A Level Chemistry 9701 Paper 41, May/June 2025

Question paper and mark scheme100 marks10 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.

Opening the question paper

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

Question paper

9701/41 · May/June 2025

Download PDF
1

Both calcium carbonate, CaCO3, and barium carbonate, BaCO3, decompose when heated to form the metal oxide and a gas.

a

Write an equation for the thermal decomposition of CaCO3.

[1]
b

State which of CaCO3 and BaCO3 decomposes at a lower temperature. Explain your answer. The compound that decomposes at a lower temperature is ................................................... explanation ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

[2]
c(i)

Calcium oxide, CaO, reacts with water to form compound A. Barium oxide, BaO, reacts with water to form compound B. Identify A.

[1]
c(ii)

Explain why A is less soluble than B.

[3]
2

Three experiments are carried out to investigate the reaction of nitrogen oxide, NO, with chlorine. 2NO+Cl2→2NOCl The rate equation for this reaction is shown. rate=k[NO]2[Cl2]

a(i)

Under the conditions used in experiments 1 and 2, the value of k is 26.4. The rate of the reaction is measured in mol dm−3 s−1. State the units of k. units of k = ..............................

[1]
a(ii)

In experiment 1, the initial concentrations of NO and Cl2 are equal. The initial rate of the reaction in experiment 1 is 2.57×10−6mol dm−3 s−1. Calculate the initial concentration of NO. Show your working. initial concentration of NO = .............................. mol dm−3

[2]
a(iii)

In experiment 2, the initial concentrations of NO and Cl2 are both ten times greater than the initial concentrations used in experiment 1. Calculate the initial rate of the reaction in experiment 2. initial rate of reaction in experiment 2 = .............................. mol dm−3 s−1

[1]
b(i)

Experiment 3 uses a large excess of NO. The initial concentration of Cl2 is 2.00×10−4mol dm−3. The graph of [Cl2] against time shows that the reaction has a constant half-life, t1/2. Explain this observation.

[1]
b(ii)

Under the conditions used in experiment 3, the value of the rate constant is 105.6. Show that t1/2 of [Cl2] is 6.56×10−3s under these conditions.

[1]
b(iii)

Calculate the time taken, in s, for [Cl2] to fall to 1.25×10−5mol dm−3 in experiment 3. time = .............................. s

[1]
c

Sulfur dioxide, SO2, reacts very slowly with oxygen in the atmosphere, forming sulfur trioxide, SO3. This reaction is much faster in the presence of NO. Explain the role of NO in this process. Include chemical equations in your answer.

[3]
3
a(i)

Chromium(III) hydroxide, Cr(OH)3, is only slightly soluble in water. The value of the solubility product, Ksp, of Cr(OH)3 is 1.0×10−33 at 298K. Complete the expression for Ksp of Cr(OH)3. Include the units. Ksp= units = ..............................

[2]
a(ii)

Calculate the solubility, in g dm−3, of Cr(OH)3 in pure water at 298K. Show your working. solubility = .............................. g dm−3

[3]
a(iii)

Cr(OH)3 is less soluble in 0.100mol dm−3 NaOH than it is in pure water. Explain this observation.

[1]
b(i)

The value of the acid dissociation constant, Ka, of butanoic acid, CH3CH2CH2COOH, is 1.51×10−5 at 298K. Calculate the pH of 0.100mol dm−3 CH3CH2CH2COOH at 298K. Show your working. pH = ..............................

[2]
b(ii)

Calculate the pH of 0.100mol dm−3 NaOH at 298K. pH = ..............................

[1]
b(iii)

5.00cm3 of 0.100mol dm−3 NaOH is added to 10.00cm3 of 0.100mol dm−3 CH3CH2CH2COOH. Calculate the pH of the resulting solution. Show your working. pH = ..............................

[2]
c

80.0cm3 of an aqueous solution containing 0.704g of CH3CH2CH2COOH is shaken with 100cm3 of benzene, C6H6. There is 0.556g of CH3CH2CH2COOH in the 100cm3 of C6H6 at equilibrium. Calculate the partition coefficient, Kpc, of CH3CH2CH2COOH between C6H6 and water. Show your working. Kpc = ..............................

[2]
4

Table 4.1 gives the enthalpy changes of hydration, ΔHhyd, of three ions, F−, K+ and Ca2+. Table 4.1

ionΔHhyd/kJ mol−1
F−−506
K+−322
Ca2+−1650
a(i)

Define enthalpy change of hydration.

[1]
a(ii)

Explain the relative magnitudes of the enthalpy changes of hydration of K+ and Ca2+.

