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© ACJC 2012 9647/01/Prelim/12 [Turn over

1

ANGLO-CHINESE JUNIOR COLLEGE

DEPARTMENT OF CHEMISTRY Preliminary Examination

CHEMISTRY 9647/01 Higher 2 Paper 1 Multiple Choice 12 September 2012 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, Centre number and candidate number on the answer sheet in the spaces provided unless this has been done for you. There are forty questions in 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 your 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 18 printed pages.

9647/01/Prelim/012 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2012 Department of Chemistry [Turn over


2

Section A

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

1 The relative atomic mass of antimony, which consists of the isotopes 121Sb and

123Sb, is 121.8. What is the percentage of 123Sb in the isotopic mixture?

A 40%

B 45%

C 50%

D 55%

2 20 cm3 of a gaseous unknown hydrocarbon was combusted in 100 cm3 (an

excess) of oxygen. After the combustion, the mixture was left to cool and the gaseous volume was 90 cm3. Upon treatment with potassium hydroxide, the volume was decreased to 50 cm3.

What is the molecular formula of the unknown hydrocarbon?

A CH4

B C2H2

C C2H6

D C3H6

3

Species with unpaired electrons exhibit a property known as paramagnetism, that is, it is attracted by an external magnetic field which is directly proportional to the number of unpaired electrons present in the species.

Which of the following species is not paramagnetic?

A Fe3+

B Ti2+

C Cu2+

D Ca2+


3

4

The empirical formula of a fluorocarbon is CF2. At the same temperature and pressure, 1 dm3 of the fluorocarbon weighs 8.93 g while 1 dm3 of fluorine gas weighs 1.80 g.

What is the molecular formula of the fluorocarbon?

A CF2

B C2F4

C C4F8

D C5F10

5 Which of the following molecules will not form a hydrogen bond with another of

its own molecules?

A CH3NH2

B CH3COCH3

C CH3CONH2

D CH3OH

6 Adrenaline is a hormone which, when secreted directly into the bloodstream,

acts as a stimulant. The structure of adrenaline is given below.

What are the values of the angles x, y and z in a molecule of adrenaline?

x y z

A 120º 105º 107º

B 109º 180º 90º

C 120º 105º 120º

D 180º 120º 107º

HO C

HO

O

H

C

H

H

N H

CH3

H

x

yz


4

7

Propene burns completely in oxygen to give carbon dioxide and water.

CH3CH=CH2 + 9/2 O2 3 CO2 + 3 H2O

Using the Data Booklet, calculate the enthalpy change of the reaction.

A - 890 kJ mol-1

B - 1548 kJ mol-1

C - 2569 kJ mol-1

D - 2890 kJ mol-1

8

Which graph correctly represents an auto-catalytic reaction (reaction in which one of the products catalyses the reaction)?

A

C

B

D

Rate

Time

Rate

Time

Rate

Time

Rate

Time


5

9

0.12 g of graphite was added to a vessel that contained CO2 (g) at a pressure of 0.824 atm at 400K. The total pressure reached an equilibrium value of 1.366 atm after a period of time.

C(graphite) + CO2 (g) 2 CO (g)

Given that the temperature did not change, what is the value of the equilibrium constant Kp?

A 0.239

B 0.260

C 3.84

D 4.17

10 What is the pH of the final solution formed when V cm3 of dilute sulfuric acid of

pH 2.0 is mixed with V cm3 of dilute sulfuric acid of pH 4.0 followed by the addition of 2V cm3 of water?

A 2.3

B 2.6

C 3.0

D 3.6


6

11

When 5.00 cm3 of a 0.100 mol dm−3 base is titrated against 0.100 mol dm3 of ethanoic acid, CH3COOH, the following titration curve is obtained.

Which of the following correctly identifies the base used and the point on the curve at which pH = pKa?

Base used Point where pH = pKa

A NaOH(aq) M

B Ba(OH)2(aq) N

C NaOH(aq) P

D Ba(OH)2(aq) P

12 A sparingly soluble salt, cobalt(II) hydroxide, dissociates in aqueous solution

according to the equation below.

Co(OH)2 (s) Co2+ (aq) + 2 OH- (aq)

What is the concentration of OH- at equilibrium given that the solubility product of cobalt(II) hydroxide is 1.6 x 10-15 mol3 dm-9?

A 7.37 x 10-6 mol dm-3

B 9.28 x 10-6 mol dm-3

C 1.47 x 10-5 mol dm-3

D 2.34 x 10-5 mol dm-3

Volume of 0.100 mol dm−3 Of CH3COOH / cm3

pH

14

12

10

8

6

4

2

0

0 2 4 6 8 10 12 14 16 18 20

P

M

N


7

13

The following reaction has a negative Eθ

cell so it does not occur under standard conditions.

2NO3-(aq) + 8H+(aq) + 6Cl-(aq) 2NO (g) + 4H2O (l) + 3Cl2(g)

However, the reaction may be made to proceed under non-standard conditions. Which of the following changes will not aid the reaction to proceed?

A Addition of NaCl

B Addition of HCl

C Addition of KNO3

D Addition of AgCl

14

Use of the Data Booklet is relevant to this question.

Sacrificial protection is a common method used to prevent the rusting of large steel objects such as underground pipes, oil tankers or ships.

A piece of metal that is more reactive than iron is connected to the object made of iron. Over time, the more reactive metal will corrode in place of the iron and hence “sacrificed”.

Which metal can be used in the above mentioned method to prevent rusting?

A Copper

B Zinc

C Cobalt

D Lead

15 Which of the following species has the smallest radius?

A S2-

B Cl-

C K+

D Ca2+


8

16

A mixture of the oxides of two elements in the third period is dissolved in water. The resultant solution is acidic.

What could be the constituents of the mixture?

A Na2O and MgO

B Na2O and Al2O3

C Al2O3 and SiO2

D SO3 and P4O10

17

X, Y and Z are Group II elements. They form compounds with the following properties:

Y(NO3)2 has a higher thermal decomposition temperature than Z(NO3)2.

XSO4 has a more exothermic lattice energy than ZSO4.

Which of the following statements is correct?

A On heating, Y(NO3)2 decomposes to the nitrite and oxygen while Z(NO3)2 decomposes to the oxide, nitrogen dioxide and oxygen.

B XCl2 has a more covalent character than YCl2.

C Z is more reducing than Y.

D X is Mg and Z is Ba.

18

0.03 mol of chlorine gas was bubbled into 100 cm3 of hot aqueous sodium hydroxide of concentration, 1.5 mol dm-3.

Which of the following statement is correct regarding the above reaction?

A 0.03 mol of sodium chlorate (I) was formed.

B 0.03 mol of sodium chlorate (V) was formed.

C The excess sodium hydroxide required 0.06 mol of dilute hydrochloric acid for complete neutralization.

D The chloride produced required 0.05 mol of silver nitrate for complete precipitation.


9

19

Transition metals like platinum and rhodium are found in catalytic converters fitted into cars. Which of the following statements best explains the role of transition metals in this use?

A Transition metals can exhibit variable oxidation states in their compounds as 3d and 4s electrons have similar energies.

B Transition metals have available and partially filled 3d orbitals for the adsorption of reactant molecules.

C Transition metals have very high melting points because both 3d and 4s electrons are involved in forming strong metallic bonds.

D Transition metals form coloured ions due to absorption of energy in the visible light region to promote an electron from a lower to a higher energy 3d orbitals.

20

The table shows the electronic configuration of the three d-block elements in the Periodic table.

Element Electronic Configuration

P [Ar]3d64s2

Q [Ar]3d84s2

R [Ar]3d104s1

Which one of the following statements is incorrect?

A The electronic configuration of central metal ion for [P(CN)6]4- is [Ar]3d6.

B Upon reduction from RCl2 (aq) to [RCl2]-(aq), the solution turned

colourless.

C P is likely to exist as KPO4.

D The E value of P3+/P2+ is less positive than that of Q3+/Q2+.

21 C4H8 decolourises aqueous bromine in the dark. How many isomers (including

both structural and geometric isomers) are possible for the compound C4H8?

A 1

B 2

C 3

D 4


10

22

Which type of reaction is the following compound unlikely to undergo?

NO2

CH2O C

O

CH2Br

A Nucleophilic substitution

B Reduction

C Electrophilic addition

D Free radical substitution

23

Consider the compound Y, which has the structural formula as shown.

ClH2C

Cl

COCl

CH2CH=CHClY

1 mole of compound Y is warmed with aqueous sodium hydroxide. The resulting solution is cooled and acidified with dilute nitric acid. How many moles of silver chloride will be precipitated out when excess aqueous silver nitrate is then added?

A 1

B 2

C 3

D 4

COCl

CH2CH=CHCl

ClH2C

Cl


11

24

Thyroxine and Cyproterone are compounds that are often used in hormone therapy.

HO

I

I O

I

I

CH2 C

H

NH2

COOH

Thyroxine

O

Cl

COCH3

OHCH3

CH3

Cyproterone

Which of the following reagents could not be used to distinguish between the two compounds?

A Neutral FeCl3

B Alkaline aqueous iodine

C Aqueous silver nitrate

D Sodium carbonate

25 Which process does not form CH3CO(CH2)3CO2H?

A Hot KMnO4(aq), H3O+KOH in ethanol

heat

CH3

Cl

B KOH(aq)H2NCO(CH2)3CHClCH3

Hot KMnO4(aq), H3O+

heat

C CH3CH(OH)(CH2)3CH2Cl

Hot KMnO4(aq), H3O+NaOH(alcohol)

Heat

D CH3

Hot KMnO4(aq), H3O+


12

26

Compound G gives the following observations:

Test Observation

Tollen’s Reagent Silver mirror formed

Fehling’s Reagent No visible reaction

Aqueous sodium carbonate Colourless gas evolved

What is compound G?

A

C

B

H

HOOC

O

D

27 During frozen storage, white fish develops a typical “cold storage” flavour. One

of the compounds contributing to this off-flavour has been isolated and identified as hept-4-enal.

CH3CH2CH=CHCH2CH2CHO

hept-4-enal

What is formed when hept-4-enal is reduced with either hydrogen and a nickel catalyst or lithium aluminum hydride?

Reagent Product

A With H2/Ni CH3(CH2)5CH2OH

B With H2/Ni CH3(CH2)5CH3

C With LiAlH4 CH3(CH2)5CH2OH

D With LiAlH4 CH3(CH2)5CHO


13

28

The ester CH3CH2CH2CO2CH3 is responsible for the aroma of apples.

When this ester is hydrolysed by acid in the stomach, what is the empirical formula of the organic acid produced?

A CH2O

B C2H4O

C C2H4O2

D C3H7O2

29

Which of the following is the correct order of compounds arranged in decreasing Kb values?

A

CH2ClCH2NH2 CH3CH2NH2

B

CH2ClCH2NH2 CH3CH2NH2

C CH3CH2NH2 CH2ClCH2NH2

D CH2ClCH2NH2 CH3CH2NH2


14

30 The reduction of a nitrile Q produced a compound of the formula CH3CH2NH2. The same nitrile Q was then hydrolysed in acidic medium.

What would be formed if the products from the two reactions are mixed together?

A CH3CONHCH2CH3

B CH3CH2CONHCH2CH3

C (CH3CH2CO2–)(CH3CH2NH3

+)

D (CH3CO2–)(CH3CH2NH3

+)


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 that 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 change(s) in conditions would not increase the volume of a fixed mass

of gas?

Pressure / kPa Temperature / K

1 Halved Halved

2 Doubled Halved

3 Halved Doubled

32

Sodium hypochlorite is a chemical compound with the formula NaOCl. It is frequently used in household as a disinfectant or a bleaching agent.

