Download 2008-2009 answers

Baptist Lui Ming Choi Secondary School
Form 7
Mock Examination (2008-2009)
Chemistry (Paper I)
Suggested Solutions
Date: 23-02-2009
SECTION A (60 marks)
1.
a. Octahedral
[1]
b.
complex I
complex II and complex III
[1]
trans-isomer
c.
Complex I:
OR
Complex II:
OR
[1+1]
cis-isomer and its mirror image
trans-dichlorobis(ethane-1,2-diamine)chromium(III) ion
trans-bis(ethane-1,2-diamine)dichlorochromium(III) ion
cis-dichlorobis(ethane-1,2-diamine)chromium(III) ion
cis- bis(ethane-1,2-diamine)dichlorochromium(III) ion
[1+1]
d.
e.
2
a.
b.
c.
i. They are geometrical isomers of each other (or geometric isomerism) [1]
ii. They are optical isomers (or enantiomers) of each other (or optical isomerism) [1]
Complex II and complex III [1]. Since they are asymmetric (or The two polar Cr-Cl bonds do
not cancel out each other). [1]
Two amino acids joined together by a peptide/amide/ –CONH– link [1]
IUPAC name of aspartic acid: Aminobutanedioic acid [1]
Structural formula aspartic acid HOOCCH(NH2)CH2COOH [1]
i. Heat under reflux [1/2 mark]
with HCl (or NaOH) [1/2 mark]
ii. Methanol [1 mark]
iii. Paper chromatography/ thin layer chromatography/ electrophoresis/ Mass
spectrometry/… (any one reasonable answer) [1]
d.
3D structural formulae of two optical isomers of phenylalanine. [1+1]
e.
f.
Dipolar ion of phenylalanine,
[1]
Hydrolysis of aspartame takes place in acidic medium. [1]
P.T.O.
Form 7
3
a.
b.
c.
d.
e.
f.
Chemistry (08-09)
P.2
High resistance voltmeter (or potentiometer) [1/2]
Salt bridge [1/2]
Standard hydrogen electrode (or other reference cell; e.g. Cu2+/ Cu) [1/2]
Vanadium ions half-cell (or general ion/ ion half-cell) [1/2]
Labels: Any two [1/2+1/2]
i. Stable oxidation state would be expected to be +5 [1]
ii. Both vanadium half-cells are more negative than oxygen half-cell [1].
(or V(+4) to V(+5) is still sufficiently negative to supply electrons to oxygen half-cell)
Electrode potential values depend on the ligands present in a complex [1]
[Ar] 4s23d3 (or [Ar] 3d34s2) [1]
V2+(g)  V3+(g) + e–
[1]
After the loss of 2 electrons from calcium, further ionisation requires a large energy input.
Since third ionization involves the removal of electron from argon like electronic configuration
which carries extra stability. [1]
Vanadium can lose a greater number of electrons before a large energy input is required since
electrons are being removed from the 3d-orbitals. [1]
4.
a.
i.
Energy cycle: [1]
H2
8 CO(g) + 17 H2(g)
C8H18(g) + 8 H2O(g)
H3
H4
8 C(s) + 4 O2(g) + 17 H2(g)
H2 + H3 = H4
H2 + 8 x Hf(CO) = Hf(C8H18) + 8 x Hf(H2O)
ii.
b.
i.
ii.
H2 + 8 x (-111) = (-169) + 8 x (-242)
[1M]
-1
H2 = -1217 (kJ mol )// [1A]
Temperature: 300-500oC and pressure: 1-10 atm (or high temperature and low pressure)
[1]. Since the forward reaction is endothermic and no. of gaseous products are greater
than that of reactant, hence the eqm. pos. will shift to product side at high temp and low
pressure [1+1].
Labelled axes [1/2 + 1/2]
correctly plotted data and straight line [1]
Initial rate of reaction is proportional to starting concentration of H2O2 [1], so reaction is
first order with respect to hydrogen peroxide [1].
To be continued
Form 7
5
a.
b.
c.
d.
Chemistry (08-09)
P.3
Water ligands around the copper ions are replaced by new ligands (L) [1]
The chemical bonds between copper(II) ion and ligand L are dative covalent bond (or
coordinate bond) [1]. Where copper(II) ion is an acceptor of electron pairs, while ligand L is
the donor of electron pairs [1].
Kc = [CuL2(org)][H+(aq)]2/ [Cu2+(aq)][HL(org)]2
[1M]
= [(0.045)(10-2)2]/ [(0.002)(0.1)2]
(or 1M)
Kc = 0.225 [1A]
i. Awareness of Le Chatelier's Principle or reversibility of process
Cu2+(aq) + 2 HL(organic)
CuL2(organic) + 2 H+(aq) [1]
Correct discussion of effect of [H+] on position of equilibrium
e.g. Increase in [H+] shifts eqm. pos. to L.H.S. [1]
ii. The process can be reversed by then shaking the organic solution with moderately
concentrated acid. This pushes Cu2+ ions back into aqueous solution and, again, an
increase in concentration can be achieved. [1]
iii. Separating funnel [1].
iv. Good ventilation/ release pressure inside the separating funnel/ restrict the use of naked
flame/… (any reasonable answer) [1].
6.
a.
i.
C6H5NH2 + H2SO4  C6H5NH3+HSO4- (or C6H5NH4SO4) [1]
ii.
b.
Curly arrow: 0.5 each (max 2 marks)
Intermediate and product: [0.5 + 0.5]
Bonus mark for resonance structures of intermediate: 0.5 each (max 1)
[max 3]
P.T.O.
Form 7
c.
Chemistry (08-09)
P.4
Steric effect between the very bulky sulphonic acid group at the 1-position and the hydrogen
atom at the adjacent 8-position.
(or The difference in thermodynamic stability of the two isomers can be attributed largely to
the destabilising repulsion between the very bulky sulphonic acid group at the 1-position and
the hydrogen atom at the adjacent 8-position.) [1]
d.



