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NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2

CHEMISTRY 9647/01Paper 1 Multiple Choice 23 Sep 2013

1 hour

Additional Materials: Multiple Choice Answer Sheet Data Booklet

READ THESE INSTRUCTIONS FIRST

Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, class and tutor’s name on the Answer Sheet in the spaces provided unless this has been done for you. There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record you choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet.

This document consists of 19 printed pages.

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2

H2 Chemistry 9647/01 NYJC J2/2013 Prelim

Section A For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct.

1 A 15 cm3 mixture of carbon monoxide and methane was mixed with excess oxygen and exploded. There was a contraction in volume of 15 cm3 at room temperature and pressure. What is the percentage by volume of methane in the mixture? A 33 B 50 C 60 D 75

2 Nitrogen is essential in plants for the synthesis of proteins and nuclei acids. A farmer intends to switch his nitrogen fertiliser from potassium nitrate to urea. Currently, he uses 300 kg of potassium nitrate per month. Given the following information about the two fertilisers, calculate his monthly savings (to the nearest dollar) if he switches to urea as a fertiliser.

Fertiliser Potassium nitrate Urea

Chemical Formula KNO3 CO(NH2)2

Mr 101.1 60.0

Cost / tonne [1 tonne = 1000 kg]

$850 $340

A $310 B $225 C $195 D $153

3

H2 Chemistry 9647/01 NYJC J2/2013 Prelim [Turn over

+

–

R

S

T

3 The following are flight paths of charged particles when accelerated in an electric field.

Which of the following are correct positions for the particles?

R S T

A 16O+ 14C+ 14N+

B 15O⎯ 14C+ 28Si+

C 14N⎯ 15O+ 28Si2+

D 14N⎯ 14C+ 28Si4+

4 Consider the following reaction of aluminium chloride:

AlCl3 + Cl– ⇌ [AlCl4]–

Which of the following is correct regarding AlCl3 and [AlCl4]–?

A Both AlCl3 and [AlCl4]– are tetrahedral in shape.

B AlCl3 is tetrahedral in shape and [AlCl4]– is planar in shape.

C The Cl–Al–Cl bond angle in AlCl3 is larger than that in [AlCl4]–.

D There are more electron pairs around the Al atom in AlCl3 than that in [AlCl4]

–.

4


5 The graph of P / kPa against (1 / V) / m–3 was obtained for 0.400 g of an unknown gas at 25 C.

What is the relative molecular mass of the gas? A 19.3 B 21.0 C 28.0 D 47.1

6 In a calorimetric experiment conducted at standard conditions, 1.60 g of fuel underwent complete combustion. The reaction was 45 % efficient in heating 200 g of water. The temperature of the water rose to 73 C. Given that the specific heat capacity of water is 4.2 J g–1 K–1, what is the total energy released per gram of fuel burnt? A 18.1 kJ g-1 B 40.3 kJ g-1 C 56.0 kJ g-1 D 85.2 kJ g-1

7 Under standard conditions, solid iodine slowly sublimes into a violet gas. Data for each form of iodine is given in the table.

Hf / kJ mol–1 S/ J K–1 mol–1

I2 (s) 0 116

I2 (g) 62.4 261

What is the value of G in kJ K–1 mol–1, for the formation of gaseous iodine from solid iodine at 25 °C? A –15.4 B +19.2 C +27.8 D +106

P / kPa

60.8

0 1290 (1 / V) / m–3

5


0 0 0 0

8 The decomposition of hydrogen peroxide is known to be a first order reaction.

2H2O2 2H2O + O2

The rate constant is found to be 4.95 x 10–2 min–1. If the initial concentration of H2O2 is 4.0 mol dm–3, what will be the concentration of H2O2 after 42 min?

A 0.5 mol dm–3 B 1.0 mol dm–3

C 0.25 mol dm–3 D 2.0 mol dm–3

9 Which curve is obtained if the rate of reaction is plotted against time for an autocatalytic reaction (i.e. a reaction in which one of the products catalyses the reaction)?

A B C D

10 Consider the following equilibrium.

3Y2(g) + 2XY(s) ⇌ 2XY4(g)

When 3 mol of XY4 was heated in a 2 dm3 container, the equilibrium mixture contained 0.8 mol of XY. What is the numerical value of the equilibrium constant, Kc?

A 0.0286 B 0.179 C 5.60 D 35.0

6


11 Use of the Data Booklet is relevant to this question. Ammonia can be synthesised directly using a mixture of hydrogen gas and nitrogen gas in a reversible reaction as shown:

3H2(g) + N2(g) ⇌ 2NH3(g) Which one of the following describes the change in yield of ammonia at equilibrium as temperature increases?

A B

C D

12 Given that a 0.500 mol dm-3 of HNO2 solution has a Ka value of 7.1 x 10–4, calculate the percentage of undissociated HNO2 molecules.

A 1.88 % B 3.77 % C 96.2 % D 98.1 %

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8


15 During electrolysis under suitable conditions, 0.785 g of chromium is deposited on the cathode when a current of 5.0 A is passed into a chromium-containing electrolyte for 9.7 min.

What could the electrolyte be?

A CrCl2 B CrCl3 C K2CrO3 D K2CrO4

16 Which of the following statements is correct?

A The ionic radius of Na+ ion is greater than that of Cl– ion.

B The electronegativity of phosphorus is greater than that of nitrogen.

C The first ionisation energy of magnesium is less than that of aluminium.

D The energy required to remove an electron from F– is less than that for Ne atom.

17 Which graph represents the volume of gas collected when separate 1.0 g samples of powdered magnesium carbonate and powdered barium carbonate are heated at the same temperature?

A B

C D

Volume of gas / cm3

Time / s

BaCO3

MgCO3

Volume of gas / cm3

Time / s

MgCO3

BaCO3

Volume of gas / cm3

Time / s

BaCO3

MgCO3

Volume of gas / cm3

Time / s

MgCO3

BaCO3

9


18 Which of the following statements about chloride, iodide and their hydrogen halides is true?

A Hydrogen chloride is a stronger acid than hydrogen iodide.

B Chloride has a weaker oxidising strength than iodide.

C Boiling point of hydrogen chloride is lower than hydrogen iodide.

D Hydrogen chloride decomposes at a lower temperature than hydrogen iodide.

19 For which property are the data under the correct element?

property copper calcium

A density / g cm–3 8.92 1.54

B electrical conductivity / relative units 9.6 85

C melting point / °C 810 1083

D metallic radius / nm 0.197 0.117

20 What is the total number of all possible isomers for secondary alcohols with the molecular formula C5H12O? A 3 B 4 C 5 D 6

21 Which of the following statements about the reaction of Cl2 and ethane to form 1,2-dichloroethane is true?

A Cl2 acts as a catalyst.

B Ultra violet light is required for heterolytic fission to take place.

C CH3CH2CHClCH2Cl cannot be formed in the termination step. D Hybridisation about the reactive carbon centre changes from sp3 to sp2 in the

first propagation step.

10


22 In the upper atmosphere, chlorofluroalkanes (CFCs) are broken down to give chlorine radicals but not fluorine radicals. What is the best explanation for this?

A Fluorine is more electronegative than chlorine.

B The C─F bond is longer than the C─Cl bond.

C The C─F bond is stronger than the C─Cl bond.

D Fluorine radicals recombine immediately owing to their reactivity.

23 A study was conducted to measures the enthalpy change of mixing of PVC with plasticisers. As the reaction between PVC and plasticiser is too slow, low molecular weight analogues such as 2,4-dichloropentane are used instead of PVC.

CH3 CH CH2 CH CH3

Cl Cl

2,4-dichloropentane

Which of the following statements about 1 mole of 2,4-dichloropentane is false?

A It has 4 stereoisomers.

B One of its stereoisomer does not rotate plane polarised light.

C A total of 3 structural isomers are formed when it undergoes elimination.

D A maximum of 287g of white precipitate may be obtained when it is warmed with ethanolic silver nitrate.

24 When an organic compound X is shaken with aqueous silver nitrate at room temperature, there is no immediate precipitate. However, when X is boiled under reflux for some time with aqueous sodium hydroxide, cooled, acidified with dilute nitric acid and aqueous silver nitrate added, a white precipitate readily forms. What could X be? A C4H9Cl B CH3COCl C C6H5COCl D C6H5Cl

11


25 Aldehydes can undergo addition reactions with a variety of compounds of the form HX according to the following equation.

C O

H

HX COH

H X

An example of such a reaction is the formation of cyanohydrin, where X = CN. Which of the following compounds cannot be obtained by such an addition reaction to an aldehyde, followed by dehydration?

A

CH3CH NNH

B

O

H

OCH3

C

C C

H

CH3

D CH3

CH3

catalyst

12


26 Compound Q was subjected to the following tests and the results are recorded below.

Reagents and Conditions Observations

Acidified KMnO4, warm Purple solution decolourises, CO2 gas is evolved

2,4-DNPH, warm Orange precipitate observed

Cu(NO3)2, NaOH(aq), warm

No precipitate observed

Which of the following could be compound Q?

A

CHCH2

O H

B O CH3

CH3

C

OH

O H

D

CH3

H

O

27 In which of the following sequence is the value of pKa decreasing?

A C6H5OH > CH3CHClCHClCH2OH > CH3CH2CH2CH2OH > (CH3)3COH

B (CH3)3COH > CH3CH2CH2CH2OH > CH3CHClCHClCH2OH > C6H5OH

C CH3CH2CH2CH2OH > (CH3)3COH >CH3CHClCHClCH2OH >C6H5OH

D CH3CHClCHClCH2OH > (CH3)3COH > CH3CH2CH2CH2OH > C6H5OH

13


28 The structure of morphine is shown below.

NCH3

HO

HO

O

The reaction between morphine and dry HBr (g) gives

A B

NCH3

HO

HO

O

Br

Br

NHCH3Br

Br

HO

O

Br

Br

C D

NCH3

HO

HO

O

Br

Br

Br

NHCH3Br

Br

HO

O

Br

14


29 Which of the following reactions will take place under the given reagents and conditions?

A

B

C

D

30 What will be the net charge of the following part of the protein chain at pH 7.4?

–Lys–Glu–Lys–

Amino acid

C CO2H

H

NH2

(CH2)4

NH2

C CO2H

H

NH2

(CH2)2

HO2C

lysine (lys) Glutamic acid (glu)

pKa of R group

10.53 4.25

A –1 B 0 C +1 D +2

NH2 + CO2H + H2OCONHheat

conc H2SO4

NH2

CH3

Br2 (aq) NH2

CH3

Br+ HBr

rm temp

CONH2 + NH3

aq NaOHCO2H

heat, reflux

NH2I

N+

+

I-

heat, high pressure

excess

+ 2HI

15


Section B For each of the questions in this section one or more of the three numbered statements 1 to 3 may be correct. Decide whether each of the statements is or is not correct (you may find it helpful to put a tick against the statements which you consider to be correct). The responses A to D should be selected on the basis of

A B C D 1, 2 and 3

are correct

1 and 2 only are correct


1 only is

correct No other combination of statements is used as a correct response.