[2]
a(iii)

Define lattice energy.

[1]
a(iv)

The lattice energy, ΔHlatt, of calcium fluoride, CaF2, is −2602kJ mol−1. Calculate the enthalpy change of solution, ΔHsol, in kJ mol−1, of CaF2. ΔHsol of CaF2 = .............................. kJ mol−1

[2]
b

The formation of CaF2 at 298K is shown. Ca(s)+F2(g)→CaF2(s)ΔH⊖=−1214kJ mol−1,ΔG⊖=−1162kJ mol−1 Calculate the entropy change, ΔS⊖, in J K−1 mol−1, for this reaction. ΔS⊖ = .............................. J K−1 mol−1

[2]
5

Copper is a transition element.

a(i)

Complete the electronic configurations of a Cu+ ion and a Cu2+ ion. Cu+ ion: [Ar] .............................. Cu2+ ion: [Ar] ..............................

[1]
a(ii)

Explain why transition elements have variable oxidation states.

[1]
b(i)

Aqueous copper(II) sulfate, CuSO4, contains the [Cu(H2O)6]2+ complex ion. A few drops of NH3(aq) are added to CuSO4(aq). Describe any observations made.

[1]
b(ii)

Write an equation for the reaction taking place.

[1]
c(i)

An excess of NH3(aq) is added to CuSO4(aq). Describe any further observations made.

[1]
c(ii)

Write an equation for the reaction taking place.

[1]
c(iii)

State the name for the type of reaction taking place.

[1]
d(i)

Copper metal can be oxidised by acidified KMnO4. The relevant half-equations and their standard electrode potentials, E⊖, are shown. Cu2++2e−⇌CuE⊖=+0.34V MnO4−+8H++5e−⇌Mn2++4H2OE⊖=+1.52V A MnO4−/Mn2+ electrode is constructed using 0.0020mol dm−3 MnO4−, 1.0mol dm−3 Mn2+ and 1.0mol dm−3 H+. The temperature used is 298K. Use the Nernst equation to show that the E value for this MnO4−/Mn2+ electrode is +1.49V.

[2]
d(ii)

An electrochemical cell is constructed using a standard Cu2+/Cu electrode and the MnO4−/Mn2+ electrode described in (d)(i). Calculate the value of Ecell. Ecell = .............................. V

[1]
d(iii)

Write an equation for the reaction taking place in the electrochemical cell described in (d)(ii).

[1]
d(iv)

Complete the sentences for the electrochemical cell described in (d)(ii). The .............................. electrode is the negative electrode. Electrons flow from the .............................. electrode to the .............................. electrode when the cell is in use.

[1]
e

A solution containing [Cu(H2O)6]2+ is electrolysed for 5.00 hours using a constant electric current. 0.764g of copper metal is formed at the cathode. No other reduction reaction takes place. Calculate the electric current, in A, used. Give your answer to three significant figures. current = .............................. A

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

Iron forms complex ions with the monodentate ligand, CN−. Complex ion A contains one Fe3+ ion and six CN− ligands. Complex ion B contains one Fe2+ ion and six CN− ligands. State the formulae of these two complex ions. Include the overall charge of each complex ion. complex ion A ................................................. complex ion B .................................................

[1]
a(ii)

Explain why a solution containing complex ion A and a solution containing complex ion B are different colours.

[2]
a(iii)

In complex ion A, the carbon atom of each CN− ligand bonds to the Fe3+ ion. State the type of bonding involved.

[1]
b

Complex ions have different geometries. Complex ion A is octahedral. Ag+ ions form a linear complex with ammonia. Ni atoms form a tetrahedral complex with carbon monoxide molecules. The carbon atom in the monodentate carbon monoxide ligand bonds to the nickel atom. Pd2+ ions form a square planar complex with chloride ions. Complete Fig. 6.1 to show the geometry of each of these four ions, using three-dimensional bonds where necessary. Label one bond angle on each complex ion.

[4]
7

A hydrocarbon is known to be either compound D or compound E.

Figure 1 of 9 for question 7
Figure 2 of 9 for question 7
Figure 3 of 9 for question 7
Figure 4 of 9 for question 7
Figure 5 of 9 for question 7
Figure 6 of 9 for question 7
Figure 7 of 9 for question 7
Figure 8 of 9 for question 7
Figure 9 of 9 for question 7
a

Give the systematic name of E.

[1]
b(i)

The proton (1H) NMR spectra of D and E are compared. They are very similar. The proton (1H) NMR spectrum of D is shown in Fig. 7.2. Suggest a suitable solvent for obtaining the spectrum in Fig. 7.2.

