mock_test_(chemistry)_term 1_2015

YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan

NAME: CLASS: PRA U S

MOCK TEST 1 [SEPTEMBER 2015]

PAPER CHEMISTRY 1 DATE 15 September 2015

CODE 962/1 DAY Tuesday

COHORT STPM 2016 DURATION 1 hour 30 minutes

DURATION 1 hour 30 minutes TIME HR1130 – HR1300

INVIGILATOR(s) 1.

2.

SUBJECT TEACHER

MS UNG HIE HUONG

INSTRUCTION TO CANDIDATES:

1. This paper consists of Section A, Section B and Section C.

2. Answer ALL questions in Section A and Section B.Answer ANY TWO questions only in Section C.

3. For calculations, always show complete workings. Write your answer in correct significant figures and correct unit.

4. Arrange and stapler your answers in numerical order.

SECTION A (15 marks)

Answer ALL the questions in Section A. Blacken the corresponding answer on the objective answer sheet provided on page 4.

1. The mass spectrum of a gaseous element is shown below.

What can be deduced from the mass spectrum given?A The element has five isotopes.B The element exists as a diatomic gas.C The relative molecular mass of the element is 22.D The mass spectrum consists of four fragmental peaks and one molecular peak.

2. A quantity of 28 g of nitrogen is mixed with 32 g of oxygen at 298 K and 101 kPa. Which statement best describes the mixture of gases formed?

A More oxygen than nitrogen molecules are found in the mixture.B The average velocities of nitrogen and oxygen molecules are the same.C The average kinetic energies of nitrogen and oxygen molecules are the same.D There is no transfer of kinetic energy when nitrogen and oxygen molecules collide.

1


3. The percentage of iron in a haemoglobin molecule is 0.335%. If a haemoglobin molecule consists of four iron(III) ions, what is the relative molecular mass of haemoglobin?

A 6.66 × 102 B 4.16 × 103 C 1.67 × 104 D 6.66 × 104

4. The diagram below shows the electronic transitions between energy levels in the emission spectrum of atomic hydrogen.

Which electronic transition will produce spectral lines in the visible region?

5. Which of the following indicate the correct bonding in the species?

A C

B D

6. Which of the following species would have the smallest bond angle?

[ L = lone pair, A = central atom, M = terminal atom]

A AM4 C AM3L

B AM3 D AM2L2

7. Copper(I) oxide is a reddish-brown solid. In which orbitals are the valence electrons of copper(I) ion found?

A 3s B 4s C 3d D 3d and 4s

8. Which factor is the most significant in explaining the non-ideal behavior of the gases present in the reaction chamber in the Haber process?

A Presence of catalyst.

B High pressure of 150 atm.

C High temperature of 450°C.

D Strong bonds between atoms in the nitrogen molecule.

2


9. The rate equation for the reaction between X and Y is as follows:

Rate = k [X] [Y]2

When 0.20 mol gas X and 0.10 mol of gas Y are mixed in a 2.0 dm3 vessel at 300°C, the initial rate is 3.2 × 10–4 mol dm–3 s–1. Which statement is true of the reaction?

A The rate of reaction is four times lower in a 4.0 dm3 vessel.

B The numerical value of k is 1.28 at 300°C.

C The rate determining step is bimolecular.

D The unit of k is dm3 mol–1 s–1.

10. Which statement explains why catalysts are often used in chemical reactions?

A Catalysts increase the activation energies of reactions.

B Catalysts increase the yield of reaction products.

C Catalysts increase the enthalpy of reactions.

D Catalysts increase the rate of reactions.

11. Gas X decomposes when heated under a constant pressure P and temperature T to form an equilibrium mixture as shown below.

X(g) 2Y(g) + Z(g)

If the partial pressure of X is 14

P, what is the equilibrium constant, Kp,of the system at

temperature T?

A14

P2 B34

P2 C 4P2 D 8P2

12. Which is the correct observation when CaCO3 is heated at 800°C in a closed vessel?

A All the CaCO3 completely decomposed.

B The number of moles of CaO and CO2 differs.

