1(a) Hydrogen peroxide reacts both as an oxidising agent and as a reducing agent, depending on the conditions. Hydrogen peroxide reduces potassium dichromate(VI) in acidic solution to chromium(III) ions. Hydrogen peroxide will oxidise chromium(III) ions back to chromate(VI) ions in alkaline solution.

Cr2O72-(aq) + 14H+(aq) + 6e- 2Cr3+(aq) + 7H2O(l) E = +1.33 V

O2(g) + 2H+(aq) + 2e- H2O2(aq) E = +0.68 V

Deduce the overall equation for the reaction between hydrogen peroxide and dichromate(VI) ions using the data above.

Cr2O72- + 3H2O2 + 8H+ 2Cr3+ + 3O2 + 7H2O

.................................................................................................................................................

(b) In alkaline solution the reaction of hydrogen peroxide with chromium(III) ions is

2Cr3+(aq) + 10OH-(aq) + 3H2O2(aq) → 2CrO42-(aq) + 8H2O(l)

This reaction is used to prepare potassium dichromate(VI) from chromium(III) chloride and hydrogen peroxide in potassium hydroxide solution. The mixture is boiled until it is bright yellow. Boiling is continued until excess hydrogen peroxide has been destroyed.The solution is then cooled, and acidified with ethanoic acid.

(i) The reaction occurring on acidification with ethanoic acid is:

2CrO42-(aq) + 2H+(aq) Cr2O7

2-(aq) + H2O(l)

Show that this is not a redox reaction.

CrO42– and Cr2O7

2– both contain Cr(+6), statement that no change implies not a redox reaction.................................................................................................................................................

(ii) Hydrogen peroxide on heating reacts as follows:

2H2O2(aq) 2H2O(l) + O2(g)

Suggest how you would know when all the hydrogen peroxide has been destroyed in the reaction mixture.

no more bubbles/fizzing/effervescence or glowing splint does not relight.................................................................................................................................................

(iii) Why is it essential to destroy all the hydrogen peroxide in the mixture before it is acidified?

Class Reg Number

Candidate Name .......................................................................

Chemistry H2 9746Tutor TuteeRevision Exercise 17: Prelims Paper 2 Revision

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acid converts chromate to dichromate which oxidises H2O2 / reacts with H2O2 / reduced by H2O2. Cr3+ formed / reduces yield.................................................................................................................................................

(iv) Derive the two half equations which together give the overall equation for the reaction between hydrogen peroxide and chromium(III) ions in alkaline solution.

CrO42- + 4H2O + 3e- Cr3+ + 8OH- or Cr3+ + 8OH- CrO4

2- + 4H2O + 3e- H2O2 + 2e- 2OH- or 2OH- H2O2 + 2e-

.................................................................................................................................................

Equilibria a2 Q18 pdf2 Methane reacts with steam in a reversible reaction. In industry this reaction, carried out at a

pressure of 30 atm, is used to produce hydrogen for the manufacture of ammonia

CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ΔH = +210 kJ mol–1

(a)(i) Define the term partial pressure as applied to a gas mixture.

fraction of the total pressure generated by a gas or pressure gas would generate if it alone occupied the volume or Ptotal × mole fraction.................................................................................................................................................

(ii) Write an expression for the equilibrium constant, Kp, for this reaction.

(iii) State and explain the effect of increasing the total pressure on the position of this equilibrium;

Increase in total pressure will result in less product molecules in the equilibrium mixture / equilibrium moves to left by LCP because more molecules on product side (4) of the equilibrium than on left (2) .................................................................................................................................................

(b) State the effect on the value of Kp for this equilibrium of the following.

(i) Increasing the total pressure

No change..................................................................................................................................................

(ii) Increasing the temperature.

Kp increases..................................................................................................................................................

(iii) Adding a catalyst

No change..................................................................................................................................................

(c) There is a theory that methane, CH4, constantly leaks from the earth’s crust. This is not

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noticeable on land but at the bottom of a cold sea, such as off the Canadian coast, the methane is trapped in a solid cage of water molecules.

