A Level Chemistry 9701 Paper 42, May/June 2025

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9701/42 · May/June 2025

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1
a(i)

Calcium nitrate, Ca(NO3)2, decomposes on heating. Write an equation for the decomposition of calcium nitrate.

[1]
a(ii)

Describe the trend in the decomposition temperature of the Group 2 nitrates. Explain your answer.

[3]
b

A sample of 0.333 g of strontium oxide, SrO, is completely dissolved in distilled water to form a solution of strontium hydroxide, Sr(OH)2. The resulting solution is added to a volumetric flask and made up to 250.0 cm3 with distilled water. Calculate the pH of this solution at 298 K. Give your answer to two decimal places.

[4]
2
Figure 1 of 2 for question 2
Figure 2 of 2 for question 2
a(i)

When cobalt(II) sulfate, CoSO4, is dissolved in distilled water, solution A is formed. The reaction scheme in Fig. 2.1 shows some reactions of solution A. Complete Table 2.1 to show the formula and colour of each of the cobalt-containing species present in A, B and C. Identify the type of reaction forming each of B and C. Table 2.1

[4]
a(ii)

Suggest a suitable reagent for the formation of [CoCl4]2− from solution A.

[1]
b(i)

The complex ion [CoCl4]2− has tetrahedral geometry. The 3d orbitals in an isolated Co2+ ion are degenerate. Complete Fig. 2.2 to show the relative energies of the 3d orbitals in an isolated Co2+ ion and in Co2+ in a tetrahedral complex.

[2]
b(ii)

Draw a three-dimensional diagram to show the structure of the complex ion [CoCl4]2−.

[1]
c

H2O2 can act as an oxidising agent or a reducing agent when reacting with species that contain manganese. Table 2.2 shows electrode potentials, E⊖, for some electrode reactions. Table 2.2 Use only the species listed in Table 2.2 to suggest: • one reaction in which H2O2 acts as an oxidising agent and • one reaction in which H2O2 acts as a reducing agent. Include the value of the standard cell potential, Ecell⊖, and an overall equation for each reaction.

[4]
d(i)

Acidified manganate(VII) ions, MnO4−, can be used to analyse the content of iron tablets by titration. Two identical iron tablets are crushed and dissolved in distilled water. The resulting solution is made up to 150.0 cm3 with distilled water. 25.0 cm3 of this solution requires 18.60 cm3 of 0.0500 mol dm−3 acidified MnO4− to reach the end-point. All the Fe2+ ions are oxidised. The relevant half-equations are shown. MnO4−+8H++5e−⇌Mn2++4H2O Fe2+⇌Fe3++e− Describe the colour change observed at the end-point of this titration.

[1]
d(ii)

Calculate the mass, in mg, of iron in one tablet. Assume that all the iron in the tablets is Fe2+. Show your working.

[4]
3
Figure 1 of 1 for question 3
a

Define entropy.

[1]
b(i)

Fig. 3.1 shows how the entropy, S, of a pure substance changes with temperature, T. Identify the process occurring at each of the temperatures T1 and T2.

[1]
b(ii)

Explain why the entropy change, ΔS, at T2 is bigger than the entropy change at T1.

[1]
c

The equation for the reduction of iron(III) oxide by carbon monoxide at 450 ∘C is shown. Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)ΔG⊖=−36.2 kJ mol−1 Table 3.1 shows the enthalpy of formation, ΔHf⊖, and the entropy, S⊖, for some substances. Table 3.1 Use the data in Table 3.1 to calculate the entropy, S⊖, of carbon monoxide at 450 ∘C. Show your working.

[3]
d

Iron(II) oxide can also be reduced to iron by carbon monoxide, as shown. FeO(s)+CO(g)→Fe(s)+CO2(g)ΔH⊖=−11.1 kJ mol−1, ΔS⊖=−15.2 J K−1 mol−1 State the effect of increasing temperature on the feasibility of this reaction. Explain your answer.

