2
Section A
For each question there are four possible answers, A, B, C, and D. Choose the one you
consider to be correct.
1
Which one of the following statements about 20.3 g of Co2(SO4)3 is incorrect?
[Molar mass = 406 g mol-1, Avogadro’s constant = 6.0 x 1023 mol-1]
2
A
It contains 0.10 mol of Co3+ ions.
B
It contains 0.15 mol of SO42- ions.
C
It contains 47.3% of oxygen by mass.
D
It contains 1.5 x 1023 atoms.
In theory, 3 mol of NH3OH+(aq) is required to convert 1 mol of BrO3–(aq) to Br–(aq).
Which one of the following statements about this reaction is correct?
3
4
A
NH3OH+ is acting as an oxidising agent.
B
The change in oxidation number of nitrogen is +4.
C
N2O is the nitrogen–containing product.
D
The reaction is likely to have a high activation energy.
Which one of the following is a possible configuration of a stable M3+ ion in the ground
state?
A
1s2 2s2 2p3
B
1s2 2s2 2p6 3s2 3p1
C
1s2 2s2 2p6 3s2 3p6 3d2
D
1s2 2s2 2p6 3s2 3p6 3d2 4s2
The pressure rating of a car tire is typically given to be 210 kPa. This is based on an
internal air temperature of 25C.
A driver decides to inflate the tires of his car to the rated pressure after driving at a high
speed for some time, during which the internal air temperature has risen to 40C.
What will be the drop in pressure in each tire when the temperature has returned to
25C?
A
 VJC 2015
9.3 kPa
B
10.1 kPa
C
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78.8 kPa
D
200 kPa
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5
LiAlH4 and NaBH4 are used as a source of hydride ions, H–, in many organic reactions.
Both chemical formulae consist of a metal cation and an anion made up of a group of
atoms.
Solid LiAlH4 decomposes upon heating to 150 C while solid NaBH4 melts at a
temperature of 400 C.
Which one of the following statements about LiAlH4 and NaBH4 is correct?
6
A
The H– from LiAlH4 is more available for donation due to the higher polarity of the
Al–H bond compared to B–H bond in NaBH4.
B
The release of H– from LiAlH4 and NaBH4 involves homolytic fission of a covalent
bond.
C
The lower stability of LiAlH4 to heat can be explained by the higher charge density of
the cation and anion present compared to NaBH4.
D
The presence of a distinct melting point for NaBH4 means that the covalent bonds in
the anions are weaker than the ionic bonds between the cations and anions.
The enthalpy change of the following half equation (ΔHro) is determined
using Hess’s Law:
ΔHro
–
½ I2(s) + e
I–(aq)
ΔH1o
ΔH3o
I(g) + e–
ΔH2o
I–(g)
Which one of the following shows the correct signs for the different
enthalpy changes?
7
ΔH1o
ΔH2o
ΔH3o
A
+
+
–
B
+
–
–
C
+
–
+
D
–
+
+
Hydrogen peroxide is thermodynamically unstable and decomposes over time to form
water and oxygen gas.
H2O2(l)  H2O(l) + ½O2(g)
The following data are provided:

O2(g) + 2H2(g)  2H2O(l)

O2(g) + H2(g)  H2O2(l)
ΔH1o = –572 kJ mol–1; ΔS1o = –325 J mol–1 K–1
ΔH2o = –188 kJ mol–1; ΔS2o = –225 J mol–1 K–1
What is the value of ΔGo, in kJ mol–1, for the decomposition of hydrogen peroxide?
A
 VJC 2015
–117
B
–354
C
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–359
D
–596
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8
Artificial pacemakers are used to regulate the heartbeat of cardiac patients. The device is
powered by a water–free lithium iodide battery as shown below.
(–) electrode
(+) electrode
LiI crystal
Li(s) anode
Ni(s) mesh
I2 complex cathode
Adapted from UC Davis ChemWiki
Which one of the following statements about the lithium iodide battery is correct?
9
A
The I2 complex cathode carries a negative charge.
B
The cell potential of the lithium iodide battery has a value of +3.58 V at the
conditions prevailing in the battery.
C
The Ni mesh provides a medium for electrons to flow from Li(s) to I2 complex.
D
The pacemaker will be able to last for more than 5 years if it contains 6 g of reactive
Li(s) and operates at an average of 0.8 mA.
A gaseous dimer, X, is introduced into an empty gas syringe which has a movable, tightly–
fitting plunger. The gas is allowed to expand until equilibrium is reached at a controlled
temperature at which 20 % of X dissociates into its monomer W.
X(g)
self–sealing
cap for
introducing
sample
2W(g)
gas
plunger
syringe oven for controlling
equilibrium temperature
Which one of the following statements is correct?
A
The pressure inside the syringe at equilibrium will be higher than the atmospheric
pressure.
B
The forward reaction is exothermic.
C
The value of the equilibrium constant, Kp is 0.167 atm
D
The dissociation of dimer X will be favoured when the plunger is pushed back into
the equilibrium mixture.
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10
The table below gives the concentration and pH values of the aqueous solutions of two
compounds S and T. Either compound could be a monoprotic acid or base.
S
concentration
T
2 mol dm
pH
–3
2 mol dm–3
2
11
Student P concluded that S(aq) is a strong acid.
Student Q concluded that the extent of dissociation is lower in S(aq) than in T(aq).
Student R concluded that no suitable indicator exists for the titration of S(aq) and T(aq).
Which of the students are correct?
11
A
P and Q only
C
R only
B
Q and R only
D
none of the students
The reaction between Pb2+(aq) and different halides is investigated.
When solid NaCl or NaI is added in increasing amounts to a solution of Pb2+(aq) in
separate experiments, the following graphs are obtained.
total concentration of
lead(II)–containing species
in solution / mol dm–3
Cl
–
I–
0
V
amount of halide / mol
Which one of the following deductions is correct?
A
The increase in both graphs after V cm3 is due to common ion effect.
B
The concentration of Pb2+(aq) for both graphs is at its minimum at V cm3.
C
The ionic product is higher than the solubility product for both salts between 0 and
V cm3.
D
The precipitate that will be formed first if Pb2+(aq) is added to a solution containing
equal amounts of Cl–(aq) and I–(aq) is lead(II) chloride.
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12
The energy profile of the reaction between M and N is shown below.
energy
E1
MN + N
E2
MN2
M + 2N
progress of reaction
Which one of the following is a correct statement?
A
The rate equation for the forward reaction is rate = k[M][N]
B
The activation energy for the reverse reaction is calculated by E1.
C
The equilibrium concentrations of M and N decrease when temperature decreases.
D
Only the rate of the forward reaction increases when temperature increases.
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13
The graph below shows the ionic radius of five consecutive elements in the third period of
the Periodic Table.
ionic radius
Y
Z
V
W
X
0
proton number
Three corresponding graphs of the same elements are provided below, each with a point
that is wrongly represented.
1st ionisation energy
electrical conductivity
W
melting point
W
X
Z
Z
V
V
X
X
V
Y
Y
0
proton number
0
Y
W
Z
proton number
0
proton number
Which of the following correctly identifies the error point in each of the graphs?
14
melting point
electrical conductivity
1st ionisation energy
A
X
X
W
B
X
Y
Y
C
Z
X
Y
D
Z
Y
W
Which one of the following statements of Group II elements (magnesium to barium) is
correct?
A
The tendency to form complex ions decreases down the Group.
B
The reactivity with cold water decreases down the Group.
C
The acidity of an aqueous solution of the Group II chlorides increases.
D
The minimum temperature required for the decomposition of the Group II nitrates
decreases.
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15
16
Which one of the following best explains why barium hydroxide has a higher solubility than
magnesium hydroxide?
A
The hydroxide ion is a very small anion.
B
The charge density of barium ions is smaller than that of magnesium ions.
C
The hydration of barium ions is more exothermic than that of magnesium ions.
D
The lattice energy of barium hydroxide is less exothermic than that of magnesium
hydroxide.
The Group VII element astatine, At2, is radioactive with a short half-life.
Which one of the following statements is most likely to be correct for astatine or its
compounds?
17
A
At2 reacts with S2O32– to produce SO42–.
B
AgAt is soluble in concentrated aqueous NH3.
C
KAtO can be obtained by bubbling At2 into hot, aqueous KOH.
D
HAt cannot be prepared by heating KAt with concentrated H2SO4.
Which set of data correctly illustrates copper as a typical transition element and calcium as
an s-block element?
property
copper
calcium
A
density /g cm−3
8.92
1.54
B
electrical conductivity /relative units
9.6
85
C
melting point /C
810
1083
D
metallic radius /nm
0.197
0.117
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18
Living systems have difficulty assimilating iron because most iron compounds found in
nature are not very soluble in water. Microorganisms have adapted to this problem by
secreting an iron-binding compound called a siderophore that forms a water soluble
complex with iron(III). One such complex is ferrichrome.
Ferrichrome
Which one of the following statements about ferrichrome is not correct?
A
The siderphore is a hexadentate ligand.
B
Iron(III) ions only form ionic bonds in the ferrichrome.
C
The electronic configuration of the central metal ion is [Ar] 3d5.
D
The overall charge of ferrichrome is zero.
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19
Compound P, a derivative of a cholesterol lowering drug, is shown below.
P
What is the correct number of chiral centres in P and the number of carbon-containing
compounds produced in its reaction with hot H2SO4(aq)?
20
number of chiral
centres in P
number of carboncontaining compounds
after reaction
A
8
3
B
8
6
C
9
3
D
9
6
The following compound is reacted with NaBH4 in dry ethanol.
How many sp2 and sp3 hybridised carbon atoms does a molecule of the product contain?
21
sp2
sp3
A
0
6
B
2
4
C
3
3
D
4
2
A compound of molecular formula C8H10 is allowed to irradiate with bromine under uv
light to produce a mixture of monosubstituted products.
What is the total possible number of isomeric products, including stereoisomers, that
can be obtained?
A
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3
B
5
C
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6
D
9
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22
Aromatic compounds such as benzene can undergo Friedel Crafts acylation using an acyl
chloride, RCOCl in the presence of a halogen carrier such as AlCl3 as shown below.
To obtain compound A, the following three steps are required:
compound A
Step 1: acylation
Step 2: bromination
Step 3: nitration
Which order of the above steps would give the best yield of compound A?
23
A
1
2
3
B
1
3
2
C
2
1
3
D
2
3
1
A student synthesised the following compound in the laboratory and wrote four statements
about this compound.
Which one of the statements below is incorrect?
A
On heating with acidified K2Cr2O7(aq), the colour of the solution turns green.
B
On warming with 2,4-dinitrophenylhydrazine, no orange crystals will be formed.
C
On heating with alkaline aqueous iodine, a pale yellow precipitate will be formed.
D
On adding Br2(aq), one molecule of the compound will incorporate one bromine
atom in the major product.
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24
An organic compound, C10H11NO2 has the following physical and chemical properties:
 It gives a brick red precipitate when warmed with Fehling’s solution.
 It gives a purple colouration when warmed with neutral FeCl3(aq).
 It is moderately soluble in water.
Which one of the following is a possible structure of this compound?
25
A
C
B
D
The following compound is derived from DDT, an insecticide.
The compound is boiled with NaOH(aq), cooled and the resultant solution is acidified with
dilute HNO3(aq). AgNO3(aq) is then added to the mixture before concentrated NH3(aq) is
added dropwise, until present in excess.
Which observation is made when NH3(aq) was added?
A
All the precipitate remains.
B
All the precipitate dissolves.
C
The precipitate appears less yellow.
D
The precipitate appears more yellow.
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26
The aldol reaction between two carbonyl compounds involves the removal of a H+ from a
carbon atom adjacent to the carbonyl group by NaOH, and subsequent formation of the
product. An example is shown below.
The following compound is a diketone.
Which one of the following products cannot possibly be formed from the above diketone
via the aldol reaction?
27
A
C
B
D
Below are four nitrogen-containing compounds, labelled 1 - 4.
CH3CHClNH2
1
CH3CH2NH2
2
CH3CONH2
3
C6H5NH2
4
Which one of the following gives the correct order of decreasing pKb values of the above
compounds?
A
2
3
4
1
B
3
4
1
2
C
4
1
2
3
D
4
3
1
2
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28
1 mol of the tetrapeptide tyr-ser-gln-gly is boiled with NaOH(aq) until no further reaction
occurs.
amino acid
R group
gln
CH2CH2CONH2
gly
H
ser
CH2OH
tyr
How many moles of NaOH will react?
A
29
4
B
C
5
6
D
7
Peptide P contains seven amino acid residues. When P is partially hydrolysed, the
following dipeptide and tripeptide fragments are produced.
gly-ala
ser-lys
ala-gly-ser
ala-met
lys-gly
What could be the structure of peptide P?
A
gly-ser-lys-ala-met-gly-ala
B
ala-met-gly-ala-gly-ser-lys
C
ala-gly-ser-lys-gly-ala-met
D
lys-gly-ala-met-ala-gly-ser
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30
The amino acid threonine has the following structure:
OH
OH
H2N
O
In alkaline solution, it exists as
O-
OH
A
O-
H2N
B
H2N
O
O
OH
C
O-
+
H3N
OH
D
OH
+
H3N
O
 VJC 2015
O
O
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Section B
For each of the questions in this section, one or more of the three numbered statements 1 to 3
may be correct.
Decide whether each of the statements is or is not correct (you may find it helpful to put a tick
against the statements that you consider to be correct).
The responses A to D should be selected on the basis of
A
B
C
D
1, 2 and 3
are correct
1 and 2
only are correct
2 and 3
only are correct
1 only
is correct
No other combination of statements is used as a correct response.
31
Z is an organic compound containing only carbon, hydrogen and chlorine.
When a given sample of Z is burnt completely in air, 2.75 g of CO2 and 0.281 g of H2O are
produced.
Which of the following are possible structures of Z?
1
2
3
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32
33
In which of the following pairs of compounds would the first compound have a higher
melting point than the second compound?
1
Na2O, K2O
2
AlF3, AlCl3
3
NH2CH2CO2H, HOCH2CO2H
The graph below shows how the fraction of a substance, X, produced in an equilibrium
mixture varies with temperature at pressures of 2 x 107 Pa and 5 x 107 Pa respectively.
Fraction of X
2 x 107 Pa
5 x 107 Pa
Temperature
In which of the following equilibria would X represent the underlined species?
34
1
C(s) + H2O(g) ⇌ H2(g) + CO(g)
ΔH = +131 kJ mol-1
2
C2F4(g) + 2HCl(g) ⇌ 2CHClF2(g)
ΔH = –128 kJ mol-1
3
2N2(g) + 6H2O(g) ⇌ 4NH3(g) + 3O2(g)
ΔH = +1267 kJ mol-1
The Kolbe reaction involves decarboxylative dimerisation of carboxyate ions by
electrolysis. When an aqueous solution of sodium carboxylate is electrolysed, the overall
equation is as follows:
2RCO2Na + 2H2O  R–R + 2CO2 + 2NaOH + H2
where R = CH3, C2H5 etc.
Which of the following statements about the anode and cathode are correct?
anode
cathode
1
The solution around the anode turns
moist blue litmus red.
The solution around the cathode turns
moist red litmus blue.
2
RCO2Na is oxidised to R–R at the (+)
terminal.
H2O is reduced to H2 at the (–)
terminal.
3
The redox potential involving H2O is the
least positive.
The redox potential involving H2O is the
most positive.
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35
36
Which of the following incorrectly describes the reactivity of Period 3 elements?
1
P burns with a bright flame in air to form solid P4O6 only.
2
Al reacts rapidly in the presence of steam to give solid Al(OH)3 and H2 gas.
3
Si reacts in excess Cl2 to give a chloride which gives a weakly acidic solution in
water.
A manganese complex formed from a solution containing potassium bromide and oxalate
ion, C2O42-, is purified and analysed. It is found to have an empirical formula of
MnK4C4Br2O8. An aqueous solution of the complex has similar electrical conductivity as an
equimolar solution of K4[Fe(CN)6].
What can be deduced from the above information?
37
1
The complex is Mn(C2O4)2Br2.
2
The manganese ion in the complex has a +2 charge.
3
The anionic portion of the complex has a coordination number of 6.
Two female sex hormones are oestrone and oestradiol.
CH3OH
CH3 O
HO
HO
oestrone
oestradiol
Which of the following reagents could be used to distinguish between the two hormones?
1
SOCl2
2
KMnO4/H+
3
LiAlH4 in dry ether
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38
Compound B is related to methyl cinnamate, used as a flavouring agent.
Compound B
A student suggested the following three methods to synthesise compound B.
Which of the following synthetic routes is not correct?
1
2
3
39
The synthesis of physostigmine, a reversible cholinesterase inhibitor that was found in
Calabar beans, was first achieved in 1935. Part of the synthetic route is shown below.
Which type of reaction is shown by at least one step in the above route?
40
1
reduction
2
nucleophilic substitution
3
electrophilic substitution
What is a possible type of reaction that occurs when a protein undergoes denaturation?
1
neutralisation
2
precipitation
3
reduction
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1
VICTORIA JUNIOR COLLEGE
2015 JC2 PRELIMINARY EXAMINATIONS
H2 CHEMISTRY PAPER 1 ANSWERS
 VJC 2015
1
D
6
B
11
C
16
D
21
C
26
A
31
D
36
C
2
C
7
A
12
A
17
A
22
B
27
B
32
A
37
D
3
C
8
C
13
B
18
B
23
D
28
C
33
C
38
A
4
B
9
C
14
A
19
C
24
D
29
C
34
D
39
B
5
A
10
C
15
D
20
D
25
B
30
A
35
A
40
A
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1
Planning
You are provided with 5 unlabelled bottles containing pure natural acids from the
extracts of roasted coffee.
Each bottle contains one of the following natural acids:

