2011 JC1 H2 Chemistry Promotional Examination Paper 2 Corrections
SECTION A
1 In car accidents, the activation of an airbag is often able to save a life when it cushions impact on the
passenger’s head. The expansion of the airbag is caused by the electronic detonation ofasodium
compound,NaNx, that triggersa decomposition reaction, producing only sodium metal and nitrogen
gas.
A student was given 1.00 g of the sodium compound. He heated the solid on a weighing balance until
a constant mass reading of 0.354 g was obtained. The gas produced was collected in a container
under roomconditions.
(a) Find the empirical formula of the sodium compound, NaNx.
(Assume that all products formed do not take part in further reactions.)
[2]
Na N
Mass / g 0.354 0.646
Amount / mol �.�����.� = 1.539 × 10
-2
�.�����.� = 4.614 × 10
-2
Simplest mole ratio 1 3
The empirical formula of the compound is NaN3.
(b) Write a balanced chemical equation for the decomposition and hence calculate the volume (in
cm3) of nitrogen gas collected.
(Assume that molar volume under room conditions is 24.0 dm3.)
[2]
2NaN3 (s) → 2Na (s) + 3N2 (g)
Amount of N2 = �� × 1.539 × 10
-2 = 2.309 × 10
-2mol
Volume of N2 = 2.309 × 10-2
× 24.0 = 0.554 dm3 = 554 cm
3
(c) The nitrogen gas collected is compressed to halfthe volume calculated in (b) and heated from
20 oCto 40
oC. Calculate the pressure exerted by the gas on the container.
[2]
��
= � �
���������������� =
��������
p2 = 216000 Pa
(d) In reality, the pressure exerted by the gas on the container was measured to be lower than the
calculated value in (c).State two reasons why this discrepancy was observed.
[2]
The particles of nitrogen gas, a non-ideal gas, have significant volume compared to that of the
container
and there are significant intermolecular forces of attractions between the particlesand between
the particles and the wall of the container.
Total 8 marks
Comment [LHYC1]: Question is poorly
done in general.
Students show a lack of understanding of
mole concept and ideal gas theories.
Comment [LHYC2]: A common
mistake made was to find the amount of N2,
compare it to the amount of Na and obtain
a mole ratio of 2:3, and using that mole
ratio for the empirical formula.
Some students simply wrote an equation
assuming one mole of N2 is formed, then
balanced the equation.
Comment [LHYC3]: Students
commonly wrote the Na (s) as Na+ (s)
instead.
Comment [LHYC4]: Students found
the amount of N atoms instead of the
amount of N2.
Comment [LHYC5]: A very common
mistake was to leave the answer in dm3.
Question asks for answer in cm3.
Comment [LHYC6]: Students gave a
wide range of irrelevant answers including
heat loss to surroundings, average values
used, not a pure ionic compound, gas
leaked from container, LCP, etc.
Many students stated that the assumption
for the calculations above is that the gas is
ideal, but did not proceed to elaborate on
how it is non-ideal.
Students who caught on to the correct
concepts commonly left out the key word,
“particles” (“gas has significant volume
compared to container” not acceptable),
and the key phrase, “and between particles
and the walls of the container”.
2 The table below shows some thermochemical data:
∆Hθ / kJ mol
-1
Ca (s)→ Ca2+
(aq) + 2e- +347
Ca(s) + S(s) + 2O2(g) → CaSO4(s) -1435
S(s) +2O2(g) + 2e-→ SO4
2-(aq) -907
(a)
Use the data in the table above to construct an energy cycle and hence calculate the standard
enthalpy change of the following reaction:
Ca(NO3)2 (aq) + H2SO4 (aq) → CaSO4 (s) + 2HNO3 (aq)
[3]
By Hess’ Law,
∆Hθ
rxn = -(-907) – (+347)+ (- 1435) = – 875 kJmol-1
(b) (i) Explainwhat is meant by the lattice energy of calcium sulfate.
Lattice energy is the energy evolved when 1 mol of solid CaSO4 is formed from its
constituent gaseous ionsOR
Ca2+
(g) + SO42-
(g) → CaSO4 (s)
(ii) Calcium sulfate is sparingly soluble in water while magnesium sulfate is soluble. Explain,
in terms of bonding and structure whymagnesiumsulfate is soluble in water.
[2]
Magnesiumsulfate is soluble in water as favourable ion-dipole interactions can be
formed, which releases energy to break the giant ionic lattice structure.
