CADANGAN JAWAPAN PSPM 20192020

CADANGAN JAWAPAN PSPM 2019/2020
CHEMISTRY 2 SK 025

SOALAN 1

a) i.

order of reaction:

rate= k[CH3COOCH3]m[OH-]n

12.5.08 1100−−33= (0.05) (0.154)
(0.05) (0.120)

n = 1, order of reaction for [OH-] is First order

23.0.43 1100−−33= (0.084) (0.154)
(0.05) (0.154)

n = 1, order of reaction for [CH3COOCH3] is First order

ii. rate= k[CH3COOCH3][OH-]
iii. 1.58 x 10-3 = k (0.05)(0.120)

k = 0.263 M-1s-1
iv. rate of reaction increase 4 times faster.

b) i. Tungsten because its has lower Ea value.

ii. 1 = ( 1 − 1 )

2 2 1

1 = 1.63 10−3 1 1
2 8.314 (823 − 723)

1 = 0.037 Thus, 2 = 27
2 1

The reaction is 27 times faster.

SOALAN 2 2CO2(g) + 3 H2O (l)

a) i.
CH3CH2OH (g) + 3O2(g)

ii. ℎ = 1.0
46 /

= 0.0217 mol

-q rxn = q cal

= Cc ∆T
= 11.0 kJoC-1 (2.7oC)
= -29.7 kJ

0.0217 mol CH3COOCH3 combusted = -29.7 kJ
1.0 mol CH3COOCH3 combusted = -1368.66 kJmol-1

∆H combustion = -1368.66 kJmol-1

b) i. ∆H7 = Lattice energy CaS
∆H8 = Enthalpy of formation CaS

ii. S-
iii. ∆H6 = + 539 kJmol-1
iv. - size of anion O2- is smaller than S2-.

- The attraction between O2- toward Ca2+ is stronger than S2- toward Ca2+.
-Thus lattice energy CaO is larger than CaS.
c)

CS2(l) C(s) + 2S(s) + 87.9 kJ
C(s) + O2(g)
2S(s) + 3O2(g) CO2(g) -393.5 kJ

2SO2 (g) -296.8 kJ x 2

Target eq : CS2 (l) + 3O2(g) CO2(g) + 2SO2(g)

∆H = + 87.9 kJ + (-393.15 kJ) + (-296.8 kJ x 2)
= -899.2 kJ

SOALAN 3
a) i. Strength of reducing agent increased: Cu < Ni < Cr
*Reducing agent is a species that can undergoes oxidation.
ii. Eo cell = Eocathode – Eoanode
= + 0.34V – (-0.74V)
= +1.08V
Since Eo cell > 0, spontaneous reaction

iii.

Cr is a reducing agent. The wall of the container will corrode/thicker because Cr metal will
dissolve in CuSO4 solution.

Cr(s) Cr3+ (aq) + 3e

b) Zn(s)

Zn2+ (aq) + 2e

Q = It
= 1.2 x 3600
= 43200 C

1 mol Zn = 2 mol e = 2x96500C
? mol = 43200 C
n= 0.0223 mol

mass Zn = 0.0223 mol x 64.5 gmol-1
= 1.46 g

SOALAN 4
a) Intermediate species (steps that produce a free radical)

hν

Br Br 2Br

CH3CHCH3 + Br CH3CHCH3 + HBr
H

CH3CHCH3 + Br Br CH3CH2CH3 + Br

b) B : CH3CH2CH2CH2CH=CH2
C : CH3CH2CH2CH2CH2CH2
D : CH3CH2CH2CH2CH2COOH
Test to differentiate B from C : Baeyers’ test

SOALAN 5

a) E : conc HNO3, conc H2SO4 , 50-55oC
F : CH3CHClCH3, AlCl3
G : KMnO4, H+, ∆

H:
Cl

H3C C CH3

b) + + AlCl4-

CH3CHCH3 + AlCl3 CH3CHCH3
Cl
CH3 CH3 CH3
H + HC CH3 HC CH3 HC CH3

+ H H H

+ CH3CHCH3 + +

CH3 CH3
HC CH3 HC CH3

H + Cl AlCl3 + HCl + AlCl3

+

SOALAN 6
a) i. SN2 mechanism. Bimolecular Nucleophilic Substitution reaction.
Mechanism

