Chemistry 9701/22 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Level Structured Questions, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Complete Table 1.1 using relevant information from the Periodic Table.
Table 1.1
nucleon number proton number number of electrons
Mg2+ 24
Al 3+ 27
[2]
(b) State and explain the difference in the ionic radius of Al 3+ compared to Mg2+.
…… [3]
(c) Draw a labelled diagram to show the structure and bonding in sodium.
[1]
(d) Fig. 1.1 shows the variation in melting point of some Period 3 elements in their standard
states at room temperature and pressure.
2500
2000 melting point / K 1500
1000
500
0
Na Mg Al Si P S Cl
element
Fig. 1.1
(i) Explain why Si has a high melting point.
…… [1]
(ii) Complete Fig. 1.1 to show the variation in the melting points of the elements P, S and Cl.
[2]
(e) Two Period 3 elements react with an excess of oxygen at room pressure.
(i) Complete Table 1.2.
Table 1.2
1 2 3 state of oxide at approximate pH of
Period 3 element room temperature and solution made when pressure oxide is added to water
Na
S
[2]
(ii) The solutions made in column 3 of Table 1.2 are mixed together.
Name the type of reaction that occurs.
…… [1]
(iii) Write an equation to describe the reaction between P4O10 and an excess of water.
…… [1]
(f) Aluminium hydroxide is amphoteric.
(i) Explain what is meant by amphoteric.
…… [1]
(ii) Write an equation to describe the reaction that occurs when aluminium hydroxide,
Al (OH)3, reacts with NaOH(aq).
…… [1]
[Total: 15]
Mark scheme
1(a) 2
Mg2+ 24 12 10
Al3+ 27 13 10
4 correct = 2 marks
2 or 3 correct = 1 mark
0 or 1 correct = 0 marks
1(b) M1 it / Al 3+ is smaller (compared to Mg2+) 3
M2 greater nuclear attraction for remaining electrons
M3 same shielding effect AND greater nuclear charge
1(c) positively charged ions 1
delocalised electrons
labelled diagram shows:
regular arrangement of minimum of 2 × 2 circles containing + for positive ions
AND surrounded by (sea of) delocalised electrons
1(d)(i) large amount of energy required to break bonds in giant covalent structure 1
1(d)(ii) 2
M1 all 3 melting points lower than Mg
M2 show melting point Cl < P < S
1(e)(i) 2
Na solid 10–14
S gas 0–4
4 correct = 2 marks
2 or 3 correct = 1 mark
0 or 1 correct = 0 marks
1(e)(ii) acid–base OR neutralisation 1
1(e)(iii) P4O10 + 6H2O → 4H3PO4 1
1(f)(i) reacts with both acids and bases OR shows both acidic and basic behaviour 1
1(f)(ii) Al (OH)3 + NaOH → NaAl (OH)4 1
Separate samples of Na2CO3 and NaHCO3 react with HCl (aq) to produce the same products, as shown in Table 2.1.
Table 2.1
reaction equation ∆H / kJ mol–1
1 Na2CO3 + 2HCl 2NaCl + H2O + CO2 ∆H1
2 NaHCO3 + HCl NaCl + H2O + CO2 ∆H2 = +27.2
(a) Complete the reaction pathway diagram in Fig. 2.1 for reaction 2.
Label the diagram to show the enthalpy change, ∆H2, and the activation energy, EA.
energy
/ kJ mol–1
NaHCO3 + HCl
progress of reaction
Fig. 2.1
[2]
(b) The value for ∆H1 is determined by experiment using the following method.
• 50.0 cm3 of 2.00 mol dm–3 HCl (aq) is added to a polystyrene cup.
• The initial temperature of the acid is recorded as 19.6 °C.
• 0.0400 mol of Na2CO3 is added and the mixture is stirred.
• All the solid Na2CO3 disappears and a colourless solution is produced.
The maximum temperature recorded during the reaction is 26.2 °C.
(i) Describe one other observation that shows the reaction is complete.
…… [1]
(ii) Calculate the value of ∆H1 in kJ mol–1.
Assume the specific heat capacity of the reaction mixture is the same as for water and no heat is lost to the surroundings.
Show your working.
