Model Answer
Grade 12 Preliminary Examination
Paper II
2013
Question 1:
1.1
1.6
B
C
Multiple choice
1.2
1.7
D
B
Question 2:
1.3
1.8
B
B
1.4
1.9
B
B
1.5
1.10
B
C
[20]
Quantitative Chemistry
2.1.1
The amount of substance expressed in grams which contains as many elementary
particles as there are carbon atoms in exactly 12g of the carbon-12 isotope.
(3)
____________________________________________________________________________
2.1.2 The volume occupied by 1 mol of any gas at S.T.P.
(2)
____________________________________________________________________________
2.2.1
2CuFeS2(s) + 3O2(g)  2FeO(s) + 2Cu2S(s) + 2SO2(g)
CuFeS2:
𝑀 = 63,5 + 56 + 64 = 183,5 𝑔. 𝑚𝑜𝑙 −1
𝑚
42,3
𝑛 = 𝑀 = 183,5 = 0,231 𝑚𝑜𝑙 
Mol ratio: CuFeS2
:
O2
2 mol
:
3 mol 
1 mol
:
1,5 mol
0,231 mol
:
0,346 mol (= 0,35 mol)
(4)
____________________________________________________________________________
2.2.2
Mol ratio: CuFeS2
2 mol
1 mol
0,231 mol
Cu2S:
:
:
:
:
Cu2S
2 mol
1 mol
0,231 mol (= 0,23 mol)
c.o.e from 2.2.1
𝑀 = 2(63,5) + 32 = 159 𝑔. 𝑚𝑜𝑙 −1 
𝑚 = 𝑛. 𝑀 = (0,231). (159) = 𝟑𝟔, 𝟕𝟑 𝒈
(= 36,57𝑔)
(3)
____________________________________________________________________________
1
2.2.3
Mol ratio: CuFeS2
0,231 mol
:
:
SO2
0,231 mol
(= 0,23 mol)
SO2:
𝑉0 = 𝑛. 𝑉𝑚 = (0,231). (22,4) = 𝟓, 𝟏𝟕 𝒅𝒎𝟑 
(= 5,15 𝑑𝑚3 )
(2)
____________________________________________________________________________
2.2.4
𝑛 = 0,231 𝑚𝑜𝑙
(= 0,23 𝑚𝑜𝑙)
Reaction 1:
Cu2S:
Reaction 2:
Cu2S(s) + O2(g)  2Cu(s) + SO2(g)
Mol ratio: Cu2S
1 mol
0,231 mol
:
:
:
Cu
2 mol
0,462 mol
c.o.e from 2.2.2
(= 0,46 mol)
𝑀 = 63,5 𝑔. 𝑚𝑜𝑙 −1
𝑚 = 𝑛. 𝑀 = (0,462). (63,5) = 𝟐𝟗, 𝟑𝟒 𝒈
(= 29,21𝑔)
(4)
____________________________________________________________________________
Cu:
2.2.5
Two relevant reasons:
High levels of unemployment lead to theft.
Copper is valuable (fetches a good price at scrap yards)
Cash on delivery.
Copper has many uses.
Dangerous (electrocution)
Disrupts communications
Unnecessary costs in replacement, etc.
(4)
____________________________________________________________________________
2.3.1
Li2SO4:
𝑀 = 2(7) + 32 + 4(16) = 110 𝑔. 𝑚𝑜𝑙 −1 
𝑚
7,1
𝑛 = 𝑀 = 110 = 0,064545 𝑚𝑜𝑙
𝑛
𝑐=𝑉=
0,064545
0,5
 = 0,13 𝑚𝑜𝑙. 𝑑𝑚−3
(3)
____________________________________________________________________________
2.3.2 Li2SO4  2Li+  + SO42- 
(2)
____________________________________________________________________________
2.3.3
[Li+] = 2×(0,13) = 0,26 mol.dm-3 
c.o.e from 2.3.1
(1)
____________________________________________________________________________
[28]
2
Question 3:
Organic Chemistry
3.1
C3H8 + 5O2  3CO2 + 4H2O
Balanced
(3)
____________________________________________________________________________
3.2
Aldehydes
(1)
____________________________________________________________________________
3.3




