CADANGAN JAWAPAN PSPM 2 SESI 20212022

Marziah binti Mohamad

CADANGAN JAWAPAN PSPM 2 SESI 2021/2022
CHEMISTRY 2 (SK 025)

1. a) i. rate = k[NO]x[O3]y
1.98 10−4 (1.0 10−6 ) (9.0 10−6)
1.32 10−4 = (1.0 10−6) (6.0 10−6)

Y=1
Order of reaction respect to O3 is first order

3.96 10−4 (2.0 10−6 ) (9.0 10−6)
1.98 10−4 = (1.0 10−6) (9.0 10−6)

X=1
Order of reaction respect to NO is first order

Rate law:
rate = k[NO][O3]

ii. 1.32 x 10-4 = k (1.0 x 10-6)(6.0 x 10-6)
k = 2.2 x 107 M-1s-1

b) i. Rate of decomposition = 2.18 x 10-2 Ms-1
Rate of decomposition = − 1 [ 2 5]

2

= 0.0436 Ms-1

ii.

1 = 1 1
2 ( 2 − 2)

= + 175.16 kJmol-1

iii. when temperature increase:
- The frequency of collision increase, thus the frequency of effective collision
increase.
- More molecules possesses average kinetic energy equal or greater than Ea.
- Thus, the rate of reaction increase.

2. a) i.
-qrxn = qc + qw
= Cc∆T + mwcw∆T
= (320 x 7.5) + (1100 x 4.18 x 7.5)
= -36 885 J
Amount of heat released is : 36 885 J @ 33.885 kJ

ii.
0.25

ℎ = 128 −1

= 1.953 x 10-3 mol

Marziah binti Mohamad

1.953 x 10-3 mol ≡ -36.885 kJ
1 mol C10H8 ≡ -18886.33 kJmol-1

∆Hc = -18886.33 kJmol-1

b) i. Thermochemical equation:

6 C (s) + H2 (g) C6H14 (l) ΔHof = -199.0 kJmol-1

H2 (g) + 1/2 O2 (g) H2O (l) ΔHof = -241.8 kJmol-1

C (s) + O2 (g) CO2 (g) ΔHof = -393.5 kJmol-1

ii. ∆Hoc hexane?

C6H14 (l) + 19/2 O2 (g) 6CO2 (g) + 7H2O (l)

∆H = [(∆Hof CO2 x 6) + (∆Hof O2)]- [(∆Hof CO2) + (∆Hof C6H14)]

= -3854.6 kJmol-1

3. a) i.

Eo cell = Eocathode - Eoanode

= + 0.80 – (-0.13 V)

= + 0.93 V

Ecell = Eocell -0.0 5 92 log [ 2+]
[ +]+

Ecell = +0.93 v - 0.0592 log (0.1)
2 (0.01)2

= + 0.8412 V

ii. voltage of cell will increase when concentration of [Ag+] increased.

b) i. product at cathode: copper solid

ii.

Cu2+ (aq) + 2e Cu (s)