H H

C Cl slow CH3O C Cl H
H CH2CH2CH3
CH3O- + H Fast C

CH2CH2CH3 CH3O H CH2CH2CH3

ii. rate =k[CH3CH2CH2CH2Cl][NaOCH3]

b) J : HCl
K : CH3CHClCH3
L : CH3CH(COOH)CH3

SOALAN 7

a) M : CH3CH2CH2CH2CH2OH
N : CH3C(CH3)OH CH3

P : CH3CH3CH3CH3COOH
b)

CH3CH2CH2CH2CH2OH + Na CH3CH2CH2CH2CH2O-Na+ + H2

CH3 CH3 H2 gas compulsory to write since its already
H3C C CH2 CH3 stated in the question as a buble gas formed.
H3C C CH2 CH3 + Na
O-Na+ + H2
HO

CH3CH2CH2CH2CH2OH Na2Cr2O7, H+ CH3CH2CH2CH2CH2OOH

Δ

c) Ester P : O CH3
C CH2 CH3
H3C CH2 CH2 CH2 C O CH3

Catalyst : concentrated H2SO4

SOALAN 8 OH OH
H3C CH2 C H H3C CH2 CH2
a) OH
CN S
H3C CH2 C H
R

Q

b)
i. T C4H8O

-Orange precipitate with Bradys’ reagent: contain carbonyl group.
-react with I2 in NaOH (yellow precipitate): contains methyl carbonyl group.

O

T H3C C CH2 CH3

ii. U : CH3CH2C(CH3)=C(CH3)CH2CH3

Reagent for ozonolysis:

i. O3 @ i. O3
ii. Zn, H2O
ii. (CH3)2S

SOALAN 9

a) Acidity increase, phenol< propanoic acid<2-chloropropanoic acid

-propanoic acid and 2-chloropropanoic acid are stronger acid compared to phenol due to
resonance stabilisation on their conjugate base.

- For the carboxylate ions, the electrons are delocalised between two oxygen atoms, while
for phenoxide ion the electrons are delocalised within carbon atoms in benzene ring.

-the presence of Cl atom in 2-chloropropanoic acid will stabilised the carboxylate ion
through inductive effect by withdrawing the electrons density from carboxylate ion.

b) i.

CH3CH2CH2CH2Br Mg, dry ether CH3CH2CH2CH2MgBr i. CO2 CH3CH2CH2CH2COOH
ii. H3O+

ii. i. O3 CH3CH2CH2COOH
CH3CH2CH2CH=CHCH2CH2CH3 ii. Zn, H2O SOCl2

CH3CH2CH2COCl

SOALAN 10

a) 3 isomers C3H9N (different class of amine)

CH3CH2CH2NH2 CH3CH2NH(CH3) CH3N(CH3)2
V W X

b) Chemical test : Reaction with nitrous acid.

CH3CH2CH2NH2 NaNO2, HCl CH3CH=CH2 + CH3CH(OH)CH3 + CH3CH(Cl)CH3 + N2

< 5°C

1o aliphatic amine produce an unstable diazonium salt at a temp < 5oC.
Observation buble gas released, indicate N2 gas.

CH3CH2NH(CH3) NaNO2, HCl CH3
< 5°C H3C CH2 N N O

2o amine. Yellow oil formed.

CH3N(CH3)2 NaNO2, HCl CH3 OCl- CH3
< 5°C H3C N+ N
+ H3C N+ H
H3C
H3C

3o amine. Clear solution.

SOALAN 11 O Na O O
a) CH2 CH C OH NaOH(aq) CH2 CH C O Na

HO H2N H2N

HO O HCl HO O
CH2 CH C OH CH2 CH C OH
HO
H2N NH3+Cl-

O H+ O
HO CH2 CH C O CH3
CH2 CH C OH + CH3OH Δ
H3N+
H2N

Tyrosine can react as acid and base because its contain 2 functional groups which are amino
group and carboxyl group. Thus, it can acts as acid and base.

SOALAN 12
Monomer CH2=C(CH3)CH=CH2

Polymer

Prepared by;
Marziah binti Mohamad
Kolej Matrikulasi Melaka