Total 5 marks
Ca(NO3)2 (aq) + H2SO4 (aq) CaSO4 (s) + 2HNO3 (aq)
Ca(s) + S(s) + 2O2(g) + 2HNO3 (aq) Ca2+
(aq) + 2e- + S (s) + 2O2 (g) + 2HNO3 (aq)
∆Hθ
rxn
-1435
+347
-907
Comment [LHYC7]: Students who
attempted this question with memory of
definitions and explanations for solubility
managed to obtain the latter 2 marks.
Overall, surprisingly better done than
expected.
Comment [LHYC8]: Students are
unable to form an energy cycle that is
balanced. HNO3 and number of electrons
are stumbling blocks.
A common careless mistake is the leaving
out of state symbols.
Comment [LHYC9]: “energy change”
was accepted in place of “energy evolved”,
but students should be told to use the
latter in future.
Students commonly left out the word
“solid” and other parts of the definition.
Comment [LHYC10]: Second marking
point commonly left out.
Some students cited magnesium sulphate
as having “intermolecular forces of
attraction”, a major misconception.
3 Ethene reacts with bromine in tetrachloromethane to form 1,2-dibromoethane (CH2BrCH2Br)as shown
by the equation:
CH2=CH2 + Br2→ CH2BrCH2Br
To determine the orders of reaction with respect to ethene and bromine, ethene and bromine were
first dissolved separately in tetrachloromethane. Various volumes of these solutions and
tetrachloromethane were then mixed and the time taken for the brown colour of bromine to
disappear was recorded. The results are shown in the table below:
Experiment
Volume of
ethene
solution/ cm3
Volume of
bromine
solution/ cm3
Volume of
tetrachloromethane/
cm3
Time taken for
colour of bromine
to disappear/ s
1 20 20 0 15
2 12 20 8 25
3 20 10 10 25
4 40 20 20 t4
(a) With reference to Experiments 1 to 3, explain why varying volumes of tetrachloromethane
were used.
[1]
To keep the total volume constant so that the concentration of ethene or bromine used is
directly proportional to the volume used.
Alternatively:
[ethene] α volume of ethene and [bromine] α volume of bromine.
(b) The relationship between the rate of reaction with the time taken for the colour of bromine to
disappearandthe volume of bromine used is given as shown:Rate∝ 2BrV
t
Using the experimental data, deduce the order of reaction with respect to ethane and bromine.
[2]
Comparing Experiments 1 and 3,
���������
= �[������]��[��� !��]�"�[������]�[��� !��]"
# $ #$
= �[��]�[��]"�[��]�[��]"
��� = ����
%�y ≈ 2
Order of reaction w.r.t. bromine is 2.
Comparing Experiments 1 and 2,
���� ����
= �[������] �[��� !��] "�[������]�[��� !��]"
# $ #$
= �[��]�[��]"�[��]�[��]"
�� = ����
&�x = 1
Order of reaction w.r.t. ethene is 1.
Comment [KSH11]: Question is very
badly done. Students did not understand
that tetrachloromethane is a solvent and
that it’s role is to keep the total volume
constant such that concentration is
proportional to volume and mentioned
that it was to determine the order of
reaction wrt tetrachloromethane. In
addition, many students mentioned that
the total volume was constant but did not
elaborate on why total volume was
constant.
Comment [KSH12]: Many students
who attempted the question were unable
to use the given relationship to determine
the order of reaction wrt ethane and
bromine. Students used 1/t to calculate the
rate instead. Some students concluded the
order of reaction without relevant working
and marks were not awarded.
(c) Suggest a value for t4, the time taken for the brown colour of bromine to disappear in
Experiment 4.
[2]
Comparing Experiments 1 and 4,
����'����
= �[������]'[��� !��]' �[������][��� !��]
#/)#*' #/'#$
= �+'#)#,+
#)#,
�+ #'#,+ #'#,
t4= 30s
(d) The following graph shows the second ionisation energy trend of successive elements from
Period 3.
(i) With the aid of the Data Booklet , account for the decrease in 2nd
ionisation energy from
element B to element C.
Element B is Al, electronic configuration: 1s22s
22p
63s
23p
1;
Element C is Si, electronic configuration: 1s22s
22p
63s
23p
2
For 2nd
IE: Al+� Al
2++e ; Si
+� Si
2+ +e
electronic configuration of Al+ : 1s
22s
22p
63s
2electronic configuration of Si
+: 1s
22s
22p
63s
23p
1
The 3p electron in Si or Si+ (C or C
+) is further away from the nucleus.
(or the 3p1 electron experience a weaker electrostatic force of attraction from the nucleus.)