What type(s) of bonding occur in NaOCl?

1 covalent bond

2 ionic bond

3 Van der Waals forces

33 Which of the following reaction(s) has/have a positive enthalpy change of

reaction?

1 H2O (l) H+ (aq) + OH- (aq)

2 Br2 (g) 2 Br (g)

3 F (g) + e F- (g)


16

34

In a chemical reaction, X reacts with Y to form Z. The initial rates of the reaction are shown for the following experiments:

Experiment [X]/ mol dm–3 [Y]/ mol dm–3 Initial rate/ mol dm–3

1 0.150 0.250 2.80 x 10–5

2 0.150 0.500 5.60 x 10–5

3 0.075 0.500 2.80 x 10–5

4 0.075 0.250 1.40 x 10–5

The energy profile diagram for the reaction is as shown:

Which of the following is/are possible overall equation(s) of the above reaction?

1 X + Y Z

2 X+ 2Y Z

3 2X + Y Z

A B C D

1, 2 and 3 are

correct



1 only is

correct

Energy

Progress of reaction


17

A B C D

1, 2 and 3 are

correct



1 only is

correct

35

At 200 oC, the value of Kc for the following reaction is 1.00 x 1010.

PCl5 PCl3 + Cl2

Which factor(s) below can affect the magnitude of Kc?

1 Temperature

2 Pressure

3 Concentration of Cl2 present

36 Which of the following statement(s) is/are true for period 3 elements and their

compounds?

1 The electronegativity decreases from sodium to chlorine.

2 The pH of the chlorides decreases from sodium to chlorine

3 The atomic radius decreases from sodium to chlorine.

37

The element astatine, At2, is radioactive with a short half-life, hence its chemistry is not easily investigated. However, astatine is below iodine in the Periodic Table.

Which of the following prediction(s) is/are correct for astatine?

1 Chlorine is able to displace astatide from its aqueous solution.

2 Astatine reacts less readily with hydrogen than bromine.

3 Astatine oxidises thiosulfate to sulfate.


18

A B C D

1, 2 and 3 are

correct



1 only is

correct

38 Which of the following reaction(s) will form a racemic mixture?

1 CH3CHO with HCN in trace amounts of NaOH at a temperature of 10 ºC to 20 ºC.

2

3

CH3CH2C CCH2CH3

CH3CH3

39 Thiols are organic compounds containing the −SH functional group. They are

sulphur analogue of alcohols. Some common reactions undergone by thiols are shown as follows.

I CH3CH2SH + CH3CH2Br (CH3CH2)2S + HBr

II CH3CH2SH + NaOH CH3CH2S−Na+ + H2O

III 2CH3CH2SH + Br2 CH3CH2−S−S−CH2CH3 + 2HBr

Which of the following statement(s) comparing thiols with alcohols is/are true?

1 Thiols are stronger nucleophiles than alcohols.

2 Thiols are stronger acids than alcohols.

3 Thiols are stronger reducing agents than alcohols.

with aqueous sodium hydroxide, heat under reflux.

with Br2 dissolved in CCl4.

CH3CH2CH2C

CH3

C6H5

Cl


19

A B C D

1, 2 and 3 are

correct



1 only is

correct

40

Fumaric acid is generally used as a substitute for tartaric acid and occasionally in place of citric acid to add sourness to foods.

Which of the following statement(s) is/are correct?

1 It reacts with Br2 (aq) to give 2 chiral centres.

2 It reacts with PCl5 to give white fumes.

3 It reacts with two equivalents of phenol to form a sweet smelling compound.

END OF PAPER

ACJC H2 Prelim 2012 Paper 1 Answers

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


Index No. Name

Form Class

Tutorial Class

Subject Tutor

ANGLO-CHINESE JUNIOR COLLEGE DEPARTMENT OF CHEMISTRY

Preliminary Examination

CHEMISTRY 9647/02Higher 2 Paper 2 Structured Questions 22 August 2012 2 hours Candidates answer on the Question Paper Additional Materials: Data Booklet

READ THESE INSTRUCTIONS FIRST Write your name, index number, form class, tutorial class and subject tutor’s name on all the work you hand in. 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. Answer all questions. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together including the cover page.

For Examiner's Use

Question no. Marks

1

2

3

4

5

6

7

TOTAL

This document consists of 22 printed pages and 1 blank page

9647/02/Prelim/12 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2012 Department of Chemistry [Turn over

2


1 Planning (P)

A student is interested to determine the Faraday constant. He proposed that the value of the Faraday constant can be determined experimentally using the electrolysis of aqueous nickel (II) sulfate.

The student did some research and made the following notes.

General information

Reaction at the cathode: Ni2+ (aq) + 2e → Ni (s)

Reaction at the anode: Ni (s) → Ni2+ (aq) + 2e

1 mol of Ni (s) is deposited at the cathode by 2 mol of electrons.

The Faraday constant is the charge in coulombs, C, carries by 1 mol of electrons.

The Faraday constant = 96500 C mol-1

Experimental setup

The experimental setup for the electrolysis of nickel (II) sulfate solution is shown above. The student will electrolyse some nickel (II) sulfate solution using nickel electrodes. Before the nickel cathode is placed into the electrolyte, it will be cleaned and weighed.

The current that passes through the electrolyte will be kept constant at 0.3 A for 40 min by adjusting a variable resistor in the experiment. At the end of 40 min, the nickel cathode will be weighed.

The experiment will be repeated for another 7 times at 40 min intervals.

For Examiner’s

use

3


(a) Based on the information given under experimental setup, give a full description of the procedures you would use in Step A (Start of experiment) in the space provided. Step B (Rinsing and reweighing of cathode) is provided.

Experimental method

Step A (Start of experiment)

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

………………………………………………………………………………………….

[2]

Step B (Rinsing and reweighing of cathode)

1. The cathode is removed from the solution and carefully washed with distilled water to remove any nickel (II) sulfate solution.

2. Distilled water is removed from the cathode by rinsing it with propanone in which the water dissolves.

3. The cathode is finally dried by allowing the propanone to evaporate from the surface.

4. The cathode is reweighed and placed back to the solution.

5. A constant current of 0.3 A is passed for a further 40 mins when the rinsing, drying and weighing are repeated.

This procedure (Step A and B) is repeated for a further 6 times.

For Examiner’s

use

4


(b) The student performed the experiment using the experimental setup and method above.

The results of his experiment are recorded below.

Time/ min Mass of cathode/ g

Charge passed/ C

Mass of Ni (s) deposited on

the cathode/ g

0 115.74 0 0

40 115.97 720 0.23

80 116.22 1440 0.48

120 116.46 2160 0.72

160 116.70

200 116.94 3600 1.20

240 117.19 4320 1.45

280 118.01 5040 2.27

320 117.67 5760 1.93

Calculate and record the charge passed and the mass of nickel deposited on the cathode at 160 min in the space given above.

[1]

For Examiner’s

use

5


(c) The student plotted the data on the graph of mass of Ni (s) deposited on the cathode against charge passed.

Draw a best-fit line on the graph below.

[1]

For Examiner’s

use

(d) With reference to part (b) and part (c), state the anomalous point. Based on the procedures for the experiment, explain for the anomaly.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

[2]

(e) The weighing balance used by the student weighed to 2 decimal places. With reference to the results of the experiment, explain why it is not appropriate to use a weighing balance accurate to 2 decimal places. ……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

[1]

Charge passed/ C

Mass of Ni(s) deposited on the cathode/ g

0

x

x

x

x

x

x

xx

x

6


(f) The student calculated the value of the gradient in part (c). Gradient of the line = 3.05 x 10-4 g C-1 Using the information provided, calculate a numerical value of the Faraday constant.

[2]

For Examiner’s

use

(g) What other measurements could be made during the course of the experiment to provide alternative data to confirm the determined value of the Faraday constant? ……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

[1]

7


(h) The same experiment was performed by another student. The plotted mass of nickel deposited against charge passed was obtained below. Suggest an explanation for the shape of this graph. Hence, comment on the accuracy of the value of the Faraday constant obtained by this student.

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

……………………………………………………………………………………………

[2]

For Examiner’s

use

[Total: 12 marks]

Charge passed/ C


0

8


2 Period 3 elements vary in their melting points and electrical conductivities. For

Examiner’s use

(a) (i) On the axes below, sketch these trends for the stated elements.

[4]

(ii) Phosphorus reacts with chlorine to form mixtures of PCl3 and PCl5 which are commonly used in organic reaction synthesis. Draw and name the shapes of PCl3 and PCl5, and state clearly the bond angles.

[4]

Melting point

Na Mg Al Si P S Cl

Electrical Conductivities

Na Mg Al Si P S Cl

9


(b) The second ionisation energies of Period 3 elements from sodium to phosphorus are given in the table below.

Element Second Ionisation Energy / kJ mol-1 Sodium 4560

Magnesium 1450 Aluminium 1820

Silicon 1580 Phosphorus 1900

For Examiner’s

use

(i) Explain why the second ionisation energy of sodium is the highest. …………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

(ii) Explain why the second ionisation energy of silicon is lower than that of aluminium. …………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

…………………………………………………………………………………..

(iii) Generally, second ionisation energy of an element is more endothermic than its first ionisation energy. Explain why this is so. Include the equations for the first and second ionisation energy of an element in your answer. You may use element ‘X’ to write the equations. …………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

…………………………………………………………………………………..

…………………………………………………………………………………..

[4]

[Total: 12 marks]

10


3 (a) Aluminium reacts with various non-metals to form simple compounds.

For Examiner’s

use

(i) Aluminium reacts with phosphorus to form aluminium phosphide, AlP which is used as a rodenticide.

Aluminium phosphide is hydrolysed by water to generate the highly toxic gas phosphine, PH3. NH3 is a polar molecule with a dipole moment of 1.47 D while PH3 has a dipole moment of 0.58 D. Predict whether PH3 is acidic, neutral or basic.

…………………………………………………………………………………….

Solutions of aluminium chloride and lithium hydride in ethoxyethane are mixed together and the resultant white precipitate filtered off. The filtrate is carefully evaporated to dryness and white crystals, compound A, are obtained. They contain:

Li,18.2%; Al,71.2%; H,10.6% by mass

(ii) Suggest the identity of compound A and write a balanced equation for the formation of compound A.

(iii) Compound A reacts violently with water, producing hydrogen gas and a white precipitate. Write a balanced equation for the reaction of A with water. …………………………………………………………………………………..

11


(iv) Draw the structural formula of compound A, clearly illustrating its shape and bondings.

For Examiner’s

use

(v) Compound A is one of the most useful reducing agents in organic chemistry. It serves generally as a source of H-, the hydride ion. Reduction of ethanoic acid to ethanol by compound A occurs in two steps as shown:

CH3C OH

O

CH3C H

O

CH3C H

OH

H

Step I Step II

In step II, the aldehyde is rapidly reduced further to the primary alcohol and cannot be isolated. State the types of reaction mechanism that occur in Step I and Step II.

Type of reaction mechanism

Step I

Step II

[7]

12


(b) Magnalium is an alloy of aluminium and magnesium which is used in boat-building. The diagram below shows some reactions of magnalium.

colourless solution

white precipitate

white residue, B colourless filtrate, C

White solid, D

HCl(aq)

excess NaOH(aq)

f ilter

sample of magnalium

HCl(aq) added dropwise

For Examiner’s

use

(i) Identify B,C and D …………………………………………………………………………………..

…………………………………………………………………………………..

…………………………………………………………………………………..

13


(ii) When 1.75 g sample of magnalium is used in the above reactions, it is found that 0.18 g of compound B is obtained. Determine the percentage by mass of magnesium and aluminium in the magnalium sample.

