H2 Mock Paper 2 Answers (1)

CATHOLIC JUNIOR COLLEGE

JC2 Mock Examination 2010

Higher 2

CANDIDATENAME

CLASS

CHEMISTRY 9647/02

Paper 2 Structured 08 October 2010

Candidates answer on the Question paper 2 hrAdditional Materials: Data Booklet

READ THESE INSTRUCTIONS CAREFULLY

Write your name and class in the spaces provided above.Write in dark blue or black pen.You may use a soft pencil for any diagrams, graphs or rough working.Do not use staples, paper clips, highlighters, glue or correction fluid.

You are reminded of the need for good English and clear presentation in your answers.The number of marks is given in brackets [ ] at the end of each question or part question.

Answer all questions.

A Data Booklet is provided.

This document consists of 12 printed pages and 0 blank page.

CJC Chemistry Department

© CJC 2010

2 T

For Examiner’s Use

Q 1

Q 2

Q 3

Q 4

Q 5

Q 6

72

1 Planning (P)

Some iron tablets, containing iron (II) sulfate, were left on the shelf for too long such that some of the iron (II) sulfate was oxidised into iron (III) sulfate.

You are to carry out a titrimetric analysis to determine the percentage by mass of Fe3+ ions present in the sample of oxidised iron tablets.

The following reagents are provided to enable the experiment to be carried out: Sample of oxidised iron tablets 1.00 mol dm-3 sulfuric acid, H2SO4

0.0110 mol dm-3 potassium dichromate (VI), K2Cr2O7

N-phenylanthranilic acid as indicator (1 cm3 of indicator produces a violet colour change at end-point)

Zinc powder

(a) The oxidised iron tablets must first be dissolved in water to liberate the Fe2+ and Fe3+

ions. A suitable quantity of sulfuric acid is then added to the solution.

Suggest a reason why sulfuric acid is used in the preparation of the mixture of Fe2+

and Fe3+ solution.

Fe2+ is stabilised in an acidic medium ORTo prevent oxidation of Fe2+ to Fe3+

[1]

(b) Write the ionic equation for the reaction between Fe2+ ions and potassium dichromate (VI) solution.

6Fe2+ + Cr2O72- + 14H+ 6Fe3+ + 2Cr3+ + 7H2O

[1]

(c) An unoxidised sample of iron tablet was dissolved and made up to 250 cm3. 25.0 cm3

of the Fe2+ ions solution required 15.00 cm3 of potassium dichromate (VI) to reach an end-point in a titration.Calculate the mass of FeSO4 in the unoxidised sample of iron tablet that was used.

No. of mol of Cr2O72- used =

15.000.0110

1000

= 1.65 × 10-4 mol6Fe2+ Cr2O7

2-

No. of mol of Fe2+ present in 25.0 cm3 = 6 × 1.65 × 10-4

= 9.91 × 10-4 mol [1, c.a.o.]

No. of mol of Fe2+ present in 250 cm3 =

4 2509.91 10

25

= 9.91 × 10-3 mol

Mass of iron tablet (which contains FeSO4) = 9.91 × 10-3 × (55.8+32.1+4(16.0)) = 1.51 g [1, e.c.f.]

[2]

CJC H2 JC2 Mock Examination 2010 2

(d) A sample of oxidised iron tablet is equally divided and dissolved to give two portions of 250 cm3 acidified Fe2+ and Fe3+ ions solutions for analysis.

Procedure 1

To one portion, the Fe3+ (aq) should be converted into Fe2+ (aq) first using excess zinc metal, before titration against potassium dichromate (VI) solution.

Procedure 2

To the other portion, the Fe2+ (aq) can be directly titrated against potassium dichromate (VI) solution.

Outline in order, the steps you would carry out for Procedure 1. In your description, you should include details of the apparatus and chemicals used and measure(s) taken to ensure reliability of results.

1) Add Zn powder to the beaker containing the 250 cm3 solution of Fe2+/Fe3+ ions with stirring until no more Zn dissolves.