[1]
b(ii)

The proton (1H) NMR spectrum of E is obtained twice, once before and once after shaking with D2O. Describe any differences between these two spectra. Explain your answer.

[1]
b(iii)

Complete Table 7.1 for the proton (1H) NMR spectrum of D. Table 7.1

chemical shift δ/ppmnumber of 1H atoms responsible for the peakgroup responsible for the peaksplitting pattern
1.3
2.7
7.1
[3]
b(iv)

Compounds D and E can be distinguished by carbon-13 NMR spectroscopy. State the number of peaks in each spectrum. The carbon-13 NMR spectrum of D has .............................. peaks. The carbon-13 NMR spectrum of E has .............................. peaks.

[1]
c(i)

Compound D can be oxidised to compound F by alkaline KMnO4 followed by dilute acid. Write an equation for this reaction using molecular formulae for D and F. The products of this reaction are F, water and carbon dioxide. Use [O] to represent one atom of oxygen from the oxidising agent.

[1]
c(ii)

F reacts with an excess of SOCl2 to form compound G. The molecular formula of G is C8H4Cl2O2. G reacts with ethane-1,2-diol, HOCH2CH2OH, to form a mixture of products that includes compounds J, molecular formula C10H8O4, and K, molecular formula C18H12Cl2O6. Draw the structures of compounds G, J and K in Fig. 7.4.

[3]
8

Bromine reacts with methylbenzene in the dark in the presence of a suitable catalyst to form HBr and compound L, C7H7Br. L is one of three isomers that can form in this reaction.

Figure 1 of 1 for question 8
a(i)

The mechanism for the reaction involves methylbenzene reacting with a Br+ ion. This ion is produced when bromine reacts with the catalyst. Complete the equation for the reaction of Br2 with the catalyst. Br2+..............................→Br++..............................

[1]
a(ii)

One of the isomers of L forms in much smaller amounts than the other two isomers. Draw the structure of this isomer and explain why it forms in the smallest amount.

[2]
a(iii)

Complete the mechanism in Fig. 8.1 for the reaction between methylbenzene and the Br+ ion. Include all relevant curly arrows and charges.

[3]
b(i)

Chlorobutane and chlorobenzene are added separately to samples of warm aqueous AgNO3. One of the chloro-compounds reacts slowly and the other does not react. Identify the chloro-compound that reacts and describe any observations.

[1]
b(ii)

Write two equations to explain any observations in (b)(i).

[2]
b(iii)

Explain the difference in reactivity of chlorobutane and chlorobenzene with warm aqueous AgNO3.

[2]
9
a(i)

Phenylamine, C6H5NH2, and propylamine, CH3CH2CH2NH2, can be produced by different reduction reactions. Identify an organic compound that can be converted into C6H5NH2 by a reduction reaction. State the reagents and conditions for this reaction. organic compound ............................................................................................................. reagents ............................................................................................................................ conditions ..........................................................................................................................

[2]
a(ii)

Identify an organic compound that can be converted into CH3CH2CH2NH2 by a reduction reaction. State the reagent for this reaction. organic compound ............................................................................................................. reagent ..............................................................................................................................

[2]
a(iii)

Identify a single test that will distinguish between C6H5NH2 and CH3CH2CH2NH2 by producing a white precipitate with only one of these amines. Draw the structure of the compound that is precipitated. testing reagent .................................................................................................................. amine that gives a precipitate ........................................................................................... structure of the compound that is precipitated:

[2]
b

Describe the relative basicities of C6H5NH2, CH3CH2CH2NH2 and NH3. Explain your answer. .............................................. < .............................................. < .............................................. least basic most basic

[4]
10

Propanoic acid, methanoic acid and ethanedioic acid are all weak acids.

a

Draw the displayed formula of ethanedioic acid.

[1]
b

The three acids, propanoic acid, methanoic acid and ethanedioic acid, can be distinguished using a combination of two chemical tests. Neither testing reagent is an acid–base indicator. Identify two suitable testing reagents and complete Table 10.1 to show the observations from each test. reagent 1 ........................................................ reagent 2 ........................................................ Table 10.1

observation when treated with reagent 1observation when treated with reagent 2
propanoic acid
methanoic acid
ethanedioic acid
[4]
c

When propanoic acid is treated with chlorine gas in the presence of ultraviolet light, a mixture of products is formed. One of these products is 2,2-dichloropropanoic acid. Explain why 2,2-dichloropropanoic acid is stronger than propanoic acid. Refer to the structure of each compound in your answer.

