CIE A-Level Chemistry 9701 Paper 4 Practice Paper (A Level Structured) - 100 marks, 2 hours, answers and mark scheme included

CIE A-Level Chemistry 9701 Paper 4 Practice Paper (A Level Structured) – 100 marks, 2 hours, answers and mark scheme included

2026-10-30

TutorZone original practice paper – CIE A-Level Chemistry 9701 Paper 4 (A Level structured)

7 questions, 41 parts – 100 marks – 2 hours – calculator allowed

Total 100 marks · 2 hours · 41 parts

Mark distribution 16 + 14 + 16 + 14 + 12 + 16 + 12 = 100

Level AS (first year of A-Level)

Paper structure

Paper	Time	Marks	Role
Paper 4 (A Level structured)	2 hours	100 marks	this paper
Paper 5 (planning, analysis and evaluation)	1 h 15 min	30 marks	other A Level paper
Paper 2 (AS structured)	1 h 15 min	60 marks	AS paper
Paper 1 (multiple choice)	1 h	40 marks	AS paper

Questions

Total: 100 marks · 2 hours · calculator allowed

Q 1 · Lattice energy and Born-Haber cycles [16]

(a) Define the lattice energy of an ionic compound. [2]

(b) Calculate the lattice energy of sodium chloride from: dHf[NaCl(s)] = -411 kJ/mol, dHatom[Na] = +107, 1/2 dHdiss[Cl2] = +122, IE1[Na] = +496, EA[Cl] = -349. [3]

(c) Data for magnesium oxide: dHf[MgO] = -602, dHatom[Mg] = +148, 1/2 dHdiss[O2] = +249, IE1+IE2[Mg] = 738 + 1451, EA1+EA2[O] = -141 + 798. Calculate the lattice energy of MgO. [3]

(d) Explain why the lattice energy of MgO is far more negative than that of NaCl. [3]

(e) The hydration enthalpies of Na+(g) and Cl-(g) are -406 and -364 kJ/mol. Calculate the standard enthalpy of solution of NaCl(s). [3]

(f) The standard entropy change on dissolving NaCl is +43 J K-1 mol-1. Calculate dG at 298 K in kJ/mol and explain why NaCl still dissolves readily in water. [2]

Q 2 · Electrochemistry and cells [14]

(a) Define the standard electrode potential of a half-cell. [2]

(b) E for Zn2+/Zn is -0.76 V and E for Cu2+/Cu is +0.34 V. Calculate the standard emf of the cell Zn | Zn2+ || Cu2+ | Cu. [2]

(c) Use dG = -nFE to calculate the standard free energy change of this cell (F = 96,500 C/mol), in kJ/mol. [3]

(d) Write the two half-equations and state the direction of electron flow and which electrode is the positive electrode. [3]

(e) If the concentration of Cu2+ is raised from 1.0 to 2.0 mol/dm3 with everything else unchanged, how does the emf change? Explain. [2]

(f) If [Zn2+] = 1.0 mol/dm3 and [Cu2+] = 0.010 mol/dm3 at 298 K, calculate the emf of the cell. [2]

Q 3 · Acid-base equilibria and buffers [16]

(a) Define an acid according to the Bronsted-Lowry theory. [1]

(b) Calculate the pH of 0.100 mol/dm3 ethanoic acid (Ka = 1.80 x 10-5). [3]

(c) A weak acid solution of concentration 0.050 mol/dm3 has pH 3.10. Calculate its Ka. [3]

(d) A solution contains 0.200 mol/dm3 ethanoic acid and 0.100 mol/dm3 sodium ethanoate. Calculate its pH (Ka = 1.80 x 10-5). [3]

(e) Adding a small amount of HCl to this buffer changes the pH only slightly. Explain why. [2]

(f) 25.0 cm3 of 0.100 mol/dm3 ethanoic acid is titrated with 0.100 mol/dm3 NaOH. Calculate the pH at the equivalence point (Kw = 1.00 x 10-14, Ka = 1.80 x 10-5). [4]

Q 4 · Reaction kinetics [14]

(a) Initial rate data for A + B -> products: experiment 1 [A] = 0.100, [B] = 0.100, rate 2.40×10-3 mol dm-3 s-1; experiment 2 [A] = 0.200, [B] = 0.100, rate 4.80×10-3; experiment 3 [A] = 0.100, [B] = 0.200, rate 9.60×10-3. Deduce the order with respect to A and to B and write the rate equation. [4]