[2]
4
Figure 1 of 3 for question 4
Figure 2 of 3 for question 4
Figure 3 of 3 for question 4
a(i)

In aqueous solution, iodide ions react with acidified hydrogen peroxide, as shown in reaction 1. reaction 12I−+H2O2+2H+→I2+2H2O The rate equation for reaction 1 is shown. rate=k[I−][H2O2] Explain what is meant by order of reaction.

[1]
a(ii)

Complete Table 4.1. Table 4.1

[2]
a(iii)

Sketch a line on Fig. 4.1 to show the relationship between [I−] and time.

[1]
b(i)

Nitrogen dioxide, NO2, reacts with ozone, O3, as shown in reaction 2. reaction 22NO2+O3→O2+N2O5 The rate equation for reaction 2 is shown. rate=k[NO2][O3] Two experiments are carried out to measure the rate of reaction 2. In the first experiment, the initial rate is measured starting with known concentrations of NO2 and O3. In the second experiment, the concentrations of NO2 and O3 are both increased by a factor of four. Predict how the initial rate for reaction 2 would change.

[1]
b(ii)

The reaction mechanism for reaction 2 has two steps. Define rate-determining step.

[1]
b(iii)

Suggest equations for the two steps of the reaction mechanism for reaction 2.

[2]
c(i)

Dinitrogen pentoxide, N2O5, can decompose to NO2 and O2. The rate equation for this decomposition is shown. rate=k[N2O5] Fig. 4.2 shows the graph of rate against [N2O5] for this decomposition. Use Fig. 4.2 to calculate a value for the rate constant, k, for this decomposition.

[1]
c(ii)

Use your answer to (c)(i) to calculate the half-life, t12, in seconds, for this decomposition.

[1]
5
a(i)

Chloric(I) acid, HClO, is a weak Brønsted–Lowry acid. When chloric(I) acid is added to aqueous sodium hydroxide, an acid–base reaction takes place, as shown. HClO+NaOH→NaClO+H2O Identify the two conjugate acid–base pairs in this reaction.

[1]
a(ii)

The value for the acid dissociation constant, Ka, of HClO(aq) is 3.70×10−8. Calculate the concentration, in mol dm−3, of HClO(aq) at a pH of 4.51.

[2]
a(iii)

When a solution of HClO(aq) is heated, chloric(V) acid and a strong acid not containing oxygen are formed as the only products. Write an equation for this reaction.

[1]
b(i)

Define a buffer solution.

[1]
b(ii)

Suggest a substance that could be added to aqueous ethanoic acid to form a buffer solution. Explain your answer.

[1]
c

Some fertilisers contain calcium dihydrogenphosphate, Ca(H2PO4)2. An aqueous solution containing dihydrogenphosphate ions, H2PO4−, can act as a buffer solution. Write two equations to show how H2PO4− ions can act as a buffer.

[2]
d(i)

The solubility of calcium phosphate, Ca3(PO4)2, is 1.14×10−7 mol dm−3 at 25 ∘C. The expression for the solubility product, Ksp, of Ca3(PO4)2 is shown. Ksp=[Ca2+]3[PO43−]2 Calculate the value of Ksp for Ca3(PO4)2. Include units.

[3]
d(ii)

Some solid sodium phosphate is added to a saturated solution of Ca3(PO4)2. Predict the effect, if any, on the solubility of Ca3(PO4)2. Explain your answer.

[1]
6
Figure 1 of 5 for question 6
Figure 2 of 5 for question 6
Figure 3 of 5 for question 6
Figure 4 of 5 for question 6
Figure 5 of 5 for question 6
a(i)

Methylbenzene reacts readily with nitronium ions, NO2+. NO2+ ions are generated by the reaction between concentrated nitric acid and concentrated sulfuric acid. Write an equation for the formation of the NO2+ ion.

[1]
a(ii)

Complete the mechanism in Fig. 6.1 for the nitration of methylbenzene to form 1-methyl-2-nitrobenzene. Include all relevant curly arrows and charges.

[3]
b

Phenol can be nitrated with dilute nitric acid. Explain why the nitration of phenol occurs under milder conditions than the nitration of benzene.

