year 6 preliminary examination (ii) -...

River Valley High School 9647/01/PRELIM/14 2014 Preliminary Examination II

RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION (II)

CANDIDATE NAME

CLASS 6

CENTRE NUMBER

S INDEX NUMBER

H2 CHEMISTRY 9647/01

Paper 1 Multiple Choice 26 September 2014

1 hour

Additional Materials: Multiple Choice Answer Sheet

Data Booklet

READ THESE INSTRUCTIONS FIRST

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

Write your name, class, centre number and index number on the Answer Sheet in the spaces provided.

There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D.

Choose the one you consider correct and record 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 19 printed pages and 1 blank page.

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 Wines often contain a small amount of sulfur dioxide that is added as a preservative. The sulfur dioxide content of a wine is found by the following method:

A 50 cm3 sample of white wine is reacted with 40.0 cm3 of 0.01 mol dm3 of excess aqueous iodine. The sulfur dioxide in the wine is oxidised to sulfate, SO4

2−, in the process. The unreacted iodine requires exactly 23.60 cm3 of 0.02 mol dm−3 sodium thiosulfate, Na2S2O3, for complete reaction.

Determine the concentration of sulfur dioxide, in mol dm−3, in the wine.

A 1.64 104

B 3.28 103

C 4.72 103

D 9.44 103

2 An ion X2+ contains 24 protons.

What is the electronic configuration of X3+?

A 1s2 2s2 2p6 3s2 3p6 3d3

B 1s2 2s2 2p6 3s2 3p6 3d4

C 1s2 2s2 2p6 3s2 3p6 3d1 4s2

D 1s2 2s2 2p6 3s2 3p6 3d5 4s1

3


3 Hydroxylamine (NH2OH) can react with propanone via a nucleophilic addition in the following manner:

What are the values of the bond angles y and z?

y z

A 109.5 118

B 90 118

C 109.5 180

D 90 180

4 Which graph correctly describes the behaviour of fixed masses of the ideal gases L and M where the number of moles of L is greater than number of moles of M?

A

B

C

D

y

z

4


6 A typical protein forms hundreds of hydrogen bonds and thousands of van der Waals’ forces in folding from primary to tertiary structures.

Which of the following thermodynamic state functions of the protein best represents the folding process?

G / kJ mol–1

H / kJ mol–1

S / J K–1

mol–1

A

B + +

C +

D + +

7 When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 723 K until equilibrium is established, the equilibrium mixture is found to contain 0.26 mol of hydrogen iodide. The equation for the reaction is as follows:

H2(g) + I2(g) ⇌ 2HI(g)

What is the correct numerical value for the equilibrium constant, Kc?

A 2.25 B 48.3 C 51.8 D 185.7

5 Tetrachloroethene is commonly used as degreasing solvent. The enthalpy change

for the following reaction is 878.5 kJ mol1.

C2H4(g) + 4HCl(g) + 2O2(g) C2Cl4(l) + 4H2O(l)

Hf (C2H4(g)) = +52.3 kJ mol1

Hf (HCl(g)) = 92.3 kJ mol1

Hf (H2O(l)) = 285.8 kJ mol1

Which of the following is the enthalpy change of formation of tetrachloroethene given the information above?

A 52.2 kJ mol1

B 582 kJ mol1

C 608 kJ mol1

D 633 kJ mol1

5


8 A system at equilibrium is subjected to the following changes separately:

(i) The pressure is reduced at constant temperature

(ii) The temperature is increased at constant pressure

For which equilibrium will both of these changes result in an increase in the proportion of products?

A H2(g) + I2(g) ⇌ 2HI(g) H = +53 kJ mol1

B 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) H = 950 kJ mol1

C N2(g) + 3H2(g) ⇌ 2NH3(g) H = 92 kJ mol1

D N2O4(g) ⇌ 2NO2(g) H = +57 kJ mol1

9 Solutions P, R and S contain a strong acid, a weak acid and a strong base, but not necessarily in the same order. The concentration and pH of each solution are shown below.

Solution Concentration / mol dm3 pH

P 1.00 2.4

R 0.01 12

S 0.001 3

Which one of the following statements is correct?

A P contains a strong acid while S is contains a weak acid.

B The pH at equivalence point when R is titrated against S is less than 7.

C Mixing 500 cm3 of P and 500 cm3 of R produces a buffer solution.

D Mixing 100 cm3 of R and 100 cm3 of S produces a buffer solution.

6


11 When a dilute sulfate solution of a metal J is electrolysed, the metal J and a diatomic gas K are produced at the cathode and the anode respectively in the molar ratio 2 : 1. In another experiment, the same quantity of electricity is used to electrolyse a saturated sodium chloride solution and a gas L is evolved at the anode.

What is the molar ratio of J : K : L?

A 2 : 1 : 1 B 2 : 1 : 2 C 4 : 2 : 1 D 4 : 2 : 3

10 The mechanism involved in the formation of XY2 is shown below:

Step I : X + Y XY (fast)

Step II : XY + Y XY2 (slow)

What is the rate equation for this reaction?

A Rate = k[XY][Y]

B Rate = k[X][Y]

C Rate = k[X][Y]2

D Rate = k[XY][Y]2

12 Use of the Data Booklet is relevant to this question.

Iodine clock reaction can be used to study how the concentration of iodide ions affect the rate of oxidation of iodide ions with peroxodisulfate ions.

Peroxodisulfate ions convert iodide ions into iodine slowly.

S2O82−(aq) + 2I−(aq) 2SO4

2−(aq) + I2(aq)

The rate of the reaction can be increased by the addition of catalysts such as aqueous iron(III) ions.

A possible catalysed reaction pathway involves the following steps:

Step 1: 2Fe3+(aq) + 2I−(aq) I2(aq) + 2Fe2+(aq)

Step 2: S2O82−(aq) + 2Fe2+(aq) 2SO4

2−(aq) + 2Fe3+(aq)

Which of the following statements is incorrect?

A Fe2+ is a stronger reducing agent than I.

B S2O82 is a stronger oxidising agent than Fe3+.

C The Ecell⦵for step 2 is more positive than step 1.

D Aqueous cobalt(II) ions can be used as a catalyst in this reaction.

7


13 Which of the following oxides is unlikely to dissolve in concentrated sodium hydroxide?

A Al2O3 B MgO C P4O10 D SiO2

14 The ionic radii of P3, S2 and Cl are 0.212 nm, 0.184 nm and 0.181 nm

respectively.

Which of the following statements correctly explains the decrease in radius from P3

to Cl?

A an increase in the nuclear charge and a constant total number of electrons

B an increase in the nuclear charge and total number of electrons

C an increase in total number of electrons while nuclear charge remains constant

D a decrease in nuclear charge while total number of electrons remains constant

15 Which one of the following statements about Group II elements is correct?

A The reactivity with cold water decreases down the group.

B The charge density of the cations increases down the group.

C The magnitude of the lattice energy of their oxides increases down the group.

D The thermal stability of their carbonates increases down the group.

16 Which one of the following statements is most likely to be true for astatine, the element below iodine in Group VII of the periodic table?

A Silver astatide and dilute aqueous ammonia will react to form a soluble complex.

B Sodium astatide react with hot concentrated sulfuric acid to form hydrogen astatide.

C When sparked, hydrogen reacts explosively and completely with astatine.

D Aqueous potassium astatide and chlorine will produce black solids of astatine.

8


18 Gibberelic acid is a plant hormone that controls various aspects of plant growth.

gibberelic acid

What is the number of chiral centres in the product after gibberelic acid reacts with HCl?

A 7 B 8 C 9 D 11

19 The organic molecule cumene is derived from cumin, a spice well-known for its distinctive flavour and aroma and an essential flavouring in South Asian dishes.

Cumene

When cumene undergoes free radical reaction with bromine, x mol of a non-chiral monosubstituted product is obtained.

What is the amount of chiral monosubstituted product formed?

A 6

x B x C 3x D 6x

17 Which of the following statements concerning transition metals is correct?

A They are the only metals which have more than one oxidation state.

B They are the only metals which form complex ions.

C They are the only metals with high melting points.

D They are the only metals which give coloured ions in an aqueous solution.