lactic acid,
CH3CH(OH)CO2H

maleic acid,
HO2CCH=CHCO2H

oxalacetic acid,
HO2CCOCH2CO2H

pyruvic acid,
CH3COCO2H

quinic acid,
HO
OH
HO
CO2H
HO
All the acids above are soluble in water.
You are also provided with any other common laboratory reagents and apparatus.
(a)
All natural acids contain one or more carboxylic acid functional groups.
Other than the carboxylic acid functional group, what other functional groups
are also present in these natural acids?
lactic acid:
………………………………………………………..…
maleic acid:
………………………………………………………..…
oxalacetic acid:
………………………………………………………..…
pyruvic acid:
………………………………………………………..…
quinic acid:
………………………………………………………..…
[2]
(b)
Two of these natural acids are colourless liquids and the rest are white
crystalline solids.
Using relevant chemical knowledge, identify the two liquids. Explain your
choices.
The two liquids are ………………………………….……………………………..…
Explanation: ……………………………………………………..……………………..
……………………………………………………………………….………………..…
……………………………………………………………………….………………..…
[2]
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(c)
Suggest a reagent that could be used to carry out a test-tube experiment to
distinguish the two liquids.
Reagent: ………………………………….……………………………..…
Describe what would be observed for each compound in the experiment.
Observation: ………………………………….……………………….……………..…
……………………………………………………………………………………………
[2]
(d)
Outline a logical sequence of chemical tests that would enable you to identify
the remaining 3 solids. You should aim for a minimum number of reactions.
Your plan should include
 a positive test to confirm the identity of each compound;
 detailed procedure (including quantities of chemicals and conditions used);
 expected observations for each compound in each test.
[2]
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(e)
How would you ensure the reliability of the test result for quinic acid?
……………………………………………………………………….………………..…
……………………………………………………………………….………………..…
……………………………………………………………………….………………..…
[1]
(f)
Suggest a safety measure that you would consider in carrying out your plan.
……………………………………………………………………….………………..…
……………………………………………………………………….………………..…
……………………………………………………………………….………………..…
[1]
(g)
Draw a set-up of the apparatus for the synthesis of pyruvic acid from lactic
acid. State the required reagents and conditions.
Reagents and conditions: ………………………………….…..…………………..…
Diagram of the set-up:
[2]
[Total: 12]
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2
Use of the Data Booklet is required for this question.
(a)
A 2.85 g sample of haematite iron ore, Fe2O3, was dissolved in hydrochloric acid and
the solution diluted to 250 cm3 in a standard flask. A 25.0 cm3 of this solution was
completely reduced with excess tin (II) chloride to form a solution of iron(II) ions.
After t After the remaining tin(II) ions were removed with a suitable reagent, the solution of
iron(II) ions was titrated against an acidified solution of 0.020 mol dm-3 potassium
dichromate(VI) and required 26.40 cm3 for complete oxidation back to iron(III) ions.
(i)
Give the balanced
dichromate(VI) ions.
equation for
the reaction
between
iron(II)
and
………………………………….……………………………………………………..…
[1]
(ii)
Calculate the percentage of iron(III) oxide, Fe2O3, in the ore.
[3]
(b)
(i)
Suggest whether the acidified potassium dichromate(VI) can be replaced by
potassium manganate(VII) for oxidising iron(II) back to iron(III). Explain your
answer.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
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(ii)
Explain why the excess tin(II) ions have to be removed before titration with
potassium dichromate(VI).
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
(c)
The compounds of manganese catalyse a wide variety of reactions, one instance
being manganese dioxide, MnO2, which catalyses the decomposition of hydrogen
peroxide:
(i)
With the aid of a sketch of the Boltzmann distribution curve, explain how the
presence of a catalyst increases the rate of reaction.
[2]
(ii)
The relationship between the rate constant and temperature is generally
governed by the Arrhenius equation:
k=A
where
k = rate constant
A = Arrhenius constant
Ea = activation energy in J mol-1
R = molar gas constant = = 8.31 J K-1 mol-1
T = temperature in K
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7
To determine the activation energy of the above reaction, the rate constant
was determined at various temperatures and the experimental results were
then processed by plotting a graph of ln k vs 1/T (K-1):
Use the given information to determine a value for the activation energy of the
reaction.
[2]
(d)
Manganese compounds are also involved in many redox reactions. For example,
MnO4− ions oxidize C2O42− ions in an acidic medium according to the following
equation:
2MnO4– + 5C2O42− + 16H+  2Mn2+ + 10CO2 + 8H2O
A student decided to investigate the rate of the above reaction. He placed the
reaction flask in a thermostatically controlled water bath to keep the temperature
constant and measured how the conductivity of the solution changed with time.
 VJC 2015
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8
He observed that the conductivity changes slowly at first but in the course of time,
the conductivity changes at a progressively faster rate. Towards the end of the
reaction, the conductivity is again detected to change slowly.
(i)
Deduce whether the conductivity readings increased or decreased as the
reaction progressed.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
(ii)
Suggest reasons why the conductivity changes at different rates as observed
by the student.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
(iii)
Sketch on the axes below to show how the amount of CO2 formed varies with
time.
Amount of CO2
0
time
[1]
[Total: 15]
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3
Hydrogen halides (or hydrohalic acids) are inorganic compounds with the formula HX
where X is one of the halogens. The hydrogen halides are colourless gases at
s.t.p., except for hydrogen fluoride, which boils at 19°C.
(a)
The hydrogen halides are diatomic molecules with no tendency to ionize in the gas
phase. However, in the liquid state, HF exhibits both the Bronsted-Lowry properties
of an acid and a base. The conjugate base that is formed is especially stable in
liquid HF.
(i)
Write an equation to show how HF can act both as a Bronsted-Lowry acid and
base in the liquid phase. Hence, explain with a diagram why the conjugate
base that is formed is especially stable in liquid HF.
[3]
(ii)
HF is a weak acid in aqueous solution with a pKa > 1 whereas the pKa values
become increasingly negative from HCl to HI. Explain briefly why the pKa
values become increasingly negative from HCl to HI.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
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(iii)
A 20.0 cm3 solution of 0.10 mol dm3 NaOH(aq) is titrated with 0.15 mol dm3
HF(aq). When 20.0 cm3 of HF(aq) is added, the pH of the solution is 3.44.
Determine the pKa of HF.
[2]
(iv)
Suggest a suitable indicator for the titration of NaF(aq) with HI(aq). Explain
your choice.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
(b)
Hydrogen and chlorine are both obtained by electrolyzing concentrated sodium
chloride solution. The direct reaction of hydrogen with chlorine gives hydrogen
chloride. However, hydrogen chloride is produced industrially by treatment of halide
salts with sulfuric acid. On the other hand, the least stable hydrogen halide, HI, is
produced by the reaction of iodine with hydrogen sulfide or with hydrazine.
(i)
Explain briefly why it is not industrially viable to produce HCl through the direct
reaction of hydrogen with chlorine.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
 VJC 2015
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11
(ii)
Write a balanced equation showing the formation of HI from iodine and
hydrogen sulfide. Hence, with reference to the Data Booklet, explain why the
reaction is thermodynamically spontaneous.
(Given that the Eo = +0.14 V for S + 2H+ + 2e ⇌ H2S)
[3]
(iii)
Determine the amount of time that is needed to produce 1.0 dm 3 of chlorine
gas from electrolyzing concentrated sodium chloride solution with a current of
5.0 A at r.t.p.
[2]
(iv)
In addition to hydrogen and chlorine, state an important by-product that is
produced during the electrolysis of concentrated sodium chloride solution.
………………………………….……………………………………………………..…
[1]
[Total: 15]
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12
4
(a)
Phosphorus trichloride, PCl3, is a colourless fuming liquid which boils at 76°C.
(i)
Draw a ‘dot-and-cross’ diagram of PCl3.
[1]
(ii)
Hence, suggest the hybridisation state of phosphorus in PCl3.
………………………………….……………………………………………………..…
[1]
(b)
Ethanol and PCl3 reacts via the nucleophilic substitution mechanism, whereby PCl3
is reacting as an electrophile.
By studying the structure of PCl3, explain why PCl3 is able to react as an electrophile.
……………………………………..….……………………………………………………..…
………………………………………...……………………………………………………..…
………………………………………………………………………………………………….
[1]
(c)
Phosphonic acid, H3PO3, is most commonly used in the production of phosphites. It
is produced when PCl3 is reacted with steam according to the equation below:
PCl3(g) + 3H2O(g)  H3PO3(l) + 3HCl(g)
(i)
 VJC 2015
Two possible structures can be drawn for phosphonic acid.
9647/02/PRELIM/15
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Make use of the following bond energies to suggest which structure is more
likely to be formed.
bond
bond energy /kJ mol−1
O–H
460
P–H
322
P–O
335
P=O
544
[3]
(ii)
10.0 cm3 of 0.50 mol dm−3 H3PO3(aq) was mixed with 10.0 cm3 of excess
NaOH(aq). It was found that the temperature of the solution after mixing rose
by 6.4°C.
Specific heat capacity of water = 4.2 J g−1 K−1; density of water = 1.0 g cm−3.
Calculate the enthalpy change of neutralisation per mole of H3PO3(aq).
[2]
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14
(iii)
Given that the enthalpy change of neutralization of HCl(aq) by NaOH(aq) is
−55.8 kJ mol−1; comment on how the answer in (c)(ii) confirms the structure of
H3PO3 that you determined previously in (c)(i).
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
(d)
PCl5 can be formed from the reaction between PCl3 and Cl2.
Below 163°C, PCl5 condenses directly from the gaseous phase to the solid phase. In
the process, a chloride ion is transferred from one of the PCl5 molecule to the other,
leaving a positive ion, [PCl4]+ and negative ion, [PCl6]− which gives rise to strong
electrostatic forces of attraction between the oppositely charged ions.
State the shapes of [PCl4]+ and [PCl6]−.
(i)
[PCl4]+
………………………………………………………..…
(ii)
[PCl6]−
………………………………………………………..…
[2]
(e)
(i)
Name and describe the mechanism of the reaction between benzene and
1-chloropropane, CH3CH2CH2Cl, in the presence of AlCl3. Your answer should
include curly arrows showing the movement of electrons and all charges.
Name of mechanism: ………………………………………………………..…
[2]
 VJC 2015
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(ii)
In
the
reaction
between
benzene
and
1-chloropropane,
(1-methylethyl)benzene is formed as a by-product of the reaction.
(1-methylethyl)benzene
Carefully examine the structure and stability of the reactive carbocation
intermediate in your mechanism.
Hence, explain the formation of (1-methylethyl)benzene from the reaction
between benzene and 1-chloropropane.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
(iii)
Explain why BeCl2 can be used in place of AlCl3 as the catalyst in the
formation of chlorobenzene from benzene.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[1]
[Total: 15]
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16
5
Bovine serum albumin (BSA) is a protein derived from cows.
The common amino acid residues in BSA are listed below:
(a)
amino acid
3-letter
abbreviation
formula of side chain
alanine
ala
–CH3
cysteine
cys
–CH2SH
glutamic acid
glu
–(CH2)2CO2H
glycine
gly
–H
leucine
leu
–CH2CH(CH3)2
lysine
lys
–(CH2)4NH2
serine
ser
–CH2OH
valine
val
–CH(CH3)2
The following shows part of the sequence of amino acids in BSA.
N
terminus
val – ala – cys – gly – val
lys – gly – glu – val – ala
gly – leu – val – cys – ser
C
terminus
A proteolytic enzyme Z is capable of hydrolysing peptide bonds at the carboxylic end
of glycine and the amino end of valine.
Draw the structural formula of the tripeptide formed when enzyme Z is added to the
sequence of amino acids in BSA.
[2]
 VJC 2015
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17
(b)
By considering the R–groups of common amino acids present in BSA, suggest the
predominant secondary protein structure present in BSA. Draw a diagram to
illustrate the secondary structure.
[3]
(c)
The interactions of small anions, such as halides, with proteins have important
significance in the transportation and distribution process of biological systems.
A solution containing excess Br– of known concentration is added to a fixed
concentration of BSA, forming a BSA complex at equilibrium as shown:
BSA + Br–
BSA complex
The concentration of Br– that binds to BSA is calculated by finding the difference
between the original concentration of Br– and the concentration of free Br– at
equilibrium. The affinity for binding is then reflected in the value of the binding
equilibrium constant, K, calculated.
The data obtained at three different temperatures are given below:
 VJC 2015
T/K
K for Br–
288
1.85
298
2.62
308
3.82
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(i)
Account for the trend in binding constants in terms of the structural changes
that BSA undergo at elevated temperatures.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
(ii)
A student suggested measuring the concentration of free Br– by adding a
known excess of aqueous silver nitrate, and measuring the mass of the
precipitate formed. With reference to the types of side chain given in the
question, explain why this suggestion is not feasible.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
(d)
When 10 cm3 of 0.100 mol dm–3 of a fully protonated amino acid in BSA is
completely neutralised with 0.100 mol dm–3 of aqueous sodium hydroxide, the
following pH curve is obtained
pH
X
isoelectric point
W
9.
7
0
 VJC 2015
10
9647/02/PRELIM/15
20
30
VNaOH /
cm3
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(i)
Identify, with reasoning, the amino acid.
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
………………………………….……………………………………………………..…
[2]
(ii)
Draw the structures of the amino acid at point W and X.
[2]
(iii)
Using a relevant equation, account for the relatively small change in pH in the
region between W and X.
[2]
[Total: 15]
 VJC 2015
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1
Victoria Junior College
2015 H2 Chemistry Prelim Exam 9647/2
Suggested Answers
1
Planning
You are provided with 5 unlabelled bottles containing pure natural acids from the
extracts of roasted coffee.
Each bottle contains one of the following natural acids:

lactic acid,
CH3CH(OH)CO2H

maleic acid,
HO2CCH=CHCO2H

oxalacetic acid,
HO2CCOCH2CO2H

pyruvic acid,
CH3COCO2H

quinic acid,
HO
OH
HO
CO2H
HO
All the acids above are soluble in water.
You are also provided with any other common laboratory reagents and apparatus.
(a)
All natural acids contain one or more carboxylic acid functional groups.
Other than the carboxylic acid functional group, what other functional groups
are also present in these natural acids?
lactic acid:
secondary alcohol
maleic acid:
alkene
oxalacetic acid: ketone
pyruvic acid:
ketone
quinic acid:
Secondary alcohol and tertiary alcohol
[2]
(b)
Two of these natural acids are colourless liquids and the rest are white
crystalline solids.
Using relevant chemical knowledge, identify the two liquids. Explain your
choices.
The two liquids are lactic acid and pyruvic acid.
Explanation: These two compounds have a relatively smaller Mr and
thus weaker dispersion forces between their respective molecules
than the other 3 compounds (due to smaller number of electrons
present).
[2]
 VJC 2015
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2
(c)
Suggest a reagent that could be used to carry out a test-tube experiment to
distinguish the two liquids.
Reagent: Brady’s reagent [or 2,4-dinitrophenylhydrazine]
Describe what would be observed for each compound in the experiment.
Observation: Pyruvic acid will give an orange ppt. but lactic acid will not.
[2]
(d)
Outline a logical sequence of chemical tests that would enable you to identify
the remaining 3 solids. You should aim for a minimum number of reactions.
Your plan should include
 a positive test to confirm the identity of each compound;
 detailed procedure (including quantities of chemicals and conditions used);
 expected observations for each compound in each test.
Step 1: Prepare aqueous solutions of each of the 3 solid samples by
dissolving 1-cm depth of solid in about 5 cm depth of deionised
water in each test-tube.
Step 2: To 1-cm depth of each of the 3 samples in separate test-tubes,
add equal volume of Brady’s reagent.
Oxalacetic acid will give an orange ppt. while the other 2 samples
will not.
Step 3: To 1-cm depth of each of the 2 remaining samples in separate
test-tubes, add a few drops of Br2(aq).
Maleic acid will decolourise reddish-brown Br2 while the
remaining sample will not.
Step 4: To 1-cm depth of the last sample in a test-tube, add a few drops of
acidified KMnO4(aq). Heat the mixture in a hot water-bath. Quinic
acid will turn purple KMnO4 colourless.
[2]
 VJC 2015
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3
(e)
How would you ensure the reliability of the test result for quinic acid?
To ensure a colour change for KMnO4, the acidified KMnO4(aq) must be
added slowly and dropwise. It should not be added in excess. The
mixture must also be heated to prevent incomplete reduction to black
MnO2 solid.
[Or hydrochloric acid cannot be used to acidify KMnO4 as the chloride
ions can be oxidised by KMnO4. As a result, there may be a colour
change of KMnO4 due to oxidation of chloride to Cl2.]
[1]
(f)
Suggest a safety measure that you would consider in carrying out your plan.
Use a hot water bath for heating instead of using a direct naked flame
from the bunsen burner as most organic compounds are highly
flammable.
[1]
Draw a set-up of the apparatus for the synthesis of pyruvic acid from lactic
acid. State the required reagents and conditions.
(g)
Reagents and conditions: KMnO4(aq) / H2SO4(aq), reflux
[OR K2Cr2O7(aq) / H2SO4(aq), reflux]
[2]
[Total: 12]
 VJC 2015
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2
Use of the Data Booklet is required for this question.
(a)
A 2.85 g sample of haematite iron ore, Fe2O3, was dissolved in hydrochloric acid and
the solution diluted to 250 cm3 in a standard flask. A 25.0 cm3 of this solution was
completely reduced with excess tin (II) chloride to form a solution of iron(II) ions.
After t After the remaining tin(II) ions were removed with a suitable reagent, the solution of
iron(II) ions was titrated against an acidified solution of 0.020 mol dm-3 potassium
dichromate(VI) and required 26.40 cm3 for complete oxidation back to iron(III) ions.
(i)
Give the balanced
dichromate(VI) ions.
equation for
the reaction
between
iron(II)
and
Cr2O72–(aq) + 14H+(aq) + 6Fe2+(aq)  2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
[1]
(ii)
Calculate the percentage of iron(III) oxide, Fe2O3, in the ore.
Amount of Cr2O72- used = 0.0200 x
= 5.28 x 10-4 mol
Amount of Fe2+ present in 25.0 cm3 solution = 6 x 5.28 x 10-4 mol
Total amount of Fe2+ present = 6 x 5.28 x 10-4 x
= 3.168 x 10-2 mol
Hence amount of Fe2O3 present = 3.168 x 10-2 x
mol
Mass of Fe2O3 present = 3.168 x 10-2 x
% of Fe2O3 present =
x 159.6 = 2.528 g
x 100 = 88.7%
[3]
(b)
 VJC 2015
(i)
Suggest whether the acidified potassium dichromate(VI) can be replaced by
potassium manganate(VII) for oxidising iron(II) back to iron(III). Explain your
answer.
KMnO4 cannot be used as it is a stronger oxidising agent than K2Cr2O7
(more positive E⍬). Thus chloride ions in both hydrochloric acid and
SnCl2 would also be oxidised leading to inaccurate titration results.
[2]
9647/02/PRELIM/15
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5
(ii)
Explain why the excess tin(II) ions have to be removed before titration with
potassium dichromate(VI).
The excess tin(II) ions would also be oxidised to tin(IV) by Cr2O72- ions
hence again leading to inaccurate titration results.
[1]
(c)
The compounds of manganese catalyse a wide variety of reactions, one instance
being manganese dioxide, MnO2, which catalyses the decomposition of hydrogen
peroxide:
(i)
With the aid of a sketch of the Boltzmann distribution curve, explain how the
presence of a catalyst increases the rate of reaction.
Correct shape of Boltzman distribution curve and axis labels
When a catalyst is used in a reaction, it provides an alternative reaction
pathway with lower activation energy. Hence, number of reacting
particles with energy ≥ Ea increases as shown by the shaded area. This
leads to higher frequency of effective collisions, thereby increasing the
rate of reaction.
[2]
 VJC 2015
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6
(ii)
The relationship between the rate constant and temperature is generally
governed by the Arrhenius equation:
k=A
where
k = rate constant
A = Arrhenius constant
Ea = activation energy in J mol-1
R = molar gas constant = = 8.31 J K-1 mol-1
T = temperature in K
To determine the activation energy of the above reaction, the rate constant
was determined at various temperatures and the experimental results were
then processed by plotting a graph of ln k vs 1/T (K-1):
Use the given information to determine a value for the activation energy of the
reaction.
Taking ln on both sides of the Arrhenius equation gives:
ln k = ln A – (Ea / R) × (1/T)
Hence in the graph of ln k vs 1/T, gradient = – Ea / R
Gradient
=
= – 6250
– 6250 = – Ea / R
Ea
= 6250 × 8.31
= 51900 J mol-1
[2]
 VJC 2015
9647/02/PRELIM/15
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7
(d)
Manganese compounds are also involved in many redox reactions. For example,
MnO4− ions oxidize C2O42− ions in an acidic medium according to the following
equation:
2MnO4– + 5C2O42− + 16H+  2Mn2+ + 10CO2 + 8H2O
A student decided to investigate the rate of the above reaction. He placed the
reaction flask in a thermostatically controlled water bath to keep the temperature
constant and measured how the conductivity of the solution changed with time.
He observed that the conductivity changes slowly at first but in the course of time,
the conductivity changes at a progressively faster rate. Towards the end of the
reaction, the conductivity is again detected to change slowly.
(i)
Deduce whether the conductivity readings increased or decreased as the
reaction progressed.
Since the reaction results in less number of ions formed, the conductivity
should have decreased.
[1]
(ii)
Suggest reasons why the conductivity changes at different rates as observed
by the student.
The reaction is slow initially as it has a high activation energy. As time
progresses, Mn2+, the autocatalyst, is formed in increasing amounts,
hence reaction speeds up. Towards the end, reactant concentration is
low, hence rate decreases again.
[2]
(iii)
Sketch on the axes below to show how the amount of CO2 formed varies with
time.
Amount of CO2
0
time
Sketch [-½ m for each mistake]
- start from origin
- gradient increases then decreases
- constant max volume reached at the end
[1]
[Total: 15]
 VJC 2015
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8
3
Hydrogen halides (or hydrohalic acids) are inorganic compounds with the formula HX
where X is one of the halogens. The hydrogen halides are colourless gases at
s.t.p., except for hydrogen fluoride, which boils at 19°C.
(a)
The hydrogen halides are diatomic molecules with no tendency to ionize in the gas
phase. However, in the liquid state, HF exhibits both the Bronsted-Lowry properties
of an acid and a base. The conjugate base that is formed is especially stable in
liquid HF.
(i)
Write an equation to show how HF can act both as a Bronsted-Lowry acid and
base in the liquid phase. Hence, explain with a diagram why the conjugate
base that is formed is especially stable in liquid HF.
HF + HF  F− + H2F+
H bond
−
+
F−•• ||||||||| H
F
In liquid HF, as HF has H bonded to highly electronegative F, hence, H is
very electron deficient and it acquires a partial positive charge. It is
therefore bonded to the lone pair electrons on F− through a hydrogen
bond. Hence, this makes the conjugate base formed, F−, especially stable
in liquid HF.
[3]
(ii)
HF is a weak acid in aqueous solution with a pKa > 1 whereas the pKa values
become increasingly negative from HCl to HI. Explain briefly why the pKa
values become increasingly negative from HCl to HI.
Down the group, size of halogen atoms increases. Hence, the extent of
overlapping of atomic orbitals between that of H and halogen (X) atoms
decreases, H-X bond strength decreases. It is therefore easier to break
the H-X bond to dissociate H+ and halide ions in aqueous solution. Acid
strength increases from HCl to HI, Ka values therefore increases
numerically and pKa decreases numerically (or becomes increasingly
negative).
[2]
(iii)
 VJC 2015
A 20.0 cm3 solution of 0.10 mol dm3 NaOH(aq) is titrated with 0.15 mol dm3
HF(aq). When 20.0 cm3 of HF(aq) is added, the pH of the solution is 3.44.
Determine the pKa of HF.
NaOH + HF  NaF + H2O
Excess HF and NaF formed make up a weak acid buffer with pH of 3.44.
[HF] left = (20x0.15 – 20x0.10) / 40 = 0.0250 mol dm−3
[F−] formed = 20x0.10 / 40 = 0.0500 mol dm-3
Ka = [H+][F−]/[HF] = (10−3.44)(0.050) / 0.025 = 7.26 x 10−4 mol dm−3
pKa = −log Ka = 3.14
[2]
9647/02/PRELIM/15
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(iv)
Suggest a suitable indicator for the titration of NaF(aq) with HI(aq). Explain
your choice.
As NaF is a weak base while HI is a strong acid, pH < 7 at equivalence
point. Suitable indicator is methyl orange.
[1]
(b)
Hydrogen and chlorine are both obtained by electrolyzing concentrated sodium
chloride solution. The direct reaction of hydrogen with chlorine gives hydrogen
chloride. However, hydrogen chloride is produced industrially by treatment of halide
salts with sulfuric acid. On the other hand, the least stable hydrogen halide, HI, is
produced by the reaction of iodine with hydrogen sulfide or with hydrazine.
(i)
Explain briefly why it is not industrially viable to produce HCl through the direct
reaction of hydrogen with chlorine.
Reaction of hydrogen and chlorine is explosive to handle.
[1]
(ii)
Write a balanced equation showing the formation of HI from iodine and
hydrogen sulfide. Hence, with reference to the Data Booklet, explain why the
reaction is thermodynamically spontaneous.
(Given that the Eo = +0.14 V for S + 2H+ + 2e ⇌ H2S)
I2 + H2S  2HI + S
S + 2H+ + 2e ⇌ H2S
