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SEKOLAH -SEKOLAH MENENGAH KINTA UTARA, PERAK

GERAK GEMPAR 2012

Instructions to candidates:

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

There are fifty questions in this paper. For each question, four suggested answers are given.

Choose one correct answer and indicate it on the Multiple Choice Answer Sheet provided.

Read the instruction on the Multiple Choice Answer Sheet very carefully.

Answer all questions. Marks will not be deducted for wrong answers.

This question paper consists of 12 printed pages.

STPM 962/1* This question paper is CONFIDENTIAL until the examination is over. CONFIDENTIAL

K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A



K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA

KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A KINTA UTARA K K K I I I N N N T T T A A A U U U T T T A A A R R R A A A




962/1 STPM CHEMISTRY

PAPER 1

MULTIPLE-CHOICE

One hour and forty-five minutes

http://edu.joshuatly.com/ http://fb.me/edu.joshuatly



CONFIDENTIAL* 2

962/1*This question paper is confidential until the examination is over. CONFIDENTIAL*

Section A

Four suggested answers labelled A, B, C and D are given for each question. Choose one correct

answer.

1. Which of the following peaks is not found in the mass spectrum of butanoic acid?

e

m

value Ion responsible for peak

A. 15 CH3+

B. 73 CH2CH2CO2H+

C. 74 CH3CH2CO2H+

D. 88 CH3CH2CH2CO2H+

2. At 273 K, 50 % of 1 mole of PCl5 in a container of 22.4 dm capacity decomposes

according to the following equation. PCl5 (g) PCl3 (g) + Cl2 (g)

What is the total pressure in the container after the dissociation at s.t.p. (273 K)?

A. 0.5 atm B. 1.0 atm C. 1.5 atm D. 2.0 atm

3. Which of the following phase diagrams is consistent with the sublimation of a solid

at room temperature (22 oC) and atmospheric pressure?

4. A covalent molecule, XYn , has the following properties.

(i) It is polar (ii) It has a pyramidal shape

(iii) It can form coordinate bond with boron trifluoride molecule.

Which of the following statements is not true regarding the XYn molecule?

A. X element is more electronegative than Y element.

B. X element is in Group 15 of the Periodic Table.

C. X atom undergoes sp hybridisation to form bonding with Y atom.

D. The XYn molecule can act as a Lewis acid.

A

B

C

D




CONFIDENTIAL* 3


5. The proton number of chlorine, potassium, and calcium are given below :

Cl = 17, K = 19 and Ca = 20.

Thus, we can conclude that the ionic radii of Cl – , K + and Ca2+ ions increases in the

order

A. Cl – < K + < Ca + C. K + < Ca + < Cl –

B. Ca+

< K +

< Cl –

D. K +

< Cl –

< Ca+

6. The boiling points of water and ammonia are 373 K and 240 K, respectively. Which

of the following statements best explains their difference in the boiling points?

A. Oxygen is more electronegative than nitrogen.

B. The covalent bond between oxygen and hydrogen is stronger than the covalent

bond between nitrogen and hydrogen.

C. The hydrogen bond between water molecules is stronger than the hydrogen

bond between ammonia molecules.

D. There are Van der Waals’ forces between water molecules but no Van der

Waals’ forces between ammonia molecules.

7. Decomposition of benzenediazonium chloride at 320 K is a first order reaction.

N2+ Cl – OH

+ H2O + HCl + N2

Which of the following graphs of the rate of reaction, R, against concentration of

benzenediazonium chloride, [A] is correct?

A. R C. R

0 [A] 0 [A]

B. R D. R

0 [A] 0 [A]

8. Which of the following equilibrium systems has the value of K c equal to the value of

K p?

A. Cl2 (g) 2 Cl (g)

B. 2 NO (g) + Cl2 (g) 2 NOCl (g)

C. 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g)

D. CH3COOH (l) + CH3CH2OH (l) CH3COOCH2CH3 (l) + H2O (l)




CONFIDENTIAL* 4


9. The industrial process for the manufacturing of ammonia is represented by the

equation below :

N2 (g) + 3 H2 (g) 2 NH3 (g) H = –92 kJWhich of the following can be deduced regarding the reaction below :

2 NH3 (g) N2 (g) + 3 H2 (g)

A. The reaction is exothermic.B. Catalyst will increase the value of the equilibrium constant.

C. The equilibrium constant increases as temperature decreases.

D. The production of nitrogen increases with decreasing pressure.

10. Which of the following indicators is most suitable to be used in a titration between

0.1 mol dm –3 hydrochloric acid and 0.1 mol dm –3 aqueous ammonia?

Indicator pH rangeA. Thymol blue 1.2 – 2.8

B. Methyl orange 3.2 – 4.4

C. Bromothymol blue 6.0 – 7.6D. Phenolphthalein 8.2 – 10.0

11. Which of the following graphs correctly represents the variation of pH against the

volume of nitric acid added when 20.0 cm3 of a 0.2 mol dm –3 sodium carbonate

solution is titrated against a 0.2 mol dm –3 nitric acid using phenolphthalein and

methyl orange indicators?

A. C.

20 40 20 40

Volume of HNO3 / cm Volume of HNO3 / cm

B. D.

20 40 20 40

Volume of HNO3 / cm Volume of HNO3 / cm

12. The alum, KAl(SO4)2.2H2O, is used in the purification of water. The precipitation of

Al(OH)3 helps to eliminate suspended particles in water. At what pH will

precipitation of Al(OH)3 begins if 3.30 kg of alum is dissolved in 5000 dm3 of

water? [Relative molecular mass of alum = 474 and the solubility product, K sp for

Al(OH)3 = 3.70 x 10 –15

mol4

dm –12

]

A. 3.86 B. 9.98 C. 10.0 D. 10.1http://edu.joshuatly.com/ http://fb.me/edu.joshuatly



CONFIDENTIAL* 5


13. The boiling point–composition curve for a mixture of two liquids, X and Y is shown

below.