Energy barrier (E1) for the formation of isomer 1 is lower than that of isomer 2 (E2).
Fewer molecules exceed E2 at low temperature than that exceeding E1.
Isomer 1 is formed more rapidly than isomer 2 at lower temperature.




Sulphonation is reversible.
At high temperature more molecules exceed E2 (and E1).
Both isomers are formed.
Fall in energy for isomer 2 is more negative than for isomer 1 (or isomer 2 is more stable).
Isomer 1 will convert to isomer 2.
(1 mark each, any 4 points)
Section B (20 marks)
7.
a.
c.
C6H5COOCH3 + H2O
C6H5COOH + CH3OH
(or C6H5COOCH3 + NaOH  C6H5COO-Na+ + CH3OH)
i. No. of mole of methyl benzoate used = 2.70/ 136 = 0.01985 mole//
ii. % yield of benzoic acid = [(1.5 / 122) / 0.01985] x 100% = 61.94%//
iii. Hydrolysis is not complete/ C6H5COOH slightly soluble in water/ Lost of product during
manipulation (1 each, any TWO)
i. To remove methanol from the reaction mixture.
d.
ii. Since C6H5COONa+ is water soluble/ to precipitate C6H5COOH
iii. to remove any impurities/ to purify the product.
Benzoic acid: 1680 – 1750 cm-1 C=O and 2500 – 3300 cm-1 O-H [0.5+0.5]
b.
Methanol: 3230 – 3550 cm-1 OH [0.5]
O-H absorption peak of benzoic acid is boarder than that of methanol [0.5].
To be continued
Form 7
8.
a.
b.
c.
Chemistry (08-09)
P.5
Si + O2 
[1]
2
SiO2 + CaO 
[1]
3
ii. Otherwise it would be removed as an oxide. [1]
Four marks for any four of the following points:
 Select a suitable filter
 Use a blank/ reference cell (to set zero absorbance or 100% transmittance)
 Make up solutions of known concentration
 Measure the absorptions of the solutions
i.