(b) Use experiment 1 to calculate the rate constant k, with units. [2]

(c) Calculate the initial rate when [A] = 0.150 and [B] = 0.0500 mol/dm3. [2]

(d) Deduce which species take part in the rate-determining step and justify your answer. [2]

(e) k is 2.40×10-3 dm6 mol-2 s-1 at 300 K and 9.60×10-3 at 320 K. Calculate the activation energy (R = 8.314 J K-1 mol-1). [3]

(f) Use collision theory to explain why the rate constant increases with temperature. [1]

Q 5 · Transition elements and redox titrations [12]

(a) Define a transition element. [2]

(b) Explain why most transition metal complexes are coloured. [2]

(c) Write the ionic equation for the reaction of [Cu(H2O)6]2+ with excess NH3 and describe the colour change. [2]

(d) 25.0 cm3 of a solution containing Fe2+ needs 24.50 cm3 of 0.0200 mol/dm3 KMnO4 in acid. Given MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O, calculate the concentration of Fe2+. [3]

(e) The Fe2+ solution has a total volume of 250 cm3. Calculate the mass of iron it contains (Ar: Fe = 55.8). [2]

(f) Why is no added indicator needed in this titration? [1]

Q 6 · Organic synthesis and structure determination [16]

(a) Starting from ethanol, state the reagents and conditions needed to make ethanal and to make ethanoic acid. [3]

(b) Write the equation for the reaction of ethanol with ethanoic acid to form ethyl ethanoate and state the catalyst and conditions. [2]

(c) 23.0 g of ethanol reacts with excess ethanoic acid to give 30.8 g of ethyl ethanoate. Calculate the percentage yield (Mr: ethanol 46.0, ester 88.0). [3]

(d) Why is the yield below 100%? Suggest one way of improving it. [2]

(e) Compare the products of the acid hydrolysis and the alkaline hydrolysis of ethyl ethanoate. [2]

(f) Compound X contains only C, H and O, has Mr = 74.0, and contains 48.6% carbon and 8.1% hydrogen. Its IR spectrum has absorptions at 1710 cm-1 and a broad band from 2500 to 3300 cm-1; its 1H NMR spectrum shows a triplet, a quartet and one exchangeable singlet. Deduce the structure of X and justify your answer. [4]

Q 7 · Chemistry of nitrogen and sulfur [12]

(a) Write the equation for the catalytic oxidation of ammonia to nitrogen monoxide and state the industrial conditions. [2]

(b) 100 kg of ammonia is completely oxidised to NO. Calculate the mass of NO obtained (Mr: NH3 17.0, NO 30.0). [3]

(c) The NO is oxidised to NO2, which is absorbed in water: 3NO2 + H2O -> 2HNO3 + NO. Calculate the maximum mass of nitric acid obtainable from that NO (Mr: HNO3 63.0). [3]

(d) Explain why the NO leaving the absorption tower is returned to the oxidation tower. [2]

(e) Flue gas contains 640 kg of SO2, removed with CaO: CaO + SO2 -> CaSO3. Calculate the minimum mass of CaO required (Mr: SO2 64.1, CaO 56.1). [2]