31 Which of the following molecules are bent?

1 SCl2

2 NO2+

3 XeF2

32 Which of the following statements are incorrect?

1 All substances with covalent bonding cannot conduct electricity.

2 Ionic compounds can be distinguished from metals by their electrical conductivity in the liquid states.

3 Metals are conductors of electricity in all states.

16


The responses A to D should be selected on the basis of

A B C D 1, 2 and 3

are correct



1 only is


33 The energy profile for a reaction, P R is shown below.

Which of the following statements are true?

1 The mechanism for the reaction is

P Q slow

Q R fast

2 The addition of a catalyst does not affect H. 3 Increasing the temperature decreases both the values of E1 and E2.

17



A B C D 1, 2 and 3

are correct



1 only is


34 The table below shows the numerical value of the solubility product, in mol dm–3 for three metal sulfides. In an acidic solution, [S2–] in a saturated solution = 10–18 mol dm–3.

Metal ion Mn2+ Ni2+ Ag+

Ksp of sulfide 10–16 10–21 10–36

Which of the following statements are correct?

1 NiS and Ag2S would be precipitated from the acidic solution containing 0.010 mol dm–3 of the metal ion when the solution is saturated with hydrogen sulfide.

2 Ag2S is the least soluble salt because its Ksp is the smallest amongst the three sulfides.

3 When the concentration of sulfide is increased, the value of the solubility product for the three metal sulfides will increase.

35 Which of the reagents can be used to differentiate the pair of compounds given?

Compound 1 Compound 2 Reagent

1 AlCl3 SiCl4 H2O

2 Al2O3 SiO2 dil HCl

3 NaCl MgO dil H2SO4

36 Which transition metal ions possess unpaired electrons?

1 TiO2

2 Ni(CO)4

3 [Fe(CN)6]4–

18



A B C D 1, 2 and 3

are correct



1 only is


37 Which of the following organic compounds will produce 1,2–benzene–dicarboxylic acid when subjected to the reagents and conditions as specified?

1 , KMnO4/H+, heat

2

COCl

COCl

, AgNO3 (aq)

3

O

O

O

O, NaOH (aq), heat

38 Which of the following reagents will result in different observations when reacted with the following two compounds?

CH2CH2OH OH

CH2CH3

1 Br2 (aq)

2 K2Cr2O7 / H+ (aq)

3 NaOH (aq)

19



A B C D 1, 2 and 3

are correct



1 only is


39 Which products are formed when phenylpropanoate undergoes acidic hydrolysis in the presence of water labelled with 18O isotope?

1 CH3CH2CO18OH

2 C6H518OH

3 C6H5CO18OH

40 A tripeptide, thr-ala-arg, was analysed using electrophoresis. The tripeptide was hydrolysed and the resulting solution was then placed at the centre of the plate in a buffer solution of pH 7.0. A potential difference was then applied across the plate.

Amino acid

C CO2H

H

NH2

C

H

OHCH3

C CO2H

H

NH2

CH3

C CO2H

H

NH2

(CH2)3

NH

CNH NH2

threonine (thr) alanine (ala) arginine (arg)

Isoelectric point

5.60 6.00 10.76

With reference to the above table, which of the following statements is correct?

1 Alanine would migrate towards the anode while arginine would migrate towards the cathode.

2 Threonine would migrate a lesser distance from the centre of the plate as compared to alanine.

3 The above amino acids will exist as H3N+CHRCO2

– at their isoelectric point.

2013 NYJC H2 Chem Prelim Paper 1 Answer Key

1 A 11 A 21 D 31 D 2 B 12 C 22 C 32 B 3 D 13 D 23 A 33 B 4 C 14 A 24 A 34 D 5 B 15 A 25 D 35 B 6 C 16 D 26 A 36 C 7 B 17 A 27 B 37 B 8 A 18 C 28 D 38 A 9 C 19 A 29 D 39 D 10 C 20 C 30 C 40 B

H2 Chemistry 9647/02 NYJC J2/2013 Prelim [Turn Over


CANDIDATE NAME

CLASS

TUTOR’S NAME

CHEMISTRY 9647/02Paper 2 Structured Questions 23 September 2013

2 hoursCandidates answer on the Question Paper. Additional Materials: Data Booklet.


Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions in the spaces provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use

1 /12

2 /22

3 /10

4 /18

5 /10

Total /72

This document consists of 17 printed pages.

2


For Examiner's

Use

Answer all questions in the spaces provided. 1 Planning (P)

Acids can be classified in terms of their basicity depending on the number of protons that can be donated to a base. A student is investigating the basicity of a solution of 0.80 mol dm-3 acid, A, using just the acid and sodium hydroxide of the same concentration. He conducted two experiments by mixing different volumes of A and sodium hydroxide, and taking the temperature rise for each experiment.

(a) Based on the definition of the enthalpy change of neutralisation (∆Hn) and his results, deduce the basicity of acid A.

Experiment Volume of acid Volume of NaOH ∆T /oC

1 40 20 3.6

2 20 40 7.2

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

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[2]

(b) (i) The temperature rise of the reaction can be determined graphically by measuring how the temperature changes with time.

You are to plan an experiment to determine the ∆Hn between a solid sample of A and sodium hydroxide.

You are provided with a solid sample of A (Mr = 174) and sodium hydroxide solution of 0.80 mol dm-3.

In your plan, you should give:

details including calculations to determine the quantities of the reactants to be used;

choice of apparatus (You may use apparatus normally found in the school laboratory.);

the essential details for obtaining the graph of temperature against time in order to determine the temperature rise;

table(s) of readings to taken.

3


For Examiner's

Use

Calculations:

Procedure:

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Table of Results

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For Examiner's

Use

(ii) Sketch the graph that you would expect, showing clearly how you can determine the temperature rise.

[8]

(c) Explain why temperature rise in the experiment should not be too low.

…………………………………………………………………………………………….…

……………………………………………………………………………………………….

[1] (d) An alternative method to determine the temperature rise is to measure the highest

temperature reached. Explain if this is a better method.

…………………………………………………………………………………………….…

……………………………………………………………………………………………….

[1]

[Total: 12]

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For Examiner's

Use

2 One of the most important features of the transition elements is that they exhibit variable oxidation states. This question illustrates the various oxidation states shown by iron in its compounds.

(a)

The reaction between iodide ions, I-, and peroxidisulfate ions, S2O82-, is slow.

The reaction can be catalysed by adding a small amount of Fe2+ ions. The initial rate of the slow reaction between iodide ions and peroxidisulfate ions can be studied by using thiosulfate ions. The equations for the reactions are as follows. 2I- + S2O8

2- I2 + 2SO42- (slow) reaction I

I2 + 2S2O3

2- 2I- + S4O62- (fast) reaction II

In the presence of a constant amount of thiosulfate ions, the iodine being slowly produced by reaction l will immediately react in reaction ll until all the thiosulfate ions has been used up. At that point, free iodine will be present in the solution, which will cause a sudden appearance of a deep blue colour if starch is present. A series of experiments was carried out using different volumes of the five reagents. The following results were obtained.

Expt Volume of

S2O82-

/cm3

Volume of I-

/cm3

Volume of

S2O32-

/cm3

Volume of

distilled water /cm3

Volume of

Starch /cm3

Time for the

appearance of deep blue

colour/s 1 20 20 10 5 5 30 2 20 15 10 10 5 40 3 5 25 10 15 5 t34 10 15 10 20 5 80

(i) If the orders of reaction with respect to peroxidisulfate ions and iodide ions are both one respectively, deduce an expression relating the volume of these two reactants and time taken for the appearance of deep blue colour. Explain your reasoning.

6


For Examiner's

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(ii) Hence, predict the time, t3, required for the appearance of deep blue colour in experiment 3. ……………….…………………………………………………………………………

…………………….……………………………………………………………………

………………………….………………………………………………………………

……………………………….…………………………………………………………

[3]

(b) (i) In the Haber process, the rate of formation of ammonia is increased by using an iron catalyst. State the type of catalysis occurring here and describe using the Maxwell-Boltzmann Distribution curve, how the iron catalyst increases the rate of reaction.

…………………….……………………………………………………………………

………………………….………………………………………………………………

…………………….……………………………………………………………………

………………………….………………………………………………………………

(ii) Ammonia shows significant deviations from the ideal behaviour that is predicted by the kinetic theory of gases. State two assumptions of the kinetic theory of gases.

……………….…………………………………………………………………………

…………………….……………………………………………………………………

………………………….………………………………………………………………

…………………….……………………………………………………………………

7


For Examiner's

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(iii) A sample of ammonia can be liquefied at room temperature just by increasing the pressure of the sample. Why does the application of pressure cause the gas to liquefy? …………………….……………………………………………………………………

………………………….………………………………………………………………

…………………….……………………………………………………………………

………………………….………………………………………………………………

[7]

(c) Iron(III) chloride is a dark brown solid which dissolves in water to give an acidic solution. This solution is often used, in the electronics industry, to etch (dissolve) the copper used in printed circuit boards.

(i) Explain, with the aid of a chemical equation, why aqueous iron(III) chloride is acidic. …………………….……………………………………………………………………

………………………….………………………………………………………………

…………………….……………………………………………………………………

(ii) By selecting appropriate E values from the Data Booklet, explain why aqueous iron(III) chloride etches (dissolves) copper. Construct a balanced equation for this reaction. …………………….……………………………………………………………………

…………………….……………………………………………………………………

………………………….………………………………………………………………

…………………….……………………………………………………………………

[4]

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For Examiner's

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(d) Iron(III) iodide has been successfully made under non-aqueous conditions by following method. Iron pentacarbonyl, Fe(CO)5, reacted with iodine in hexane solution to form a solution of compound B, whose molecular formula is FeC4O4I2. A further calculated amount of iodine is added, and the solution exposed to uv light. A black precipitate of iron(III) iodide results, and a colourless gas is evolved.