[2]

Mark scheme

9701/41 · May/June 2025

Download PDF

1a · 1 mark

  • 1 mark

    CaCO3 → CaO + CO2

1b · 2 marks

  • M1

    calcium carbonate AND calcium ion / Ca2+ smaller / ionic radius bigger down group

  • M2

    carbonate ion / anion / CO32– is more distorted/polarised by Ca2+

1c(i) · 1 mark

  • 1 mark

    calcium hydroxide / Ca(OH)2

1c(ii) · 3 marks

  • M1

    A / calcium hydroxide has more exothermic ΔHlatt AND ΔHhyd than B / barium hydroxide

  • M2

    difference in ΔHlatt is greater for A / calcium hydroxide than for B / barium hydroxide

  • M3

    ΔHsol is less exothermic for A / calcium hydroxide than for B / barium hydroxide

2a(i) · 1 mark

  • 1 mark

    mol–2 dm6 s–1

2a(ii) · 2 marks

  • M1

    2.57 × 10–6 = 26.4 × [concentration]3

  • M2

    [concentration] = 4.60 × 10–3

2a(iii) · 1 mark

  • 1 mark

    2.57 × 10–3

2b(i) · 1 mark

  • 1 mark

    overall 1st order / pseudo 1st order OR 1st order with respect to Cl2 AND concentration of NO doesn’t change

2b(ii) · 1 mark

  • 1 mark

    ln2 / 105.6 or 0.693 / 105.6

2b(iii) · 1 mark

  • 1 mark

    0.0262 / 2.62 × 10–2

2c · 3 marks

  • M1

    homogeneous catalyst OR catalyst because it is regenerated

  • M2

    2NO + O2 → 2NO2

  • M3

    NO2 + SO2 → NO + SO3

3a(i) · 2 marks

  • M1

    Ksp = [Cr3+] [OH–]3

  • M2

    mol4 dm–12

3a(ii) · 3 marks

  • M1

    [OH–] = 3[Cr3+]

  • M2

    2.47 × 10–9mol dm–3

  • M3

    2.54 × 10–7 g dm–3

3a(iii) · 1 mark

  • 1 mark

    common ion effect

3b(i) · 2 marks

  • M1

    [H+] = 1.23 × 10–3

  • M2

    pH = 2.91

3b(ii) · 1 mark

  • 1 mark

    13.0

3b(iii) · 2 marks

  • M1

    [CH3CH2CH2COOH] = [CH3CH2CH2COO–]

  • M2

    pH = 4.82

3c · 2 marks

  • M1

    0.148 g is left in water

  • M2

    Kpc = 3.01 OR 0.333

4a(i) · 1 mark

  • 1 mark

    energy change when one mole of gaseous ions dissolves in water

4a(ii) · 2 marks

  • M1

    K+ has larger radius AND K+ has smaller charge

  • M2

    Ca2+ has larger attraction for H2O

4a(iii) · 1 mark

  • 1 mark

    energy change when one mole of ionic solid forms from gaseous ions

4a(iv) · 2 marks

  • M1

    –1650–(2 × 506) + 2602

  • M2

    –60

4b · 2 marks

  • M1

    ΔG = ΔH –TΔS AND T = 298

  • M2

    –174.5

5a(i) · 1 mark

  • 1 mark

    (Ar) 3d10 AND (Ar) 3d9

5a(ii) · 1 mark

  • 1 mark

    similar energy of 3d and 4s subshells

5b(i) · 1 mark

  • 1 mark

    pale blue precipitate

5b(ii) · 1 mark

  • 1 mark

    [Cu(H2O)6]2+ + 2OH– → Cu(OH)2(H2O)4 + 2H2O

5c(i) · 1 mark

  • 1 mark

    deep blue solution

5c(ii) · 1 mark

  • 1 mark

    [Cu(H2O)6]2+ + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 4H2O

5c(iii) · 1 mark

  • 1 mark

    ligand exchange

5d(i) · 2 marks

  • M1

    E = Eo +(0.059 / z) log(ox/red)

  • M2

    E = 1.52 + (0.059 / 5) × log (0.002)

5d(ii) · 1 mark

  • 1 mark

    1.15

5d(iii) · 1 mark

  • 1 mark

    5Cu + 2MnO4– + 16H+ → 5Cu2+ + 2Mn2+ + 8H2O

5d(iv) · 1 mark

  • 1 mark

    Cu2+ / Cu, Cu2+ / Cu, MnO4– / Mn2+ / platinum

5e · 3 marks

  • M1

    moles of copper = 0.0120 mol

  • M2

    coulombs required = 2322 C

  • M3

    current required = 0.129 A

6a(i) · 1 mark

  • 1 mark

    [Fe(CN)6]3– AND [Fe(CN)6]4–

6a(ii) · 2 marks

  • M1

    different ΔE

  • M2

    absorption of different frequency of visible light OR wavelength of visible light