i.
Plot a calibration curve
Measure absorption of the green solution of unknown concentration
Read off its concentration from calibration curve
Amount of MnO4– used in the titration = 0.0200 x 24.2/ 1000 [1 M]
(= 4.84 x 10–4 moles)
Mass of iron in steel sample = 5 x 4.84 x 10–4 x 10 x 56.0 = 1.36 g [1A]
Percentage by mass of iron in steel = 1.36/ 1.40 x 100% = 96.8% [1A]
Section C (20 marks)
9.
Introduction
A definition of polymer: a very large molecule obtained from monomers.
Types pf polymers (max 4)
This essay requires that some kind of classification be given. There are many ways of doing this.
This part also is accompanied by examples of each type.
Natural polymers and synthetic polymers
Definitions and examples of natural polymers (e.g. proteins, polysaccharides and nucleic acids)
and synthetic polymers [1+1]
Addition polymers
Definition: Formed from intermolecular reactions in which monomers joined through without
elimination of small molecules [0.5]
Examples: Polythene, PVC, Polystyrene, …[0.5]
Chemical equation of Addition polymerization (a suitable example) [1]
Condensation polymers
Definition: Formed from intermolecular reactions in which monomers joined through
elimination of small molecules e.g. H2O, ROH [0.5]
Examples: nylon, PET, urea methanal, …[0.5]
Chemical equation of Condensation polymerization [1]
For example, equations of Nylon (or terylene)
P.T.O.
Form 7
Chemistry (08-09)
Mechanisms of their formation (max 4)
Addition polymers:- Free radical addition polymerization [0.5]
Mechanisms are generally chain reaction involving initiation, propagation and termination. [2]
Illustrate with polythene (or any suitable polymer) as an example [0.5]
Or
Condensation Polymer mechanisms: Nucleophilic addition followed elimination [0.5]
Mechanism [2]
Suitable example [0.5]
Effect of bonding and structures on properties of polymers (max 2)
Compare ANY ONE of the following pair of polymers
 LDPE and HDPE
 Conditions of polymerization [0.5+0.5]
 Crystalline structures in solid state [0.5+0.5]
 Different in properties [0.5+0.5]
 Nylon and Kelvar
 Crystalline structures in solid state [0.5+0.5]
 Extent of intermolecular hydrogen bond [0.5+0.5]

Thermoplastic and thermosetting plastic
Definitions [0.5 each]
Examples [0.5 each]
OR
Describe the effect of bonding and structure on properties of …[1+1]
Vulcanizied plastics [e.g. poly(2-methylbuta-1,3-diene)]
Biopolymer [e.g. poly(lactic acid)]
Photodegradable plastics.
Synthetic biodegradable plastics
P.6
10.
A.
1.
2.
3.
B.
1.
Bonding, structures and physical properties of the oxides (max 5)
The oxides of period 3 elements are classified into two types according to the nature of their
bonding.
a. Ionic oxides
b. Covalent oxides
Formulae of oxides:
Na2O, MgO, Al2O3, SiO2, P4O6 (or P4O10), SO2 (or SO3), Cl2O [0.5]
Correctly matched with …
Bonding. [0.5]
Structures. [0.5]
States (or melting points). [0.5]
Electron diagrams of an ionic oxide and a covalent oxide. [0.5 + 0.5]
[max 3]
Going across period 3 from left to right, the nature of bonding of the oxides changes from ionic
to covalent bonding. [1]
Expansion of octet for period 3 oxides due to the presence of low lying vacant 3d orbitals. [1]
Use phosphorus (or sulphur) as example. [0.5]
Chemical properties of the oxides (max 5)
The general periodic patterns in the reactions of period 3 oxides with water, dilute acids and
alkalis.
a. Ioinc oxides react with water to form hydroxides. [0.5]
Example (or equation) [0.5]
b. Ionic oxide with covalent character,
i. aluminium oxide does not react with water but it is amphoteric. [0.5]
Examples (or equations) [0.5 + 0.5]
ii. silicon(IV) oxide does not reacts with water but it reacts with hot strong alkalis. [0.5]
Example (or equation) [0.5]
c.
2.
Covalent oxides reacts with water to form acids [0.5]
Example [0.5]
Going across period 3 from left to right, the acid base nature of the oxides changes from acidic
to alkaline. [1]