H
H
H
O
C
C
C
H
H
H
OH
Propanoic acid
+
HO
C
H
H
methanol

H
H
H
O
C
C
C
H
H
H
O
C
H
+
H
H
methyl propanoate
O
H
water
(7)
____________________________________________________________________________
3.4.1
Addition
(1)
Hydrohalogenation
____________________________________________________________________________
3.4.2
Hydrogen bromide
(1)
(Don’t accept HBr)
____________________________________________________________________________
3.5
Carboxyl group
(1)
____________________________________________________________________________
3.6
2 – methylpropane (or methyl propane)
(2)
____________________________________________________________________________
3.7
Hydrogen bond
(1)
____________________________________________________________________________
3.8
Higher boiling and melting points
(2)
Higher viscosity
Lower vapour pressure
____________________________________________________________________________
3.9
CnH2n – 2 
(1)
____________________________________________________________________________
3.10
The double bond in the alkene consists of a strong 𝜎 bond and a weaker 𝜋 bond.
An alkane only contains only strong 𝜎 bonds.
Less energy is required to break the weaker 𝜋 bond in the alkene.
(2)
____________________________________________________________________________
[22]
3
Question 4:
Organic Chemistry
4.1.1
A bond, an atom or a group of atoms which identifies to which homologous series a
molecule belongs and is responsible for the chemical properties of that compound.
(3)
____________________________________________________________________________
4.1.2
A carbon based side chain which is attached to the longest continuous chain in the
molecule.
(2)
____________________________________________________________________________
4.2.1
CH3 – CBr = CH – CHBr – CH2 – CH3
Double bond at 2nd C
Br in correct position (2,4)
(2)
____________________________________________________________________________
4.2.2
CH3CH(CH3)CH(CH2CH3)CH2CH3
5 C’s in main chain
(CH3) and (CH2CH3) in correct positions
(2)
____________________________________________________________________________
4.3.1
van der Waals forces (London forces)
(1)
4.3.2
The larger the molecule the greater the electron cloud density giving rise to stronger
London forces.
(2)
____________________________________________________________________________
4.4.1
A: Alkylhalides or Haloalkanes
B: Alkenes
(2)
____________________________________________________________________________
4.4.2 Elimination
(1)
____________________________________________________________________________
4.4.3 Water (don’t accept H2O)
(1)
____________________________________________________________________________
Concentrated sulphuric acid, H2SO4
Pentanoic acid.
(2)
____________________________________________________________________________
4.4.4
4
4.4.5
Can be a structural or semi-structural formula.
H
H
H
H
H
H
H
H
O
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
OH
7 carbon atoms
carboxyl group
Name:
heptanoic acid
(3)
You must consider the many branched carboxylic acid possibilities:
For example:
CH3
CH3
CH2
CH
O
CH2
CH2
C
OH
Check the name:
4-methyl hexanoic acid
____________________________________________________________________________
4.4.6 Bromination 
(1)
____________________________________________________________________________
4.4.7 Decolourisation of bromine.
(2)
____________________________________________________________________________
[24]
5
Question 5:
Polymers and Plastics
5.1
A molecular fragment with an unpaired electron.
(2)
____________________________________________________________________________
5.2
A bond between C atoms in monomer is broken.
+
R
H
CH3
C
C
H
COOCH3

Free radical bonds with monomer, creating another free radical.
R
H
CH3
C
C

(2)
H
COOCH3
____________________________________________________________________________
5.3
Propagation
(1)
____________________________________________________________________________
5.4
1 mark for the unit:
H
CH3 
C
C
H
COOCH3
Another 2 marks for three repeated units.
H
CH3
H
CH3
H
CH3
C
C
C
C
C
C
•
H
COOCH3 H
COOCH3
COOCH3 H
(4)
____________________________________________________________________________
5.5
Safety.
Doesn’t break as easily as glass.
Not produced at very high temperatures.
(4)
____________________________________________________________________________
6
5.6
5.6.1
5.6.2
5.6.3
Thermoset
Cross-links between molecules
Char and burn
Cannot be recycled
Thermoplastic
None
melt
Can be recycled
(6)
____________________________________________________________________________
5.7
Give two marks for each well described problem.
Give one mark if you’re not quite convinced.
Fine pieces could be ingested by marine life, resulting in death.
These deaths could have a negative effect on the ecosystem.
Marine life could become entangled in larger pieces.
Not biodegradable, etc.
(4)
____________________________________________________________________________
[23]
7
Question 6:
Electrochemistry
6.1.1 The gain of electrons.
(2)
____________________________________________________________________________
6.1.2 Fe3+ 
(2)
____________________________________________________________________________
6.1.3 Fe3+ + e-  Fe2+ 
(1)
____________________________________________________________________________
6.1.4
(H2O2) is a stronger reducing agent than the oxidising agent (Fe3+) in the Table of
Standard Reduction Potentials.
Or comment on the different electrode potential values.
(3)
____________________________________________________________________________
6.2.1
Zn
It is highest on the right hand side of the table.
Or Zn has the more negative electrode potential, meaning it is more easily oxidised.
(2)
____________________________________________________________________________
6.2.2
(i) Fe  Fe2+ + 2e- 
(1)
(ii) Au3+ + 3e-  Au 
(1)
(iii) 3Fe + 2Au3+  3Fe2+ + 2Au