(i) State the co-ordination number of iron in the complex, Fe(CO)5 and hence predict the shape of the complex, Fe(CO)5. Co-ordination number: ………………………….………………...…………………

Shape: ………………………...………………………………………………………

(ii) Describe the bonding between iodine and iron in compound B and write a balanced equation for the reaction between compound B and iodine. …………………….……………………………………………………………………

…………………….……………………………………………………………………

…………………….……………………………………………………………………

[4]

(e) When a solution of NaClO is added to a strongly alkaline suspension of Fe2O3, a purple solution results. When BaCl2(aq) is added to the solution, a red solid is precipitated with the composition by mass: Ba, 53.4%; Fe, 21.7%; O,24.9%.

(i) Calculate the empirical formula of the red solid, and hence determine the oxidation number of iron in the red solid.

(ii) Construct a balanced ionic equation for the reaction between Fe2O3, ClO- and OH-. …………………….……………………………………………………………………

[4]

[Total: 22]

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For Examiner's

Use

3 Aromatic sulfonic acid is an important precursor to industrial materials and has the following structure:

Aromatic sulfonic acid

(a) Sulfonation of benzene is the reaction mechanism between benzene and sulfuric acid. The first step of the reaction mechanism is given by the following equation.

H2SO4 H2O + SO3 Sulfur trioxide is an important intermediate in the reaction mechanism.

(i) Draw a dot-and-cross diagram of sulfur trioxide.

(ii) With reference to the structure of the molecule, state and explain the function of sulphur trioxide in the mechanism. …………………….……………………………………………………………………

…………………….……………………………………………………………………

…………………….……………………………………………………………………

…………………….……………………………………………………………………

…………………….……………………………………………………………………

CHCH

CCH

CHCH

S

O

O

OH

10


For Examiner's

Use

(iii) The final step of the mechanism is different from other benzene reactions because the hydrogen atom is not removed from the ring by a separate negative ion. Instead it is removed by a lone pair of electrons on a negative oxygen atom in the sigma complex. The sigma complex formed has the following structure:

Sigma complex With the above information, complete the mechanism for the reaction between benzene and sulfuric acid.

(iv) Sulfonation of aniline produces sulfanic acid compounds with an unusually high melting point. With reference to the structure of sulfanic acid, suggest why this is so.

Sulfanic acid

…………………….……………………………………………………………………

…………………….……………………………………………………………………

…………………….……………………………………………………………………

[7]

SHO

OO-

+

CHCH

CC

CHCH

S

O

O

OH

NH2

11


For Examiner's

Use

(b) The key reaction during the Contact process is as follows:

2SO2 (g) + O2 (g) 2SO3 (g); ∆H = - 98 kJ mol-1 Industrially, the gases are compressed to a pressure of 3 atm before it is passed over a bed of vanadium(V) oxide at 430 ○C. Explain the conditions used.

…………………….……………………………………………………………………….…

……………….……………………………………………………………………………….

…….…………………….……………………………………………………………………

……………….…………………………………………………………………………….…

…….…………………….……………………………………………………………………

…….…………………….……………………………………………………………………

[3]

[Total: 10]

12


For Examiner's

Use

4 Magnesium is the fourth most abundant element in the earth. Scientists are working on developing rechargeable magnesium-ion batteries for use in electric cars, as they offer a cheaper and more energy efficient alternative to lithium-ion batteries. A simplified representation of a magnesium-ion battery is shown below. When the battery discharges, magnesium ions are formed, which move across the cation exchange membrane to the Sn electrode, forming solid Mg2Sn.

(a) (i) Write half equations for the reactions occurring at the cathode and the anode when the battery discharges. Cathode: ……………….…………………………………………………………..…

Anode: ….…………………….………….……………………………………………

(ii) The magnesium-ion battery provides a voltage of 3.01 V. With reference to the Data Booklet, calculate E (Sn/Mg2Sn), assuming standard conditions.

13


For Examiner's

Use

(iii) The car battery has to be recharged after use. After clocking a mileage of 500 km, 83.8 g of Mg2Sn was deposited on the electrode of the battery. How many hours will it take for the car battery to be recharged, using a current of 10.0 A? What is the polarity of each electrode during the charging process?

Mg electrode: ………….…………………………………………………………..…

Sn electrode: ..……………….………….……………………………………………

(iv) Besides cost, explain another advantage the magnesium-ion battery has over the lithium-ion battery. ….……….…………………….…………………………………………………….…

………………….…………………….………………………………………………..

[8]

(b) Magnesium is also used in the synthesis of Grignard reagents, RMgX (R = alkyl or aryl groups, X = Cl or Br), from organic halides. The organic halide is added to a suspension of magnesium in dry ether, forming the Grignard reagent

RX + Mg RMgX Grignard reagents are examples of organometallic compounds as they contain a carbon-metal bond. They are useful for organic synthesis as they can form new carbon-carbon bonds with unsaturated compounds such as carbonyls. The reaction of methanal and a general Grignard reagent is illustrated below:

O

H

H

RMgX+1. dry ether

2. H+ (aq)

OH

H

R

H

(i) Draw the displayed formula of the product of the reaction between butanone and C6H5MgCl.

14


For Examiner's

Use

(ii) How would you determine if the reaction in b(i) has gone to completion? ……………….…………………………………………………………………………

….……….…………………….…………………………………………………….…

………………….…………………….……………………………………….……….

(iii) The Grignard reagent can be considered as a good source of a carbanion, “R–”. Using this information, explain whether the product mixture formed will be optically active. Explain your answer with the aid of a diagram. ……………….…………………………………………………………………………

….……….…………………….…………………………………………………….…

………………….…………………….……………………………………….……….

[5]

15


For Examiner's

Use

(c) Refer to the reaction scheme involving a Grignard reagent below.

(i) Suggest the identities of C, D and E.

(ii) Suggest reagents and conditions for Step 1 and 2. Step 1: …………………………………………… Step 2: ……………….……………………………

[5]

[Total: 18]

Step 1 C

Mg

dry ether Grignard reagent

1. E

2. H+ (aq)

D

Na2CO

3 (aq)

Step 2

Effervescence

OH

HO Br

16


For Examiner's

Use

5 Some chemistry students were doing an organic chemistry practical.

(a) Student X heated compound F with hydrochloric acid to form two organic products. In order to isolate both products, the reaction mixture was placed in a separating funnel and shaken with water.

Compound F

(i) Draw the products of the above reaction.

(ii) Explain why the separation will not be successful.

……………….…………………………………………………………………………

….……….…………………….………………………………………………….….…

………………….…………………….……………………………………….….…….

……………….…………………………………………………………………………

(iii) Describe the correct procedure to separate the hydrolysis products.

……………….…………………………………………………………………………

….……….…………………….……………………………………………………..…

………………….…………………….……………………………………….………..

……………….…………………………………………………………………………

CH

CH

CH

C

CH

CH

NH C

O

CH3

17


For Examiner's

Use

(b) Under acidic conditions, Student Y treated ethylamine with excess iodomethane in an attempt to form a quarternary amine salt. Explain why this synthesis will not be successful.

……………….…………………………………………………………………………….…

…….…………………….…………………………………………………….…………......

.………….…………………….……………………………………….…………………….

[2]

(c) Student Z predicted that compound G was formed from the reaction of propene and aqueous bromine in the presence of concentrated sodium chloride.

C C

H

H

OH

H

Br

CH3

Compound G

(i) Explain why his prediction is wrong.

……………….…………………………………………………………………………

….……….…………………….…………………………………………………….....

.………….......………….…………………….………………………………………..

.………….......………….…………………….………………………………………..

(ii) Hence, suggest the structure of the major product formed.

[3]

[Total: 10]


CHEMISTRY 9647/03Paper 3 Free Response 19 Sep 2013

2 hours Candidates answer Section A on the Question Paper Additional Materials: Answer Paper Data Booklet


Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers.

At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Section B Answer all questions on separate answer paper.

For Examiner’s Use

Section A

1

2

3

4

5

Total

This document consists of 11 printed pages

2

Answer any four questions.

1 The elements of Group VII of the Periodic Table are called halogens, which mean "salt formers". They lack only one electron to form a complete shell, and are extremely active chemically. This question examines the chemistry of two reactions of halogens.

(a) In the first reaction, a 1:2 mixture of H2 and I2 gases was mixed at a constant temperature of 430 oC and a constant pressure of 2 atm. The following equilibrium was set up.

H2 (g) + I2 (g) 2HI (g) ∆H = −11 kJ mol1

(i) Given that degree of dissociation of H2 is 86 %, calculate the equilibrium

constant, Kp for the above reaction under these conditions.

(ii) Suggest, with reasoning, a way to increase degree of dissociation of H2 for the above reaction.

(iii) Deduce how Kp for the following system would be compared to your answer in (i).

H2 (g) + Br2 (g) 2HBr (g) [7]

In another reaction, halogens can react with unreactive alkanes to form mono-

substituted product. An example is the reaction of chlorine with 2-methylpentane shown below.

2-methylpentane

(b) (i) Explain why alkanes are generally unreactive.

(ii)

This reaction is seldom used for synthesis as there are many associated problems. Firstly, several isomeric products are formed. The relative ratio of the isomeric products may be more accurately determined if relative rates of abstraction of H atoms are taken into account.

The relative rates of abstraction of H atoms are

Type of H atoms Relative rate of abstraction Primary 1

Secondary 4 Tertiary 6

By examining the difference in stability of the intermediates formed when different types of H atom are abstracted, explain the trend in relative rate of abstraction.

3


(iii)

Predict the ratio of the following two products A and B, taking into account the relative rates of abstraction given in (ii). Explain your reasoning.

Cland

AB

Cl

(iv) Describe suitable chemical test(s) to distinguish compounds A and B in (iii). [7] (c) Another problem of the reaction is multi-substitution.

(i) Suggest the conditions that will give rise to formation of multi-substituted

products.

(ii)

A di-substituted product C with molecular formula C6H12Cl2 was obtained in the reaction of chlorine with 2-methylpentane. C does not exhibit optical activity. On warming with alcoholic KOH, a major product D is formed. When D is treated with hot acidified potassium manganate(VII) solution, effervescence was observed and the only organic product remaining gives a yellow precipitate with alkaline aqueous iodine. Use this information to deduce the structure of C and D, explaining your reasoning.

[6]

[Total: 20]

4 2 Precipitation plays an important function in growth of bones and teeth in our bodies.

Teeth and bones are largely composed of calcium phosphate salt Ca3(PO4)2. In order for this precipitation to occur, the concentrations of the ions in blood must exceed the solubility product in the immediate region of deposition.