[5]

For Examiner’s

use

(c) Across Period 3, the types of oxides vary from basic oxide to amphoteric oxide to acidic oxide. In Group IV, the most typical oxides of tin and lead are SnO, SnO2, PbO and PbO2. The following two generalisations can be made about the oxides of the elements in group IV.

As the metallic character of the elements increases down the Group, the oxides become more basic.

The oxides of the elements in their higher oxidation states are more acidic than the oxides of the elements in their lower oxidation states.

Use these generalisations to suggest which of the above oxides of tin or lead is most likely to react with each of the following reagents. (i)

With NaOH(aq)

Formula of oxide:

(ii)

With HCl(aq)

Formula of oxide:

[2]

[Total: 14 marks]

14


4 (a) Fibrinogen is produced by the liver and it is used for blood coagulation. This protein is converted by thrombin into fibrin, which is cross-linked for a blood clot to be formed. One of the amino acids in fibrinogen is glutamic acid. Glutamic acid can be synthesized from cyclopentene (Mr of 68) by the following route.

Br NH2

glutamic acid

I II III

cyclopentene K L

C COOH

(CH2)2

H

NH2

COOH

For Examiner’s

use

(i) Suggest the reagents and conditions for steps I, II and III. Step I: …………………………………….

Step II: ……………………………………

Step III: ………………………………….. [3]

(ii) Suggest a simple chemical test by which compounds K and L could be distinguished from each other. Write any appropriate equation.

[3]

15


(iii) State the name of the reaction in step I. The reaction in step I also produces small quantities of compound M which has a Mr of 134. Draw the structure of M.

[2]

For Examiner’s

use

(iv) State two reasons to explain why the yield of step I may be small. …………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

[2]

(b) Vitamin K is a fat-soluble vitamin that is essential to allow blood clotting to occur. There are three basic forms of vitamin K and they are K1, K2 and K3. All the three basic forms contain two rings fused together. The following is the structure of Vitamin K1.

O

O

CH3

R

CH3

R is an alkyl group and it contains 16 carbons. You can assume that R remains inert for the reactions below.

16


Draw the structures of organic compounds formed when vitamin K1 reacts with each of the following reagents: (i) conc. H2SO4, conc. HNO3, heat

(ii) KMnO4/H+, heat

(iii) LiAlH4, in dry ether.

(iv) HCN with traces of NaCN, followed by heating with aqueous sodium hydroxide.

[5]

[Total: 15 marks]

For Examiner’s

use

17


5 (a) Hydrogen is made from methane via a process known as steam reforming.

CH4(g) + H2O(g) CO(g) + 3 H2(g) H = +206 kJ mol-1

The value of the equilibrium constant, Kp, for this reaction is 1.80 x 10-7 at 600 K.

For Examiner’s

use

(i) Gaseous CH4, H2O and CO are introduced into an evacuated container at 600 K and their initial partial pressures (before reaction) are 1.40 atm, 2.30 atm and 1.60 atm respectively.

Determine the partial pressure of H2 (g) when equilibrium is reached. (You may assume that the extent of the forward reaction is small.)

(ii) Should the temperature of the reaction be raised or lowered to increase the yield of H2? Explain your answer.

…………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

…………………………………………………………………………………..

[5]

18


(b) Additional hydrogen can be recovered using the carbon monoxide produced in another reaction known as the water-gas shift reaction.

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

For Examiner’s

use

Given that the above reaction was conducted at 400 oC and 1 atm, calculate the volume of H2 that can be recovered from 10 kg of CO. (You may assume that H2 behaves as an ideal gas)

[3]

[Total: 8 marks]

19


6 Calcium (Ca2+) ions play an important role in the clotting of blood as well as other cellular processes. As such, an abnormal Ca2+ concentration is of concern.

To determine the Ca2+ concentration, 1.00 cm3 of human blood is treated with aqueous Na2C2O4 solution. The resulting CaC2O4 precipitate is filtered and then treated with H2SO4 to release the C2O4

2- ions into solution. This solution is then titrated with acidified KMnO4.

2.05 cm3 of 4.88 x 10-4 mol dm-3 KMnO4 was required to reach the end-point for a particular blood sample.

For Examiner’s

use

(a) Given that C2O42- is oxidised to CO2, write a balanced ionic equation for the

reaction between MnO4- and C2O4

2-.

………………………………………………………………………………………….

………………………………………………………………………………………….

[1]

(b) Calculate the amount of Ca2+ present in the blood sample.

[2]

(c) Given that the normal concentration of Ca2+ is 90 – 110 mg / L blood, show

whether the concentration of Ca2+ in the sample is acceptable.

[2]

[Total: 5 marks]

20


7 The acid dissociation constants, Ka, of three acids are shown as follows:

compound Ka/ mol dm-3

3-hydroxybenzoic acid

COOH

OH

8.7 x 10-5

Benzoic acid COOH

6.3 x 10-5

4-hydroxybenzoic acid COOH

OH

3.3 x 10-5

For Examiner’s

use

(a) (i) Explain the trend of decreasing Ka values of the three acids.

…………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

…………………………………………………………………………………..

…………………………………………………………………………………..

……………………………………………………………………………………

…………………………………………………………………………………..

21


(ii) Calculate the pH of 0.0180 mol dm-3 solution of benzoic acid?

[3]

For Examiner’s

use

(b) (i) State the reagents and condition required to convert salicyclic acid to aspirin.

COOH

OH

COOH

OCOCH3

salicyclic acid aspirin ……………………………………………………………………………………

(ii) The purity of the aspirin obtained is often estimated by hydrolysis with an excess of aqueous sodium hydroxide. The equation for the hydrolysis is

COO-Na+

OH

COOH

OCOCH3

+ 2 NaOH + CH3COO-Na+

The amount of excess sodium hydroxide is found by titration with hydrochloric acid. Although phenol reacts with sodium hydroxide, the hydrolysis equation indicates that the hydroxyl group in salicyclic acid does not. With the aid of a diagram, suggest a reason for this behaviour.

22


(iii) When 0.10 mol of aspirin was treated with 0.30 mol of aqueous sodium hydroxide, the excess sodium hydroxide reacts with 0.15 mol of hydrochloric acid. Calculate the percentage purity of the aspirin prepared.

[3]

[Total: 6 marks]

For Examiner’s

use

23


BLANK PAGE

1


ACJC H2 Prelim 2012 Paper 2 Answers 1 Planning (P)

A student is interested to determine the Faraday constant. He proposed that the value of the Faraday constant can be determined experimentally using the electrolysis of aqueous nickel (II) sulfate.

The student did some research and made the following notes.

General information

Reaction at the cathode: Ni2+ (aq) + 2e → Ni (s)

Reaction at the anode: Ni (s) → Ni2+ (aq) + 2e

1 mol of Ni (s) is deposited at the cathode by 2 mol of electrons.

The Faraday constant is the charge in coulombs, C, carries by 1 mol of electrons.

The Faraday constant = 96500 C mol-1

Experimental setup

The experimental setup for the electrolysis of nickel (II) sulfate solution is shown above. The student will electrolyse some nickel (II) sulfate solution using nickel electrodes. Before the nickel cathode is placed into the electrolyte, it will be cleaned and weighed.

The current that passes through the electrolyte will be kept constant at 0.3 A for 40 min by adjusting a variable resistor in the experiment. At the end of 40 min, the nickel cathode will be weighed.

The experiment will be repeated for another 7 times at 40 min intervals.

2


(a) Based on the information given under experimental setup, give a full description of the procedures you would use in Step A (Start of experiment) in the space provided. Step B (Rinsing and reweighing of cathode) is provided.

[2]

Experimental method

Step A (Start of experiment)

1. The cathode is cleaned and weighed before being placed in the nickel (II) sulfate solution.

2. The circuit is switched on and the current is maintained at 0.3 A by adjusting the variable resistor.

3. The current is maintained at 0.3 A for exactly 40 mins and the circuit is switched off.

Step B (Rinsing and reweighing of cathode)

1. The cathode is removed from the solution and carefully washed with distilled water to remove any nickel (II) sulfate solution.

2. Distilled water is removed from the cathode by rinsing it with propanone in which the water dissolves.

3. The cathode is finally dried by allowing the propanone to evaporate from the surface.

4. The cathode is reweighed and placed back to the solution.

5. A constant current of 0.3 A is passed for a further 40 mins when the rinsing, drying and weighing are repeated.

This procedure (Step A and B) is repeated for a further 6 times.

3


(b) The student performed the experiment using the experimental setup and method above.

The results of his experiment are recorded below.

Time/ min Mass of cathode/ g

Charge passed/ C

Mass of Ni (s) deposited on

the cathode/ g

0 115.74 0 0

40 115.97 720 0.23

80 116.22 1440 0.48

120 116.46 2160 0.72

160 116.70 2880 0.96

200 116.94 3600 1.20

240 117.19 4320 1.45

280 118.01 5040 2.27

320 117.67 5760 1.93

Calculate and record the charge passed and the mass of nickel deposited on the cathode at 160 min in the space given above.

[1]

4


(c) The student plotted the data on the graph of mass of Ni (s) deposited on the cathode against charge passed.

Draw a best-fit line on the graph below. [1]

(d) With reference to part (b) and part (c), state the anomalous point. Based on the procedures for the experiment, explain for the anomaly. [2] (5040C, 2.27g) There could be some residual liquid on the cathode when it is weighed.

(e) The weighing balance used by the student weighed to 2 decimal places. With reference to the results of the experiment, explain why it is not appropriate to use a weighing balance accurate to 2 decimal places. [1] As the mass of Ni (s) deposited is very small, the % error of the calculated value is large.

(f) The student calculated the value of the gradient in part (c). Gradient of the line = 3.05 x 10-4 g C-1 Using the information provided, calculate a numerical value of the Faraday constant. [2] Ni2+ (aq) + 2e → Ni (s) Mass of Ni (s) deposited = 3.05 x 10-4 (charge passed) Mass of Ni (s) deposited by 1 mol of electron = 0.5 x 58.7 = 29.35g Charge passed to deposit 29.35 g of Ni (s) = 29.35/3.05 x 10-4 = 96230 C Faraday constant = 96230 C mol-1


0 Charge passed/ C

x

x

x

x

x

x

xx

x

5


(g) What other measurements could be made during the course of the experiment to provide alternative data to confirm the determined value of the Faraday constant? [1] The loss in mass of the Ni anode can also be measured and recorded to confirm the determined value of the Faraday constant.

(h) The same experiment was performed by another student. The plotted mass of nickel deposited against charge passed was obtained below. Suggest an explanation for the shape of this graph. Hence, comment on the accuracy of the value of the Faraday constant obtained by this student. [2] The NiSO4 solution is impure. It could be acidified NiSO4 solution. Thus, the evolution of H2 (g) could compete with the deposition of Ni (s) as H+ and Ni2+ could be simultaneously discharged. [Note: Eo (H+/H2) = 0.00V and Eo (Ni2+/Ni) = -0.25V). Simultaneous reduction of H+ and Ni2+ can occur although Eo (H+/H2) is more positive than Eo (Ni2+/Ni) ]. Given the amount of charges pass through the circuit, less Ni(s) is deposited initially. The gradient becomes steeper when all the H+ are completed reduced and only Ni2+ is being discharged. The F constant is accurate if the student use the steeper gradient in his calculation. OR The F constant is not accurate if the student use the gentle gradient in his calculation.

[Total: 12 marks]

Charge passed/ C


0

6


2 Period 3 elements vary in their melting points and electrical conductivities.

(a) (i) On the axes below, sketch these trends for the stated elements.