[2]
c

A sample of 2-nitrophenol is reacted with sodium. Complete the equation in Fig. 6.2 for the reaction of 2-nitrophenol with sodium.

[1]
d(i)

2-nitrophenol can undergo different reactions as shown in Fig. 6.3. Suggest reagents and conditions for reaction 1.

[1]
d(ii)

Name the type of reaction for reaction 1.

[1]
d(iii)

Reaction 2 is carried out at room temperature. Draw the structure of the organic product from reaction 2 in Fig. 6.3.

[1]
d(iv)

Name the mechanism for reaction 2.

[1]
e

The NO2 group in 2-nitrophenol is electron withdrawing. Suggest the relative acidities of ethanol, 2-nitrophenol, phenol and water. Explain your answer.

[4]
f(i)

Salbutamol is a pharmaceutical drug that contains a phenol functional group. Name and classify the three other functional groups in salbutamol in Table 6.1. Table 6.1

[2]
f(ii)

Salbutamol reacts with Br2(aq) to form organic product X. Draw the structure of X.

[1]
f(iii)

Salbutamol reacts with an excess of SOCl2 to form organic product Y. The molecular formula of Y is C13H19Cl2NO. Draw the structure of Y.

[1]
7
Figure 1 of 1 for question 7
a

2-aminobutane, CH3CH(NH2)CH2CH3, exists as a mixture of two enantiomers. Define enantiomers.

[1]
b(i)

Explain why an aqueous solution of 2-aminobutane has a pH greater than 7. Include an equation in your answer.

[2]
b(ii)

A 0.10 mol dm−3 solution of diethylamine, (CH3CH2)2NH, has a higher pH than a 0.10 mol dm−3 solution of 2-aminobutane, CH3CH(NH2)CH2CH3. Suggest why.

[2]
b(iii)

(CH3CH2)2NH reacts with ethanoic acid, CH3COOH. Complete the equation for this reaction. (CH3CH2)2NH+CH3COOH→..................................................................................

[1]
b(iv)

(CH3CH2)2NH reacts with ethanoyl chloride, CH3COCl. Complete the equation for this reaction. (CH3CH2)2NH+CH3COCl→....................................................................................

[1]
c(i)

Table 7.1 shows monomers that can undergo polymerisation. Complete Table 7.1. Table 7.1

[1]
c(ii)

Ethanedioic acid, HOOCCOOH, can react with propane-1,3-diamine, H2NCH2CH2CH2NH2, to form polymer W. Draw a section of polymer W showing only one repeat unit. The new functional group formed should be displayed.

[2]
c(iii)

Poly(alkenes) biodegrade very slowly. Explain why.

[1]
8
Figure 1 of 1 for question 8
a

Compound A is analysed by carbon-13 NMR and proton (1H) NMR spectroscopy. State the reference substance and a solvent that can be used in NMR spectroscopy.

[1]
b

Predict the number of peaks in the carbon-13 NMR spectrum of A.

[1]
c

The proton (1H) NMR spectrum of A shows peaks in four different chemical environments. Complete Table 8.1 for the proton (1H) NMR spectrum of A. Table 8.1

[4]
9
Figure 1 of 4 for question 9
Figure 2 of 4 for question 9
Figure 3 of 4 for question 9
Figure 4 of 4 for question 9
a

The structures of the amino acids serine and lysine are shown in Fig. 9.1. Draw the structure for the dipeptide, ser–lys, with molecular formula C9H19N3O4. The peptide functional group formed should be displayed.

[2]
b(i)

The isoelectric point of serine is 5.7 and of lysine is 9.7. State what is meant by isoelectric point.

[1]
b(ii)

A mixture of serine, lysine and ser–lys is analysed by electrophoresis using a buffer at pH 5.7. Draw and label three spots on Fig. 9.2 to indicate the predicted position of each of these three species, serine, lysine and ser–lys, after electrophoresis. Explain your answer.

[3]
c(i)

Fig. 9.3 shows the synthesis of compound R from compound P. Draw the structure of Q in Fig. 9.3.

[1]
c(ii)

State the reagents and conditions for steps 1 and 2 in Fig. 9.3.