(a) Write an ionic equation with state symbols to show the formation of teeth and bones at the deposition region and hence an expression for the solubility product of the Ca3(PO4)2 deposits. [2]

(b) (i) At a temperature of 37 oC, the solubility product for calcium phosphate Ca3(PO4)2 has a numerical value of 4 × 10-27. In a sample of blood at a pH of 7.4, the concentrations of calcium and phosphate ions are found to be 1.2 × 10-3 mol dm-3 and 1.6 × 10-8 mol dm-3 respectively. State and explain if the above composition in blood will result in the growth or degeneration of teeth and bones.

(ii) Hence, state and explain if the pH environment of 7.4 in (i) is that for a growing child, a fully grown man or an old person. [3]

(c) The pH at the area of growth may change due to metabolic processes in the body. The hydrogen ions that are present combine phosphate ions as hydrogen phosphate ions. This prevents growth of strong teeth and bones.

PO43- (aq) + H+ (aq) HPO4

2- (aq)

Assuming that the [HPO42-

] is maintained at 1.7 × 10-4 mol dm-3 by cell functions,

(i) calculate the pH that will result in neither growth nor degeneration at the teeth and bones deposition region when the concentration of calcium ions in blood is maintained at 1.2 × 10-3 mol dm-3. [Ka of HPO4

2- = 4.4 × 10-13 mol dm-3]

(ii) Suggest another acid–conjugate base equation that accounts for the constant supply of HPO4

2- ions from cell functions. [3]

5


(d)

(i) Construct an energy level diagram for the formation of Ca3(PO4)2 from its elements using relevant data from the table below and other appropriate values in the Data Booklet. Hence calculate the lattice energy of Ca3(PO4)2. Enthalpy ∆H / kJ mol-1

Standard enthalpy change of formation of Ca3(PO4)2 −4132 Standard enthalpy change of atomisation of Ca +178 Standard enthalpy change of atomisation of P +315 P(g) + 4O(g) + 3e → PO4

3-(g) −2810

(ii) Using your answers to (i) and other enthalpy change value if any, explain why Ca3(PO4)2 is insoluble.

[6]

(e) Use of the Data Booklet is relevant to this question. Group VII elements exist as ions and play a vital role in bodily functions. The following observations are made about Group VII elements and their reactions. Explain the observations made.

(i) F2 cannot be prepared by electrolytic oxidation of aqueous F- solutions.

(ii) HI can be prepared by treating NaI with phosphoric acid, H3PO4 but not with concentrated sulfuric acid.

(iii) Interhalogen compounds such as IF3, IF5 and IF7, BrF3, BrF5, ClF3, ICl3 exists. On the other hand, ClF5 is not easily formed and BrCl3 does not exists.

[6]

[Total: 20]

6 3 (a) Period 3 elements react with oxygen to form the corresponding oxides of varying

acid-base nature. Using magnesium, aluminium and phosphorus as examples, describe the acid-base nature of the oxides across Period 3 with suitable equations. [4]

(b) Aluminium is a reactive metal but its reactivity is limited due to a layer of inert aluminium oxide being formed. Both aluminium and its oxide react readily with aqueous sodium hydroxide to give similar products. Aluminium, however, produces another product which is a combustible gas.

(i) Write a balanced equation for the reaction when aluminium is heated in aqueous sodium hydroxide.

(ii) Explain why a homogenous solution is obtained in (i). [3]

(c) Using 2,6−dichloroheptane as the starting compound, suggest the reagents and conditions for steps I, II and IV and the products J to M in the synthetic scheme below. [7]

Cl Cl

PCl5

O O

O O

IV

I

II

III

VJ

K L

M

N

7


(d) (i) Propyne, C3H4, like ethyne, C2H2, is commonly used for blow torches to cut metals in heavy metal industries. By drawing an appropriate energy cycle, determine the enthalpy of combustion of propyne, given the following enthalpy values: Hf(C3H4) = +184 kJ mol⎯1

Hc(C) = −394 kJ mol⎯1

Hc(H2) = −286 kJ mol⎯1

(ii) Determine the enthalpy of combustion of propyne using bond energy values. Explain why the use of bond energy values to calculate the enthalpy of combustion is significantly less exothermic that those obtained in (i).

[6]

[Total: 20]

8 4 Nickel is a typical transition element which is commonly used as a catalyst and metal

for electroplating. Nickel also forms complex ions with ligands such as H2O, NH3 and NH2CH2CH2NH2 to give various coloured complexes.

(a) What do you understand by the terms transition element, ligand and complex ion? [2] [3]

(b) (i) Explain why transition metal complexes are coloured.

(ii) Complex ions of nickel such as [Ni(H2O)6]2+, [Ni(NH3)6]

2+ and [Ni(NH2CH2CH2NH2)3]

2+ are green, blue and violet colour respectively. The visible absorption spectra of the three complex ions of nickel are shown below: Suggest an explanation for the difference in colour of the three complexes and using the colours of the complexes given, assign the peaks A, B and C to the three complex ions of nickel. [3] [5]

wavelength / nm

relative absorbance

A

B

C

700 600 500

colour of visible light

red orange green

650 550

yellow

9


(c) When Ni2+ (aq) is mixed with an excess of NH3 (aq), each H2O is replaced by a NH3 at a time. The stepwise formation of [Ni(NH3)6]

2+ from [Ni(H2O)6]2+ can be

summarised by the following equations with their associated equilibrium constants: Step 1: [Ni(H2O)6]

2+ + NH3 [Ni(H2O)5(NH3)]2+ + H2O K1 = 630 mol–1dm3

Step 2: [Ni(H2O)5(NH3)]

2+ + NH3 [Ni(H2O)4(NH3)2]2+ + H2O K2 = 160 mol–1dm3

and so on for steps 3, 4 and 5. Step 6: [Ni(H2O)(NH3)5]

2+ + NH3 [Ni(NH3)6]2+ + H2O K6 = 1.1 mol–1dm3

(i) The stepwise formation of [Ni(NH3)6]

2+ from [Ni(H2O)6]2+ undergoes a

dissociative mechanism which resembles a SN1 mechanism in organic chemistry. Suggest a possible mechanism for Step 1 and show clearly how the shape of the complex ion changes.

(ii) Sketch an energy profile diagram for the suggested mechanism in (i) given the reaction is exothermic.

(iii)

Suggest a possible reason for the decrease in value of the equilibrium constant for each successive replacement of water by ammonia ligands. [6]

(d) [Ni(H2O)6]2+ can also undergo ligand exchange reaction with NH2CH2CH2NH2 to

form [Ni(NH2CH2CH2NH2)3]2+. Using the signs of G, H and S, explain how the

formation of [Ni(NH2CH2CH2NH2)3]2+ would compare to that of [Ni(NH3)6]

2+. [2] [3]

(e) With the aid of a clearly labelled diagram, show the calculations you would carry out to determine a value for the Avogadro constant by the electroplating of nickel from nickel(II) ions. You may assume that you are provided with the following data. • mass of nickel deposited, m • current used, I • time taken, t [3]

[Total: 20]

10 5 (a) Amino acids play central roles both as building blocks of proteins and as

intermediates in metabolism. The 20 amino acids that are found within proteins convey a vast array of chemical versatility and they have relatively high melting points and are soluble in water.

(i) Define the secondary structure of proteins.

(ii) What is meant by the term denaturation of proteins?

(iii) Suggest a brief outline of one method by which proteins may be denatured. [4]

(b) An octapeptide A was analyzed and contained the following amino acids:

Amino acid abbreviation R-group Number of residues

Aspartic acid Asp – CH2CO2H 1

Glycine Gly – H 2

Histidine His

NH

N

CH2

CH3

1

Phenylalanine Phe CH2 1

Proline Pro

CH2

CH2 CH2

(cyclic)

2

Valine Val – CH(CH3)2 1

(i) Analysis showed that the C-terminus contained an amino acid which is

optically inactive and partial hydrolysis gives the following fragments:

Val-Pro-His, Gly-Asp-Pro-Phe, Pro-Phe-Val Use the above results to deduce the amino acid sequence of the octapeptide A and explain your reasoning.

(ii) Draw the displayed formula of Gly-Asp-Pro which is part of the octapeptide chain in A.

[4]

11


(c) Histidine is utilized by your body to develop and maintain healthy tissues. It is especially important in the myelin sheath that coat nervous cells to ensure the transmission of messages from your brain to organs throughout your body. Histidine also stimulates the secretion of the digestive enzymes gastrin and is also required to manufacture both red and white blood cells.

NHN

COOH

NH2

1

2

Structure of histidine

Histidine contains a total of 3 nitrogen atoms and each N atom exhibit different properties in aqueous solution.

(i) Considering the orbital arrangement of the atoms, explain why 2N atom in histidine is less basic than 1N atom.

(ii) There are three pKa values associated with the histidine: 1.8, 6.0, 9.2.

Assuming that the 2N atom does not exhibit any basic properties, make use of the three pKa values to suggest the major species present in solutions of histidine with the following pH values. • pH 1 • pH 5 • pH 7 • pH 11

(iii) Consider the buffer solution at pH = 5, calculate what fraction of the histidine side chains will carry a positive charge at pH 5.

[8]

(d) 25.0 cm3 of 0.150 mol dm3־ of diethylamine solution, (CH3CH2)2NH, was placed in a conical flask and titrated against a solution of 0.200 mol dm-3 hydrochloric acid from a burette. Given that the Kb of (CH3CH2)2NH is 8.60 x 10-4 mol dm-3,

(i) explain what is meant by the term base dissociation constant, Kb of diethylamine.

(ii) Calculate the pH at equivalence point.

[4]

[Total: 20]

1


For Examiner's

Use

Answer all questions in the spaces provided. 1 Planning (P)

Acids can be classified in terms of their basicity depending on the number of protons that can be donated to a base. A student is investigating the basicity of a solution of 0.80 mol dm-3 acid, A, using just the acid and sodium hydroxide of the same concentration. He conducted two experiments by mixing different volumes of A and sodium hydroxide, and taking the temperature rise for each experiment.

(a) Based on the definition of the enthalpy change of neutralisation (∆Hn) and his results, deduce the basicity of acid A.

Experiment Volume of acid Volume of NaOH ∆T /oC

1 40 20 3.6

2 20 40 7.2

If it is a dibasic acid, n(H2O) formed in experiment 2 is twice that of experiment 1. Since total volume is constant, ∆T for experiment 2 is doubled. Hence it is dibasic. If it is a monobasic acid, n(H2O) formed in expt 1 = n(H2O) formed in expt 2 Since total volume is constant, ∆T must be the same for both experiments.