9


20 Which of the following isomers of C5H11OH gives, on dehydration, the greatest

number of different alkenes (including stereoisomers)?

A

B

C

D

21 Catechin is a type of natural phenol and antioxidant found in green tea, chocolate and peaches.

Catechin

The functional group is inert.

Which statement about catechin is correct?

A 1 mol of catechin reacts with 5 mol of NaOH.

B 1 mol of catechin reacts with Na to form 5 mol of H2.

C It can undergo electrophilic substitution with FeCl3(aq).

D It has 2 chiral centres.

10


22 Cyclohexane undergoes substitution with chlorine gas.

Which of the following is a possible by-product during the termination stage?

A

B

C

D

23 Naphthalene is an organic aromatic compound with the formula C10H8. It undergoes similar reactions as benzene and is best known as the main ingredient in traditional mothballs.

Naphthalene

Which of the following statements about naphthalene is correct?

A Naphthalene is a planar molecule.

B Naphthalene has 12 electrons.

C Naphthalene undergoes electrophilic substitution with aqueous bromine.

D

Naphthalene undergoes free radical substitution to form two isomeric monobromo products.

11


24 Saccharin is a commonly used artificial sweetener. The original synthetic route proposed by Remsen and Fahlberg started with methylbenzene.

Which of the following best describes reactions I, II and III?

Reaction I Reaction II Reaction III

A Electrophilic Substitution Condensation Oxidation

B Electrophilic Substitution Elimination Condensation

C Electrophilic Addition Condensation Condensation

D Electrophilic Addition Elimination Oxidation

25 In which of the following reactions is the inorganic reagent acting as a nucleophile?

A C6H6 + H2SO4 C6H5SO3H + H2O

B CH3CH=CH2 + HBr CH3CHBrCH3

C CH3CH2NH2 + HCl CH3CH2NH3Cl

D CH3CH2Br + H2O CH3CH2OH + HBr

ClSO3H

NH3

Saccharin

I

II

III

12


26 The reaction conditions of four different transformations are given below.

Which transformation has a set of conditions that is correct?

A

B

C

D

27 Which of the following reagents and conditions can be used to distinguish between the two compounds X and Y below?

X Y

A Fehling’s reagent, warm

B Aqueous bromine, r.t.p.

C P4 and Cl2, heat

D Acidified K2Cr2O7, heat

13


28 After the reduction of nitrobenzene to phenylamine, using tin and concentrated hydrochloric acid, an excess of sodium hydroxide is added.

What is the purpose of the sodium hydroxide?

A To lower the boiling point for subsequent distillation.

B To remove water and dry the product as it is hygroscopic.

C To liberate phenylamine.

D To precipitate tin(II) hydroxide.

29 The planet Jupiter is known to have a reducing atmosphere consisting mainly of methane and ammonia, with some ethene and water in the form of ice crystals. Astronomers used the impact of fragments of Comet Shoemaker-Levy-9 with Jupiter to identify more complex molecules present in the planet’s atmosphere.

Which molecule is unlikely to have been observed?

A CH3CH2NH2

B CH3CH2OH

C CH3CH2CH2CH3

D CH3CO2H

30 GABA is a neuro-transmitter released by red algae which will encourage shellfish larvae to settle on the ocean bed.

H2NCH2CH2CH2CO2H

GABA

In what way does GABA differ from amino acids derived from the hydrolysis of proteins?

A It does not form zwitterions.

B It is not soluble in water.

C It is not a 2-aminocarboxylic acid.

D It does not form a polyamide linkage.

14


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.

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

A B C D

1, 2 and 3 are correct

1 and 2 only are correct


1 only is

correct

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

31 Which of the following particles have a single unpaired electron?

1 The copper ion in CuI

2 The methyl free radical

3 A molecule of NO

32 After an oil spillage at sea, liquid hydrocarbon layer floats on the surface of the sea water. Which of the following statements help to explain why liquid hydrocarbons both float on, and are less dense than water?

1 Hydrocarbons molecules are not solvated by water.

2 There are only van der Waals’ interactions between hydrocarbon molecules.

3

Hydrogen bonds between the molecules in liquid water cause them to pack closer together.


An electrochemical cell contains a Mn2+/Mn half−cell and a Q3+/Q half−cell. The following reactions take place at the two electrodes.

Mn → Mn2+ + 2e−

Q3+ + 3e− → Q

Which of the following statements are correct?

1 The Mn electrode is the negative electrode.

2 Q can be chromium metal.

3

The number of moles of Q deposited is 1.5 times the number of moles of Mn2+ formed.

15



A B C D




1 only is

correct


34 The table gives data for the reaction between X and Y at constant temperature.

X + Y → Z

Experiment [X] / mol dm−3 [Y] / mol dm−3 Initial rate / mol dm−3 s−1

1 0.4 0.1 2.5 10−4

2 0.2 0.2 5.0 10−4

3 0.2 0.4 1.0 10−3

Which statements follow from these results?

1 The half-life of this reaction is 277 s.

2 The mechanism involves more than 1 step.

3 The rate equation can be written as: rate = k[X][Y].

16



A B C D




1 only is

correct


35 A gas P can react to give another gas R according to this equation:

2P(g) ⇌ R(g)

The graph below represents the decomposition of 1.0 mol of P in a container of a fixed volume at various temperatures.

Which of the following statements about the above system are correct?

1

The equilibrium constant, Kp, for the decomposition reaction increases with decreasing temperature.

2 At T1 K, the mole fraction of P is greater than that of R at equilibrium.

3 The decomposition is spontaneous at all three temperatures.

Amount of P present / mol

Time / min

T1 K

T2 K

T3 K

17



A B C D




1 only is

correct



Which of the following statements are true about vanadium and its compounds?

1 Lead is able to reduce vanadium from the +5 to the +3 oxidation state.

2 The maximum oxidation state of vanadium is found in the oxoanion VO2+.

3 The density of vanadium is smaller than that of calcium.

37 Which of the following reactions will lead to a racemic mixture of products?

1 CH3CHO with HCN and a small amount of NaOH

2 CH3CH2CH2 C

CH3

Cl

CH3CH2 with aqueous sodium hydroxide, heat under reflux

3 CH3CH2CH2 C

CH3

C

H

CH3 with HBr dissolved in CCl4

18



A B C D




1 only is

correct


38 Enalapril and carvedilol are two drugs that are commonly used in the treatment of heart diseases.

Enalapril

Carvedilol

Which of the following can be used to distinguish between the two compounds?

1 sodium carbonate

2 aqueous chlorine

3 hot acidified potassium manganate(VII)

19



A B C D




1 only is

correct


39 Surgicryl is a synthetic polymer that can be used to stitch wounds together during a surgery. It degrades into smaller molecules over two to three months. Hence, patients do not have to return to the hospital to remove the stitches.

Which of the following statements explain Surgicryl’s suitability for this function?

1 Surgicryl can be hydrolysed in the body.

2 The small molecules produced are soluble in water.

3 Surgicryl is degraded slowly due to its large size.

40 Glutamic acid and ornithine are both amino acids that can be found in the human body.

Glutamic acid Ornithine

Which of the following statements about the two amino acids are correct?

1

Both amino acids are likely to be found on the outer surface of a globular protein.

2

During an electrophoresis, glutamic acid will be nearer to the cathode than ornithine.

3 The isoelectric point of glutamic acid is higher than that of ornithine.

surgicryl

20


[BLANK PAGE]

Answers for RVHS Preliminary Examination II Paper 1

1 B 11 B 21 D 31 C

2 A 12 A 22 B 32 A

3 A 13 B 23 A 33 B

4 B 14 A 24 A 34 B

5 A 15 D 25 D 35 A

6 A 16 D 26 D 36 D

7 B 17 D 27 B 37 A

8 D 18 C 28 C 38 D

9 C 19 D 29 D 39 A

10 C 20 B 30 C 40 D



CANDIDATE NAME

CLASS 6

CENTRE NUMBER

S INDEX NUMBER


Paper 2 Structured Questions 15 September 2014

2 hours

Candidates answer on the Question Paper.

Additional Materials: Data Booklet


Write your name, class, centre number and index number on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.

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


Answer all questions in the space provided. A Data Booklet is provided. Do NOT write anything on it.

