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2020-DSE-CHEM|MOCK EXAM|Paper 2

MOCK EXAMINATION
CHEMISTRY
PAPER 2
Section A – Industrial Chemistry
1.
(a)
(i)
Nitric acid is manufactured by the catalytic oxidation of ammonia in the Ostwald process.
Pt
4NO(g) + 6H2O(g) H = –970 kJ mol–1
Stage 1: 4NH3(g) + 5O2(g)
900C, 8 atm
Stage 2: 2NO(g) + O2(g)  2NO2(g) H = –117 kJ mol–1
Stage 3:
(ii)
1
4NO2(g) + O2(g) + 2H2O(l)  4HNO3(aq)
 flowing mercury cell:
 membrane cell:
sodium amalgam, Na/Hg(l)
1
hydrogen, H2(g) / sodium hydroxide solution, NaOH(aq)
H+(aq) ions from water are preferentially discharged (reduced) in the cathode compartment,
forming hydrogen gas. OH–(aq) ions accumulate in the cathode compartment. The ionpermeable membrane in the membrane cell allows the passage of Na+(aq) ions but not that of
Cl–(aq) or OH–(aq) ions. The used brine is simply removed from the anode compartment. The
resultant solution in the cathode compartment is sodium hydroxide solution containing
no sodium chloride impurity. It is composed of about 30–35% sodium hydroxide by mass
(iii)
The by-product is water, which is harmless. [Green Chemistry  Principle #3]
1
The cyclohexene used in the new synthetic pathway is less hazardous than the benzene
1
(toxic/carcinogenic) used in the traditional synthetic pathway.
(Accept other reasonable answers)
(iv)
HI CH COOH
CH3OH + CO 
3
1
Note: Monsanto process uses rhodium (Rh) as catalyst while CATIVA process uses Iridium (Ir),
in addition to HI.
[5]
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1.
(b)
(i)
Set B (250C, 50100 atm, Cu/ZnO/Al2O3) should be adopted.
As the forward reaction is exothermic, a lower temperature shifts the equilibrium position to
1
the right (so that the yield of methanol is higher). A lower temperature is favoured.
Since there is greater number of moles of gas at the left-hand side of the equation, a high pressure
1
shifts the equilibrium position to the right. A higher pressure is favoured.
(ii)
Number of particles
Correct drawing + Correct labelling of axes
Correct labelling of the activation energy of
the catalysed conversion (Ea’) and that of the
uncatalysed conversion (Ea)
Ea’
Ea
1+1
Kinetic energy
With reference to the diagram, in the catalysed reaction, the number of particles having energy
equal to or greater than the activation energy is larger than that in the uncatalysed reaction.
1
Hence, there is an increased number of effective collisions per unit time, increasing the rate of
reaction.
1.
(b)
(iii)
The statement is incorrect. A catalyst does not lower the activation energy of the original
1
reaction pathway. A catalyst works by providing an alternative reaction pathway with lower
activation energy for the reaction to proceed. Can you draw an energy profile to illustrate the
role of a catalyst for a reaction?
(iv)
The use of catalyst can reduce the use of energy (allowing the reaction to take place at a lower
1
temperature) / the catalysed reaction is more efficient as more products can be obtained in a
relatively shorter period of time. [Green Chemistry  Principle #6 & #9]
[7]
Further question to think:
1.
(b)
(v)
In recent years, direct oxidation of methane to methanol has been the focus of many researches.
However, methanol is often further oxidized to methanal in the process. Explain, in view of green
chemistry, why it is worthwhile to develop this new technique.
(1 mark)

In the single-step direct oxidation of methane, no derivatives or intermediates form.
This helps minimize the use of energy and reagents. [Green Chemistry  Principle #8]
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1
 Mark Scheme


2020-DSE-CHEM|MOCK EXAM|Paper 2
Green chemistry is the utilization of a set of principles that reduce or eliminate the use or generation of hazardous
substances in the design, development and application of chemical products and processes. Green chemistry can be
regarded as a reduction process that aims at reducing the cost, waste, materials, energy, risk and hazard.


Answer to Q.1 (a) (ii):
Further information to the supplementary question of Q.1 (b) (iii):
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1.
(c) (i)
From experiments 1 and 2, at the same initial [NO2(g)], the initial rate tripled when [F2(g)] was tripled.
1
Hence, the order of reaction with respect to F2(g) is 1.
From experiments 2 and 3, at the same initial [F2(g)], the initial rate increased by a factor of 5 when
1
[NO2(g)] was increased by a factor of 5. Hence, the order of reaction with respect to NO2(g) is 1.
∴
(ii)
(iii)
rate = k[NO2(g)][F2(g)]
4.0  10 5 mol dm 3 s 1
= 40.0 mol1 dm3 s1
k=
0.0010 mol dm 3  0.0010 mol dm 3
1
The initial concentrations of reactants are known.
(iv)
Arrhenius equation: log
k = constant 
k1 = 40.0, T1 = (25 + 273) K = 298 K
log k1  log k2 =
1
Ea
2.3RT
and
k2 = 80.0, T2 = (50 + 273) K = 323 K
Ea
Ea 1 1
1
1
(  )  log 40.0  log 80.0 =
(

