Chemistry 2022 Oct/Nov Paper 42 | Past Paper

Q2(a)

Short Answer 1 mark

Nitrogen monoxide, NO, reacts with ozone, O NO(g) + O

Q2(g)

Short Answer 1 mark

This reaction is first order with respect to both NO and O

Q2(i)

Short Answer 1 mark

Complete the rate equation for this reaction. rate =

Q2(ii)

Short Answer 2 marks

A reaction is carried out in which the initial concentrations of NO and O are both 1.20 × 10 Calculate the initial rate of the reaction. State its units. rate of reaction = units =

Q2(iii)

Short Answer 1 mark

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

Q2(b)

Short Answer 1 mark

Nitrous oxide, N O, decomposes into its elements.

Q2(g)

Short Answer 1 mark

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.

Q2(i)

Short Answer 1 mark

Sketch a graph, on the axes below, of reaction rate against time for the catalysed decomposition of N O under these conditions. reaction rate time

Q2(ii)

Short Answer 1 mark

Sketch a graph, on the axes below, of [N O] against time for this reaction. time

Q2(iii)

Short Answer 1 mark

Platinum behaves as a heterogeneous catalyst in this reaction. Describe the mode of action of a heterogeneous catalyst.

Q2(iv)

Short Answer 1 mark

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. [Total: 10] Question 3 starts on the next page.

Q3

Short Answer 1 mark

Data should be selected from Table 3.1 in order to answer some parts of this question. Table 3.1 + 2e Mg -2.38 + 2e Mn -1.18 + e +1.49MnO + 4H + 2e + 2H O +1.23MnO + e MnO +0.56MnO + 4H + 3e MnO + 2H O +1.67MnO + 8H + 5e + 4H O +1.52

Q3(a)

Short Answer 1 mark

An electrochemical cell can be constructed from a Mg potentials of the two half-cells.

Q3(i)

Short Answer 1 mark

Define standard electrode potential. Include details of the standard conditions used.

Q3(ii)

Short Answer 3 marks

Complete the diagram below to show an electrochemical cell constructed from a Mg half-cell. Label your diagram.

Q3(iii)

Short Answer 1 mark

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.

Q3(iv)

Short Answer 1 mark

Calculate the standard cell potential, , of this cell. = V

Q3(v)

Short Answer 1 mark

Construct an equation for the cell reaction.

Q3(vi)

Short Answer 1 mark

Predict how the cell reaction will change, if at all, when the solution in the Mg is diluted by the addition of a large volume of water. Explain your answer.

Q3(b)

Short Answer 1 mark

A molten magnesium salt is electrolysed for 15.0 minutes by a constant current. 4.75 × 10 magnesium atoms are produced at the cathode. Calculate the value of the current used. current = A [2] [Total: 11]

Q4(a)

Short Answer 1 mark

The value of the solubility product, K , of iron(III) hydroxide, Fe(OH) , is given by the following expression. = [Fe ][OH = 2.0 × 10-39-12

Q4(i)

Short Answer 1 mark

Calculate the solubility of Fe(OH) in water.

Q4(ii)

Short Answer 1 mark

Calculate the solubility of Fe(OH) barium hydroxide, Ba(OH)

Q4(iii)

Short Answer 1 mark

Fe(OH) is less soluble in Ba(OH)

Q4(aq)

Short Answer 1 mark

than it is in pure water. Name this effect.

Q4(b)

Short Answer 1 mark

The numerical value of the K of HBrO is 2.00 × 10 X is a solution of HBrO which contains 4.00 × 10 of solution. In this solution the following equilibrium is established in which there are two conjugate acid-base pairs. HBrO + H BrO + H

Q4(i)

Short Answer 1 mark

Define conjugate acid-base pair.

Q4(ii)

Short Answer 1 mark

Identify the two conjugate acid-base pairs shown in the equation above. pair one acid base pair two acid base

Q4(iii)

Short Answer 2 marks

Calculate the pH of solution X. Show all your working. pH =

Q4(iv)

Short Answer 1 mark

A solution containing 2.00 × 10 formed. Calculate the pH of this buffer solution. pH = [1] [Total: 9]

Q5

Short Answer 1 mark

Copper is a transition element. It forms a wide variety of compounds.

Q5(a)

Short Answer 1 mark

Define transition element.

Q5(b)

Short Answer 1 mark

An aqueous solution of copper( II) sulfate, CuSO , contains [Cu(H complex ions. If an excess of concentrated hydrochloric acid is added to this solution a ligand exchange reaction complex ions are formed.

Q5(i)

Short Answer 3 marks

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(H

Q5(ii)

Short Answer 1 mark

In an isolated Cu ion the d-orbitals are all degenerate. In a complex ion such as [Cu(H the d-orbitals are non-degenerate. Define degenerate and non-degenerate in this context. degenerate non-degenerate

Q5(iii)

Short Answer 1 mark

Explain why the solutions of the two complex ions in Table 5.1 are different colours.