Temperature / C

100 % X Z 100 % Y

Composition

Which of the following statements can explain the curve above?

A. The mixture shows positive deviation from Raoult’s law.

B. When X and Y are mixed, heat is absorbed.

C. Fractional distillation of a mixture containing 50 % X will yield a residual with

the Z composition in the distillation flask.

D. The total vapour pressure of the mixture is the total vapour pressure of pure Xand pure Y.

14. The relationship between the electrode potential and concentration is given by the

Nernst equation.

E = E – 0.06

lg[oxidised ion]

n [reduced ion]

The standard e.m.f. of the cell

Zn (s) / Zn2+ (aq, 1 mol dm –3) || Cu2+ (aq, 1 mol dm –3) / Cu (s) is + 1.10 V. If the

concentration of the Cu2+ ions is reduced to 0.10 mol dm –3 and the concentration of

the Zn2+ ions is fixed, what will be the e.m.f. of the cell?

A. 1.04 V B. 1.07 V C. 1.13 V D. 1.16 V

15. Consider the half-cells below :

Half-cell reaction E

/ V

Ba2+ + 2e Ba – 2.90

Cr 3+ + 3e Cr – 0.74

Co2+ + 2e Co – 0.28

PbO2 + 4H+ + 2e Pb2+ + 2H2O +1.47

Co3+ + e Co2+ +1.82

Which of the following reactions will give the highest e.m.f.?

A. 2 Cr 3+ + 3 Ba 2 Cr + 3 Ba2+

B. 2 Co3+ + Ba Ba2+ + 2 Co2+ C. 3 Co2+ + 2 Cr 3 Co + 2 Cr 3+

D. 2 Co3+ + Pb2+ + 2 H2O 2 Co2+ + PbO2 + 4 H+




CONFIDENTIAL* 6


16. Which of the following represents an ion, Z, which will be discharged at the anode

when electrolysis of a molten compound contains Z ions?

Proton number Electronic configurationA. 10 1s 2s 2p

B. 11 1s 2s 2p

C. 13 1s 2s 2p

D. 17 1s 2s 2p 3s 3p

17. Which of the following is not required in the calculation of the lattice energy of

calcium oxide using the Born–Haber cycle?

A. Enthalpy of hydration C. Enthalpy of atomisation

B. Enthalpy of ionisation D. Electron affinity

18. Consider the neutralisation reactions below :

NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) H1 = –52.7 kJ

NaOH (aq) + ½ H2SO4 (aq) ½ Na2SO4 (aq) + H2O (l) H2 = –68.0 kJ

H2 is more negative than H1 because

A. H2SO4 is a stronger acid than HCl

B. H2SO4 dissociates completely in water but HCl only dissociates partially

C. the heat of dilution of H2SO4 is larger than that of HCl

D. the lattice energy of HCl is higher than that of H2SO4

19. Which of the following statements is true regarding the oxides of the elements in

Period 3 (sodium to chlorine) of the Periodic Table?

A. Sodium oxide is the strongest base because sodium is the most electropositive

element in Period 3.

B. Across Period 3, the properties of the oxides change from base to acid because

the bond between the element and oxygen gets stronger.

C. Aluminium oxide is amphoteric because aluminium is in Group 13.

D. Silicon dioxide does not react with aqueous sodium hydroxide because it is

neutral.

20. When aqueous X is added to aqueous Y, a white precipitate is formed immediately.

What is X and Y?

X YA. magnesium chloride sodium sulphate

B. barium nitrate potassium nitrate

C. barium chloride sodium sulphate

D. barium nitrate ammonia

21. Which of the following about the property and use of aluminium is correct?

Property UseA. Good heat conductor Aluminium paint

B. Good light reflector Cooking pot

C. Resistance to corrosion Electrical cableD. Low density Aircraft body




CONFIDENTIAL* 7


22. Which of the following represents the correct set of conditions used in the extraction

of aluminium by electrolysis?

Anode Cathode Temperature / CA. Graphite Mercury 900

B. Platinum Platinum 600

C. Mercury Graphite 600D. Graphite Graphite 859

23. Below are the structural formulae of three compounds, X, Y and Z.

H

C H H

X Y Z

Which of the following shows the position of the atoms in the respective molecules?

All atoms in the molecule

are located in the same plane

Only carbon atoms

are located in the same plane

A. X, Y X

B. X, Z X, Y

C. X, Y X. Y, Z

D. X, Z X, Y, Z

24. Why is the halogenation of alkanes considered a chain reaction?

A. it occurs with the presence of ultraviolet light

B. it occurs without the generation of intermediates

C. each step generates the reactive intermediate that causes the next step to occur

D. the reaction allows long chains of halogenated alkanes to be formed

25. Benzene reacts with a mixture of concentrated nitric acid and concentrated sulphuric

acid to produce nitrobenzene according to the equation :

C6H6 + HNO3 4SO2H

C6H5 NO2 + H2O

What is the function of sulphuric acid in the above reaction?

A. To protonate the nitric acid

B. To eliminate the water produced

C. To prevent di–substitution from occurring

D. To prevent oxidation of benzene by nitric acid

26. What is the best choice of reagent to perform the following transformation?

A. Br 2, H2O B. HBr C. Br 2, CCl4 D. HBr, H2O2




CONFIDENTIAL* 8


27. What is the organic product obtained when o–chloroethylbenzene is refluxed with

alkaline potassium manganate (VII) for several hours, followed by dilute sulphuric

acid?

