MOCK_TEST_(chemistry)_Term 1_2015

YEE 6301 SMK MERADONG

Kecergasan Untuk Kemajuan

NAME:

CLASS: PRA U S MOCK TEST 1 [SEPTEMBER 2015]

PAPER CODE COHORT DURATION

CHEMISTRY 1 962/1 STPM 2016 1 hour 30 minutes

INVIGILATOR(s)

1.

DATE DAY DURATION TIME SUBJECT TEACHER

2.

15 September 2015 Tuesday 1 hour 30 minutes HR 1130 – HR1300

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. Relative intensity

10

11

20

21

22

m e

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

YEE 6301 SMK MERADONG

Kecergasan Untuk Kemajuan

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

× 102

6.66

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? n=5 n=4

Energy

n=3 n=2

A

B

C

n=1

D

5. Which of the following indicate the correct bonding in the species? H O A C H O N H N H O H H

B H

C

H2O

D C

H2 O

N

H2O

H

Cl

M

6. Which of the following species would have the smallest [ L = lone pair, A = central atom, M = terminal atom] A AM4 C AM3L B AM3 D AM2L2

Cu H2O

A

2+

OH2 OH2

M

bond angle?

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

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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 dm 3 vessel at 300°C, the initial rate is 3.2 A B C D

× 10–4 mol dm–3 s–1. Which statement is true of the reaction?

The rate of reaction is four times lower in a 4.0 dm3 vessel. The numerical value of k is 1.28 at 300°C. The rate determining step is bimolecular. 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

1 4 P, what is the equilibrium constant, Kp,of the system at

temperature T? A

1 4 P2

B

3 4 P2

C

4P 2

D

8P 2

12. Which is the correct observation when CaCO3 is heated at 800°C in a closed vessel? All the CaCO3 completely decomposed. A The number of moles of CaO and CO2 differs. B Only part of the CaCO3 decomposed even after prolonged heating. C The pressure in the vessel will increase until no more CaCO3 is left. D 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 that A 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 C 1 mol dm–3 H2SO4 1 mol dm–3 H2CO3 3

B

1 mol dm– H3PO4 3

YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan D 1 mol dm–3 HClO4 OBJECTIVE ANSWER SHEET

SECTION A (Question 1 – 15): 1 2 3 4 5

A

B

A

B

A

B

A

B

A

B

 

 

 

 

 

 

 

 

 

 

C C C C C

 

 

 

 

 

D D D D D

6 7 8 9 10

A

B

A

B

A

B

A

B

A

B

 

 

 

 

 

 

 

 

 

 

C C C C C

 

 

 

 

 

D D D

11 B

A

B

A

B

A

B

A

B

 

 

C

 

D

12  

 

C

 

D

13

D D

A

 

 

C

 

D

14  

 

C

 

D

15  

 

C

 

D

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) Colourless

2NO2(g)

∆H = positive

Brown

(a) 1.00 g of the above mixture occupies a volume of 380 cm 3 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.

4

[3 marks]

YEE 6301 SMK MERADONG

Kecergasan Untuk Kemajuan

(c) Determine the equilibrium constant, Kp, for the reaction.

[3 marks]

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 ]

SECTION C (30 marks) Answer ANY TWO questions only in this section. Write your answers on the answer sheets on page 7-9.

5

[3 marks]

YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan 18. (a) The figure below shows spectrum lines of the Balmer series in the emission spectrum of atomic hydrogen. L

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

Initial concentration/ mol dm–3 6

Initial rate

YEE 6301 SMK MERADONG I II III

[HgCl2]

[C2O42–]

0.068 0.068 0.102

0.035 0.14 0.035

Kecergasan Untuk Kemajuan (

× 10–3 / mol dm–3 min–1 ) 0.230 4.16 0.345

(i) Determine the rate equation for the reaction. (ii) Suggest a reaction mechanism that is consistent with the rate equation.