Q5(c)

Short Answer 1 mark

Cu forms a complex ion containing water molecules and ethanedioate ions, C , as ligands. The formula of the complex is [Cu(C . The ethanedioate ion is a bidentate ligand.

Q5(i)

Short Answer 1 mark

Explain what is meant by bidentate.

Q5(ii)

Short Answer 2 marks

There are three stereoisomers with the formula [Cu(C Complete the three-dimensional diagrams to show these three stereoisomers. stereoisomer 1 stereoisomer 2 stereoisomer 3

Q5(iii)

Short Answer 3 marks

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 stereoisomersandand [2] [Cu(C of solution. A acidified potassium manganate( VII). The equation for the redox reaction is shown. + 2MnO + 16H + 2Mn + 8H acidified potassium manganate( VII) needed to oxidise all of the ethanedioate ions, C sample. Show all your working. : Na [Cu(C ], 321.5] minimum volume = cm

Q5(e)

Short Answer 1 mark

Copper( II) nitrate, Cu(NO , and barium nitrate, Ba(NO , both decompose when heated. Copper( II) nitrate decomposes at a lower temperature than barium nitrate. Suggest a reason for this difference. Explain your answer. [Total: 16]

Q6

Short Answer 1 mark

An aqueous solution of cobalt(II) chloride is a pink colour due to the presence of [Co(H complex ions.

Q6(a)

Short Answer 2 marks

(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 reagent colour of cobalt-containing product state of cobalt-containing product NaOH(aq) an excess of conc. HCl

Q6(ii)

Short Answer 1 mark

Write an equation for the reaction between [Co(H ions and NaOH(aq).

Q6(iii)

Short Answer 1 mark

Write an equation for the reaction between [Co(H ions and an excess of conc. HCl.

Q6(b)

Short Answer 1 mark

(i) Define stability constant.

Q6(ii)

Short Answer 1 mark

Write an expression for the stability constant, K stab , of the [Co(NH complex ion. stab

Q6(iii)

Short Answer 1 mark

Give the units of the stability constant, K stab , of the [Co(NH complex ion. units =

Q6(iv)

Short Answer 1 mark

The numerical value of the stability constant, K stab , of the [Co(NH complex ion is 7.7 × 10 In an aqueous solution the concentration of the [Co(NH and the concentration of NH at equilibrium. Calculate the concentration of [Co(H in this solution. [1] [Total: 8]

Q7(a)

Short Answer 1 mark

The structural and displayed formulae of three aromatic compounds, A, B and C, are shown in Fig. 7.1. COOH HO O CHCl CO HO O Compare the relative acidities of A, B and C. most acidic least acidic Explain your answer.

Q7(b)

Short Answer 1 mark

Methylbenzene, C , can be made from benzene by an electrophilic substitution reaction.

Q7(i)

Short Answer 1 mark

Identify a compound that reacts with benzene to form methylbenzene. Identify the catalyst used. compound catalyst

Q7(ii)

Short Answer 1 mark

The first step in the reaction is the generation of the CH electrophile. Write an equation for the reaction that generates this electrophile.

Q7(iii)

Short Answer 3 marks

Describe the mechanism for the reaction between benzene and the CH electrophile. Include all relevant curly arrows and charges.

Q7(iv)

Short Answer 2 marks

Identify a suitable reagent to oxidise methylbenzene to benzoic acid, C COOH. Write an equation for this reaction using [O] to represent one atom of oxygen from the oxidising agent. reagent equation

Q7(v)

Short Answer 1 mark

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 CH and HOOCC respectively. Draw the structures of these two products.

Q7(c)

Short Answer 1 mark

A reaction scheme is shown in Fig. 7.2. reaction 1 reaction 2 OH CH dye molecule Y reaction 3 reaction 4

Q7(i)

Short Answer 1 mark

Describe the reagents and conditions to produce CH from CH reaction 2. reagents conditions

Q7(ii)

Short Answer 1 mark

Describe how CH OH can be produced from CH in reaction 3.

Q7(iii)

Short Answer 1 mark

Draw the structure of the dye molecule Y formed when CH and CH react together in reaction 4. Describe the conditions for this reaction. structure conditions [2] [Total: 15]

Q8

Short Answer 1 mark

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 ( H) NMR spectrum. The structures of five isomeric ketones, P, Q, R, S and T are given. P C COCH(CHS C COCH Q C COCHT C CH(CH )COCH R C COCH

Q8(a)

Short Answer 1 mark

Identify all the chiral carbon atoms on the structures above. Label each chiral carbon atom with an asterisk (*).

Q8(b)

Short Answer 1 mark

The proton ( H) NMR spectrum of one of the five isomers, P, Q, R, S or T, is shown in Fig. 8.1. 1110 9 8 7 6 5 / ppm 4 3 2 1 0

Q8(i)

Short Answer 2 marks

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.