A. COOH

B. COOH

Cl

C. CH2COOH

Cl

D. CHO

Cl

28. Conversion of ethene to ethanol can be carried out in two stages as below :

I CH2 = CH2 (g) + HI (g) CH3CH2I (l)II CH3CH2I (g) + KOH (aq) CH3CH2OH (aq) + KI (aq)

Which of the following is true about reaction I and II?

I II

A. Addition EliminationB. Substitution Addition

C. Addition Substitution

D. Substitution Substitution

29. Menthol is one of the substances added into a cough mixture. The

structure of its molecule is as given. Which statement is true of

menthol?

CH3

OH

CH(CH3)2

A. It has two chiral centres.

B. It has two functional groups.

C. It reacts with an aqueous solution of sodium hydroxide.

D. It decolourises an acidified solution of potassium manganate (VII).

30. Which of the following reagents will react with ethanol and phenol?

A. Ethanoic acid C. Ethanoyl chloride

B. Aqueous bromine D. Aqueous sodium hydroxide

31. For which one of the following pairs of compounds can the members be

distinguished by means of Tollen’s test?

A. HCHO and CH3CHO C. CH3COCH3 and CH3CO2CH3

B. CH3CHO and CH3COCH3 D. CH3CO2H and CH3CO2CH3

32. Oxidation of an organic compound, X, produces an organic liquid which has the

following properties.

(i) Gives an orange precipitate with 2,4–dinitrophenylhydrazine

(ii) Do not form precipitate with warm Fehling’s solution

(iii) Do not form precipitate with warm alkaline iodine aqueous

X compound might be

A. CH3CH2CH2CH2CH2OH C. C6H5CH(OH)CH3

B. CH3CH2CH(OH)CH2CH3 D. C6H5CH2CH2CH2OHhttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly



CONFIDENTIAL* 9


33. Consider the following reaction scheme. K is an organic compound whereas L is an

inorganic compound.

K + L M(aq)HCl

boil (CH3)2C(OH)COOH

K and L may be

K LA. CH3CHO HCN

B. CH3COCH3 HCN

C. (CH3)2CHCHO HCN

D. (CH3)2CHCHO KMnO4 / H +

34. Chloroethane is converted into a carboxylic acid containing one more carbon atom

through a two–stage process. Which of the following compounds could be the

intermediate in the synthesis of the carboxylic acid?

A. CH3CH2OH C. CH3CH2COOCH3

B. CH3CH2CH2CN D. CH3CH2CN

35. The reaction scheme for the synthesis of the organic compound, Y, is shown below.

CH3CH2OH

H/4KMnO

refluxX

h/2Cl

boilY

The pK a values of ethanol, X and Y decrease in the order

A. pK a (CH3CH2OH), pK a (X), pK a (Y)

B. pK a (CH3CH2OH), pK a (Y), pK a (X)

C. pK a (X), pK a (CH3CH2OH), pK a (Y)

D. pK a (X), pK a (Y), pK a (CH3CH2OH)

36. Cold water is added to the following compound.

CH2Cl

Cl CH2COCl

Which of the following compounds is the product obtained?

A.

CH2OH

Cl CH2COCl C.

CH2Cl

Cl CH2COOH

B.

CH2OH

HO CH2COOH D.

CH2OH

HO CH2COCl




CONFIDENTIAL* 10


37. X and Y compounds are produced by the following route.

CN

H2 , Pt

X COCl

Y

What is Y compound?

A. C CH2 N O H

C. C N CH2 O H

B. CH2 N H

C Cl

O

D. CH2 N CH2 H

38. Which of the following reagents does not react with phenylamine?

A. Hydrogen bromide C. Ethanoyl chloride

B. Bromine water D. Concentrated ammonia solution

39. When 1 mole of compound X is heated with a concentrated solution of potassium

manganate (VII), 2 moles of compound Y with the molecular formula C2H4O2 is

produced. Compound X undergoes polymerisation in the presence of Al(CH3)3 and

TiCl4 to form compound Z. The structure of Z could be

A.

CH3 CH3 CH3 CH3 CH3

C C C C C H H H H H

B.

CH3 H CH3 H CH3

C C C C C H H H H H

C.

CH3 H CH3 H CH3

C C C C C CH3 H CH3 H CH3

D.

CH3 H CH3 H CH3

C C C C C

H CH3 H H H




CONFIDENTIAL* 11


40. A Ziegler–Natta catalyst is used to produce high density polyethene because the

titanium (IV) ion in the catalyst

A. has empty d orbitals. C. has many valence electrons.

B. has a low activation energy. D. can change its oxidation states.

Section B

For each question in this section, one or more of the three numbered statements 1 to 3 may be

correct. The responses A to D should be selected as follows:

A B C D

1 only

is correct

1 and 2 only

are correct

2 and 3 only

are correct

1, 2 and 3

are correct

41. The ideal gas equation is pV = nRT. It can be concluded that, for an ideal gas

1 the volume of a fixed mass of gas will be doubled when its temperature isincreased from 25 C to 50 C.

2 the density of a gas, at constant pressure, is inversely proportional to the

temperature.

3 one mole of any gas will occupy the same volume under the same conditions.

42. Solid carbon dioxide is used as ‘dry ice’ especially in the food industry because

1 it is less dense than ice

2 it sublimes at room temperature

3 it does not contaminate

43. Consider the reaction A + B C + D

Two experiments are carried out and the results are shown in the graph below.

pressure of C experiment II

experiment I

time

Which of the following separate changes in the conditions of experiment I wouldgive the results as shown in experiment II?

1 Increasing the concentration of A and B

2 Adding a suitable catalyst

3 Removing D from the reacting vessel as it is formed

44. Carbonic acid, H2CO3 , and hydrogen carbonate ion, HCO3 – , are the agents of the

buffer system in blood. Which reaction/s occur when the level of acidity in blood

increases?

