

Q5(c)

Short Answer 1 mark

Ozone reacts with nitrogen dioxide, as shown. + 2NO + O The rate of reaction is first order with respect to O and first order with respect to NO step 1 step 2 [2] [Total: 9] * 0019655314911 *

Q6(a)

Short Answer 1 mark

Aqueous solutions of methanoic acid, HCOOH, and propanoic acid, CH COOH, are mixed together. An equilibrium is set up between two conjugate acid±base pairs.

Q6(i)

Short Answer 1 mark

Define conjugate acid±base pair.

Q6(ii)

Short Answer 1 mark

The pK of HCOOH is 3.75 and of CH COOH is 4.87. Complete the equation for the Brùnsted±Lowry equilibrium between the stronger of these two acids and water.

Q6(b)

Short Answer 1 mark

(i) Write an expression for the acid dissociation constant, K , for butanoic acid, COOH.

Q6(ii)

Short Answer 2 marks

The pK of CH COOH is 4.82. A solution of CH COOH(aq) has a pH of 3.25. Calculate the concentration, in mol dm , of CH COOH in this solution. concentration of CH COOH = m ol dm

Q6(c)

Short Answer 1 mark

(i) Define buffer solution.

Q6(ii)

MCQ 3 marks

A buffer solution containing a mixture of CH COOH and CH COONa is prepared as follows. A solution of 600 cm of CH COOH is mixed with 400 cm of 0.125 mol dm COONa. The buffer solution has pH 5.70. The K of CH COOH is 1.78 × 10 mol dm Calculate the initial concentration, in mol dm , of CH COOH used. concentration of CH COOH = m ol dm

A

buffer solution containing a mixture of CH

A

solution of 600 cm

Q6(d)

Short Answer 1 mark

A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Methanoic acid, HCOOH, is being investigated as a fuel in fuel cells. When the cell operates, HCOOH is oxidised to carbon diox ide. + 4H + 4e In this fuel cell, the overall cell reaction is the same as that for the complete combustion of HCOOH.

Q6(ii)

Short Answer 1 mark

Calculate the volume, in cm , of oxygen used when a current of 3.75 A is delivered by the cell for 40.0 minutes. Assume the cell operates at room conditions. volume of oxygen = cm [2] [Total: 13] * 0019655315013 *

Q7

Short Answer 1 mark

Methyl red can be synthesised as shown in Fig. 7.1. COOH step 1 step 2 step 3 methyl redR S COOH N N N Fig. 7.1 * 0019655315014 *

Q7(a)

Short Answer 1 mark

(i) Give the systematic name of P.

Q7(ii)

Short Answer 1 mark

P can be synthesised as shown in Fig. 7.2. COOH Suggest reagents and conditions for this reaction.

Q7(iii)

Short Answer 1 mark

A student attempts to synthesise P by an alternative route, as shown in Fig. 7.3. Compound T is the major product in this reaction rather than P. COOH conc. HNO conc. H COOH Explain why T is the major product in this reaction.

Q7(b)

Short Answer 1 mark

S reacts in a similar way to phenol in step 3.

Q7(i)

Short Answer 3 marks

Draw the structures of Q, R and S in the boxes in Fig. 7.1.

Q7(ii)

Short Answer 1 mark

Suggest reagents and conditions for steps 1 and 2 in Fig. 7.1. step 1 step 2 [3] [Total: 9] * 0019655315015 *

Q8(a)

Short Answer 3 marks

State the relative basicities of phenylamine, C , benzylamine, C , and ammonia, NH , in aqueous solution. Explain your answer. most basic least basic

Q8(b)

Short Answer 1 mark

An excess of Br

Q8(aq)

Short Answer 1 mark

is added to separate samples of C and benzene, C

Q8(i)

Short Answer 1 mark

C reacts readily with Br

Q8(aq)

Short Answer 2 marks

to form organic product M. State the expected observations for this reaction. Draw the structure of M. observations structure of M

Q8(ii)

Short Answer 1 mark

C does not react with Br

Q8(aq)

Short Answer 1 mark

Suggest why Br

Q8(aq)

Short Answer 1 mark

reacts with C but not with C

Q8(c)

Short Answer 1 mark

Explain why benzamide, C CONH , is a much weaker base than ammonia, NH

Q8(d)

Short Answer 1 mark

C CONH is formed by reacting benzoyl chloride, C COCl, with NH Complete the mechanism in Fig. 8.1 for the reaction of C COCl with NH Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the organic intermediate. organic intermediate HCl+ [4] * 0019655315017 *

Q8(e)

Short Answer 1 mark

Phenylalanine, C CH(NH )COOH, is an amino acid with an isoelectric point of 5.5.

Q8(i)

Short Answer 1 mark

State what is meant by isoelectric point.

Q8(ii)

Short Answer 1 mark

Draw the structure of C CH(NH )COOH at pH 10.