Q=It
= 5 x (100x60)
= 30 000 C

1 mol Cu ≡ 2 mol e ≡ 2 x 96500 C
1 mol Cu ≡ 193 000 C
? mol ≡ 30 000 C

nCu = 0.1554 mol

mass = n x Mr
= 0.1554 mol x 63.6 gmol-1

= 9.88 g

Marziah binti Mohamad

4. OH

CH3

I = H2SO4 (aq) @ dilute H2SO4
D

OO KMnO4, H+ HBr CH3
H3C C CH2 CH2 CH2 C OH CH3 Br
Δ H
II = H2, Pt

III= Br2, uv

CH3
B

5. a) O
CH
H3C CH CH3
F
+ H3C CH CH3 AlBr3 KMnO4, H+
Br Δ

b) E contain benzylic hydrogen E
c)
H3C CH CH3

+ H3C CH CH3 AlBr3 + HBr
Br
+
d) Step 1 CH3CHCH3 + AlBr4-

H3C CH CH3 + Br H3C CH Br+ AlBr3
Br Br Al Br CH3

Step 2 +H H H
+ +
H
+

+ CH3CHCH3

Step 3 + Br AlBr3- + HBr + AlBr3
H

+

6. a,b H i. CO2 J
CH3CH2CH2CH2MgBr ii. H3O+ CH3CH2CH2CH2COOH
Mg dry ether

CH3CH2CH2CH2Br

Marziah binti Mohamad

c) HH H
HO C Br H
H
HO C
H slow fast
C Br CH2CH2CH3
-OH +

CH3CH2CH2 CH3CH2CH2

7. a) Alcohol K optically active : has chiral carbon
Yellow precipitate when treated with alkaline iodine : has methyl carbinol group

Alcohol K conc H2SO4 L i. O3 OO
H3C ii. Zn, H2O +
OH alkenes
C CHCH3 H3C C H H3C C CH3
CH3 CH3
K H3C CH C CH3
H3C

L

b) Chemical test to differentiate K from L

Baeyers' test

H3C OH KMnO4, OH- purple colour of KMnO4 remain
C CHCH3 cold and no brown precipitate formed
CH3 CH3

H3C CH C CH3 KMnO4, OH- CH3 + MnO2
H3C cold H3C CH C CH3

OH HO

purple colour of KMnO4 decolourised
and brown precipitate formed

Or Br2 (aq) reddish brown colour of
bromine remain unchange
Bromine water
OH

H3C C CHCH3
CH3 CH3

H3C CH C CH3 Br2 (aq) CH3 + MnO2
H3C H3C CH C CH3

Br HO

reddish brown colour of bromine
decolourised

8. a) O Marziah binti Mohamad
C CH3 H3C C N OH
H
+ H N OH N
OH
O HC CH3
C CH3

NaBH4, CH3OH

O P
C CH3 OH

+ CH3CH2OH H+ H3C C O CH2 CH3

O R
C CH3 HO

+ H3C CH MgBr H3O+ H3C C CH CH3
CH3 CH3

b) Q
O
CH3 C CH3

C i. O3 2
C ii. Zn, H2O

CH3

S

or
oxidation using KMnO4 acidified heat
9. a) Phenol< hexanoic acid < benzoic acid

Phenol least acidic because in phenoxide ion electrons are delocalised between O and
C atom of benzene ring.
In hexanoic ion, electrons are delocalised between two oxygen atoms.
Benzoate ion is more stable than hexanoate ion due to the present of pentyl group
which is an electron donating group in hexanoate ion @ due to the presence of
benzene ring as electron withdrawing group in benzoate ion.

Marziah binti Mohamad

b) H3C OH conc H2SO4 HBr, H O H3C CH2 CH2 KCN, H+
CH CH3 22 Br
Δ CH3CH2CH2CN
H3C CH CH2
H O+
3

CH3CH2CH2COOH

10. a) N,N-dimethylethanamine< n-methylethanamine < propanamine <ethanol

Ethanol has the highest boiling point because hydrogen bond in ethanol is stronger
than hydrogen bond in amine.

Propanamine has higher boiling point than N-methylethanamine because propanamine
can formed more hydrogen bond than N-methylethanamine.

N,N-dimethylethanamine has the lowest boiling point because it has van der Waals
forces. Van der Waals forces is weaker than hydrogen bond.

b) NaNO2, HCl + + OH
N Cl
NH2 N

< 5°C diazonium salt
stable salt
11. a) pH 12 (basic)

O O

HO CH2 CH C O- @ HO CH2 CH C O-Na+
NH2 NH2

b) O
H+
O Δ HO CH2 CH C O CH2 CH3

HO CH2 CH C OH + CH3CH2OH +NH3
NH2
T

V

OH O OH O
HO CH2 CH C OH + H N
CH C OH HO CH2 CH C N CH C OH
NH2 CH2
OH NH2 CH2
OH

peptide bond

12. a) HH HH
CC CC
H CN CH3 Cl n

Marziah binti Mohamad

b)

+ H2O