1 HCO3 –

H+

+ CO32–

2 HCO3

– + H+ H2CO3

3 H2CO3 CO2 + H2Ohttp://edu.joshuatly.com/ http://fb.me/edu.joshuatly



CONFIDENTIAL* 12


A B C D

1 only

is correct

1 and 2 only

are correct

2 and 3 only

are correct

1, 2 and 3

are correct

45. In general, the first ionisation energy of the Period 3 elements (from sodium to

argon) increases because

1 the nuclear charge increases

2 the screening effect decreases

3 the number of electrons increases

46. Beryllium and aluminium has similar properties in that

1 both their oxides are amphoteric

2 both elements can form complex ions

3 both their chlorides are covalent

47. When excess chlorine gas is passed through 0.1 mole of a warm X compound, all the

hydrogen atoms of 2.24 dm3 of X are substituted by chlorine gas and 13.44 dm3 of

hydrochloric acid are produced. What is the possible structural formula of X?

[Assume that all the compounds are at gas condition at the same temperature and

pressure.]

1 CH4 2 CH3CH3 3 CH3COCH3

48. Which of the following substances does not react with chlorobenzene under suitable

conditions?

1 ammonia 2 sodium hydroxide 3 sodium metal

49. Which of the following give benzoic acid on boiling with dilute sulphuric acid ?

1 CN 2 COCl 3 OOC

50. Which of the following compounds dissolves in acidic aqueous sodium nitrite at

70 C to release nitrogen gas?

1 NH2 2 NO2

H2 N

3 N2+ Cl –




CONFIDENTIAL* 13


Answers for Paper 1 Chemistry STPM

Program Gerak Gempur PPD Kinta Utara 2012

1 C 26 B2 C 27 B

3 C 28 C

4 D 29 D

5 B 30 C

6 C 31 B

7 B 32 B

8 C 33 B

9 D 34 D

10 B 35 A

11 C 36 D

12 D 37 C

13 C 38 D

14 B 39 A

15 B 40 A

16 D 41 C

17 A 42 C

18 C 43 A

19 A 44 C

20 C 45 A

21 D 46 D

22 D 47 C

23 D 48 A

24 C 49 D

25 A 50 D




Instructions to candidates :

This paper consists of 10 printed pages

Section A[ 40 marks ]

Answer all the questions in this section.

This question paper consists of 8 printed pages.

For Examiner’s Use

Section A

1

2

3

4

Section B

Total

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962 / 2

Chemistry

Gerak Gempur ( 2012 )

Chemistry Paper 2

( Structure & Essay )

DO NOT OPEN THIS QUESTION PAPER UNTIL YOU ARE

TOLD TO DO SO.

Answer all the questions in Section A. Write your answers in

the spaces provided.

Answer any four questions from Section B. Write your

answers on papers provided. Begin each answer on a fresh

sheet of paper, and arrange your answers in numerical order.

Tie your answer sheets to this booklet.

All working must be shown. Numerical answers should begiven to an appropriate number of significant figures ; units

should be quoted wherever they are appropriate.

A Data Booklet is provided.

STPMName: ______________________________

Class: ______________________________




Section A [ 40 marks ]

Answer ALL questions in this section.

1. Biodiesel makes use of plants’ ability to fix atmospheric carbon by photosynthesis. Many companies and

individuals are now using biodiesel as a fuel in order to reduce their carbon footprint. Biodiesel can be

synthesized from vegetable oil according to the following reaction.

(a) Identify the organic functional group present in both vegetable oil and biodiesel.

………………………………………………………………………………………………………………[1]

(b) For part of her extended essay investigation into the efficiency of the process, a student reacted a pure

sample of a vegetable oil (where R = C17H33) with methanol.

The raw data recorded for the reaction is below.

The relative molecular mass of the oil used by the student is 885.6. Calculate the amount (in moles) of

the oil and the methanol used, and hence the amount (in moles) of excess methanol.

[2]

(c) The reversible arrows in the equation indicate that the production of biodiesel is an equilibrium process.

(i) State what is meant by the term dynamic equilibrium. [1]

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

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

(ii) Using the abbreviations [vegetable oil], [methanol], [glycerol] and [biodiesel], deduce the equilibrium

constant expression ( K c ) for this reaction. [1]

- 2 -

Mass of oil = 1013.0 g

Mass of methanol = 200.0 g

Mass of sodium hydroxide = 3.500 gMass of biodiesel produced = 811.0 g




(iii) Suggest a reason why excess methanol is used in this process. [1]

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

(iv) State and explain the effect that the addition of the sodium hydroxide catalyst will have on the position

of equilibrium. [2]

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

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

(d) Calculate the percentage yield of biodiesel obtained in this process. [2]

2 (a) State the meaning of the term first ionisation energy of an atom. [1]

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

(b) Explain the trend in variation of the first ionisation energies of the Period 3 elements Na to Ar. [2]

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

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

(c) Compare the first ionisation of phosphorus and sulphur . Explain your answer. [3]

…………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………

(d) (i) Draw the shape of a BF3 molecule and the shape of a H2S molecule. In each case show any lone pairs

of electrons. [2]

BF3 H2S

(ii) Hydrogen sulphide, H2S, reacts with boron trifluoride, BF3, to form compound A.

Explain the formation of compound A with the help of its Lewis structure to indicate the type of

bonding involved . [2]

- 3 -http://edu.joshuatly.com/ http://fb.me/edu.joshuatly



3. Nitrogen(II) oxide reacts with hydrogen as shown by the following equation.

2NO(g) + 2H2 (g) → N2 (g) + 2H2O(g)

The table below shows how the rate of reaction varies as the reactant concentrations vary.