At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

For Examiner’s Use

Paper 2

Question Number

1 2 3 4 5 Total

Marks 12 16 15 15 14 72

Paper 1 40

Paper 3 80

Total 192

This document consists of 20 printed pages.

2


1 Planning (P)

On a field trip to Bukit Timah Hill in Singapore, a new metal carbonate called rvium carbonate, RvCO3, was discovered by Susie. It was found that the dissolving of rvium carbonate in water is endothermic, and that the salt is only slightly soluble. The solubility of rvium carbonate is approximately 0.25 mol dm−3 at the laboratory temperature.

You are to plan an experiment to determine the accurate values of both the solubility and enthalpy change of solution of rvium carbonate at the laboratory temperature.

This can be done by preparing a saturated solution of rvium carbonate and measuring the temperature change. The concentration of carbonate ions in the saturated solution can then be determined through volumetric analysis.

You may assume that you are provided with:

Solid rvium carbonate, RvCO3;

Aqueous hydrochloric acid, HCl;

Acid−base indicators;

The apparatus and chemicals normally found in a school or college laboratory.

(a) Write the balanced equation for the reaction between rvium carbonate and hydrochloric acid.

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

.................................................................................................................... [1]

(b) Draw a well-labelled diagram to show the experimental set-up you would use to determine the temperature change during the preparation of the saturated rvium carbonate solution.

[2]

3


(c) Using the information given above, you are required to write a plan to determine the solubility and enthalpy change of solution of rvium carbonate.

You plan should include details of:

the preparation of a saturated rvium carbonate solution

the measurement of the temperature change during the dissolution process

the intended concentration of the standard hydrochloric acid solution to be used

the titration of the saturated rvium carbonate solution with the standard hydrochloric acid solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4


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

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

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

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

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

.................................................................................................................... [6]

(d) Assume that the temperature drop during the preparation of V cm3 of saturated rvium carbonate solution was ΔT °C. When a 25.0 cm3 sample of the saturated rvium carbonate solution was titrated against a standard hydrochloric acid solution of C mol dm−3, the average titre was W cm3.

Outline how you would use these results to determine the

(i) solubility, in mol dm−3, of rvium carbonate;

(ii) enthalpy change of solution, ΔHsoln, for rvium carbonate.

[3]

[Total: 12]

5


2 The halogens are a group of elements found in Group VII of the periodic table.

(a) Bromine was discovered in the 1820s by Antoine-Jérôme Balard and is commonly used in halogen lamps. The halogen lamp has a tungsten filament similar to standard incandescent lamps. However, in addition, the halogen lamp contains a small amount of bromine gas in the bulb that slows down the thinning of the tungsten filament, increasing the lamp’s lifetime.

When the halogen lamp is switched on, all the bromine molecules dissociate at the high temperatures and exist as single bromine atoms.

Br2(g) → 2Br(g)

The hot tungsten filament also evaporates and condenses on the relatively cool walls of the bulb, where it reacts with the bromine atoms to give gaseous tungsten bromide, WBr4.

W(s) + 4Br(g) ⇌ WBr4(g)

(i) A halogen lamp was constructed using a 50 cm3 bulb containing 0.107 g of Br2(g). When the halogen lamp was switched on, the temperature at the walls of the bulb reached 900 K and the equilibrium gas density was measured to be 0.003 g cm−3.

Calculate the number of moles of Br(g) produced initially when the halogen lamp was switched on.

(ii) Calculate the number of moles of Br(g) and WBr4(g) that were present at equilibrium at 900 K.

6


(iii) Calculate the equilibrium partial pressures of Br(g) and WBr4(g).

Hence, calculate the value for Kp for the reaction between tungsten and bromine atoms at 900 K, giving the units in terms of Pa−3.

(iv) Near the tungsten filament, where the temperature is about 2800 K, the Kp for the reaction between tungsten and bromine atoms has a smaller value of 1 x 10−16 Pa−3.

Explain whether the forward or backward reaction is favoured near the tungsten filament.

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

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

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

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

(v) In standard incandescent lamps, the hot tungsten filament that evaporates will collect on the walls of the bulb. As more and more tungsten evaporates, the filament starts to thin, reducing the lifespan of the incandescent lamp considerably.

Using your answer to 2(a)(iv), suggest how the addition of bromine gas into the bulb helps to slow down the thinning of the tungsten filament in halogen lamps.

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

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

........................................................................................................... [9]

7


(b) Chlorine gas was first produced by Carl Wilhelm Scheele in 1774. It can be synthesised into phosphorus pentachloride, PCl5, a colourless and moisture−sensitive solid.

(i) Write an equation for the reaction of phosphorus pentachloride with excess water.

Suggest the pH of the resultant solution.

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

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

(ii) While the reaction between phosphorus pentachloride and water takes place vigorously at room temperature, nitrogen trichloride, NCl3, is stable in the presence of water at room temperature.

Suggest why these two chlorides differ in their reactions with water.

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

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

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

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

8


(iii) Phosphorus pentachloride can react with phosphorus pentaoxide to produce phosphoryl chloride.

6PCl5(g) + P4O10(s) → 10POCl3(g)

Given the following data, construct an energy cycle and determine the enthalpy change of the reaction above.

P4(s) + 6Cl2(g) → 4PCl3(g) ΔHr = −1226 kJ mol−1

P4(s) + 5O2(g) → P4O10(s) ΔHr = −2967 kJ mol−1

PCl3(g) + Cl2(g) → PCl5(g) ΔHr = −84.2 kJ mol−1

PCl3(g) + ½O2(g) → POCl3(g) ΔHr = −286 kJ mol−1

[7]

[Total: 16]

9


3 Copper is a rare element that makes up only 6.8 10–3 percent of the Earth’s crust by mass. It is used in alloys, electrical cables, plumbing pipes and coins.

Some properties of copper and its compounds are given below.

Cu CuO Cu(OH)2 CuCO3

Melting point / C 1085 1326

80 (decomposes)

200

Electrical conductivity in solid state

conductor non-

conductor non-conductor

non-conductor

Solubility at 25 C / mol dm–3 - negligible 1.77 10–7 1.18 10–5

(a) Explain, in terms of bonding and structure, the following observations.

(i) Copper has a higher melting point than calcium (melting point:

843 C).

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

(ii) Copper conducts electricity in solid state while CuO does not.

………………………………………………………………………………

………………………………………………………………………………

………………………………………………………………………………

……………………………………………………………………………… [4]

10


(b) Copper(II) hydroxide, Cu(OH)2, is a sparingly soluble salt.

(i) Write an expression for the solubility product, Ksp, of copper(II)

hydroxide, and calculate its value at 25 C.

(ii) State what is meant by the common ion effect.

Illustrate your answer by determining the solubility of copper(II)

hydroxide in 0.050 mol dm–3 of NaOH at 25 C.

………………………………………………………………………………

………………………………………………………………………………

(iii) Describe the observations you would expect when excess aqueous ammonia is added to a solid sample of Cu(OH)2.

Write an equation for any reaction that occurs.

………………………………………………………………………………

……………………………………………………………………………… [6]

11


(c) Copper(II) carbonate, CuCO3, undergoes the follow reactions.

The three gases evolved (Gas G) have the same identity.

(i) Suggest the identities of X, Y and Z, and the colours of X and Y.

Ppt X

Identity : ……………………………………………………….................

Colour : …………………………………………………………………...

Solution Y

Identity : ……………………………………………………………….....

Colour : …………………………………………………………………...

Solid Z

Identity : ……………………………………………………………….....

(ii) Suggest the role of carbon in formation of solid Z.

……………………………………………………………………………… [4]

CuCO3

H+(aq)

pale blue solution (+ Gas G)

KI(aq)

ppt X in solution Y

heat stronglyblack solid (+ Gas G)

carbon

pink solid Z (+ Gas G)

12


(d) The Statue of Liberty is a landmark monument in New York and was a gift from France to the United States to mark her 100 years of independence. It is made of a copper alloy and has a distinct dull green coating, which consists of an equimolar mixture of Cu(OH)2 and CuCO3.

Copper reacts with moist air to form the dull green coating.

Given that three gases are involved in the reaction with copper, write an equation for the formation of the dull green coating.

