CADANGAN JAWAPAN PSPM 2 20182019

CHEMISTRY SK025 PSPM II 2018/2019

Answer all questions.

1. a) On heating, dinitrogen pentoxide, N2O5 decomposes as follow:

2N2O5 (g) 4NO2 (g) + O2 (g)

The reaction is first order with respect to N2O5.

i. Determine the rate of formation of O2 if the rate of disapperance of
N2O5 is 1.25 x 10-2 M min-1.

Given : − [ ] = 1.25 x 10-2 M min-1

+ [ ] = − [ ]



= ( . − )



= 6.25 x 10-3 M min-1

ii. Calculate the rate constant for this reaction if 25% of N2O5 is
decomposed within 10 minutes.

ln [N2O5]o - ln [N2O5] = kt

ln (100) – ln (75) = k (10)
k = 0.0287 min-1

iii. Determine the half-life for this decomposition reaction.

[7 marks]

/ =


=
.

= 24.15 min

b) TABLE 1 shows the rate constants for the decomposition of N2O5 at two
different temperatures.

TABLE 1

Temperature (oC) Rate constant (s-1)
25.0 1.74 x 10-5
35.0 6.61 x 10-5

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CHEMISTRY SK025 PSPM II 2018/2019

i. Determine the activation energy, Ea for this reaction in kJmol-1.

ln = ( − )



. − ( - )
. . .
In . − =

Ea = 101950 J mol-1

= 101.95 kJ mol-1

ii. Activation energy can also be determined graphically. Sketch the graph
and explain briefly.
[7 marks]

In k

m= -Ea/R

1 / T (K-1)

From Arrhenius equation : k = A −
In k = ln A -



y = c + mx

Thus, by plotting graph ln k against 1/T, a straight line graph with

negative slope form. Activation energy can be determined from

gradient of the graph.

Gradient, m = −


Ea = - m x R

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CHEMISTRY SK025 PSPM II 2018/2019

2. a) In an experiment to measure the enthalpy change of neutralisation, 25.0 ml
of 1.0 M HCl is placed in a polystyrene cup. At the same temperature, 25.0 ml
of 1.0 M NaOH is added and stirred. The temperature of the resulting solution
increases by 7.0oC. Calculate the enthaply of neutralisation for this reaction.

[Assume the density and specific heat capacity of the final solution are the

same as water].

[6 marks]

NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)

-qreleased = +qabsorbed

-qrxn = qsoln

= mscs∆T

= (50) (4.18) (7.0)

= 1463 J

No. of mole HCl and NaOH = MV/1000
= (1.0)(25) /1000
= 0.025 mol

Thus, mole of H2O = 0.025 mol
0.025 mol H2O = - 1463 J
1 mol H2O = - 58.52 k J
∆Hn = - 58.52 k J mol-1

b) Thermochemical data for the formation of sodium oxide, Na2O is given in
TABLE 2.

TABLE 2

Enthalpy ∆Ho (kJmol-1)

Ionisation energy of sodium +496

Enthalpy of formation of sodium oxide -414

First electron affinity of oxygen -141

Second electron affinity of oxygen +750

Enthalpy of atomisation of oxygen +249

Lattice energy of sodium oxide -2478

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CHEMISTRY SK025 PSPM II 2018/2019

Based on TABLE 2,
i. construct a Born-Haber cycle for sodium oxide.

2Na (s) + ½ O2 ∆Hf = -414 kJ Na2O (s)
(g)

∆Ha= ? x 2 ∆Ha= 249 kJ

2Na (g) O (g)

IE = 496kJ x 2 EA1 = -141 kJ ∆Hlattice = -2478 kJ
O- (g)

EA2 = 750 kJ

2Na+ (g) + O2- (g)
ii. calculate the enthalpy of atomisation of sodium.

[8 marks]

∆Hf Na2O = (∆Ha Na x 2) + (IE Na x 2) + ∆Ha O + EA1 O + EA2 O + ∆Hlattice
-414 = (∆Ha Na x 2) + (496 x 2) + 249 -141 + 750 -2478

∆Ha Na = 107 kJmol-1

3. a) A galvanic cell is constructed based on the notation below:

Zn(s) | Zn2+(aq) || Ag+ (aq) | Ag(s)

The cell potential is 1.58 V when concentrations of Zn2+ and Ag+ are x M and

0.15 M, respectively. Calculate the value of x.

[Given: Eocell = + 1.56 V]

[4 marks]

Anode : Zn (s) → Zn2+ (aq) + 2e-

Cathode : 2Ag+ (aq) + 2e- → Ag (s)_______

Overall : Zn (s) + 2 Ag+ (aq) → Zn2+ (aq) + Ag (s)

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CHEMISTRY SK025 PSPM II 2018/2019

b) A charge of 0.025 F was passed through an aqueous solution of Mn+ using a
current of 2.00 A. An amount of 0.79 g of M was deposited at the cathode.