(a) Determine the order of reaction with respect to (i) NO (ii) H2 [2]

(b) Write the rate law for the above reaction. [1]

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

(c) Calculate the value for the rate constant, including its units. [2]

(d) A suggested mechanism for this reaction is as follows.

H2 + NO Ý X fast

X + NO ↓ Y + H2O slow

Y + H2 ↓ N2 + H2O fast

State and explain whether this mechanism agrees with the experimental rate expression in (b). [3]

…………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………

e) Deduce the initial rate of formation of H2O(g) as compared to that of N2(g) for experiment 1. Explain

your answer. [2]

…………………………………………………………………………………………………………………

…………………………………………………………………………………………………………………

- 4 -




4. (a) 2 – butene is a straight-chain alkene with formula C4H8. The molecule contains both σ and π bonds.

H H H H

H C C = C C H

H H

(i) Name the type of hybridisation shown by the C2 carbon atom . [1]

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

(ii) Explain the formation of the π bond in terms of overlapping of orbitals. [1]

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

(b) 2 – butene shows structural isomerism and also stereoisomerism.

(i) Draw and name one other structural isomer of 2 – butene . [1]

(ii) Draw and name 2 stereoisomers of 2 – butene . [2]

(c) Identify the structural formula of an isomer of 2 – butene which does not decolourise bromine water.

[1]

(d) The polymerisation of alkenes is one of the most significant reactions of the twentieth century.

(i) Give two reasons why the polymers of alkenes are of economic importance. [2]

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

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

(ii) Deduce the structure of the polymer for 2 – butene showing 3 repeating units . [1]

(iii) Explain why monomers are often gases or volatile liquids but polymers are solids. [1]

…………………………………………………………………………………………………………………

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




Section B [ 60 marks ]

Answer any FOUR questions in this section.

5. (a) X, Y and Z represent elements of proton number 9 , 19 and 34 .

Predict the type of bonding and draw ‘ dot – cross ’ diagrams for the compounds formed

( showing only the electrons in the outermost shell for each atom ) between

(i) X and Y (ii) X and Z [6]

(b) Predict, giving reasons, the relative (i) volatility (ii) electrical conductivity

of the compound formed between X and Y compared with that formed between X and Z.

[4]

(c) (i) Define the term ‘ activation energy ’.

(ii) The endothermic reaction between substances P and Q can be represented by the following

equation.

P(g) + Q(g) R(g) + S(g)

Draw the energy profile for this reaction.

Indicate and label clearly the activation energy and the enthalpy change for the reaction.

(iii) Explain how a catalyst affects the rate of a chemical reaction. [5]

6. The table shows the variation of pV against p for 65.0 g of a gas M at 298 K.

p ( x 103 ) kPa 0.20 0.40 0.60 0.80 1.00

pV kPa m3 6.82 6.81 6.80 6.79 6.78

(a) (i) Plot a graph of pV against p.

(ii) Add to your graph the line expected if gas M is an ideal gas.

(iii) Comment on the shape of your graph.

(iv) Use the data in the graph to calculate the relative molecular mass of M . [8]

(b) Sketch the phase diagram of carbon dioxide.

Label the areas and explain why solid carbon dioxide can sublime under room conditions. [4]

(c) (i) Write an expression for K w, the ionic product of water.

(ii) Use the expression above, to determine the pH value for 0.200 moldm –3 NaOH (aq ) . [3]

7. (a) Describe the reactions, if any , that will occur when separate samples of sodium and phosphorus are

(i) added to water

(ii) burned in excess oxygen and then water is added to the resulting oxide.

Write equations wherever appropriate and suggest pH of any aqueous solution formed. [8]




(b) (i) Aluminium is extracted by the electrolysis of its mineral bauxite.

Identify some important aspects of this process, including equations for the reactions at the

electrodes.

(ii) When heated both aluminium fluoride and aluminium chloride sublime at 1270 oC and

178 oC respectively . Explain differences in these values based on the bonding in both compounds.

[7]

8. Hydrogen cyanide, HCN, is a highly toxic substance. 50.0 mg of it will cause death in a few seconds.

(a) Calculate the lethal dose (50.0 mg) of hydrogen cyanide in moles. [2]

(b) Hydrogen cyanide is manufactured by passing a mixture of ammonia and methane over a platinum

catalyst. The reaction is endothermic.

NH3(g) + CH4(g) Ý HCN(g) + 3H2(g)

(i) Suggest why the reaction is carried out at 1000 °C.

(ii) Explain if a high pressure should be used in the manufacture of hydrogen cyanide. [4]

(c) If ammonia (0.200 mol) and methane (0.200 mol) are placed in a 1.00 dm3 container and heated

to 500°C, it is found that 0.100 mol of hydrogen cyanide and 0.300 mol of hydrogen are produced at

1.00 atmosphere pressure. Calculate the equilibrium constant, K c for the reaction under these

conditions and state its units. [3]

(d) Hydrogen cyanide dissolves in water to form a weakly acidic solution. It has a dissociation constant of

4.90 × 10 –10 at 25 °C. Alkalis react with hydrogen cyanide to form salts known as cyanides.

(i) Explain why an aqueous solution of sodium cyanide is alkaline.(ii) Using equations, explain how a mixture of sodium cyanide and hydrogen cyanide is able to act as a

buffer.

(iii) Calculate the pH of the buffer solution formed when 200 cm3 of a 0.500 mol dm –3 solution of

hydrogen cyanide is added to 200 cm3 of a 1.00 mol dm –3 solution of sodium cyanide. [6]

9. Carefully study the steps involved in the conversion of the starting reagent to the final product.

(a) With the help of equations, describe the mechanism for step 1. [4]

(b) (i) Suggest why SnCl2 is used as the reducing agent for step 3 instead of lithium

tetrahydridoaluminate(III), LiAlH4 .

