Chemistry Paper 2 Marking Scheme
MBOONI WEST SUB – COUNTY JOINT EVALUATION TEST
Kenya Certificate of Secondary Education
233/2
CHEMISTRY
PAPER 2(THEORY)
JULY/AUGUST 2014
MARKING SCHEME
1. (a) P removes carbon (IV) oxide  1 mark
Q removes oxygen  1 mark
(b) P – Potassium hydroxide/Sodium hydroxide  1 mark
Q – Heated copper metal  1 mark
(c) P – 2KOH(aq) + CO2(g)
K2CO3(aq) + H2O (l)
or 2NaOH(aq) + CO2(g)
Na2CO3(aq) + H2O(l)
Q – 2Cu(s) + O2 (g)
2CuO(s)  1 mark
(d) - Oxyacetylene flame for welding  ½ mark
- In hospitals for patients with breathing difficulties  ½ mark
- In respiration  ½ mark
- When mixed helium it is used by deep sea divers and mountain climbers  ½ mark
(e) Noble gases, at least one mentioned
2. (a) (i) Element S  1 mark
(ii) E(s) + Y2(g)
EY2 (s)  1 mark
(iii) Transition metals  1 mark
(iv) Y  1 mark It has the highest ability to attract electrons  1 mark
(v) Y is smaller than T  1 mark since Y has a greater nuclear charge than T 1 mark//Y has many
protons than T.
(b) (i) F:  ½ mark Atomic no.13
G:  ½ mark Atomic no.20
(ii) F ½ mark and H½ mark// F and I// H and I
(iii) 2I(s) + 2H2O(l)
2IOH(aq) + H2(aq)
(iv) The ion has one energy  1 mark level less than H. G loses  1 mark its valency electrons with
much ease than H.
(v) G is more reactive than H. G loses its valences electrons with much ease than H.  1
3. (a) Hydrogen is obtained from methane and water  1 mark. Methane is reacted with water to give
carbon (IV) oxide and hydrogen.  1 mark
(b) N2(g) + 3H2(g)
2NH3(g)  1 mark
(c) (i) More yield  1 mark because the forward reaction is accompanied by decrease in volume
 1 mark
(ii) More yield  1 mark because the forward reaction is exothermic and is thus favoured by low
temperature  1 mark
(d) 3CuO(s) + 2NH3(g)
3Cu(s) + 3H2O(g) + N2(g)  1 mark
(e) The catalyst increases the rates of both forward and backward reactions  1 mark thus allowing the
equilibrium to be attained faster  1 mark
(f) 22.4dm3 of NH3 of NH3 = 1 mole
1 𝑚𝑜𝑙𝑒 𝑥 0.34𝑑𝑚3
034dm3 =
 ½ mark
22.4𝑑𝑚3
= 0.015 moles
0.015
From the rations in the equation the number of moles of (NH4)2SO4 = 2  ½ mark
= 0.0075 moles
1 mole of (NH4)2SO4 = 132g  ½ mark
0.0075 moles = 132g mol-1 x 0.0075 mol
= 1.0g  ½ mark
Mbooni West Sub-County Joint Evaluation 2014
Page 1
Chemistry Paper 2 Marking Scheme
4. (a) - Lubricating oil
- Fuel oil
- Diesel
- Kerosene
Mark atleast four components each ½ mark
- Petrol
- Bitumen
- Gasoline
- Naptha
(b) Thermal cracking is the breaking down 1 mark chosen alkaline using high temperature only, while
catalytic cracking involves breaking of long alkaline at lower temperature in the presence of a
catalyst  1 mark
(c) (i) W – Fermentation  ½ mark
Y = Oxidation  ½ mark
(ii) B – Ethane  ½ mark
C – Sodium ethanote  ½ mark
(iii) C2H5OH
Conc.
H2SO4
C2H4(g) + H2O(l)  1 mark
(iv) 2C2H6(g) + 5O2(g)
4CO2(g) + 6H2O(l)  1 mark
(v) Brown bromine  1 mark is decolourised in the presence of sunlight, substitution  1 mark
reaction takes place.