[2]

Mark scheme

9701/42 · May/June 2025

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1a(i) · 1 mark

  • 1 mark

    Ca(NO3)2 → CaO + 2NO2 + ½O2

1a(ii) · 3 marks

  • M1

    temperature increases down the group

  • M2

    ionic radius OR size of cation / M(2)+ AND increases

  • M3

    less polarisation / less distortion AND of anion / nitrate ion OR less weakening of NO bond / harder to break NO bond

1b · 4 marks

  • M1

    moles of SrO = 0.333 ÷ 103.6 = 3.214 × 10-3 AND moles of Sr(OH)2 = 3.214 × 10-3 OR 9 / 2800

  • M2

    moles of OH– in 250 cm3 = 3.214 × 10-3 × 2 = 6.429 × 10-3 AND moles of OH– in 1 dm3 = 6.429 × 10-3 × 4 = 0.0257

  • M3

    [H+] = 1.00 × 10–14 ÷ 0.0257 = 3.89 × 10–13 OR pOH = –log[OH–] = 1.59

  • M4

    pH = –log(3.89 × 10–13) = 12.41 must be to 2dp OR pH = 14 –1.59 = 12.41 must be 2 dp

2a(i) · 4 marks

  • 1 mark

    A formula: [Co(H2O)6]2+

  • 1 mark

    A colour: pink

  • 1 mark

    B formula: [Co(H2O)4(OH)2] / Co(OH)2 and colour: blue

  • 1 mark

    C formula: [Co(NH3)6]2+, colour: straw / yellow-brown, type of reaction for B and C

2a(ii) · 1 mark

  • 1 mark

    concentrated HCl

2b(i) · 2 marks

  • 1 mark

    five (3d) orbitals (lines, boxes) in the isolated Co2+ ion of same energy

  • 1 mark

    splitting three higher and two lower d orbitals AND energy of all d orbitals in the complex clearly higher than all d orbitals in isolated ion

2b(ii) · 1 mark

  • 1 mark

    correct 3D diagram showing tetrahedral geometry of [CoCl4]2–

2c · 4 marks

  • M1

    H2O2 acting as oxidising agent with Mn(OH)2 / Mn2+

  • M2

    Mn2+ + H2O2 → MnO2 + 2H+ AND E⦵cell = +0.56 OR Mn(OH)2 + H2O2 → MnO2 + 2H2O AND E⦵cell = +0.92

  • M3

    H2O2 acting as reducing agent with MnO2

  • M4

    MnO2 + 2H+ + H2O2 → Mn2+ + 2H2O + O2 AND E⦵cell = +0.54

2d(i) · 1 mark

  • 1 mark

    colourless to pale pink

2d(ii) · 4 marks

  • M1

    moles MnO4– in 25.0 cm3 and moles Fe2+ in 25.0 cm3

  • M2

    moles Fe2+ in 150.0 cm3 = 2.79 × 10-2

  • M3

    mass Fe = 55.8 × 2.79 × 10-2 = 1.55682 (g) in two tablets

  • M4

    mass Fe = 0.5 × 1.55682 = 0.77841 (g) in one tablet AND mass Fe = 7.7841 × 10–2 × 1000 = 778.4

3a · 1 mark

  • 1 mark

    number of possible arrangements of particles/molecules AND energy in a system

3b(i) · 1 mark

  • 1 mark

    T1 melting OR solid to liquid AND T2 boiling OR liquid to gas

3b(ii) · 1 mark

  • 1 mark

    change in disorder from a liquid to a gas is much bigger than solid to liquid OR change in intermolecular distance between liquid & gas is much bigger than solid to liquid

3c · 3 marks

  • M1

    ΔH⦵r = (3 × –393.5) – ((–824.2) + (3 × –110.5)) = –24.8 (kJ mol–1)

  • M2

    ΔG⦵ = ΔH⦵ – TΔS⦵ AND T= 723 used

  • M3

    (24800 – 36200) ÷ 723 = – [3 × 213.8 + 2 × 27.3) – (87.4) – (3x)] –15.8 = –608.6 + 3x, so 3x = 592.8, so x = 197.6 / 198 (J K–1 mol–1) min 3sf