- 7 -

CH3 CH3 COOH COOH COOCH2CH3

conc. HNO3 KMnO4 / H

+

SnCl2

conc. H2SO4

Step 1 Step 2 Step 3 Step 4

NO2 NO2 NH2 NH2




(ii) SnCl2 is formed by the following reaction.

100oC

SnCl4 SnCl2 + Cl2

Show the oxidation state of each element in the equation and name the type of reaction that

has taken place in the above reaction.

(iii) Despite the metallic nature of tin, SnCl4 has a molecular structure.

Draw the molecular structure of SnCl4 .

What will be observed when a little water is added to SnCl4 ? Give reasons for your answer. [6]

(c) Describe a chemical test to confirm that step 3 has produced the amino group. [3]

(d) (i) Give the reagents and conditions for step 4.

(ii) Write a balanced equation for step 4 . [2]

10. (a) Draw all structural formulae for all the monobrominated product formed when 2 – methylpropane istreated with bromine gas in the presence of sunlight.

Predict the major product. [3]

(b) Compound X, C7H14 is optically active. On catalytic reduction of X over nickel, 1 mole of hydrogen

gas is absorbed giving 1 mole of compound Y, C7H16. Reaction of X with hot acidified potassium

manganate(VII) produces ethanoic acid and compound Z , C5H10O2 which is a carboxylic acid and is

also optically active.

(i) Draw the structures of X, Y and Z . Explain your answers.

(ii) Identify a chemical test to distinguish X from Y . [9]

(c) Give structures of the organic products, A, B and C formed in the following reaction scheme.

Compound Z

LiAlH4, dry ether

H3O+ SOCl2 Mg

A B C [3]

***********************************




Marking Scheme - GERAK GEMPUR – CHEMISTRY 2 ( 2012 )

SECTION A ( STRUCTURE ) SCORE REMARKS

1 a ester 1 mk

b No. of moles of oil = mass of oil = 1013.0 = 1.144

formula mass 885.6

No. of moles of CH3OH = 200.0 = 6.25

32.0

From equation, 1 mole oil ≡ 3 moles CH3OH

1.144 moles oil ≡ 3(1.144) = 3.432 moles CH3OH

Excess CH3OH = 6.25 – 3.432 = 2.818 = 2.82 moles ( 3 sf )

1 mk

1 mk

c (i) Dynamic equilibrium – a condition in a reversible reaction

when the rate of forward reaction = rate of reversereaction

1 mk

(ii) Kc = [glycerol] [biodiesel]3

[vegetable oil] [methanol]3

1 mk

(iii) -so that equilibrium shifts to the right according to

Le Chatelier’s Principle OR

-to increase the yield of biodiesel

1 mk either answer

(iv) - Catalyst NaOH has no effect on the position of

equilibrium i.e. it has no effect on the yield of biodiesel- as a catalyst increase the rate of both the forward and

reverse reactions to the same extent

1 mk

1 mk

d 3.432 moles CH3OH 3.432 moles biodiesel

= 3.432 x formula mass

= 3.432 x 296.0 = 1015.9 g ( theory)

% yield of biodiesel = 811.0 g x 100 = 79.83 % ( 4sf )

1015.9 g

1 mk

1 mk

Total = 10 mk

2 a Minimum energy required to remove 1 mole electron from1 mole atom of the element in gaseous state

1 mk

b - IE1 increases across Period 3 from Na to Ar

- Effective nuclear charge increases due to increase in no.

of protons but screening effect is constant

1 mk

1 mk

c - P : 1s2

2s2

2p6

3s2

3p3

S : 1s2

2s2

2p6

3s2

3p4

P has extra stability due to half-filled p orbitals

- Repulsion between e- pair in 3p of S enables the first mole

of e- to be easily removed

- Hence IE1 of P > IE1 of S

1 mk

1 mk

1 mk

-1-http://edu.joshuatly.com/ http://fb.me/edu.joshuatly




d (i) F

B 120o S

H

F F 105o

H

trigonal planar V-shaped / bent

1 mk

1 mk

(ii) xx

xx F xx

xx xo

xx F xo B S x H

xx xo x

xx F xx H

xx

- B atom is e- deficient, having an empty orbital

while S has an e- pair to donate

A coordinate / dative bond forms between BF3 and H2S

1 mk

1 mk

Total = 10 mk

3 a Rate2 = 5.05 x 10 –6

= ( 0.200 )n

Rate1 2.53 x 10 –6

( 0.100 )n

2 = 2n

, n=1 first order w.r.t. H2

Rate3 = 10.10 x 10 –6

= ( 0.200 )n

Rate1 2.53 x 10 –6

( 0.100 )n

3.99 ≈ 4 = 22

= 2n

, n=2 second order w.r.t. NO

1 mk

1 mk

1 mk

1 mk Max 3 mk

b Rate = k[NO]2

[H2] 1 mk

c Using expt 1 : k = Rate1 = 2.53 x 10 –6

mol dm –3

s-1

[NO]2

[H2] (0.100)2(0.100)( mol dm

–3)

3

= 2.53 x 10 –3

mol –2

dm6

s –1

( 3 sf )

1 mk

1 mk

d - At equilibrium for step 1 K = [X]___

[NO][H2]

- For step 2, slow or rate determining step ,- Rate = k [X][NO] = kK[NO] [H2] [NO] = kK[NO]2[H2]

- Mechanism agrees with experimental rate expression.