[4]

(ii) Phosphorus reacts with chlorine to form mixtures of PCl3 and PCl5 which are commonly used in organic reaction synthesis. Draw and name the shapes of PCl3 and PCl5, and state clearly the bond angles. [4]

PCl

ClCl

PCl

ClCl

Cl

Cl Trigonal pyramidal Trigonal bipyramidal

Electrical Conductivities

Na Mg Al Si P S Cl

107º

120º

90º

Melting point

Na Mg Al Si P S Cl

7


(b) The second ionisation energies of Period 3 elements from sodium to phosphorus are given in the table below.

Element Second Ionisation Energy / kJ mol-1 Sodium 4560

Magnesium 1450 Aluminium 1820

Silicon 1580 Phosphorus 1900

(i) Explain why the second ionisation energy of sodium is the highest. The removal of the second electron from sodium is from inner quantum shell which experience greater nuclear attraction whereas the rest are removing from valence shell, thus it has the highest second ionization energy.

(ii) Explain why the second ionisation energy of silicon is lower than that of aluminium. The removal of second electron from silicon is from 3p orbital which is further away from the nucleus compared to that from 3s orbital of aluminium.

(iii) Generally, second ionisation energy of an element is more endothermic than its first ionization energy. Explain why this is so. Include the equations for the first and second ionisation energy of an element in your answer. You may use element ‘X’ to write the equations. X (g) X+(g) + e X+ (g) X2+(g) + e After the first electron is removed from a neutral element, a unipositively charged ion is formed. The second electron will be removed from this positively charge ion. Due to electrostatic attraction, it is more difficult to pull an electron away from a positively charge ion than a neutral atom hence second electron needs more energy to be removed.

[4]

[Total: 12 marks]

8


3 (a) Aluminium reacts with various non-metals to form simple compounds.

(i) Aluminium reacts with phosphorus to form aluminium phosphide, AlP which is used as a rodenticide.

Aluminium phosphide is hydrolysed by water to generate the highly toxic gas phosphine, PH3. NH3 is a polar molecule with a dipole moment of 1.47 D while PH3 has a dipole moment of 0.58 D. Predict whether PH3 is acidic, neutral or basic.

Neutral

Solutions of aluminium chloride and lithium hydride in ethoxyethane are mixed together and the resultant white precipitate filtered off. The filtrate is carefully evaporated to dryness and white crystals, compound A, are obtained. They contain:

Li,18.2%; Al,71.2%; H,10.6% by mass

(ii) Suggest the identity of compound A and write a balanced equation for the formation of compound A. Li Al H mass 18.2 71.2 10.6 n 18.2/6.9

=2.64 71.2/27 =2.64

10.6/1 =10.6

Simplest ratio 1 1 4 Compound A is LiAlH4 4LiH + AlCl3 → LiAlH4 + 3LiCl

(iii) Compound A reacts violently with water, producing hydrogen gas and a white precipitate. Write a balanced equation for the reaction of A with water. LiAlH4 + 4H2O → LiOH + Al(OH)3 + 4H2

(iv) Draw the structural formula of compound A, clearly illustrating its shape and bondings.

Al

H

HH

H

-

Li+

9


(v) Compound A is one of the most useful reducing agents in organic chemistry. It serves generally as a source of H-, the hydride ion. Reduction of ethanoic acid to ethanol by compound A occurs in two steps as shown:

CH3C OH

O

CH3C H

O

CH3C H

OH

H

Step I Step II

In step II, the aldehyde is rapidly reduced further to the primary alcohol and cannot be isolated. State the types of reaction mechanism that occur in Step I and Step II.

Type of reaction mechanism

Step I

Nucleophilic substitution

Step II

Nucleophilic addition

[7]

10


(b) Magnalium is an alloy of aluminium and magnesium which is used in boat-building. The diagram below shows some reactions of magnalium.

colourless solution

white precipitate

white residue, B colourless filtrate, C

White solid, D

HCl(aq)

excess NaOH(aq)

f ilter

sample of magnalium

HCl(aq) added dropwise

(i) Identify B,C and D B----Mg(OH)2 C----Al(OH)4

- D----Al(OH)3

11


(ii) When 1.75 g sample of magnalium is used in the above reactions, it is found that 0.18 g of compound B is obtained. Determine the percentage by mass of magnesium and aluminium in the magnalium sample. No of moles of Mg(OH)2 = 0.18/58.3 =0.00309 Mass of Mg = 0.00309 x 24.3 = 0.0751 g % by mass of Mg = (0.0751/1.75) x 100 =4.29% % by mass of Al= 100-4.29 =95.71% (or 95.7%)

[5]

(c) Across Period 3, the types of oxides vary from basic oxide to amphoteric oxide to acidic oxide. In Group IV, the most typical oxides of tin and lead are SnO, SnO2, PbO and PbO2. The following two generalisations can be made about the oxides of the elements in group IV.

As the metallic character of the elements increases down the Group, the oxides become more basic.

The oxides of the elements in their higher oxidation states are more acidic than the oxides of the elements in their lower oxidation states.

Use these generalisations to suggest which of the above oxides of tin or lead is most likely to react with each of the following reagents. (i)

With NaOH(aq)

Formula of oxide: SnO2

(ii)

With HCl(aq)

Formula of oxide: PbO

[2]

[Total: 14 marks]

12


4 (a) Fibrinogen is produced by the liver and it is used for blood coagulation. This protein is converted by thrombin into fibrin, which is cross-linked for a blood clot to be formed. One of the amino acids in fibrinogen is glutamic acid. Glutamic acid can be synthesized from cyclopentene (Mr of 68) by the following route.

Br NH2

glutamic acid

I II III

cyclopentene K L

C COOH

(CH2)2

H

NH2

COOH

(i) Suggest the reagents and conditions for steps I, II and III. Step I: Br2, UV light Step II: NH3 (alc), heat in sealed tube Step III: KMnO4/H

+ heat under reflux.

[3]

(iii) Suggest a simple chemical test by which compounds K and L could be distinguished from each other. Write any appropriate equation. R/C: Heat with NaOH (aq), cool, then add dil HNO3. Followed by AgNO3. Sample that produces a cream ppt is K, sample that does not is L.

Br

+ NaOH

OH

+ NaBr

Ag+ + Br

- AgBr

[3]

(iii) State the name of the reaction in step I. The reaction in step I also produces small quantities of compound M which has a Mr of 134. Draw the structure of M. Free radical substitution

[2]

13


(iv) State two reasons to explain why the yield of step I may be small. Electrophilic addition across the alkene functional group serves as a competing reaction. There is more than one H atom on the cyclopentene that can be substituted/ the number of byproducts produced.

[2]

(b) Vitamin K is a fat-soluble vitamin that is essential to allow blood clotting to occur. There are three basic forms of vitamin K and they are, K1, K2 and K3. All the three basic forms contain two rings fused together. The following is the structure of Vitamin K1.

O

O

CH3

R

CH3

R is an alkyl group and it contains 16 carbons. You can assume that R remains inert for the reactions below. Draw the structures of organic compounds formed when vitamin K1 reacts with each of the following reagents: (i) conc. H2SO4, conc. HNO3, heat

O

O

CH3

R

CH3NO2

(ii) KMnO4/H+ heat

COOH

O

O

CH3

O

O

+

CH3

O

R OR

[5]

14


(iii) LiAlH4, in dry ether.

OH

OH

CH3

R

CH3

(iv) HCN with traces of NaCN. Followed by heating with aqueous sodium hydroxide.

OH

OH

CH3

R

CH3COO-Na

+

COO-Na

+

[Total: 15 marks]

5 (a) Hydrogen is made from methane via a process known as steam reforming.

CH4(g) + H2O(g) CO(g) + 3 H2(g) H = +206 kJ mol-1

The value of the equilibrium constant, Kp, for this reaction is 1.80 x 10-7 at 600 K.

(i) Gaseous CH4, H2O and CO are introduced into an evacuated container at 600 K and their initial partial pressures (before reaction) are 1.40 atm, 2.30 atm and 1.60 atm respectively.

Determine the partial pressure of H2(g) when equilibrium is reached. (You may assume that the extent of the forward reaction is small.) [3]

CH4(g) + H2O(g) CO(g) + 3H2(g)

Initial /atm 1.40 2.30 1.60 0

Change/atm -x -x +x +3x

Eqm 1.40 - x 2.30 – x 1.60 + x 3x

Kp = )P)(P(

)P)(P(

OHCH

3HCO

24

2

1.80 x 10-7 =)x30.2)(x40.1(

)x3)(x60.1( 3

Assuming x is small,

15


1.80 x 10-7 = )30.2)(40.1(

)x3)(60.1( 3

Solving for x,

x = 2.38 x 10-3

Partial pressure of H2 at equilibrium = 3 x 2.38 x 10-3 = 7.14 x 10-3 atm

(ii) Should the temperature of the reaction be raised or lowered to increase the yield of H2? Explain your answer.

[2]

The temperature should be raised to increase the yield of H2.

When the temperature is raised, the endothermic reaction is favoured since it absorbs energy. The position of equilibrium shifts to the right in favour of the forward reaction, increasing the yield of H2.

(b) Additional hydrogen can be recovered using the carbon monoxide produced in another reaction known as the water-gas shift reaction.

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

Given that the above reaction was conducted at 400 oC and 1 atm, calculate the volume of H2 that can be recovered from 10 kg of CO. (You may assume that H2 behaves as an ideal gas)

[3] Amount of H2 that can be recovered = 10000/28 = 357 mol pV = nRT

V = p

nRT

V = 51001.1

)273400(31.8357

= 19.8 m3

[Total: 8 marks]

16


6 Calcium (Ca2+) ions play an important role in the clotting of blood as well as other cellular processes. As such, an abnormal Ca2+ concentration is of concern.

To determine the Ca2+ concentration, 1.00 cm3 of human blood is treated with aqueous Na2C2O4 solution. The resulting CaC2O4 precipitate is filtered and then treated with H2SO4 to release the C2O4

2- ions into solution. This solution is then titrated with acidified KMnO4.

2.05 cm3 of 4.88 x 10-4 mol dm-3 KMnO4 was required to reach the end-point for a particular blood sample.

(a) Given that C2O42- is oxidised to CO2, write a balanced ionic equation for

the reaction between MnO4- and C2O4

2-. [1]

2 MnO4- + 5 C2O4

2- + 16 H+ 2 Mn2+ + 10 CO2 + 8 H2O

(b) Calculate the amount of Ca2+ present in the blood sample. [2]

Amount of KMnO4 reacted

= 2.05/1000 x 4.88 x 10-4

= 1.00 x 10-6 mol

Amount of Ca2+ present in the blood sample

= 1.00 x 10-6 x 5/2

= 2.50 x 10-6 mol

(c) Given that the normal concentration of Ca2+ is 90 – 110 mg / L blood,

show whether the concentration of Ca2+ in the sample is acceptable. [2] Amount of Ca2+ present in 1 L of blood sample = 2.50 x10-6 x 1000 = 2.50 x 10-3 mol Mass of Ca2+ present in 1 L of blood sample = 2.50 x10-3 x 40.1 = 0.10035g = 100 mg The concentration of Ca2+ in the sample falls within the acceptable range.

[Total: 5 marks]

17


7 The acid dissociation constants, Ka, of three acids are shown as follows:

compound Ka/ mol dm-3

3-hydroxybenzoic acid

COOH

OH

8.7 x 10-5

Benzoic acid COOH

6.3 x 10-5

4-hydroxybenzoic acid COOH

OH

3.3 x 10-5

(a) (i) Explain the trend of decreasing Ka values of the three acids.