Q8(ii)

Short Answer 1 mark

Name the splitting pattern of the peak at δ = 3.7. Explain why it has this splitting pattern.

Q8(c)

Short Answer 1 mark

Choose from the letters P, Q, R, S and T to identify:

Q8(i)

Short Answer 1 mark

the two compounds that each have a doublet peak in the proton ( H) NMR spectrum

Q8(ii)

Short Answer 1 mark

the compound with only three peaks in its proton ( H) NMR spectrum.

Q8(d)

Short Answer 1 mark

Suggest a suitable solvent that should be used for obtaining the spectrum shown in Fig. 8.1.

Q8(e)

Short Answer 1 mark

The proton ( H) NMR spectrum of compound T is compared in the presence of D O and in the absence of D Describe any difference between the two spectra. Explain your answer.

Q8(f)

Short Answer 1 mark

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]

Q9

Short Answer 1 mark

Tyrosine and lysine, shown in Fig. 9.1, are naturally occurring amino acids found in proteins. tyrosine lysine

Q9(a)

Short Answer 3 marks

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] 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

Q9(c)

Short Answer 1 mark

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 -CH CHNH COOH substituent. Table 9.1 reagent structure an excess of Br

Q9(aq)

Short Answer 1 mark

an excess of NaOH(aq) an excess of HNO

Q9(d)

Short Answer 1 mark

A mixture of tyrosine and lysine can be separated by thin-layer chromatography. Under certain conditions the R value of lysine is 0.14 and the R value of tyrosine is 0.45.

Q9(i)

Short Answer 1 mark

Explain what is meant by R value.

Q9(ii)

Short Answer 1 mark

Suggest an explanation for the difference in R values. [Total: 10] Important values, constants and standards Faraday constant F = 9.65 u 10 Avogadro constant L = 6.022 u 10 electronic charge e = ?1.60 u 10?19 molar volume of gas V at room conditions ionic product of water K = 1.00 u 10?14 Group The Periodic Table of Elements hydrogen 1.0 He helium 4.0 12131415161718 34567891011123Li 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 strontiu m 87.6 caesium 132. 9 barium 137.3 francium Ra radium boron10.8 13 Al aluminium 27.0 31 Ga gallium 69.7 49 In indium 114.8 thallium 204.4 carbon 12.0 14 Si silicon 28.1 32 Ge germanium 72.6 50 Sntin 118.7 82 Pb lead 207.2 titanium 47.9 40 Zr zirconium 91.2 72 Hf hafnium178. 5 rutherfordium vanadium 50.9 41 Nb niobium 92.9 73 Ta tantalum 180.9 105 Db dubnium chromium 52.0 42 Mo molybdenum 95.9 74 W tungste n 183.8 106 Sg seaborgium manganese 54.9 43 Tc technetium rhenium 186. 2 107 Bh bohrium iron 55.8 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium Co cobalt 58.9 45 Rh rhodium 102.9 iridium192.2 109 Mt meitnerium nicke l58.7 46 Pd palladium 106. 4 platinum 195. 1 110 Ds darmstadtium copper 63.5 47 Ag silver 107. 9 79 Au gold 197. 0 111 Rg roentgenium zinc 65.4 48 Cd cadmium 112.4 80 Hg mercury 200. 6 112 Cn copernicium flerovium livermorium nitrogen 14.0 phosphorus 31.0 33 As arsenic 74.9 51 Sb antimony 121.8 bismuth 209.0 oxygen 16.0 sulfur 32.1 34 Se selenium 79.0 52 Te tellurium 127.6 84 Po polonium fluorine 19.0 17 Cl chlorine 35.5 35 Br bromine 79.9 iodine 126.9 Ne neon 20.2 18 Ar argo n39.9 krypto n 83.8 54 Xe xenon131. 3 86 Rn radon nihonium moscovium tennessine oganesson scandium 45.0 yttriu m 88.9 57?71 lanthanoids 89?103 actinoids 57La lanthanum 138.9 89 Ac lanthanoids actinoids actinium Ce cerium 140.1 90 Th thoriu m232.0 praseodymium 140.9 91 Pa protactinium 231.0 neodymium 144.4 uranium 238.0 promethium neptunium samarium 150.4 94 Pu plutonium europium 152.0 95 Am americium gadolinium 157.3 96 Cm curium terbiu m158.9 97 Bk berkelium dysprosium 162. 5 californium holmium 164. 9 99 Es einsteinium erbium 167. 3 100 Fm fermium Tm thulium168. 9 101 Md mendelevium ytterbiu m 173. 1 102 No nobelium lutetium 175.0 lawrencium 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.

Board	CAIE
Level	A Level
Subject	Chemistry
Code	`9701`
Year	2022
Session	Oct/Nov
Paper No.	Paper 42
Q&A	78 questions

Chemistry 2022 · Oct/Nov · Paper 42

Questions & Model Answers 78 Qs

Paper Details