(b) (i) The temperature rise of the reaction can be determined graphically by measuring how the temperature changes with time.

You are to plan an experiment to determine the ∆Hn between a solid sample of A and sodium hydroxide.

You are provided with a solid sample of A (Mr = 174) and sodium hydroxide solution of 0.80 mol dm-3.

In your plan, you should give:

details including calculations to determine the quantities of the reactants to use;

choice of apparatus (You may use apparatus normally found in the school laboratory.);

the essential details for obtaining the graph of temperature against time in order to determine the temperature rise;

table(s) of readings to taken.

Calculations:

Volume of NaOH to use = 40 cm3 Amount of acid used in experiment 2 = 0.016 mol Mass of acid used in experiment 2 = 0.016 x 174 = 2.784 g To ensure acid is limiting here, we use 2.5 g. Procedure:

1. Using a weighing balance, weigh accurately about 2.5 g of A. 2. Using a 50 cm3 measuring cylinder, measure 50 cm3 of NaOH and add into a

Styrofoam cup and cover with a lid. 3. Using a 0.2 oC division thermometer, note the temperature of the solution at 1 min

interval for the first 3 minutes. 4. At 3.5 min, rapidly add sample A into the cup, close the lid. 5. Stir the mixture. 6. Measure the temperature of the mixture at the 4th minute, and at every 1 minute

interval until the 10th minute.

Apparatus: weighing balance, 50 cm3 measuring cylinder, Styrofoam cup, 0.2 oC division thermometer. (2 items missing, 0) Timing: take initial temp for at least 3 min, time intervals (at most 1 min), read minimum of 5 readings after mixing. (2 items missing, 0) General procedure: rapidly pour, stir, close lid. (2 missing, 0) Quantities: vol. of NaOH, mass of X (no penalty since calculation was done)

Table of readings: Mass of X and weighing bottle /g Mass of weighing bottle and residue /g Mass of empty weighing bottle /g Mass of X used /g Time /min Temperature /oC

3


For Examiner's

Use

(ii) Sketch the graph that you would expect, showing clearly how you can determine the temperature rise.

∆T 3.5 (or point of mixing)

[8]

(c) Explain why temperature rise in the experiment should not be too low. Too low, result in high percentage error.

[1] (d) An alternative method to determine the temperature rise is to measure the highest

temperature reached. Explain if this is a better method.

Poorer method as it does not take into account heat loss to the surroundings.

[1] [Total: 12]

T /°C

Time / s

2 One of the most important features of the transition elements is that they exhibit variable oxidation states. This question illustrates the various oxidation states shown by iron in its compounds.

(a)

The reaction between iodide ions, I-, and peroxidisulfate ions, S2O82-, is slow.

The reaction can be catalysed by adding a small amount of Fe2+ ions. The initial rate of the slow reaction between iodide ions and peroxidisulfate ions can be studied by using thiosulfate ions. The equations for the reactions are as follows.

2I- + S2O82- I2 + 2SO4

2- (slow) reaction I

I2 + 2S2O32- 2I- + S4O6

2- (fast) reaction II In the presence of a constant amount of thiosulfate ions, the iodine being slowly produced by reaction l will immediately react in reaction ll until all the thiosulfate ions has been used up. At that point, free iodine will be present in the solution, which will cause a sudden appearance of a deep blue colour if starch is present. A series of experiments was carried out using different volumes of the five reagents. The following results were obtained.

Expt Volume of

S2O82-

/cm3

Volume of I-

/cm3

Volume of

S2O32-

/cm3

Volume of

distilled water /cm3

Volume of

Starch /cm3

Time for the

appearance of deep blue

colour/s 1 20 20 10 5 5 30 2 20 15 10 10 5 40 3 5 25 10 15 5 t34 10 15 10 20 5 80

(i) If the orders of reaction with respect to peroxidisulfate ions and iodide ions are both one respectively, deduce an expression relating the volume of these two reactants and time taken for the appearance of deep blue colour. Explain your reasoning. Rate = k [S2O8

2-][ I-] Since total volume of solution is constant, concentration of a reactant volume of reactant rate 1/time 1/time = k (volume of S2O8

2-)(volume of I-) (volume of S2O8

2-)(volume of I-) x time = constant

5


For Examiner's

Use

(ii) Hence, predict the time, t3, required for the appearance of deep blue colour in experiment 3. time for expt 3, t3 = 96 s

[3](b) (i) In the Haber process, the rate of formation of ammonia is increased by using

an iron catalyst. State the type of catalysis occurring here and describe using the Maxwell-Boltzmann Distribution curve, how the iron catalyst increases the rate of reaction.

Heterogeneous catalysis Boltzmann Distribution Diagram: In the presence of a catalyst, a reaction has a different mechanism with a lower activation energy

compared to the uncatalysed reaction. More molecules will possess energy greater than this lowered activation

energy, hence frequency of effective collisions will increase. Hence, rate of reaction increases.

(ii) Ammonia shows significant deviations from the ideal behaviour that is predicted by the kinetic theory of gases. State two assumptions of the kinetic theory of gases.

2 assumptions for ideal gas behaviour are: - Particles in gaseous state do not exert any force or negligible forces of

attraction; - volume of particles is negligibly small compared with that of the container.

Key: No of molecules with energy ≥ Ea No of molecules with energy ≥ Eac (catalysed)

0

No of molecules

Eac Catalysed rxn

Ea uncatalysed rxn

Energy

(iii) A sample of ammonia can be liquefied at room temperature just by increasing the pressure of the sample. Why does the application of pressure cause the gas to liquefy? At high pressure, volume of gas decreases, and the particles are much closer to one another. They are able to form more significant intermolecular forces (hydrogen bond between ammonia molecules) and hence, gases condense into liquid.

[7](c) Iron(III) chloride is a dark brown solid which dissolves in water to give an acidic

solution. This solution is often used, in the electronics industry, to etch (dissolve) the copper used in printed circuit boards.

(i) Explain, with the aid of a chemical equation, why aqueous iron(III) chloride is acidic. Fe3+(aq) can undergo hydrolysis with water due to its high charge density and hence it’s able to polarize one of the water ligands to produce H3O

+ [Fe(H2O)6]

3+ + H2O [Fe(OH)(H2O)5]2+ + H3O

+

(ii) By selecting appropriate E values from the Data Booklet, explain why aqueous iron(III) chloride etches (dissolves) copper. Construct a balanced equation for this reaction. Cu2+ + 2e Cu E = +0.34V 2Fe3+ + 2e 2Fe2+ E = +0.77V Since E

cell = +0.43V 0, reaction is feasible 2Fe3+ + Cu Cu2+ + 2Fe2+

[4]

(d) Iron(III) iodide has been successfully made under non-aqueous conditions by following method. Iron pentacarbonyl, Fe(CO)5, reacted with iodine in hexane solution to form a solution of compound E, whose molecular formula is FeC4O4I2. A further calculated amount of iodine is added, and the solution exposed to uv light. A black precipitate of iron(III) iodide results, and a colourless gas is evolved.

(i) State the co-ordination number of iron in the complex, Fe(CO)5 and hence predict the shape of the complex, Fe(CO)5. Co-ordination number: 5 Shape: trigonal bipyramidal

7


For Examiner's

Use

(ii) Describe the bonding between iodine and iron in compound E and write a balanced equation for the reaction between compound E and iodine. Co-ordinate bond / Dative bond / Dative covalent bond [1] 2FeC4O4I2 + I2 2FeI3 + 8CO

[4](e) When a solution of NaClO is added to a strongly alkaline suspension of Fe2O3, a

purple solution results. When BaCl2(aq) is added to the solution, a red solid is precipitated with the composition by mass: Ba, 53.4%; Fe, 21.7%; O,24.9%.

(i) Calculate the empirical formula of the red solid, and hence determine the oxidation number of iron in the red solid. Ba Fe O Mass/g 53.4 21.7 24.9 No of mole/mol 53.4/137 21.7/55.8 24.9/16 = 0.390 = 0.390 = 1.56 simplest ratio 1 1 4 Empirical formula is BaFeO4 Oxidation number = +6

(ii) Construct a balanced ionic equation for the reaction between Fe2O3, ClO- and OH-. [O] Fe2O3 + 10OH- 2FeO4

2- + 5H2O + 6e [R] ClO- + H2O + 2e Cl- + 2OH- Overall: Fe2O3 + 3ClO- + 4OH- 2FeO4

2- + 3Cl- + 2H2O [4]

[Total: 22]

3 Aromatic sulfonic acid is an important precursor to industrial materials and has the following structure:

Aromatic sulfonic acid

(a) Sulfonation of benzene is the reaction mechanism between benzene and sulfuric acid. The first step of the reaction mechanism is given by the following equation.

H2SO4 H2O + SO3 Sulfur trioxide is an important intermediate in the reaction mechanism.

CHCH

CCH

CHCH

S

O

O

OH

(i) Draw a dot-and-cross diagram of sulfur trioxide.

S

O

OOx

xxx

x x

xx

x x

x x

xx

xx

x x

(ii) With reference to the structure of the molecule, state and explain the function of sulphur trioxide in the mechanism. Electrophile The three oxygen atoms are more electronegative than the sulphur [1] and so draw electrons towards themselves. The sulphur atom is electron deficient and is attacked by the benzene ring.

(iii) The final step of the mechanism is different from other benzene reactions because the hydrogen atom is not removed from the ring by a separate negative ion. Instead it is removed by a lone pair of electrons on a negative oxygen atom in the sigma complex. The sigma complex formed has the following structure:

Sigma complex With the above information, complete the mechanism for the reaction between benzene and sulfuric acid.

Electrophilic Substitution Step 2: Formation of the sigma complex (Slow step)

Step 3: Loss of proton to form the substitution product

(iv) Sulfonation of aniline produces sulfanic acid compounds with an unusually

SHO

OO-

+

slowSH

O

OO-

+CH

CH

CH

CH

CH

CH

S

O

O

O

+-

SHO

OO-

+CH

CH

C

CH

CH

CH

S O

O

OH

fast

9


For Examiner's

Use

high melting point. With reference to the structure of sulfanic acid, suggest why this is so.

Sulfanic acid

Sulfanic acid forms zwitterions, hence the electrostatic forces of attraction between oppositely charge ions on neighboring give rise to the high melting point.