2) Filter the mixture using a filter funnel and collect the filtrate containing Fe2+ ions.3) Pipette 25.0 cm3 of the Fe2+ solution (filtrate) into a conical flask and to it, add 1 cm3 of

N-phenylanthranilic acid using a 10 cm3 measuring cylinder / graduated dropper.4) Fill a burette with the given standard K2Cr2O7 solution and note the initial reading on

the burette.5) Titrate the Fe2+ solution in the conical flask until the solution turns (from pale green to

yellow to) violet; note the final reading on the burette.6) Repeat the titration (steps 3 – 5) for consistent results (ie. titre values are within 0.10

cm3 of each other).

Marking pointsProcedure should include appropriate apparatus commonly found in a college laboratory:

Addition of excess Zn powder into Fe2+/Fe3+ solution until no more Zn dissolves Filtration and collection of filtrate Proper preparation of aliquot Use of correct volume of indicator added Titration of aliquot against standard K2Cr2O7 with end-point colour change to violet

noted Initial and final burette readings noted with repeat of titration for consistent results

[5]


(e) Using the following information, calculate the percentage by mass of Fe3+ ions in the oxidised iron tablet sample.

No. of mol of Fe2+ (in 250 cm3) from Procedure 1 = pNo. of mol of Fe2+ (in 250 cm3) from Procedure 2 = qAr: O, 16.0; S, 32.1; Fe, 55.8

No. of mol of Fe3+ present in 250 cm3 = p - q No. of mol of Fe3+ present in sample = 2 × (p – q)

Mass of Fe3+ in actual sample = 2(p – q) × 55.8 [1]

Mass of oxidised iron tablet (containing FeSO4 + Fe2(SO4)3)= (2q × 151.9) + [½ × 2(p – q) × 399.9]= 399.9p – 96.1q

Percentage by mass of iron(III) in sample =

= [1]

[2]

(f) By modifying Procedure 1, suggest another means by which the amount of Fe3+ in an oxidised iron tablet sample can be determined.

Add a known mass of excess Zn powder to the solution containing Fe2+/Fe3+ ions.

Filter the resultant solution, recover and dry the residue containing the unreacted Zn powder.

The difference in mass of Zn can be used to calculate the no. of mol of Fe3+ present.

Alternative test: Iodometric titration (Fe3+ / I-)

[1]

[Total: 12]


2 (a) The melting points of the chlorides of the Period 3 elements magnesium to phosphorous, are given below.

(i)Account for

the difference in the melting points of magnesium chloride and aluminium chloride, in terms of structure and bonding.

MgCl2 has a giant ionic lattice structure held by strong electrostatic forces of attraction between the oppositely charged Mg 2+ and C l - ions . A large amount of energy is required to overcome the strong electrostatic forces of attraction, hence the melting point is high.

Aluminium chloride has a simple molecular structure. The molecules are non-polar and held by which can be easily broken during melting weak induced dipole-induced dipole attractions/ Van der Waal’s forces of attraction. A small amount of energy is required to overcome the weaker forces of attraction between the molecules, hence the melting point is low. [1: Correct structures, 1: Correct bonding, 1: relative energy mentioned]

(ii) Explain the difference in the pH values when AlCl3 and PCl5 are separately dissolved in water. Write equations to support your answer.

Al2Cl6 undergoes both hydration and hydrolysis when dissolved in water. The high charge density of hydrated Al3+ ion enables it to attract electrons away from one of its surrounding water molecules, thereby polarising and weakening the O-H bond to a greater extent which results in the release of a proton. pH = 3.0

AlCl3 + 6H2O [Al(H2O)6]3+ + 3Cl-

[Al(H2O)6]3+ [Al(H2O)5(OH)]2+ + H+

PCl5 undergoes complete hydrolysis to form a strongly acidic solution with pH=1.0/2.0

PCl5 + 4 H2O H3PO4 + 5 HCl [1: Correct description for Aluminium, 1: Correct pH, 1: Correct Equations]

[6]

CJC H2 JC2 Mock Examination 2010

CompoundMagnesium

chlorideAluminium

chloridePhosphorous pentachloride

Melting point / oC 714 178 162

5

(b) Elements K, L, M and N are four consecutive Period 3 elements. The table below shows the first four ionisation energies of elements K, L, M and N.