CH4(g) + 6H2O(s) [CH4(H2O)6](s) methane hydrate

At –29 °C the equilibrium pressure of the methane is 101.3 kPa.

(i) Write an expression for Kp for this equilibrium.

(ii) Deduce the value of Kp at –29°C, stating its units.

9.87 × 10–3 kPa–1/ 9.87 × 10−6 Pa−1

(iii) At 0°C the equilibrium pressure of methane rises to 2600 kPa. What does this tell you about the effect of temperature change on the position of equilibrium and about the enthalpy change for this reaction?

equilibrium has moved left in favour of gas. exothermic going left to right/in the forward direction .................................................................................................................................................

(iv) Some people have suggested collecting the methane hydrate from the bottom of the sea and allowing it to warm up to 0 °C on board a ship. Comment on whether this would be a useful method for collecting methane.

Answer yes or no with some sensible justificatione.g. No the costs would not justify the amount produced.................................................................................................................................................

Equilibria A2 Q31 pdf3(a) The bombardier beetle Metrius contractus persuades potential predators to disappear by

firing a boiling mixture of irritants at them. The reaction producing this ammunition is a redox reaction, H2O2 being the oxidising agent.

The two half-reactions involved are:

H2O2 + 2H+ + 2e- 2H2O E = +1.77 V

(i) Write the overall equation for the reaction and show that the reaction is feasible.

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E = + 2.47V. This is positive, therefore reaction feasible

(ii) The beetle makes use of an enzyme catalyst in the reaction. Explain in general terms how catalysts increase the rate of a chemical reaction using a graph of the Maxwell-Boltzmann distribution of molecular energies.

area under graph to right of Ecat> area to right of Ea.greater number/ fraction of molecules or particles have enough energy to react on collision or greater number of effective / successful collisions

(iii) The reaction is highly exothermic; in principle its enthalpy of reaction could be found by using average bond enthalpies. By a consideration of the structure and bonding in the compounds involved, suggest why the use of the average bond enthalpies for C=O, C–C, C=C and O–H would give a highly inaccurate answer for the enthalpy of reaction.

The reactant/ 1,4-dihydroxybenzene (and the product/ quinone both) have delocalised ring (or resonance) systems/ or described delocalisation (or resonance) these average bond energies are for localised bonds / do not apply to benzene ring compounds / compounds with delocalised or resonance systems.

(b) On heating hydrogen peroxide decomposes according to the equation.

2H2O2 2H2O + O2

Hydrogen peroxide is marketed as an aqueous solution of a given ‘volume strength’. The common 20-volume solution gives 20 dm3 of oxygen from 1 dm3 of solution. What is the concentration in g dm–3 of such a solution?

(Molar volume of any gas at the temperature and pressure of the experiment is 24 dm3.)

20 dm3 oxygen is 20/24 = 0.833 mol amount peroxide in 1 dm3 = 0.833 mol × 2 (1) = 1.67 mol mass 1.67 mol × 34 g mol−1 = 57 / 56.6 / 56.7 / 56.8 g (dm–3)

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(c) Hydrogen peroxide, H2O2, can also act as a reducing agent.

The rapid oxidation of hydrogen peroxide was used in World War II to generate steam to launch the V1 ‘flying bomb’. H2O2 (100 volume) was reacted with acidified potassium manganate(VII) solution.

(i) Write the half-equation for the oxidation of hydrogen peroxide to oxygen, O2.

H2O2 O2+ 2H+ + 2e−

.................................................................................................................................................

(ii) The MnO4- ions are reduced to Mn2+ during the reaction. Derive the overall equation for the

reaction between H2O2 and acidified KMnO4.

2MnO4− + 6H+ + 5 H2O2 2Mn2+ + 8H2O + 5O2

.................................................................................................................................................

(iii) Suggest in terms of the collision theory of chemical kinetics why 100-volume hydrogen peroxide (this gives 100 dm3 of oxygen from 1 dm3 of hydrogen peroxide when it decomposes to water and oxygen) was used rather than the more common 20-volume solution.

Higher concentration increases collision frequency / more collisions per unit of time therefore causes increase in reaction rate.................................................................................................................................................