C Only part of the CaCO3 decomposed even after prolonged heating.

D The pressure in the vessel will increase until no more CaCO3 is left.

13. Solid silver chloride is soluble in aqueous ammonia via the following equilibrium:

AgCl(s) + 2NH3(aq) [Ag(NH3)2]+(aq) + Cl–(aq)

Which of the following would most likely cause the reappearance of silver chloride?A Adding more ammonia.B Adding excess of dilute hydrochloric acid.C Adding ammonium nitrate.D Warming the mixture.

14. Pure water is a weak electrolyte. This indicates thatA Water is neutral.B Water is an amphoteric solvent.C Water undergoes partial dissociation.D The concentrations of H+ and OH– ions in water are the same.

15. Which of the following solutions has the highest concentration of H+ ions?A 1 mol dm–3 H2SO4 C 1 mol dm–3 H2CO3

3


B 1 mol dm–3 H3PO4 D 1 mol dm–3 HClO4

OBJECTIVE ANSWER SHEET

SECTION A (Question 1 – 15):

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

15

SECTION B (15 marks)

Answer ALL the questions in Section B. Write your answers in the spaces provided.

16. Dinitrogen tetroxide, N2O4 gas and nitrogen dioxide, NO2 gas exists in equilibrium as follows.

N2O4(g) 2NO2(g) ∆H = positive Colourless Brown

(a) 1.00 g of the above mixture occupies a volume of 380 cm3 at 60°C and 100 kPa. Calculate the average relative molecular mass of the mixture. [Gas constant, R = 8.31 J g–1 °C–1] [2 marks]

(b) Determine the mole fractions of N2O4 and NO2 in the mixture. [3 marks]

(c) Determine the equilibrium constant, Kp, for the reaction. [3 marks]

4

A B C D A B C D A B C D A B C D A B C D




17. A buffer solution is a solution that can resist change in pH when a small amount of acid or base is added.

(a) One of the buffer systems in the human blood is the carbonic acid/ bicarbonate ion system. The equilibrium is given as shown below:

H2CO3(aq) H+(aq) + HCO3–(aq)By using suitable equations, explain how the mixture acts as a buffer. [3 marks]

(b) The normal pH of human blood is maintained between 7.35 and 7.45 by buffer systems. State another buffer system in human blood. [1 mark]

(c) Calculate the pH of a buffer solution formed by mixing 100 cm3 of 0.050 mol dm–3

ethanoic acid and 50 cm3 of 0.20 mol dm–3 sodium ethanoate.[Ka for ethanoic acid is 1.7 × 10–5 mol dm–3 ] [3 marks]

SECTION C (30 marks)

Answer ANY TWO questions only in this section.Write your answers on the answer sheets on page 7-9.

18. (a) The figure below shows spectrum lines of the Balmer series in the emission spectrum of atomic hydrogen.

Using a labelled energy diagram, show how the line marked L on the spectrum is formed.[2 marks]

5


(b) (i) State two conditions when real gases behave almost like ideal gases. [2 marks]

(ii) Explain why xenon exhibits the greatest deviation from ideal behavior compared with other elements in Group 18. [2 marks]

(c) Helium has a triple point temperature of 1.0 K and critical point temperature of 5.0 K. Solid helium has the same density as liquid helium.

(i) Draw a labelled phase diagram for helium. [3 marks]

(ii) State and explain the effect on the melting point of helium when the pressure is increased. [2 marks]

(iii) Helium-5 is an unstable isotope of helium. The rate constant, k, for its radioactive disintegration is 9.12 × 1020 s–1. Determine the half-life of helium-5 and state why helium-5 is rarely found in nature. [4 marks]

19. (a) State Le Catelier’s principle. [1 mark]

(b) The Haber process for the manufacture of ammonia involves the following equilibrium.

N2(g) + 3H2(g) 2NH3(g) ∆H = –95 kJ mol–1

State and explain how the equilibrium composition of ammonia would change (if any) with the following alterations:

(i) Lowering the temperature, [3 marks]

(ii) Decreasing the pressure, [3 marks]

(iii) Addition of a suitable catalyst. [3 marks]

(c) A mixture containing 1 mol of nitrogen and 3 mol of hydrogen were allowed to achieve equilibrium at 180°C, 2000 atm and in the presence of a catalyst. The equilibrium mixture was found to contain 1.8 mol of ammonia.