……………………………………………………………………………………. [1]

[Total: 15]

13


4 Group II and Group VII elements are not found naturally in their elemental state. Many of these elements were first isolated by means of electrolysis.

(a) A major method of extracting magnesium is by electrolysis of molten magnesium chloride.

(i) Write the ion-electron half equations, including state symbols, for the reactions that would occur at the anode and cathode during the electrolysis of magnesium chloride.

Anode : ………………………………………………………………….

Cathode: ………………………………………………………………….

(ii) Calculate the current required to produced 20 tonnes of magnesium per day.

(1 tonne = 1000 kg)

[4]

(b) Seawater contains a significant quantity of magnesium ion. Seawater is treated with alkali to precipitate the magnesium in the form of magnesium hydroxide which is then heated to produce magnesium oxide. Magnesium oxide is then reacted with chlorine gas to produce solid magnesium chloride. Magnesium is isolated by electrolysing molten magnesium chloride.

The following table shows the three most common ions in seawater.

Species % composition in seawater

chloride ion 1.94

sodium ion 1.08

magnesium ion 0.129

14


The following table shows melting points and boiling points for two magnesium compounds used industrially.

Compound Melting point / C Boiling point / C

magnesium oxide 2852 3600

magnesium chloride 714 1412

Using the information provided and quoting any relevant additional information from the Data Booklet, suggest an explanation for the following.

(i) A solid magnesium compound is isolated for electrolysis rather than directly electrolysing seawater to obtain magnesium.

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

(ii) The magnesium oxide is converted to magnesium chloride before electrolysis.

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

…………………………………………………………………………….. [4]

15


(c) A fireworks specialist was tasked by the National Day Parade organiser to design fireworks that showcased the colours of the Singapore flag. Suggest which Group II elements he should use in the fireworks and their corresponding colours.

…………………………………………………………………………………….

……………………………………………………………………………………. [2]

(d) Three separate experiments in which aqueous sodium chloride was electrolysed were performed. The experimental conditions are shown below.

Experiment Conditions Observations at anode

1

dilute aqueous sodium chloride

30 C

Solution is unstirred.

Colourless gas evolved.

2

concentrated aqueous sodium chloride

30 C

Solution is unstirred.

Greenish yellow gas evolved.

3

concentrated aqueous sodium chloride

70 C

Solution is stirred continuously.

No gas collected.

16


(i) Using relevant information from the Data Booklet, identify the gases evolved in experiments 1 and 2 and explain the difference in the observations between experiments 1 and 2.

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

……………………………………………………………………………..

(ii) Write an equation to account for the observation made in experiment 3.

…………………………………………………………………………….. [5]

[Total: 15]

17


5 Ibuprofen is an over-the-counter remedy for mild to moderate pain and is sold under the brand name Nurofen. In the United Kingdom, ibuprofen was originally synthesised and first patented in 1961 by the research team in Boots Company. The starting material for the synthesis of ibuprofen is 2-methylpropylbenzene, which has a similar carbon skeleton to that of ibuprofen. The reaction scheme below shows Boots’ synthesis of ibuprofen:

(a) Step I proceeds via an electrophilic substitution mechanism where the aluminium chloride acts as a Lewis acid catalyst. The aluminium atom is electron deficient, allowing it to accept a lone pair of electrons from the

oxygen on the molecule. The CO bond breaks, generating the electrophile and another singly-charged anion as the only products.

(i) Draw the structure of the electrophile.

(ii) From the information given and your answer in (i), suggest a balanced equation to illustrate the step involving the generation of the electrophile.

step I step II

step III

step IV step V

step VI

2-methylpropylbenzene

ibuprofen

A B

C D E

18


(iii) Describe the mechanism for the formation of intermediate A from 2-methylpropylbenzene, including curly arrows showing the movement of electrons, and all charges.

[4]

(b) The three-membered ring in intermediate B is known as an epoxide functional group.

State the hybridisation of the carbon atom in the epoxide ring and suggest why epoxides are highly reactive.

…………………………………………………………………………………….

…………………………………………………………………………………….

……………………………………………………………………………………. [2]

19


(c) The conversion of intermediate C to D in step IV involves the following sub-steps:

1. nucleophilic addition of NH2OH to the carbonyl group in C to form a zwitterion

2. intramolecular transfer of proton, forming an OH bond and breaking

an NH bond in the zwitterion to give intermediate X

Draw the zwitterion formed from the nucleophilic addition step in the box. Complete the reaction scheme below for sub-steps 1 and 2, using curly arrows to show movement of electrons and label all charges.

where R =

[3]

(d) A dose of Nurofen contains more than just ibuprofen, the active ingredient. The science of blending other ingredients into the dose is known as formulation and Nurofen is available in a number of formulations. They include tablets to be taken orally or gels to be applied onto the skin. The oral variant is often formulated as the free acid or a lysine salt while the gel variant is often formulated with propan-2-ol.

(i) The ibuprofen in the lysine salt variant is found to be absorbed by the body nearly twice as quickly as the free acid variant.

Suggest why this is so.

……………………………………………………………………………...

C X D

20


(ii) The gel variant is rubbed into painful parts of the body so that the active ingredient is applied directly to the site of pain. The inner layer of the skin consist mainly of polar molecules.

With reference to the polarities of ibuprofen and propan-2-ol, explain why ibuprofen is formulated with propan-2-ol instead of being applied directly on the skin as the free acid.

……………………………………………………………………………...

……………………………………………………………………………...

……………………………………………………………………………... [3]

(e) Describe a simple chemical test that can be used to distinguish between 2-methylpropylbenzene and ibuprofen.

State the observation for each compound.

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

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

……………………………………………………………………………............ [2]

[Total: 14]

- End of Paper -

Answer key to Paper 2

1 (a) RvCO3 + 2HCl → RvCl2 + CO2 + H2O [1]

(b)

[2]

(c) Preparation of saturated RvCO3 solution

1. Using a 100 cm3 measuring cylinder, place about 100 cm3 of deionised water (accept any volume between 80 – 250 cm3) into a polystyrene cup.

2. Measure and record the initial temperature of the water using a thermometer.

3. Using a spatula, add solid RvCO3 into the polystyrene cup and stir/swirl gently with the thermometer / stirrer. Continue to add more solid and stir, until some solid RvCO3 remains undissolved to ensure a saturated solution is obtained.

4. Record the lowest temperature reached on the thermometer. 5. Filter the saturated solution using a dry filter funnel and dry filter paper

into a dry 250 cm3 conical flask / beaker to ensure the saturated solution is not diluted.

Justification of concentration of standard HCl solution

For a 25.0 cm3 aliquot of saturated RvCO3(aq), a reasonable HCl titre value should be 25.00cm3 (accept any value between 20−30 cm3).

RvCO3 + 2HCl → RvCl2 + CO2 + H2O

RvCO3 ≡ 2HCl

Given that solubility of RvCO3 is approximately 0.25 mol dm−3,

(250 cm3) beaker

polystyrene cup

thermometer

lid

(100 cm3) H2O + RvCO3

2

[HCl] required =

= 0.500 mol dm−3

Titration of saturated RvCO3 solution against HCl

1. Pipette 25.0 cm3 of saturated RvCO3 solution into a 250 cm3 conical flask.

2. Add 2-3 drops of methyl orange indicator.

3. Titrate with the standard 0.500 mol dm−3 HCl from the burette until the colour of the solution changes from yellow to orange.