Equilibria A2 Q34 pdf4(a) A compound D, CH3CH(OH)COOH, may be prepared from C3H7OH by the following series

of reactions.

(i) Identify compounds A, B and C.

(ii) Outline the reaction mechanism with chlorine.

Free radical substitution.

Initiation, propagation and termination stages (with relevant equations)

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(b) Assuming the percentage yield for each step in sequence to be 80%, calculate the mass of D that could be made from 60 g of C3H7OH.

No of moles of D = 0.8 × 0.8 × 0.8 = 0.512 1 mol propanol = 60g 1 mol D = 90g

therefore 0.512 × 90 = 46.1 g

(c) An aqueous solution of D of concentration 0.100 mol dm–3 has a pH value of 2.04.

(i) Calculate the value of the dissociation constant, Ka, for D.

pH = 2.04 [H+] = 10-2.04 = 9.12 ×10−3

Ka = (9.12 × 10−3)2/0.1 = 8.32 x 10-4 mol dm-3

(ii) Suggest, with reasoning, whether D or propanoic acid, CH3CH2COOH, Ka = 1.3 × 10–5 mol dm–3, would be more exothermic in reaction with aqueous sodium hydroxide solution of concentration 0.1 mol dm–3.

Propanoic acid is a weaker acid than D and would produce less energy than D in its reaction with NaOH as more energy would be needed to dissociate the acid into ions. (extension: link to stability of carboxylate anion in the presence of electron-withdrawing inductive –OH group on D as compared to no –OH group on propanoic acid, thereby favouring dissociation to produce H+ and hence a stronger acid).................................................................................................................................................

Equilibria A2 pdf Q395 Hard water contains dissolved calcium or magnesium ions, which come from the rocks over

which the water has flowed. The concentration of these ions can be found in several ways.

(a) One method is to titrate a known volume of the water with a standard solution of the compound edta.

Edta complexes with Ca2+ ions in a 1:1 ratio. To ensure that the edta complexes satisfactorily with the Ca2+ ions, the solution must be buffered at about pH 10. At this pH the indicator used will change colour when all the Ca2+ ions have been complexed.

(i) The buffer which is usually used in this titration is a mixture of aqueous ammonia and ammonium chloride solution. Explain how this mixture behaves as a buffer solution.

NH3 + H2O ⇌ NH4+ + OH– NH4Cl NH4

+ + Cl–

1st equilibrium suppressed by 2nd reaction or with NH3 NH4+ addition of H+ : reacts

with OH– NH4+ + OH– reacts with OH– \ more NH3 dissolves to restore equilibrium

addition of OH– : reacts with NH4+ NH3 + H2O since very large concentration of NH4

+. Hence, the solution is one which can maintain a fairly constant pH when a small amount of acid or base is added to it due to presence of a large reservoir of NH3 and NH4

+ respectively.

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.................................................................................................................................................(ii) A 50.0 cm3 sample of tap water was titrated with edta solution of concentration 0.0100 mol

dm–3. In the titration 31.2 cm3 of the edta solution was needed before the indicator changed colour. What is the calcium ion concentration in the water in mol dm–3?

moles edta moles Ca2+

No. mols of edta solution used = 31.2 × 0.01/1000 = 3.12 × 10–4

No. mols of calcium ion in sample used = 3.12 × 10–4

Concentration of calcium in sample = 3.12 × 10–4 × 1000/50 = 6.24 × 10–3 mol dm–3

(b) A second method for the determination of Ca2+ uses the precipitation of the salt calcium ethanedioate, CaC2O4 The precipitate is filtered off, dissolved in warm dilute nitric acid, and the ethanedioate, now in solution as ethanedioic acid, is determined by titration with standard potassium manganate(VII) solution.