Less energy is required to remove the electron resulting in the decrease of 2nd
IE from B to C
(ii) Sketch the trend of electrical conductivity for element A to D. [3]
Electrical Conductivity
element
A B C D
Total 8 marks
Comment [KSH13]: Students did not
recognise that [ethene] and [Br2] is
affected by total volume and that they
need to make use of the given relationship
of rate ∝ VBr/t to determine the time taken.
Comment [KSH14]: Students need to
identify the elements or electronic
configuration correctly in order to explain
the decrease in 2nd
IE. Many students
identified the wrong elements or gave the
wrong electronic configuration and
mentioned inter-electron repulsion
between electrons in the same orbitals led
to the decrease in 2nd
IE.
Comment [KSH15]: Most students
were unable to sketch the correct trend of
electrical conductivity. This may be
because they did not identify the elements
A to D and thus unable to know the
electrical conductivity of the elements.
4 Nitrates are commonly used in the making of pyrotechnics. Hydroxylamine nitrate, NH3OHNO3, is one
of them and it can also be used as a rocket propellant. Hydroxylamine nitrate undergoes catalytic
decomposition according to the following equation:
4NH3OHNO3 (s) → 3N2O (g) + 7H2O (l) + 2HNO3 (l)
The standard enthalpy change of reaction, ΔHrxnθ, is -526.6 kJ mol
-1 and the standard entropy change,
ΔSθ, is +180 J mol
– 1K
-1.
(a) Explain the significance of the sign of ΔSθ. [1]
The positive sign for ΔSθ implies that there is an increase in the number of gaseous
molecules/particles, ie. Δn = +3.
Or A change in state from solid reactants to liquid and gaseous products.
(b) Determine the value of the standard Gibbs free energy, ΔGθ, for the reaction. [1]
ΔGθ = (-526.6) – 298 (180/1000)= -580 kJ mol
–1
(c) Comment on the effect of high temperatures on the feasibility of the reaction. [2]
The reaction will be feasible at all temperatures./ Temperature has no effect on feasibility of
reaction.
ΔGθ = ΔH
θ - TΔS
θ = -ve – (+ve)( +ve) = -ve
ΔGθwill be negative at all temperatures and the reaction will be spontaneous.
(d) Draw the dot-and cross diagram of N2O.[1]
Comment [ECH(C16]: Most students
give the definition of ΔSθ instead of
explaining the sign in reference to the
chemical reaction in the question.
Comment [ECH(C17]: Most students
forget to convert ΔSθto kJ mol
-1 K
-1 and
substitute temperature as ‘293 K’ and ‘ 273
K’ instead.
Comment [ECH(C18]: Some students
interpret as ‘kinetic feasibility’ instead
of ’thermal feasibility’, hence use Ea to
explain. The above reaction is not
reversible so do not cite LCP! Most
students compare the magnitude of
ΔHθvs.TΔS
θ when it is not necessary.
Comment [ECH(C19]: Do not draw
circles to represent electron shelves, use
other shapes (besides dots and crosses) to
represent electrons and ‘→’ to represent
dative bond.
(e) The plots of pV/RT against p for one mole of an ideal gas at 300K is given below.
(i) Show on the same axes how one mole of N2O gaswill behave at the same temperature of 300K.
(ii) On the same axes, draw a curve to representN2O gas when it is cooled to 50K.
(iii) On the same axes show how one mole of N2 gaswill behave at the same temperature of 300K. Explain your answer.
[4]
pV/RT
1.0 Ideal Gas
p
Ideal Gas
(300K)
pV/RT
1.0
(i)N2O (300 K)
p
(iii) N2 (300 K)
(ii) N2O (50 K)
Or
(iii) N2O and N2 both have simple molecular structure. N2 has weaker temporary dipole-dipole interaction as compared to the permanent dipole-dipole interaction in N2O Or N2O has more electrons/ higher Mr than N2 hence experiences more extensive Van der Waals’ forces of attractions. N2moleculeswill tend towards ideal gas behaviour.
Total 9 marks
pV/RT
1.0
(i)N2O (300 K)
p
(iii) N2 (300 K)
(ii) N2O (50 K)
Comment [ECH(C20]: Many students
compare N2 at 300 K with N2O at 50 K. They
should compare with N2O at 300 K since
both gases are at the same temperature!
N2O cannot formed hydrogen bonds. Do
not compare the size of N2 and N2O in
relation to the size of the container as the
focus of the questions is about changes in
temperature and not pressure.