[6 marks] [3 marks]

(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

YEE 6301 SMK MERADONG ANSWER SHEET

Kecergasan Untuk Kemajuan

SECTION C Question

Answers

Examiner’s use only

SECTION C Question

Answers

Examiner’s use only

8

YEE 6301 SMK MERADONG

SECTION C Question

Answers

9

Kecergasan Untuk Kemajuan

Examiner’s use only

YEE 6301 SMK MERADONG

10

Kecergasan Untuk Kemajuan

Periodic Table (Jadual Berkala) 1 (I) 1.0 H

2 (II)

6.9 Li

9.0 Be

3

4

5

6

7

8

Group (Kumpulan) 9 10

11

12

13 (III)

14 (IV)

15 (V)

16 (VI)

17 (VII)

5

6

7

8

9

27.0 Al 13 69.7 Ga 31 115 In 49 204 Tl 81

28.1 Si 14 72.6 Ge 32 119 Sn 50 207 Pb 82

31.0 P 15 74.9 As 33 122 Sb 51 209 Bi 83

32.1 S 16 79.0 Se 34 128 Te 52 [209] Po 84

35.5 Cl 17 79.9 Br 35 127 I 53 [210] At 85

18 (VIII) 4.0 He 2 20.2 Ne 10 40.0 Ar 18 83.8 Kr 36 131 Xe 54 [222] Rn 86

10.8 B

12.0 C

14.0 N

16.0 O

19.0 F

163 Dy 66 [251] Cf 98

165 Ho 67 [252] Es 99

167 Er 68 [257] Fm 100

169 Tm 69 [258] Md 101

173 Yb 70 [259] No 102

175 Lu 71 [262] Lr 103

1

3

4

23.0 Na 11 39.1 K 19 85.5 Rb 37 133 Cs 55 [223] Fr 87

24.3 Mg 12 40.1 Ca 20 87.6 Sr 38 137 Ba 56 [226] Ra 88



a X b

45.0 Sc 21 88.9 Y 39 139 La 57 227 Ac 89

47.9 Ti 22 91.2 Zr 40 178 Hf 72 [261] Rf 104

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

50.9 V 23 92.9 Nb 41 181 Ta 73 [262] Db 105

52.0 Cr 24 95.9 Mo 42 184 W 74 [266] Sg 106

54.9 Mn 25 [98] Tc 43 186 Re 75 [264] Bh 107

55.8 Fe 26 101 Ru 44 190 Os 76 [269] Hs 108

58.9 Co 27 103 Rh 45 192 Ir 77 [268] Mt 109

58.7 Ni 28 106 Pd 46 195 Pt 78 [281] Ds 110

63.5 Cu 29 108 Ag 47 197 Au 79 [272] Rg 111

65.4 Zn 30 112 Cd 48 201 Hg 80 [285] Cn 112

140 Ce 58 232 Th 90

141 Pr 59 231 Pa 91

144 Nd 60 238 U 92

[145] Pm 61 237 Np 93

150 Sm 62 [244] Pu 94

152 Eu 63 [243] Am 95

157 Gd 64 [247] Cm 96

159 Tb 65 [247] Bk 97

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.

11



(Nombor proton dan anggaran jisim atom relatif yang ditunjukkan dalam jadual adalah untuk digunakan dalam peperiksaan kecuali yang sebaliknya dinyatakan dalam soalan yang tertentu.)

12

YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan 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] 1 6

a

pV =nRT @ pV =¿

( Mm ) r

M r=¿

RT @ M r=¿

mRT pV

1

( 1.00 )( 8.31 )( 273+60)

( 100× 103 ) (380 ×10−6 )

1

M r=72.8 b

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

1

46 x +92 ( 1−x )=72.8 x=0.417

1

∴

Mole fraction of N2O4 = 0.417

Mole fraction of NO2 = 0.583

13

1

YEE 6301 SMK MERADONG Q

Kecergasan Untuk Kemajuan

RUBRIC c

M

2

K p=¿

(PNO ) PN O 2

2

1

4

1

2

(0.417 ×100) ( 0.583× 100)

¿

1

¿ 29.8 kPa 1 7

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 H 2CO3. 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)

1

1

1

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

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

c

1

WCR

[CH3COOH] =

0.050× 100 (100+50)

=

1 30

mol dm–3

[CH3COONa] =

0.20 × 50 (100+50)

=

1 15

mol dm–3

∴

pH= p K a +log 10

[CH 3 COONa ] [ CH 3 COOH ]

1

1 1

14

YEE 6301 SMK MERADONG Q

Kecergasan Untuk Kemajuan

RUBRIC

M

( 151 ) ( 301 )

pH=−log 10 ( 1.7 × 10−5 ) +log 10

pH of the buffer solution=5.1

SECTION C [45 marks] 1 8

a

n=5 n=4

Energy L

n=3 n=2

n=1

1 8

 

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

1

b(i)

 

Low pressure High temperature

1 1

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.