1 mk

1 mk1 mk

1 mk Max 3 mk

e Rate of formation of H2O(g) = 2 x rate of formation of N 2(g)

= 2 (2.53 x 10 –6

) mol dm –3

s-1

= 5.06 x 10 –6 mol dm

–3s-1 (3sf)

1 mk

1 mk

4 a (i)

(ii)

sp2

hybridisation

Side-by-side / sideways overlapping of 2 unhybridised 2p

orbitals

1 mk

1 mk

-2-http://edu.joshuatly.com/ http://fb.me/edu.joshuatly




4 b(i) Any one

H2C=CHCH2CH3 1 – butene

H2C=C(CH3)2 methylpropene

cyclobutane

methylcyclopropane

CH3

1 mk

(ii) CH3 CH3 CH3 H

C = C C = C

H H H CH3

cis 2 – butene trans 2 – butene

2 mk

c

or ( both are saturated hydrocarbons )

CH3

1 mk

d (i) - plastics have a wide range and versatile uses

- cheap

- chemically inert products

1 mk

1 mk

Any 2

(ii) CH3 CH3 CH3 CH3 CH3 CH3

C C C C C C

H H H H H H

1 mk

(iii) - Monomers are small monomers with with weak

intermolecular forces of attraction e.g. VDW or H-bonds

1 mk

- 3 -




SECTION B ( ESSAY ) SCORE REMARKS

5 a(i)

(ii)

- X : 1s2

2s22p

5 Group 17, non-metal , electronegative,

accepts electron

- Y : [Ar] 4s1 Group 1, metal, electropositive , donates

electron

•• ••

- Y○ + • X •• [ Y ]+

[ ○• X •• ] –

•• ••

- Involves transfer of electron from metal to non-metal.

YX is an ionic compound.

- Z : [Ar] 3d10

4s2

4p4 Group 16, non-metal ,

electronegative

- Both X and Z combine through sharing of electron pairs .

X2Z covalent compound.

•• ○○ ••

- •• X •○ Z ○• X ••

•• ○○ ••

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

Any 6

Max 6 mk

b(i)

(ii)

- Strong electrostatic forces of attraction exist between

oppositely-charged ions, Y+

and X –

, difficult for solid YX to

vapourise, making YX non – volatile.

- X2Z exists as simple covalent molecules. with weak VDW

forces between them.

- Thus X2Z have lower b.p. than YX , making it more volatile

than YX.

- In molten state or in aqueous solution, the ions in YX aremobile, making it an electrical conductor.

- X2Z being neutral molecules, are unable to conduct

electricity.

2 mk

1 mk

1 mk

Any 2

c(i) - Activation energy - minimum energy that must be

overcome by the reactant particles before a reaction can

occur, forming products

1 mk

(ii) Energy E a = activation energy

ΔH = enthalpy of reaction

Ea

R(g)+S(g)

ΔH = positive

P(g)+Q(g)

Reaction

2 mk

(iii) - Catalyst increases the rate of a reaction by providing an

alternative pathway, with a lower activation energy, Ea

- More reactant particles have energy equal or greater than

the lower activation energy,increasing the no. of effective

collisions.

1 mk

1 mk




6 a(i) 3 mk axes + correct

units

– 1 mk

correct plot

- 1 mk

balanced graph

- 1 mk

(ii) - straight line parallel to x-axis ( as shown ) 1 mk

(iii) - Gas M is a non-ideal gas, it does not obey Boyle’s law as

pV is not constant for all pressures of p.

1 mk

(iv) - Non-ideal gas shows ideal behaviour under low pressure.

At p = 0 kPa, pV = 6.83 x 103

kPa m3

- Assuming gas M is ideal, pV = mRT

Mr

Mr = mRT = ( 65.0 g ) ( 8.31 J K –1

mol –1

) ( 298 K )

pV 6.83 x 103

N m –2

m3

Mr = 23.6 ( 3sf )

1 mk

1 mk

1 mk Total 8 mk

b Pressure ( atm )

solid liquid gas

5.1 atm triplepoint

1 atmTemp.(

oC )

3 mk correct axes

- 1 mk

correct curves

- 1 mk

correct areas

- 1 mk

a(ii)




- Triple point for solid CO2 has a pressure above 1 atm., so

when temperature of a sample of solid CO2 increases, it

turns to gas without melting.

1 mk

Total 4 mk

c(i)

(ii)

- H2O Ý H+

+ OH –

Kw = [H+

][OH –

]

- At 25oC, Kw = 10

–14mol

2dm

–6

0.200 mol dm –3

NaOH [OH –

] = 0.200 mol dm –3

[H+] = 10

–14= 10

–14= 5.00 x 10

– 14mol dm

–3

[OH –

] 0.200

- pH = - log10 [H+] = 13.3 ( 3 sf )

1 mk

1 mk

1 mk Total 3 mk

7 a(i)

(ii)

- 2 Na + 2 H2O 2 NaOH + H2 (g)

Sodium burns & fizzes on the surface of the water.

- pH of strong alkali NaOH about 13

- Phosphorus has no reaction with water.

- 4 Na + O2 2 Na2O

Sodium burns in air vigorously.

- Na2O + H2O 2 NaOH

- 4 P + 5 O2 P4O10 OR 4 P + 3 O2 P4O6

- P4O10 + 6 H2O 4 H3PO4 OR P4O6 + 6 H2O 4 H3PO3

- pH for a weak acid, H3PO4 or H3PO3 about 3

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk Total 8 mk

b(i) - Electrolyte used is molten aluminium oxide.- Cryolite, Na3AlF6 is added to lower the melting point of

Al2O3 from 2050oC to 900

oC .

- Graphites are used as electrodes.

- At cathode : Al3+

(l) + 3e Al (s)

- At anode : 2O2 –

(l) O2 (g) + 4e

1 mkeach

Total 5 mk

(ii) - AlF3 is an ionic compound with covalent character, with

strong electrostatic forces of attraction between its Al3+

and F –

ions

- Cl –

ion bigger than F –

ion, polarised by Al3+

ion making

- aluminium chloride exists as simple covalent dimer Al2Cl6 molecules with weak intermolecular Van der Waals forces.