)
2.3  8.31 323 298
2.3R T2 T1
Solving for Ea = 22.2 kJ mol1
1
1
(v)
1+1
Potential energy
Transition state (activated complex formed)
Correct drawing + Correct labelling of axes
Ea
2NO2(g) + O3(g)
Correct labelling of enthalpy change
ΔH = 198 kJ mol1
N2O5(g) + O2(g)
Reaction coordinate
[8]
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 Mark Scheme

2020-DSE-CHEM|MOCK EXAM|Paper 2
Section C – Analytical Chemistry
3.
(a)
(i)
Chlorophyll b
1
 Chlorophyll b is most polar so that it does not tend to dissolve in non-polar hexane mobile phase
1
and moves with the mobile phase the least.
Note:
In paper chromatography, the cellulose fibres in paper hold a thin film of water, which forms
the stationary phase. Components that are more soluble in the developing solvent move
up the paper more quickly. As a result, different components are separated.

Paper Chromatography and Thin-Layer Chromatography (chemistry edb)
https://www.youtube.com/watch?v=KLp7J51rGm4
(ii)
1
[3]
Note:
3.
(b)
(i)
anhydrous calcium chloride / magnesium sulphate / sodium sulphate
1
ALLOW silica gel
(ii)
distillation / fractional distillation
1
ALLOW evaporate under reduced pressure / rotary evaporation
 Basic Laboratory Techniques http://www.cuhk.edu.hk/chem/en/resources/basic.html
(iii)
Recrystallization
1
ALLOW column chromatography NOT distillation
 Chromatographic Techniques http://www.cuhk.edu.hk/chem/en/resources/chromatographic.html
 How to extract Caffeine from Tea (Classic DCM Method)
https://www.youtube.com/watch?v=RIbff5iD0GQ
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[3]
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3.
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
ClO3-(aq) + 6Fe2+(aq) + 6H+(aq)  Cl-(aq) + 3H2O(l) + 6Fe3+(aq)
(c)
… (1)
(in excess)
5Fe (aq) + MnO4-(aq) + 8H+(aq)  5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
2+
… (2)
(unreacted in (1))
No. of moles of MnO4-(aq) used =
14.93
1000
dm3 0.1672M
2.496  10-3 mol
No. of moles of excess Fe2+(aq) = 2.49610-3 mol  5 = 0.01248mol
Original no. of moles of Fe2+(aq) =
50.00
1000
1
dm3 0.8930M = 0.04465mol
No. of moles of Fe2+(aq) reacting with ClO3-(aq) = 0.04465  0.01248 = 0.03217 mol
1
No. of moles of ClO3-(aq) = 0.03217mol = 5.36210-3 mol
Mass of KClO3 = 5.362  10-3 mol  (39.1+35.5+16.03) g mol-1 = 0.6574 g
% by mass of KClO3 =
0.6574 g
1.279 g
100%
1
51.40%
[3]
3.
(d)
(i)
The absorbance/value was out with the calibration range / line.
OR
1
To bring the absorbance/value to within the calibration range / line
[Must make reference to the calibration range or line]
(ii)
From the calibration curve, concentration of Cu2+(aq) ion the diluted solution = 0.032 M
Concentration of Cu2+(aq) ion in the undiluted solution with absorbance of 0.71
= 0.032 M  2 = 0.064 M
1
2+
3
No. of moles of Cu (aq) in 20 cm of conc. HNO3
= 0.064M
250
1000
dm3
0.016 mol
1
Mass of Cu (aq) = 0.016 mol  63.5 g mol = 1.016 g
2+
% by mass of Cu in brass =
(iii)
-1
1.016 g
1.43 g
100% = 71.05% OR 71.0%
The presence of zinc ion will not affect the result ⸪ Zinc ions are colourless
1
1
[5]
3.
(e)
(i)
A strong absorption from 1680 to 1800 cm-1 indicates the presence of C=O.
1
-1
A strong, broad absorption from 2500 to 3300 cm indicates the presence of OH of carboxylic acid.
(ii)
 Carboxyl group / COOH is present.
1
Add Na2CO3 solution / NaHCO3 solution to compound X. (∵ water should be present for showing acidic properties of the acid)
1
Evolution of colourless gas bubbles confirms the presence of COOH group.
(iii)
/
m/z = 43 indicates the presence of
m/z = 88 indicates the presence of
1
/
1
 Compound X is
/
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 Mark Scheme
3.
(e)
2020-DSE-CHEM|MOCK EXAM|Paper 2