4. Repeat the titration to obtain two consistent results within 0.10 cm3 in difference. [6]

(d) (i) nHCl used =

= 1.00 x 10−3 WC mol

RvCO3 ≡ 2HCl

nRvCO3 in 25.0 cm3 of saturated solution = 1.00 x 10−3 WC

= 5.00 x 10−4 WC mol

solubility of RvCO3 = 5.00 x 10−4 WC

= 0.0200 WC mol dm−3

(ii) Heat taken in by V cm3 solution = VCΔT

= 4.18 V ΔT J

nRvCO3 dissolved in V cm3 solution = 0.0200 WC

= 2.00 x 10−5 WCV mol

ΔHsoln =

J mol−1 [3]

[Total: 12]

3

2 (a) (i) initial nBr2 =

= 6.70 x 10−4 mol

At 900 K,

Br2(g) → 2Br(g)

initial nBr = 2 x 6.70 x 10−4 = 1.34 x 10−3 mol

(ii) W(s) + 4Br(g) ⇌ WBr4(g)

initial amount / mol − 1.34 x 10−3 0

change in amount / mol − −4y +y

equilibrium amount / mol − 1.34 x 10−3 −4y y

mass of gas at equilibrium

= (1.34 x 10−3 −4y)(79.9) + (y)(184 + 4(79.9))

= 0.107 + 184y

0.107 + 184y = 0.003 x 50 g

y = 2.34 x 10−4 mol

equilibrium nWBr4 = 2.34 x 10−4 mol

equilibrium nBr = 1.34 x 10−3 −4(2.34 x 10−4) = 4.04 x 10−4 mol

(iii) equilibrium pWBr4 =

= 3.50 x 104 Pa

equilibrium pBr =

= 6.04 x 104 Pa

Kp at 900 K =

= 2.63 x 10−15 Pa−3

(iv) When the temperature increases from 900 K to 2800 K, the Kp decreases. This implies that the equilibrium position shifts to the left to favour the backward reaction near the tungsten filament.

(v) The addition of bromine gas allows the formation of WBr4(g), which is converted back to tungsten at the filament, slowing down the thinning of the tungsten filament. [9]

(b) (i) PCl5 + 4H2O → H3PO4 + 5HCl

pH of resultant solution = 2

4

(ii) PCl5 is able to undergo hydrolysis in water because P atom is able to use its energetically accessible vacant 3d orbitals for dative bonding with water molecules.

NCl3 does not undergo hydrolysis because N atom does not have energetically accessible vacant orbitals/ N atom is in Period 2 and not Period 3.

(iii)

6PCl5(g) + P4O10(s) 10POCl3(g)

Using Hess’s law,

ΔHr = 6(+84.2) +2967 + (−1226) + 10(−286)

= −614 kJ mol−1 (3sf) [7]

[Total: 16]

10PCl3(g) + 5O2(g)

6(+84.2)

6PCl3(g) + 6Cl2(g) + P4O10(s)

+2967

6PCl3(g) + 6Cl2(g) + P4(s) + 5O2(g)

−1226

10(−286)

ΔHr

5

3 (a) (i) Both copper and calcium have giant metallic structure. In Cu, both the 4s and 3d electrons can be contributed to the sea of delocalised valence electrons as they are very close in energy. The resulting copper ions have a smaller ionic radius than Ca2+. A larger amount of energy is required to overcome the stronger electrostatic attraction between the copper ions and the delocalised/mobile valence electrons compared to that between Ca2+ and delocalised/mobile valence.

(ii) Cu conducts electricity in the solid state due to the presence of mobile (delocalised) valence electrons that can migrate freely through the giant metallic structure when a potential difference is applied.

For CuO in solid state, the Cu2+ and O2– ions are held in fixed positions in a giant ionic lattice structure and are not mobile. [4]

(b) (i) Ksp = [Cu2+][OH–]2

= (1.77 10–7)(2 1.77 10–7)2

= 2.22 10–20 mol3 dm–9

(ii) Common ion effect refers to the reduced solubility of a sparingly soluble salt (when compared to its solubility in water) in an aqueous solution containing an ion common to that salt.

Let x be the solubility of Cu(OH)2 in 0.050 mol dm–3 of NaOH

Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH–(aq)

Initial [ ] / mol dm–3 0 0.050

Change in [ ] / mol dm–3

+ x +2x

Eqm [ ] / mol dm–3 x 2x +

0.050

x (2x + 0.050)2 = 2.22 10–20

Assume x is small such that 2x + 0.050 ≈ 0.05

x = 8.88 10–18 mol dm–3 << 1.77 10–7

(iii) (Pale) Blue precipitate dissolves to give a dark blue solution.

Cu(OH)2 + 4NH3 + 2H2O [Cu(NH3)4(H2O)2]2+ + 2OH– [6]

6

(c) (i) Ppt X

Identity : CuI

Colour : off-white/cream

Solution Y

Identity : I2

Colour : (reddish) brown

Solid Z

Identity : Cu

(ii) Reducing agent [4]

(d) 2Cu(s) + CO2(g) + H2O(g) + O2(g) Cu(OH)2(s) + CuCO3(s) [1]

[Total: 15]

7

4 (a) (i) Anode: 2Cl(l) → Cl2(g) + 2e

Cathode: Mg2+(l) + 2e → Mg(l)

(ii) Amount of Mg per day = (20 1000 1000) 24.3

= 8.23 105 mol

Mg 2e

Amount of electrons per day = 1.65 106 mol

I = nF/t

= (1.65 106 96500) (24 60 60)

= 1.84 106 A [4]

(b) (i) E⦵

(H2O/H2) = 0.83 V

E⦵

(Mg2+/Mg) = 2.38 V

Electrolyis of seawater would produce hydrogen gas as water is

preferentially reduced since the E⦵

(H2O/H2) is more positive than

E⦵

(Mg2+/Mg) .

(ii) The melting point of magnesium oxide is much higher than magnesium chloride. Melting magnesium oxide would require significantly more energy and thus incur higher cost.

(iii) Magnesium – White

Strontium/calcium – crimson/(brick) red [6]

8

(c) (i) Experiment 1: Oxygen

Experiment 2: Chlorine

E⦵

(O2/H2O) = +1.23 V

E⦵

(Cl2/Cl) = +1.36 V

In experiment 1 water is preferentially oxidised to oxygen gas as

E⦵

(O2/H2O) is more negative than E⦵

(Cl2/Cl).

In experiment 2 chloride anion is preferentially oxidised to chlorine gas as the increased concentration of choride ions shifts the

position of equilibrium Cl2 +2e ⇌ 2Cl to the left, making

E(Cl2/Cl) more negative than E(O2/H2O).

(ii) 3Cl2(g) + 6OH–(aq) → 5Cl −(aq) + ClO3

−(aq) + 3H2O(l) [5]

[Total: 15]

9

5 (a) (i)

(ii)

+ AlCl3 +

(iii)

[4]

(b) The carbon is sp3 hybridised. The 60° bond angle makes the three-membered ring highly strained hence epoxides react in ring-opening reactions to relief the ring strain/to attain the (preferred) 109.5°. [2]

(c)

[3]

(d) (i) The solubility of ibuprofen lysine salt is higher so it is more easily absorbed by the body.

(ii) The non-polar part of ibuprofen makes it difficult to penetrate the polar inner layer of the skin. When mixed with the more polar propan-2-ol, the mixture becomes sufficiently polar to penetrate the inner layer of the skin. [3]

C D

X

+

10

(e) Any one of the following tests and observations:

1) Add Na2CO3

Observation: Effervescence of carbon dioxide gas that formed white precipitate with limewater observed for ibuprofen but no gas evolved for 2-methylpropylbenzene.

2) Add PCl5 or SOCl2

Observation: White fumes of HCl observed for ibuprofen but no fumes evolved for 2-methylpropylbenzene. [2]

[Total: 14]



CANDIDATE NAME

CLASS 6

CENTRE NUMBER

S INDEX NUMBER


Paper 3 Free Response 23 September 2014

2 hours

Candidates answer on separate paper.

Additional Materials: Answer Paper

Graph Paper

Cover Page

Data Booklet


Write your name, class, centre number and index number on all the work you hand in.

Write in dark blue or black pen on both sides of paper.

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


Answer any four questions.

Begin each question on a fresh sheet of paper.

A Data Booklet is provided. Do not write anything on it.

You are reminded of the need for good English and clear presentation in your answers.

The number of marks is given in brackets [ ] at the end of each question or part question.

At the end of the examination, fasten all your work securely together, with the cover page on

top.

This document consists of 16 printed pages.

2


Answer any four questions.

1 (a) TRIS is an organic compound with the formula (HOCH2)3CNH2. It is extensively used in biochemistry and molecular biology as a component of buffer solutions.

TRIS

The table below summarises key information about TRIS and other amines.

Compound m.p./C b.p./C pKb Odour

CH3NH2 93 6 3.43 Fishy

CH3CH2NH2 80 20 3.19 Fishy

C6H5NH2 6 184 9.37 Fishy

TRIS 176 219 5.93 Characteristic

(i) What is meant by the term buffer?