Any 2

Total 2 mk

8 a - 50.0 mg = 50.0 × 10 –3 g = 0.0500 g

1 mole HCN = 1.0 + 12.0 + 14.0 = 27.0 g

- No. of moles HCN = 0.0500/27.0 = 0.00185 ( 3 sf )

1 mk

1 mk

b(i) - Forward reaction is endothermic, according to Le

Chatelier’s principle

- high temperature will shift equilibrium to the right

increasing the yield of HCN.

1 mk

1 mk

(ii) - Forward reaction involves an increase in volume of gases- so high pressure will not favour production of HCN as

1 mk1 mk




equilibrium will shift to the left.

c NH3(g) + CH4(g) Ý HCN(g) + 3H2(g)

Initial/mol 0.200 0.200 0 0

Change/mol__- 0.100 - 0.100 + 0.100__ +3(0.100)

Eqm/mol 0.100 0.100 0.100 0.300

Kc = [HCN][H2]3

= (0.100)(0.300)3

= 0.270 mol2

dm –6

[NH3][CH4] (0.100)(0.100)

1 mk

2 mk

d (i) - NaCN is salt of weak acid HCN, under hydrolysis in water

CN –

+ H2O Ý HCN + OH –

giving an alkaline solution due to presence of OH –

ions

1 mk

(ii) - HCN + H2OÝ

H3O

+

+ CN

–

HCN undergoes partial dissociation in water.

NaCN Na+

+ CN –

( salt complete dissociates )

- Concentration of HCN and CN –

high in mixture.

- Addition of an acid, H+

+ CN – HCN

Addition of an alkali, OH –

+ HCN H2O + CN –

1 mk

1 mk

1 mk

(iii) - For a buffer, pH = pKa + log10 [ Salt ]

[ Acid ]

- pH = - log10 (4.90 × 10 –10

) + log10 (1.00 x 200 /400)

(0.500 x 200/400)= 9.61 ( 3 sf )

1 mk

1 mk

9 a - CH3 CH3

+ HNO3 + H2O

NO2

Mechanism for nitration – electrophilic substitution

- Formation of electrophile, nitronium ion

2 H2SO4 + HNO3 Ý

2 HSO4

–

+ NO2

+

+ H3O

+

fast

CH3 CH3

- slow

+ NO2+ intermediate

arenium / carbonium ion

NO2

CH3 CH3

-

+ HSO4 –

+ H2SO4 fast

catalyst regenerated

H NO2 NO2

1 mk

1 mk

1 mk

1 mk

+

+

- 7 -




b (i) SnCl2 is a weak reducing agent, but LiAlH4, is a very strong

reducing agent which can reduce - COOH group to –CH2OH

1 mk

(ii) - SnCl4 SnCl2 + Cl2(+4)4(-1) (+2)2(-1) 0

- Sn undergoes reduction, while Cl undergoes oxidation

so reaction is a redox reaction

1 mk

1 mk

(iii) - Cl

Sn Cl tetrahedral molecule

Cl Cl

- White fumes and a white solid are observed

- due to hydrolysis reaction.or SnCl4 + 2 H2O SnO2 (s) + 4 HCl (g)

1 mk

1 mk

1 mk

c - Sodium nitrite and dilute HCl are added to mixture.

- Effervescence of a colourless gas observed.

- HOOC – C6H4 – NH2 + O=N –OH HOOC – C6H4 – OH

+ N2(g) + H2O

OR

- A little aqueous copper(II) sulphate solution is added tothe mixture.

- A dark blue solution is observed.

- 2 H2N –C6H4 –COOH + Cu2+ H2N - OOC

C6H4 Cu2+

C6H4

COO- NH2

+ 2 H+

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

d (i)

(ii)

- Ethanol, concentrated sulphuric acid , reflux

- COOH + HOCH2CH3 COOCH2CH3

Ý + H2O

NH2 NH2 (esterification)

1 mk

1 mk

- 8 -




10 a CH3 CH3

CH3 – C – CH3 , CH3 – C – CH2Br

Br H

- 2-bromo-2-methylpropane is major product as its

formation involves stable 3o

free radical (CH3)3C•

1 mk

1 mk

1 mk

b(i)

b(ii)

H H H

X = CH3 – C = C – C* – CH2CH3

CH3

H2(g)

H H H

Y = CH3 – C – C – C* – CH2CH3

H H CH3

H H H

X = CH3 – C = C – C* – CH2CH3

CH3 Oxidation

H

CH3COOH + HOOC – C* – CH2CH3

CH3 ( Z )

- Both X and Y, each has a chiral C atom bonded to 4

different groups making it optically active.

- X is an alkene with one double bond as only 1 mole of

H2 can be added.

- X is unsaturated hydrocarbon but Y is saturated

- Bromine water OR acidified potassium manganate(VII)

is added to liquid X and liquid Y in separate test tubes.

- X decolourises reddish-brown bromine water OR

purple colour of acidified KMnO4.

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk

1 mk




c

- CH3CH=CHCH(CH3)CH2CH3 + Br2

CH3CH(Br)CH(Br)CH(CH3)CH2CH3

OR

KMnO4/H+

- CH3CH=CHCH(CH3)CH2CH3 CH3CH(OH)CH(OH)CH(CH3)CH2CH3

H

A = HOCH2 – C* – CH2CH3 ( reduction of RCOOH to

1o

ROH )

CH3

O H

B = Cl – C – C* – CH2CH3 ( - OH Cl )

CH3

H O

C = CH3CH2 – C – C – O –

Mg2+

CH3 2

1 mk

1 mk

1 mk

1 mk

stpm-trial-2012-chemistry-qa-perak.pdf

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