[Given molar mass of M is 63.2 g mol-1]

i. Calculate the time taken in minutes for the deposition of M.

n=2

Ecell = Eocell - . [ + ]
[ +]

1.58 = 1.56 - . log
( . )

x = 4.75 x 10-3 M

1 F ≡ 96500 C
0.025 F = 2412.5 C

Q = It
2412.5 = 2.00 t

t = 1206.25 s
= 20.1 min

ii. Calculate the volume of oxygen gas liberated at the anode at STP.

[6 marks]

2H2O (l) → O2 + 4H+ (aq) + 4e-

1 mol O2 ≡ 4 F ≡ 4 (96500) C

Thus, 1 mol O2 ≡ 386 000 C
6.25 x 10-3 mol O2 = 2412.5 C

Volume of O2 = 6.25 x 10-3 (22.4)
= 0.14 L

4. Compound A is a four-membered ring alkane with molecular formula C5H10.

a) Draw the structural formulae for all possible monosubstitued products for the
reaction between C5H10 and chlorine in the presence of light.

CH3 CH2 Cl CH3 CH3

Cl

Cl Cl

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CHEMISTRY SK025 PSPM II 2018/2019

b) Write the reaction mechanism for the formation of the most stable reaction

intermediate in 4(a).

[7 marks]

Cl Cl +uv Cl
Cl

CH3 Cl +CH3 HCl

H+

5. Compounds B and C are hydrocarbons with the structural formulae as shown below.

CH3 CH3

BC

a) Name compounds B and C according to IUPAC nomenclature.
B: 1-methylcyclohexene @ methylcyclohexene
C: toluene @ methylbenzene

b) Both B and C can undergo oxidation reaction with the same oxidising agent.
Write chemical equations involved and explain the differences between these
two reactions.

CH3 O

i. KMnO4 , OH-,  OH
O
ii. H3O+ H3C COOH

B

CH3
i. KMnO4 , OH-, 

ii. H3O+

C

B undergoes oxidation that occur cleavage of double bond and produce
carboxylic acid and ketone.

C undergoes oxidation that occur on side chain of benzene since contains of
benzylic hydrogen and produce benzoic acid.

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CHEMISTRY SK025 PSPM II 2018/2019

c) Name one reaction that converts B to methylcyclohexane. [8 marks]
Hydrogenation of alkene @ Reduction

6. Two organic compounds D and F with molecular formula C4H9Br react with suitable
reagents to form E and G via SN 1 and SN 2 reactions respectively. Both E and G have a
molecular formula of C4H10O.

a) Draw the structural formulae for D, E, F and G.

D: (CH3)3CBr

E: (CH3)3COH

F: CH3CH2CH2CH2Br

G: CH3CH2CH2CH2OH

b) Suggest reagents for the formation of E and G.

Reagent for formation of E: H2O
Reagent for formation of G: NaOH (aq)

c) Write the mechanism for the formation of E.

[8 marks]

Step 1: Formation of Carbocation

CH3 H3C CH3 + Br-
H3C C Br C+

CH3 CH3

Step 2: Nucleophilic attack

CH3 .. CH3
+H3C C+ :O H C O+ H
H3C
H CH3 H
CH3

Loss of H+

CH3 .. CH3
:O H
C O+ H
CH3 H H
+H3C +H3C C O H3O+

CH3 H

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CHEMISTRY SK025 PSPM II 2018/2019

7. The structural formula of compound H is shown below.

HO

Compound H

a) Name the functional group of compound H.
Hydroxyl

b) Suggest a synthetic pathway to prepare H via Grignard reagent and propanal.

+H3C CH2 CH2CH2 Br Mg dry ether H3C CH2 CH2CH2 MgBr

H3C CH2 CH2CH2 MgBr i. CH3CH2CHO , dry ether H3C CH2CH2CH2 CH CH2 CH3
ii. H3O+ OH

c) Compound H reacts with J to form K, C10H20O2, in the presence of a few drops
8. a) of sulphuric acid. Draw the structural formula of J. State the function of
sulphuric acid.

[7 marks]
J: CH3CH2COOH
Function of sulphuric acid : acts as catalyst.

Arrange the following compounds in ascending order of boiling point. Explain
your answer.