(2)
____________________________________________________________________________
6.3
+6 
(1)
____________________________________________________________________________
6.4.1 zinc (Zn) or chromium (Cr)
(1)
____________________________________________________________________________
6.4.2
It is a stronger reducing agent and is therefore more easily oxidised than Fe.
(2)
____________________________________________________________________________
[18]
8
Question 7:
Rates of Reactions
7.1
Increasing the temperature will increase the reaction rate.
(2)
Decreasing the temperature will decrease the reaction rate.
____________________________________________________________________________
Two marks
–1 for each mistake.
(2)
st
nd
Temperature (°C) Time for 1 trial (s) Time for 2 trial (s) Average time (s)
10
48,2
47,8
48,0
20
27,5
28,5
28,0
30
15
15,4
15,2
40
9,6
9,4
9,5
50
6,78
6,28
6,5
____________________________________________________________________________
7.2
7.3
Concentration of sodium thiosulphate.
Concentration of acid.
Volume of solutions.
(2)
____________________________________________________________________________
7.4
Time
(2)
____________________________________________________________________________
9
7.5
Temperature of reactants versus average time for reaction
50
40
Time (s)
30
20
10
0
0
10
20
Title
Both axes labelled with units
Good choice of scales on both axes
Plotting 5 points (–1 for each mistake)
Line of best fit
30
40
Temperature (°C)
50
(1)
(2)
(1)
(3)
(1)
If variables are plotted on the wrong axes, then award a maximum of 5 marks.
(8)
____________________________________________________________________________
7.6
As the temperature increases, average time for the reaction decreases.
(2)
____________________________________________________________________________
[18]
10
Question 8:
Rates of Reactions
8.1.1 Decrease
(1)
____________________________________________________________________________
8.1.2 Increase
(1)
____________________________________________________________________________
8.1.3 No change
(1)
____________________________________________________________________________
8.2
Decrease in surface area. therefore less particles available
Fewer collisions occur (or collision frequency decreases).
Leading to fewer effective collisions.
(3)
____________________________________________________________________________
8.3.1 The number of particles with enough energy to take part in a reaction.
(1)
____________________________________________________________________________
8.3.2
Particles must collide.
Particles must collide with enough EK.
(greater than or equal to EA)
Particles must collide with the correct orientation.
(3)
____________________________________________________________________________
8.3.3
Must show a lower EA value.
Number of particles
with kinetic energy EK
EA
EA
Kinetic energy EK
(2)
____________________________________________________________________________
8.3.4
Increasing [HCl], increases the number of particles per unit volume.
This increases the collision frequency.
This increases the number of effective collisions per unit time.
Increasing the reaction rate.
(3)
____________________________________________________________________________
[15]
11
Question 9:
9.1
Chemical Equilibrium
The concentration of the N2 and O2 (reactants) at equilibrium is very high.
Or the concentration of NO (product) at equilibrium is very low.
(2)
Simply stating that the equilibrium lies to the left gets only one mark.
____________________________________________________________________________
9.2.1
Reverse reaction
Only N2 and O2 are present in car engine.
Forward reaction rate would therefore be high.
(3)
Or
Reverse reaction
There is no product (NO) present at the start of the reaction.
Reverse reaction rate would be zero and increasing as NO is formed.
____________________________________________________________________________
9.2.2
5 minutes
Rate of forward and reverse reactions are equal.
(2)
____________________________________________________________________________
9.3.1


Reaction
rate


0
5
10
17
Time (minutes)
 sudden increase in rate of forward reaction.
 gradual decrease in rate of forward reaction.
 gradual increase in rate of reverse reaction.
 rates are equal at 17 minutes.
(4)
____________________________________________________________________________
9.3.2 Equal to
(2)
____________________________________________________________________________
[13]
12
Question 10:
Chemical Equilibrium
10.1
For a closed system, the rate of the forward reaction is equal to the rate of the
reverse reaction.
(2)
____________________________________________________________________________
10.2
n
[AB3 ] =
V
6
[AB2 ] =
[B2 ] =
Kc =
Kc =
2
3
2
=
4
2
= 2 mol.dm-3 
= 3 mol.dm-3 
= 1,5 mol.dm-3 
[AB2 ]2 ×[B2 ]
[AB3 ]2
(3)2 ×(1,5)
(2)2


K c = 3,375 
(6)
____________________________________________________________________________
10.3.1 According to LCP, a temperature increase will favour the endothermic reaction as it
uses up the heat energy.
The graph shows that the reverse reaction is being favoured (amount of AB3 is
increasing).
It follows then that the forward reaction is exothermic.
(4)
____________________________________________________________________________
10.3.2 Greater than.
(2)
____________________________________________________________________________
10.4.1 Increases
(1)
____________________________________________________________________________
10.4.2 Increases
(1)
____________________________________________________________________________
10.5
Decreasing the volume will increase the pressure.
According to LCP, an increase in pressure will favour the reverse reaction, since it
produces fewer mols of gas thus decreasing the pressure.
(3)
____________________________________________________________________________
[19]
[200]
13