harris federation year 13b spring chemistry internal ... · 02/02/2018 · english o organise...

Harris Federation

Year 13B

Spring Chemistry Internal Assessment

Student Name ..........................................................

Time allowed for this test is 1 hour 40 minutes

Answer all the questions

Use a black pen

Maximum mark for this test is 86

Ensure that throughout your test, you: o use good

English o organise

information clearly

o use specialist vocabulary where appropriate

Inorganic and Physical Questions

Question Topic Mark

Marks available

1-2 3.1.8 Thermodynamics

14

3-4 3.1.9 Rate equations

8

5 3.1.11 Electrode potentials

14

6 3.1.12 Acids and bases

12

7 3.2.5 Transition metals

15

8 3.2.6 Reactions of ions in Aqueous solutions

11

9 3.2.4 Properties of Period 3

12

TOTAL 86 Grade

Harris Federation

Q1.This question is about the reaction given below.

CO(g) + H2O(g) CO2(g) + H2(g)

Enthalpy data for the reacting species are given in the table below.

Substance CO(g) H2O(g) CO2(g) H2(g)

ΔH / kJ mol−1 −110 −242 −394 0

Which one of the following statements is not correct?

A The value of Kp changes when the temperature changes.

B The activation energy decreases when the temperature is increased.

C The entropy change is more positive when the water is liquid rather than gaseous.

D The enthalpy change is more positive when the water is liquid rather than gaseous. (Total 1 mark)

Q2.The enthalpy of hydration for the chloride ion is −364 kJ mol−1 and that for the bromide ion is −335 kJ mol−1.

(a) By describing the nature of the attractive forces involved, explain why the value for the enthalpy of hydration for the chloride ion is more negative than that for the bromide ion.

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Harris Federation

(b) The enthalpy of hydration for the potassium ion is −322 kJ mol−1. The lattice enthalpy of dissociation for potassium bromide is +670 kJ mol−1.

Calculate the enthalpy of solution for potassium bromide.

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(c) The enthalpy of solution for potassium chloride is +17.2 kJ mol−1.

(i) Explain why the free-energy change for the dissolving of potassium chloride in water is negative, even though the enthalpy change is positive.

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Harris Federation

(ii) A solution is formed when 5.00 g of potassium chloride are dissolved in 20.0 g of water. The initial temperature of the water is 298 K.

Calculate the final temperature of the solution.

In your calculation, assume that only the 20.0 g of water changes in temperature and that the specific heat capacity of water is 4.18 J K−1 g−1.

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(Total 13 marks)

Harris Federation

Q3.The equation and rate law for the reaction of substance P with substance Q are given below.

2P + Q → R + S

rate = k[P]2[H+]

Under which one of the following conditions, all at the same temperature, would the rate of reaction be slowest?

[P] / mol dm−3 pH

A 0.1 0

B 1 2

C 3 3

D 10 4 (Total 1 mark)

Q4. The hydrolysis of methyl propanoate was studied in acidic conditions at 25°C and the rate equation was found to be

rate = k[CH3CH2COOCH3][H+]

(a) Use the data below to calculate the value of the rate constant, k, at this temperature. Deduce its units.

Initial rate of reaction / mol dm–3 s–1

Initial concentration of methyl propanoate / mol dm–3

Initial concentration of hydrochloric acid / mol dm–3

1.15 × 10–4 0.150 0.555

Rate constant ...............................................................................................

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Units .............................................................................................................

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Harris Federation

(b) The reaction in part (a) was repeated at the same temperature, but water was added so that the volume of the reaction mixture was doubled. Calculate the initial rate of reaction under these conditions.

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(c) A third experiment was carried out at a different temperature. Some data from this experiment are shown in the table below.

Initial rate of reaction / mol dm–3 s–1

Value of rate constant at this different temperature

Initial methyl propanoate / mol dm–3

4.56 × 10–5 8.94 × 10–4 0.123

Calculate the initial pH of the reaction mixture. Give your answer to two decimal places.

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(Total 7 marks)

Q5. The electrons transferred in redox reactions can be used by electrochemical cells to provide energy.

Some electrode half-equations and their standard electrode potentials are shown in the table below.

