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].