Element First I.E Second I.E Third I.E Fourth I.EK 577 1820 2740 11600L 786 1580 3230 4360M 1060 1900 2920 4960N 1000 2260 3390 4540

(i) For element K, briefly explain why the energy difference between the removal of the 2nd and 3rd electrons is smaller than that between the removal of the 3rd and 4th electrons.

2nd and 3rd electrons occupy the same quantum shell while 3rd and 4th electrons are removed from the different quantum shell. [1]

(ii) Explain why the first ionisation energy of element N is lower than that of element M.

N N+ + e1s22s22p63s23p4 1s22s22p63s23p3

M M+ + e1s22s22p63s23p3 1s22s22p63s23p2

OR

Element N is Group VI and Element M is Group V. There is inter-electron repulsion between the paired electrons in the 3p orbital of element N. Hence less energy is required to remove one valence 3p electron from N. 1st ionisation energy of M > N.[1: correct electronic configuration or group, 1: correct reason]

(iii) Element T is a Period 2 element in the same group as element M. Suggest with reasoning why element M can form an oxo-anion of formula MO4

3- but element T cannot.

M, being in Period 3, has empty and energetically accessible d orbitals to expand the octet configuration. Hence MO4

3- can exist. [1]

T, being in Period 2, does not have empty and energetically accessible d orbitals to accommodate the extra electrons and cannot expand the octet configuration. Hence TO4

3- cannot exist. [1]

[5](c) Many interhalogen compounds are very active oxidising agents as well as strong

acids. Examples are oxoacids of chlorine, chloric(V) acid, HClO3 and chloric(VII) acid, HClO4.

(i) With an aid of an equation, show HClO3 as a strong Bronsted-Lowry monoprotic acid in an aqueous solution.

HClO3 + H2O → H3O+ + ClO3- [1: only full arrow accepted]


(ii) Explain why chloric(VII) is a stronger acid than chloric(V) acid in terms of their structures and bonding.

ClO4- more stable as the negative charge is delocalised over a larger number of

oxygen atoms. Hence chloric(VII) is a stronger acid.[1]

(iii) Draw the dot-cross diagram of HClO3. Using VSEPR model, predict the shape with respect to Cl and indicate a possible O-Cl-O bond angle.

Shape – trigonal pyramidal ; bond angle - 1070

[1: Correct dot and cross, 1: correct shape, 1: correct angle] [5][Total: 16]

3 (a) Consider an equilibrium mixture,

4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g) ΔH = −1267 kJmol−1

The graph below shows how the fraction of N2(g) in the equilibrium mixture varies with temperature and pressure.

(i) Comment on the shape of the curve.

As temperature increases, amount of N2 decreases[1]. Therefore the equilibrium position has shifted to the left[1] since ΔH is exothermic.

(ii) On the same graph, sketch a curve to show how the fraction of N2 changes at a higher pressure, P2 . [1]

[3]

(b) An initial 5 moles of NH3 and 3 moles of O2 are mixed in a 2 dm3 vessel. At 4500C, an equilibrium mixture shows 80% of ammonia remains. Calculate a value for the equilibrium constant, Kc and give its unit.

(0.5/2) 2 (1.5/2) 6 Kc = (4/2)4 (2.25/2)3 [1 correct equation used, correct values substituted] = 4.88 X 10-4 mol dm-3 [1 e.c.f. with units]

[3]


Fraction of N2 in equilibrium mixture

temperature

P1

P2

(c) Hydrogen gas is used in the industry to convert N2 into ammonia for the use in fertilizers. One method of preparing H2 is to pass a mixture of carbon monoxide and steam over a Cu/Zn catalyst.

CO(g) + H2O(g) CO2(g) + H2(g) ΔH = −41 kJmol−1

(i) Use the given ΔH and bond energies values from the data booklet, construct an energy cycle and hence calculate the bond energy in the CO molecule.

[1: energy cycle with correct values]B.E.(CO) = –2(460) –41+ 2(740) + 436 = 955 kJmol-1 [1 e.c.f. with units]

(ii) With the aid of the Boltzman-maxwell distribution curve, explain how Cu/Zn affects the rate of the above reaction.