(i) 25.0 cm3 of a solution of Ca2+ ions containing 0.0500 mol dm–3 of Ca2+ ions was treated with excess ammonium ethanedioate solution, and the precipitate of calcium ethanedioate was filtered off. Find the mass of the salt which would be precipitated.

molar mass = (2 × 12) + (4 × 16) + 40 = 128 (g mol–1) therefore mass precipitated = 25 × 0.05 × 28 /1000 g = 0.160 g

(ii) The precipitate was washed with warm dilute nitric acid until completely dissolved, and the washings made up to 250 cm3 with pure water. 25.0 cm3 portions of this solution were added to about 25 cm3 of dilute sulphuric acid, and the mixture was titrated at 60ºC with 0.00200 mol dm–3 potassium manganate(VII) solution. Write the equation for the reactions occurring during the titration and calculate the volume of potassium manganate(VII) solution which would be required.

2MnO4- + 5C2O4

2– + 16H+ 2Mn2+ + 10CO2 + 8H2O

mol ethanedioic acid = 1.25 x 10-4

mol MnO4- = 5 x 10-5 (1) therefore

vol MnO4- = 5 x 10-5 (mol)/0.002 (mol dm-3) = 0.025 dm3 = 25cm3

Quantitative Chemistry Q18 pdf6 Spider silk is a natural polymer which has an exceptional strength for its weight. Kevlar is a

man-made polymer designed to have similar properties. It has a wide variety of uses from sporting equipment to bullet-proof vests.

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correct diagram showing at least one monomer unit, and at least one N-H and C=O.i.e. –NH-C6H2-NH-CO– or –CO-C6H4-CO-NH–

one H-bond between N-H of original chain and C=O group of new chain one H-bond between C=O of original chain and N-H group of new chainlabelled hydrogen bonds or H-bonds

(a) In Kevlar, the polymer strands line up to form strong sheets with bonds between the strands.

On the diagram above, draw a labelled part of a second polymer chain showing how bonds could be formed between the chains.

(b) The transport of oil by sea has resulted in a number of oil spills in recent years. As well as a waste of a valuable resource, these have caused major environmental problems. Traditional sorbent materials absorb water and sink. Researchers have developed new sorbent materials to help collect the spilled oil. The sorbent consists of a material called ‘hydrophobic aerogels’. This is a network of silicon(IV) oxide with some of the silicon atoms attached to fluorine-containing groups.

—O—Si—CH2—CF3

The introduction of these fluorine-containing groups allows the oil to be absorbed but not the water. Tests show that these materials can absorb more than 200 times their mass of oil without sinking.

(i) Suggest what the word hydrophobic means.

Water-hating/fearing/repelling/resistant or can’t form bonds with water (molecules).................................................................................................................................................

(ii) Suggest why the fluorine-containing groups allow oil to pass through but not water molecules.

Fluorine-containing groups form van der Waals bonds (with the oil molecules) but cannot form hydrogen bonds (with the water molecules).................................................................................................................................................

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(iii) Suggest another important fluorine-containing polymer that repels water-containing materials.

Teflon/PTFE (under CFCs sections in lecture notes)

.................................................................................................................................................

9701_s09_qp4 Q97 Silver bromide, AgBr, is widely used in photography. In a photographic film, AgBr crystals

are precipitated into a gelatine base as ‘grains’ of diameter about 1 x 10–6 m.

(a) Calculate the approximate number of silver ions contained in a grain of AgBr of mass 2.5 x 10–12 g.

Mr(AgBr) = 108 + 79.9 = 187.9

moles = 2.5 x 10-12/187.9 = 1.33 x 10-14

no. of ions = 1.33 x 10-14 x 6.02 x 1023 = 8.01 x 109 ions (to 3sf)

(b) AgBr is only sparingly soluble in water. The [Ag+] in a saturated solution of AgBr can be estimated by measuring the E

cell of a particular cell.

(i) Draw a labelled diagram to show how the E(Ag+|Ag) can be measured.

Labelled diagram showing standard hydrogen electrode (with [H+(aq)] = 1 mol dm-3, H2 at 1 atm and 298 K), salt bridge, voltmeter, Ag electrode immersed into AgBr(aq) at 1 mol dm-3 and 298K

(ii) Explain briefly the effect on the e.m.f of the cell if [Ag+] decreases.

(As [Ag+] decreases), the Ereduced will decrease since by LCP, a decrease in [Ag+]

causes equilibrium position to shift left, favouring the backward reaction, hence overall emf will decrease/become more negative..................................................................................................................................................