In 3-hydroxybenzoic acid, the negative charge of the carboxylate anion is stabilized by the electron-withdrawing inductive effect of OH group, making it more acidic.

For 4-hydroxybenzoic acid, the negative charge on the anion is destabilized by the electron-donating resonance effect of the OH that acts over the electron system of the ring. Hence it is less acidic than benzoic acid

(ii) Calculate the pH of 0.0180 mol dm-3 solution of benzoic acid?

[[H+] =√(6.3 x 10-5)(0.0180) = 1.06 x 10-3 mol dm-3

pH= -lg{H+] = -lg(1.06 x 10-3) = 2.97

[3]

18


(b) (i) State the reagent and condition required to convert salicyclic acid to Aspirin.

COOH

OH

COOH

OCOCH3

aspirin salicyclic acid NaOH(aq), CH3COCl, room temperature

(ii) The purity of the Aspirin obtained is often estimated by hydrolysis with an excess of aqueous sodium hydroxide. The equation for the hydrolysis is

COO-Na+

OH

COOH

OCOCH3

+ 2NaOH + CH3COO-Na+

The amount of excess sodium hydroxide is found by titration with hydrochloric acid. Although phenol reacts with sodium hydroxide, the hydrolysis equation indicates that the hydroxyl group in salicyclic acid does not. With the aid of a diagram, suggest a reason for this behavior.

C

O

H

O O-

Intramolecular hydrogen bond stabilizes the anion.

19


(iii) When 0.10 mol of aspirin was treated with 0.30 mol of aqueous sodium hydroxide, the excess sodium hydroxide reacts with 0.15 mol of hydrochloric acid. Calculate the percentage purity of the aspirin prepared. No. of moles of excess NaOH =0.15 No. of moles of NaOH reacted with aspirin = 0.30-0.15 = 0.15 No of moles of aspirin = 0.15/2 = 0.0750 %purity = (0.075/0.10) x 100 = 75.0%

[3]

[Total: 6 marks]

1


ANGLO-CHINESE JUNIOR COLLEGE

DEPARTMENT OF CHEMISTRY Preliminary Examination

CHEMISTRY H2 9647/03 Paper 3 Free Response 29 August 2012 2 hours Candidates answer on separate paper.

Additional Materials: Data Booklet Answer Paper Cover Page

READ THESE INSTRUCTIONS FIRST Write your index number, name, form class and tutorial class on all the work you hand in.

Write in dark blue or black pen.

You may use a pencil for any diagrams, graphs, or rough working.

Do not use staples, paper clips, highlighters, glue or correction fluids.

Answer any four questions.

A Data Booklet is provided.

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 of part question.

At the end of the examination, fasten all your work securely together.

This document consists of 12 printed pages and 1 blank page.

9647/Prelim/12 ANGLO-CHINESE JUNIOR COLLEGE © ACJC 2012 Department of Chemistry [Turn over

2


Answer any four questions.

1 (a) When hydrochloric acid is added to sodium thiosulfate, Na2S2O3, a fine precipitate of sulfur appears. This gradually makes the solution opaque.

Na2S2O3(aq) + 2 HCl(aq) S(s) + SO2(g) + 2 NaCl (aq) + H2O(l)

The rate of the reaction may be determined by measuring how long it takes for the solution to become opaque such that a cross marked on a piece of paper placed under the beaker containing the reaction mixture just becomes obscured.

A series of experiments were carried out to study the order of reaction with respect to Na2S2O3 and the following results were obtained.

Experiment Number

Volume / cm3 Time / s

HCl Na2S2O3 H2O

1 20.0 40.0 40.0 39

2 20.0 30.0 50.0 50

3 20.0 20.0 60.0 83

4 20.0 10.0 70.0 170

(i) Explain why the total volume of the reaction mixture is kept constant for all

the four experiments.

(ii) Calculate the values of 1s/)

Time

1( for the various volumes of Na2S2O3 used.

(iii) Hence or otherwise, state how the rate changes with the volume of Na2S2O3 used.

(iv) Explain why the rate changes with the volume of Na2S2O3 used.

[5]

(b) Sulfur reacts with chlorine gas to form a compound A containing 47.4% sulfur by mass. When compound A is reacted with water, a yellow precipitate was formed together with a solution containing a mixture of sulfurous acid, H2SO3 and hydrochloric acid, HCl.

(i) Given that the relative molecular mass of A is 135, determine the empirical formula and the molecular formula of A.

(ii) Explain why compound A can react with water.

(iii) If 1.00 g of compound A was dissolved in water, calculate the volume of 1.00 mol dm-3 NaOH that would be required to completely neutralise the resultant solution.

[5]

3


(c) The Wittig reaction was discovered by German chemist Georg Wittig. This reaction is significant because it uses a phosphorous-containing compound to produce an alkene molecule. In 1979, Georg Wittg was awarded the Nobel Prize for his work in discovering the Wittig reaction. The Wittig reaction is thought to proceed through the following stages.

CH3Cl P+

CH3

Ph

Ph

Ph P CH2

Ph

Ph

Ph C C

H

H

C2H5

CH3

P(Ph)3

I

PhLi

II

CH3COC2H5

III

where Ph refers to phenyl substituent.

(i) What type of reaction takes place in stage I?

(ii) Describe the mechanism for stage I. In your answer, show any relevant charges, lone pair of electrons and movement of electrons.

(iii) State the role of PhLi in stage II.

(iv) Suggest the other product of stage III and hence construct the balanced equation for this stage.

(v) When butane reacts with chlorine under ultraviolet light, a mixture of monochlorobutane is produced. P(Ph)3, PhLi and ethanal are then added in succession to the resultant mixture. This results in the formation of two pairs of geometric isomeric alkenes. Suggest the structures of the four alkenes, and the ratio in which they might be produced. Explain your suggested ratio. [10]

[Total: 20 marks]

4


2 (a) Silver chloride is a white crystalline solid well known for its low solubility in water.

It is often used in photography as it darkens quickly when it is exposed to light.

(i) Explain what is meant by the lattice energy of silver chloride. Write an equation to represent the lattice energy of silver chloride.

(ii) Using the following data, and relevant data from the Data Booklet, construct a thermochemical cycle to calculate the lattice energy of silver chloride. Include state symbols in your cycle. Value / kJ mol-1 Standard enthalpy change of formation of AgCl -127 Standard enthalpy change of atomization of Ag +285 First ionization energy of Ag +731 First electron affinity of Cl -349

(iii) Explain how you would expect the numerical magnitude of the lattice energy of silver chloride to compare with the lattice energy of silver iodide.

[7]

(b) Equal volumes of 0.020 mol dm-3 of aqueous silver nitrate and an unknown concentration of aqueous sodium chloride were mixed in a beaker. A white precipitate was observed. The precipitate formed was filtered and washed. A colourless solution was seen upon addition of aqueous ammonia to the white precipitate. Solubility product, Ksp, of silver chloride is 1.77 x 10-10 mol2 dm-6.

(i) Write an expression for the solubility product of silver chloride.

(ii) Calculate the maximum concentration of aqueous sodium chloride such that no precipitate of silver chloride is formed upon mixing the two solutions.

(iii) Explain why a colourless solution is obtained when aqueous ammonia is added to the white precipitate.

[4]

(c) Iodine, a member of Group VII element, reacts with alkali at room temperature to form iodide and iodate(V) ions. Group II metals form iodate(V) compounds which can undergo decomposition to give metal oxide, iodine vapour and oxygen gas.

(i) Name the reaction between iodine and alkali and write balanced ionic equation for the reaction.

A student conducted an experiment to find out the thermal stability of Ca(IO3)2, Sr(IO3)2 and Ba(IO3)2. He took 2.0 g of each compound and heated them separately in a test-tube at a temperature T ºC. He recorded the time of the appearance of purple iodine vapour.

(ii) Briefly explain the relationship between thermal stability of iodate compounds with the time of appearance of purple iodine vapour.

5


(iii) Arrange Ca(IO3)2, Sr(IO3)2 and Ba(IO3)2 in the order of increasing thermal stability. Explain the trend as fully as you can.

(iv) Solid sodium iodide undergoes reaction with concentrated sulfuric acid State the observations and write balanced equations for the reactions that take place.

[9]

[Total: 20 marks]

3 (a) Histidine is an essential α-amino acid in humans and other mammals. Histidine is

utilized by the 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 the brain to organs throughout the body. Therefore, adequate histidine levels are essential to good mental and physical health. Food such as yellowfin tuna, mackerel and cauliflowers are excellent sources of histidine.

The structure of the fully protonated form of histidine is given below.

+H3N C C

CH2

OH

O

HN

NH+

H

The pKa values of the respective functional groups are provided in the following table.

Functional group pKa value

-COOH 1.82

-NH3+ 9.17

CH2

HN

NH+

6.00

6


(i) Make use of the pKa values to suggest the major species present in solutions of histidine with the following pH values.

pH 1 pH 4 pH 8 pH12

(ii) Explain, with the aid of relevant chemical equation(s), how histidine acts as a buffer at pH of 1.82.

(iii)

+H3N C C

CH2

OH

O

HNA

NBH+

H

As shown in the structure of the fully protonated histidine above, the side chain of the fully protonated histidine has 2 different N atoms, NA and NB atom. NB atom is protonated but not NA atom. The heterocyclic ring of the side chain is known to be as stable as benzene ring. Explain why NA atom cannot be protonated.

[7]

(b) Histidine is also important for our digestion as it helps to produce gastric juices in the stomach. However, high level of histidine causes acid indigestion which is also known as heartburn. Heartburn results in burning sensation at the upper stomach as the stomach acid rises into the esophagus. The pH of the stomach of a patient who suffers from heartburn is 4. A stomach juice sample is extracted from a patient to determine the concentration of histidine using sodium hydroxide, NaOH (aq).

(i) Draw a labeled titration curve of pH against amount of NaOH (aq) added when 1 mole of fully protonated histidine is titrated with NaOH. You should clearly label the following in your titration curve.

Equivalence point(s). Maximum buffer capacities.

(ii) Write a balanced equation of the reaction between histidine in the patient’s stomach juice sample and NaOH.

[5]

7


(c) Histidine is converted to histamine by an enzyme known as histidine decarboxylase. Histamine increases the permeability of the capillaries to white blood cells, to allow them to engage harmful pathogens in infected body tissues. Histidine decarboxylase has a quaternary protein structure. The enzyme-catalysed decarboxylation reaction is shown below.

H2N CX CY

CH2

OH

O

HN

N

H

H2N CX

CH2

HN

N

H

H

+ CYO2

Histidine decarboxylase

Histidine Histamine The decarboxylation of histidine occurs in the active site of histidine decarboxylase. In one part of the enzyme’s active site, there is a section of 11 amino acid residues which is critical in the catalysis of decarboxylation of histidine. In addition, the active site also causes some of the bond angles in histidine molecule to deviate from its ideal bond angles. Consequently, the catalysis of decarboxylation of histidine is further increased.

(i) Explain what is meant by the term quaternary structure of protein.

(ii) Histidine decarboxylase undergoes enzymatic partial hydrolysis. Upon hydrolysis, the polypeptide fragments of the critical section of the enzyme’s active site are isolated as shown below.

Ala-Cys-Phe Gly-Gly Lys-Asp-Asp-Gly-Gly Phe-Arg-Lys Ala-Cys-Phe-Phe-Arg-Lys Asp-Asp-Gly

Deduce the sequence of the critical section of 11 amino acid residues of the enzyme’s active site. Show your working clearly.

(iii) The CX-CY bond in histidine is weakened during the enzyme-catalysed decarboxylation reaction. Explain why the CX-CY bond in histidine is weakened.