∆H1 = …… kJ mol–1 [3]
(iii) Thermal decomposition occurs when NaHCO3 is heated.
reaction 3 2NaHCO3 Na2CO3 + H2O + CO2
Calculate the enthalpy change for reaction 3, ∆Hr , using the data in Table 2.1 and the value of ∆H1 calculated in (b)(ii).
(If you were unable to calculate a value for ∆H1 in (b)(ii), assume the enthalpy change is
–38.4 kJ mol–1. This is not the correct value.)
∆Hr = …… kJ mol–1 [2]
(c) Z is a salt that contains a Period 4 element from Group 2. When Z is heated brown gas forms.
Identify the formula of Z and use it to write an equation for the reaction.
…… [2]
[Total: 10]
Mark scheme
2(a) 2
NaCl + H2O + CO2
All three correct (2 marks) or two correct (1 mark):
profile for endothermic reaction (include ‘hump’)
arrow representing EA
arrow representing ∆H OR (+) 27.2
2(b)(i) no more fizzing / effervescence stops 1
2(b)(ii) M1 calculate Q from = 50 × 4.18 × (26.2 – 19.6)) = 1379.(4) 3
M2 Use M1 to find energy change when 1 mol Na2CO3 / 2 mol HCl reacts
M1 ÷ 0.04 = 34485
M3 express Q as ∆H(1) in kJ mol–1 = – (value for M2) 1000
= –34.5 kJ mol–1
Alternative method
M1 evidence of use Q = mc∆T = 50 × 4.18 / 1000 × (26.2 – 19.6) = 1.3794
M2 Use M1 to find energy change when 1 mol Na2CO3 / 2 mol HCl reacts
M1 ÷ 0.04 = 34.485
M3 express Q as ∆H(1) in kJ mol –1 = – (value for M2) (expressed in kJ)
= –34.5 kJ mol–1
2(b)(iii) Using answer from 2(b)(ii) 2
(=) 2∆H2 – (∆H1)
(=) 2 × 27.2 – (2b(ii))
2 × 27.2 – (–34.50) = (+) 88.9 kJ mol–1
All three correct (2 marks) or two correct (1 mark):
(27.2) – (–34.5) OR (27.2) – (2b(ii))
2 × 27.2 and 1 × 2b(ii)
calculate correctly their expression
Using alternative value of ∆H1 = –38.4 kJ mol-1
∆Hr = (+)92.8 kJ mol–1 = 2 marks
2(c) Ca(NO3)2 → CaO + 2NO2 + ½O2 2
M1 correct formula for calcium nitrate: Ca(NO3)2
M2 correct balanced equation
(a) Describe what is meant by dynamic equilibrium.
…… [2]
(b) Reaction 4 describes the reversible reaction between yellow Fe3+(aq) and colourless
SCN–(aq) to produce red FeSCN2+(aq).
reaction 4 Fe3+(aq) + SCN–(aq) FeSCN2+(aq) yellow colourless red
An equilibrium mixture contains Fe3+(aq), SCN–(aq) and FeSCN2+(aq). A few colourless crystals of soluble KSCN(s) are added. The mixture is then left until it reaches equilibrium again. The temperature of both equilibrium mixtures is the same.
(i) Deduce the changes that occur, if any, in the equilibrium mixture after KSCN(s) is added compared to the original equilibrium mixture.
• change in appearance
• change in relative concentration of Fe3+(aq)
• change in value of the equilibrium constant, Kc
[3]
(ii) The expression for the equilibrium constant, Kc, for reaction 4 is shown.
[FeSCN2+(aq)]
Kc =
[Fe3+(aq)] × [SCN–(aq)]
5.00 × 10–5 mol of Fe3+(aq) and 5.00 × 10–5 mol of SCN–(aq) are added together and allowed to reach equilibrium. The total volume of the mixture is 25.0 cm3.
At equilibrium the concentration of FeSCN2+(aq) is 4.23 × 10–4 mol dm–3.
Calculate the equilibrium constant, Kc, for reaction 4.
Include the units in your answer.
Kc = ……
units ……
[4]
(c) Determine the full electronic configuration of Fe3+.
…… [1]
(d) SCN–(aq) is colourless.