5 (a) Formic acid, also known as methanoic acid, is the simplest carboxylic acid. It occurs naturally in the venom of bee and ant stings. Formic acid dissociates in water as shown:
HCOOH (aq) HCOO- (aq) + H+ (aq)
In an experiment, 25.0 cm3 of 0.1 mol dm-3 formic acid was titrated against 0.05 mol dm-
3 of aqueous sodium hydroxide. [Ka of formic acid = 1.8 x 10-4 mol dm-3]
(i) At a certain point in the titration, a solution of maximum buffering capacity can be
obtained. Calculate the pH of this point.
At half-equivalence point,
pH = pKa
pH = -log10(1.8 x 10-4
)= 3.74
(ii) Calculate the concentration of the formic acid when 15.00 cm3 of aqueous sodium
hydroxide has been added. Hence, calculate the pH of the mixture. [4]
HCOOH (aq) HCOO–(aq) + H
+ (aq)
HCOO–Na
+(aq) →HCOO
–(aq) + Na
+(aq)
Amount of HCOOH before mixing = ������ × 0.1 = 2.5 x 10
-3 mol
Amount of NaOH added = ������ × 0.05 = 7.5 x 10
-4 mol
HCOOH (aq) + NaOH (aq) → HCOO-Na
+ (aq) + H2O (l)
Amount of HCOOH reacted = 7.5 x 10-4
mol
Amount of HCOO-Na
+ formed = 7.5 x 10
-4 mol
New volume after mixing = 25 + 15 = 40 cm3
New [HCOOH] = -�.�×��0�1–-�.�×��0'1
�.�� = 0.0438mol dm-3
New [HCOO-Na
+] =�.���0'�.�� = 0.01875 mol dm
-3
pH = pKa + log10
[34550][34553]= –log10(1.8 × 10
-4) + log10��.��6���.������ = 3.38
Comment [TEY21]: This part is well
done. Most students are able to recognise
that at maximum buffering capacity, pH =
pKa. However, there are some students
who think that at this point, the solution is
that of a weak acid.
Comment [TEY22]: Many students do
notknow that the mixture is a buffer. The
weak acid is in the conical flask and as it
has only been partially neutralised, some
weak acid remains and some salt would
have been formed.
A number of students mistook the mixture
to be a salt solution or a solution of weak
acid.
Comment [TEY23]: Some students did
not realise that some of the formic acid has
reacted with the sodium hydroxide added
and used the original amount of acid /
amount of acid reacted to find its
concentration.
Comment [TEY24]: Note that since the
question asked for boththe concentration
of the weak acid and the pH of the mixture,
both answers have to be calculated and
left to 3 sig. fig.
Comment [TEY25]: Note the formula!
Some students remembered the buffer
formula wrongly.
(b) A saturated solution of iron(II) hydroxide has a pH of 8.76 at 25°C.
(i) Write an expression for the solubility product of iron(II) hydroxide.
Fe(OH)2(s) Fe2+
(aq) + 2OH- (aq)
Ksp = [Fe2+
][OH-]
2
(ii) Calculate the value of the Ksp of iron(II) hydroxide, stating its units.
pOH = 14 – 8.76 = 5.24
pOH = -lg [OH-]
[OH-] = 10
-5.24 =5.754 x 10
-6 mol dm
-3
Let the solubility of Fe(OH)2 be x mol dm–3
.
Fe(OH)2(s) Fe2+
(aq) + 2OH- (aq)
∆ in [ ] : -x +x +2x
Ksp = [Fe2+
][OH-]
2
=(5.754 x 10-6
÷2)(5.754 x 10-6
)2
= 9.53 x10-17
mol3 dm
-9
(iii) Explain, with the aid of an equation, the decrease in solubility of iron(II)hydroxide in the
presence of 0.10 mol dm-3
of aqueous iron(II) sulphate. [5]
Fe(OH)2 (s) Fe2+
(aq) + 2OH– (aq) ----- (1)
FeSO4 (aq) →Fe2+
(aq)+SO42–
(aq) ----- (2)
Due to thecommon ion effect, by Le Chatelier’s Principle, the position of equilibrium in (1)
will shift left to reduce [Fe2+
].
Hence, solubility of Fe(OH)2 will be reduced.
Total 10 marks
TOTAL for SECTION A: 40 Marks
Comment [TEY26]: This is a very easy
question. However, some students do not
know that solubility product refers to Ksp
and wrote the dissociation equation
instead! Some students included the solid
in the expression!
Comment [TEY27]: Most students
know that they need to find [OH-] from the
pH given. However, many students do not
know that they need to divide by 2 in order
to find the [Fe2+
] at saturation.