Eo = +0.14 V
I2 + 2e ⇌ 2I
Eo = +0.54 V
o
o
o
E cell = E Red  E Ox = 0.54  0.14 = +0.40 V
Since Eocell > 0, the redox reaction is thermodynamically feasible under
standard conditions.
[3]
(iii)
Determine the amount of time that is needed to produce 1.0 dm 3 of chlorine
gas from electrolyzing concentrated sodium chloride solution with a current of
5.0 A at r.t.p.
At anode, 2Cl  Cl2 + 2e
No. of moles of electrons transferred = 2 x 1000/24000 = 0.0833 mol
Quantity of charge transferred = 0.0833 x 96500 = 8.04 x 103 C
Hence, Q = I x t
8.04 x 103 = 5.0 x t
t = 1.61 x 103 s (or 26.8 min)
[2]
(iv)
In addition to hydrogen and chlorine, state an important by-product that is
produced during the electrolysis of concentrated sodium chloride solution.
sodium hydroxide
[1]
[Total: 15]
 VJC 2015
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[Turn over
10
4
(a)
Phosphorus trichloride, PCl3, is a colourless fuming liquid which boils at 76°C.
(i)
Draw a ‘dot-and-cross’ diagram of PCl3.
[1]
(ii)
Hence, suggest the hybridisation state of phosphorus in PCl3.
sp3 (because four electron domains around phosphorus)
[1]
(b)
Ethanol and PCl3 reacts via the nucleophilic substitution mechanism, whereby PCl3
is reacting as an electrophile.
By studying the structure of PCl3, explain why PCl3 is able to react as an electrophile.
Phosphorus is bonded to three electronegative chlorine atoms making the
former electron deficient and susceptible to attack by nucleophile (lone pair of
electron on oxygen in ethanol)
[1]
(c)
Phosphonic acid, H3PO3, is most commonly used in the production of phosphites. It
is produced when PCl3 is reacted with steam according to the equation below:
PCl3(g) + 3H2O(g)  H3PO3(l) + 3HCl(g)
(i)
Two possible structures can be drawn for phosphonic acid.
Make use of the following bond energies to suggest which structure is more
likely to be formed.
 VJC 2015
bond
bond energy /kJ mol−1
O–H
460
P–H
322
P–O
335
P=O
544
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Total bond energy in (structure on left)
= 3 × 460 + 3 × 335
= 2385 kJ mol−1
Total bond energy in (structure on right)
= 2 × 460 + 2 × 335 + 322 + 544
= 2456 kJ mol−1
Structure on right is more likely to be formed since its formation will lead
to a greater decrease in enthalpy (and, therefore, leading to greater
stability) for the chemical system.
[3]
(ii)
10.0 cm3 of 0.50 mol dm−3 H3PO3(aq) was mixed with 10.0 cm3 of excess
NaOH(aq). It was found that the temperature of the solution after mixing rose
by 6.4°C.
Specific heat capacity of water = 4.2 J g−1 K−1; density of water = 1.0 g cm−3.
Calculate the enthalpy change of reaction per mole of H3PO3(aq).
No. of moles of H3PO3 used = 5.0 × 10−3 mol
∆Q = mc∆T = 20.0 g × 4.2 J g−1 K−1 × 6.4 K = 537.6 J
Enthalpy of reaction = −
537.6
= −108 kJ mol−1
3
5.0  10
[2]
(iii)
Given that the enthalpy change of neutralization of HCl(aq) by NaOH(aq) is
−55.8 kJ mol−1; comment on how the answer in (c)(ii) confirms the structure of
H3PO3 that you determined previously in (c)(i).
The enthalpy change of reaction per mole of phosphonic acid is about
twice that of hydrochloric acid (monoprotic acid) which means that the
former is a diprotic acid and this confirms the structure determined
previously.
[1]
(d)
PCl5 can be formed from the reaction between PCl3 and Cl2.
Below 163°C, PCl5 condenses directly from the gaseous phase to the solid phase. In
the process, a chloride ion is transferred from one of the PCl5 molecule to the other,
leaving a positive ion, [PCl4]+ and negative ion, [PCl6]− which gives rise to strong
electrostatic forces of attraction between the oppositely charged ions.
State the shapes of [PCl4]+ and [PCl6]−.
(i)
[PCl4]+
tetrahedral
(ii)
[PCl6]−
octahedral
[2]
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(e)
(i)
Name and describe the mechanism of the reaction between benzene and
1-chloropropane, CH3CH2CH2Cl, in the presence of AlCl3. Your answer should
include curly arrows showing the movement of electrons and all charges.
Name of mechanism: electrophilic subsititution
Generation of electrophile
AlCl3 + CH3CH2CH2Cl
AlCl4− + CH3CH2CH2+
Electrophilic attack on benzene ring by carbocation
Restoration of resonance in benzene ring by proton abstraction
[2]
(ii)
In
the
reaction
between
benzene
and
1-chloropropane,
(1-methylethyl)benzene is formed as a by-product of the reaction.
(1-methylethyl)benzene
Carefully examine the structure and stability of the reactive carbocation
intermediate in your mechanism.
Hence, explain the formation of (1-methylethyl)benzene from the reaction
between benzene and 1-chloropropane.
In the original mechanism, the carbocation is a primary carbocation
whereas to form (1-methylethyl)benzene, the carbocation is a secondary
carbocation.
The original carbocation was transformed/rearranged to give the
secondary carbocation which is more stable than the primary
carbocation.
[1]
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(iii)
Explain why BeCl2 can be used in place of AlCl3 as the catalyst in the
formation of chlorobenzene from benzene.
Beryllium in BeCl2 is electron-deficient and can react as a halogen carrier
(or form a bond with chlorine atom).
[1]
[Total: 15]
5
Bovine serum albumin (BSA) is a protein derived from cows.
The common amino acid residues in BSA are listed below:
(a)
amino acid
3-letter
abbreviation
formula of side chain
alanine
ala
–CH3
cysteine
cys
–CH2SH
glutamic acid
glu
–(CH2)2CO2H
glycine
gly
–H
leucine
leu
–CH2CH(CH3)2
lysine
lys
–(CH2)4NH2
serine
ser
–CH2OH
valine
val
–CH(CH3)2
The following shows part of the sequence of amino acids in BSA.
N
terminus
val – ala – cys – gly – val
lys – gly – glu – val – ala
gly – leu – val – cys – ser
C
terminus
A proteolytic enzyme Z is capable of hydrolysing peptide bonds at the carboxylic end
of glycine and the amino end of valine.
Draw the structural formula of the tripeptide formed when enzyme Z is added to the
sequence of amino acids in BSA.
[2]
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(b)
By considering the R–groups of common amino acids present in BSA, suggest the
predominant secondary protein structure present in BSA. Draw a diagram to
illustrate the secondary structure.
The R–groups are generally non–bulky. Thus, alpha helix is likely to be the
predominant secondary protein structure.
R
√R–groups
pointing outwards
–
–
+
R
+
√hydrogen bonds
R
–
–
R
[3]
(c)
The interactions of small anions, such as halides, with proteins have important
significance in the transportation and distribution process of biological systems.
A solution containing excess Br– of known concentration is added to a fixed
concentration of BSA, forming a BSA complex at equilibrium as shown:
BSA + Br–
BSA complex
The concentration of Br– that binds to BSA is calculated by finding the difference
between the original concentration of Br– and the concentration of free Br– at
equilibrium. The affinity for binding is then reflected in the value of the binding
equilibrium constant, K, calculated.
The data obtained at three different temperatures are given below:
(i)
 VJC 2015
T/K
K for Br–
288
1.85
298
2.62
308
3.82
Account for the trend in binding constants in terms of the structural changes
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that BSA undergo at elevated temperatures.
As temperature increases, binding constant increases. Disruption of
dispersion forces, hydrogen bonds, ionic interactions and disulfide linkages
results in the denaturation of 2°, 3°, and 4° structures. The unfolding of the
protein results in more (positively charged) regions for the Br– to bind to.
[2]
(ii)
A student suggested measuring the concentration of free Br– by adding a
known excess of aqueous silver nitrate, and measuring the mass of the
precipitate formed. With reference to the types of side chain given in the
question, explain why this suggestion is not feasible.
Ag+ from AgNO3 will disrupt ionic interactions between glutamic acid and
lysine residues as well as disulfide linkages between cysteine residues,
resulting in the denaturation of 3° and 4° structures. This will not only result in
less AgBr formed, but also result in the probable precipitation of BSA.
Hence suggestion is not feasible.
[2]
(d)
When 10 cm3 of 0.100 mol dm–3 of a fully protonated amino acid in BSA is
completely neutralised with 0.100 mol dm–3 of aqueous sodium hydroxide, the
following pH curve is obtained
pH
X
isoelectric point
W
9.
7
0
(i)
10
20
30
VNaOH /
cm3
Identify, with reasoning, the amino acid.
3 points of neutralisation means that the amino acid have 3 acidic groups in
its fully protonated form. pI > 7 means that that amino acid has more NH2
groups compared to CO2H groups. Hence amino acid is lysine.
[2]
(ii)
 VJC 2015
Draw the structures of the amino acid at point W and X.
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[2]
(iii)
Using a relevant equation, account for the relatively small change in pH in the
region between W and X.
The region between W and X is a buffer region.
When small amount of base is added, the large reservoir of W will remove
the additional amount of OH– added. Thus pH remains relatively constant.
[2]
[Total: 15]
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1
Magnesium ethanoate is commonly used as a source of magnesium or ethanoate ions in
chemistry experiments. One of the more prevalent uses of magnesium ethanoate is in the
mixture called calcium magnesium ethanoate (CMA). It is a mixture of calcium ethanoate
and magnesium ethanoate. CMA acts as a powerful SO2, NOx, and toxic particulate
emission control agent in coal combustion processes to reduce acid rain, and as an
effective catalyst for the facilitation of coal combustion.
(a) Samples of anhydrous magnesium ethanoate, Mg(CH3COO)2 and barium ethanoate,
Ba(CH3COO)2, were heated to a temperature of 600oC causing them to thermally
decompose. White residues were formed in both cases and a common gaseous product,
X, with the molecular formula of C3H6O was also given off. X gives a yellow precipitate
upon warming with aqueous alkaline iodine.
In addition, the decomposition of magnesium ethanoate also produced a second
gaseous product which formed a white precipitate with calcium hydroxide solution.
(i)
Identify the gaseous product, X.
[1]
(ii)
Write a balanced equation with state symbols, for each decomposition reaction.
[2]
(iii)
Account for the difference in the decomposition products.
[2]
(b) Magnesium, is an extremely important light weight structural metal which can be
produced by the electrolysis of magnesium chloride. Magnesium chloride can be
prepared from magnesium oxide which is obtained from sea-water containing a
significant amount of Mg2+ and Ca2+. The steps involved are shown below:
Step 1
Controlled
addition of
CO32-
Sea water
(containing
Mg2+ and Ca2+)
Step 2
addition of
OH-(aq)
Mg(OH)2
filtrate
filter
Step 3
heat
Mg
MgO
MgCl2
Step 5
Step 4
The numerical values of the relevant solubility products are given below.
Magnesium carbonate
Calcium carbonate
Magnesium hydroxide
Calcium hydroxide
 VJC 2015
1.0 x 10-5
8.7 x 10-9
1.1 x 10-11
5.5 x 10-6
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(i)
Calculate and compare the solubility between magnesium carbonate and
magnesium hydroxide.
[3]
(ii)
Assuming the concentration of magnesium ions in the filtrate in step 2 is
3.0 x 10-5 mol dm-3, calculate the concentration of hydroxide ions present in the
filtrate when trace of solid magnesium hydroxide first appears.
[1]
(iii)
Explain why the addition of carbonate ions in step 1 has to be controlled.
[1]
(iv)
Give a reason why the electrolysis of magnesium chloride is preferred to that of
magnesium oxide.
[1]
(c) Some organic and inorganic compounds are classified under a category called ‘nonexistent compounds’. They are called non-existent because so far chemists had been
unable to synthesise them. Some of the reasons why these compounds are unstable
are:



unfavourable bond energy terms
a redox incompatibility of the ions making up the compound
reaction occurring between the ions
For the following cases, suggest an explanation for each observation. You may use data
from the Data Booklet to assist you in your answers. Give relevant equations to support
your answers.
(i)
When aqueous sodium carbonate is added to aqueous aluminium chloride, a
colourless gas is liberated and the precipitate formed is not aluminium carbonate,
Al2(CO3)3.
[4]
(ii)
Caprolactum, a monomer of the polymer nylon 6, exists as an alicyclic ring
structure and not as 6-aminohexanoic acid, a straight chain aliphatic structure as
most monomers do.
O
H
C
N
H2NCH2CH2CH2CH2CH2CO2H
Caprolactum
6-aminohexanoic acid
[3]
(iii)
It is possible to find MnCl2 in the laboratory but not MnCl3.
[2]
[Total: 20]
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2
(a)
The Contact Process is an industrial process for manufacturing sulfuric acid. The
key stage in this process is the reaction between sulfur dioxide and oxygen.
2SO2(g) + O2(g) ⇌ 2SO3(g)
H = –197 kJ mol–1
Vanadium(V) oxide, V2O5, is used as a heterogeneous catalyst for this process.
Describe the mode of action by which V2O5 fulfils this role.
[3]
(b)
In an experiment to determine the Kp of the above equilibrium, a mixture containing
0.200 mol of SO2 and 0.100 mol of O2 was heated in a closed flask and allowed to
reach equilibrium at 550oC and 3.5 atm.
The flask was then rapidly cooled to liquefy the SO3 so that it can be separated from
the gaseous SO2 and O2. Excess water was carefully added to the liquid SO3,
causing the following reaction to occur:
SO3(l) + H2O(l)  H2SO4(aq)
The resulting solution was made up to 250 cm3 in a standard volumetric flask.
25.0 cm3 of this solution was titrated with 1.00 mol dm−3 NaOH and required
36.0 cm3 for complete neutralisation.
(i)
Calculate the equilibrium amount of SO2, O2 and SO3 at 550oC.
[2]
(ii)
Hence, calculate a value for Kp, including its correct units.
[1]
(iii) In liquefying the SO3, it was mentioned that the cooling was performed rapidly.
Explain how the titration volume would be affected if the cooling had been
performed slowly.
[2]
(c)
One important use of vanadium compounds is the vanadium redox battery, a
rechargeable battery which utilizes graphite electrodes. The electrolytes are pumped
from separate storage tanks A and B into compartments X and Y of the cell
respectively, where they are separated by a semi–permeable proton exchange
membrane which allows only passage of protons.
electrical load
Tank A
acidified
VO2+(aq) and
VO2+(aq)
X
Y
e–
Tank B
acidified
V3+(aq) and
V2+(aq)
membrane
The direction of the electron flow during the discharging process is indicated in the
diagram of the battery above.
 VJC 2015
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(i)
By means of the letters X and Y, state and briefly justify the direction of proton
flow across the membrane during the discharging process.
[1]
In a particular setup, compartment X contained 5 dm3 of the electrolyte, with VO2+
and VO2+ each having an original concentration of 1.5 mol dm-3. The cell discharged
until 90% of the reactants was consumed, after which the cell was recharged by
connecting an electricity source across the electrodes.
(ii)
With the aid of the Data Booklet, write half–equations for the reactions that
occur at the electrodes in X and Y during the charging process.
[1]
(iii) If the current used for the charging process is 14 A, calculate the time taken, in
hours, needed to restore the concentration of the vanadium-containing ions in
compartment X to their original value.
[2]
(d)
Chromium is another transition metal adjacent to vanadium in the Periodic Table.
Some of its reactions are described below:
When chromium metal is dissolved in dilute sulfuric acid, it produces a blue solution
containing Cr2+(aq). The blue solution slowly turns green even in the absence of air.
This mixture is then filtered to remove the unreacted metal. To the green filtrate V,
sodium hydroxide solution was added, forming a grey−green precipitate, W. In
excess sodium hydroxide, the precipitate dissolves to form a dark green solution
containing an anion, X. The dark green solution turns into a yellow solution, Y, upon
warming with hydrogen peroxide solution.
(i)
By quoting and using relevant Eo values from the Data Booklet, explain why
the blue solution turns green.
[2]
(ii)
Identify the precipitate W and the anion X and suggest ionic equations, with
state symbols, for the formation of W, and for the formation of X.
[2]
(iii) State the role of hydrogen peroxide in converting the dark green solution into
Y.
[1]
(e)
A particular chromium compound has the formula Cr(H2O)6Cl3. It contains a complex
Z in which the coordination number of chromium is 6.
When a solution containing 0.02 mol of the compound, Cr(H2O)6Cl3, is treated with
excess aqueous Pb(NO3)2, a white precipitate is formed. After filtration and drying,
the mass of the precipitate is found to be 2.78 g.
(i)
Use the given information to determine the formula of the complex, Z.
[1]
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(ii)
Z is known to exist as two stereoisomers. Draw the structures of these two
stereoisomers, showing clearly the spatial arrangement of the ligands around
the central metal ion.
[2]
[Total: 20]
3
Phosphorus belongs to Period 3, an element that is essential for life. The elemental form
of phosphorus consists of tetrahedral P4 molecules, in which each phosphorus atom forms
three single bonds with three other phosphorus atoms. Phosphates, compounds
containing the PO43 ion, are a component of DNA, RNA, ATP and also the phospholipids,
which form all cell membranes.
(a)
Each of the Period 3 elements, Na to S, reacts at a different rate with oxygen to form
oxides of different oxidation states.
(i)
Describe what you see when magnesium and sulfur are separately burned in
air or oxygen. Write equations for the reactions that occur.
[3]
(ii)
Explain briefly why the bonding of the oxides of Period 3 elements vary from
ionic to covalent.
[1]
(iii) X and Y are Period 3 elements.
Element X forms a white oxide that is slightly soluble in cold water. Its chloride
dissolves in water to form a weakly acidic solution.
Element Y forms two oxides. 0.03 mol of one of the two oxides produces
6.99 g of white precipitate when shaken with excess barium chloride solution.
Another solution containing 0.05 mol of this oxide of element Y forms a neutral
solution when the same amount of the oxide of the element X is added to it.
Identify the elements X, Y, the oxide of Y and the white precipitate that is
formed in the above reaction. Explain why the chloride of element X is acidic.
[3]
(b)
 VJC 2015
Sulfur, another Period 3 element, forms a wide range of sulfides with phosphorus.
One of the most well-known examples is the three-fold symmetric P4S3 used in
strike-anywhere matches.
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(i)
State the two oxidation states of phosphorus that exist in P4S3.
[1]
(ii)
P4O3 has a structure that is similar to P4S3. Explain why the OPO bond angle
is smaller than the SPS bond angle.
[2]
(iii) The bond energy for the PP single bond is 197 kJ mol1.
Given the enthalpy change for the following transformation
P4S3(g)  P4(g) + 3S(g)
H = +1029 kJ mol1,
calculate the bond energy for the PS bond.
[2]
(iv) In the presence of excess sulfur, P4S3 can be converted to P4S10, which is
structurally and chemically similar to P4O10. Give a balanced equation for the
reaction between P4S10 and H2S which does not involve any change in
oxidation state. Hence, state the role that P4S10 plays in the reaction.
[2]
(c)
Aluminium is an important element in the reducing agent, LiAlH4, for organic
synthesis. When benzoic acid is reduced by LiAlH4 in dry ether followed by addition
of heavy water (D2O), the product, benzyl alcohol-OD, is formed:
C6H5CH2OD
(i)
Give the structure of the product formed when 2-hydroxybenzaldehyde
, is reduced by LiAlD4 in dry ether followed by H2O.
[1]
(ii)
LiAlH4 reacts with H2O to give two insoluble hydroxides and a gas. Give a
balanced equation for the reaction.
[1]
(iii) When 2-hydroxybenzaldehyde reacts with aqueous BrCl, an insoluble white
precipitate is formed. The Mr of the precipitate is 157.8 units more than that of
2-hydroxybenzaldehyde.
Write a balanced equation to show the reaction.
[1]
(iv) Give a 2-step synthesis to convert methylbenzene to C6H5CH2OD.
[3]
[Total: 20]
 VJC 2015
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4(a)
Nitrous oxide is a chemical compound with formula N2O. It is commonly known as
laughing gas due to the euphoric effects of inhaling it, a property that had led to its
recreational use as a hallucinogenic anaesthetic.
At a high temperature of 1200 K, in the presence of a gold surface, N2O decomposes to
its elements according to the following equation.
The rate of decomposition of pure N2O may be followed by measuring the total pressure
of the system. In such an experiment, the total pressure, p, increased as follows:
Total pressure, p /kPa
Time /s
Partial pressure of N2O /kPa
25.0
27.5
30.0
32.5
34.0
35.0
0
1030
2360
4230
5870
7420
25.0
x
15
10
7
5
The data were plotted on the following graph.
(i)
Explain mathematically, or otherwise, that the partial pressure of N2O at 1030 s is
20 kPa.
[1]
(ii)
By using a graphical method, determine the order of the reaction with respect to
N2O.
[1]
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(iii)
Calculate the rate constant for the reaction.
[1]
(iv)
Suggest a reaction mechanism consisting of two elementary steps for the
decomposition of nitrous oxide that is consistent with the rate equation.
Label the slow step of the reaction clearly.
[2]
(b)
Alkene metathesis is a reaction that breaks the double bond in an alkene and then
rejoins the fragments. When the fragments are joined, each new double bond is formed
between two sp2 carbons that were not previously bonded. Metathesis is a Greek word
that means “transposition.”
Compound A (C8H12O2), is a sweet-smelling compound that undergoes metathesis
reaction to give ethene and compound B (C6H8O2), which possesses a five-membered
ring.
When reacted with hot H2SO4(aq), B gave compound C (C6H10O3).
One mole of compound C gave two moles of carbon dioxide gas and one mole of
compound D (C4H6O3) when treated with hot acidified concentrated KMnO4.
Compound D
Suggest structures for A, B and C and explain the reactions.
[5]
(c)
Geraniol (C10H18O) appears as a clear to pale-yellow oil that is insoluble in water but
soluble in most common organic solvents. It has a rose-like scent and is commonly used
in perfumes.
Geraniol
 VJC 2015
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On treatment with phosphoric acid, H3PO4 at 200°C, geraniol dehydrates to give
dipentene, C10H16, which reacts with an excess of bromine to give a tetrabromide,
C10H16Br4.
The dehydration reaction can be broken into the following steps:
Step 1
The primary alcohol is protonated by phosphoric acid.
Intermediate X
Step 2
In a concerted manner, the C–O bond breaks resulting in the expulsion of a water
molecule and the carbon atom is attacked by an electron-rich double bond leading to the
formation of a carbocation with a six-membered ring.
Step 3
The octet electronic configuration is restored to the positively charged carbon in the
carbocation by a transfer of electrons from a C–H bond. The transfer of electron is
facilitated by H2PO4− acting as a base.
Dipentene, with the following structure, is formed as the product:
Dipentene
 VJC 2015
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Copy the structure of the organic intermediate X formed in step 1 onto your answer
script and complete the remaining steps 2 and 3 of the mechanism.
In the description of the mechanism, you should clearly show curly arrows, charges,
dipoles and any relevant lone pair.
[2]
(d)
(i)
Name the mechanism of the reaction between dipentene and bromine.
[1]
(e)
(ii)
Draw the structure of the tetrabromide that is produced from the reaction between
dipentene and bromine.
[1]
(iii)
State the type of stereoisomerism exhibited by the tetrabromide compound and the
number of isomers present.
[2]
Nerol, which is present in the essential oil, bergamot, has the same formula as geraniol
and undergoes the same dehydration reaction to give the same dipentene. The main
difference between geraniol and nerol is the melting point. Geraniol has a melting point
of 15°C while nerol has a melting point of around −10°C.
(i)
Draw the structure of nerol and state the type of isomerism that exists between
geraniol and nerol.
[2]
(ii)
Hence, explain why there is a difference in melting point between the two isomers.
[2]
[Total: 20]
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5 This question is about the chemistry of some nitrogen and oxygen containing compounds.
Nitrogen and oxygen are two very electronegative elements which make up about 78%
and 20% of the atmosphere by volume respectively. Nitrogen containing compounds are
vital components of foods, fertilizers, and explosives while oxygen is the third most
abundant element found in the sun.
(a)
Procaine is a nitrogen containing compound which is used primarily to reduce the
pain of intramuscular injection of penicillin.
CO2CH2CH2N(C2H5)2
NH2
Procaine
Draw the structure of the compound(s) formed when procaine reacts with:
(i)
Hot, dilute sulfuric acid
[2]
(ii)
Aqueous bromine
[1]
(b)
Compare and explain the relative basicity between the two nitrogen-containing
compounds below in terms of their structures.
CO2CH2CH2N(C2H5)2
CO2CH2CH2N(C2H5)2
Cl
Cl
NH2
Procaine
NH2
Compound L
[2]
(c)
Suggest a chemical test that allows you to distinguish procaine from compound L.
[2]
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(d)
Dinitrogen tetroxide, commonly referred to as nitrogen tetroxide, N2O4, is a useful
reagent in chemical synthesis. It is also a powerful oxidizer that spontaneously
reacts upon contact with various forms of hydrazine, which makes the pair a
popular bipropellant for rockets.
(i)
Given the following reactions and their standard enthalpy changes, construct
an energy cycle to calculate the enthalpy change for the reaction below
involving the formation of gaseous N2O4.
N2O3(g) + N2O5(s)  2N2O4(g)
NO(g) + NO2(g)  N2O3(g)
NO(g) + NO2(g) + O2(g)  N2O5(g)
2NO2(g)  N2O4(g)
2NO(g) + O2(g)  2NO2(g)
N2O5(g)  N2O5(s)
H1 = –39.8 kJ mol–1
H2 = –113 kJ mol–1
H3 = –57.2 kJ mol–1
H4 = –114 kJ mol–1
H5 = –54.1 kJ mol–1
[3]
(ii)
The standard molar entropy, S is the entropy content of one mole of
substance under standard conditions of 298 K and 1 atm. The S values of
some nitrogen oxides are given in the table below:
Nitrogen oxides
S /J K−1 mol−1
N2O4(g)
304.29
N2O3(g)
314.63
N2O5(s)
178.20
Given that Sreaction = ∑S(products) – ∑S(reactants) and using the data
above, calculate the ∆G value for the reaction below at 298 K:
N2O3(g) + N2O5(s)  2N2O4(g)
[2]
(iii)
 VJC 2015
Hence, calculate the minimum temperature at which the reaction will take
place.
[1]
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(e)
The graph below is an Ellingham diagram, which shows the variation in the Gibbs
free energy change of formation, ∆Gf, with temperature, T, for some oxides. This
graph is used to evaluate the ease of reduction of metal oxides.
∆Gf
/ kJ mol−1
0
4Ag + O2  2Ag2O
2Zn + O2  2ZnO
−200
−400
2C + O2  2CO
−600
500
(i)
1000
1500
2000
2500
Temp./ K
Which oxide is the least stable to heat at 1250 K? Explain your choice.
[2]
(ii)
What does the gradient of the graph in the Ellingham diagram represent?
Hence, explain why the sign of the gradient for 2Zn + O2  2ZnO is as
reflected on the graph.
[2]
(iii)
Calculate the ∆G value for the reduction of zinc oxide by carbon at 2000K.
Hence, comment on the feasibility of the reaction at 2000 K.
[3]
[Total: 20]
 VJC 2015
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Victoria Junior College
2015 H2 Chemistry Prelim Exam 9647/3
Suggested Answers
1
(a)
Magnesium ethanoate is commonly used as a source of magnesium or ethanoate ions in
chemistry experiments. One of the more prevalent uses of magnesium ethanoate is in
the mixture called calcium magnesium ethanoate (CMA). It is a mixture of calcium
ethanoate and magnesium ethanoate. CMA acts as a powerful SO2, NOx, and toxic
particulate emission control agent in coal combustion processes to reduce acid rain, and
as an effective catalyst for the facilitation of coal combustion.
Samples of anhydrous magnesium ethanoate, Mg(CH3COO)2 and barium ethanoate,
Ba(CH3COO)2, were heated to a temperature of 600oC causing them to thermally
decompose. White residues were formed in both cases and a common gaseous product,
X, with molecular formula of C3H6O was also given off. X gives a yellow precipitate upon
warming with aqueous alkaline iodine.
In addition, the decomposition of magnesium ethanoate also produced a second
gaseous product which formed a white precipitate with calcium hydroxide solution.
(i)
Identify the gaseous product, X.
Propanone, CH3COCH3
[1]
(ii)
Write a balanced equation with state symbols, for each decomposition reaction.
Mg(CH3COO)2 (s)
MgO(s) + CO2(g) + CH3COCH3(g)
Ba(CH3COO)2 (s)
BaCO3(s) + CH3COCH3(g)
[2]
(iii)
Account for the difference in the decomposition products.
The Mg2+ ion is much smaller than Ba2+ ion, hence its higher charge density
enables it to polarise the CH3COO- ion and distort the C-O bond to a greater
extent resulting in complete decomposition.
[2]
 VJC 2015
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(b)
Magnesium, is an extremely important light weight structural metal which can be
produced by the electrolysis of magnesium chloride. Magnesium chloride can be
prepared from magnesium oxide which is obtained from sea-water containing a
significant amount of Mg2+ and Ca2+. The steps involved are shown below:
Step 1
Controlled
addition of
CO32-
Sea water
(containing
Mg2+ and Ca2+)
Step 2
addition of
OH-(aq)
Mg(OH)2
filtrate
filter
Step 3
heat
Mg
MgO
MgCl2
Step 5
Step 4
The numerical values of the relevant solubility products are given below.
Magnesium carbonate
Calcium carbonate
Magnesium hydroxide
Calcium hydroxide
(i)
1.0 x 10-5
8.7 x 10-9
1.1 x 10-11
5.5 x 10-6
Calculate and compare the solubility between magnesium carbonate and
magnesium hydroxide.
Ksp (Magnesium carbonate) = [Mg2+][CO32-]
1.0 x 10-5 = (x)(x)
x = 3.16 x 10-3 mol dm-3
Ksp (Magnesium hydroxide) = [Mg2+][OH-]2
1.1 x 10-11 = (y) (2y)2
y = 1.40 x 10-4 mol dm-3
Solubility of MgCO3 is higher than that of Mg(OH)2
[3]
(ii)
Assuming the concentration of magnesium ions in the filtrate in step 2 is
3.0 x 10-5 mol dm-3, calculate the concentration of hydroxide ions present in the
filtrate when trace of solid magnesium hydroxide first appears.
First trace of ppt appears when
Ionic product = Ksp
[Mg2+][OH-]2 = 1.1 x 10-11
(3.0 x 10-5)(z)2 = 1.1 x 10-11
z = 6.06 x 10-4 mol dm-3
[1]
 VJC 2015
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(iii)
Explain why the addition of carbonate ions in step 1 has to be controlled.
To ensure maximum precipitation of Ca2+ in the form of CaCO3 which is
then removed while preventing the precipitation of Mg2+ in Step 1.
[1]
(iv)
Give a reason why the electrolysis of magnesium chloride is preferred to that of
magnesium oxide.
The melting point of MgCl2 is lower than MgO and hence less energy will be
required to melt MgCl2.
[1]
(c) Some organic and inorganic compounds are classified under a category called ‘nonexistent compounds’. They are called non-existent because so far chemists had been
unable to synthesise them. Some of the reasons why these compounds are unstable
are:



unfavourable bond energy terms
a redox incompatibility of the ions making up the compound
reaction occurring between the ions
For the following cases, suggest an explanation for each observation. You may use data
from the Data Booklet to assist you in your answers. Give relevant equations to support
your answers.
(i)
When aqueous sodium carbonate is added to aqueous aluminium chloride, a
colourless gas is liberated and the precipitate formed is not aluminium carbonate,
Al2(CO3)3.
Aluminium carbonate is not precipitated because Al3+(aq) is acidic and will
react with the carbonate, liberating carbon dioxide gas.
When aluminum chloride dissolves in water, it undergoes hydrolysis to
produce an acidic solution as follows:
Al(H2O)63+ + H2O
Al(H2O)5(OH)2+ + H3O+
This acidic solution will react with sodium carbonate to produce carbon
dioxide:
2H3O+ + CO32-  CO2 + 3H2O
[OR 2H+ + CO32-  CO2 + H2O]
The precipitate is aluminium hydroxide formed due to hydrolysis of sodium
carbonate.
CO32- + H2O
HCO3- + OH3+
Al (aq) + 3OH (aq)  Al(OH)3(s)
[4]
 VJC 2015
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(ii)
Caprolactum, a monomer of the polymer nylon 6, exists as an alicyclic ring
structure and not as 6-aminohexanoic acid, a straight chain aliphatic structure as
most monomers do.
O
H
C
N
H2NCH2CH2CH2CH2CH2CO2H
Caprolactum
6-aminohexanoic acid
6-aminohexanoic acid can undergo internal nucleophilic substitution [OR
cyclisation] to form caprolactum.
H
O
O
NH HOC
CH2
H
C
CH
N