Half-equation Eο/V

Cr2O72–(aq) + 14H+(aq) + 6e– → 2Cr3+(aq) + 7H2O(l) +1.33

Fe3+(aq) + e– → Fe2+(aq) +0.77

2H+(aq) + 2e– → H2(g) 0.00

Fe2+(aq) + 2e– → Fe(s) –0.44

Li+(aq) + e– → Li(s) –3.04

Harris Federation

(a) Describe a standard hydrogen electrode.

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(b) A conventional representation of a lithium cell is given below. This cell has an e.m.f. of +2.91 V

Li(s) | Li+(aq) || Li+(aq) | MnO2(s) , LiMnO2(s) | Pt(s)

Write a half-equation for the reaction that occurs at the positive electrode of this cell.

Calculate the standard electrode potential of this positive electrode.

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(c) Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer.

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Harris Federation

(d) A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution. The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions. A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid.

Calculate the percentage of iron(II) ions that had been oxidised by the air.

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(Total 14 marks)

Harris Federation

Q6.Ammonium chloride, when dissolved in water, can act as a weak acid as shown by the following equation.

NH4+(aq) NH3(aq) + H+(aq)

The following figure shows a graph of data obtained by a student when a solution of sodium hydroxide was added to a solution of ammonium chloride. The pH of the reaction mixture was measured initially and after each addition of the sodium hydroxide solution.

(a) Suggest a suitable piece of apparatus that could be used to measure out the sodium hydroxide solution. Explain why this apparatus is more suitable than a pipette for this purpose.

Apparatus ......................................................................................................

Explanation ....................................................................................................

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(b) Use information from the curve in the figure above to explain why the end point of this reaction would be difficult to judge accurately using an indicator.

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Harris Federation

(c) The pH at the end point of this reaction is 11.8.

Use this pH value and the ionic product of water, Kw = 1.0 × 10−14 mol2 dm−6, to calculate the concentration of hydroxide ions at the end point of the reaction.

Concentration = ............................... mol dm−3

(3)

(d) The expression for the acid dissociation constant for aqueous ammonium ions is

The initial concentration of the ammonium chloride solution was 2.00 mol dm−3.

Use the pH of this solution, before any sodium hydroxide had been added, to calculate a value for Ka

Ka = ............................... mol dm−3

(3)

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(e) A solution contains equal concentrations of ammonia and ammonium ions.

Use your value of Ka from part (d) to calculate the pH of this solution. Explain your working.

(If you were unable to calculate a value for Ka you may assume that it has the value 4.75 × 10−9 mol dm−3. This is not the correct value.)

pH= ............................... (2)

(Total 12 marks)

Q7.Complexes containing transition elements have a wide variety of uses including acting as dyestuffs like Prussian Blue.

Cisplatin is a platinum-based chemotherapy drug used to treat various types of cancers. It was the first member of a class of anti-cancer drugs that react with DNA in tumour cells.

Cisplatin is prepared from K2PtCl4 according to the following scheme.

All the reactions shown are reversible.

(a) Name the type of reaction occurring in all four steps of the scheme.

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(b) Explain why an excess of potassium iodide is used in Reaction 1.

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(c) (i) Write an equation for Reaction 1.

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(ii) Calculate the percentage atom economy for the formation of K2PtI4 in Reaction 1. Show your working.

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(d) In Reaction 3, silver nitrate solution is added to improve the yield of product.

(i) Write the simplest ionic equation for the reaction of iodide ions with silver nitrate.

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(ii) Suggest why addition of silver nitrate improves the yield of product from Reaction 3.

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(e) Suggest two reasons, other than poor practical technique, why the overall yield of cisplatin in this synthesis may be low.

Reason 1 .......................................................................................................

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(f) The cisplatin formed in Reaction 4 is impure. Outline how the impure solid is purified by recrystallisation.

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(g) Platinum compounds are highly toxic.

(i) State why cisplatin is used in cancer treatment despite its toxicity.

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(ii) Suggest a suitable precaution that should be taken by medical staff when using cisplatin.

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(Total 15 marks)

Harris Federation

Q8.(a) A sample of solid chromium(III) hydroxide displays amphoteric character when treated separately with dilute hydrochloric acid and with dilute aqueous sodium hydroxide.