8


(iv) The decarboxylation reaction of histidine can be disrupted due to fever and lead poisoning. Explain how the following conditions disrupt the decarboxylation reaction of histidine.

Fever Lead Poisoning

[8]

[Total: 20 marks]

4 (a) Statistics show a correlation between road accidents and drivers who drink too much.

One of the first chemical breathalysers was based on the oxidation process of ethanol using acidified potassium dichromate(VI).

(i) Derive the half equation for the oxidation of ethanol to ethanoic acid.

(ii) Hence, using relevant information from the Data Booklet, write a balanced ionic equation for the oxidation of ethanol using acidified potassium dichromate(VI).

The breathalyser kit consists of an inflated 1000 cm3 plastic bag connected to a glass tube containing the dichromate crystals. When the breath is expelled through the tube the crystals change colour as they are reduced. The proportion of the crystals that change colour indicates the amount of alcohol present. The current legal maximum blood alcohol concentration when driving is 80 mg per 100 cm3 of blood. Alcohol concentration in the blood can be estimated by analysing the alcohol in the breath because an equilibrium is set up between the alcohol in the blood and the alcohol in the breath: CH3CH2OH(blood) CH3CH2OH(breath) At body temperature, the concentration of alcohol in the blood is about 2300 times that in the breath.

(iii) What is the corresponding breath alcohol concentration in mg per 1000 cm3 of breath?

(iv) What colour change would you expect for a positive result? [4]

(b) Modern breathalysers use fuel-cell technology. The fuel cell has two platinum electrodes with a porous acid-electrolyte material sandwiched between them. As the exhaled air from the suspect flows past one side of the fuel cell, the platinum anode oxidizes any alcohol in the air to produce ethanoic acid. The electrons produced flow through a wire from the platinum anode to the cathode. The protons move through the lower portion of the fuel cell to the cathode and combine with oxygen to form water. It is the size of this electric current which determines the amount of alcohol in the blood stream.

(i) Construct an equation for the cathode reaction.

(ii) Hence write a balanced equation for the overall reaction.

9


(iii) ∆GѲ and EcellѲ are related by the following equation, ∆GѲ = -nFEcell

Ѳ where ∆GѲ is the standard free energy change in J mol-1, n is the number of electrons transferred during the redox reaction and F is the Faraday constant. ∆GѲ, for the oxidation of ethanol to ethanoic acid at 298 K, is -374 kJ mol-1. Use the equation you have written in (b)(ii) to decide on a suitable value for n and hence calculate the Ecell

Ѳ for the fuel cell breathalyser. [4]

(c) EDTA (compound B) and the ‘crown thioether’ (compound C) are polydentate ligands that are used to remove harmful metals from the environment.

NCH2CH2N

CH2COO-

CH2COO-

-OOCH2C

-OOCH2C

B

Compound B complexes with Cu2+ and can be used to remove Cu2+ from contaminated water.

Compound C complexes with Ni2+ and can be used to extract Ni2+ from solutions of industrial waste.

(i) Explain what is meant by a polydentate ligand and state the features present in both compounds B and C which make them suitable to act as ligands.

(ii) For both complexes formed between compound B and Cu2+ ions and between compound C and Ni2+ ions respectively, suggest the co-ordination numbers and shapes of both complexes.

[4]

(d) When water ligands in a hydrated metal ion are substituted by other ligands, the equilibrium constant for the reaction is referred to as the stability constant, Kstab, of the new complex. For example, [Ni(H2O)6]

2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l)

[Ni(NH3)6]

2+ Kc = Kstab = [Ni(H2O)6]

2+[NH3]6

C

10


The table below lists the colours and stability constants (in logarithm form) of some nickel(II) complexes.

Complex lg Kstab Colour

D [Ni(H2O)6]2+(aq)

- green

E [Ni(NH3)6]2+(aq)

7.7 blue

F [Ni(H2NCH2CH2NH2)3]2+(aq)

18.3 purple

G [Ni(EDTA)]2-(aq)

19.0 blue

H [Ni(CN)4]2-(aq)

31.1 Pale yellow

(i)

The standard free energy change, ∆GѲ, and the stability constant, Kstab, are also related by the following equation, ∆GѲ = -2.3RT lg Kstab …….(1) where R is the molar gas constant, T is the temperature in Kelvin and Kstab is the stability constant. Using this equation [∆GѲ = ∆HѲ-T∆SѲ ] and the above equation (1), explain why [Ni(H2NCH2CH2NH2)3]

2+ has a higher lg Kstab than that of [Ni(NH3)6]2+ even though

both complexes have six Ni2+-N dative bonds. [ You may represent [Ni(H2NCH2CH2NH2)3]

2+ as [Ni(en)3]2+]

(ii) Use the data in the table to predict and explain what you will observe when a

solution of pale yellow [Ni(CN)4]2- is treated with EDTA.

(iii) The following table lists the colours of the photons of light of certain wavelengths. Colour of photon violet blue green yellow red

Wavelength/nm 400 450 500 600 650

Based on the observed colours of the complexes, deduce which complex, D or H, will have a larger energy gap between the two groups of 3d orbitals? Explain your answer. Predict the colour of [Ni(H2O)4(H2NCH2CH2NH2)]

2+(aq). [8]

[Total: 20 marks]

11


5 (a) The molecules of ethanoic acid and methyl methanoate contain the same number of electrons. These molecules have the same mass and size but their boiling points differ widely.

Compound Boiling point /oC CH3COOH 112 HCOOCH3 32

(i)

Suggest an explanation for the difference in boiling point.

(ii) Draw a diagram showing why ethanoic acid can dimerise in methyl methanoate. [3]

(b) The classic synthesis of an ester is done via the Fischer Esterification, which involves treating a carboxylic acid with an alcohol in the presence of a catalyst. The reaction between butanoic acid and ethanol produces ethyl butanoate, an ester which smells like pineapples.

CH3CH2CH2COOH + CH3CH2OH CH3CH2CH2COOCH2CH3 + OH2

H+

An alternative way of producing ethyl butanoate is to mix ethanol with the acid chloride, butanoyl chloride.

(i) State the name of the reaction between butanoyl chloride and ethanol.

(ii) Using relevant data from the Data Booklet, explain why the acid chlorides are more reactive than the carboxylic acid in the reaction to form esters.

(iii) Suggest and explain one safety precaution you should employ when using acid chlorides in the laboratory.

[5] (c) In 1899 the German chemists Adolf von Baeyer and Victor Viliger discovered that

by treating ketones with peroxyacids, such as peroxybenzoic acid (C6H5COOOH), gives an ester by “insertion” of one atom of oxygen. The Baeyer-Viliger reaction is thought to proceed through the following stages.

C

O

R1 R2

C

O

OOH

CO

OO

C

O

R1 R2

H

C

O

R1 OR2I II

12


For asymmetrical ketones the oxygen atom is inserted in the order: tertiary alkyl > secondary alkyl > primary alkyl > CH3.

(i) Peroxybenzoic acid can be formed by reacting benzoic acid and hydrogen peroxide. Write the balanced equation of this reaction.

(ii) State the role of peroxybenzoic acid in the Baeyer-Viliger reaction.

(iii) What type of reaction takes place in stage I?

(iv) Suggest the other product of stage II.

(v) Molecules of compound J, C6H12O are chiral. The bond angles about all the carbon atoms are approximately 109.5o. When H2O is eliminated from J, a mixture of three isomeric alkenes, C6H10, is produced. Only two of which are optical isomers of each other. Heating compound J with acidified potassium permanganate produces K, C6H10O. Mixing compound K with peroxybenzoic acid produces one compound L as the product. Suggest the structures of J, K, L and the three alkenes. [9]

(d) Using your answers in (c)(v), suggest a simple chemical test by which one of the alkenes and compound J could be distinguished from each other. Write any appropriate equation. [3]

[Total: 20 marks]

13


BLANK PAGE

1


ACJC H2 Prelim Paper 3 Answers

1 (a) (i) When hydrochloric acid is added to sodium thiosulfate, Na2S2O3, a fine precipitate of sulfur appears. This gradually makes the solution opaque.

Na2S2O3(aq) + 2 HCl(aq) S(s) + SO2(g) + 2 NaCl (aq) + H2O(l)

The rate of the reaction may be determined by measuring how long it takes for the solution to become opaque such that a cross marked on a piece of paper placed under the beaker containing the reaction mixture just becomes obscured.

A series of experiments were carried out to study the order of reaction with respect to Na2S2O3 and the following results were obtained.

Experiment Number

Volume / cm3 Time / s

HCl Na2S2O3 H2O

1 20.0 40.0 40.0 39

2 20.0 30.0 50.0 50

3 20.0 20.0 60.0 83

4 20.0 10.0 70.0 170

Explain why the total volume of the reaction mixture is kept constant for all the four experiments.

This is so that the concentration of Na2S2O3 is directly proportional to the volume used.

[1]

(ii) Calculate the values of 1s/)

Time

1( for the various volumes of Na2S2O3 used.

Volume / cm3 Time / s 1s/)Time

1(

Na2S2O3

40.0 39 0.0256

30.0 50 0.0200

20.0 83 0.0120

10.0 170 0.00588

[1]

(iii) Hence or otherwise, state how the rate changes with the volume of Na2S2O3 used. The rate increases with the increase in the volume of thiosulfate used.

[1]

(iv) Explain why the rate changes with the volume of Na2S2O3 used.

When the volume of sodium thiosulfate used increases, the number of

[2]

2


sodium thiosulfate particles having energy ≥ Ea increases.

This results in an increase in the frequency of collision and number of effective collisions. Therefore the rate increases.

(b) Sulfur reacts with chlorine gas to form a compound A containing 47.4% sulfur by mass. When compound A is reacted with water, a yellow precipitate was formed together with a solution containing a mixture of sulfurous acid, H2SO3 and hydrochloric acid, HCl.

(i) Given that the relative molecular mass of A is 135, determine the empirical formula and the molecular formula of A.

S Cl % by mass 47.4 53.6 No of moles 47.4/32.1 =1.48 53.6/35.5 = 1.51 Simplest ratio 1 1

Empirical formula: SCl Let molecular formula of X be (SCl)y. y(32 + 35.5) = 135 y = 2 Molecular formula of X is S2Cl2

[2]

(ii) Explain why compound A can react with water. S in S2Cl2 has vacant low-lying 3d orbitals that can accept lone pair of electrons from the oxygen on H2O and undergoes hydrolysis.

[1]

(iii) If 1.00 g of compound A was dissolved in water, calculate the volume of 1.00 mol dm-3 NaOH that would be required to completely neutralise the resultant solution. 2S2Cl2 + 3H2O 3S +H2SO3 + 4HCl Amount of S2Cl2 = 1/135 = 7.407 x 10-3 mol Amount of NaOH required = 7.407 x 10-3 + 2 (7.407 x 10-3) = 0.0222 mol Volume of NaOH required = 0.0222 ÷ 1/1000 = 22.20 cm3

[2]

3


(c) The Wittig reaction was discovered by German chemist Georg Wittig. This reaction is significant because it uses a phosphorous-containing compound to produce an alkene molecule. In 1979, Georg Wittg was awarded the Nobel Prize for his work in discovering the Wittig reaction. The Wittig reaction is thought to proceed through the following stages.

CH3Cl P+

CH3

Ph

Ph

Ph P CH2

Ph

Ph

Ph C C

H

H

C2H5

CH3

P(Ph)3

I

PhLi

II

CH3COC2H5

III

where Ph refers to phenyl substituent.

(i) What type of reaction takes place in stage I?