+ H2O
CH2CH2CH2
Energy requirements:
To break NH and CO bonds = + 390 + 360 = +750 kJ mol¯1
To form OH and CN bonds = – 460 – 305 = –765 kJ mol¯1
Overall H = –15 kJ mol-1.
Reaction is exothermic and feasible.
[Note: Overall there is also an increase in entropy of the system.]
[3]
(iii)
It is possible to find MnCl2 in the laboratory but not MnCl3.
Cl2 + 2eMn3+ + e-
2ClMn2+
Eo = +1.36 V
Eo = +1.49 V
2Mn3+ + 2Cl-  2Mn2+ + Cl2
= 1.49 – (1.36)
= +0.13 V
Since
> 0, a redox reaction will occur in which Mn3+ is reduced to Mn2+
and Cl- is oxidized to chlorine gas when Mn3+ is combined with Cl-.
[2]
[Total: 20]
 VJC 2015
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2
(a)
The Contact Process is an industrial process for manufacturing sulfuric acid. The
key stage in this process is the reaction between sulfur dioxide and oxygen.
2SO2(g) + O2(g) ⇌ 2SO3(g)
H = –197 kJ mol–1
Vanadium(V) oxide, V2O5, is used as a heterogeneous catalyst for this process.
Describe the mode of action by which V2O5 fulfils this role.
Availability of partially filled 3d orbitals in vanadium allow reactant molecules
to be adsorbed onto metal surface.
Reactant molecules are brought closer together and bonds within them are
weakened. Activation energy is lowered.
Subsequently, products formed desorb from the catalyst surface.
[3]
(b)
In an experiment to determine the Kp of the above equilibrium, a mixture containing
0.200 mol of SO2 and 0.100 mol of O2 was heated in a closed flask and allowed to
reach equilibrium at 550oC and 3.5 atm.
The flask was then rapidly cooled to liquefy the SO3 so that it can be separated from
the gaseous SO2 and O2. Excess water was carefully added to the liquid SO3,
causing the following reaction to occur:
SO3(l) + H2O(l)  H2SO4(aq)
The resulting solution was made up to 250 cm3 in a standard volumetric flask.
25.0 cm3 of this solution was titrated with 1.00 mol dm−3 NaOH and required
36.0 cm3 for complete neutralisation.
(i)
Calculate the equilibrium amount of SO2, O2 and SO3 at 550oC.
2NaOH + H2SO4  Na2SO4 + 2H2O
No. of moles of H2SO4 in 25.0 cm3 = ½ × 36.0 / 1000 × 1.00
= 0.0180 mol
No. of moles of H2SO4 in 250 cm3
Initial mol
Change in mol
2SO2(g)
0.200
–0.180
Eqm mol
0.020
+
= (250 / 25.0) × 0.0180
= 0.180 mol
= no. of moles of SO3
O2(g)
0.100
–½ (0.180)
0.01
⇌ 2SO3(g)
0
+0.180
0.180
[2]
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(ii)
Hence, calculate a value for Kp, including its correct units.
Total number of moles
Kp
= 0.020 + 0.010 + 0.180
= 0.210 mol
=
=
= 486 atm-1
[1]
(iii) In liquefying the SO3, it was mentioned that the cooling was performed rapidly.
Explain how the titration volume would be affected if the cooling had been
performed slowly.
If the cooling had been performed slowly, equilibrium position will be
affected. By Le Chatelier’s Principle, equilibrium position will shift to the
right since the forward reaction is exothermic, releasing heat to
counteract the decrease in temperature. Hence, more SO3 will be formed,
causing titration volume to be larger.
[2]
(c)
One important use of vanadium compounds is the vanadium redox battery, a
rechargeable battery which utilizes graphite electrodes. The electrolytes are pumped
from separate storage tanks A and B into compartments X and Y of the cell
respectively, where they are separated by a semi–permeable proton exchange
membrane which allows only passage of protons.
electrical load
Tank A
acidified
VO2+(aq) and
VO2+(aq)
X
Y
e–
Tank B
acidified
V3+(aq) and
V2+(aq)
membrane
The direction of the electron flow during the discharging process is indicated in the
diagram of the battery above.
(i)
By means of the letters X and Y, state and briefly justify the direction of proton
flow across the membrane during the discharging process.
Protons flow from Y to X to maintain electrical neutrality of the
electrolytes.
[1]
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In a particular setup, compartment X contained 5 dm3 of the electrolyte, with VO2+
and VO2+ each having an original concentration of 1.5 mol dm-3. The cell discharged
until 90% of the reactants was consumed, after which the cell was recharged by
connecting an electricity source across the electrodes.
(ii)
With the aid of the Data Booklet, write half–equations for the reactions that
occur at the electrodes in X and Y during the charging process.
X: VO2+ + H2O  VO2+ + 2H+ + e
Y: V3+ + e  V2+
[1]
(iii) If the current used for the charging process is 14 A, calculate the time taken, in
hours, needed to restore the concentration of the vanadium-containing ions in
compartment X to their original value.
Moles of reactant (VO2+) consumed during discharging = 0.90 × 1.5 × 5
= 6.75 mol
Since nreactant : nelectrons = 1:1,
Moles of electrons = 6.75 mol
Moles of electrons = I × t / 96500
6.75 = 14 × t / 96500
t = 46500 s = 12.9 h
[2]
(d)
Chromium is another transition metal adjacent to vanadium in the Periodic Table.
Some of its reactions are described below:
When chromium metal is dissolved in dilute sulfuric acid, it produces a blue solution
containing Cr2+(aq). The blue solution slowly turns green even in the absence of air.
This mixture is then filtered to remove the unreacted metal. To the green filtrate V,
sodium hydroxide solution was added, forming a grey−green precipitate, W. In
excess sodium hydroxide, the precipitate dissolves to form a dark green solution
containing an anion, X. The dark green solution turns into a yellow solution, Y, upon
warming with hydrogen peroxide solution.
(i)
By quoting and using relevant Eo values from the Data Booklet, explain why
the blue solution turns green.
Cr3+ + e−
2H+ + 2e−
Cr2+
H2
Eo = −0.41 V
Eo = 0.00 V
= 0 – (−0.41)
= +0.41 V
Since
Cr3+.
> 0, reaction is feasible and the blue Cr2+ is oxidised to green
[2]
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(ii)
Identify the precipitate W and the anion X and suggest ionic equations, with
state symbols, for the formation of W, and for the formation of X.
W is Cr(OH)3
X is [Cr(OH)6]3Cr3+(aq) + 3OH-(aq) ⇌ Cr(OH)3(s)
Cr(OH)3(s) + 3OH-(aq) ⇌ [Cr(OH)6]3-(aq)
[OR Cr(H2O)63+(aq) + 6OH-(aq) ⇌ [Cr(OH)6]3-(aq) + 6H2O(l)]
[2]
(iii) State the role of hydrogen peroxide in converting the dark green solution into
Y.
Oxidising agent
[1]
(e)
A particular chromium compound has the formula Cr(H2O)6Cl3. It contains a complex
Z in which the coordination number of chromium is 6.
When a solution containing 0.02 mol of the compound, Cr(H2O)6Cl3, is treated with
excess aqueous Pb(NO3)2, a white precipitate is formed. After filtration and drying,
the mass of the precipitate is found to be 2.78 g.
(i)
Use the given information to determine the formula of the complex, Z.
The precipitate formed is PbCl2. Hence,
Amount of free Cl- anions = 2.78 / (207 + 71.0) × 2
= 0.02 mol
=1
Hence, 2 Cl- ligands are present in the complex.
Formula of the complex is [Cr(H2O)4Cl2]+.
[1]
 VJC 2015
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(ii)
Z is known to exist as two stereoisomers. Draw the structures of these two
stereoisomers, showing clearly the spatial arrangement of the ligands around
the central metal ion.
(cis)
(trans)
[2]
[Total: 20]
3
Phosphorus belongs to Period 3, an element that is essential for life. The elemental form
of phosphorus consists of tetrahedral P4 molecules, in which each Phosphorus atom forms
three single bonds with three other Phosphorus atoms. Phosphates, compounds
containing the PO43 ion, are a component of DNA, RNA, ATP and also the phospholipids,
which form all cell membranes.
(a)
Each of the Period 3 elements, Na to S, reacts at a different rate with oxygen to form
oxides of different oxidation states.
(i)
Describe what you see when magnesium and sulphur are separately burned in
air or oxygen. Write equations for the reactions that occur.
2Mg(s) + O2(g)  2 MgO(s)
Mg burns with a brilliant white flame to give a white solid.
The reaction is vigorous.
S(s) + O2(g)  SO2(g)
S burns with a pale blue flame to give a pungent gas (with burning S
smell).
The reaction is slow.
[3]
(ii)
Explain briefly why the bonding of the oxides of Period 3 elements vary from
ionic to covalent.
The difference between the electronegativity values of the Period 3
elements and oxygen decreases across the period.
[1]
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(iii) X and Y are Period 3 elements.
Element X forms a white oxide that is slightly soluble in cold water. Its chloride
dissolves in water to form a weakly acidic solution.
Element Y forms two oxides. 0.03 mol of one of the two oxides produces
6.99 g of white precipitate when shaken with excess barium chloride solution.
Another solution containing 0.05 mol of this oxide of element Y forms a neutral
solution when the same amount of the oxide of the element X is added to it.
Identify the elements X, Y, the oxide of Y and the white precipitate that is
formed in the above reaction. Explain why the chloride of element X is acidic.
Element X : Magnesium
Element Y: Sulfur
Formula of the oxide of Element Y: SO3
6.99g of white ppt correspond to the mass of 0.03 mol of BaSO4.
Upon addition of BaCl2, white ppt formed is BaSO4.
Due to the high charge density of the Mg2+ ion, MgCl2 hydrolyses slightly to
form a weakly acidic solution.
[3]
(b)
Sulfur, another Period 3 element, forms a wide range of sulfides with phosphorus.
One of the most well-known examples is the three-fold symmetric P4S3 used in
strike-anywhere matches.
(i)
State the two oxidation states of phosphorus that exist in P4S3.
The oxidation states are +1 and +3.
[1]
(ii)
P4O3 has a structure that is similar to P4S3. Explain why the OPO bond angle
is smaller than the SPS bond angle.
As O atom is more electronegative than S atom, the shared electrons are
attracted more towards the O atom in the PO bond than for the PS bond.
Hence, the smaller bond pairbond pair repulsion causes the OPO bond
angle to be smaller than the SPS bond angle.
[2]
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(iii) The bond energy for the PP single bond is 197 kJ mol1.
Given the enthalpy change for the following transformation
P4S3(g)  P4(g) + 3S(g)
H = +1029 kJ mol1,
calculate the bond energy for the PS bond.
6 PS bonds are broken and 3 PP bonds are formed.
H


