MEMO: GRADE 12 MOCK MATRIC SEPTEMBER 2012
Question One:
1.1
catalyst

1.2
endothermic

1.3
alkyl-chlorides/halides  or halo-alkanes
1.4
electrolyte
1.5
eutrophication 

5
Question Two:
2.1 C
2.2 C
2.3 B
2.4 D 2.5 D 2.6 B
2.7 A 2.8 A 2.9 A 2.10 C  x 10 = 20
Question Three:
3.1
Einput
= 2 x 3 x (C-C) + 2 x 10 x (C-H) + 13 x (O=O) 
= 6 x 348 + 20 x 412 + 13 x 496
= 16 776kJ
3.2
Eoutput
4
= 8 x 2 x (C=O) + 10 x 2 x (O-H) 
= 16 x 743 + 20 x 463
= 21 148kJ
3.3
∆E
3
= 16 776 – 21 148
= -4 372kJ
Since ∆E is negative – reaction is exothermic
2
Any correctly explained answer without the calculation may also score 2
Question Four:
4.1.1
The rate at which reactants are used up or products are produced. 
2
4.1.2
nCaCO3

3
= 2g/(40 + 12 + 48) = 0,02 moles of CaCO3
4.1.3 According to the mole ratio of 2:1 HCl to CaCO3 the number of moles of HCl required to
react with the CaCO3 is 2 x 0,02moles or 0,04 moles. Since we have 0,3 moles of HCl - it is in
excess and CaCO3 is the limiting reagent.

2

4.2.1
On graph paper.
5
4.2.2
Aprox 0,6 grams (mark off graph allowing some leeway)
4.2.3
Since only 0,04 moles of HCl react according to the balanced chemical equation,  its
concentration only drops to 0,026 mol•dm-3 which is still close to 0,3 mol•dm-3. 2
4.2.4
The graph flattens out as the CaCO3 gets used up. / Slope gets less / any reasonable
description and explanation gets 2.

2
4.2.5
See graph.
4.2.6
Volume of acid used / Temperature / state of division / each x 2
4.2.7
Greater concentration gives greater number of collisions between particles this results
in more effective collisions per unit time and this gives rise to an increase in the rate
of reaction. 
3

2
2
Question Five:
[NO]2/[N2]•[O2]

5.1.1
kc =
2
5.1.2
The equilibrium lies almost completely to the left of this equation: i.e. very little product
is formed and almost all the reactants remain unreacted.

2
5.2.1
For any closed system which is at equilibrium, a disturbance to that equilibrium will
result in the system reacting in such a way as to overcome that disturbance. 2
5.2.2
kc = (2 x 10-6)2 / (2 x 10-2) x (2.5 x 10-3)  = 8 x 10-8 
3
5.3.1* YES - NO is a greenhouse gas and by removing NO from exhaust fumes we are
removing a potentially harmful gas from the environment. 
3
5.3.2
Replace fossil fuel driven cars with electric cars / hydrogen driven cars / steam driven
cars etc. (Any reasonable suggestion) 
2
*
If the answer to 5.3.1 is NO then they must give a plausible reason as to why the device
is not cost effective.
Question Six:
6.1.1
Mn3+ because it has the most positive / highest E value 
6.1.2
Co2+
6.1.3
Co2+ → Co 3+  and Mn3+ → Mn2+ 
6.1.4
Ecell = Ecathode - Eanode

2
1
If they choose the wrong equations then mark
2
positively in 6.1.4

= 1,49 – (-0,28)

= 1,77V

4
6.2.1
Fe has a greater oxidizing potential than Cu and so will be oxidized in preference to
Cu.  (boys may say that Fe lies above Cu in the table 4B and so will act as the
anode with respect to Cu. )
3
6.2.2
It completes the circuit by allowing ions to flow in the solutions. 
6.2.3
The damp soil which would naturally carry some salts and hence form an electrolyte. 1
1
Question Seven:
7.1.1
A homologous series is one which is characterized by a general formula. 
2
7.1.2
Unsaturated hydrocarbons contain at least one double or triple bond. 
2
7.1.3
a)
alkyl-halides or halo-alkanes.

1
b)
carboxylic acids

1
c)
ketones

1
d)
alcohols / alkanols.

1

2
7.1.4
(boys will need to put in the C-H bonds – I have
left them out because I couldn’t find good pics!)
7.1.5

When marking assume 2 marks and subtract for each mistake – 2 mistakes – no marks!
2
7.1.6
pent-1-ene

2
7.1.7
1-chlorobutane

2
7.2.1
A molecule with the same molecular formula as another but with a different structural
formula.

2
7.2.2
propan-2-ol or 2-propanol.
7.2.3
Heated alcohol reacted with hot conc H2SO4 or Heated alcohol passed over a
catalyst - Al2O3

7.2.4

1
Dehydration/ elimination of water/ dehydrolysis

2
1
7.2.5

2-bromopropane
7.3.1
CH3COOH + C4H9OH ⇌ CH3COOC4H9 + H2O
7.3.2

7.3.3
4

3
2
Concentrated sulphuric acid. / H2SO4(conc) if formula given must have (conc)
1
Question Eight:
8.1.1
Electrolytic cells convert electrical energy into chemical change/ chemical potential
energy

2
8.1.2
Concentrated sodium chloride solution / salt solution.
8.1.3
Chlorine gas – used in plastics and the paper /pulp industry / insecticides. 
2
Sodium hydroxide – soap manufacture / bleaching / tanning industry 
2
8.2.1
2Cl- → Cl2 + 2e-
2
8.2.2
Mercury is a poisonous metal which can leak into the environment and poison
ecological systems. 
3


1
Question Nine:

9.1.1
Nitrogen
1
9.1.2
1:
The production of H2 which is a primary feedstock for the process is generally
produced from fossil fuels.

2:
Fossil fuels are used to heat the gasses prior to reaction in the Haber process.
Or used in the generation of electricity to power the plant
2
9.1.3
N2 + 3H2 ⇌ 2NH3 
9.1.4
Increased yields of crops per hectare of land / quicker crop rotation in fields without
them having to lie fallow / better quality of crops as a result of better nutrients. 1
9.1.5
It could poison underground water resources which may be scarce in farming areas. It
could leak into natural water supply and result in eutrophication / algal bloom / blue
baby syndrome etc.

2
(-1 for each mistake)
3
Question Ten:
10.1.1 +2

2
10.1.2 Pb(s) + PbO2(s) + 2H2SO4 ⇌ 2PbSO4 + 2H2O

3
10.1.3 Pb - it is the only element oxidized in the process.

2
10.2.1 2H2O ⇌ O2 + 4H+ + 4e-

2
10.2.2 If both the H2 and the O2 are produced at the time of charging they could potentially
form an explosive mixture which could be detonated by a spark. 
2
Graph for Question Four:
0.8
Mass lost in grams
0.7
4.2.2
0.6
4.2.1
0.5
4.2.5
0.4
0.3
0.2
0.1
0
0
100
200
300
Time in seconds
400
500