02112020/CAPE/MS 2015
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
MARK SCHEME
MAY/JUNE 2015
-202112020/CAPE/MS/2015
CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 1
Specific Objectives: — 6.3, 6.4, 6.5, 6.6, 6.9
KC
(a)
(b)
(i)
Bond energy — energy required to break the covalent
bonds in one mole of a gaseous substance/ molecule. Or
the energy given out when one mole of a substance/
molecule is formed.
1
(ii)
As the bond length decreases, the bond strength
increases. Or shorter bond length , stronger bond
1
(i)
Eqn:
CH4(g) + Cl2(g)
CH3 C1
+ HCl(g)
(g)
(Must have BOTH correct formulae (1 UK) and state
symbols) (1 UK)
(ii)
UK
Hrxn    (bond breaking) -   (bond forming) (1 KC)
( or if 1st statement is implied
get 1 mark)
2
1
Hrxn = (410 + 244) kJ mol-1 - (340 + 431) kJ mol-1
-1
-1
= 654 kJ mol - 771 kJ mol
-1
= -117 kJ mol
(1 UK)
Alternative calculation
( 4x 410) + 244 – [(3x410) + 340 +431]
= 1884 – 2001
= -117 kJ mol -1
(Must have both units and value) based on students’
eqn in (b) (i),)
(1 UK)
(iii) Exothermic
(based on students’ answer in (b)(ii))
If student got a positive response for (b)ii then
their answer must be endothermic
1
1
1
(iv)





shape of curve [1 UK]
labels
reactants Any 2 correct
products
[1 UK]
∆Hrxn

If (b)ii and (b)iii responses are omitted no marks can
be awarded for (b)iv unless the student states
exothermic on a properly labelled diagram. 1 mark is
awarded.
1
1
XS
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 1 cont’d
KC
(c)
UK
XS
Steps
Steps 1& 2 1mk
1. Measure temperature of 75 mL, 1.0 M KOH solution. (T1)
2. Measure temperature of 75 mL, 1.00 M HCl. (T2)
3. Place thermometer in KOH solution.
steps3&4 1mk
4. Add HCl, stirring to ensure complete mixing.
5. Record highest temperature observed. (T3)
6. Calculate
step5 1mk
average initial temp [(T1 + T2) / 2] (T4)
Step6 1mk
7. Calculate T = T3 – T4
8. Calculate H neutralization using; H = mc T
Where m = 150 g = .150 kg
c = 4.18 kJ kg-1 K-1
T = T3 – T4
[5 XS]
Must show M and
T
Total 15 marks
steps 1mk
3
7
5
5
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 2
Specific Objectives Module 2: — 6.1, 6.3, 6.4, 6.5
KC
(a)
(i)
(ii)
(b)
(i)
The potential of that half-cell relative to a standard
hydrogen (1 KC) electrode under standard conditions 1KC
or the potential difference of the half cell and the
standard hydrogen electrode 1KC under standard
conditions.
(1 KC)
2
The maximum potential difference/ electromotive
force(EMF)/ electrode potential between two half cell(1
KC) between two half-cells connected under standard
conditions (1 KC)
2
ANODE: A
(S)
⇌ A
3+(aq)+
3e-/
Al(s) – 3e⇌ A
3+(aq)
UK
XS
(1 UK)
1
CATHODE: Sn2+(aq) + 2e- ⇌ Sn(s) (1 UK) (ignore if the arrow
is single or reversible, both arrow should be of the same type)
1
If the the anode and the cathode are in reversed positions give 1
mk
A (s)| A 3+(aq) || Sn2+(aq) | Sn(s) (1 UK)
State symbols must be shown
(ii)
1
(iii)




Or 25оC
direction of electron flow
indication of temperature
concentrations of solutions 1 m
indication of salt bridge
(1
(1
(1
(1
UK)
XS)
XS)
XS)
1
1
1
1
1
1
ensure salt bridge in
is solution