Write an ionic equation for each of these reactions. Include the formula of each complex ion formed. Describe the changes that you would observe in each reaction.

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(b) Aqueous solutions of copper(II) sulfate and cobalt(II) sulfate undergo ligand substitution reactions when treated separately with an excess of dilute aqueous ammonia.

Write equations for these reactions. Include the formulae for any complex ions. Describe the changes that you would observe in each reaction.

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(Total 11 marks)

Harris Federation

Q9.Magnesium oxide, silicon dioxide and phosphorus(V) oxide are white solids but each oxide has a different type of structure and bonding.

(a) State the type of bonding in magnesium oxide. Outline a simple experiment to demonstrate that magnesium oxide has this type of bonding.

Type of bonding ............................................................................................

Experiment ....................................................................................................

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(b) By reference to the structure of, and the bonding in, silicon dioxide, suggest why it is insoluble in water.

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(c) State how the melting point of phosphorus(V) oxide compares with that of silicon dioxide. Explain your answer in terms of the structure of, and the bonding in, phosphorus(V) oxide.

Melting point in comparison to silicon dioxide ...............................................

Explanation ....................................................................................................

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Harris Federation

(d) Magnesium oxide is classified as a basic oxide.

Write an equation for a reaction that shows magnesium oxide acting as a base with another reagent.

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(e) Phosphorus(V) oxide is classified as an acidic oxide.

Write an equation for its reaction with sodium hydroxide.

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(Total 12 marks)

END OF EXAM

Harris Federation

M1.B [1]

M2.(a) Chloride (ions) are smaller (than bromide ions)

Must state or imply ions.

Allow chloride has greater charge density (than bromide).

Penalise chlorine ions once only (max 2 / 3). 1

So the force of attraction between chloride ions and water is stronger

This can be implied from M1 and M3 but do not allow intermolecular forces.

1

Chloride ions attract the δ+ on H of water / electron deficient H on water

Allow attraction between ions and polar / dipole water.

Penalise H+ (ions) and mention of hydrogen bonding for M3

Ignore any reference to electronegativity.

Note: If water not mentioned can score M1 only. 1

(b) ΔHsolution = ΔHL + ΔHhyd K+ ions + ΔHhyd Br − ions / = 670 − 322 − 335

Allow ΔHsolution= ΔHL + ΣΔHhyd

1

= (+)13 (kJ mol−1)

Ignore units even if incorrect.

+13 scores M1 and M2

−13 scores 0

−16 scores M2 only (transcription error). 1

(c) (i) The entropy change is positive / entropy increases

ΔS is negative loses M1 and M3 1

Harris Federation

Because 1 mol (solid) → 2 mol (aqueous ions) / no of particles increases

Allow the aqueous ions are more disordered (than the solid).

Mention of atoms / molecules loses M2 1

Therefore TΔS > ΔH 1

(ii) Amount of KCl = 5/Mr = 5/74.6 = 0.067(0) mol

If moles of KCl not worked out can score M3, M4 only (answer to M4 likely to be 205.7 K)

1

Heat absorbed = 17.2 × 0.0670 = 1.153 kJ

Process mark for M1 × 17.2 1

Heat absorbed = mass × sp ht × ΔT

(1.153 × 1000) = 20 × 4.18 × ΔT

If calculation uses 25 g not 20, lose M3 only (M4 = 11.04, M5 = 287)

1

ΔT = 1.153 × 1000 / (20 × 4.18) = 13.8 K

If 1000 not used, can only score M1, M2, M3

M4 is for a correct ΔT

Note that 311.8 K scores 4 (M1, M2, M3, M4). 1

T = 298 − 13.8 = 284(.2) K

If final temperature is negative, M5 = 0

Allow no units for final temp, penalise wrong units. 1

[13]

Harris Federation

M3.C [1]

M4. (a) k = rate/[CH3CH2COOCH3][H+] 1

or

=

= 1.38 × 10–3 to 1.4 × 10–3

1

mol–1 dm3 s–1

1

(b) ans = rate constant × (½ × 0.150) × (½ × 0.555)

ignore units

= rate constant × 0.0208

2.88 × 10–5 (1.38 × 10–3 gives 2.87 × 10–5)