(ii) 25.0 cm3 of 0.10 mol dm–3 TRIS is titrated against 0.20 mol dm–3 hydrochloric acid.

Calculate the pH of the solution when the following volumes of hydrochloric acid are added.

1. 0 cm3

2. To reach equivalence point

3. 30 cm3

(iii) Apart from its relatively low cost, suggest why TRIS is suitable for use in biochemistry and molecular biology experiments. [6]

3


(b) C6H5NH2, also known as aniline, played a key role in the proliferation of synthetic dyes. Many of these compounds involve azo dyes, which contain two aromatic fragments connected by a N=N double bond. Azo dyes are prepared in a two-step reaction, the first being the synthesis of an aromatic diazonium ion from an aniline derivative (note: alkyl diazonium ions are very unstable). The next step is the coupling of the diazonium salt with an aromatic compound. The reaction scheme is shown below.

(i) A student who was interested in studying the rate of reaction of the diazotisation step found the following mechanism on the internet.

HONO ⇌ H+ + ONO

4


Based on the mechanism, state the rate equation for the reaction between C6H5NH2 and HONO.

(ii) State the units for the rate constant for this reaction. [2]

(c) Compound G has the molecular formula C12H15NO3. G does not react with PCl5 to give white steamy fumes.

On boiling G with aqueous sodium hydroxide, H and a salt of compound I are formed. Compound H has a fishy smell and does not have a stable reaction with HONO.

Compound I, C11H12O4, rotates plane-polarised light and produces a yellow solid on warming with alkaline aqueous iodine. I gives a yellow precipitate with 2,4-dinitrophenylhydrazine. On addition of aqueous bromine, compound J is formed with the formula, C11H11O4Br.

Suggest structures for compounds G to J and explain the observations. [9]

(d) By means of equations, describe two reactions in which iron or one of its compounds behaves as a catalyst, and state whether the catalyst is homogenous or heterogenous. [3]

[Total: 20]

5


2 Nitrogen is an element that is essential to life on earth. Nitrogen is found in nucleic acids such as DNA and RNA which store genetic information. It is also found in proteins.

In spite of nitrogen’s abundance in the atmosphere, the quantity of nitrogen containing compounds that were available for human use was limited. Prior to the 20th century, nitrogenous compounds were either obtained from biological sources such as manure and guano (seabird or bat droppings) or by mining nitrate deposits. Nitrate deposits were very rare and the main deposit that was exploited was in the Atacama Desert, Chile.

The Haber process for the manufacture of ammonia and the Ostwald process for the conversion of ammonia to nitric acid were developed in the early 20th century.

The First World War began one hundred years ago in July 1914. Almost all explosives contain nitrogen and nitric acid is an essential reagent in their manufacture. The ability to convert atmospheric nitrogen to ammonia and hence other nitrogenous compounds freed Germany and her allies from depending on the Chilean nitrate deposits which had to be transported across the Atlantic by ship. Historians estimate that the First World War would have ended in 1916 instead of 1918 as Germany would have run out of sources of nitrogen.

(a) Ammonium salts and nitrates are sources of nitrogen.

Suggest two reasons why deposits of ammonium and nitrate compounds are rare and found only in deserts. [2]

(b) Ammonia is described as a ‘real’ gas.

State and explain the conditions under which ammonia deviates the most from ideal gas behaviour. Relate your answer to the bonding in ammonia. [3]

(c) In an industrial accident, a large quantity of ammonia was released. Ammonia is lethal if the concentration of the gas exceeds 5000 ppm.

(1 ppm is equivalent to 1 mg dm3)

Assuming that the pressure of ammonia was 1 atm and that ammonia

behaves as an ideal gas, calculate the density of ammonia at 30 C.

Hence, determine if the concentration of the ammonia was lethal. [3]

6


(d) The Van’t Hoff equation relates the equilibrium constant of a reaction with temperature. The linear form of the Van’t Hoff equation is given below.

ln c H

T

where R is 8.31 J mol1 K1

The following table gives equilibrium constant values measured at different temperatures for the Haber process.

2

1N2(g) +

2

3H2(g) ⇌ NH3(g)

T / K Kc / mol1 dm3 ln Kc T

1/ K1

300 4.34 103 5.44 3.33 103

400 1.64 104 8.72 2.50 103

(i) Determine the enthalpy change of the reaction.

(ii) Determine the entropy change of the reaction.

(iii) Hence, determine the temperature at which the reaction becomes spontaneous. [6]

(e) Proteins are an important class of biological molecules.

(i) State three functions of proteins.

(ii) Secondary structures are repeated arrangements of amino acids that can be found in proteins. One such structure is termed the

-pleated sheet.

Describe the -pleated sheet structure of protein. Your answer should include a suitable diagram which shows the bonding between the strands and at least two amino acid residues per strand.

7


(iii) Globular proteins are a class of water-soluble proteins. Many techniques have been developed to extract proteins from natural sources. One simple technique is termed ‘salting-out’. In salting out an inorganic salt such as ammonium sulfate is gradually added to a solution containing proteins. Different proteins will be selectively precipitated out as concentration of the salt in the solution increases.

Using your knowledge of the tertiary structure of proteins, explain why globular proteins can be water soluble and why the addition of ammonium sulfate may cause proteins to be precipitated out. [6]

[Total: 20]

8


3 Ethyl diazoethanoate is an important organic chemistry building block, but its use as an industrial intermediate is limited due in part to the safety concerns associated with its instability and high reactivity.

Ethyl diazoethanoate

(a) At 298 K, ethyl diazoethanoate reacts with water to give ethyl hydroxyethanoate and nitrogen gas.

N2CHCO2CH2CH3(aq) + H2O(l) HOCH2CO2CH2CH3(aq) + N2(g)

The rate of the reaction was followed by measuring the volume of nitrogen gas evolved at various time intervals after the start of the reaction. After 50 minutes, when all of the ethyl diazoethanoate had been consumed, the total volume of nitrogen gas produced was 60.0 cm3. The experimental results are given in the table below:

Time / min Volume of nitrogen gas / cm3

0.0 0.0

10.0 45.0

20.0 55.7

30.0 58.1

(i) Using 2 cm to represent 10.0 min and 10.0 cm3 on the x-axis and y-axis respectively, plot a graph of volume of nitrogen produced against time.

(ii) Use your graph to determine the order of the reaction with respect to ethyl diazoethanoate and explain how you deduced your answer.

(iii) What is the value of the rate constant for this reaction at 298 K? [5]

9


(b) Ethyl diazoethanoate also undergoes the Büchner-Curtius-Schlotterbeck reaction involving the reaction with an aldehyde to form a β-keto ester, A. The general reaction scheme is as follows:

(i) State the type of reaction involved in step I.

(ii) Suggest why step II is spontaneous. [2]

step I

step II

A

resonance

rearrangement

10


(c) The β-keto ester formed, A, undergoes reactions in the following scheme. The products formed in the boxes are subjected to further reactions.

(i) State the types of reactions in steps I and III.

(ii) Give the reagents and conditions required in step II.

(iii) Draw the structures of the organic products C to H. [9]

A

B

C

D

E

F + G + H

step I

step II

LiAlH4 in dry ether

step III

11


(d) Compound B, formed in the above scheme, undergoes condensation reaction with methanol. The reaction is exothermic.

CH3COCH2CO2H(l) + CH3OH(l) ⇌ CH3COCH2CO2CH3(l) + H2O(l)

A mixture of 0.25 mol of B and 0.34 mol of methanol was allowed to reach equilibrium. The equilibrium mixture was found to contain 0.13 mol of the ester.

(i) Calculate the value of the equilibrium constant, Kc, for the condensation reaction.

(ii) Explain the effects of the following actions on the equilibrium position of the reaction:

addition of a small amount of concentrated sulfuric acid

increasing the temperature at which the reaction was carried out [4]

[Total: 20]

12


4 Iron is one of the most abundant element found on earth. It comprises about 5.6% of the earth's crust and almost all of the earth's core. It is also commonly found in both humans and plants.

Iron is a typical transition element. Transition elements typically form complexes. Complexes having different ligands exhibit different properties even when the central metal ions are the same.

The common oxidation states of iron are +2 and +3. Both iron(II) and iron(III) ions form complexes with H2O and CN–.