Ecat Euncat

[1: correct diagram]

Catalyst provides an alternative pathway for the reaction with a lower activation energy. [1].Hence more reactants have energy greater than or equal to the activation energy[1]. as shown by bigger area. Hence effective collisions increase and rate increases[1].

[6]

[Total: 12]


Fraction of particles with energy

energy

Area under curve withEcat > Euncat

4 The elements of Group VII were discovered in the 17 th century by different European scientists. Due to their high reactivity, they are found in the environment only in compounds such as halides in minerals and halogenated organic compounds in living organisms.

(a) The plots of PV/RT against P for one mole of an ideal gas and one mole of HCl at 100 K are given below.

(i) Show clearly, on the same axes, how one mole of HF will behave at the same temperature of 100 K.

(ii) Explain the difference in behaviour between HF and HCl at 100 K.

HF is less ideal than HCl as HF has stronger intermolecular hydrogen bonding whereas HCl has weaker intermolecular Van der Waals forces of attraction. [1]

(iii) Explain what happens when the HCl gas is heated to 400 K.Illustrate your answer clearly on the same axes in a(i).

At high temperature, there are negligible forces of attraction between the gas particles. Hence, HCl gas will deviate less from ideal gas behaviour. [1]

[4]


Ideal Gas

HF (100 K) PV/RT

1.0

HCl (100 K)

P

1 mark

1 mark

HCl (400 K)

(b) 2-chloropropane can be made by the chlorination of propane.

(i) Outline the reaction mechanism of its formation, labelling each step in the mechanism appropriately.

Free radical substitution [1]

Initiation stage

Cl2 2 Cl

Propagation stageCH3CH2CH3 + Cl CH3CHCH3 + HCl

CH3CHCH3 + Cl2 CH3CH(Cl)CH3 + Cl

Termination stage2 Cl Cl2

CH3CHCH3 + Cl CH3CH(Cl)CH3

CH3CHCH3 + CH3CHCH3 CH3CH(CH3)CH(CH3)CH3 or C6H14

(ii) Explain why the reaction requires only a flash of ultraviolet light rather than prolonged radiation.

Condition of heat or uv light applies only in initiation step as chlorine free radicals are regenerated in propagation stage to keep reaction going. [1]

[4]

(c) Bromine and fluorine can react to form an interhalogen liquid compound BrF3, a common fluorinating agent. This compound can react with silicon (IV) oxide according to the following equation.

4BrF3 + 3SiO2 3SiF4 + 2Br2 + 3O2

The amount of bromine produced can be determined by reacting it with sodium thiosulfate solution, Na2S2O3 (aq).

(i)Write the equation for the reaction between bromine and thiosulfate ions in acidic medium, given S2O3

2- + 5H2O ⇌ 2SO42- + 10H+ + 8e-.

4Br2 + S2O32- + 5H2O → 2SO4

2- + 10H+ + 8Br - [1]


Ultraviolet light

[1]

[1]

(ii) Determine the mass of SiF4 produced, to one decimal place, if 30.0 cm3 of 0.100 mol dm3 Na2S2O3 solution was required to completely react with the bromine liberated from the reaction between BrF3 and SiO2.

No. of moles of S2O32- = 30/1000 x 0.100

= 0.00300 mol [1]S2O3

2- 4Br2 6 SiF4

No. of moles of SiF4 = 6 x 0.003 = 0.018

Mass of SiF4 = 0.036 x (28.1 +19.0 x 4) = 1.9 g [1]

(iii) The bromine produced could also be reacted with cold aqueous sodium hydroxide to form a mixture of products.

State the type of reaction that it has undergone and write a balanced equation for the reaction involved.

Disproportionation [1]

3 Br2 + 6 OH- 5Br

- + BrO3

- + 3 H2O [1]

[5][Total: 13]

5 The structural formula of a compound R is shown below.

CH2 – C – CH3

(a) Compound R undergoes a 2 step reaction to produce Compound Z. Draw the intermediate product and state the reagents and conditions used for both steps. [3]


O

O

OH

O

(b) Draw the structural formula(e) of the organic product(s) formed when compound R reacts with each of the following reagents.