Comment [TEY28]: Some students
forgot to write down the units which can
easily earn them one mark!
Comment [TEY29]: This question does
not require a quantitative explanation.
Students who attempted to give a
quantitative explanation failed to fully
understand what is happening and to
correctly prove that the solubility of
Fe(OH)2 is decreased in the presence of
FeSO4.
Comment [TEY30]: Many students did
not get the mark. This was due to the
following reasons: (i) no/wrong state
symbols (ii) wrong type of arrow (iii) wrong
/no equation.
Comment [TEY31]: For the
explanation, it is important for students to
include that the reason the position of
equilibrium will shift left is to decrease
[Fe2+
]. Many answers omitted this point
and was hence, not awarded the mark. A
number of students also missed out the
word ‘position’ despite numerous
reminders.
Some students used Ksp to explain. Students
need to take note that Ksp will not change
as temperature did not change!
For this examination, the explanation mark
was awarded as long as the equation is
written in parts (i) – (iii). However, it should
be noted that in the future, the correct
equation should be written together with
the explanation before the explanation can
be awarded any marks.
Mark Scheme for SECTION B
6 (a) Brass is a mixture of copper and zinc. It dissolves in nitric acid to give a mixture of Cu2+
(aq) and
Zn2+
(aq) ions.
3Cu (s) + 2NO3- (aq) + 8H
+ (aq) → 3Cu
2+ (aq) + 2NO (g) + 4H2O (l)
The copper ions, Cu2+
, may be analysed by means of iodide and sodium thiosulfate. The zinc
ions do not react during this analysis.
1.00 g of brass was dissolved in nitric acid and after boiling off oxides of nitrogen and
neutralisation, excess potassium iodide, KI, was added to the Cu2+
(aq) ions and white
precipitate of copper(I) iodide was formed in iodine solution, I2 (aq).
The iodine formed then reacted with 0.0100 mol of sodium thiosulfate.
I2 (aq) + 2S2O32-
(aq) → 2I- (aq) + S4O6
2- (aq)
Construct an equation between the copper(II) ions and iodide. Hence, calculate the percentage
by mass of copper in the brass.
[2]
2Cu2+
(aq) + 4I- (aq) →2CuI (s) + I2 (aq)
Amt of I2 produced = 0.0100 ÷ 2 = 0.00500 mol
Amt of Cu2+
= 0.005 x 2 = 0.0100 mol
Amt of Cu = 0.0100 mol
Mass of Cu in sample = 0.0100 x 63.5 = 0.635 g
% by mass of Cu in brass = (0.635 ÷ 1) x 100% = 63.5%
(b) The sulfur dioxide and carbon dioxide mixture was subsequently passed through an industrial
scrubber to separate the two gases. The sulfur dioxide gas obtained was then passed through
excess oxygen, using vanadium(V) oxide as the catalyst at a temperature of 500°C in a reactor
of 2 dm3. This is the key stage in the Contact process to produce sulfuric acid.
2SO2 (g) + O2 (g) 2SO3(g) ∆H = −197 kJ mol-1
(i) Explain the considerations which lead to the temperature of 500oC being used.
At low temperature, according to Le Chatelier’s Principle, the position of equilibrium
shifts to the right to release heat, favouring the exothermic reaction. Hence, a high yield
of SO3 is obtained.
However, too low a temperature will cause the reaction to be too slow, which makes the
process uneconomical.
Thus a moderate temperature of 500oC is adopted.
Comment [GXYS32]: Most students
were not able to provide the correct
equation (i.e. to form Cu+ (aq), instead of
CuI (s)). However, it should not be the case
since a similar question is found in Redox
tutorial Q5.
Some students also placed Cu2+
and I- as
the products formed. This shows poor
understanding of the question.
Comment [GXYS33]: No e.c.f. was
given as long as overall redox equation was
not justifiable.
Comment [ec34]: A lot of students did
not take into consideration the rate of
reaction. They only wrote down
equilibrium considerations.
(ii) Write an expression for Kc for this reaction.
Kc = [75�]
[75 ] [5 ]
(iii) Assuming a 95% conversion of SO2 (g) into SO3 (g) was achieved, use your expression in
(b)(i) to calculate the value for Kc when 4 mol of SO2 and 3 mol of O2 were allowed to
reach equilibrium at 500°C.