(vi) Rmm of ethane = 28  ½ mark
28n = 112000  ½ mark
112000
n = 28  ½ mark
= 4000  ½ mark
5. (a) E  1 mark
- Has the highest standard electrode potential  1 mark
(b) (i) K and E half-cells  1 mark
−
+
1
(ii) K(s)/K (𝑎𝑞) 2E2(g) / E(𝑎𝑞) pt E = + 4.28V  1 mark
(c) E.M.F = Ered – Eox
= -2.92 – (-0.44)
= -2.92 + 0.44  ½ mark
= -2.48V  ½ mark
Overall e.m.f is negative ½ mark, the reaction does not take place  1 mark
OR
+
2A(𝑎𝑞) + 2eA (s) E = -2.92V
2+
D(s)
D
+ 2e- E = +0.44V
(𝑎𝑞)
+
2+
2A(𝑎𝑞) + D(s)
A(s) + D
, E = -2.48V  1 mark
(𝑎𝑞)
E.M.F is negative  ½ mark, No reaction  ½ mark
II. (a) H: Anode  ½ mark
J: Cathode  ½ mark
(b) A burning splint is introduced in the mouth  1 mark of a test tube containing gas F. A “pop” sound
is produced  1 mark. (Reject: It burns with a “pop” sound)
(c) Q = 5 x [3 𝑥 60 + 21] ½ mark
= 5 x 180 + 21
= 5 x 201  ½ mark
= 1005C
−
4OH(𝑎𝑞)
2H2O(l) + O2(g) + 4e-  1 mark
4 x 96500C
24,000cm3
Mbooni West Sub-County Joint Evaluation 2014
Page 2
Chemistry Paper 2 Marking Scheme
1005 𝑋 24,000
1005C ⟨ 4 𝑋 96500 ⟩  ½ mark
= 62.49cm3  ½ mark
6. (a) DH = McDT
100
= 1000 x 4.2 x 8.5  1 mark
= 3.57Kj  ½ mark
0.32
Moles of CH3OH = 32 = 0.01moles  ½ mark
If
0.01 moles = 3.57Kj  ½ mark
1 mole = x
1 𝑥 3.57
x = 0.01  ½ mark
= -357 kJ/mol  ½ mark
(b) Exothermic  1 mark
(c) The experiment value -357kg/mole is lower than the accurate (theoretical) value  1 mark
– 638kJ/mole. This is due to: (i) Heat lost to the container and the atmosphere during the experiment.
 1 mark
3
(d) CH3OH (l) + 2 O2 (g)
CO2(g) + 2H2O (l) DH2 = -357kJ/mol  1 mark
(e)
Activation energy  ½
½
CH3OH(l) + O2 (g)
2X2Y(g)
Energy
kJ/mol
 ½ mark
DHc = -357kJ/mol
 ½ mark
CO2(g) + 2H2O ½ mark
Reaction progress  ½
7. (i) (a) Zinc blende  1 Mark
(b) Limestone  1 Mark
(ii) P – Carbon (IV) oxide  1 Mark
Q – Sulphur (II) oxide  1 Mark
(iii) R – Calcium oxide  1 Mark
S – Zinc (II) Oxide  1 Mark
(iv) ZnO (s) + C(s)
Zn(g) + CO(g)  1 Mark OR
ZnO(s) + CO(g)
Zn (g) + CO(g)  1 Mark
(v) Decomposes to give carbon (IV) oxide which is reduced by coke to form carbon (II) oxide
 ½ Mark. Carbon (II) oxide is a reducing agent  ½ Mark
(vi) - Marking galvanized iron sheets  1 Mark
- Make brass  1 Mark
- Casing in dry cells  1 Mark
Any two correct answers award 2
(vii) (i) Leclanche cells industries  1 Mark
(ii) Industries for manufacturing medals  1 Mark
Mbooni West Sub-County Joint Evaluation 2014
Page 3