3d · 2 marks

  • M1

    (as temperature increases) the reaction is less feasible

  • M2

    as ΔG⦵ becomes less negative / (more) positive AND due to TΔS becoming more negative / –TΔS (becoming more) positive

4a(i) · 1 mark

  • 1 mark

    the power/exponent to which a concentration AND of a reactant is raised in the rate equation

4a(ii) · 2 marks

  • 1 mark

    any two correct from the table

  • 1 mark

    all four entries correct

4a(iii) · 1 mark

  • 1 mark

    negative gradient curve

4b(i) · 1 mark

  • 1 mark

    rate = 4 × 4 = 16 times

4b(ii) · 1 mark

  • 1 mark

    slowest step / slow step in the mechanism / multi-step reaction

4b(iii) · 2 marks

  • M1

    step 1 O3 + NO2 → NO3 + O2

  • M2

    step 2 NO3 + NO2 → N2O5 AND sum of step 1 and 2 is consistent with equation for reaction 2

4c(i) · 1 mark

  • 1 mark

    Use of graph k = 12 × 10–4 / 0.04 = 0.03

4c(ii) · 1 mark

  • 1 mark

    k = 0.693 / t1/2 and t1/2 = 0.693 / 0.030 = 23.1 (s) min 2sf

5a(i) · 1 mark

  • 1 mark

    acid I HClO conjugate base of acid I ClO– AND acid II H2O conjugate base of acid II OH–

5a(ii) · 2 marks

  • M1

    [H+] = 10–4.51 = 3.09 × 10–5

  • M2

    [HClO] = (3.09 × 10–5)2 ÷ (3.70 × 10–8) = 0.0258 min 2sf

5a(iii) · 1 mark

  • 1 mark

    3 HClO → HClO3 + 2 HCl

5b(i) · 1 mark

  • 1 mark

    (a solution that) resists / minimises changes pH AND when small amounts of acid / H+ and base / OH– are added to it

5b(ii) · 1 mark

  • 1 mark

    hydroxide ions / NaOH / OH– AND it reacts with the acid to form the conjugate base / ethanoate ions OR ethanoate ions / sodium ethanoate AND as it is the conjugate base (of ethanoic acid)

5c · 2 marks

No marking points are listed for this part.

5d(i) · 3 marks

  • M1

    Use of [Ca2+] = [PO43–] = 1.14 × 10–7 in the expression OR Ksp = (3.42 × 10–7)3 × (2.28 × 10–7)2

  • M2

    Ksp = (3 × 1.14 × 10–7)3 × (2 × 1.14 × 10–7)2 = 2.079 × 10–33 OR 2.08 × 10–33 OR 2.1 × 10–33 min 2sf

  • M3

    units = mol5 dm–15

5d(ii) · 1 mark

  • 1 mark

    solubility decreases AND due to the common ion effect

6a(i) · 1 mark

  • 1 mark

    HNO3 + H2SO4 → NO2+ + HSO4– + H2O OR HNO3 + 2H2SO4 → NO2+ + 2HSO4– + H3O+

6a(ii) · 3 marks

  • M1

    curly arrow 1 from inside the hexagon to nitrogen of NO2+

  • M2

    intermediate, usual rules for horseshoe and + charge

  • M3

    curly arrow 2 from C-H bond into hexagon AND H+

6b · 2 marks

  • M1

    p-orbital/lone pair from O / oxygen AND is delocalised / overlaps AND into the ring / π system

  • M2

    so electron density of the ring is increased OR polarises electrophiles / NO2+ better

6c · 1 mark

  • 1 mark

    correct structure of sodium 2-nitrophenolate (with ONa) + 1/2 H2

6d(i) · 1 mark

  • 1 mark

    (hot) tin / Sn AND concentrated AND HCl

6d(ii) · 1 mark

  • 1 mark

    reduction

6d(iii) · 1 mark

  • 1 mark

    structure of 2-nitrophenyl ethanoate

6d(iv) · 1 mark

  • 1 mark

    (nucleophilic) addition – elimination

6e · 4 marks

  • M1

    2-nitrophenol > phenol > water > ethanol

  • M2

    weakens O—H bond / anion stabilised AND proton/H+ (more easily) lost / proton / H+ (more easily) donated