Bimolecular nucleophilic substitution reaction / SN2

[1]

(ii) Describe the mechanism for stage I. In your answer, show any relevant charges, lone pair of electrons and movement of electrons.

C

H

H

HCl

d+ d-

P(Ph)3

C

H

H

H

Cl(Ph)3P C

H

H

H(Ph)3P+ Cl x

x

transition state

d+ d-

[2]

(iii) State the role of PhLi in stage II.

It functions as a base.

[1]

(iv) Suggest the other product of stage III and hence construct the balanced equation for this stage. (Ph)3PO (or (Ph)3P=O) (Ph)3P=CH2 + CH3COC2H5 →(Ph)3PO + CH3C(C2H5)CH2

[2]

(v) When butane reacts with chlorine under ultraviolet light, a mixture of monochlorobutane is produced. P(Ph)3, PhLi and ethanal are then added in succession to the resultant mixture. This results in the formation of two pairs of geometric isomeric alkenes. Suggest the structures of the four alkenes, and the ratio in which they might be produced. Explain your suggested ratio.

[4]

4


C C

CH3

CH2CH3

CH3

H

C C

CH3

CH2CH3

H

CH3

C C

H

CH2CH2CH3

CH3

H

C C

H

CH2CH2CH3

H

CH3

1A 1B

2A 2B

Ratio of 1A:1B:2A:2B = 2:2:3:3 Both the cis and trans isomer are equally likely to be produced. 1A and 1B came from CH3CHClCH2CH3 while 2A and 2B came from CH2ClCH2CH2CH3. The ratio of the two alkyl halides are 2:3 (due to number of hydrogen atoms) Alternate answer: Ratio of 1A:1B:2A:2B ≈ 2:2:3:3 (or all approximately equal amounts) – This ratio is dependent on student’s explanation. Both the cis and trans isomer are equally likely to be produced. 1A and 1B came from CH3CHClCH2CH3 while 2A and 2B came from CH2ClCH2CH2CH3. The ratio of the two alkyl halides are nearly 1:1 (despite the different number of hydrogen atom, the 2o alkyl radical is more stable than the 1o hence more likely to be produced)

5


2 (a) Silver chloride is a white crystalline solid well known for its low solubility in water. It is often used in photography as it darkens quickly when it is exposed to light.

(i) Explain what is meant by the lattice energy of silver chloride. Write an equation to represent the lattice energy of silver chloride.

Lattice energy is the heat evolved when one mole of solid silver chloride is formed from its component gaseous ions. Ag+ (g) + Cl- (g) AgCl (s)

[2]

(ii) Using the following data, and relevant data from the Data Booklet, construct a thermochemical cycle to calculate the lattice energy of silver chloride.

Include state symbols in your cycle.

Value / kJ mol-1 Standard enthalpy change of formation of AgCl -127 Standard enthalpy change of atomization of Ag +285 First ionization energy of Ag +731 First electron affinity of Cl -349

Ag (s) + ½ Cl2 (g) AgCl (s)

+285 ½ (+244)

Ag (g) + Cl (g) LE

+731 -349

Ag+(g) + Cl- (g)

(+285) + (+731) + ½ (+244) +(-349) + LE = -127

LE = -916 kJ mol-1

[4]

(iii) Explain how you would expect the numerical magnitude of the lattice energy of silver chloride to compare with the lattice energy of silver iodide. The lattice energy magnitude of AgCl is larger than AgI as the ionic radius of Cl- is smaller than I-.

[1]

(b) Equal volumes of 0.020 mol dm-3 of aqueous silver nitrate and an unknown concentration of aqueous sodium chloride were mixed in a beaker. A white precipitate was observed. The precipitate formed was filtered and washed. A

-127

6


colorless solution was seen upon addition of aqueous ammonia to the white precipitate. Solubility product, Ksp, of silver chloride is 1.77 x 10-10 mol2 dm-6.

(i) Write an expression for the solubility product of silver chloride. Ksp = [Ag+][Cl-]

[1]

(ii) Calculate the maximum concentration of aqueous sodium chloride such that no precipitate of silver chloride is formed upon mixing the two solutions. [Ag+][Cl-] ≤ Ksp (0.02/2)(x/2) ≤ 1.77 x 10-10 x ≤ 3.5 x 10-8 Max conc of NaCl(aq) = 3.5 x 10-8 mol dm-3

[2]

(iii) Explain why a colorless solution is obtained when aqueous ammonia is added to the white precipitate. AgCl(s) dissolves readily in dilute NH3(aq) due to the formation of [Ag(NH3)2]

+.

Or Ag+Cl-(s) + 2NH3(aq) [Ag(NH3)2]+(aq) + Cl-(aq)

Or When excess NH3 is added, the concentration of Ag+ decreases and the ionic product becomes numerically smaller than the Ksp. Hence, AgCl dissolves in excess NH3(aq).

[1]

(c) Iodine, a member of Group VII element, reacts with alkali at room temperature to form iodide and iodate(V) ions. Group II metals form iodate(V) compounds which can undergo decomposition to give metal oxide, iodine vapour and oxygen gas.

(i) Name the reaction between iodine and alkali and write balanced ionic equation for the reaction. Disproportionation 3I2 + 6OH- 5I- + IO3

- + 3H2O

[2]

A student conducted an experiment to find out the thermal stability of Ca(IO3)2, Sr(IO3)2 and Ba(IO3)2. He took 2.0 g of each compound and heated them separately in a test-tube at a temperature T ºC. He recorded the time of the appearance of purple iodine vapour.

(ii) Briefly explain the relationship between thermal stability of iodate compounds with the time of appearance of purple iodine vapour. The more stable the iodate compounds, the longer is the time for iodine to appear.

[1]

7


(iii) Arrange Ca(IO3)2, Sr(IO3)2 and Ba(IO3)2 in the order of increasing thermal stability. Explain the trend as fully as you can. Thermal stability: Ca(IO3)2 < Sr(IO3)2 < Ba(IO3)2 Cationic size increases from Ca2+ to Sr2+ to Ba2+, resulting in decreasing charge density and hence decreasing polarizing power. Thus the electron cloud of the anion is less distorted and thus less weakening effect on the intramolecular bonds. The result is thermal decomposition becomes more difficult.

[3]

(iv) Solid sodium iodide undergoes reaction with concentrated sulfuric acid State the observations and write balanced equations for the reactions that take place. Purple fumes of I2, pungent gas/rotten egg smell H2S and black solid I2.

NaI (s) + H2SO4 (l) NaHSO4 (s) + HI (g)

2HI (g) + H2SO4 (l) I2 (g) + 2H2O (l) + SO2(g)

6HI (g) + H2SO4 (l) 3I2 (g) + 4H2O (l) + S (s)

8HI (g) + H2SO4 (l) 4I2 (g) + H2S(g) + 4H2O (l)

[3]

[Total: 20 marks]

8


3 (a) Histidine is an essential α-amino acid in humans and other mammals. Histidine is utilized by the 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 the brain to organs throughout the body. Therefore, adequate histidine levels are essential to good mental and physical health. Food such as yellowfin tuna, mackerel and cauliflowers are excellent sources of histidine.

The structure of the fully protonated form of histidine is given below.

+H3N C C

CH2

OH

O

HN

NH+

H

The pKa values of the respective functional groups are provided in the following table.

Functional group pKa value

-COOH 1.82

-NH3+ 9.17

CH2

HN

NH+

6.00

(i)

Make use of the pKa values to suggest the major species present in solutions of histidine with the following pH values.

pH 1

pH 4

pH 8

pH12

[4]

9


pH 1

+H3N C C

CH2

OH

O

HN

NH+

H

pH 4

+H3N C C

CH2

O-

O

HN

NH+

H

pH 8

+H3N C C

CH2

O-

O

HN

N

H

pH 12

H2N C C

CH2

O-

O

HN

N

H

(ii) Explain, with the aid of relevant chemical equation(s), how histidine acts as a buffer at pH of 1.82.

When small amount of alkali is added,

+H3N C C

CH2

OH

O

HN

NH+

H

+ OH-

+H3N C C

CH2

O

HN

NH+

H

O-

+ H2O

[2]

10


When small amount of acid is added,

+H3N C C

CH2

OH

O

HN

NH+

H

+ H+

+H3N C C

CH2

O

HN

NH+

H

O-

(iii)

+H3N C C

CH2

OH

O

HNA

NBH+

H

As shown in the structure of the fully protonated histidine above, the side chain of the fully protonated histidine has 2 different N atoms, NA and NB atom. NB atom is protonated but not NA atom. The heterocyclic ring of the side chain is known to be as stable as benzene ring.

Explain why NA atom cannot be protonated.

The lone pair of electron on NA atom is not available to accept a proton as the lone pair of electron is involved in the delocalization of 4n +2 = 6 electrons around the heterocyclic ring which results in the stability of the heterocyclic ring/ the heterocyclic ring displays aromaticity.

[1]

(b) Histidine is also important for our digestion as it helps to produce gastric juices in the stomach. However, high level of histidine causes acid indigestion which is also known as heartburn. Heartburn results in burning sensation at the upper stomach as the stomach acid rises into the esophagus. The pH of the stomach of a patient who suffers from heartburn is 4.

A stomach juice sample is extracted from a patient to determine the concentration of histidine using sodium hydroxide, NaOH (aq).

(i) Draw a labeled titration curve of pH against amount of NaOH (aq) added when 1 mole of fully protonated histidine is titrated with NaOH. You should clearly label the following in your titration curve.

Equivalence point(s). Maximum buffer capacities.

[4]

11


(ii) Write a balanced equation of the reaction between histidine in the patient’s stomach juice sample and NaOH.

+ 2OH-+ 2H2O

+H3N C C

CH2

O-

O

HN

NH+

H

H2N C C

CH2

O-

O

HN

N

H

[1]

(c) Histidine is converted to histamine by an enzyme known as histidine decarboxylase. Histamine increases the permeability of the capillaries to white blood cells, to allow them to engage harmful pathogens in infected body tissues. Histidine decarboxylase has a quaternary protein structure. The enzyme-catalysed decarboxylation reaction is shown below.

12


H2N CX CY

CH2

OH

O

HN

N

H

H2N CX

CH2

HN

N

H

H

+ CYO2

Histidine decarboxylase

Histidine Histamine The decarboxylation of histidine occurs in the active site of histidine decarboxylase. In one part of the enzyme’s active site, there is a section of 11 amino acid residues which is critical in the catalysis of decarboxylation of histidine. In addition, the active site also causes some of the bond angles in histidine molecule to deviate from its ideal bond angles. Consequently, the catalysis of decarboxylation of histidine is further increased.

(i) Explain what is meant by the term quaternary structure of protein. Quaternary structure of protein refers to individually folded tertiary polypeptide/protein subunits are grouped to form a large aggregated structure.

[1]

(ii) Histidine decarboxylase undergoes enzymatic partial hydrolysis. Upon hydrolysis, the polypeptide fragments of the critical section of the enzyme’s active site are isolated as shown below.

Ala-Cys-Phe Gly-Gly Lys-Asp-Asp-Gly-Gly Phe-Arg-Lys Ala-Cys-Phe-Phe-Arg-Lys Asp-Asp-Gly

Deduce the sequence of the critical section of 11 amino acid residues of the enzyme’s active site. Show your working clearly. Working Ala-Cys-Phe Phe-Arg-Lys Ala-Cys-Phe-Phe-Arg-Lys Lys-Asp-Asp-Gly-Gly Asp-Asp-Gly Gly-Gly Ala-Cys-Phe-Phe-Arg-Lys-Asp-Asp-Gly-Gly-Gly

[2]

13


(iii) The CX-CY bond in histidine is weakened during the enzyme-catalysed decarboxylation reaction. Explain why the CX-CY bond in histidine is weakened. The CX-CY bond in histidine is weakened in the active site because the bond angles involving the CX-CY bond deviates from the ideal bond angles.