= BE(bond breaking)  BE(bond forming) = +1029
6 x BE(PS)  3 x BE(PP) = 1029
BE(PS) =
= 270 kJ mol1.
[2]
(iv) In the presence of excess sulfur, P4S3 can be converted to P4S10, which is
structurally and chemically similar to P4O10. Give a balanced equation for the
reaction between P4S10 and H2S which does not involve any change in
oxidation state. Hence, state the role that P4S10 plays in the reaction.
P4S10 + 6H2S  4H3PS4
P4S10 is acting as a base.
[2]
(c)
Aluminium is an important element in the reducing agent, LiAlH4, for organic
synthesis. When benzoic acid is reduced by LiAlH4 in dry ether followed by addition
of heavy water (D2O), the product, benzyl alcohol-OD, is formed:
C6H5CH2OD
(i)
Give the structure of the product formed when 2-hydroxybenzaldehyde
, is reduced by LiAlD4 in dry ether followed by H2O.
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[1]
(ii)
LiAlH4 reacts with H2O to give two insoluble hydroxides and a gas. Give a
balanced equation for the reaction.
LiAlH4(s) + 4H2O(l)  LiOH(s) + Al(OH)3(s) + 4H2(g)
[1]
(iii) When 2-hydroxybenzaldehyde reacts with aqueous BrCl, an insoluble white
precipitate is formed. The Mr of the precipitate is 157.8 units more than that of
2-hydroxybenzaldehyde.
Write a balanced equation to show the reaction.
[1]
(iv) Give a 2-step synthesis to convert methylbenzene to C6H5CH2OD.
[3]
[Total: 20]
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4(a)
Nitrous oxide is a chemical compound with formula N2O. It is commonly known as
laughing gas due to the euphoric effects of inhaling it, a property that had led to its
recreational use as a hallucinogenic anaesthetic.
At a high temperature of 1200 K, in the presence of a gold surface, N2O decomposes to
its elements according to the following equation.
The rate of decomposition of pure N2O may be followed by measuring the total pressure
of the system. In such an experiment, the total pressure, p, increased as follows:
Total pressure, p /kPa
Time /s
Partial pressure of N2O /kPa
25.0
27.5
30.0
32.5
34.0
35.0
0
1030
2360
4230
5870
7420
25.0
x
15
10
7
5
The data were plotted on the following graph.
half life of reaction ≈ 3200
s
(i)
 VJC 2015
Explain mathematically, or otherwise, that the partial pressure of N2O at 1030 s is
20 kPa.
9647/03/PRELIM/15
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The reaction results in an overall increase in number of moles of gas. From
two moles of gaseous reactant to a total of three moles of gaseous products.
The increase in total pressure from 25 kPa to 27.5 kPa is due to the difference
between the number of moles of gaseous reactant and products. Since two
moles of gaseous reactants are consumed for every one mole increase in
gaseous products therefore,
partial pressure of N2O at 1030 s = 25 – 2(2.5) = 20 kPa
[1]
(ii)
By using a graphical method, determine the order of the reaction with respect to
N2O.
Since half-life is constant at 3200 s, reaction is first order wrt N2O.
[1]
(iii)
Calculate the rate constant for the reaction.
k=
ln2 ln2
=
=2.17×10-4 s-1
t 1 3200
2
[1]
(iv)
Suggest a reaction mechanism consisting of two elementary steps for the
decomposition of nitrous oxide that is consistent with the rate equation.
Label the slow step of the reaction clearly.
slow
• N2O 
 N2 +O
fast
• N2O+O 
 N2 +O2
[2]
(b)
Alkene metathesis is a reaction that breaks the double bond in an alkene and then rejoins
the fragments. When the fragments are joined, each new double bond is formed between
two sp2 carbons that were not previously bonded. Metathesis is a Greek word that means
“transposition.”
Compound A (C8H12O2), is a sweet-smelling compound that undergoes metathesis
reaction to give ethene and compound B (C6H8O2), which possesses a five-membered
ring.
When reacted with hot H2SO4(aq), B gave compound C (C6H10O3).
One mole of compound C gave two moles of carbon dioxide gas and one mole of
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compound D (C4H6O3) when treated with hot acidified concentrated KMnO4.
Compound D
Suggest structures for A, B and C and explain the reactions.
If a single molecule of A can undergo metathesis, this means A contains two C=C
double bonds.
A is sweet-smelling, therefore, A is an ester.
The number of C=C double is conserved during a metathesis reaction, B must
contain a C=C double bond since the other metathesis product is ethene.
B still contains the original ester group since metathesis does not affect ester.
B undergoes ester hydrolysis to give C.
C contains one carboxylic acid FG, one alcohol FG and one alkene FG.
C undergoes oxidative cleavage to give D and two moles of CO2 gas.
The ketone carbon in D must have come from a cleavage of a C=C.
A
B
C
[5]
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(c)
Geraniol (C10H18O) appears as a clear to pale-yellow oil that is insoluble in water but
soluble in most common organic solvents. It has a rose-like scent and is commonly used
in perfumes.
Geraniol
On treatment with phosphoric acid, H3PO4 at 200°C, geraniol dehydrates to give
dipentene, C10H16, which reacts with an excess of bromine to give a tetrabromide,
C10H16Br4.
The dehydration reaction can be broken into the following steps:
Step 1
The primary alcohol is protonated by phosphoric acid.
Intermediate X
Step 2
In a concerted manner, the C–O bond breaks resulting in the expulsion of a water
molecule and the carbon atom is attacked by an electron-rich double bond leading
to the formation of a carbocation with a six-membered ring.
Step 3
The octet electronic configuration is restored to the positively charged carbon in the
carbocation by a transfer of electrons from a C–H bond. The transfer of electron is
facilitated by H2PO4− acting as a base.
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Dipentene, with the following structure, is formed as the product:
Dipentene
Copy the structure of the organic intermediate X formed in step 1 onto your answer script
and complete the remaining steps 2 and 3 of the mechanism.
In the description of the mechanism, you should clearly show curly arrows, charges,
dipoles and any relevant lone pair.
+ H3PO4
two arrows indicating flow of electrons in first step
two arrows indicating flow of electron in second step
[2]
 VJC 2015
9647/03/PRELIM/15
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18
(d)
(i)
Name the mechanism of the reaction between dipentene and bromine.
electrophilic addition
[1]
(ii)
Draw the structure of the tetrabromide that is produced from the reaction between
dipentene and bromine.
[1]
(iii)
State the type of stereoisomerism exhibited by the tetrabromide compound and the
number of isomers present.
optical isomerism
number of isomers = 24 = 16
[2]
(e)
Nerol, which is present in the essential oil, bergamot, has the same formula as geraniol
and undergoes the same dehydration reaction to give the same dipentene. The main
difference between geraniol and nerol is the melting point: geraniol has a melting point of
15°C while nerol has a melting pointing of around −10°C.
(i)
Draw the structure of nerol and state the type of isomerism that exists between
geraniol and nerol.
structure of nerol (with 120° bond angle about double bond)
cis-trans isomerism
[2]
 VJC 2015
9647/03/PRELIM/15
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19
(ii)
Hence, explain why there is a difference in melting point between the two isomers.
The trans isomers pack more compactly / efficiently resulting in stronger
dispersion forces of attraction between molecules. More energy is required to
break the dispersion forces between the trans isomers than cis isomers
resulting in nerol (cis isomer) having lower melting point than geraniol.
[2]
[Total: 20]
5 This question is about the chemistry of some nitrogen and oxygen containing compounds.
Nitrogen and oxygen are two very electronegative elements which make up about 78%
and 20% of the atmosphere by volume respectively. Nitrogen containing compounds are
vital components of foods, fertilizers, and explosives while oxygen is the third most
abundant element found in the sun.
(a)
Procaine is a nitrogen containing compound which is used primarily to reduce the
pain of intramuscular injection of penicillin.
CO2CH2CH2N(C2H5)2
NH2
Procaine
Draw the structure of the compound(s) formed when procaine reacts with:
(i)
Hot, dilute sulfuric acid
H3N+
COOH
SO422
H
HOCH2CH2N+(C2H5)2
2
SO42[2]
(ii)
Aqueous bromine
CO2CH2CH2N(C2H5)2
Br
Br
NH2
[1]
 VJC 2015
9647/03/PRELIM/15
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20
(b)
Compare and explain the relative basicity between the two nitrogen-containing
compounds below in terms of their structures.
CO2CH2CH2N(C2H5)2
CO2CH2CH2N(C2H5)2
Cl
Cl
NH2
Procaine
NH2
Compound L
Procaine is more basic.
Cl in compound L is electronegative. It exerts electron-withdrawing inductive
effect which reduces the availability of the lone pair electrons on N for
donation to acid.
[OR Cl in compound L is electronegative. It exerts electron-withdrawing
inductive effect which intensifies the positive charge on the conjugate acid
ion and hence destabilises it.]
[2]
(c)
Suggest a chemical test that allows you to distinguish procaine from compound L.
Add Br2(aq) to each compound. There will be decolourisation of reddishbrown Br2 for procaine but no colour change for L.
[2]
(d)
Dinitrogen tetroxide, commonly referred to as nitrogen tetroxide, N2O4, is a useful
reagent in chemical synthesis. It is also a powerful oxidizer that spontaneously
reacts upon contact with various forms of hydrazine, which makes the pair a
popular bipropellant for rockets.
(i)
Given the following reactions and their standard enthalpy changes, construct
an energy cycle to calculate the enthalpy change for the reaction below
involving the formation of gaseous N2O4.
N2O3(g) + N2O5(s)  2N2O4(g)
NO(g) + NO2(g)  N2O3(g)
NO(g) + NO2(g) + O2(g)  N2O5(g)
2NO2(g)  N2O4(g)
2NO(g) + O2(g)  2NO2(g)
N2O5(g)  N2O5(s)
 VJC 2015
9647/03/PRELIM/15
H1 = –39.8 kJ mol–1
H2 = –113 kJ mol–1
H3 = –57.2 kJ mol–1
H4 = –114 kJ mol–1
H5 = –54.1 kJ mol–1
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Hreaction
N2O3(g)
+
39.8

N2O5(s)
2N2O4(g)
54.1
NO(g) + NO2(g)
N2O5(g)
2(–57.2)
113
NO(g) + NO2(g)
NO(g) + NO2(g) + O2(g)
–114
2NO2(g) + 2NO2(g)
[2 m for energy cycle, −1/2m for every mistake]
By Hess’s Law:
Hreaction = 39.8 + 54.1 + 113 + (–114) + 2(–57.2) = –21.5 kJ mol–1
[3]
(ii)
The standard molar entropy, S is the entropy content of one mole of
substance under standard conditions of 298 K and 1 atm. The S values of
some nitrogen oxides are given in the table below:
Nitrogen oxides
S /J K−1 mol−1
N2O4(g)
304.29
N2O3(g)
314.63
N2O5(s)
178.20
Given that Sreaction = ∑S(products) – ∑S(reactants) and using the data
above, calculate the ∆G value for the reaction below at 298 K:
N2O3(g) + N2O5(s)  2N2O4(g)
Sreaction = 2(304.29) – 178.20 – 314.63
= 116 J K-1 mol-1
Greaction = Hreaction – TSreaction
= –21.5 x 103 – 298(116)
= –56.1 kJ mol-1
[2]
(iii)
Hence, calculate the minimum temperature at which the reaction will take
place.
Using ∆G = ∆H − T∆S and noting that reaction will take place when
∆G = 0,
T = −21.5 x 103 / 116
= −185 K
[1]
 VJC 2015
9647/03/PRELIM/15
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22
(e)
The graph below is an Ellingham diagram, which shows the variation in the Gibbs
free energy change of formation, ∆Gf, with temperature, T, for some oxides. This
graph is used to evaluate the ease of reduction of metal oxides.
∆Gf
/ kJ mol−1
0
4Ag + O2  2Ag2O
2Zn + O2  2ZnO
−200
−400
2C + O2  2CO
−600
500
(i)
1000
1500
2000
2500
Temp./ K
Which oxide is the least stable to heat at 1250 K? Explain your choice.
AgO
∆G < 0 for the decomposition of AgO, 2Ag2O  4Ag + O2
It means that the decomposition of AgO is spontaneous at 1250 K.
[2]
(ii)
What does the gradient of the graph in the Ellingham diagram represent?
Hence, explain why the sign of the gradient for 2Zn + O2  2ZnO is as
reflected on the graph.
Gradient represents –S (equation of line is G = –TS + H).
A positive gradient implies that S is negative. A decrease in
disorderliness of the system (Or S < 0) is due to a decrease in amount
of gases (from 1 to 0 mol) as the reaction proceeds.
[2]
(iii)
Calculate the ∆G value for the reduction of zinc oxide by carbon at 2000K.
Hence, comment on the feasibility of the reaction at 2000 K.
At 2000 K,
2C + O2  2CO
G = – 600 kJ mol-1 ………. (1)
2Zn + O2  2ZnO
G = – 100 kJ mol-1 ........... (2)
Hence, 2ZnO  2Zn + O2
G = + 100 kJ mol-1 ........... (3)
Reduction of ZnO by C:
(1) + (3) gives 2ZnO + 2C  2Zn + 2CO
Hence, G = –600 + 100 = –500 kJ mol-1
Since G < 0, the reaction is therefore thermodynamically feasible at
2000 K.
[3]
[Total: 20]
 VJC 2015
9647/03/PRELIM/15
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