2SO2 (g) + O2 (g) 2SO3 (g)
Initial amount
/mol 4 3 0
Change in
amount/mol -3.8
–(3.8/2)
= -1.9 +3.8
Equilibrium
amount / mol 0.2 1.1 3.8
Kc = ��.) �
8#. 9 �. �
= 656.4 ≈656 mol-1
dm3
(iv) Some of the SO3 (g) formed was immediately removed from the reactor once equilibrium
was established. Calculate the newequilibrium amount of SO3 (g) at 500°C if the amount
of O2 (g) at the new equilibrium was 1.01 mol.
2SO2 (g) + O2 (g) 2SO3 (g)
Initial amount
/mol 0.2 1.1
3.8-y
Change in
amount/mol
-(2 x 0.09)
= -0.18 -0.09 +0.18
Equilibrium
amount / mol 0.02 1.01 3.98-y
Since temperature remains constant, Kc = 656.4 mol-1
dm3.
−
=
2
2
3.98
2K
0.02 1.01
2 2
c
y
( ) ( )
−
=
2
2
3.98
2656.4
0.01 0.505
y
0.3641 = 3.98 – y
y = 3.6159
SO3 at equilibrium = 3.98 – 3.6159 = 0.364 mol
Comment [ec35]: Some students
cannot differentiate between initial and
final amount. Therefore, they are unable to
draw the correct ICE table.
A lot of students did not substitute
concentrations in Kc expression. Instead
they substitute amount (mol). DO
REMEMBER to divide amount by volume
before calculating Kc.
Comment [ec36]: A lot of students
cannot draw the correct ICE table. They did
not minus “y” from the initial amount.
(v) State and explain the impact on the equilibrium yield of SO3 (g) if more vanadium(V)
oxide was added to the reacting system.
No change on the equilibrium yield of SO3(g).
Catalyst merely speeds up the rate at which equilibrium is achieved.
(vi) State and explain the impact on the equilibrium yield of SO3 (g) if the process was carried
out at the temperature of 800oC.
[11]
The equilibrium yield of SO3(g) will decrease.
When temperature is increased, the endothermic reaction is favoured so as to absorb the
excess heat and position of equilibrium shifts left in order reduce the temperature.
(c) Sulfuric acid, H2SO4,can behave as an acid, as an oxidising agent and as a dehydrating agent and
is a central substance in the chemical industry. Its principal uses include lead-acid batteries for
cars and other vehicles, ore processing, fertiliser manufacturing, oil refining, wastewater
processing, and chemical synthesis.
(i) 50 cm3 of 1 mol dm
-3 of ethanoic acid, CH3COOH (aq), was mixed together with 25 cm
3
of 1 mol dm-3
of sodium hydroxide, NaOH (aq).
A small amount of sulfuric acid was then added to the mixture. With an aid of an
equation, comment if there is any pH change.
With 50 cm3 of 1 mol dm
-3 of ethanoic acid, CH3COOH (aq) and 25 cm
3 of 1 mol dm
-3 of
sodium hydroxide, NaOH (aq) � Presence of Buffer Solution.
The small amount of H2SO4 will be removed by the buffer system
H+ + CH3COO
-���� CH3COOH
Thus, pH will remain fairly constant.
(ii) The boiling point of pure sulfuric acid, at 270oC, is higher than that of SCl2. Explain, in
terms of structure and bonding, why the boiling point of sulfuric acid is higher than that
of SCl2. [5]
Both have simple molecular structure.
H2SO4 has intermolecular hydrogen bonds and SCl2 has intermolecular van der
Waals’forces of attractions.
More energyis required to overcome the stronger hydrogen bonds in H2SO4 than the
weaker van der Waals’ forces of attraction in SCl2.
Hence H2SO4 has a higher boiling point.
(d) Give the symbols (showing the proton number, nucleon number and charges) of the
following two particles
particle protons neutrons electrons
Q 17 17 20
R 18 17 17
[2]
−34 3
17Q
+35
18R
Total 20 marks
Comment [ec37]: Some students has
the misconception that when rate increase,
yield must increase which is wrong. Rate
change but equilibrium yield is the same.
Comment [GXYS38]: Students were
generally able to identify SCl2 as a simple
molecule. However, many students
identified H2SO4 as a giant ionic compound
due to the conception of H+ and SO4
2- ions
present during dissociation. Students
should be able to identify that H2SO4 is in
fact a simple molecule since it consists of
non-metal atoms only.
Comment [GXYS39]: Most students
were not able to identify “intermolecular
H-bonds” present in H2SO4. Most
mentioned about the greater
extensiveness of VDW forces due to
polarity, or greater number of electrons.