Complete the dot-and-cross diagram in Fig. 3.1 to show the arrangement of outer electrons in an SCN– ion.
–
S C N
Fig. 3.1
[2]
[Total: 12]
Mark scheme
3(a) M1 rates of forward and reverse / backward reactions are equal 2
M2 no change in measurable properties OR concentration of reactants AND products remain constant
3(b)(i) M1 change to appearance of the mixture: (goes) darker red / more red 3
M2 change to relative concentration of Fe3+(aq): decreases
M3 change to the value of equilibrium constant, Kc: constant / none
3(b)(ii) Method 1 4
M1 conc of initial conc of Fe3+ and SCN– = 0.0020 (mol dm-3)
M2 at equilibrium the conc of Fe3+ and SCN– are (M1 – 0.000423)
M3 use 0.000423 (M1 – 0.000423)2 to calculate Kc = 170(.089)
M4 units = mol–1 dm3 OR dm3 mol-1
Method 2
M1 no mol FeSCN2+ = 4.23 × 10–4 × 25 / 1000 = 1.0575 × 10–5
M2 no mol Fe3+ and SCN– are (5 × 10–5 – M1) = 3.9425 × 10–5
M3 0.000423 (M2 × 1000 / 25)2 = 170(.089)
M4 units = mol–1 dm3 OR dm3 mol–1
3(c) full electronic configuration of Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 1
3(d) 2
M1 6 electrons between C and N AND 2 electrons between C and S
M2 3 lone pairs around S AND none around C AND 1 lone pair around N
CH3(CH2)5CHBrCH3 exists as a pair of stereoisomers.
(a) Draw the three-dimensional structures of the two stereoisomers of CH3(CH2)5CHBrCH3.
R can be used to represent CH3(CH2)5.
[2]
(b) A sample of CH3(CH2)5CHBrCH3 reacts with NaOH to make CH3(CH2)5CH(OH)CH3 in an
SN1 mechanism.
Complete Fig. 4.1 to show the mechanism for the reaction of CH3(CH2)5CHBrCH3 and NaOH.
Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.
H H
CH3(CH2)5 C Br CH3(CH2)5 C OH
CH3 CH3
Fig. 4.1
[3]
(c) Separate samples of CH3(CH2)5CHBrCH3, CH3(CH2)5CH(OH)CH3 and CH3(CH2)5CHCH2 are tested with different reagents.
Complete Table 4.1. If no reaction occurs, write × in the relevant box.
Table 4.1
observation with observation with observation with reagent added
CH3(CH2)5CHBrCH3 CH3(CH2)5CH(OH)CH3 CH3(CH2)5CHCH2
Br2(l) in the dark
PCl 5(s)
AgNO3(aq)
[3]
(d) CH3(CH2)5CHBrCH3 is heated with D to produce three different molecules, E, F and G.
E F G
Fig. 4.2
(i) Name the type of reaction.
…… [1]
(ii) Identify D and the conditions used.
…… [1]
(e) (i) Both σ and π bonds are present in a molecule of E as a result of different types of hybridisation in the carbon atoms.
Complete Table 4.2 to show the number of carbon atoms with each type of hybridisation in a molecule of E.
Table 4.2
number of carbon atoms sp hybridised sp2 hybridised sp3 hybridised
E F G
[2]
(ii) Describe the essential feature of an unbranched hydrocarbon that causes its molecules to show stereoisomerism. Explain how this feature leads to stereoisomerism.
…… [3]
[Total: 15]
5 Compound W has molecular formula C4H10O. It contains only one functional group.
(a) Table 5.1 shows the two peaks with the greatest m / e values in the mass spectrum of W.
Table 5.1
m/e relative abundance
74 50
75 x
(i) Calculate the relative abundance, x, of the peak at m / e = 75 using the information from
Table 5.1.
x = …… [1]
(ii) The mass spectrum of W also shows peaks at m / e = 29 and m / e = 59.
Suggest the molecular formulae of these fragments.
m / e = 29 ……
m / e = 59 ……
[2]
(b) A sample of W, C4H10O, is heated under reflux with an excess of acidified K2Cr2O7 until there is no further reaction. Only one organic product, X, is present in the mixture at the end of the reaction.