In its saturated solution, [AgBr(aq)] = 7.1 x 10–7 mol dm–3.

(iii) Write an expression for the solubility product of AgBr, and calculate its value, including units.

Ksp = [Ag+][Br-] = (7.1 x 10-7)2 = 5.04 x 10-13 mol2dm-6

.................................................................................................................................................

(c)(i) Write a chemical equation representing the lattice energy of AgBr.

Ag+(g) + Br-(g) AgBr(s).................................................................................................................................................

(ii) Use the following data to calculate a value for the lattice energy of AgBr(s).

first ionisation energy of silver = +731 kJ mol–1

electron affinity of bromine = –325 kJ mol–1

enthalpy change of atomisation of silver = +285 kJ mol–1

enthalpy change of atomisation of bromine = +112 kJ mol–1

enthalpy change of formation of AgBr(s) = –100 kJ mol–1

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LE = ΔHf - (all the rest)= -100 – (731 + 285 + 112 – 325)(= -100 - 731 - 285 - 112 + 325)= -903 kJ mol-1

OR show Born-Haber cycle as well and calculate LE using Hess’ Law, where

285 + 112 + 731 + (-325) + LE = -100

LE = -903 kJ mol-1

(iii) How might the lattice energy of AgCl compare to that of AgBr? Explain your answer.

LE(AgCl) should be higher/more negative, due to size/radius of Cl- being less than that of Br-. LE is proportional to (q+q- / r+ + r-).................................................................................................................................................

(d) In photography a bromide ion absorbs a photon and releases an electron which reduces a silver ion to a silver atom.

Br- Br + e-

Ag+ + e- Ag

Predict whether it would require more energy or less energy to initiate this process in a AgCl emulsion, compared to a AgBr emulsion. Explain your answer.

more energy needed, since rCl- < rBr- or ionised electron nearer to nucleus or less shielding etc. or in terms of I.E.(Cl) > I.E.(Br).................................................................................................................................................

9701_w05_qp4 Q18 Rodinol is used as a photographic developer. In alkaline solution it is a mild reducing agent,

providing electrons according to the following half equation.

Rodinol ‘develops’ a latent photographic image by reducing activated silver bromide grains to silver metal and bromide ions.

(a) Construct a balanced equation for the reaction between rodinol and AgBr.

HO-C6H4-NH2 + 2AgBr + 2OH- O=C6H4=O + H2O + NH3 + 2Ag + 2Br-

(or C6H7NO) (or C6H4O2).................................................................................................................................................

(b) Suggest, with a reason, how the basicity of rodinol might compare to that of ammonia.

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Rodinol should be less basic than NH3 because the lone pair on N is delocalised over/overlaps with the aryl ring/interacts with the p-orbitals of the carbon atoms of the benzene ring, forming a partial double bond character, reducing the availability of the lone pair of electrons on N for donation..................................................................................................................................................

(c) Suggest structural formulae for the compounds E, F and G in the following chart of the reactions of rodinol.

E is H2N-C6H4-O- Na+ or H2N-C6H4-ONa F is HO-C6H4NH3

+ Cl- or HO-C6H4NH3ClG is HO-C6H2Br2-NH2 up to HO-C6Br4-NH2 (ignore orientation)

(d) Rodinol can be synthesised from phenol by the following route.

(i) Suggest reagents and conditions for step I.

HNO3(aq) or dil HNO3 (NOT conc., and NOT + conc. H2SO4).................................................................................................................................................

(ii) State the type of reaction in step II and suggest reagents and conditions for it.

Reduction, Sn + excess concentrated HCl, reflux followed by NaOH(aq).................................................................................................................................................

(e) Rodinol is also an important intermediate in the commercial production of the analgesic

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drug paracetamol.

(i) Name two functional groups in paracetamol.

phenol, amide.................................................................................................................................................

(ii) State the reagent and condition to convert rodinol into paracetamol.

CH3COCl, room temperature .................................................................................................................................................

9701_w05_qp4 Q49 The phenol 1-naphthol is a starting point for the manufacture of carbaryl, an insecticide and

a plant growth inhibitor.