Allow 2.87 – 2.91 × 10–5 (1.4 × 10–3 gives 2.91 × 10–5) 1

(c) [H+] = rate/ k[CH3COOCH2CH3] 1

=

= 0.415 (0.4146) 1

pH = 0.38 mark independently

[H+] = 0.41 gives pH = 0.39 1

[7]

Harris Federation

M5. (a) Hydrogen/H2 gas/bubbles 1

1.0 mol dm–3 HCl/H+

1

At 298K and 100kPa

Allow 1 bar instead of 100 kPa Do not allow 1 atm

1

Pt (electrode) 1

(b) Li+ + MnO2 + e– → LiMnO2

Ignore state symbols 1

–0.13(V) 1

(c) Fe3+ ions reduced to Fe2+

Can score from equation/scheme 1

Because E(Fe3+(/Fe2+)) > E(H+/H2)/E(hydrogen)

Allow emf/Ecell +ve/0.77V Allow Fe3+ better oxidising agent than H+

Allow H2 better reducing agent than Fe2+

Only award this explanation mark if previous mark given 1

(d) Moles Cr2O72– =23.7 × 0.01/1000 = 2.37 × 10–4

1

1 mol Cr2O72– reacts with 6 mol Fe2+ so moles

Fe2+ in 25 cm3 = 6 × 2.37 × 10–4 = 1.422 × 10–3

1

M1 × 6

Moles Fe2+ in 250 cm3 = 1.422 × 10–2

M2 × 10 or M4/10 1

Original moles Fe2+ = 10.00/277.9 = 0.0360

Independent mark 1

Moles Fe2+ oxidised = 0.0360 – 0.0142 = 0.0218

Harris Federation

M4 – M3 1

% oxidised = (0.0218 × 100)/0.0360 = 60.5%

(M5 × 100)/M4 Allow 60 to 61 Note Max 3 if mol ratio for M2 wrong eg 1:5 gives 67.1% 1:1 gives 93.4%

Note also, 39.5% (39-40) scores M1, M2, M3 and M4 (4 marks)

1

[14]

M6.(a) Burette 1

Because it can deliver variable volumes 1

(b) The change in pH is gradual / not rapid at the end point 1

An indicator would change colour over a range of volumes of sodium hydroxide

Allow indicator would not change colour rapidly / with a few drops of NaOH

1

(c) [H+] = 10–pH = 1.58 × 10–12

1

Kw = [H+] [OH–] therefore [OH–] = Kw / [H+] 1

Harris Federation

Therefore, [OH–] = 1 × 10–14 / 1.58 × 10–12 = 6.33 × 10–3 (mol dm–3)

Allow 6.31–6.33 × 10–3 (mol dm–3) 1

(d) At this point, [NH3] = [H+]

Therefore Ka 1

[H+] = 10–4.6 = 2.51 × 10–5

1

Ka = (2.51 × 10–5)2 / 2 = 3.15 × 10–10 (mol dm–3)

Allow 3.15 – 3.16 × 10–10 (mol dm–3) 1

(e) When [NH3] = [NH4+], Ka = [H+] therefore – log Ka= – log [H+]

Answer using alternative value 1

Therefore pH = –log10(3.15 × 10–10) = 9.50

M2 pH = – log10(4.75 × 10–9) = 8.32

Allow consequential marking based on answer from part (d) 1

[12]

M7.(a) (ligand) substitution

Allow ‘ligand exchange’. 1

(b) To displace the equilibrium to the right

To ensure reaction goes to completion. 1

To improve the yield

Harris Federation

Allow ‘to replace all chlorines’. 1

(c) (i) K2PtCl4 + 4Kl → K2Ptl4 + 4KCl

Allow correct ionic equations PtCl42− + 4l− → Ptl42− + 4Cl−

Allow multiples and fractions. 1

(ii) = (780.9) ×100 / (415.3 + 664)

Working must be clearly shown.

Allow one mark for correct relationship even if Mr values are incorrect eg using values from ionic equation.

1

= 72.4

Allow 72% 1

(d) (i) Ag+ + I− → AgI

Ignore state symbols even if incorrect.

