02112020/CAPE/KMS 2018
C A R I B B E A N
E X A M I N A T I O N S
C O U N C I L
CARIBBEAN ADVANCED PROFICIENCY EXAMINATION®
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
MAY/JUNE 2018
– 2 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
SECTION A
MODULE 1
FUNDAMENTALS IN CHEMISTRY
Question 1.
Specific Objectives: Module 1. 6.6 – 6.9
KC
(a)
The total/standard enthalpy/energy/heat change for a chemical
reaction is independent of the route by which the reaction
proceeds/occurs. (1)OR
UK
1
The total/standard enthalpy/energy/heat change for a chemical
reaction is the same whatever route is taken. OR
The total/standard enthalpy/energy/heat change for a chemical
reaction is the same whether the reaction occurs in one
step/stage or through steps/stages
(b)
(i)
H1
–
Enthalpy of
potassium
(1)
atomization
(or
sublimation)
for
4
H2 – Enthalpy of atomization (or bond dissociation
energy) for fluorine
(1)
Students must indicate each element for marks to be
awarded
H3 – Enthalpy of first ionization energy for potassium
(1)
H4 – Enthalpy of electron affinity for fluorine
(ii)
(1)
4
H6 = Hlatt  (H1 + H2 + H3 + H4)
H5 = Hlatt = H6  (H1 + H2 + H3 + H4)
(1)
= 562.6  (89.6 + 419.0 + 79.1  332.6) (1)
= 562.6  255.1
= 817.7 (1) kJ mol1 with units
(1)
If student had no statements but only uses numerical values, in
the correct order,the candidates will receive the two marks.
If the student has ONE incorrect numerical value, they lose ONE
mark. If their numerical answer corresponds to what they have
2
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02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
written, they will be awarded that mark.
More than one incorrect numerical value, the student will ONLY
have access to the mark for the UNITS.
(iii) The theoretical and experimental lattice energy values
are close therefore, suggesting an ionic lattice with
ionic bonding (1) due to the large difference in
electronegativity of the ions. (1)
If the value calculated in (b)(ii) is much lower than it should
be and the candidate writes that “ the theoretical and
experimental lattice energy values are far apart suggesting
that the structure has covalent bonding or covalency. (Assign 1
mark)
Question 1. (continued)
KC
(c)



Pipette 50 cm3 of potassium hydroxide (or acid) solution
in a styrofoam cup and note the temperature, T1.
(1)
The temperature of the acid (or alkali) is noted and
acid placed in a burette, T2.
Add small portions (5 cm3) of acid (or alkali) from burette
to styrofoam cup,
mixing and
recording the highest
temperature, T3. (1)[Note addition can also be done all at
once]
T1 + T2
2
=
T4
(1)
T = T3 – T4
H = 100 g  4.18  T kJ
(1)[All terms must be correct]
UK
XS
4
– 4 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Total 15 marks
5
6
4
– 5 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
MODULE 2
KINETICS AND EQUILIBRIA
Question 2.
Specific Objectives: Module 2. 1.1 – 1.8
KC
(a)





Surface area
Concentration of reactants
Temperature of reactants
Catalysts
Light
UK
2
Any 2 for 2 marks
(b)
(i)
2
Change in concentration of nitrogen (1)
Time taken (1)
OR
Change in amount/concentration of nitrogen (1) per unit
time/per second (1)
OR
Δ[N2]/Δt
(ii)
Hydrogen:
First order with respect to hydrogen
1
(1)
Nitrogen oxide:
Second order with respect to nitrogen oxide
1
(1)
(iii) Rate law:
Rate = k [NO]2 [H2] (1)
If student has lost a mark for an incorrect order of the
reaction in part (b)(ii) and their rate law matches that
error, the student will be awarded this mark.
1
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02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
– 7 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 2. (continued)
KC
(b)
(iv)
UK
XS
Choosing values from experiment 2 or any other:
[NO] = 6  103 mol dm3
[H2] = 2  103 mol dm3
3
Rate = 6  103 mol dm3 s1
6  103 = k (6  103)2 (2  103)
Ensure that student values ACTUALLY match an experiment. If
values DO NOT match an experiment,the ONLY mark available to
the student is the one for UNITS. Note that the units must
match with their rate law written in part (b)(iii).
6
-3
k =
6 × 10
6 × 10  × 2 × 10
-3 2
-3
=
10
12
(1)
= 8.33  104 (1)mol2 dm6 s1 units (1)
Note – if student has incorrect rate law but all expressions
and calculations are consistent with incorrect rate law, NO
MARKS ARE DEDUCTED i.e use student rate law from b(iii).
(c)
(i)
The
reactants
and
products
are
all
gases
so
a
measurement of pressure at suitable time intervals would
be appropriate. (1)
(ii) 