6a(iii) · 1 mark

  • 1 mark

    dative covalent / coordinate

6b · 4 marks

  • 1 mark

    correct structure 1 (octahedral Fe(CN)6 with 90° angle)

  • 1 mark

    correct structure 2 (linear Ag(NH3)2 with 180° angle)

  • 1 mark

    correct structure 3 (tetrahedral Ni(CO)4 with 109.5° angle)

  • 1 mark

    correct structure 4 (square planar PdCl4 with 90° angle)

7a · 1 mark

  • 1 mark

    1,3-diethylbenzene

7b(i) · 1 mark

  • 1 mark

    CDCl3

7b(ii) · 1 mark

  • 1 mark

    no difference AND no H that can be exchanged with D

7b(iii) · 3 marks

  • 1 mark

    number of protons column: 6, 4, 4

  • 1 mark

    group responsible column: CH3, CH2

  • 1 mark

    name of splitting pattern column: triplet, quartet / quadruplet

7b(iv) · 1 mark

  • 1 mark

    5 AND 6

7c(i) · 1 mark

  • 1 mark

    C10H14 + 12[O] → C8H6O4 + 2CO2 + 4H2O

7c(ii) · 3 marks

  • 1 mark

    compound G correct structure

  • 1 mark

    compound J correct structure

  • 1 mark

    compound K correct structure

8a(i) · 1 mark

  • 1 mark

    Br2 + AlBr3 → Br+ + AlBr4–

8a(ii) · 2 marks

  • 1 mark

    reference to methyl group having an effect

  • 1 mark

    methyl group is 2,4,6 directing (along with correct structure)

8a(iii) · 3 marks

  • M1

    first curly arrow from inside hexagon towards Br+

  • M2

    correct intermediate

  • M3

    second curly arrow from bond into hexagon AND organic product, H+

8b(i) · 1 mark

  • 1 mark

    chlorobutane AND white precipitate

8b(ii) · 2 marks

  • M1

    C4H9Cl + H2O → C4H9OH + H+ + Cl–

  • M2

    Cl– + Ag+ → AgCl

8b(iii) · 2 marks

  • M1

    delocalisation of lone pair of Cl into benzene ring / π system

  • M2

    strengthens C–Cl bond / causes C–Cl bond to have partial double bond nature

9a(i) · 2 marks

  • 1 mark

    nitrobenzene / C6H5NO2 and Sn and HCl

  • 1 mark

    concentrated HCl and heat / boil / reflux

9a(ii) · 2 marks

  • M1

    propylamide / propenamide / C2H5CONH2 OR propanenitrile / C2H5CN

  • M2

    LiAlH4

9a(iii) · 2 marks

  • M1

    bromine (aq) AND phenylamine / C6H5NH2

  • M2

    correct structure

9b · 4 marks

  • M1

    phenylamine ammonia propylamine

  • M2

    a base has a lone pair that can accept a proton / H+

  • M3

    the lone pair on the nitrogen atom of phenylamine is delocalised into the benzene ring / π-system

  • M4

    propylamine has an electron donating alkyl group

10a · 1 mark

  • 1 mark

    correct structure of ethanedioic acid

10b · 4 marks

  • M1

    reagent 1:Fehling’s OR Tollens’ reagent

  • M2

    acidified MnO4–

  • M3

    One completely correct column

  • M4

    both completely correct columns

10c · 2 marks

  • M1

    chlorine atoms are electron withdrawing / electronegative

  • M2

    weakening O–H bond OR stabilising conjugate base

Questions

  1. 1
    Question 17 marks
    Mark scheme ›
  2. 2
    Question 210 marks
    Mark scheme ›
  3. 3
    Question 313 marks
    Mark scheme ›
  4. 4
    Question 48 marks
    Mark scheme ›
  5. 5
    Question 515 marks
    Mark scheme ›
  6. 6
    Question 68 marks
    Mark scheme ›
  7. 7
    Question 711 marks
    Mark scheme ›
  8. 8
    Question 811 marks
    Mark scheme ›
  9. 9
    Question 910 marks
    Mark scheme ›
  10. 10
    Question 107 marks
    Mark scheme ›

More 9701 papers

Other papers from May/June 2025

Solve it on ExamPalStart