A solution of Fe3+(aq) contain the ion [Fe(H2O)6]3+.

(a) The oxidising power of a complex may be deduced from its standard

reduction potential, E⦵

.

(i) Draw a fully labelled diagram to describe how the standard reduction potential of the [Fe(CN)6]

3–/[Fe(CN)6]4– system can be

measured using a standard hydrogen electrode.

(ii) By quoting E⦵

values from the Data Booklet, show that [Fe(H2O)6]

3+ has a stronger oxidising power than [Fe(CN)6]3–.

(iii) Suggest a reason for the difference in oxidising power of the two complexes in (a)(ii). [6]

(b) In an octahedral complex such as hexacyanoferrate(II) ions, [Fe(CN)6]4–,

the d-orbitals are split into two levels, as shown in Figure 1 below.

d-orbitals of an isolated Fe2+ ion without ligands

d-orbitals of Fe2+ in an octahedral complex

Figure 1

Energy gap ( E)

13


Complexes exist in two spin states.

In a high spin state, the electrons of the central metal occupy all the d-orbitals singly, before starting to pair up in the lower energy d-orbitals.

In a low spin state, the lower energy d-orbitals are filled first, with pairing if necessary, before the higher energy d-orbitals are used.

(i) Write the full electronic configuration of the iron ion in [Fe(CN)6]4–.

(ii) Electrons usually prefer to occupy orbitals singly because of repulsion that would occur between two electrons occupying the same orbital.

In light of this, suggest why a complex might still adopt a low spin state.

The value of the energy gap ( E) is dependent on the strength of ligand. For a given metal ion, the series below shows how the ligand affects E:

CO > CN– > NO2– > NH3 > edta4– > H2O > OH– > F–

larger E smaller E

(iii) Predict whether [Fe(CN)6]4– will adopt a high spin or low spin

state. Briefly explain your answer.

A paramagnetic substance is one that will be attracted by an externally applied magnetic field. Chemical species with at least one unpaired electrons will be paramagnetic.

(iv) By using your answers to (b)(i) and (b)(iii), deduce whether [Fe(CN)6]

4– will be paramagnetic. Explain your answer. You may find it useful to include a diagram in your answer. [7]

(c) Iron also exists in other oxidation states.

An oxoanion of iron has the formula FeOx2–. In acidic conditions, FeOx

2– is a strong oxidising agent and is reduced to Fe3+. In an experiment, a 10 cm3 of 0.200 mol dm–3 of FeOx

2– completely oxidised 30 cm3 of 0.200 mol dm–3 of Fe2+ to Fe3+.

Determine the value of x. [3]

14


(d) Iron(III) chloride can be used in qualitative analysis for both inorganic and organic chemistry.

(i) A label has fallen off a bottle containing a colourless solution suspected to be either NH4SCN or NH4NO3.

Outline how a solution of iron(III) chloride can be used to determine the identity of the solution.

A neutral solution of iron(III) chloride can be used to test for a particular organic functional group.

(ii) Describe the appearance of a positive result of this test.

(iii) Draw the structure of one compound with the molecular formula C7H8O that would show a positive result in this test, and the structure of one compound with the same molecular formula that would not. Label your structures clearly. [4]

[Total: 20]

15


5 Chlorine was identified as an element in 1810 by the English Chemist Sir Humphry Davy. The reactions of chlorine and chlorine-containing compounds span all fields of chemistry.

(a) When concentrated sulfuric acid is warmed with solid sodium chloride, white fumes are produced. However, when concentrated sulfuric acid is warmed with solid sodium iodide, purple fumes are produced.

(i) Identify the fumes that are produced.

(ii) Write equations for the reactions that occurred for NaI.

(iii) Briefly explain the differences in the reactions of solid sodium chloride and sodium iodide with concentrated sulfuric acid. [4]

(b) (i) Halogens can also act as oxidising agents.

Describe and explain the trend in oxidising power of the halogens down the group.

(ii) Describe and explain how the thermal stabilities of the hydrogen halides vary down the group. [5]

(c) Methylmalonic acid, MMA, is a compound that reacts with vitamin B12 to produce coenzyme A (CoA). Coenzyme A is essential to normal cellular function. When vitamin B12 deficiencies occur, MMA levels increase. Measurement of MMA levels can provide the doctor with information about an existing vitamin B12 deficiency.

The following table compares the pKa values of methylmalonic acid with malonic acid.

HO2C-(CH2)n-COOH HO2C-(CH2)n-COO O2C-(CH2)n-COO

Acid Formula pK1 pK2

Malonic acid HO2CCH2COOH 2.85 5.70

Methylmalonic acid HO2CCH(CH3)COOH 3.07 5.76

(i) Suggest a reason why the pK1 value of methylmalonic acid is higher than that of malonic acid.

(ii) Suggest a reason why the pK2 values of methylmalonic acid and malonic acid are higher than their corresponding pK1 values.

pK1 pK2

16


(iii) High concentration of calcium malonate, C3H2O4Ca, is found in beetroot juice. A cup of 100 cm3 of beetroot juice is found to have a pH of 8.

Given that calcium malonate is the only component in beetroot juice, calculate

1. the concentration of OH in the beetroot juice;

2. the initial concentration of calcium malonate.

You may use X2 to represent the malonate ion in this calculation. [7]

(d) State and explain how the acidities of phenol, ethanoic acid and propan-1-ol compare with each other. [4]

[Total: 20]

- End of Paper -

1

Answer key to Paper 3

1 (a) (i) A buffer solution is one that can resist a change in pH when a small amount of acid or base is added to it.

(ii) 1.

pKb = 5.93

Let [OH] be x

x2 (0.10 x) = 105.93

Since x is small,

x2 0.10 = 105.93

x = 3.43 × 104 mol dm3

pOH = 3.47

pH = 10.5

2.

At equivalence point, 12.5 cm3 of HCl solution is required.

pKa of conjugate acid = 8.07

Amt of RNH3+ = (25/1000) × 0.1 = 2.50 × 103 mol dm3

[RNH3+] = 2.50 × 103 (37.5/1000)

= 6.67 × 102 mol dm3

Let the concentration of [H+] be y

y2 (6.67 × 102 –y) = 108.07

Since y is small,

y = 2.38 × 105 mol dm3

pH = 4.62

3.

Amt of HCl remaining = (30/1000) × 0.20 - 2.50 × 103

= 3.50 × 103 mol

[HCl] = 3.50 × 103 (55/1000)

= 6.36 × 102 mol dm3

pH = 1.20

2

(iii) Maximum buffering capacity occurs at 8.07, which is close to physiological pH of 7.4/ The buffering action occurs near physiological pH of 7.4/ Non-toxic [6]

(b) (i) Rate = k[HONO]2[C6H5NH2]0

(ii) Units for k mol1 dm3 s1 or mol1 dm3 min1 [2]

(c) Observations/Reactions Deduction

G does not undergo (nucleophilic) substitution with PCl5 to give white fumes

G does not contain carboxylic nor alcohol functional group.

G undergoes alkaline hydrolysis to give H and salt of I.

G is an amide

I is a carboxylic acid

G has a fishy smell and does not react with HONO.

G is an alkyl amine

I rotates plane-polarised light I is chiral

I undergoes positive iodoform test I contains –COCH3 or –C(OH)CH3 groups

I undergoes condensation reaction with 2,4-DNPH

I does not have -C(OH)CH3 groups / I must contain –COCH3 functional group

I undergoes electrophilic substitution with Br2(aq)

I undergoes 1 electrophilic substitution on benzene ring containing phenol functional group /other ortho & para positions are blocked from ES

G

3

H

I

J

[9]

(d) N2 + 3H2 ⇌ 2NH3

Heterogeneous catalyst: Fe

S2O82 + 2I

2SO42 + I2

Homogeneous catalyst: Fe2+ or Fe3+

Or

C6H6 + Br2 C6H5Br + HBr (or C6H6 + Cl2 C6H5Cl + HCl)

Homogeneous catalyst: FeBr3 (or FeCl3) [3]

[Total: 20]

4

2 (a) Nitrogen is inert and nitrogen compounds are hard to form.

Ammonium salts and nitrates are soluble in water and thus would only accumulate in dry areas. [2]

(b) Low temperature and high pressure.