(i) Determine the equilibrium constant Kc. [2 marks]

(ii) What can be said of the magnitude of Kc? [1 mark]

(iii) Such conditions are not practically used. Explain why. [2 marks]

20. (a) Mercury(II) chloride reacts with oxalate ion, C2O42– according to the equation:

2HgCl2(aq) + C2O42–(aq) 2HgCl(s) + 2CO2(g) + 2Cl–(aq)

A kinetic study of the reaction gives the following results:

ExperimentInitial concentration/ mol dm–3 Initial rate

( × 10–3 / mol dm–3 min–1 )[HgCl2] [C2O42–]

I 0.068 0.035 0.230II 0.068 0.14 4.16III 0.102 0.035 0.345

(i) Determine the rate equation for the reaction. [6 marks]

(ii) Suggest a reaction mechanism that is consistent with the rate equation. [3 marks]

6


(b) Phenol, C6H5OH, is a weak organic acid. A solution containing 0.385 g of phenol in 2.00 dm3 solution has a pH of 6.29 at 25°C.

Calculate:

(i) the molar concentration of the phenol solution, [2 marks]

(ii) the acid dissociation constant of phenol at 25°C. [4 marks]

7


ANSWER SHEETSECTION C Examiner’s

use onlyQuestion Answers

8


SECTION C Examiner’s use onlyQuestion Answers

9


SECTION C Examiner’s use onlyQuestion Answers

10

Periodic Table (Jadual Berkala)

Group (Kumpulan)1(I)

2(II)

3 4 5 6 7 8 9 10 11 12 13(III)

14(IV)

15(V)

16(VI)

17(VII)

18(VIII)

1.0H

1

4.0He

26.9Li

3

9.0Be

4

aX

b

a = relative atomic mass (jisim atom relatif)X = atomic symbol (symbol atom)b = atomic number (nombor atom)

10.8B

5

12.0C

6

14.0N

7

16.0O

8

19.0F

9

20.2Ne

1023.0Na

11

24.3Mg

12

27.0Al

13

28.1Si

14

31.0P

15

32.1S

16

35.5Cl

17

40.0Ar

1839.1

K19

40.1Ca

20

45.0Sc

21

47.9Ti

22

50.9V

23

52.0Cr

24

54.9Mn

25

55.8Fe

26

58.9Co

27

58.7Ni

28

63.5Cu

29

65.4Zn

30

69.7Ga

31

72.6Ge

32

74.9As

33

79.0Se

34

79.9Br

35

83.8Kr

3685.5Rb

37

87.6Sr

38

88.9Y

39

91.2Zr

40

92.9Nb

41

95.9Mo

42

[98]Tc

43

101Ru

44

103Rh

45

106Pd

46

108Ag

47

112Cd

48

115In

49

119Sn

50

122Sb

51

128Te

52

127I

53

131Xe

54133Cs

55

137Ba

56

139La

57

178Hf

72

181Ta

73

184W

74

186Re

75

190Os

76

192Ir

77

195Pt

78

197Au

79

201Hg

80

204Tl

81

207Pb

82

209Bi

83

[209]Po

84

[210]At

85

[222]Rn

86[223]

Fr87

[226]Ra

88

227Ac

89

[261]Rf

104

[262]Db

105

[266]Sg

106

[264]Bh

107

[269]Hs

108

[268]Mt

109

[281]Ds

110

[272]Rg

111

[285]Cn

112

140Ce

58

141Pr

59

144Nd

60

[145]Pm

61

150Sm

62

152Eu

63

157Gd

64

159Tb

65

163Dy

66

165Ho

67

167Er

68

169Tm

69

173Yb

70

175Lu

71232Th

90

231Pa

91

238U

92

237Np

93

[244]Pu

94

[243]Am

95

[247]Cm

96

[247]Bk

97

[251]Cf

98

[252]Es

99

[257]Fm

100

[258]Md

101

[259]No

102

[262]Lr

103

The proton numbers and approximate relative atomic masses shown in the table are for use in the examination unless stated otherwise in an individual question. (Nombor proton dan anggaran jisim atom relatif yang ditunjukkan dalam jadual adalah untuk digunakan dalam peperiksaan kecuali yang sebaliknya dinyatakan dalam soalan

yang tertentu.)