(iv)
labelling of electrodes
labelling of ions in solution
A
(s)
⇌ A 3+ (aq) + 3e-,
(1 XS)
(1 XS)
1
E⊖ = + 1.66 v
(1 UK)
Sn2+ (aq)+2e ⇌ Sn(s),
E⊖ = - 0.14 V
 E⊖ cell = + 1.66 V – 0.14 V
= + 1.52 V
(1 UK)
If half equations were not stated and it is implied in
the calculation give full marks
Total 15 marks
1
4
6
5
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 3
Specific Objectives Module 3: — 3.3, 3.4, 6.2, 6.3
KC
UK
XS
(a)
Formula of
Compounds
CO
Acid/Base
Nature
Thermal
Stability
Oxidation
State of
Group IV
Element

(b)
(i)








CO2
Pbo
acidic
amphoteric
unstable
+2
stable
+4
Pbo2
unstable
+4
(1)
(1)
(1)
(1)
Must have both answers for each compound to be awarded
(1 mark).
4
+2 state in PbO is more stable than +2 state in CO. (1)
The bonding in PbO is ionic and its structure is giant
ionic. (1)
The loss of p electrons in lead result in more stable
unreactive s electrons or inert pair effect. (1)
The bonding in CO is molecular and its structure is
simple
molecular
and
this
is
maintained
in
a
tetravalent covalent compound. (1)
OR
+2 state in PbO is more stable than +2 state in CO. (1)
The loss of p electrons in lead result in more stable
unreactive s electrons or inert pair effect. (1)
This as a result of the poor shielding of the s
electrons by the d orbital in descending the group. (1)
So electrons are difficult to remove which results in
stability. (1)
4
(ii)


(c)
(d)
The +2 state of the lead ion is more stable than the +4
state. (1)
In moving from the +4 state to +2 oxidation state the
E⊖ value is more positive. (1)
2
(i)

The solid dissolves / forms a colourless solution. (1)
1
(ii)

The solid dissolves to form a white precipitate (1) and
a green/yellow pungent gas evolves. (1) (yellow
gas/pungent gas acceptable).
2


Aqueous potassium iodide/ chromate added to lead ions,
Pb2+, produces (1) a yellow precipitate. (1)
Any aqueous chloride or sulphate (1) added to form
white ppt.(1)
2
Total 15 marks
4
6
5
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 4
Specific Objectives Module 1: — 1.4, 1.8, 1.9
KC
(a)
(i)
UK
Three types of radiation



alpha particles
beta particles
gamma radiation
3
(1 x 3 KC)
Symbols
4 He  alpha particles
2
0e
-1

- beta particles
α
β or
(3 KC)
e-
3
if number are written
on right hand side
- gamma ray
accept.
Penetrating power was removed from mark scheme
(ii)
Nuclear equations
241 Am 
95
4 He + 237 Np (1 UK)
2
93
237 NP 
93 NP
4 He + 233 Pa (1 UK)
2
91
2
Note : NP is really Np
NP
241
Am
95
(b)
4
2
He
2
+
233
Pa
91
(1UK)
(i)
(1 UK)
XS
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
If all four orbitals are on the same axes award 2mks.
KC
UK
XS
2
All orientations correct (1 UK)
(b)
(ii)
)Similarity

1
Identical arrangement for the first 18 electrons in
each species. (1 UK)
Difference



The single electron in the valence (1 UK) shell of K
is in a 4s orbital, whilst the single electron for Sc
resides in an 3d orbital.
There are no 4s electrons in Zn2+
There are no d electrons in K
1
Comment




After 18 electrons, the electrons which follow fill
the 4s before the 3d. (1 UK)
Once the 3d are filled, 4s electrons (1 UK)
are
removed
before
the
3d,
hence
the
Zn2+
arrangement.
(1 UK)
All three species have identical filled orbitals up
to 3p6.
(1 UK)
3
Zn2+ has no 4s electrons [due to their loss in the
formation of Zn2+].
(1 UK)
Any 3 — 3 x 1 mark
Total 15 marks
6
9
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 5
Specific Objectives Module 2: — 4.1, 4.2, 4.3, 4.5
KC
(a)
H2PO4- (aq) ⇌ HPO42-(aq)+ H+ (aq) or HPO42-(aq)+ H+ (aq)
⇌ H2PO4- (aq)
(2UK)