The [HCOOH] is constant due to amount being in
large excess.
Bromine (aqueous) is reddish brown and [Br2]
falls during course of reaction.
The intensity of the reddish brown colour can be
followed by a colorimeter at suitable time
intervals.
A calibration curve of known [Br2] against the
colorimeter readings plotted and [Br2] deduced.
The [Br2] is plotted against time.
Tangent is drawn to the curve at given times to
obtain reaction rate.
The reaction rate is plotted against the [Br2].
6-7 steps 4 marks; 4-5 steps 3 marks;
2-3 steps 2 marks;
1 step 1 mark
1
(1)
(1)
4
(1)
(1)
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02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
– 9 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 2. (continued)
(c) (ii) 




Carbon dioxide gas is produced. (1)
The volume of carbon dioxide is measured at
intervals. (1)
The volume of CO2 is plotted against time. (1)
The tangent is drawn to the curve to determine
reaction rate.
The reaction rate is plotted against volume of
CO2.
KC
UK
XS
4
6
5
time
(1)
5 steps
4 marks
4 steps
3 marks
3 steps
2 marks
1-2 steps 1 mark
Total 15 marks
– 10 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
MODULE 3
CHEMISTRY OF THE ELEMENTS
Question 3.
Specific Objectives: Module 3. 3.3 – 3.4
KC
(a)
Properties
Group IV Elements in
the +2 oxidation state
CO
Thermal
Stability
SiO
PbO
UK
Group IV Elements in the
+4 oxidation state
CO2
SiO2
PbO2
Readily
oxidized
to
dioxide
Stable
Stable
Changes
to PbO
(on
warming)
(1)
(1)
(1)
(1)
4
1 mark for each
(b)
Melting point of CO2 is 56 °C.
CO2 is a gas at room temperature with simple molecular
structure and weak /van der walls or intermolecular forces.
(1)
Melting point of SiO2 is 1610 °C.
SiO2 has a giant molecular/covalent structure/macromolecular in
its solid
state. (1)
Melting point of PbO2 is 290 °C.
PbO2 has a giant covalent/molecular structure with ionic
character OR some degree of covalency in ionic structure. (1)
3
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– 11 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
– 12 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 3. (continued)
KC
(c)





(d)
The stability of the +2 oxidation state decreases in moving
from PbO, SiO to CO. (1)
The bonding in the PbO compound is ionic and PbO is more
stable due to unreactive ‘s’ electrons or inert pair effect,
which is not present in carbon in CO. (1)
OR The electrode potential value, Eө is more positive for
Pb4+(aq)/Pb2+(aq) system.
The stability of the +4 oxidation state increases in moving
from PbO2, SiO2 to CO2. (1)
CO2 possesses the lowest oxidizing power. (1)
OR PbO2 accepts electrons readily. (1)OR PbO2 exhibits strong
oxidizing power. (1)
(i)
UK
XS
4
The solid dissolves.
A white precipitate (1) is formed
and a greenish yellow, (pungent gas) (1) evolves. (color
of gas MUST be noted for student to be awarded mark)
2
1
(ii) The solid dissolves and colourless solution forms.
(1)
1
(iii) The pale green solution
brown solution. (1)
changes
to
yellow
or
reddish
Total 15 marks
4
7
4
– 13 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
– 14 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
SECTION B
MODULE 1
FUNDAMENTALS IN CHEMISTRY
Question 4.
Specific Objectives: Module 1. 5.1 – 5.4
KC
(a)
The kinetic theory for an ideal gas:




(b)
UK
Particles move randomly. (1)
Particles do not attract each other/no intermolecular forces
(1)
Particles have no volume/negligible volume (1)
Collision between particles are elastic, there is no loss of
energy. (1)
(i)
Real gases deviate from ideal gas behavior
conditions of high pressures and low temperatures.
4
under
(1)
Under these conditions:


Attractive forces between gas particles pull particles
closer/ intermolecular forces are now present (1)
Volume of particles is not negligible compared to
overall volume. (1)
3
(ii)
3
OR atm
Labeled axes (1) (unit only required for x-axis
Reference ideal gas (1)
Illustration of one real gas (1)
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– 15 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 4. (continued)
KC
(c)
(i)
UK
The molecules in the liquid state are:


Close together/In clusters and not arranged in regular
order.
Have more kinetic energy than particles in the solid
state OR particles move less randomly than in gaseous
state.
2
1 mark for a statement on arrangement
1 mark for a statement on motion
(ii) Volume of vaporized liquid = (54.6  18.4) cm3
or implied)
(1)(seen
36.2 cm3 = 36.2  106 m3
[If no subtraction is performed OR if no conversion to m3
is done, student loses one mark]
p = 1.01  105 Pa
t = 273 + 57 = 330 K
r = 8.314 JK1 mol1
pv =
m
M
= 1.01  105  36.2  106
rt
(1)
0.187
 8.314  330 (1)
M
[all values must be correct for allocation of third mark]
=
M = 140
Alternatively,
calculate M:
student
may
solve
for
n,
then
use
Pv =nRT = n 8.314 x 330
n = (1.01 x 105 x 36.2 x 10-6)/ (8.314 x 330)
n = 3.62 / 2743.62
(1)
= 0.00132
0.00132 moles has a mass of .187g
Therefore 1 mole has a mass of 141.6g
(1)
m
and
3
XS
– 16 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Total 15 marks
6
9
-
– 17 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
MODULE 2
KINETICS AND EQUILIBRIA
Question 5.
Specific Objectives: Module 2. 6.1 – 6.7
KC
(a)
(i)
The potential of that half-cell relative to a standard
hydrogen electrode (1) under standard conditions. (1)
UK
2
(ii)
4
Fe2+/Fe3+ (aq) 1 mol dm-3
8 – 9 labels
4 marks
6 – 7 labels
3 marks
4 – 5 labels
2 marks
2 – 3 labels
1 mark
1 label
0 mark
(iii) Standard electrode potentials can be used to:




Predict the feasibility of a reaction/if the reaction
can occur.
Indicate the oxidizing/reducing ability of species.
Determine the direction of electron flow in a cell.
Calculate standard cell potentials
Any two
2
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– 18 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 5. (continued)
KC
(b)
(i)
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
state symbols)
(ii)
Cu(s)  Cu2+(aq) + 2e  0.34V
UK
1
(1) (balanced with
2

2Ag (aq) + 2e  2Ag(s) + 0.80 V
(1)

2+
Cu(s) + 2Ag (aq)  Cu (aq) + 2Ag(s)
OR Eθ cell = ERHS – ELHS OR Eθ = Ecathode- Eanode = 0.80V - 0.34V (1)
Eө = +0.46V (WITH UNIT)
(1)
2
(iii) The positive electrode potential value of emf value (1)
indicates that the reaction is feasible. (1)
If the candidate has a negative EӨ value, then accept the
appropriate answer as: the negative electrode potential
value (1) indicates that the reaction is not feasible (1)
(iv) If the concentration Ag+ increases, according to Le
Chatelier’s Principle, the equilibrium will shift to
the right (1) and the cell potential will become more
positive. (1)
2
XS
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02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Total 15 marks
6
9
-
KC
UK
XS
MODULE 3
CHEMISTRY OF THE ELEMENTS
Question 6.
Specific Objectives: Module 3. 4.1 – 4.6
(a)
(b)
The volatility of the elements decreases down the group.
(1)
This is due to the increasing strength of Van der Waal forces
as the molecular mass increases. (1)
2
The Eө values for the halogens become less positive in moving
down the group from fluorine to iodine and this is an
indication of the decreasing oxidizing (1) power or ability of
the halogens. OR Based on the values, fluorine and chlorine are
strong oxidizing agents (1) and iodine is the weakest oxidizing
agent.
This is illustrated in their reactions with sodium thiosulfate.
Fluorine, chlorine and bromine can all oxidize (1) the
thiosulfate ion, S2O32 to the sulfate ion, SO42. (1)
4
Iodine oxidizes the thiosulfate ion to the tetrathionate ion,
S4O62. (1)
(c)
(i)




Fluorine explodes with hydrogen in the dark.
OR Fluorine reacts with hydrogen even under cool
conditions/low temperature. (1)
Chlorine reacts slowly with hydrogen in the dark.
OR Chlorine explodes with hydrogen in sunlight. (1)
Bromine reacts with hydrogen on heating (and in the
presence of platinum catalyst). (1)
Iodine
reacts
partially
and
slowly
even
when
heated. (1)
4
– 20 –
02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Question 6. (continued)
KC
(c)
(ii)
•The bond energies for the hydrides:
HF > HCl > HBr > HI
HF is the most stable hydride and HI the least stable
hydride. (1) OR The length of the H-X bond increases
as we descend group VII. Longer bonds are weaker bonds
and hence the strength of the H-X bond decreases.
•Fluorine is the strongest oxidizing agent and the most
reactive and forms
strong
covalent bonds with
hydrogen.
(1)
OR
Flourine
having
the
highest
electronegativity forms the most polar bonds with
hydrogen giving rise to strong intermolecular hydrogen
bonding between molecules.
 The reactivity of chlorine, bromine and iodine
decreases with hydrogen and the strength of the
covalent bonds in the hydrides decreases down the
group. (1)
(iii) H2(g) + F2(g)  2 HF(g)
OR
(balanced with state symbols)
H2(g) + X2(g)  2 HX(g)
(2)(ALL reagents are gaseous)
X2 = Cl2, Br2, I2
1 mark – balanced equation (correct formula)
1 mark – state symbols
UK
3
2
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02112020/CAPE/KMS 2018
CHEMISTRY
UNIT 1 – PAPER 02
KEY AND MARK SCHEME
Total 15 marks
6
9
-