Chemistry 9701/22 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Chemistry (9701), Paper 2 AS Level Structured Questions, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Magnesium has a melting point of 650 °C and high electrical conductivity.
Explain these properties of magnesium by referring to its structure and bonding.
. ……
. …… [2]
(b) When magnesium is heated in air, magnesium oxide, MgO, is the major product. Smaller amounts of magnesium nitride, Mg3N2, are also made.
(i) Calculate the oxidation number for magnesium and for the nitrogen species in Mg3N2 to complete Table 1.1.
Table 1.1
species magnesium in Mg3N2 nitrogen in Mg3N2
oxidation number
[1]
(ii) Identify the type of reaction which takes place between magnesium and nitrogen.
Explain your answer.
. ……
. …… [1]
(iii) Define enthalpy change of formation.
. ……
. …… [2]
(iv) When 3.645 g of Mg(s) burns in excess N2(g) to form Mg3N2(s), 23.05 kJ of energy is released.
Calculate the enthalpy change of formation, ∆Hf, of Mg3N2. Show your working.
∆Hf (Mg3N2) = …… [3]
[Total: 9]
Question 2 starts on the next page.
Mark scheme
1(a) M1 comment explaining high melting point of Mg in terms of 2
many strong metallic bonds
OR
many strong (electrostatic) attractions between cations and delocalised electrons
OR
strong bonds in giant metallic structure.
M2 comment explaining electrical conductivity of Mg in terms of movement of delocalised electrons
delocalised electrons can move through the structure
1(b)(i) 1
species magnesium in Mg3N2 nitrogen in Mg3N2
oxidation number 2(+) / (+)II 3– / –III
1(b)(ii) redox 1
Mg loses electrons / Mg increases ox no / Mg is oxidised
N gains electrons / N reduces ox no / N is reduced
1(b)(iii) M1 (enthalpy change / energy change) when one mole of a compound / substance is formed 2
M2 from its elements in their standard states
1(b)(iv) M1 correct method to find amount Mg3N2 formed in reaction 3
3.645 ÷ 24.3 mol Mg produces 0.15 ÷ 3 mol of Mg3N2 OR 0.05 mol of Mg3N2
M2 use M1 in correct expression to find energy released by 1 mol Mg3N2
23.05 ÷ M1 OR 23.05 ÷ mol Mg3N2
OR 23 050 ÷ M1 OR 23 050 ÷ Mg3N2
M3 calculate value from M2 expression correctly
∆H = –(M2) = –461(.0) kJ mol–1 OR –461 000 J mol–1
Radium, Ra, is an element found in Group 2 of the Periodic Table. It is a crystalline solid at room temperature and conducts electricity.
Radium chloride, RaCl 2, has a melting point of 900 °C and is soluble in water.
(a) Predict the lattice structure of RaCl 2(s) based on the properties described.
. …… [1]
(b) Draw a dot-and-cross diagram to show the arrangement of outer electrons in RaCl 2.
[1]
(c) Solid Ra and Ca show similar reactions with H2O, but the reactions occur at different rates.
Separate samples, each containing a single piece of solid Ra or Ca, are added to equal volumes of cold water.
Each sample contains equal numbers of moles of solid and the H2O is in excess.
(i) Construct an equation for the reaction of Ra with H2O.
. …… [1]
(ii) Identify which element, Ra or Ca, reacts with H2O at a faster rate. Suggest how the observations of each reaction would differ.
. ……
. …… [1]
(iii) Suggest why these reactions occur at different rates.
. ……
. ……
. …… [2]
(iv) One of the solutions is cloudy when the reaction has finished.
At the end of each reaction, universal indicator is added to each reaction mixture.
Suggest pH values of the solutions made in both reactions. Explain your answer.
. ……
. ……
. …… [2]
(d) A sample of aqueous calcium halide, CaX2(aq), contains either chloride, bromide or iodide ions.
Complete Table 2.1 to describe a two-step process that could be used to identify the halide ion present.