Comment [GXYS40]: Some students
identified SCl2 to consist of temporary
dipole-dipole interactions, which is not
true. This is because they failed to identify
that SCl2 is a polar molecule.
Comment [GXYS41]: Phrasing for
intermolecular forces needs to be
improved. Many students failed to include
keywords: “intermolecular”, or “between
molecules”. Some students also had
misconceptions about IMF – “between
atoms” was mentioned.
Comment [GXYS42]: Most students
were able to identify the accurate nucleon
and proton number, and charges. However,
some failed to read the question carefully
to express the answer in symbols.
7 Tea light candles are a popular form of decoration during the Christmas season and are
commonly made from alkanes. A student performed an experiment using tea light candles
purchased from IKEA and The Body Shop.
The calorimeter used in the experiment involving the IKEA candle was calibrated using the
following formula:
Q = C ΔT
where Q is the heat energy transferred, C is the heat capacity of the calorimeter and ΔT is the
temperature change.
The student used the calibrated calorimeter to determine the standard enthalpy change of
combustion,ΔHcθ,of The Body Shop candle.
The information below shows the results of the experiment.
Calibration of calorimeter using IKEA candle:
ΔHcθ -12 x 10
6 J mol
-1
Formula C20H42
Initial mass of candle 3.000 g
Final mass of candle 2.500 g
Temperature change 5 oC
Determination of ΔHcθ of The Body Shop candle:
ΔHcθ ?
Formula C22H46
Initial mass of candle 3.000 g
Final mass of candle 2.400 g
Temperature change 6 oC
After the experiment, the student realised that he had forgotten to record the standard enthalpy
change of combustion of The Body Shop candle.
(a) Define standard enthalpy change of combustion.
[1]
The standard enthalpy change of combustion is the heat evolved when 1 mol of the
substance is completely combusted under standard conditions.
Comment [YWC43]: Students wrote
“heat” required which is wrong.
Comment [YWC44]: Students left out
the “standard condition” in their answer.
(b) Using your practical knowledge, illustrate,with a clearly labelled diagram, the set-up for this
experiment. You should include in your diagram the apparatus mentioned in this experiment
and those commonly found in a school laboratory.
[2]
(c) (i) Using the instructions and the data collected from the experiment with the IKEA candle,
calculate the heat capacity of the calorimeter. Leave your answer to 3 significant
figures.
ΔHc
θ =
Q
n−
-12 x 106 = −
−3 2.5
282
Q
Q = 2.1277 x 104 J
Using Q = C ΔT
2.1277 x 104= C (5)
C = 4255.3 J K-1
= 4260 J K-1
Tea Light Candle
Calorimeter
with water
Thermometer
Comment [YWC45]: Students did not
mentioned or draw the water in the
calorimeter.
Students used Styrofoam cup to contain
the water, not allowed as it will burn.
Thermometer is not included in the
diagram.
Comment [YWC46]: Students got this
formula wrong.
Comment [YWC47]: Students did not
subtract the mass.
Comment [YWC48]: Students added
273 to the temperature change factor.
Comment [YWC49]: Unit is wrong.
(ii) Hence, calculate the standard enthalpy change of combustion of The Body Shop candle.
State one assumption you made in your calculations in c(i) and c(ii).
[5]
Q =C ΔT= 4255.32 x 6 = 25531.91 J
ΔHcθ =
25531.91
0.60
12(22) 46
−
+
= -13.2 x 106 J mol
-1
Assumption: 100% efficiency / no heat lost to surrounding.
(d) Iron(III) oxide, Fe2O3, is produced from the oxidation of iron metal. It is one of the three main
oxides of iron, the other two being FeO, which is rare and Fe3O4 which occurs naturally as the
mineral magnetite.
Standard enthalpy change of atomisation of iron = +414 kJ mol-1
First and second electron affinity of oxygen = +650 kJ mol-1
Standard enthalpy change of formation of iron(III) oxide = -823 kJ mol-1
(i) With reference to the Data Booklet and given information, construct and label a Born-
Habercycle for the formation of iron(III) oxide from its elements.
2Fe (s) + 3/2 O2 (g)
Fe2O3 (s)
2Fe (g) + 3/2 O2 (g)
2Fe (g) + 3 O (g)
2Fe+ (g) + 3 O (g) +2e
2Fe2+ (g) + 3 O (g) +4e
2Fe3+ (g) + 3 O (g) + 6e
0
2Fe3+ (g) + 3 O2- (g)
∆Hlatt
∆Hf
2x∆Hat (Fe)
2x 1st IE Fe
2x 2nd IE Fe
2x 3rd IE Fe
3x 1st& 2nd
EA O
3/2 x BE O
Energy
Comment [YWC50]:
Energy and zero label is not present
Electrons not balanced.
