A Level Chemistry 9701 Paper 22, May/June 2024

Question paper and mark scheme60 marks4 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Papers open in

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9701/22 · May/June 2024

Download PDF
1
Figure 1 of 1 for question 1
a

Complete Table 1.1 using relevant information from the Periodic Table. Table 1.1

nucleon numberproton numbernumber of electrons
Mg2+24
Al3+27
[2]
b

State and explain the difference in the ionic radius of Al3+ compared to Mg2+.

[3]
c

Draw a labelled diagram to show the structure and bonding in sodium.

[1]
d(i)

Fig. 1.1 shows the variation in melting point of some Period 3 elements in their standard states at room temperature and pressure. Explain why Si has a high melting point.

[1]
d(ii)

Complete Fig. 1.1 to show the variation in the melting points of the elements P, S and Cl.

[2]
e(i)

Two Period 3 elements react with an excess of oxygen at room pressure. Complete Table 1.2. Table 1.2

123
Period 3 elementstate of oxide at room temperature and pressureapproximate pH of solution made when oxide is added to water
Na
S
[2]
e(ii)

The solutions made in column 3 of Table 1.2 are mixed together. Name the type of reaction that occurs.

[1]
e(iii)

Write an equation to describe the reaction between P4O10 and an excess of water.

[1]
f(i)

Aluminium hydroxide is amphoteric. Explain what is meant by amphoteric.

[1]
f(ii)

Write an equation to describe the reaction that occurs when aluminium hydroxide, Al(OH)3, reacts with NaOH(aq).

[1]
2

Separate samples of Na2CO3 and NaHCO3 react with HCl(aq) to produce the same products, as shown in Table 2.1. Table 2.1

reactionequationΔH/kJ mol−1
1Na2CO3+2HCl→2NaCl+H2O+CO2ΔH1
2NaHCO3+HCl→NaCl+H2O+CO2ΔH2=+27.2
Figure 1 of 1 for question 2
a

Complete the reaction pathway diagram in Fig. 2.1 for reaction 2. Label the diagram to show the enthalpy change, ΔH2, and the activation energy, EA.

[2]
b(i)

The value for ΔH1 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. Describe one other observation that shows the reaction is complete.

[1]
b(ii)

Calculate the value of ΔH1 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. ΔH1=.............................. kJ mol−1

[3]
b(iii)

Thermal decomposition occurs when NaHCO3 is heated. reaction 3 2NaHCO3→Na2CO3+H2O+CO2 Calculate the enthalpy change for reaction 3, ΔHr, using the data in Table 2.1 and the value of ΔH1 calculated in (b)(ii). (If you were unable to calculate a value for ΔH1 in (b)(ii), assume the enthalpy change is −38.4 kJ mol−1. This is not the correct value.) ΔHr=.............................. 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]
3
Figure 1 of 1 for question 3
a

Describe what is meant by dynamic equilibrium.

[2]
b(i)

Reaction 4 describes the reversible reaction between yellow Fe3+(aq) and colourless SCN−(aq) to produce red FeSCN2+(aq).

reaction 4Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)
yellowcolourlessred

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. 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]
b(ii)

The expression for the equilibrium constant, Kc, for reaction 4 is shown. Kc=[FeSCN2+(aq)][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.

[2]
4

CH3(CH2)5CHBrCH3 exists as a pair of stereoisomers.

Figure 1 of 3 for question 4
Figure 2 of 3 for question 4
Figure 3 of 3 for question 4
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.

[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

reagent addedobservation with CH3(CH2)5CHBrCH3observation with CH3(CH2)5CH(OH)CH3observation with CH3(CH2)5CHCH2
Br2(l) in the dark
PCl5(s)
AgNO3(aq)
[3]
d(i)

CH3(CH2)5CHBrCH3 is heated with D to produce three different molecules, E, F and G. Name the type of reaction.

[1]
d(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 hybridisedsp2 hybridisedsp3 hybridised
E
[2]
e(ii)

Describe the essential feature of an unbranched hydrocarbon that causes its molecules to show stereoisomerism. Explain how this feature leads to stereoisomerism.