[7]

(b) The key reaction during the Contact process is as follows:

2SO2 (g) + O2 (g) 2SO3 (g); ∆H = - 98 kJ mol-1 Industrially, the gases are compressed to a pressure of 3 atm before it is passed over a bed of vanadium(V) oxide at 430 ○C. Explain the conditions used. By Le Chatelier’s principle, high pressure and low temperature are conditions favorable for the production of the product, SO3. At high pressure, the foward reaction is favoured and POE shifts to the right as there are less gas molecules produced, hence helping to remove the stress of high pressure. However it is costly to operate the reaction at too high a pressure as stronger pipes and vessels are needed. Hence an optimal pressure of 3 atm is used. At low temperature, the forward reaction is favoured and POE shifts to the right as the forward reaction is exothermic, hence helping to remove the stress of low temperature. However rate of reaction would be affected by the low temperature. Hence an optimal temperature of 450oC and catalyst is used.

[3][Total: 10]

4 Magnesium is the fourth most abundant element in the earth. Scientists are

CHCH

CC

CHCH

S

O

O

OH

NH2

working on developing rechargeable magnesium-ion batteries for use in electric cars, as they offer a cheaper and more energy efficient alternative to lithium-ion batteries. A simplified representation of a magnesium-ion battery is shown below. When the battery discharges, magnesium ions are formed, which move across the cation exchange membrane to the Sn electrode, forming solid Mg2Sn.

(a) (i) Write half equations for the reactions occurring at the cathode and the anode when the battery discharges. Cathode: 2Mg2+ + Sn + 4e Mg2Sn Anode: Mg Mg2+ + 2e

(ii) The magnesium-ion battery provides a voltage of 3.01 V. With reference to the Data Booklet, calculate E (Sn/Mg2Sn), assuming standard conditions. E

cell = E[R] – E[O] 3.01 = E[R] – (-2.38) E[R] = +0.63 V

(iii) The car battery has to be recharged after use. After clocking a mileage of 500 km, 83.8 g of Mg2Sn was deposited on the electrode of the battery. How many hours will it take for the car battery to be recharged, using a current of 10.0 A? What is the polarity of each electrode during the charging process? n(Mg2Sn) = 0.500 mol 1 mol of Mg2Sn requires 4 mol of electrons It = nzF 10.0t = (0.500)(4)(96500) t = 5.36 hours (3sf) Sn electrode: +ve, Mg electrode: -ve

(iv) Besides cost, explain another advantage the magnesium-ion battery has over

11


For Examiner's

Use

the lithium-ion battery. 1 mol of Mg can transfer 2 mol of electrons as compared to 1 mol for Li, so the process is more efficient. OR Mg reacts less vigorously with water than Li, so it poses less of a safety hazard if the battery is accidentally exposed to moisture.

[8]

(b) Magnesium is also used in the synthesis of Grignard reagents, RMgX (R = alkyl or aryl groups, X = Cl or Br), from organic halides. The organic halide is added to a suspension of magnesium in dry ether, forming the Grignard reagent

RX + Mg RMgX Grignard reagents are examples of organometallic compounds as they contain a carbon-metal bond. They are useful for organic synthesis as they can form new carbon-carbon bonds with unsaturated compounds such as carbonyls. The reaction of methanal and a general Grignard reagent is illustrated below:

O

H

H

RMgX+1. dry ether

2. H+ (aq)

OH

H

R

H

(i) Draw the displayed formula of the product of the reaction between butanone and C6H5MgCl.

H C

H

H

C

O

C

H

H

C

H

H

H

H

[1]

(ii) How would you determine if the reaction in b(i) has gone to completion?

To a small portion of reaction mixture, add 2,4 dinitrophenylhydrazine, warm. If there is no orange ppt, the reaction is complete.

(iii) The Grignard reagent can be considered as a good source of a carbanion, “R–”. Using this information, explain whether the product mixture formed will be optically active. Explain your answer with the aid of a diagram.

+- O

Product is not optically active as the nucleophile has equal probability of attack from above or below the planar carbonyl centre.

[5]

(c) Refer to the reaction scheme involving a Grignard reagent below.

13


For Examiner's

Use

(i) Suggest the identities of C, D and E.

C: Br

D: OH

O

E: CO2

(ii) Suggest reagents and conditions for Step 1 and 2. Step 1: concentrated H2SO4, 170 ºC or Al2O3, 350 ºC Step 2: LiAlH4 / dry ether

[5][Total: 18]

5 Some chemistry students were doing an organic chemistry practical.

(a) Student X heated Compound F with hydrochloric acid to form two organic products. In order to isolate both products, the reaction mixture was placed in a separating funnel and shaken with water.

Step 1 C

Mg

dry ether Grignard Reagent

1. E

2. H+ (aq)

D

Na2CO

3 (aq)

Step 2

Effervescence

OH

HO Br

NCH

CH3

O

Compound F

(i) Draw the products of the above reaction.

N HH

H

CCH3

O

O

(ii) Explain why the separation will not be successful.

The amine will form ion-dipole interactions with water molecules. The acid will form hydrogen bonds with water molecules.

(iii) Describe the correct procedure to separate the hydrolysis products.

Add NaOH to hydrolysis products until moist litmus paper turns blue. Shake the reaction mixture with water in the separating funnel to isolate the () in the aqueous medium, leaving behind the amine.

(b) Under acidic conditions, Student Y treated ethylamine with excess iodomethane in an attempt to form a quarternary amine salt. Explain why this synthesis will not be successful.

Under acidic conditions, NH2 is protonated to NH3. The lone pair on the nitrogen is not available for the nucleophilic substitution to form the quarternary amine salt.

[2](c)

15


For Examiner's

Use

C C

H

H

OH

H

Br

CH3

Compound G Student Z predicted that Compound G was formed from the reaction of propene and aqueous bromine in the presence of concentrated sodium chloride.

(i) Explain why his prediction is wrong.

Propene undergoes an electrophilic addition reaction with Br2(aq) in conc. NaCl. The secondary carbocation is more stable than the primary carbocation as it has one more electron donating alkyl groups help to stabilise the carbocation intermediate by increasing the electron density

C C

H

H

Br

H

CH3

There is a high concentration of Cl-, thus it will form a bond with the carbocation.

(ii) Hence, suggest the structure of the major product formed.

C C

H

H

Br

H

Cl

CH3

[3]

[Total: 10]

1


Answer for Paper 3

1 The elements of Group VII of the Periodic Table are called halogens, which mean "salt formers". They lack only one electron to form a complete shell, and are extremely active chemically. This question examines the chemistry of two reactions of halogens.

(a) In the first reaction, a 1:2 mixture of H2 and I2 gases was mixed at a constant temperature of 430 oC and a constant pressure of 2 atm. The following equilibrium was set up.

H2 (g) + I2 (g) 2HI (g) ∆H = −11 kJ mol1

(i) Given that degree of dissociation of H2 is 86 %, calculate the equilibrium

constant, Kp for the above reaction under these conditions. H2 (g) + I2 (g) 2 HI (g) Initial amt / mol

X 2x 0

Change / mol 0.86x 0.86x +1.72x

Equilbrium / mol

0.14x 1.14x 1.72x

. n(total) = 1.14x + 0.14x + 1.72x = 3x

Kp = 2 2

22

1.72( 2)( ) 3

0.14 0.14( )( ) ( 2)( 2)3 3

HI

H I

xP x

x xP Px x

= 18.5

(ii) Suggest, with reasoning, a way to increase degree of dissociation of H2 for the above reaction. Decrease temperature. By Le Chatelier’s Principle, when temperature is decreased, The forward exothermic reaction is favoured. Hence POE shift to the right, and degree of dissociation of H2 increases.

(iii) Deduce how Kp for the following system would be compared to your answer in (i).

H2 (g) + Br2 (g) 2HBr (g)

It will be larger since a stronger HBr bond is formed. Extent of equilibrium lies more to the right.

[7]

In another reaction, halogens can react with unreactive alkanes to form mono-substituted product. An example is the reaction of chlorine with 2-methylpentane shown below.

2-methylpentane

(b) (i) Explain why alkanes are generally unreactive. The C-H bonds are non-polar and very strong

(ii)

This reaction is seldom used for synthesis as there are many associated problems. Firstly, several isomeric products are formed. The relative ratio of the isomeric products may be more accurately determined if relative rates of abstraction of H atoms are taken into account.

The relative rates of abstraction of H atoms are

Type of H atoms Relative rate of abstraction Primary 1

Secondary 4 Tertiary 6

By examining the difference in stability of the intermediates formed when different types of H atom are abstracted, explain the trend in relative rate of abstraction. The tertiary radical is the most stable as it contains the most electron-donating alkyl groups which help to stabilise the electron-deficient radical centre. Hence, it is abstracted more easily/faster.

(iii)

Predict the ratio of the following two products A and B, taking into account the relative rates of abstraction given in (ii). Explain your reasoning.

Cland

AB

Cl

3


C C

CH a3

C C CHa

Ha

Ha

Hb

H

H H

H

H

H

H

6 possible (primary) hydrogens (Ha) can be substituted to form A 1 possible (tertiary) hydrogens (Hb) can be substituted to form B Assuming equal probability of abstraction, Ratio of A : B = 6:1 However since a tertiary H is abstracted 6 times faster than a primary H, Ratio of A : B = 6:1x6 = 1:1

(iv) Describe suitable chemical test(s) to distinguish compounds A and B in (iii). Add NaOH(aq) to both unknowns, heat (to form alcohols) [1] followed by KMnO4, NaOH(aq), heat. Observation for A: Purple KMnO4 decolourised, Brown ppt of MnO2 formed. (as 1o alcohol formed is oxidised). Observation for B: Purple KMnO4 remains (as 3o alcohol formed cannot be oxidised).

[7] (c) Another problem of the reaction is multi-substitution.

(i) Suggest the conditions that will give rise to formation of multi-substituted

products. Excess Cl2 or limited alkane. [1]

(ii)

A di-substituted product C with molecular formula C6H12Cl2 was obtained in the reaction of chlorine with 2-methylpentane. C does not exhibit optical activity. On warming with alcoholic KOH, a major product D is formed. When D is treated with hot acidified potassium manganate(VII) solution, effervescence was observed and the only organic product remaining gives a yellow precipitate with alkaline aqueous iodine. Use this information to deduce the structure of C and D, explaining your reasoning.