[6]

Reagent Structural formula(e) of the organic product(s)

Aqueous alkaline I2

[2]

Acidified potassium dichromate(VI), reflux

[1]

Ethanoyl Chloride

[1]

Aqueous NaOH, reflux

[1]

Thionyl chloride

[1]

[Total: 9]


CH2COCH3

O

OH OH

O

O

O

OCOCH3

O IICH2 – C – CH3

OH

CH2COCH3

O

O- Na+ O- Na+

O

O

Cl

O IICH2 – C – CH3

O

O

OH

O IICH2 – C – O– + CHI3

6 Chymotrypsin is a digestive enzyme found in the small intestines that can hydrolyze peptide bonds within the protein. It is also used as a form of treatment for sports injuries to reduce pain and inflammation.

(a) Using chymotrypsin, a small peptide Y was broken down to form three different amino acids according to the following reaction.

(i) Calculate the Mr of peptide Y.

Hydrolysis of peptide Y gain in 1 H2O for every peptide bond broken. No. of peptide bonds broken = 6 [1]

Mr of peptide Y = 2 x 181 + 3 x 165 + 174 x 2 - 6 x 18 = 1097 [1]


A 2

CH2

OH

+

NH2CHCO2H

CH2+3

Tyrosine Phenylalanine

NH2CHCO2H

Mr= 181 Mr = 165 Mr = 174

NH2CHCO2H

(CH2)3

HN

H2N

NH

2

Arginine

Y

13

(ii) The graph shows the results of an investigation of the initial rate of hydrolysis of peptide Y by the enzyme chymotrypsin. In the experiments, the initial concentration of peptide Y was varied but that of chymotrypsin was kept constant.

Explain the difference in the rate of hydrolysis at high and low concentrations of peptide Y.

At low concentration of peptide Y, Rate of reaction increases linearly as active sites of the enzyme are

not fully occupied Reaction is approximately first order wrt the concentration of

peptide Y. [1]However, at high concentration of peptide Y, Rate of reaction is constant Rate is independent of concentration of peptide Y as all active sites

occupied Reaction is zero order wrt the concentration of peptide Y. [1]

[4](b) The table below shows some of the amino acid residues in chymotrypsin.

Name of amino acid residues

Formula of side chain (R in RCH(NH2)CO2H)

Serine ―CH2OH

Histidine

Aspartic Acid ―CH2COOH

Valine ―CH(CH3)2

Lysine ―(CH2)4NH2

Phenylalanine ―CH2(C6H5)

(i) The amino acids which are on the outside of the chymotrypsin are hydrophilic. Suggest, with reasoning, which two of the amino acids in chymotrypsin are likely to have this property.

Any 2 of the following: Serine, Aspartic acid, Lysine, Histidine. [1]This is due to the formation of hydrogen bonding between the amino acids and water molecules. [1]


CH2

N

NH

―

14

[Peptide Y]

Rate

(ii) Explain, in terms of R group interactions, how a low pH might affect the enzymatic activity of chymotrypsin.

At low pH,

Excess H+ will react with basic amine group in histidine/lysine of the chymotrypsin.

Hence, low pH will disrupt ionic bonds and hydrogen bonds formed between basic amine group in histidine/lysine and the acidic carboxylic group in aspartic acid which are critical to the tertiary and quaternary structure of the enzyme.

This will thus lead to changes in enzyme shape and shape of active site and hence loss of enzyme activity and denaturation. [1]

[3]

(b) Part of the chain of polypeptide Z is shown below.

CONHCHCONHCHCON

(CH2)4NH2

CH2CONH2

CHCONH

Draw the structure of the products formed when polypeptide Z is treated with excess aqueous NaOH under prolonged heating.

[1 mark each, must show carboxylate ion]

[3][Total: 10]


NH2CHCO2-

(CH2)4NH2

NH2CHCO2-

CH2COO-

HN

COO-

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H2 Mock Paper 2 Answers (1)