This equation only. 1

(ii) Stops the reverse reaction / equilibrium displaced to the right 1

(e) Number of steps in the process

Allow ‘equilibrium may lie on the reactant side’ / side reactions / isomer formation.

1

Losses at each stage of the synthesis

Equilibrium losses or practical losses or yield not 100% for each step.

1

(f) Minimum amount of hot solvent

Harris Federation

Accept ‘small’ for minimum.

Accept water. 1

Cool / crystallise 1

Filter 1

(g) (i) Small amounts are more likely to kill cancer cells rather than the patient 1

(ii) Wear gloves / wash hands after use

Ignore masks.

Apply the list principle if more than one answer. 1

[15]

M8.(a) Cr(OH)3 + 3H2O + 3H+ → [Cr(H2O)6]3+

Can start with Cr(H2O)3(OH)3 for each equation

Ignore any unnecessary preliminary preparation of Cr(OH)3

1

Green / grey−green solid

Mark colours independently from equations

Allow green ppt. 1

Forms green / purple / ruby / violet solution

ignore shades of colours 1

Cr(OH)3 + 2H2O + OH− → [Cr(H2O)2(OH)4]−

Allow with 5 or 6 OH− provided complex has co−ordination number of 6

Penalise complex ions with incorrect charges overall or if shown on ligand.

1

Harris Federation

Forms green solution

Note that for each equation final complex must be 6 co−ordinate

1

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

Allow two correct equations via intermediate hydroxide in both cases even if first equation uses OH− instead of NH3

1

Blue (solution)

Mark colours independently from equations 1

Dark / deep / royal blue solution 1

[Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ +6H2O 1

pink / red (solution) 1

Brown / straw / yellow solution

ignore darkens in air / with time 1

[11]

M9.(a) MgO is ionic

If not ionic, CE = 0 1

Melt it

If solution mentioned, cannot score M2 or M3 1

(Molten oxide) conducts electricity

Allow acts as an electrolyte.

Cannot score M3 unless M2 is correct. 1

(b) Macromolecular

CE = 0 if ionic, metallic or molecular.

Harris Federation

Allow giant molecule. 1

Covalent bonding

Giant covalent scores M1 and M2 1

Water cannot (supply enough energy to) break the covalent bonds / lattice

Hydration enthalpy < bond enthalpy. 1

(c) (Phosphorus pentoxide’s melting point is) lower

If M1 is incorrect, can only score M2 1

Molecular with covalent bonding

M2 can be awarded if molecular mentioned in M3 1

Weak / easily broken / not much energy to break intermolecular forces OR weak vdW / dipole-dipole forces of attraction between molecules

Intermolecular / IMF means same as between molecules. 1

(d) Reagent (water or acid)

Can be awarded in the equation. 1

Equation eg MgO + 2HCl → MgCl2 + H2O

MgO + H2O → Mg(OH)2

Equations can be ionic but must show all of the reagent eg H+ + Cl−

Simplified ionic equation without full reagent can score M2 only.

Allow 6MgO + P4O10 → 2Mg3(PO4 )2

1

(e) P4O10 + 12NaOH → 4Na3PO4 + 6H2O

Harris Federation

Allow P2O5 and acid salts.

Must be NaOH not just hydroxide ions. 1

[12]

Harris Federation

E2.This question proved to be difficult. In part (a), most students scored one mark for mentioning that the ionic radius of the chloride ion is less than that for the bromide ion. However, many weaker students lost this mark because they did not make reference to a chloride ion; often, just chlorine was mentioned. Only the best students went on to discuss how a chloride ion is attracted to the partially positive hydrogen atom in a water molecule. Others referred incorrectly to intermolecular forces or to hydrogen bonding. About 60% of students gained both marks for the calculation in part (b). In part (c)(i), most students stated that the entropy would increase but many did not clearly explain why and also that the TΔS term would outweigh the positive enthalpy change. Although about a quarter of all students scored at least 4 marks in part (c)(ii), less than 10% went on to score full marks by subtracting the temperature change from 298 K. Marks were generally low because many students did not recognise that the heat absorbed could be calculated by multiplying the enthalpy of solution by the amount in moles of potassium chloride. Weak students also used a wrong value for the mass of water heated – usually 25 g or 5.00 g.