Ammonia gas has hydrogen bonding between its molecules and thus significant intermolecular forces are present at low temperatures.

At high pressure, the ammonia molecules would be very close to each other and the volume of the gas molecules would be significant when compared to the volume of the container / gas. [3]

(c) pV = nRT

pV = mRT/Mr

density = m/V = pMr/RT

= (1.01 105 17.0) / (8.31 303)

= 682 g m3

= 682 mg dm3

Concentration is below lethal value. [3]

(d) (i) Gradient

= [(5.44 (8.72)] / (3.33 103 2.50 103)

= 3.95 103 K

H = 3.95 103 8.31 = 32825 J mol1 = 32.8 kJ mol1

(ii)

5.44 = (3.95 103)(3.33 103) + S/8.31

S = 155 J mol1 K1

(iii) G = H – TS

T = 32825/155 = 212 K [6]

5

(e) (i) Transport, structure, catalysts, defence, signalling / control (any 3)

(ii) β-pleated sheet consists of adjacent polypeptide strands stabilised by hydrogen bonds between the backbone C=O group of one strand and the backbone N-H group of the adjacent strand.

N

CH

C

N

CH

C

N

CH

C

N

CH

O

H O

H O

H

C

O

H

N

CH

C

N

CH

C

N

CH

C

N

CH

O

HO

HO

H

C

O

H

hydrogen bond

R

R

R

R

R

R

R

R

(iii) Charged and polar R groups / side chains of amino acid residues will form ion-dipole and hydrogen bonds with water molecules.

Addition of ammonium ions and sulfate ions can disrupt these interactions which will cause the proteins to precipitate out. [6]

[Total: 20]

+

+

+

+

6

3 (a) (i)

(ii) From the graph, t1/2 is (approximately) constant at 5.0 min. (Since the reaction occurs in aqueous medium, the concentration of water (the other reactant) can be assumed to be constant.) Therefore the reaction is first order with respect to ethyl diazoethanoate.

(iii) Rate constant =

2/1t

2ln = 0.139 (min1)

[5]

(b) (i) Nucleophilic Addition

(ii) Formation of gaseous molecule results in an increase in entropy. [2]

(c) (i) Reduction

Oxidation

(ii) Dilute HCl/H2SO4(aq), heat under reflux

0.0

10.0

20.0

30.0

40.0

50.0

60.0

0.0 10.0 20.0 30.0 40.0 50.0

Volume of nitrogen / cm3

time / min

7

(iii) C D E

F G H

[9]

(d) (i) B + Methanol ⇌ Ester + Water

Initial amount / mol

0.25

0.34

0

0

Change in amount / mol

0.13

0.13

+ 0.13

+ 0.13

Eqm amount / mol

0.12

0.21

0.13

0.13

Kc = ]OHCH][HCOCOCHCH[

]OH][CHCOCOCHCH[

3223

23223

= )V/21.0)(V/12.0(

)V/13.0)(V/13.0(

= 0.671

8

(ii) Concentrated sulfuric acid removes the water formed / is a dehydrating agent, causing the equilibrium position to shift to the right to produce more water.

Or concentrated sulfuric acid acts as a catalyst and does not change the equilibrium position.

Increasing the temperature favours the endothermic reaction to absorb some of the added heat. Since the forward reaction is exothermic, the equilibrium position shifts to the left to favour the backward endothermic reaction. [4]

[Total: 20]

9

4 (a) (i)

(ii) E⦵

/ V

Fe3+ + e– ⇌ Fe2+ + 0.77

[Fe(CN)6]3– + e– ⇌ [Fe(CN)6]

4– + 0.36

Since E⦵

(Fe3+/Fe2+) is more positive than E⦵

([Fe(CN)6]3–/

[Fe(CN)6]4–), Fe3+/[Fe(H2O)6]

3+ is more easily reduced than [Fe(CN)6]

3–.

Hence, [Fe(H2O)6]3+ is a stronger oxidising agent.

(iii) There will be repulsion between the negatively charged

[Fe(CN)6]3– and the electron to be added. / CN is a stronger

ligand than H2O, hence it stabilises the complex ion to a greater extent.

Hence, [Fe(CN)6]3– is less easily reduced and a weaker oxidising

agent. [6]

(b) (i) 1s2 2s2 2p6 3s2 3p6 3d6

(ii) The energy gap between the two sets of d-orbitals is greater than the repulsion energy, favouring the pairing of electrons.

(iii) [Fe(CN)6]4– will adopt a low spin state. since CN– ligands will result

in a large ∆E so electrons will pair up in the lower energy orbital.

Voltmeter

10

(iv)

electron arrangement in [Fe(CN)6]4–

The 6 d-electrons are all paired up in the lower energy orbitals/There are no unpaired electrons in [Fe(CN)6]

4–. Hence [Fe(CN)6]

4– will not be paramagnetic. [7]

(c) Amount of Fe2+ reacted = (30/1000) 0.200 = 6.00 10–3 mol

Amount of FeOx2– reacted = (10/1000) 0.200 = 2.00 10–3 mol

Since Fe2+ e–

Amount of e– gained by 1 mol of FeOx2–

= 6.00 10–3 / 2.00 10–3 = 3 mol

Oxidation state of Fe in FeOx2– = +3 + 3 = +6

+6 + (–2)x = –2

x = 4 [3]

(d) (i) NH4SCN gives a blood/deep red colouration with iron(III) chloride but NH4NO3 does not give blood/deep red colouration.

(ii) A violet colouration/solution will be formed.

(iii) Positive test: Negative test:

[4]

[Total: 20]

OH

CH3

CH2OH

11

5 (a) (i) White fumes : HCl

Purple fumes : I2

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

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

(iii) I is a stronger reducing agent/undergoes oxidation more readily.

Hence, HI will be further oxidised by H2SO4 which is an oxidising

agent to I2. [4]

(b) (i) The oxidising power of the halogens decreases down the group.

Oxidising power of halogens decreases down the group due to

increasing radius of halogen atoms from F2 to I2.

Hence, the attractive force by the nucleus for electrons decreases and the tendency for the halogen atom to gain electrons decreases.

(ii) The thermal stabilities decreases down the group.

The size of the halogen atoms increases from F to I and their

valence orbitals become more diffuse. This results in less effective overlap of the orbitals between the H atom and halogen atom and less energy is required to break the weaker H–X bond. [5]

(c) (i) The methyl group has an electron-donating effect, intensifying the

negative charge on oxygen atom, making the HO2CCH(CH3)COO

ion less stable than HO2CCH2COO. Hence methylmalonic acid is a weaker acid/has a lower K1 with a higher pK1.

(ii) Both HO2CCH(CH3)COO and HO2CCH2COO can be stabilised by intramolecular hydrogen bonding. / It is difficult to remove the H+ from both anions.

12

(iii) 1. pOH = 14 8 = 6

Concentration of OH = 106

= 1.00 106 mol dm3

2. X2–(aq) + H2O(l) ⇌ HX– (aq) + OH–(aq)

Initial [ ] / mol dm–3 y 0 0

Change in [ ] / mol dm–3 1.00106 +1.00106 +1.00106

Equilibrium [ ] / mol dm–3 y 1.00106 1.00106 1.00106

Ka2 = 105.70

Ka2 = 2.00 106 mol dm3

Kb (X2–) =

6

14

1000.2

1000.1

= 5.01 109 mol dm3

Kb = ][

]][H[-2X

OHX

5.01 109 = ]1000.1y[

)1000.1(6

26

y 1.00 106 = 2.00 104 mol dm3

Initial concentration of malonate = y = 2.01 104 mol dm3 [7]

(d) Ethanoic acid is the most acidic, followed by phenol then propan-1-ol.

Ethanoic acid is the most acidic due to delocalisation of the negative

charge over the COO group, making the ethanoate anion the most stable anion.

Comparing phenol and propan-1-ol, phenoxide anion is more stable than the propanoxide anion as the negative charge on the oxygen is delocalised into the benzene ring.

On the other hand, the electron-donating propyl group intensifies the negative charge on oxygen of propanoxide anion, making the anion less stable. [4]

[Total: 20]

year 6 preliminary examination (ii) -...

Documents