11


MARKING SCHEME

MOCK TEST 1 [SEPTEMBER 2015]

CHEMISTRY 1 (962/1)

Q RUBRIC M

SECTION A [15 marks]

1 B 1

2 C 1

3 D 1

4 C 1

5 A 1

6 D 1

7 C 1

8 B 1

9 B 1

10 D 1

11 A 1

12 C 1

13 B 1

14 C 1

15 A 1

SECTION B [15 marks]

16 apV=nRT @ pV=¿ ( m

M r) RT @ M r=¿

mRTpV

M r=¿ (1.00 ) (8.31 )(273+60)

(100×103 )(380× 10−6)

M r=72.8

1

1

b Let: Mole fraction of N2O4 = x Mole fraction of NO2 = (1 – x)

Therefore, 46 x+92 (1−x )=72.8

x=0.417

∴ Mole fraction of N2O4 = 0.417

Mole fraction of NO2 = 0.583

1

1

1

cK p=¿

(PNO2)2

PN2 O4

¿ (0.417 × 100)2

(0.583 ×100)

1

1

1

12


Q RUBRIC M

¿29.8 kPa

17 a The buffer solution consists of undissociated weak acid, H2CO3 and its conjugate

base, HCO3–.

When [acid] increased,

The equilibrium position will shift to the left to form undissociated H2CO3. The added H+ is removed and pH remains constant.

H+(aq) + HCO3–(aq) H2CO3(aq)

Then, the unstable H2CO3 will decompose into CO2 and H2O. CO2 is removed from the blood via exhalation.

H2CO3(aq) CO2(g) + H2O(l)

Extra OH– is neutralized/ removed by reaction with H2CO3:

H2CO3(aq) + OH–(aq) H2O(l) + HCO3–(aq)

NOTE: [H+] and [OH–] remains constant, hence the pH of buffer solution remains unchanged.

1

1

1

b Either one: Amino acid, H2NCHRCOOH H2PO4– / HPO42– system

NOTE: WCR

1

c[CH3COOH] =

0.050 ×100(100+50)

= 1

30 mol dm–3

[CH3COONa] = 0.20 ×50(100+50)

= 1

15 mol dm–3

∴ pH=p Ka+ log10 [CH3 COONa ][CH3 COOH ]

pH=−log10 (1.7 × 10−5 )+ log10 ( 1

15 )( 1

30 )pH of thebuffer solution=5.1

1

1

1

SECTION C [45 marks]

18 a

13


Q RUBRIC M

Draw all energy levels from n1 until n5 Draw arrow from n5 to n2

1

1

b(i) Low pressure High temperature

11

18 b(ii) Xe has the largest atomic size in Group 18 and the most number of electrons. Intermolecular forces and volume of gas particles/ atoms cannot be ignored.

11

c(i) Drawn and labelled axes + boiling curve + sublimation curve Correct melting line (vertical line) + Label phases Mark and state temperatures for Triple point = 1.0 K

Critical point = 5.0 K

NOTE:1. If axes are not labelled 0 mark2. Curves must have positive gradient (upwards from left to right)3. The three phase transition lines/ curves must meet at triple point

11

1

c(ii) There is no change in volume during the phase change between solid and liquid helium.

Thus, pressure will not affect the melting point of helium.

1

1

c(iii) Radioactive disintegration is a first order reaction.

t 12

=¿ ln 2k

t 12

=¿ ln2

9.12× 1020

t 12

=7.60 ×10−22 s

Helium-5 is rarely found in nature because it has a very short half-life.