Equations not balanced.
State symbols not included.
If 1st
EA and 2nd
EA is used, students
represented it wrongly on the Born Haber
cycle.
3rd
IE of Iron not mentioned in the Born
Haber cycle.
(ii) Hence, use the cycle in d(i) to calculate the lattice energy of Fe2O3(s).
[3]
By Hess Law,
[2x ΔHat (Fe)] + [3/2 BE of O] + [2 x 1st
IE Fe] + [2 x 2nd
IE Fe]+ [2 x 3rd
IE Fe] + [3 x 1st
&2nd
BE of O] + [ΔHlatt (Fe2O3)] = ΔHf (Fe2O3)
(2x414) + (3/2 x496) + (2x762) + (2x1560) + (2x2960) + (3x650) + ΔHlatt (Fe2O3)
= - 823
ΔHlattθ(Fe2O3) = -14909 kJ mol
-1 = - 1.48 x 10
7 J mol
-1
(e) The following mechanism illustrates a reaction between reactants A and B.
Step 1: A + B ⇌ C fast
Step 2: C + B → D + F slow
Prove that the overall order of this reaction is three. [2]
From Reaction 1:
Kc = [4][:][;]
[C] = Kc[A][B] ----- (1)
From reaction 2:
Rate = k[C][B] ----- (2)
Since C is an intermediate, it should not appear in the rate law.
Substituting (1) into (2):
Rate = k{Kc[A][B]} [B]
Rate = k’[A][B]
2
Therefore, the overall order of reaction is three.
Comment [S51]:
Some serious misconceptions shown by
majority of the students resulting in zero
mark awarded:
1) deduce overall order from the overall
equation instead of the rate determining
step (slowest step);
2)express (totally meaningless) rate
equation in term of products, D/F which
cannot be controlled.
(f) With the aid of an energy profile diagram,explain the following statement:
“Diamond is energetically unstable but kinetically stable.”[2]
The process of converting diamond to graphite is an exothermic reaction/diamond has
higher energy level, hence diamond is energetically unstable, but the process required
high activation energy resulting in kinetic stability.
Diamond
Graphite
Energy
Reaction Pathway
ΔH = -ve
HighEa
Comment [S52]: This question deals with two concepts of
energetics and kinetics. 1 mark is only
awarded for correct explanation per
concept if clearly supported by relevant
labels on the energy profile diagram.
No marks were given at all for any
contradiction in concepts shown and when
diamond was not indicated in the diagram
and explanation.
1 mark is also deducted for wrong labels of
energy profile diagram: common mistake is
to label x-axis as time/s.
(g) Iodination of propanone is done in aqueous acidic solution according to the equation:
CH3COCH3 (aq) + I2 (aq) → CH3COCH2I (aq) + H+ (aq) + I
- (aq)
The rate of reaction was studied via a colorimetric method, in which the colour intensity of
iodine was measured at regular time intervals.
Three separate experiments were performed, in which the initial concentrations of iodine,
propanone and acid were varied in turn, the other two being kept constant. The results are
shown below in graphical form:
Graph 1
Graph 2
Time/min
0.002
0.004
[I2]/mol dm-3
1
[CH3COCH3]
= 0.8 mol dm-3
[CH3COCH3] = 0.4 mol dm-3
[CH3COCH3]
= 1.2 mol dm-3
2 3 4 5 6
Time/min
0.002
0.004
[I2]/mol dm-3
1
[H+]
= 0.8 mol dm-3
[H+]
= 0.4 mol dm-3
[H+] = 1.2 mol dm
-3
2 3 4 5 6
(i) Using the graphs above, deduce the orders of reaction with respect to propanone,
iodine and acid, respectively. Show all working clearly.
Order of reaction with respect to:
appropriate working
Propanone = 1
Acid = 1
Iodine = 0
(ii) Hence, write down the rate equation for this reaction.
[5]
Rate = k[H+][CH3COCH3]
Total 20 marks
END OF CORRECTIONS
Comment [S53]:
It is unfortunate that many students do not
have the time to attempt this standard
question.
Among the answers given, few students
could explain why it is zero order wrt
iodine by simply stating that a straight line
for concentration-time graph indicates
constant rate which is independent of [I2].
Missing/wrong units for rate which is
moldm-3
min-1
.
Reminder that the correct concluding
statement is zero or first order with respect
to a particular reactant, NOT [reactant].