Fig. 5.1 shows the infrared spectrum of W.
100
80
60 transmittance / %
40
20
A
0
4000 3500 3000 2500 2000 1500 1000 500 wavenumber / cm–1
Fig. 5.1
Fig. 5.2 shows the infrared spectrum of X.
100
80
60 transmittance / %
40
20
B
0
4000 3500 3000 2500 2000 1500 1000 500 wavenumber / cm–1
Fig. 5.2
Table 5.2
bond functional groups containing the bond characteristic infrared absorption range
(in wavenumbers) / cm–1
C–O hydroxy, ester 1040–1300
C=C aromatic compound, alkene 1500–1680
C=O amide 1640–1690 carbonyl, carboxyl 1670–1740 ester 1710–1750
C≡N nitrile 2200–2250
C–H alkane 2850–2950
N–H amine, amide 3300–3500
O–H carboxyl 2500–3000
hydroxy 3200–3650
(i) Absorption A is shown in Fig. 5.1.
Absorption B is shown in Fig. 5.2.
Complete Table 5.3 using the information given in Fig. 5.1, Fig. 5.2 and Table 5.2.
Table 5.3
functional group absorption bond containing the bond
A
B
[1]
(ii) Use the information in (a) and (b)(i) to draw the structure of X in the box in Fig. 5.3.
X
W + acidified
K2Cr2O7
C4H10O
Fig. 5.3
[1]
(c) Y is a structural isomer of W.
Both W and Y produce colourless bubbles when sodium is added to them.
Y does not react when heated with acidified K2Cr2O7.
Y does not react when warmed with alkaline I2(aq).
(i) Name the functional group present in Y.
…… [1]
(ii) Complete the equation to describe the reaction of W or Y with sodium.
……C4H10O + ……Na …… [1]
(iii) Draw the structure of Y.
[1]
[Total: 8]
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
3 4 5 6 7 8 9 10 11 12
23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
at www.cambridgeinternational.org after the live examination series.
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
16
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
138.9 140.1 140.9 144.2 – 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
Mark scheme
4(a) Three-dimensional stereoisomers of CH3(CH2)5CHBrCH3 2
M1 correct representation of left-hand answer as 3d structure of one of the enantiomers of CH3(CH2)5CHBrCH3
M2 correct representation of 3d structure of the second enantiomer of CH3(CH2)5CHBrCH3
4(b) 3
(+ NaBr OR Br– )
M1 correct dipole (δ+C—Brδ–) AND curly arrow from bond of C—Br to Br (or just beyond)
M2 correct intermediate ALLOW R for –(CH2)5CH3
M3 curly arrow from lone pair on OH– to the C of intermediate (or suitable position where C–OH bond is made)
If SN2 mechanism shown:
Apply SN2 mechanism mark scheme only if mechanism shows OH / NaOH species attacking the reactant and no OH
M1 correct dipole (δ+C—Brδ–) AND curly arrow from bond of C—Br to Br of reactant (or just beyond)
NOT M2
M3 curly arrow from lone pair on :OH— to the appropriate C of the reactant
4(c) 3
Chemical observation with observation observation with
test 2-bromooctane with CH3(CH2)5CHCH2
octan-2-ol
Br2 in the orange / brown to
dark colourless
PCl 5 steamy
fumes
AgNO3(aq) cream precipitate
OR
off white precipitate
6 correct answers = 3 marks
4 or 5 correct answers = 2 marks
2 or 3 correct answers = 1 mark
0 or 1 correct answer = 0 marks
4(d)(i) elimination 1
4(d)(ii) NaOH / sodium hydroxide in ethanol 1
4(e)(i) 2
0 2 6
3 correct answers = 2 marks
2 correct answers = 1 mark
0 or 1 correct answers = 0 marks
4(e)(ii) M1 presence of a 𝜋 bond 3
M2 limited rotation of C=C bond
M3 EITHER
two different groups on each / both double bonded carbon atoms
OR
two different atoms on each / both double bonded carbon atom
OR
one H / atom and one alkyl group on each / both double bonded carbon atom
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Chemistry 9701/22 for May/June 2024 is out of 60 marks, with 1 hour 15 minutes to answer it, across 5 questions.
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