  • M3

    (for 2-nitrophenol / phenol), a p-orbital / lone pair from O / oxygen AND is delocalised / overlaps AND into the ring / π system (and increases electron density in the ring)

  • M4

    (for ethanol), positive inductive effect / electron donating of alkyl group

6f(i) · 2 marks

  • 1 mark

    Any three correct

  • 1 mark

    All six correct

6f(ii) · 1 mark

  • 1 mark

    correct structure showing monobromination on the aromatic ring at the position between the two phenolic OH groups

6f(iii) · 1 mark

  • 1 mark

    correct structure showing substitution of aliphatic OH groups by Cl

7a · 1 mark

  • 1 mark

    rotate the plane of polarised light equally in the opposite direction

7b(i) · 2 marks

  • M1

    nitrogen / N can accept / bond to a proton / H+ ion OR donation of lone pair/electrons on the nitrogen / N to a proton / H+ ion

  • M2

    CH3CH(NH2)CH2CH3 + H2O → CH3CH(NH3+)CH2CH3 + OH–

7b(ii) · 2 marks

  • M1

    more electron donating alkyl / ethyl group on N / NH (in diethylamine) OR two electron donating alkyl / R / alkane / ethyl group on N / NH (in diethylamine)

  • M2

    increase electron density on N OR make lone pair on N more available (to accept H+)

7b(iii) · 1 mark

  • 1 mark

    (CH3CH2)2NH + CH3COOH → (CH3CH2)2NH2+ CH3COO–

7b(iv) · 1 mark

  • 1 mark

    (CH3CH2)2NH + CH3COCl → CH3CON(CH2CH3)2 + HCl

7c(i) · 1 mark

  • 1 mark

    CH2CHCH3COOH: addition, 4-(chlorocarbonyl)phenol: condensation, CH3CH(NH2)COOH: condensation

7c(ii) · 2 marks

  • M1

    correct displayed amide linkage with an adjacent C=O (to CO) and C(H2) to NH

  • M2

    rest of the structure correct (only one repeat unit) with continuation bonds

7c(iii) · 1 mark

  • 1 mark

    they are chemically inert / difficult to hydrolyse / C-C bonds are non-polar

8a · 1 mark

  • 1 mark

    reference: TMS / tetramethylsilane / (CH3)4Si AND solvent: D2O / CDCl3

8b · 1 mark

  • 1 mark

    six / 6 peaks

8c · 4 marks

  • 1 mark

    Any three correct entries in the table

  • 1 mark

    Any six correct entries in the table

  • 1 mark

    Any nine correct entries in the table

  • 1 mark

    All eleven correct entries in the table

9a · 2 marks

  • M1

    correct displayed peptide bond with an adjacent C(H) (to CO) and C(H) to NH

  • M2

    rest of the dipeptide correct

9b(i) · 1 mark

  • 1 mark

    pH at which a molecule has no overall charge / no net charge

9b(ii) · 3 marks

  • M1

    correct relative positions of the spots drawn

  • M2

    ser not moved AND as it is a zwitterion / neutral OR ser-lys / lys AND move towards negative (pole) as they are positively charged

  • M3

    lys moves the furthest / fastest AND as it has the lower Mr / smaller size (than ser-lys)

9c(i) · 1 mark

  • 1 mark

    correct structure of Q, C6H13NO3

9c(ii) · 2 marks

  • M1

    step 1 aqueous AND HCl / H2SO4 AND heat / reflux

  • M2

    step 2 conc. H2SO4 catalyst AND high temp / heat

Questions

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  2. 2
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  3. 3
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  4. 4
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  5. 5
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  6. 6
    Question 619 marks
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  7. 7
    Question 711 marks
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  8. 8
    Question 86 marks
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  9. 9
    Question 99 marks
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