E4. As usual with questions involving rate equations, this question was answered well, particularly parts (a) and (c). Part (b) however was poorly answered and only the very best candidates gained this mark. Many of the rest only divided the original rate by 2 instead of by 4 as the concentrations of both the ester and of If” ions were halved. Again it was disappointing to see the number of candidates who did not give the pH to two decimal places despite being asked to do so.

E5. This was the most difficult question in the paper. In part (a) most candidates knew that a standard hydrogen electrode uses hydrogen gas and a platinum electrode but only a minority went on to state the correct and necessary conditions of concentration, temperature and pressure for operation of the electrode. In part (b) most candidates gave a correct value for the electrode potential of the positive electrode but very few candidates were able to use the cell representation to deduce a half-equation for the electrode reaction. Answers to part (c) were also disappointing. Despite the given electrode potentials and half-equations, most candidates could not predict and explain the redox reaction between hydrogen and iron(III) ions. Part (d) also proved to be a difficult question. Candidates were usually able to score three marks out of six for a correct calculation of the number of moles of dichromate(VI) ions, a calculation of the original number of moles of iron(II) sulfate and for recognising the involvement of a factor of ten. Only the very best candidates realised that the number of moles of iron ions was six times the number of moles of dichromate(VI) ions and that a subtraction was required to calculate the percentage of iron(II) ions that had been oxidised by air before the titration.

Harris Federation

E7.(a) While most candidates scored the mark in part (a) only the best candidates scored both marks in part (b). Many candidates could not clearly explain the changes in an equilibrium reaction.

(b) While most candidates scored the mark in part (a) only the best candidates scored both marks in part (b). Many candidates could not clearly explain the changes in an equilibrium reaction.

(c) The equation in part (i) proved quite demanding but most candidates had a good appreciation of atom economy in part (ii), though mathematical slips were quite common.

(d) (i) Most candidates scored the mark for the ionic equation in this part, the presence of spectator ions was the usual error.

(ii) Answers to this part were disappointing. Most candidates explained what happened in the reaction, rather than addressing the improvement. Many answers that had little relevance to the question were incorrectly awarded a mark.

(e) Many candidates had little awareness of the practical consequences of a multi-stage process in this part. There was a tendency for candidates to repeat themselves and for the centre, incorrectly, to award two marks.

(f) A surprising number could not give a satisfactory description of recrystallisation in this part. Far too many candidates described completely different techniques. Centres need to exercise some common sense in this type of question. For example, a candidate cannot be awarded a mark for filtration or recrystallisation if a solution was not prepared or if the candidate melted the mixture.

(g) (i) In this part many candidates described how cisplatin works rather than addressing the balance of factors when using a toxic drug.

(ii) The safety factor in this part was well known.

Harris Federation

E8.In part (a) the observations were usually described accurately. Attempts at the equations were less successful. Again, students should be encouraged to read the question carefully before starting to answer it. Many students attempted to start from an aqueous solution of chromium(III) ions instead of the solid hydroxide. Students were asked to write ionic equations because it was thought hat this would make the equations simpler and easier to write. A significant number of students ignored this instruction and were penalised. In part (b) of this question students were asked to write equations for ligand substitution reactions. It was not possible to answer this question correctly unless it was realised that the initial complex ion, in aqueous solution, was a hexaaqua ion with six water ligands. When describing an observation it is important to give the phase to which the colour applies so that the product of reacting an excess of aqueous ammonia with copper(II) sulphate solution is a ‘deep blue solution’, not just ‘deep blue’.

E9.This question was answered well and students of all abilities were able to score reasonable marks. In part (a), the most common error was to suggest that the conductivity of a solution could be tested. Apart from the problem that magnesium oxide is only sparingly soluble, A-level students should know that many covalent substances form ions in water so that testing the conductivity of a solution is not a good method for the identification of an ionic solid. In parts (b) and (c) only the higher-scoring students scored all marks. The marks for structure and bonding were usually gained but in part (b) a suggestion as to why water does not break down the structure was not given and in part (c) students often did not state clearly that the forces of attraction between molecules are weak.

harris federation year 13b spring chemistry internal ... · 02/02/2018 · english o organise...

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