(a)(i) Suggest a structure for the intermediate C and draw it in the box above.

(ii) Name the functional groups in carbaryl.

amide, ester.................................................................................................................................................

(iii) Suggest structures for the three products formed when carbaryl is hydrolysed.

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CO2 or H2CO3 or Na2CO3, CH3NH2 or CH3NH3+Cl–

(iv) What reagents and conditions would you use for this hydrolysis?

Dilute HCl(aq) and heat >80°C or reflux or NaOH(aq) and heat >80°C or reflux.................................................................................................................................................

(b) Suggest reagents and conditions for converting 1-naphthol into each of the following compounds.

(i)

Br2(aq), room temperature(ii)

dilute/aqueous HNO3, room temperature

(c) Compound D is an isomer of 4-nitro-1-naphthol. D is formed as a by-product during the reaction in b(ii). It can be converted into 2-amino-1-naphthol, E.

(i) Suggest the structural formula of the isomer D.

(ii) Suggest reagents and conditions needed for reaction I.

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Sn, excess concentrated HCl, reflux followed by NaOH(aq)

NOT LiAlH4

.................................................................................................................................................

(iii) Suggest the structural formula of the compound formed when compound E reacts with an excess of CH3COCl.

(d) When an alkaline solution of compound E is added to a solution containing Cu2+(aq) ions, a pale green-blue precipitate F forms. Analysis of F shows that its formula is Cu(C10H8NO)2(H2O)2.

(i) Complete the following structural formula of F.

(allow any orientation of groups)

When an excess of concentrated NH3(aq) is added to F, the precipitate dissolves to form a deep blue solution.

(ii) State the formula of the ion and type of reaction responsible for the formation of the deep blue colour. Explain your answer.

[Cu(NH3)4(H2O)2]2+ NOT [Cu(NH3)6]2+

ligand substitution/exchange reaction .................................................................................................................................................

9701_w07_qp4 Q610 Compound Z, an organic compound with three functional groups, has the molecular

formula C4H6O2. The functional groups can be confirmed by the following tests.

(a) Z decolourises aqueous bromine. Z reacts with sodium to give hydrogen and a solid compound of formula C4H5O2Na. When Z is heated with ethanoic acid and a few drops of concentrated sulphuric acid, a sweet smelling liquid of molecular formula C6H8O3 is formed. A few drops of Z form a yellow/orange precipitate when added to 2,4-

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dinitrophenylhydrazine reagent. When a few drops of Z are warmed with Tollens’ reagent, a silver mirror is formed. Z does not show cis-trans isomerism.

Deduce the displayed formula of Z.

In parts (b) and (c) you may use R– to represent the part of the molecule that does not react.

(b) What is the organic compound formed by the reactions of Z in each of the tests as follows?

with sodium

RONa or R+ ONa-

with ethanoic acid

RO2CCH3

(c) With the aid of an equation, draw the structure of the organic compound formed by Z in each of the tests in (a).

with Tollens’ reagent

Z + 2Ag(NH3)2+ + 3OH- RCO2- + 2Ag + 4NH3 + 2H2O

with 2,4-dinitrophenylhydrazine,

Z + 2,4-DNPH H2O + RCH=NNHC6H3(NO2)2

(d) But-2-enoic acid is an isomer of Z which shows cis-trans isomerism.

Draw a displayed (all bonds should be shown – negative example below) formula of the cis isomer of this acid.

(e) Ethanoic acid is manufactured from methanol, CH3OH, by reacting it with carbon monoxide

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in the presence of a catalyst containing rhodium metal and iodide ions.

CH3OH + CO CH3CO2H

The reaction proceeds in a number of stages.

(i) One stage in this process is the reaction of methanol with hydrogen iodide.

What organic compound is formed in this reaction?

CH3I / iodomethane.................................................................................................................................................

(ii) A later stage involves the conversion of an intermediate compound.

State the type of reaction involved.

Nucleophilic substitution.................................................................................................................................................

9701_w07_qp2 Q5, 9701_w05_qp2 End of Paper

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