C:

C C

CH3

C C CH

H

H

Cl

H

H H

H

Cl

H

H

D:

C C

CH3

C C CH

H

H H H H

H

C does not exhibit optical activity as it doesn’t contain any chiral

centre (no carbon with 4 different groups). Alkyl halide in C undergoes elimination with alcoholic KOH to form

alkene in D. Alkene in D undergoes oxidative cleavage to form the following

products:

C C

CH3

H

H

H

OC C

OH OH

O O

CO 2

C C

CH3

H

H

H

O undergoes oxidation to form yellow precipitate with alkaline aq iodine.

The terminal alkene is oxidised to CO2. Ethanedioic acid formed is also oxidised spontaneously to CO2.

[6]

[Total: 20]

2 Precipitation plays an important function in growth of bones and teeth in our bodies.

5


Teeth and bones are largely composed of calcium phosphate salt Ca3(PO4)2. In order for this precipitation to occur, the concentrations of the ions in blood must exceed the solubility product in the immediate region of deposition.

(a) Write an ionic equation with state symbols to show the formation of teeth and bones at the deposition region and hence an expression for the solubility product of the Ca3(PO4)2 deposits. [2] 3Ca2+(aq) + 2PO4

3- (aq) Ca3(PO4)2(s)

Ksp of Ca3(PO4)2 = 3 22 3

4Ca PO

(b) (i) At a temperature of 37 oC, the solubility product for calcium phosphate Ca3(PO4)2 has a numerical value of 4 × 10-27. In a sample of blood at a pH of 7.4, the concentrations of calcium and phosphate ions are found to be 1.2 × 10-3 mol dm-3 and 1.6 × 10-8 mol dm-3 respectively. State and explain if the above composition in blood will result in the growth or degeneration of teeth and bones.

ionic product = 3 22 3

4Ca PO = (1.2 × 10-3)3 × (1.6 × 10-8)2 = 4.42 × 10-25 mol5 dm-15 Since IP > Ksp (4 × 10-27), precipitation occurs resulting in growth.

(ii) Hence, state and explain if the pH environment of 7.4 in (i) is that for a growing child, a fully grown man or an old person. At pH = 7.4, since there is deposition of calcium phosphate, it is the environment for a growing child. [3]

(c) The pH at the area of growth may change due to metabolic processes in the body. The hydrogen ions that are present combine phosphate ions as hydrogen phosphate ions. This prevents growth of strong teeth and bones.

PO43- (aq) + H+ (aq) HPO4

2- (aq)

Assuming that the [HPO42-

] is maintained at 1.7 × 10-4 mol dm-3 by cell functions,

(i) calculate the pH that will result in neither growth nor degeneration at the teeth and bones deposition region when the concentration of calcium ions in blood is maintained at 1.2 × 10-3 mol dm-3. [Ka of HPO4

2- = 4.4 × 10-13 mol dm-3] To prevent neither growth nor degeneration, the [PO4

3-(aq)] must be at saturation (or equilibrium with the Ca3(PO4)2(s),

[PO43-(aq)] =

27

3 3

4 10

(1.2 10 )

=1.52 × 10-9 moldm-3

pH = -log[H+] =-log(

13 4

9

4.4 10 1.7 10

1.52 10

) = 7.3

(ii)

Suggest another acid–conjugate base equation that accounts for the constant supply of HPO4

2- ions from cell functions. H2PO4

- (aq) HPO42-(aq) + H+ (aq) [state symbols can be ignored].

[3]

(d)

(i)

Construct an energy level diagram for the formation of Ca3(PO4)2 from its elements using relevant data from the table below and other appropriate values in the Data Booklet. Hence calculate the lattice energy of Ca3(PO4)2. Enthalpy ∆H / kJ mol-1

Standard enthalpy change of formation of Ca3(PO4)2 −4132 Standard enthalpy change of atomisation of Ca +178 Standard enthalpy change of atomisation of P +315 P(g) + 4O(g) + 3e → PO4

3-(g) −2810 By Hess’ law, ∆HLE Ca3(PO4)2 = +2810 x 2 - 3 × (589+1145) - 498 × 4 - 315 × 2 - 178×3 – 4132 = -6870 kJmol-1

3Ca(s) + 2P(s) + 4O2(g) 3×∆Hatm(Ca) = +178×3

0

kJ mol-1

3Ca2+(g) + 6e+ 2P(g) + 8O(g)

3Ca(g) + 2P(s) + 4O2(g)

3Ca(g) + 2P(g) + 4O2(g)

3Ca(g) + 2P(g) + 8O(g)

2×∆Hatm(P) = +315 × 2

4×∆Hbe(O2) = +498 × 4

3 × (1st +2nd IE (Ca)) = 3 × (589+1145)

Ca3(PO4)2 (s)

∆Hf(Ca3(PO4)2 ) = −4132

∆HLE Ca3(PO4)2

-2810 x 2

3Ca2+(g) + 2PO43-

(g)

7


(ii) Using your answers to (i) and other enthalpy change value if any, explain why Ca3(PO4)2 is insoluble. ∆Hsolution Ca3(PO4)2 = -∆HLE Ca3(PO4)2 + 3∆Hhydration Ca2+ + 2∆Hhydration PO4

3-

-∆HLE Ca3(PO4)2 is very positive as seen in part (d) and both Ca2+, PO4

3- are have large radii, resulting in less negative ∆Hhydration values. Hence ∆Hsolution Ca3(PO4)2 will likely to be positive. Making it insoluble.

[6]

(e) Use of the Data Booklet is relevant to this question. Group VII elements exist as ions and play a vital role in bodily functions. The following observations are made about Group VII elements and their reactions. Explain the observations made.

(i) F2 cannot be prepared by electrolytic oxidation of aqueous F- solutions.

The electrode potential value of F2 F- is too electropositive, hence not favouring the backward reaction. In an aqueous medium, H2O will be preferentially oxidised to oxygen gas instead.

(ii) HI can be prepared by treating NaI with phosphoric acid, H3PO4 but not with concentrated sulfuric acid. Phosphoric displaces HI from I- and it does not further oxidise HI to I2. H2SO4 on the other hand oxidises HI to I2, itself being reduced to H2S.

(iii) Interhalogen compounds such as IF3, IF5 and IF7, BrF3, BrF5, ClF3, ICl3 exists. On the other hand, ClF5 is not easily formed and BrCl3 does not exists. The large size of the I atom can accommodate up to seven small F atoms. As the central atom gets smaller from Br to Cl, it cannot accommodate as many F atoms. As the surrounding atom gets bigger, the steric hindrance prevents too many of it from surrounding the central atom as well.

[6]

[Total: 20]

3 (a) Period 3 elements react with oxygen to form the corresponding oxides of varying acid-base nature. Using magnesium, aluminium and phosphorus as examples, describe the acid-base nature of the oxides across Period 3 with suitable equations. [4]

Magnesium oxide is a basic oxide and it reacts readily with acids to form salt and water. MgO (s) + 2H+ (aq) Mg2+ (aq) + H2O (l) Aluminium oxide is an amphoteric oxide and it can react with both acids and alkali to give a salt. Al2O3 (s) + 6H+ (aq) 2Al3+ (aq) + 3H2O (l) Al2O3 (s) + 2OH⎯ (aq) + 3H2O (l) 2[Al(OH)4]⎯ (aq) Phosphorus trioxide / pentoxide is acidic and reacts with alkali to give a salt and water. P4O6 (s) + 12OH⎯ (aq) 4PO3

3⎯ (aq) + 6H2O (l) OR P4O10 (s) + 12OH⎯ (aq) 4PO4

3⎯ (aq) + 6H2O (l)

(b) Aluminium is a reactive metal but its reactivity is limited due to a layer of inert aluminium oxide being formed. Both aluminium and its oxide react readily with aqueous sodium hydroxide to give similar products. Aluminium, however, produces another product which is a combustible gas.

(i) Write a balanced equation for the reaction when aluminium is heated in aqueous sodium hydroxide.

2Al (s) + 2OH⎯ (aq) + 6H2O (l) 2[Al(OH)4]⎯ (aq) + 3H2 (g)

(ii) Explain why a homogenous solution is obtained in (i). The aluminium hydroxide formed is amphoteric and can react with hydroxide to form [Al(OH)4]⎯. The resulting product is charged and can form ion−dipole interaction with water thus becomes soluble.

[3]

9


(c) Using 2,6−dichloroheptane as the starting compound, suggest the reagents and conditions for steps I, II and IV and the products J to M in the synthetic scheme below. [7]

Answer

Reagents and Conditions:

Step I: Ethanolic NaOH, heat under reflux

Step II: aq H2SO4, aq KMnO4, heat under reflux

Step IV: aq NaOH, heat under reflux

Cl Cl

PCl5

O O

O O

IV

I

II

III

VJ

K L

M

N

Cl Cl OH OH

OH OH

O O

Cl Cl

O O

PCl5

O O

O O

IV

I

II

III

V

J

K L

M

N

(d) (i) Propyne, C3H4, like ethyne, C2H2, is commonly used for blow torches to cut metals in heavy metal industries. By drawing an appropriate energy cycle, determine the enthalpy of combustion of propyne, given the following enthalpy values: Hf(C3H4) = +184 kJ mol⎯1

Hc(C) = −394 kJ mol⎯1

Hc(H2) = −286 kJ mol⎯1 Hc (propyne) = 3(−394) + 2(−286) − 184 = −1938 kJ mol−1

(ii) Determine the enthalpy of combustion of propyne using bond energy values. Explain why the use of bond energy values to calculate the enthalpy of combustion is significantly less exothermic that those obtained in (i). BD = 4(+410) + (+350) + (+840) + 4(+496) BF = 6(+740) + 4(+460) Hc(C3H4) = +4814 + (−6280) = −1466 kJ mol⎯1 Hvap(H2O) is not taken into consideration.

[6]

[Total: 20]

3C (g) + 2H2 (g) + 4O2 (g)

C3H4 (g) + 4O2 (g) 3CO2 (g) + 2H2O (l)

3(-394) 2(-286)Hf = +184

Hc = -1938

CH3 C CH + 4O2 (g)(g) CO O HOH3 (g) (l)2+

11


4 Nickel is a typical transition element which is commonly used as a catalyst and metal for electroplating. Nickel also forms complex ions with ligands such as H2O, NH3 and NH2CH2CH2NH2 to give various coloured complexes.

(a) What do you understand by the terms transition element, ligand and complex ion? [2] [3] A transition element is a d-block element that can form one or more stable ions with a partially filled d-subshell. A ligand is a molecule or anion with at least one lone pair of electrons that it can use to form coordinate bond to the central metal atom/ion in a complex ion. A complex ion is a species that contains a central metal atom/ion surrounded by ligands which forms coordinate bonds to the metal centre.