CH3CH2CH2COH CH3CH2CH2COOH CH3CH2CH2CH2OH

L MN

Boiling point of L < N < M

L, M and N has similar molecular weight. L has lower boiling point that N
and M since L only can form weak Van der Waals forces between its
molecules. N and M can form hydrogen bonds between its molecules. Van
der Waals forces weakest than hydrogen bonds. Thus, less energy needed
to break Van der Waals forces between L molecules.

N and M can form hydrogen bonds between its molecules but M can form
stable hydrogen bonded dimer between its molecules. Thus, more energy
needed to break stable hydrogen bonded dimer of M than hydrogen bond
of N.

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CHEMISTRY SK025 PSPM II 2018/2019

b) Based on the following reaction scheme, draw the structural formulae of P
and T, and give reagents Q, R and S.

CH3CH2CH2CH2OH P CH3CH2CH2COOH
i. O3
ii. Zn, H2O

QR

CH3CH2CH2CHO

S NH2OH, H+

CH3CH2CH2CH(CH3)(OH) T

[8 marks]

P: CH3CH2CH2CH=CHCH2CH2CH3
T:
Q: CH3CH2CH2CH=NOH
R: NaBH4 in CH3OH @ i) LiAlH4 ii) H3O+
S: KMnO4, H3O+ , heat (or any oxidising agent)

9. a) i. CH3MgBr, dry ether

ii. H3O+

Treatment of 2-phenylethanoic acid with alcohol U gives an ester V with
molecular formula of C11H14O2. Alcohol U reacts with alkaline iodine solution
to form a yellow precipitate W, whereas oxidation of U yields X.

i. Draw the structures for U, V and X.

U: CH3CH(OH)CH3

V:

O CH CH3
CH2 C O CH3

X : CH3COCH3

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CHEMISTRY SK025 PSPM II 2018/2019

ii. Write all chemical equations involved.

O O CH CH3
CH2 C OH CH2 C O CH3

CH3CH(OH)CH3 , H3O+



V

OH +I2 , OH- O
H3C CH CH3 H3C C O- CHI3

U W

OH KMnO4 , H+ O

H3C CH CH3  H3C C CH3

UX

iii. Give the IUPAC name of W.
triiodomethane

b) Explain why methanoic acid gives a positive test towards Tollens’ reagent but
ethanoic acid does not. Write the equation for this reaction.

[10 marks]

Methanoic acid can react with Tollens’ reagent because of presence of
aldehyde functional group. Whereas ethanoic acid does not have aldehyde
functional group.

O Ag(NH3)2+ , OH- CO2 + H2O + Ag
H C OH

10. a) The reaction scheme below shows the formation of a dye DD from benzene.

NO2 N2Cl

AA Zn, H+ CC C6H5OH
OH-
BB DD

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CHEMISTRY SK025 PSPM II 2018/2019

i. Give the reagents AA and CC and draw the structural formulae of BB
and DD.
AA : conc. HNO3 , conc. H2SO4 , 50-55oC
CC : NaNO2 , HCl , 0-5oC

NH2

BB:

DD: NN OH

ii. Name the reaction for the formation of DD.
Coupling reaction

b) N-methylamphetamine is a potent central nervous system stimulant that is
mainly used as a recreational drug. The structure of N-methylamphetamine is
shown below.

CH3
CH2 CH NHCH3

N-methylamphetamine

Name a chemical test to distinguish N-methylamphetamine from aniline.
State the observations and write the chemical equation.

[8 marks]

Name of test: Identification test of aniline

Reagent: Br2 (aq) @ Bromine water

Observation:

N-methylamphetamine : No white precipitate formed.

Aniline : White precipitate formed

Equation:

NH2 NH2
Br Br
Br2 (aq)

Br

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CHEMISTRY SK025 PSPM II 2018/2019

11. Amino acid has amphoteric properties. By using serine below as an example,

O
C NH2
HO CH

CH2
OH

Serine

a) explain what is meant by amphoteric properties?
Amino acid such as serine are amphoteric because functional group amino
(NH2) acts as base and functional group carboxyl (COOH) acts as acid.

b) write chemical equations to show the amphoteric properties.
[4 marks]

OO
C NH2
HO CH HCl C NH3+ Cl-
HO CH
CH2
OH CH2
OH

O O
C NH2
HO CH NaOH (aq) C NH2

CH2 NaO CH
OH
CH2
OH

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CHEMISTRY SK025 PSPM II 2018/2019

12. The structure of Kevlar is as shown below. O
C
H HO
N NC n

Kevlar [2 marks]
Write the structural formulae of its monomers.

Monomers: O O
H2N C Cl
NH2 and Cl C

O @
HO C
O
C OH

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