Added H+ reacts with HPO42(1) to give H2PO4-(1).
Equilibrium shifts to which ever direction depending
on equation stated (1) to maintain the pH
(1 KC)
Or
OH- (aq) +H2PO4- (aq) ⇌ H2O(l) + HPO42-(aq) 2uk
Students can only be given the mark for how the
buffer works with H+ or OHAdded OH– reacts with H+ (1 KC)
Removal of H+ causes equilibrium to shift. (1 KC)
H2PO4- dissociates to replace H+ ions. (1 KC) to
maintain the pH
(1 KC)



UK
2
5
(b) H2PO4-(aq) ⇌ HPO42- (aq) + H+ (aq)
2-
 Ka =
+
[HPO4 ][H ]
-
[M2PO4 ]
[H2PO4-]
-
[H2PO4 ] = 0.188
0.153 MM
-
[HPO4 ] = 0.129
M
0.159 M
+
[H ] = Ka
[H2PO4-]
0.153
0.188
-8
× ..15
2- = 6.4 × 10
0.129
0.159
[HPO4 ]
35
-8
6.1 ×x 10
10-8 M
= 9.3
pH =
=
=
=
+
-log [H ]
6.1 x 10-8 -8
-log (9.3
 10 )
- (-7.03
)
-7.2
7.03
7.2
Or using Henderson-Hasslebalch equation,
pKa = -log 6.4 x 10-8
pH
= -log 6.4 x 10-8
- log 0.153
0.159
pH = 7.19
(c) Biological buffers are important because
 Enzymes function at pH ≈ 7 OR
 Enzymes function over a very narrow pH
(1 KC)
 Biological reactions involve enzymes as catalyst
(1 KC)
 Blood buffers helps to maintain blood pH at around
7.4. (1 KC)

To prevent acidosis or alkalosis (1KC)
Any 3 points — 2 KC
Any 2 points — 1 KC
5
XS
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 5 cont’d
KC
UK
2
2
7
8
Examples of blood buffers include
(i)
hydrogen carbonate buffer:
CO2(g) + H2O( ) ⇌ HCO-3 (aq) + H+(aq)
(ii)
Phosphate buffer
H2PO4-(aq) ⇌ HPO42-(aq) + H+(aq)
(iii) Protein buffers
PH ⇌ P- + H+ or PH+ ⇌ P + H+
Any example of chemical reaction of blood buffer
Balance + state symbols — 2 UK
Total 15 marks
XS
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 6
Specific Objectives Module 3: — 2.1, 2.3, 2.4
KC
(a)
(b)

A fully filled orbital/shell is added from one element to
the next. (1)

The screening effect also increases and shields the outer
electrons from increasing nuclear charge. (1)

The outer electrons are further away from the nucleus.
(1)
3
Variation:
There is a slight increase in melting point from
magnesium to calcium, (1) and then a general decrease
from calcium to barium. (1)

(c)
UK
1
As the atomic/ionic radii increase down the group:

there is less attraction between the ions and
delocalized electrons (1)

therefore less energy required to melt metals. (1)
1
2
Trend:
The solubility of the sulfates decreases down the group. (1)

The large size of the sulfate anion and the slight
increase in the size of the cations (1) results in small
decrease in lattice energy. (1)

The change in density of the cations decreases due to
slight increase in size of cations and hydration energy
decreases (1).

The decrease in hydration energy becomes more significant
than the decrease in lattice energy and the solubility of
metal sulfates decreases down the group (1).
1
4
XS
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CHEMISTRY
UNIT 1 – PAPER 02
MARK SCHEME
Question 6 cont’d
KC
(d) Metal nitrates become more thermally stable going down the
group (1) OR Ease of decomposition decreases as you go
down the group.

UK
1
Size of cations increases and Large cations cannot
(is difficult to) polarize the large N03- ion(1) to
break the bonds for decomposition/ to form (stable)
oxide compounds (1).
OR the smaller ion polarizes the nitrate easily to
form oxide compounds.
2
Total 15 marks
6
9
XS