[3]

Mark scheme

9701/22 · May/June 2024

Download PDF

1a · 2 marks

  • 2 marks

    Mg2+: 24, 12, 10; Al3+: 27, 13, 10 (4 correct = 2 marks, 2 or 3 correct = 1 mark)

1b · 3 marks

  • M1

    it / Al3+ is smaller (compared to Mg2+)

  • M2

    greater nuclear attraction for remaining electrons

  • M3

    same shielding effect AND greater nuclear charge

1c · 1 mark

  • 1 mark

    labelled diagram shows: regular arrangement of minimum of 2 × 2 circles containing + for positive ions AND surrounded by (sea of) delocalised electrons

1d(i) · 1 mark

  • 1 mark

    large amount of energy required to break bonds in giant covalent structure

1d(ii) · 2 marks

  • M1

    all 3 melting points lower than Mg

  • M2

    show melting point Cl < P < S

1e(i) · 2 marks

  • 2 marks

    Na: solid, 10–14; S: gas, 0–4 (4 correct = 2 marks, 2 or 3 correct = 1 mark)

1e(ii) · 1 mark

  • 1 mark

    acid–base OR neutralisation

1e(iii) · 1 mark

  • 1 mark

    P4O10 + 6H2O → 4H3PO4

1f(i) · 1 mark

  • 1 mark

    reacts with both acids and bases OR shows both acidic and basic behaviour

1f(ii) · 1 mark

  • 1 mark

    Al(OH)3 + NaOH → NaAl(OH)4

2a · 2 marks

  • 2 marks

    All three correct (2 marks) or two correct (1 mark): - profile for endothermic reaction (include ‘hump’) - arrow representing EA - arrow representing ΔH OR (+) 27.2

2b(i) · 1 mark

  • 1 mark

    no more fizzing / effervescence stops

2b(ii) · 3 marks

  • M1

    calculate Q from = 50 × 4.18 × (26.2 – 19.6)) = 1379.(4) OR evidence of use Q = mcΔT = 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 = 34485 (or 34.485)

  • M3

    express Q as ΔH(1) in kJ mol–1 = – (value for M2) ÷ 1000 = –34.5 kJ mol–1 OR express Q as ΔH(1) in kJ mol–1 = – (value for M2) (expressed in kJ) = –34.5 kJ mol–1

2b(iii) · 2 marks

  • 2 marks

    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

2c · 2 marks

  • M1

    correct formula for calcium nitrate: Ca(NO3)2

  • M2

    correct balanced equation: Ca(NO3)2 → CaO + 2NO2 + ½O2

3a · 2 marks

  • M1

    rates of forward and reverse / backward reactions are equal

  • M2

    no change in measurable properties OR concentration of reactants AND products remain constant

3b(i) · 3 marks

  • M1

    change to appearance of the mixture: (goes) darker red / more red

  • M2

    change to relative concentration of Fe3+(aq): decreases

  • M3

    change to the value of equilibrium constant, Kc: constant / none

3b(ii) · 4 marks

  • M1

    conc of initial conc of Fe3+ and SCN–= 0.0020 (mol dm−3) OR no mol FeSCN2+ = 4.23 × 10–4 × 25 / 1000 = 1.0575 × 10–5

  • M2

    at equilibrium the conc of Fe3+ and SCN– are (M1 – 0.000423) OR no mol Fe3+ and SCN– are (5 × 10–5 – M1) = 3.9425 × 10–5

  • M3

    use 0.000423 ÷ (M1 – 0.000423)2 to calculate Kc = 170(.089) OR 0.000423 ÷ (M2 × 1000 / 25)2 = 170(.089)

  • M4

    units = mol–1 dm3 OR dm3 mol−1

3c · 1 mark

  • 1 mark

    full electronic configuration of Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5

3d · 2 marks

  • 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

4a · 2 marks

  • 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

4b · 3 marks

  • M1

    correct dipole (δ+C—Brδ–) AND curly arrow from bond of C—Br to Br (or just beyond) [or for SN2: to Br of reactant (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) [or for SN2: curly arrow from lone pair on :OH— to the appropriate C of the reactant]

4c · 3 marks

  • 3 marks

    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)

4d(i) · 1 mark

  • 1 mark

    elimination

4d(ii) · 1 mark

  • 1 mark

    NaOH / sodium hydroxide in ethanol

4e(i) · 2 marks

  • 2 marks

    0, 2, 6 (3 correct answers = 2 marks, 2 correct answers = 1 mark)

4e(ii) · 3 marks

  • M1

    presence of a π bond

  • 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

Questions

  1. 1
    Question 115 marks
    Mark scheme ›
  2. 2
    Question 210 marks
    Mark scheme ›
  3. 3
    Question 312 marks
    Mark scheme ›
  4. 4
    Question 415 marks
    Mark scheme ›

More 9701 papers

Other papers from May/June 2024

Solve it on ExamPalStart