Table 2.1
observation observation observation step method with CaCl 2 with CaBr2 with CaI2
step 1
step 2
[3]
[Total: 11]
Mark scheme
2(a) giant ionic 1
2(b) Ra2+ and 2 x Cl – 0 electrons surrounding 2Ra AND 8 electrons surrounding Cl 1
Ra
2(c)(i) Ra + 2H2O → Ra(OH)2 + H2 1
2(c)(ii) Ra – more bubbles per unit time 1
OR
With Ra solid disappears more quickly
OR
Ra is the first to stop fizzing
ora
2(c)(iii) option 1 2
suggest why these reactions occur at different rates (collision theory)
M1 difference in activation energy / ionisation energy
M2 affects the frequency of effective collisions (between particles / molecules)
OR
affects the proportion of particles with energy greater than activation energy
option 2
suggest why the elements have different reactivity in terms of atomic structure (which results in a different rate)
M1 ionisation energies are different
M2 a decrease in nuclear attraction due to
EITHER
increase in shielding
OR
increase in distance of outer electron from nucleus
OR
increase in number of shells of electrons ora
2(c)(iv) M1 2
pH value (or values) for each solution must be greater > 7
AND
pH value (or range values) identified for solution made from Ra must be greater than pH values (or range of values)
stated for Ca
M2 any one from:
solubility of group 2 hydroxides increases down the group
Ra(OH)2 is more soluble
greater concentration of OH–(aq) in the solution involving Ra
more OH– (aq) in the solution involving Ra
2(d) 3
step method observation with CaCl2 observation with CaBr2 observation with CaI2
AgNO3(aq) white ppt cream OR off-white ppt (pale) yellow ppt
step 1
(+HNO3(aq)
NH3 (aq) (ppt) (ppt) partly soluble / slightly soluble (ppt) insoluble / solid remains
dissolves / (completely) (in dilute or conc) (in dilute or concentrated)
step 2 soluble (in dilute or conc) ALLOW (ppt) dissolves in excess
OR (forms) colourless (in dilute or conc)
solution ALLOW (ppt) soluble in conc.NH3
M1 step 1 AgNO3(aq) / silver nitrate solution AND step 2 NH3(aq) / ammonia solution / dilute NH3 / concentrated NH3
M2 ALL solid / precipitate AND correct colours described in row 1
M3 correct observations on addition of ammonia
(a) 0.025 mol of HI(g) is added to a closed vessel and left to reach dynamic equilibrium. The total pressure of the vessel is 100 kPa.
equation 1 2HI(g) H2(g) + I2(g)
(i) Explain what is meant by dynamic equilibrium.
. ……
. ……
. …… [2]
(ii) Describe one difference in the initial appearance of the reaction mixture compared to the mixture at equilibrium.
. …… [1]
(iii) Write an expression for Kp for the reaction described in equation 1.
Kp =
[1]
(iv) At equilibrium the partial pressure of HI(g) is 86.4 kPa.
Calculate the amount of HI(g) present in the mixture at equilibrium. Show your working.
amount of HI(g) = …… mol [2]
(b) Use equation 1 and the bond energy values in Table 3.1 to calculate the change in enthalpy,
∆H, for the thermal decomposition of 1 mole of HI(g). Show your working.
Table 3.1
bond bond energy / kJ mol–1
H–H 436
I–I 151
H–I 299
∆H = …… kJ mol–1 [2]
(c) Describe the effect of increasing pressure on the value of Kp for the decomposition of HI(g).
. …… [1]
(d) HCl (g) is prepared by adding NaCl (s) to concentrated H2SO4.
HI(g) is not prepared by adding NaI(s) to concentrated H2SO4 because the HI(g) produced also reacts with concentrated H2SO4.
(i) Identify the type of reaction that occurs when NaI(s) reacts with concentrated H2SO4 to form HI(g).
. …… [1]
(ii) Write an equation for the reaction of HI(g) and concentrated H2SO4.
. …… [1]
(iii) Explain why HI(g) reacts with concentrated H2SO4 whereas HCl does not.