Answers and mark scheme

Part	Answer	Marking points	marks
1(a)	The enthalpy change when 1 mol of an ionic compound is formed from its gaseous ions under standard conditions (dHlatt is negative).	1 mark: 1 mol of ionic compound formed from its gaseous ions 1 mark: exothermic, so dH is negative	2
1(b)	dHlatt = -411 – (107 + 122 + 496 – 349) = -411 – 376 = -787 kJ/mol	1 mark: signs of all Born-Haber terms correct 1 mark: 107 + 122 + 496 – 349 = 376 1 mark: -787 kJ/mol	3
1(c)	Sum = 148 + 249 + 2189 + 657 = 3243; dHlatt = -602 – 3243 = -3845 kJ/mol	1 mark: IE total 2189 and EA total 657 1 mark: 148 + 249 + 2189 + 657 = 3243 1 mark: -3845 kJ/mol	3
1(d)	The charges on Mg2+ and O2- are twice those on Na+ and Cl-, so the product of charges rises by a factor of 4; the ions are also smaller, so the inter-ionic distance is shorter and the electrostatic attraction is much stronger.	1 mark: ions carry 2+ and 2- charges 1 mark: charge product about 4 times larger 1 mark: smaller ions give a shorter distance	3
1(e)	dHsol = -dHlatt + dHhyd(Na+) + dHhyd(Cl-) = 787 – 406 – 364 = +17 kJ/mol (endothermic)	1 mark: breaking the lattice costs +787 kJ/mol 1 mark: -406 – 364 1 mark: +17 kJ/mol	3
1(f)	dG = 17 – 298 x 43/1000 = +4.2 kJ/mol, which is positive, so solution is not favoured at standard conditions; in practice the solution is unsaturated, and the large entropy of mixing drives dG negative, so NaCl still dissolves.	1 mark: dG = +4.2 kJ/mol 1 mark: positive at standard conditions, but entropy of mixing in dilute solution makes dG negative	2
2(a)	The emf of a cell made by joining the half-cell to a standard hydrogen electrode at 298 K with ion concentrations of 1 mol/dm3 (gas pressure 1 atm), the hydrogen electrode being taken as 0.00 V.	1 mark: standard conditions, 1 mol/dm3, 298 K, 1 atm 1 mark: measured against the standard hydrogen electrode taken as 0.00 V	2
2(b)	Ecell = E(cathode) – E(anode) = 0.34 – (-0.76) = +1.10 V	1 mark: Ecell = Ecathode – Eanode 1 mark: +1.10 V	2
2(c)	n = 2; dG = -2 x 96,500 x 1.10 = -212,300 J/mol = -212 kJ/mol	1 mark: n = 2 1 mark: -2 x 96,500 x 1.10 1 mark: -212 kJ/mol	3
2(d)	Negative electrode (oxidation): Zn -> Zn2+ + 2e-; positive electrode (reduction): Cu2+ + 2e- -> Cu; electrons flow from zinc to copper through the external circuit, and copper is the positive electrode.	1 mark: Zn -> Zn2+ + 2e- (oxidation) 1 mark: Cu2+ + 2e- -> Cu (reduction) 1 mark: electrons Zn -> Cu and copper is positive	3
2(e)	The emf increases. By the Nernst equation E = E – (0.059/n)log([Zn2+]/[Cu2+]), a larger [Cu2+] makes the log term more negative, so E rises as the copper half-cell potential increases.	1 mark: emf increases 1 mark: Nernst equation; the reduction potential of the copper half-cell rises	2
2(f)	E = 1.10 – (0.059/2)log(1.0/0.010) = 1.10 – 0.059 = +1.041 V	1 mark: substitution with log 100 = 2 1 mark: +1.041 V	2
3(a)	A species that donates a proton (H+) to another species.	1 mark: a proton (H+) donor	1
3(b)	[H+] = root(1.80×10-5 x 0.100) = 1.34×10-3 mol/dm3; pH = 2.87	1 mark: Ka = [H+]^2/[HA] with [H+] = [A-] 1 mark: [H+] = 1.34×10-3 mol/dm3 1 mark: pH = 2.87	3
3(c)	[H+] = 10^-3.10 = 7.94×10-4; Ka = (7.94×10-4)^2/(0.050 – 7.94×10-4) = 1.28×10-5	1 mark: [H+] = 7.94×10-4 mol/dm3 1 mark: denominator 0.050 – [H+] 1 mark: Ka about 1.28×10-5	3
3(d)	[H+] = Ka x [acid]/[salt] = 1.80×10-5 x 0.200/0.100 = 3.60×10-5 mol/dm3; pH = 4.44	1 mark: buffer equation [H+] = Ka x [HA]/[A-] 1 mark: [H+] = 3.60×10-5 mol/dm3 1 mark: pH = 4.44	3
3(e)	The added H+ is removed by combining with ethanoate ions to form the weak acid HA, so the ratio [HA]/[A-] changes only slightly and the pH hardly moves.	1 mark: H+ reacts with A- to form HA 1 mark: the ratio [HA]/[A-] is almost unchanged	2
3(f)	At the equivalence point the total volume is 50.0 cm3 so [CH3COO-] = 0.0500 mol/dm3; Kb = Kw/Ka = 5.56×10-10; [OH-] = root(5.56×10-10 x 0.0500) = 5.27×10-6 mol/dm3; pOH = 5.28 and pH = 8.72	1 mark: [CH3COO-] = 0.0500 mol/dm3 1 mark: Kb = Kw/Ka = 5.56×10-10 1 mark: [OH-] = 5.27×10-6 mol/dm3 1 mark: pH = 8.72	4
4(a)	Doubling [A] doubles the rate, so first order in A; doubling [B] increases the rate fourfold, so second order in B; rate = k[A][B]^2	1 mark: experiments 1 and 2 give first order in A 1 mark: experiments 1 and 3 give a fourfold rate increase 1 mark: second order in B 1 mark: rate = k[A][B]^2	4
4(b)	k = 2.40×10-3/(0.100 x 0.100^2) = 2.4 dm6 mol-2 s-1	1 mark: correct substitution giving 2.4 1 mark: units dm6 mol-2 s-1	2
4(c)	rate = 2.4 x 0.150 x (0.0500)^2 = 9.0×10-4 mol dm-3 s-1	1 mark: correct substitution 1 mark: 9.0×10-4 mol dm-3 s-1	2
4(d)	The rate-determining step involves one A and two B particles, since two B particles must collide; the orders equal the number of each species in the rate-determining step, one for A and two for B.	1 mark: one A and two B in the rate-determining step 1 mark: orders equal particle numbers in that step	2