1 1

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

1

1

Pressure/ atm

Liquid

Solid

Gas

1.0

5.0

NOTE: 15

Temperature/ K

1 1

YEE 6301 SMK MERADONG Q

Kecergasan Untuk Kemajuan

RUBRIC

M

1. If axes are not labelled  0 mark 2. Curves must have positive gradient (upwards from left to right) 3. The three phase transition lines/ curves must meet at triple point c(ii)

 

c(iii)

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

Radioactive disintegration is a first order reaction. 

t 1 =¿ 2

t 1 =¿ 2

ln 2 k

1

ln 2 20 9.12× 10

1 1

−22

t 1 =7.60× 10

s

2

1  1 9

1 9

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

(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



Lowering the temperature will force the equilibrium position to shift to the righthand side.

1



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

1



1



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.



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

1



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

1



[Ammonia] remain the same.

1

b(ii)



b(iii)

c(i)

N 2 ( g )+3 H 2 ( g ) 2 NH 3 ( g)



1 (1 – 0.9) 0.1

3 (3 – 2.7) 0.3

1

1

⇔

Initial/ mol : Change/ mol : Equilibrium/ mol :

1

0 1.8 1.8

2



K c =¿

[NH 3 ] 3 [N 2 ][H 2]

1 16

YEE 6301 SMK MERADONG Q

Kecergasan Untuk Kemajuan

RUBRIC

M

2

(1.8) ( 0.1 )( 0.3)3



K c =¿



K c =1200 mol dm

−2

6

1 c(ii)

Kc

N2

H2

to

1



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.

1



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.

The magnitude of

is large. There is a high conversion of

and

NH 3 . c(iii)

2 0

a(i)

1

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



Experiment II 4.16 0.14 : = Experiment I 0.230 0.035

(

b

)

1

b

18.1=4

1

b=2.1≈ 2 1



Experiment III 0.345 0.102 : = Experiment I 0.230 0.068

(

a

)

1

1.5=1.5

a

1

a=1 1  2 0



a(ii)

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

k =¿

0.230 ( 0.068 ) (0.035)2

HgCl2 + 2C2O42– Hg + HgCl2

fast →

slow →

3

−2

6 −1

¿ 2.8 ×10 mol dm s

Hg + 2Cl– + CO2 + [C2O4 • CO2] 2–

1 1

2HgCl 17

YEE 6301 SMK MERADONG Q

Kecergasan Untuk Kemajuan

RUBRIC

fast

[C2O4 • CO2] 2–

M 1

CO2 + C2O42–

→

NOTE: Show 3 steps mechanism. Slow/ Rate determining step involves 1 mol of HgCl2 and 2 moles of C2O42– The actual mechanism:

fast

HgCl2 + C2O42–

→

[HgCl2• C2O4]2–

slow

[HgCl2• C2O4]2– + C2O42– Hg + HgCl2

fast

[C2O4 • CO2] 2–

Hg + 2Cl– + CO2 + [C2O4 • CO2] 2–

→

Hg2Cl2

→

fast →

CO2 + C2O42–

0.385 6 ( 12.0 )+ 6 (1.0 )+ 16.0

b(i) Number of moles of phenol

=

=

0.385 94.0

mol

=

0.004096 mol 1 Molar concentration of phenol =

0.004096 2.00

= 2.05 a(ii)

Dissociation of phenol: C6H5OH(aq)

K a=¿

mol dm–3

1

× 10–3 mol dm–3 ⇔

C6H5O–(aq) + H+(aq)

1 1

−¿ C 6 H 5 O¿ ¿ +¿ H¿ ¿ ¿ ¿

1

1

18

YEE 6301 SMK MERADONG Q

RUBRIC

K a=¿

+¿¿ H ¿ ¿2 ¿ ¿ ¿ 2

K a=¿

( 10−6.29 )

( 2.05× 10−3 )

K a=1.28 × 10−10 mol dm −3

NOTE: Calculation of degree of dissociation [H+] =

∝=¿

c∝

@

∝=¿

+¿¿ H ¿ ¿ ¿

10−6.29 2.05 × 10−3 −4

∝=2.50 ×10 @ 0.0250

19

Kecergasan Untuk Kemajuan M