. …… [1]
[Total: 12]
Mark scheme
3(a)(i) M1 rate(s) of forward and reverse / backward reactions are equal / are the same 2
M2 no change in measurable properties OR concentration of reactants and products remain constant
3(a)(ii) (colourless gas) becomes purple (gas) 1
3(a)(iii) (pH2 )(pI2 ) 1
Kp =
(pHI)2
3(a)(iv) M1 use correct expression for finding amount of HI used up in reaction 2
86.4 = 100 × (0.025 – x) / 0.025 – x + x / 2 + x / 2) where x = HI mol used up
OR
86.4 = 100 × (0.025 – 2x) / (0.025 – 2x + x + x) where 2x = HI mol used up
M2 correct calculation to find amount of HI(mol) at equilibrium based on
HI(mol) at eq’m = 0.025 – HI(mol) used = 0.025 – 0.0034 = 0.0216 OR 0.022 (mol)
alternative method
M1 use the pp H2 and pp I2 at equilibrium (= 13.6) to calculate no mol H2 & I2 at equilibrium
(either 0.0034 in total OR 0.0017 each)
M2 use value for no mol H2 + I2 in expression to find no mol HI
0.025 – (no mol H2 + I2) = 0.0216
3(b) evaluation, to give a value, based on calculation using: 2
all three correct bond energies [HI = 299, H2 = 436, I2 = 151]
correct use of stoichiometry to calculate ∆H for 1 mol HI
x = 1 and y = ½
calculate a value for ∆H correctly
i.e. x(299) – y(436 + 151)
∆H = (+)5.5 kJ mol–1
3(c) none / no change 1
3(d)(i) acid-base 1
3(d)(ii) 8HI + H2SO4 → 4I2 + H2S + 4H2O 1
OR
8H+ + 8I– + H2SO4 → 4I2 + H2S + 4H2O
3(d)(iii) explanation in terms of comparison of reducing nature of HI to HCl 1
i.e. HI / I– is stronger reducing agent compared to Cl – / HCl
OR
explanation in terms of additional behaviour of HI (compared to HCl)
i.e. HI / I– reduces S (in sulfuric acid) / (concentrated) sulfuric acid
(a) Bromine reacts with butane in the presence of ultraviolet light to form bromobutane.
Two structural isomers with the molecular formula C4H9Br are produced during this reaction.
(i) Draw the two structural isomers and state the systematic name of each isomer.
structural isomer 1 structural isomer 2
name …… name ……
[2]
(ii) Identify the type of structural isomerism shown in (a)(i).
. …… [1]
(b) Halothane is an anaesthetic.
halothane
F Cl
F C C H
F Br
Fig. 4.1
(i) Identify the chiral centre in halothane and mark it with an asterisk (*). [1]
When halothane reacts in ultraviolet light, homolytic fission occurs and the C–Br bond is broken.
(ii) Construct an equation to show the homolytic fission of halothane, CF3CHBrCl.
. …… [1]
(iii) Complete Fig. 4.2 to show the arrangement of electrons in a bromine atom using the electrons in boxes notation.
[Ar]
3d 4s 4p
Fig. 4.2
[1]
(c)
X is an addition polymer.
X
Cl
n
Fig. 4.3
(i) Draw the monomer of X.
[1]
(ii) Suggest one reason why the disposal of items made from X is difficult.
. …… [1]
[Total: 8]
Mark scheme
4(a)(i) 2
CH3(CH2)2CH2Br CH3CH2CH(Br)CH3
OR OR OR OR
name: 1-bromobutane name: 2-bromobutane
M1 structure of isomer 1 AND name
M2 structure of isomer 2 AND name (in any order)
4(a)(ii) positional 1
4(b)(i) 1
4(b)(ii) CF3CBrClH → CF3CHCl() + Br() 1
4(b)(iii) [Ar] 1
4(c)(i) 1
4(c)(ii) non-biodegradable 1
OR
poisonous / harmful HCl made when burnt
Fig. 5.1 shows three reactions of 2-bromopropane, CH3CH(Br)CH3.
CH3CH(Br)CH3
reaction 1 reaction 3 reaction 2
CH3CH(OH)CH3 CH3CH(NH2)CH3 H2C=CHCH3
Fig. 5.1
(a) Complete Table 5.1 for each reaction, by:
● stating the reagent and conditions used
● identifying the type of reaction that occurs.
Table 5.1
reaction reagent and conditions type of reaction
1
2
3
[6]
(b) A sample of 2-iodopropane, CH3CH(I)CH3, reacts under the same conditions as reaction 1 to produce CH3CH(OH)CH3.
Explain why 2-iodopropane reacts at a faster rate than 2-bromopropane.
. ……
. ……
. …… [2]
(c) Fig. 5.2 shows how butan-1-ol can be made from 1-bromopropane in three steps.
HO
N HO
Br step 1 step 2 step 3
O
Fig. 5.2
(i) In step 1, 1-bromopropane reacts with CN– to form butanenitrile.
Complete Fig. 5.3 to show the mechanism for step 1. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate.