(b) (i) Explain why transition metal complexes are coloured. In the presence of ligands, the degenerate d orbitals of the transition metal split into 2 groups with an energy gap, E. When visible light passes through the solution of the transition metal ion, the wavelength of light corresponding to the energy gap is absorbed by the d electron in the lower energy level to promote to a vacant d orbital at the higher energy level./ for d-d transition. The complementary colour corresponding to the unabsorbed wavelength is observed.

(ii) Complex ions of nickel such as [Ni(H2O)6]2+, [Ni(NH3)6]

2+ and [Ni(NH2CH2CH2NH2)3]

2+ are green, blue and violet colour respectively. The visible absorption spectra of the three complex ions of nickel are shown below:

wavelength / nm

relative absorbance

A

B

C

700 600 500

colour of visible light

redorangegreen

650 550

yellow

Suggest an explanation for the difference in colour of the three complexes and using the colours of the complexes given, assign the peaks A, B and C to the three complex ions of nickel. Due to the presence of different ligand, the d orbitals are split to different extent/ energy gap, E. Light of different wavelength will be absorbed by the d electron of Ni(II) in the three complex ions and hence different complementary colours are observed. A: [Ni(NH2CH2CH2NH2)3]

2+ absorbs yellow and appears violet B: [Ni(NH3)6]

2+ absorbs orange and appears blue C: [Ni(H2O)6]

2+ absorbs red and appears green [3] [5]

(c) When Ni2+ (aq) is mixed with an excess of NH3 (aq), each H2O is replaced by a NH3 at a time. The stepwise formation of [Ni(NH3)6]

2+ from [Ni(H2O)6]2+ can be

summarised by the following equations with their associated equilibrium constants: Step 1: [Ni(H2O)6]

2+ + NH3 [Ni(H2O)5(NH3)]2+ + H2O K1 = 630 mol–1dm3

Step 2: [Ni(H2O)5(NH3)]

2+ + NH3 [Ni(H2O)4(NH3)2]2+ + H2O K2 = 160 mol–1dm3

and so on for steps 3, 4 and 5. Step 6: [Ni(H2O)(NH3)5]

2+ + NH3 [Ni(NH3)6]2+ + H2O K6 = 1.1 mol–1dm3

(i) The stepwise formation of [Ni(NH3)6]

2+ from [Ni(H2O)6]2+ undergoes a

dissociative mechanism which resembles a SN1 mechanism in organic chemistry. Suggest a possible mechanism for Step 1 and show clearly how the shape of the complex ion changes.

NiH2O

H2O OH2

OH2

H2O

H2O

2

slowNi

H2O

H2O OH2

OH2

H2O2

intermediate

+ H2O

NiH2O

H2O OH2

OH2

H2O2

NH3

fastNi

H2O

H2O OH2

OH2

H2O2

NH3

step 1

step 2

Trigonal bipyramidal intermediate is accepted as well.

13


(ii) Sketch an energy profile diagram for the suggested mechanism in (i) given the reaction is exothermic.

(iii)

Suggest a possible reason for the decrease in value of the equilibrium constant for each successive replacement of water by ammonia ligands. For each successive substitution, the number of sites available for the ammonia ligands to replace decreases. [6]

(d) [Ni(H2O)6]2+ can also undergo ligand exchange reaction with NH2CH2CH2NH2 to

form [Ni(NH2CH2CH2NH2)3]2+. Using the signs of G, H and S, explain how the

formation of [Ni(NH2CH2CH2NH2)3]2+ would compare to that of [Ni(NH3)6]

2+. [2] [3] H for formation of [Ni(NH3)6]

2+ and [Ni(NH2CH2CH2NH2)3]2+ is similar due to

breaking of similar Ni─O bonds and forming of similar Ni─N bonds. S for formation of [Ni(NH2CH2CH2NH2)3]

2+ would be more positive than that of [Ni(NH3)6]

2+ because there is an increase in number of aqueous particles as [Ni(NH2CH2CH2NH2)3]

2+ is formed, allowing more ways of arranging the particles. Since H for formation of [Ni(NH3)6]

2+ and [Ni(NH2CH2CH2NH2)3]2+ are similar and

S for formation of [Ni(NH2CH2CH2NH2)3]2+ is more positive than that of

[Ni(NH3)6]2+ , G for formation of [Ni(NH2CH2CH2NH2)3]

2+ would be more negative than that of [Ni(NH3)6]

2+ and hence more spontaneous.

[Ni(H2O)5NH3]2+

Energy / kJmol-1

∆H

reaction pathway

intermediate reactant

product

Ea1

Ea2

Note: Ea1 > Ea2 since step 1 is slow.

[Ni(H2O)6]2+

[Ni(H2O)5]2+

(e) With the aid of a clearly labelled diagram, show the calculations you would carry out to determine a value for the Avogadro constant by the electroplating of nickel from nickel(II) ions. You may assume that you are provided with the following data. • mass of nickel deposited, m • current used, I • time taken, t

Calculations:

amount of nickel metal deposited =m

58.7 mol

amount of charge required to deposit 1 mol of Ni = Itm

58.7

=58.7It

mC

At the cathode: Ni2+ + 2e → Ni

Quantity of charge per mol of electron passes = 58.7It

m x

1

2=

58.7It

2m

Given that the charge of an electron is 1.60 x 10–19 C,

Avogadro’s constant = 19

58.7It2m

1.6 x 10 =19

58.7It

2 x 1.6 x 10 m

[3]

[Total: 20]

aqueous nickel (II) sulfate

nickel cathode (-) nickel anode (+)

15


5 (a) Amino acids play central roles both as building blocks of proteins and as intermediates in metabolism. The 20 amino acids that are found within proteins convey a vast array of chemical versatility and they have relatively high melting points and are soluble in water.

(i) Define the secondary structure of proteins. The secondary structure refers to the repeated coiling and folding of a polypeptide chain in a regular arrangement. The two common repeating structural patterns in proteins are: -helix and -pleated sheets. Both structures are stabilised by hydrogen bonding between the oxygen atom of the C=O group of a peptide bond and the hydrogen atom of the N-H group of another peptide bond.

(ii) What is meant by the term denaturation of proteins? Denaturation is an irreversible process when the protein loses its biological activity. When proteins are denatured, the R–group interactions are destroyed, only the secondary, tertiary and quaternary structures are disrupted. This results in protein disintegrating to form random coils of polypeptide chains.

(iii) Suggest a brief outline of one method by which proteins may be denatured. Proteins may be denatured by addition of acid to it. This will disrupt the ionic bonding originally present between the R groups.

[4]

(b) An octapeptide A was analyzed and contained the following amino acids:

Amino acid abbreviation R-group Number of residues

Aspartic acid Asp – CH2CO2H 1

Glycine Gly – H 2

Histidine His

NH

N

CH2

CH3

1

Phenylalanine Phe CH2 1

Proline Pro

CH2

CH2 CH2

(cyclic)

2

Valine Val – CH(CH3)2 1

(i) Analysis showed that the C-terminus contained an amino acid which is

optically inactive and partial hydrolysis gives the following fragments:

Val-Pro-His, Gly-Asp-Pro-Phe, Pro-Phe-Val Use the above results to deduce the amino acid sequence of the octapeptide A and explain your reasoning. Complete Sequence: Gly-Asp-Pro-Phe Pro-Phe-Val Val-Pro-His Gly Octapeptide: Gly-Asp-Pro-Phe-Val-Pro-His -Gly

(ii) Draw the displayed formula of Gly-Asp-Pro which is part of the octapeptide chain in A.

C C

H

N

H

H

H

O

N

H

C

C

H

C

H H

CO O

H

O

N C C

H

O

O H

[4]

(c) Histidine is utilized by your body to develop and maintain healthy tissues. It is especially important in the myelin sheath that coat nervous cells to ensure the transmission of messages from your brain to organs throughout your body. Histidine also stimulates the secretion of the digestive enzymes gastrin and is also required to manufacture both red and white blood cells.

17


NHN

COOH

NH2

1

2

Structure of histidine

Histidine contains a total of 3 nitrogen atoms and each N atom exhibit different properties in aqueous solution.

(i) Considering the orbital arrangement of the atoms, explain why 2N atom in histidine is less basic than 1N atom. The lone pair of electrons on the 2N atom is less available for donation to the proton via dative bonding because the lone pairs of electrons occupy an sp3 orbital that can be delocalised into the ring. While on the other hand, the lone pair of electrons on the 1N atom is available for donation to the proton via dative bonding because the lone pairs of electrons occupy the sp2 orbital.

(ii) There are three pKa values associated with the histidine: 1.8, 6.0, 9.2.

Assuming that the 2N atom does not exhibit any basic properties, make use of the three pKa values to suggest the major species present in solutions of histidine with the following pH values. • pH 1 • pH 5 • pH 7 • pH 11 pH 1: pH 5:

NHHN

COOH

NH3

+

+

NHHN

COO-

NH3

+

+

pH 7: pH 11:

NHN

COO-

NH3+

NH

N

COO-

NH2

(iii) Consider the buffer solution at pH = 5, calculate what fraction of the histidine side chains will carry a positive charge at pH 5.

pH = pKa + log[salt]

[acid]

5 = 6 + log[salt]

[acid]

[salt]

[acid]= 10-1 =

1

10

Since 10[salt] = [acid] Therefore at pH 5, 10/11 of the histidine will be in the protonated form.

[8]

(d) 25.0 cm3 of 0.150 mol dm3־ of diethylamine solution, (CH3CH2)2NH, was placed in a conical flask and titrated against a solution of 0.200 mol dm-3 hydrochloric acid from a burette. Given that the Kb of (CH3CH2)2NH is 8.60 x 10-4 mol dm-3,

(i) explain what is meant by the term base dissociation constant, Kb of diethylamine.

The base dissociation constant is defined as

+

3 2 2 2b

3 2 2

[OH ][(CH CH ) NH ]K

[(CH CH ) NH]

(ii) Calculate the pH at equivalence point.

V(HCl) at neutralisation point =

25.0 0.15018.75

(0.200) cm3

At equivalence point, [salt] =

salt

T

25.0 0.150n0.08571

V (25.0 +18.75) mol dm-3

Since salt is an acidic in nature, [H+] =

-14-7

-4

1 x 10x 0.08571 = 9.983 x 10

8.6 x 10

pH = -log(9.983 x 10-7) = 6.00

[4]

[Total: 20]

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