4(e)	Ea = R ln(k2/k1)/(1/T1 – 1/T2) = 8.314 x ln 4/(1/300 – 1/320) = 55.3 kJ/mol	1 mark: two-point Arrhenius equation 1 mark: ln 4 = 1.386 and 1/300 – 1/320 = 2.08×10-4 1 mark: 55.3 kJ/mol	3
4(f)	At a higher temperature the molecules have greater average kinetic energy, so a much larger fraction has energy at or above Ea; successful collisions are therefore more frequent and k increases.	1 mark: larger fraction of molecules with energy at or above Ea	1
5(a)	A d-block element that forms at least one ion in which the d subshell is incompletely filled.	1 mark: a d-block element 1 mark: forms an ion with an incomplete d subshell	2
5(b)	Ligands split the five d orbitals into groups of different energy. Electrons absorb visible light and are promoted between the split d orbitals, a d-d transition; the light not absorbed is transmitted or reflected and gives the observed colour.	1 mark: ligand field splits the d orbitals 1 mark: d-d transition absorbs visible light, the remainder gives the colour	2
5(c)	[Cu(H2O)6]2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]2+ + 4H2O; the colour changes from pale blue to deep blue.	1 mark: correct equation with coordination number 4 1 mark: pale blue to deep blue	2
5(d)	n(MnO4-) = 0.0200 x 0.02450 = 4.90×10-4 mol; n(Fe2+) = 5 x 4.90×10-4 = 2.45×10-3 mol; [Fe2+] = 2.45×10-3/0.0250 = 0.0980 mol/dm3	1 mark: n(MnO4-) = 4.90×10-4 mol 1 mark: n(Fe2+) = 2.45×10-3 mol (factor 5) 1 mark: [Fe2+] = 0.0980 mol/dm3	3
5(e)	n(Fe2+) = 2.45×10-3 x 10 = 0.0245 mol; m = 0.0245 x 55.8 = 1.37 g	1 mark: n = 0.0245 mol 1 mark: 1.37 g	2
5(f)	The MnO4- ion is purple and is reduced to almost colourless Mn2+; at the end point one excess drop leaves a permanent pale pink colour, so the reagent acts as its own indicator.	1 mark: purple MnO4- reduced to colourless Mn2+, permanent pale pink at the end point	1
6(a)	Ethanal: acidified K2Cr2O7 with heating, distilling off the aldehyde at once to prevent further oxidation; ethanoic acid: excess acidified K2Cr2O7 heated under reflux.	1 mark: acidified dichromate, K2Cr2O7/H+ 1 mark: distillation used to remove ethanal at once 1 mark: excess oxidising agent under reflux for the acid	3
6(b)	CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O; concentrated sulfuric acid catalyst with heating.	1 mark: correct reversible equation 1 mark: concentrated H2SO4 catalyst and heat	2
6(c)	n(ethanol) = 23.0/46.0 = 0.500 mol; theoretical mass = 0.500 x 88.0 = 44.0 g; yield = 30.8/44.0 x 100 = 70.0%	1 mark: n = 0.500 mol 1 mark: theoretical mass 44.0 g 1 mark: 70.0%	3
6(d)	Esterification is reversible, so reactants remain at equilibrium. Water formed can be removed continuously with a Dean-Stark trap, or the ester distilled off, or excess ethanol used, to shift the equilibrium to the right.	1 mark: the reaction is reversible and reaches equilibrium 1 mark: remove water or ester, or use excess reactant	2
6(e)	Acid hydrolysis gives ethanoic acid and ethanol and is reversible; alkaline hydrolysis with NaOH(aq) gives sodium ethanoate and ethanol and is not reversible, being saponification.	1 mark: acid hydrolysis gives ethanoic acid plus ethanol 1 mark: alkaline hydrolysis gives the ethanoate salt plus ethanol and is not reversible	2
6(f)	C : H : O = 48.6/12.0 : 8.1/1.0 : 43.3/16.0 = 4.05 : 8.1 : 2.71, giving C3H6O2, and Mr 74.0 confirms this molecular formula; 1710 cm-1 is C=O and the broad 2500-3300 cm-1 band is O-H, so a carboxylic acid is present; the triplet and quartet show a CH3CH2 group with adjacent coupling; therefore X is propanoic acid, CH3CH2COOH.	1 mark: empirical formula C3H6O2 consistent with Mr 74.0 1 mark: 1710 cm-1 is C=O and 2500-3300 cm-1 is the acid O-H 1 mark: triplet plus quartet indicates CH3CH2 with adjacent coupling 1 mark: propanoic acid, CH3CH2COOH	4
7(a)	4NH3 + 5O2 -> 4NO + 6H2O, with a platinum-rhodium catalyst at about 900 C.	1 mark: correctly balanced equation 1 mark: Pt-Rh catalyst at about 900 C	2
7(b)	n(NH3) = 100,000/17.0 = 5,882 mol; n(NO) is the same; m = 5,882 x 30.0 = 176,471 g, about 176 kg	1 mark: n(NH3) = 5,882 mol 1 mark: n(NO) = n(NH3), 1:1 1 mark: about 176 kg	3
7(c)	n(NO2) = 5,882 mol; n(HNO3) = 2/3 x 5,882 = 3,921 mol; m = 3,921 x 63.0 = 247,023 g, about 247 kg	1 mark: n(HNO3) = 2/3 n(NO2) = 3,921 mol 1 mark: multiplied by 63.0 1 mark: about 247 kg	3
7(d)	The absorption step converts only two thirds of the NO2 into HNO3, releasing the rest as NO; recycling and re-oxidising that NO raises the utilisation of ammonia and the yield of nitric acid.	1 mark: the absorption reaction releases NO 1 mark: recycling improves utilisation and yield	2
7(e)	n(SO2) = 640,000/64.1 = 9,984 mol; m(CaO) = 9,984 x 56.1 = 560,102 g, about 560 kg	1 mark: n(SO2) = 9,984 mol 1 mark: about 560 kg of CaO	2