H H H H H H
H C C C Br H C C C C N
H H H H H H
Br –
C –
N
Fig. 5.3
[2]
(ii) In step 2, butanenitrile is heated with HCl (aq). A hydrolysis reaction occurs.
Construct an equation for the reaction in step 2.
. …… [1]
(iii) Step 3 is a reduction reaction.
Construct an equation for the reduction reaction in step 3. Use [H] to represent one atom of hydrogen from the reducing agent.
. …… [1]
(iv) State the identity of a suitable reducing agent in step 3.
. …… [1]
[Total: 13]
Mark scheme
5(a) 6
reaction reagent and conditions type of reaction
1 NaOH(aq) [M1] substitution
2 NH3 [M2] substitution
in ethanol AND heat under pressure [M3]
3 NaOH in ethanol elimination
AND heat [M4]
M5 and M6 types of reaction
[3 correct types = 2 marks and 2 correct types = 1 mark]
All reagents can be identified as (correct) formula or in words
5(b) M1 C–I (covalent bond) is weaker ora 2
M2 lower activation energy / lower Ea (with 2-iodopropane)
OR
M2 explain in terms of the SN1 mechanism (that is dominant / preferred / occurring with 2-iodopropane is fast(er))
(identical) carbocation is made (more) quickly (with 2-iodopropane)
OR
low / less energy is required to make the (same)carbocation / intermediate (from 2-iodopropane)
OR
M2 iodine / I (of C–I) has weaker attraction of nucleus to bonding / shared pair (of electrons) due to
any one of:
more / high shielding (of electrons from inner shells)
more / 5 electron shells in iodine
C-I greater distance from nuclei (to bonding pair of electrons)
5(c)(i) 2
M1 dipole Cδ+–Brδ- AND curly arrow from C–Br bond to Br δ-
M2 lone pair on C of –CN AND curly arrow from lone pair to C (on C-1 of bromopropane)
5(c)(ii) CN(CH2)2CH3 + 2H2O + HCl → HOOC(CH2)2CH3 + NH4Cl 1
OR
+ 2H2O + HCl → + NH4Cl
5(c)(iii) HOOC(CH2)2CH3 + 4[H] → HO(CH2)3CH3 + H2O 1
OR
+ 4[H] → + H2O
5(c)(iv) LiAl H4 OR lithium tetrahydridoaluminate (in ether) 1
Z is a molecule which contains the elements carbon, hydrogen and oxygen only.
Z contains only alkene and carboxyl functional groups.
(a) Complete Table 6.1 by describing the observations that occur when two different reagents are added to separate samples of Z(aq).
Table 6.1
reagent added observation to Z(aq)
Br2(aq)
Na2CO3(s)
[2]
(b) Table 6.2 shows the percentage by mass of each element present in Z.
Table 6.2
element percentage by mass / % carbon 41.38 hydrogen 3.45 oxygen 55.17
Using the data in Table 6.2, demonstrate that the empirical formula of Z is CHO.
Show your working.
[1]
(c) Fig. 6.1 shows the mass spectrum of Z.
45
100
80
relative 60 abundance
40
71
20
116
117
0
10 20 30 40 50 60 70 80 90 100 110 120 m/e
Fig. 6.1
(i) Deduce the molecular formula of Z. Explain your answer by referring to the molecular ion peak in Fig. 6.1 and the empirical formula of Z.
[1]
(ii) Use Fig. 6.1 to suggest the formulae of the fragments with m / e peaks at 45 and at 71.
m / e 45 . ……
m / e 71 . ……
[2]
(iii) Suggest the structure of Z using relevant information from Table 6.1, (b) and (c).
[1]
[Total: 7]
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
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
138.9 140.1 140.9 144.4 – 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
Mark scheme
6(a) 2
Br2(aq) orange to colourless OR orange disappears
Na2CO3(s) fizzing OR bubbles OR effervescence
6(b) C : H : O 1
41.38 / 12 3.45 / 1 55.17 / 16
3.45 3.45 3.45 (so C(1)H(1)O(1))
6(c)(i) Look for some reference to 116 as the mass of the molecular ion AND mass of C(1)H(1)O(1) = 29 to conclude molecular 1
formula is C4H4O4
116 / 29 = 4 so C4H4O4
6(c)(ii) M1 m / e 45: +COOH OR +CHO2 2
M2 m / e 71: C3H3O2+
6(c)(iii) 1
OR
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Chemistry 9701/22 for May/June 2022 is out of 60 marks, across 6 questions.
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