[1]

(iv) The decarboxylation reaction of histidine can be disrupted due to fever and lead poisoning. Explain how the following conditions disrupt the decarboxylation reaction of histidine.

Fever Lead Poisoning

The enzyme is denatured and lost its catalytic property. Increasing in body temperature disrupts hydrogen bonding/VDW interaction/Ionic interaction between R groups. Pb2+ is a heavy metal ion. It can binds to negatively charged R groups and disrupt the ionic interaction between charged groups. The disulphide bonds are also disrupted by Pb2+ due to its high affinity for sulfur.

[4]

[Total: 20 marks]

4 (a) Statistics show a correlation between road accidents and drivers who drink too much. One of the first chemical breathalysers was based on the oxidation process of ethanol using acidified potassium dichromate(VI).

(i) Derive the half equation for the oxidation of ethanol to ethanoic acid.

CH3CH2OH + H2O → CH3CO2H + 4 H+ + 4 e–

(ii) Hence, using relevant information from the Data Booklet, write a balanced ionic equation for the oxidation of ethanol using acidified potassium dichromate(VI)

2Cr2O72- + 16H+ + 3 CH3CH2OH → 4Cr3+ + 3CH3CO2H + 11H2O

The breathalyser kit consists of an inflated 1000 cm3 plastic bag connected to a glass tube containing the dichromate crystals. When the breath is expelled through the tube the crystals change colour as they are reduced. The proportion of the crystals that change colour indicates the amount of alcohol present. The current legal maximum blood alcohol concentration when driving is 80 mg per 100 cm3 of blood. Alcohol concentration in the blood can be estimated by analysing the alcohol in the breath because an equilibrium is set up between the alcohol in the blood and the alcohol in the breath: CH3CH2OH(blood) CH3CH2OH(breath) At body temperature, the concentration of alcohol in the blood is about 2300 times that in the breath.

(iii) What is the corresponding breath alcohol concentration in mg per 1000 cm3 of

14


breath? Blood alcohol = 800mg/1000cm3

Breath alcohol = 800/2300mg/1000cm3 = 0.348mg/1000cm3

(iv) What colour change would you expect for a positive result? Orange to green

[4]

(b) Modern breathalysers use fuel-cell technology. The fuel cell has two platinum electrodes with a porous acid-electrolyte material sandwiched between them. As the exhaled air from the suspect flows past one side of the fuel cell, the platinum anode oxidizes any alcohol in the air to produce ethanoic acid. The electrons produced flow through a wire from the platinum anode to the cathode. The protons move through the lower portion of the fuel cell to the cathode and combine with oxygen to form water. It is the size of this electric current which determines the amount of alcohol in the blood stream.

(i) Construct an equation for the cathode reaction. 4H+ + O2 + 4e- → 2H2O

(ii) Hence write a balanced equation for the overall reaction. CH3CH2OH + O2 → CH3CO2H + H2O

(iii) ∆GѲ and EcellѲ are related by the following equation, ∆GѲ = -nFEcell

Ѳ where ∆GѲ is the standard free energy change in J mol-1, n is the number of electrons transferred during the redox reaction and F is the Faraday constant. ∆GѲ, for the oxidation of ethanol to ethanoic acid at 298 K, is -374 kJ mol-1. Use the equation you have written in (b)( ii) to decide on a suitable value for n and hence calculate the Ecell

Ѳ for the fuel cell breathalyser. -374 x 103 = -4 x 96500 x Ecell

Ѳ

Ecell Ѳ = +0.969V

[4]

15


(c) EDTA (compound B) and the ‘crown thioether’ (compound C) are polydentate ligands that are used to remove harmful metals from the environment.

NCH2CH2N

CH2COO-

CH2COO-

-OOCH2C

-OOCH2C

B

Compound B complexes with Cu2+ and can be used to remove Cu2+ from contaminated water.

Compound C complexes with Ni2+ and can be used to extract Ni2+ from solutions of industrial waste.

(i) Explain what is meant by a polydentate ligand and state the features present in both compounds B and C which make them suitable to act as ligands. Polydentate – can form more than one (dative) bond per molecule of

ligand

Both compounds contain lone pairs of electrons on atoms such as oxygen ,nitrogen and sulphur

(ii) For both complexes formed between compound B and Cu2+ ions and between compound C and Ni2+ ions respectively, suggest the co-ordination numbers and shapes of both complexes.

[4]

Complex between compound A and Cu2+ ions

Coordination number 6 Octahedral

Complex between compound B and Ni2+ ions

Coordination number 4 Tetrahedral (or square planar)

(d) When water ligands in a hydrated metal ion are substituted by other ligands, the equilibrium constant for the reaction is referred to as the stability constant, Kstab, of the new complex. For example,

C

16


[Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]

2+(aq) + 6H2O(l) [Ni(NH3)6]

2+ Kc = Kstab = [Ni(H2O)6]

2+[NH3]6

The table below lists the colours and stability constants (in logarithm form) of some nickel(II) complexes.

Complex lg Kstab Colour

D [Ni(H2O)6]2+(aq)

- green

E [Ni(NH3)6]2+(aq)

7.7 blue

F [Ni(H2NCH2CH2NH2)3]2+(aq)

18.3 purple

G [Ni(EDTA)]2-(aq)

19.0 blue

H [Ni(CN)4]2-(aq)

31.1 Pale yellow

(i) The standard free energy change, ∆GѲ, and the stability constant, Kstab, are also related by the following equation, ∆GѲ = -2.3RT lg Kstab …….(1) where R is the molar gas constant, T is the temperature in Kelvin and Kstab is the stability constant. Using this equation [∆GѲ = ∆HѲ-T∆SѲ ] and the above equation (1), explain why [Ni(H2NCH2CH2NH2)3]

2+ has a higher lg Kstab than that of [Ni(NH3)6]2+ even though

both complexes have six Ni2+-N dative bonds. [ You may represent [Ni(H2NCH2CH2NH2)3]

2+ as [Ni(en)3]2+]

Formation of both complexes involves breaking six Ni2+-O dative bonds and formation of six Ni2+-N dative bonds, hence ∆H for both complexes are similar.

Let en represent H2NCH2CH2NH2

[Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]

2+(aq) + 6H2O(l) 7 particles 7 particles

[Ni(H2O)6]2+(aq) + 3en(aq) [Ni(en)3]

2+(aq) + 6H2O(l) 4 particles 7 particles

To form [Ni(en)3]2+(aq), three en molecules replace 6 water molecules, hence

∆S increases

Large increase in ∆S results in more negative ∆G;

more negative ∆G is reflected in higher lg Kstab

17


(ii) Use the data in the table to predict and explain what you will observe when a solution of pale yellow [Ni(CN)4]

2- is treated with EDTA.

Pale yellow colour remains as [Ni(CN)4]2- is more stable than

[Ni(EDTA)]2-(aq), hence EDTA cannot replace the CN- ligands.

(iii) The following table lists the colours of the photons of light of certain wavelengths. Colour of photon violet blue green yellow red

Wavelength/nm 400 450 500 600 650

Based on the observed colours of the complexes, deduce which complex, D or H, will have a larger energy gap between the two groups of 3d orbitals? Explain your answer. Predict the colour of [Ni(H2O)4(H2NCH2CH2NH2)]

2+(aq). Energy of light is inversely proportional to wavelength of light In complex H, the energy gap is larger as it absorbs higher energy violet light(complement of yellow). In complex D, it absorbs lower energy red light(complement of green)

The colour of [Ni(H2O)4(H2NCH2CH2NH2)]2+(aq) is blue.

[8]

[Total: 20 marks]

5 (a) The molecules of ethanoic acid and methyl methanoate contain the same number

of electrons. These molecules have the same size and mass but their boiling points differ widely.

Compound Boiling point /oC CH3COOH 112 HCOOCH3 32

(i) Suggest an explanation for the difference in boiling point.

CH3COOH exhibits hydrogen bonding while HCOOCH3 exhibits pd-pd interactions between molecules

Hydrogen bonding is stronger than pd-pd interactions hence ethanoic acid has the higher boiling point.

[2]

(ii) Draw a diagram showing why ethanoic acid can dimerise in methyl methanoate.

[1]

18


OC

O

HCH3

OC

O

HCH3

(b) The classic synthesis of an ester is done via the Fischer Esterification, which involves treating a carboxylic acid with an alcohol in the presence of a catalyst. The reaction between butanoic acid and ethanol produces ethyl butanoate, an ester which smells like pineapples.

CH3CH2CH2COOH + CH3CH2OH CH3CH2CH2COOCH2CH3 + OH2

H+

An alternative way of producing ethyl butanoate is to mix ethanol with the acid chloride, butanoyl chloride.

(i) State the name of the reaction between butanoyl chloride and ethanol.

Nucleophilic substitution

[1]

(ii) Using relevant data from the Data Booklet, explain why the acid chlorides are more reactive than the carboxylic acid in the reaction to form esters.

Bond energy of C-O bond – 360 kJ mol-1

C-Cl bond – 340 kJ mol-1

C-Cl bond is weaker than the C-O bond hence it is easier to break resulting it to be more reactive.

[2]

(iii) Suggest and explain one safety precaution you should employ when using acid chlorides in the laboratory.

Acid chloride hydrolyses rapidly with water to produce HCl which is toxic and corrosive, hence ensure that test-tube is dry when collecting a sample of acid chloride.

[2]

(c) In 1899 the German chemists Adolf von Baeyer and Victor Viliger discovered that by treating ketones with peroxyacids, such as peroxybenzoic acid (C6H5COOOH), gives an ester by “insertion” of one atom of oxygen. The Baeyer-Viliger reaction is thought to proceed through the following stages.

19


C

O

R1 R2

C

O

OOH

CO

OO

C

O

R1 R2

H

C

O

R1 OR2I II

For asymmetrical ketones the oxygen atom is inserted in the order: tertiary alkyl > secondary alkyl > primary alkyl > CH3.

(i) Peroxybenzoic acid can be formed by reacting benzoic acid and hydrogen peroxide. Write the balanced equation of this reaction.

C

O

OOHC

O

OH + H2O2 + H2O

[1]

(ii) State the role of peroxybenzoic acid in the Baeyer-Viliger reaction.

Oxidising agent

[1]

(iii) What type of reaction takes place in stage I?

(nucleophilic) Addition reaction

[1]

(iv) Suggest the other product of stage II. Benzoic acid

[1]

(v) Molecules of compound J, C6H12O are chiral. The bond angles about all the carbon atoms are approximately 109.5o. When H2O is eliminated from J, a mixture of three isomeric alkenes, C6H10, is produced. Only two of which are optical isomers of each other. Heating compound J with acidified potassium permanganate produces K, C6H10O. Mixing compound K with peroxybenzoic acid produces one compound L as the product. Suggest the structures of J, K, L and the three alkenes.

[5]

20


CH3

OH

CH3CH3

H

CH3

H

CH3

O

O

CH3

O

A B C

(d) Using your answers in (c)(v), suggest a simple chemical test by which one of the alkenes and compound J could be distinguished from each other. Write any appropriate equation. R/C: Na metal Observation: Compound J shows effervescence produces a colourless and odourless gas that produces a pop sound with a burning splint.

CH3

OH

+ Na CH3

O-

+ H2

Na+

2 2 2

[3]

[Total: 20 marks]

J K L

acjc

Documents

engage harmful

coat nervous

suspect flows

producing

maintain healthy

secondary

tertiary alkyl

r2 r1 ooh