Topic map

Question	Topic	Marks
1	Lattice energy and Born-Haber cycles	16
2	Electrochemistry and cells	14
3	Acid-base equilibria and buffers	16
4	Reaction kinetics	14
5	Transition elements and redox titrations	12
6	Organic synthesis and structure determination	16
7	Chemistry of nitrogen and sulfur	12

Common lost marks

Defining lattice energy without saying the ions are gaseous.
Reversing the sign of an ionisation energy or electron affinity term in a Born-Haber cycle.
Inverting the acid to salt ratio in a buffer, or confusing pH with pOH.
Omitting the units of a second order rate constant, dm6 mol-2 s-1.
Forgetting the 1 to 5 ratio between MnO4- and Fe2+ in the titration.

How to use this paper Sit the paper closed-book in 75 minutes, then mark point by point with the answer booklet and redo every lost mark.

Syllabus reference Format follows the CIE 9701 Chemistry syllabus (2025-2027) Paper 2: 2 h, 100 marks, A Level structured questions.

FAQ

Who is this paper for?
Students who have finished the AS year and are preparing for CIE A-Level Chemistry 9701 Paper 4, especially those losing marks on long structured questions on lattice energy, electrochemistry and organic synthesis.

Does the answer booklet include a mark scheme?
Yes. Every part lists its marking points so you can mark yourself or a tutor can mark point by point.

Can I print it?
Yes, both the question paper and the answer booklet are A4 PDFs ready to print.

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This is original TutorZone practice material, not a past examination paper.

Format follows the CIE 9701 Chemistry syllabus (2025-2027) Paper 4: 2 hours, 100 marks, A Level structured questions.