1

1

1

1

19 (a) When a system in dynamic equilibrium is subjected to change,… the system will react… to remove the effect of the change… so that equilibrium is re-established.

1

b(i) Forward reaction is exothermic. 1

14


Q RUBRIC M

Lowering the temperature will force the equilibrium position to shift to the right-hand side.

More ammonia will be produced// [Ammonia] will increase.

1

1

19 b(ii) Reverse reaction involves an increase in the number of (moles of) gaseous particles.

Lowering the pressure will force the equilibrium position to shift to the left-hand side.

Less ammonia will be produced// [Ammonia] will decrease.

1

1

1

b(iii) A catalyst does not alter the equilibrium position (and the equilibrium composition).

It just increases the rates of forward and reverse reaction (by the same factor) so that equilibrium is achieved in a shorter time.

[Ammonia] remain the same.

1

1

1

c(i) N2 (g )+3 H 2 ( g )⇔

2 NH3(g)

Initial/ mol : 1 3 0Change/ mol : (1 – 0.9) (3 – 2.7) 1.8Equilibrium/ mol : 0.1 0.3 1.8

K c=¿ [NH 3]

2

[ N2][ H 2]3

K c=¿ (1.8)2

(0.1 )(0.3)3

K c=1200 mol−2 dm6

1

1

1

c(ii) The magnitude of K c is large. There is a high conversion of N2 and H 2 to NH 3. 1

c(iii) At low temperature (180°C), the yiels of ammonia is high. However, the rate of reaction at such a low temperature would be very slow// It takes a long time for the yield to be achieved. In order to increase the rate of reaction, a catalyst (iron) is added.

High pressure (2000 atm) would ensure a high yield. However, if the pressure is too high, the cost of production would also be high because the pipes and storage tanks have to be thick enough to withstand the pressure.

1

1

20 a(i) Let the rate equation be: Rate = k [HgCl2]a [C2O42– ]b

Experiment IIExperiment I

:4.16

0.230=( 0.14

0.035 )b

18.1=4b

b=2.1 ≈ 2

Experiment IIIExperiment I

:0.3450.230

=( 0.1020.068 )

a

1.5=1.5a

1

1

1

1

15


Q RUBRIC M

20

a=1

Hence, the rate equation is: Rate = k [HgCl2] [C2O42– ]2

k=¿ 0.230

(0.068 )(0.035)2 ¿2.8 ×103 mol−2 dm6 s−1

1

1

a(ii) HgCl2 + 2C2O42– slow→

Hg + 2Cl– + CO2 + [C2O4 • CO2] 2–

Hg + HgCl2 fast→

2HgCl

[C2O4 • CO2] 2– fast→

CO2 + C2O42–

NOTE: Show 3 steps mechanism.Slow/ Rate determining step involves 1 mol of HgCl2 and 2 moles of

C2O42–

The actual mechanism:

HgCl2 + C2O42– fast→

[HgCl2• C2O4]2–

[HgCl2• C2O4]2– + C2O42– slow→

Hg + 2Cl– + CO2 + [C2O4 • CO2] 2–

Hg + HgCl2 fast→

Hg2Cl2

[C2O4 • CO2] 2– fast→

CO2 + C2O42–

1

1

1

b(i)Number of moles of phenol =

0.3856 (12.0 )+6 (1.0 )+16.0

= 0.38594.0

mol = 0.004096 mol

Molar concentration of phenol = 0.004096

2.00 mol dm–3

= 2.05 × 10–3 mol dm–3

1

1

a(ii) Dissociation of phenol: C6H5OH(aq) ⇔

C6H5O–(aq) + H+(aq)

Ka=¿ ¿¿

Ka=¿ ¿¿¿

Ka=¿ (10−6.29 )2

(2.05×10−3 )

Ka=1.28× 10−10 moldm−3

NOTE: Calculation of degree of dissociation

1

1

1

1

16


Q RUBRIC M

[H+] = c∝ @ ∝=¿ ¿¿

∝=¿ 10−6.29

2.05× 10−3

∝=2.50 ×10−4 @0.0250 %

17

mock_test_(chemistry)_term 1_2015

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