MOCK_TEST_(chemistry)_Term 1_2015

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 HR_{1130 –} 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.

Relative intensity

m e 10 11 20 21 22

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.

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?

Energy n = 5_{n = 4}

n = 3 n = 2

n = 1 A B C D

5. Which of the following indicate the correct bonding in the species? A H O N O O C N H H H H Cl B C H H H C N D Cu H₂O H2O OH2 OH₂ H₂O H₂O 2+

6. Which of the following species would have the smallest M A M 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.

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

1

₄

_{P, what is the equilibrium constant, Kp,of the system at} temperature T?

A

1

₄

P2 B

3

₄

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 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.}

      A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D       A B C D B 1 mol dm–3 _H 3PO4 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]}

(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 ] [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.

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:

(

×

10–3 / mol dm–3 min–1 )

[HgCl2] [C2O42–]

I 0.068 0.035 0.230

II 0.068 0.14 4.16

III 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]

(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]

ANSWER SHEET SECTION C Examiner’s use only Question Answers SECTION C Examiner’s use only Question Answers

SECTION C Examiner’s use only

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) 1.0 H 1 4.0 He 2 6.9 Li 3 9.0 Be 4 a X b

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

b = atomic number (nombor atom)

10.8 B 5 12.0 C 6 14.0 N 7 16.0 O 8 19.0 F 9 20.2 Ne 10 23.0 Na 11 24.3 Mg 12 27.0 Al 13 28.1 Si 14 31.0 P 15 32.1 S 16 35.5 Cl 17 40.0 Ar 18 39.1 K 19 40.1 Ca 20 45.0 Sc 21 47.9 Ti 22 50.9 V 23 52.0 Cr 24 54.9 Mn 25 55.8 Fe 26 58.9 Co 27 58.7 Ni 28 63.5 Cu 29 65.4 Zn 30 69.7 Ga 31 72.6 Ge 32 74.9 As 33 79.0 Se 34 79.9 Br 35 83.8 Kr 36 85.5 Rb 37 87.6 Sr 38 88.9 Y 39 91.2 Zr 40 92.9 Nb 41 95.9 Mo 42 [98] Tc 43 101 Ru 44 103 Rh 45 106 Pd 46 108 Ag 47 112 Cd 48 115 In 49 119 Sn 50 122 Sb 51 128 Te 52 127 I 53 131 Xe 54 133 Cs 55 137 Ba 56 139 La 57 178 Hf 72 181 Ta 73 184 W 74 186 Re 75 190 Os 76 192 Ir 77 195 Pt 78 197 Au 79 201 Hg 80 204 Tl 81 207 Pb 82 209 Bi 83 [209] Po 84 [210] At 85 [222] Rn 86 [223] Fr 87 [226] Ra 88 227 Ac 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 112 140 Ce 58 141 Pr 59 144 Nd 60 [145] Pm 61 150 Sm 62 152 Eu 63 157 Gd 64 159 Tb 65 163 Dy 66 165 Ho 67 167 Er 68 169 Tm 69 173 Yb 70 175 Lu 71 232 Th 90 231 Pa 91 238 U 92 237 Np 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.



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 =

¿

(

_M

m

r

)

RT @ M

r

=

¿

mRT

pV

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

Q RUBRIC M c

K

p

=

¿

(

P

NO2

)

2

P

_N2O4

¿

(0.417 ×100) 2 (0.583× 100)

¿29.8 kPa 1 1 1 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 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] =

(100+50)

0.20 × 50

=

1

15

mol dm–3 ∴

pH= p K

_a

+log

₁₀

[

CH

3

COONa

]

[

CH

₃

COOH

_]

1 1 1

Q RUBRIC M

pH=−log

10

(

1.7 × 10

−5

)

+log

10

(

1

15

)

(

1

30

)

pH of the buffer solution=5.1 SECTION C [45 marks] 1 8 a Energy n = 5n = 4 n = 3 n = 2 n = 1 L

 Draw all energy levels from n1 until n5

 _{Draw arrow from n5 to n2} 1₁

b(i) _ Low pressure

 High temperature

1 1 1

8

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

Temperature/ K

Pressure/ atm

5.0

1.0

Solid

Liquid

Gas

1 1 1

Q 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) _ 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 2

k

t

₁ 2

=

¿

ln 2

9.12× 10

20

t

1 2

=7.60× 10

−22

_s

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

1 1 1 1 1 9

(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.

 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 1 1

9

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 H2

(

g

)

⇔2 NH3(g) Initial/ mol : 1 3 0 Change/ mol : (1 – 0.9) (3 – 2.7) 1.8 Equilibrium/ mol : 0.1 0.3 1.8 

K

c

=

¿

[

NH

3

]

2

[

N

₂

][

H

₂

]

3 1

Q RUBRIC M 

K

c

=

¿

(1.8)

2

(

0.1)(0.3)

3 

K

c

=1200 mol

−2

dm

6 1 c(ii) _{The magnitude of}

K

_{c is large. There is a high conversion of}

N

_{2 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 2 0 2 0

a(i) _{Let the rate equation be: Rate = k [HgCl2]}

a

_[C2O42–]

b



Experiment II

Experiment I

:

4.16

0.230

=

(

0.14

0.035

)

b

18.1=4

b b=2.1≈ 2



Experiment III

Experiment I

:

0.345

0.230

=

(

0.102

0.068

)

a

1.5=1.5

a a=1

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

 k =¿

₍

_0.0680.230

₎

_(0.035)2

¿

2.8 ×10

3

mol

−2

dm

6

s

−1 1 1 1 1 1 1 a(ii)

HgCl2 + 2C2O42–

slow

→ Hg + 2Cl– + CO2 + [C2O4• CO2] 2–

Hg + HgCl2

fast

→ 2HgCl

1 1

Q RUBRIC M [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

b(i)

Number of moles of phenol =

0.385

6

(

12.0

)

+6

(

1.0

)

+16.0 =

0.385

94.0

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)

K

_a

=

¿

−

¿

C

₆

H

₅

O

¿

¿

+

¿

H

¿

¿

¿

¿

1 1 1 1

Q RUBRIC M

K

_a

=

¿

+

¿

H

¿

¿

¿

2

¿

¿

¿

K

_a

=

¿

(

10

−6.29

₎

2

(

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

∝=2.50 ×10

−4

@ 0.0250