(iv)
Measure the boiling point of compound X and check with boiling point of the structure deduced in
1
(iii) from the data book.
If the boiling point of compound X is the same as that of the structure deduced in (iii), compound X
has the same structure as that deduced in (iii).
OR
Carry out thin-layer chromatography of the compound X with an authentic sample with the structure
deduced in (iii).
If compound X has the same Rf value as that of the structure deduced in (iii), then compound X has
the same structure as that deduced in (iii).
(Accept other possible answers.)
[6]
Physical Properties of Carbonyl Compounds (for reference only)
Identification of Carbonyl Compounds (for reference only)

The derivatives (condensation products) are useful in the characterisation and identification of carbonyl
compounds because they are mostly crystalline solids and have characteristic sharp melting points.

2,4-Dinitrophenylhydrazine (sometimes abbreviated as 2,4-DNP) is most commonly used for the
characterization and identification of aldehydes and ketones. 2,4-Dinitrophenylhydrazine reacts with all
aldehydes and ketones.

Phenylhydrazine forms derivatives (phenylhydrozone) readily with aromatic aldehydes, but in general 2,4dinitrophenylhydrazine is preferred because its derivatives (2,4-dinitrophenylhydrazones) have higher
melting points and are less soluble.

A solution of 2,4-dinitrophenylhydrazine in methanol and dilute sulphuric acid is called Brady’s reagent.
When it is added to a solution of a carbonyl compound, yellow to orange precipitate of 2,4dinitrophenylhydrazone forms rapidly; the crystals (purified by recrystallisation) has a well-defined melting
point and can be identified by reference to tables of melting points of 2,4-dinitrophenylhydrine derivatives.
Equation:
O 2N
H3C
C
H3C
O
propanone
+
H2NNH
NO2
- H2O
H3C
O2N
C N NH
NO2
orange
H3C
propanone 2,4-dinitrophenylhydrazone
2,4-dinitrophenylhydrazine
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
The reaction is quantitative, and the precipitate can be dried and weighed to reveal the amount of carbonyl compound present.

The derivative may be hydrolysed back to the carbonyl compound by refluxing with dilute mineral acid.

Even for closely similar carbonyl compounds (hence they have similar m.p.), melting points of their derivatives are usually
sufficiently different to enable them to be identified. For example,
Aldehyde or ketone
Derivative
O2N
H
CH3CH2CH2CH2CHO
pentanal 2,4-dinitrophenylhydrazone
C N NH
NO2
m.p. 98°C
CH3CH2CH2CH2
Pentanal , b.p. = 104°C
O2N
CH3
CH3CH2CH2COCH3
pentan-2-one 2,4-dinitrophenylhydrazone
C N NH
NO2
m.p. 141°C
CH3CH2CH2
Pentan-2-one, b.p. = 102°C
Flow scheme to show the identification steps
Unknown compound X
Adding 2,4-dinitrophenylhydrazine solution (Brady’s reagent)
Orange precipitate obtained
Compound X is a carbonyl compound.
m.p. test for the derivative of X, 2,4-dinitrophenylhydrazone after purification
m.p. of pure crystals of the orange precipitate = 126C
Check literature data:
From data book, propanone 2,4-dinitrophenylhydrazone has a m.p. of 126C.
The unknown compound X is propanone!
Physical properties of carboxylic acids (RCOOH)
NAME
FORMULA
m.p. / °C
b.p. / °C
SOLUBILITY
methanoic acid
HCOOH
8.4
101
vs
ethanoic acid
CH3COOH
16.6
118
vs
propanoic acid
CH3CH2COOH
-20.8
141
vs
butanoic acid
CH3(CH2)2COOH
-6.5
164
vs
pentanoic acid
CH3(CH2)3COOH
-34.5
186
s
hexanoic acid
CH3(CH2)4COOH
-1.5
205
ss
benzoic acid
C6H5COOH
122
249
i (see note)
phenylethanoic acid
C6H5CH2COOH
76
166
s

Benzoic acid is a white crystalline solid, sparingly soluble in cold water but quite soluble in hot water.

Methanoic acid is a colourless liquid with a pungent smell. It irritates and destroys skin.

Ethanoic acid is a colourless liquid with a sharp smell of vinegar and a sour taste. Pure ethanoic acid freezes to an ‘ice-like’
solid below 16.6°C and is commonly known as glacial ethanoic acid.

The odours of the lower aliphatic acids progress from the sharp, irritating odors of methanoic and ethanoic acids to the
distinctly unpleasant odours (smell of rancid butter) of butanoic, pentanoic and hexanoic acids; the higher acids have little
odour because of their low volatility.
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