Q8(f)

Short Answer 1 mark

C CH(NH )COOH and alanine, CH CH(NH )COOH, react to form a dipeptide containing both amino acid residues. Draw the structure of this dipeptide. The peptide functional group formed should be displayed. [2] [Total: 16] * 0019655315018 *, ,

Q9(a)

Short Answer 1 mark

Explain why trichloroethanoic acid, CCl COOH, is more acidic than ethanoic acid, CH COOH.

Q9(b)

Short Answer 1 mark

Acyl chlorides are formed by reacting carboxylic acids with thionyl chloride, SOC l

Q9(i)

Short Answer 1 mark

Ethanedioyl chloride, (COCl ) , can be prepared by reacting ethanedioic acid, (COOH) with an excess of SOCl Write an equation for this reaction.

Q9(ii)

Short Answer 1 mark

Samples of (COCl ) are reacted separately with an excess of warm acidified KMnO

Q9(aq)

Short Answer 1 mark

and with HNCH NCH has the molecular formula C in each reaction. with warm acidified KMnO

Q9(aq)

Short Answer 2 marks

withNCH

Q9(iii)

Short Answer 1 mark

A polyester can be synthesised from the reaction of (COCl ) HOCHOH. Draw two repeat units of the polymer formed. Any functional groups should be displayed. [2] * 0019655315019 *

Q9(c)

Short Answer 1 mark

Compound H, C , reacts with alkaline I

Q9(aq)

Short Answer 1 mark

to form yellow precipitate J but does not react with Na

Q9(aq)

Short Answer 1 mark

The proton ( H) NMR spectrum of H in CDCl is shown in Fig. 9.2. 4.0 3.0 2.0 chemical shift δ / ppm 1.0 Table 9.1 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=O 2.2±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 * 0019655315020 *

Q9(i)

Short Answer 1 mark

Identify yellow precipitate J.

Q9(ii)

Structured 4 marks

Complete Table 9.2 for the proton ( H) NMR spectrum of H, C Table 9.2 chemical shift δ / ppm splitting pattern number of H atoms responsible for the peak number of protons on adjacent carbon atoms 1.15 2.25 3.60 3.95

Q9(iii)

Short Answer 1 mark

Suggest a structure for H, C [1] [Total: 12] * 0019655315021 * BLANK PAGE * 0019655315022 * 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)) speci®c heat capacity of waterc = 4.18 kJ kg (4.18 J g ) * 0019655315023 * 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 Ta ble of Elements hydrogen 1.0 helium 4.0 12 13 14 15 16 17 18 3456789 10 11 12 lithium 6.9 beryllium 9.0 atomic number atomic symbolKey name relative atomic mass sodium 23.0 magnesium 24.3 potassium 39.1 calcium 40.1 rubidium 85.5 strontium 87.6 caesium 132.9 barium 137.3 francium radium boron 10.8 aluminium 27.0 gallium 69.7 indium 114.8 thallium 204.4 carbon 12.0 silicon 28.1 germanium 72.6tin 118.7 lead 207.2 titanium 47.9 zirconium 91.2 hafnium 178.5 rutherfordium vanadium 50.9 niobium 92.9 tantalum 180.9 dubnium chromium 52.0 molybdenum 95.9 tungsten 183.8 seaborgium manganese 54.9 technetium rhenium 186.2 bohrium iron 55.8 ruthenium 101.1 osmium 190.2 hassium cobalt 58.9 rhodium 102.9 iridium 192.2 meitnerium nickel 58.7 palladium 106.4 platinum 195.1 darmstadtium copper 63.5 silver 107.9 gold 197.0 roentgenium zinc 65.4 cadmium 112.4 mercury 200.6 copernicium flerovium livermorium nitrogen 14.0 phosphorus 31.0 arsenic 74.9 antimony 121.8 bismuth 209.0 oxygen 16.0 sulfur 32.1 selenium 79.0 tellurium 127.6 polonium fluorine 19.0 chlorine 35.5 bromine 79.9 iodine 126.9 neon 20.2 argon 39.9 krypton 83.8 xenon 131.3 radon nihonium moscovium tennessine oganesson scandium 45.0 yttrium 88.9 57±71 lanthanoids 89±103 actinoids lanthanum 138.9 lanthanoids actinoids actinium cerium 140.1 thorium 232.0 praseodymium 140.9 protactinium 231.0 neodymium 144.2 uranium 238.0 promethium neptunium samarium 150.4 plutonium europium 152.0 americium gadolinium 157.3 curium terbium 158.9 berkelium dysprosium 162.5 californium holmium 164.9 einsteinium erbium 167.3 fermium thulium 168.9 mendelevium ytterbium 173.1 nobelium lutetium 175.0 lawrencium ± * 0019655315024 *

Board	CAIE
Level	A Level
Subject	Chemistry
Code	`9701`
Year	2024
Session	May/June
Q&A	85 questions

Chemistry 2024 · May/June

Questions & Model Answers 85 Qs

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