GCE “A” Level H2 Chemistry Nov 2010 Paper 3 Suggested Solutions
1
(a)
(i)
(ii)
Entropy of a system measures its amount of disorder, and it gives a measure
of the extent of which particles and energy are distributed within the chemical
system. The higher the amount of disorder, the entropy of the system is
higher. (Do not mix up “entropy” with “entropy change”.)
By adding 1 mol of Cl2(g) to 1 mol of N2(g), both at 298 K, the entropy of the
system will increase (ΔS > 0), as this is the process of mixing (no chemical
reaction occurs since N2 is relatively inert with high N≡N bond energy), and
this results in more ways of arranging the particles within the system.
By heating 1 mol of Cl2(g) from 298 K to 373 K, since temperature increases,
the entropy of system will increase (ΔS > 0), as there is broadening of the
Boltzmann energy distribution and this results in more ways of arranging
energy quanta in the hotter gas.
Considering the reaction of Cl2 with I2 to form ICl3, the entropy of system will
decrease (ΔS < 0), since there is decrease in the number of particles during
the reaction (4/3 moles of reactants compared to 2/3 moles of products). OR
compare in terms of decrease in amount of gaseous particles.
Considering the photolysis of Cl2, the entropy of system will increase (ΔS > 0),
since there is an increase in amount of gaseous particles during the reaction
(1 mole of reactants compared to 2 moles of products).
(b)
(i)
(ii)
Hydrogen and chlorine react with each other at lower temperatures than the
corresponding reaction with bromine (ie. reaction is more vigorous), and also
for reaction involving bromine, a catalyst has to be used to increase the rate
of the reaction, such as using platinum.
Comparison is on rate of reaction – not on extent of reaction.
Reaction of hydrogen with fluorine would be much more vigorous (and
explosive) than the corresponding reaction with chlorine since fluorine is a
stronger oxidizing agent than chlorine.
(c)
(i)
In presence of uv light.
(ii)
Free radical substitution (with Cl• as free radical)
Initiation:
Propagation:
C2H6 + Cl•  •C2H5 + HCl
•C2H5 + Cl2  C2H5Cl + Cl•
Termination:
2 •C2H5  C2H5-C2H5 / Cl• + •C2H5  C2H5Cl
(iii)
The first propagation step of the above mechanism would become
endothermic, and hence unlikely to occur. OR
Due to low bond energy of H-I, would easily dissociate and recombine with
ethyl radical formed in propagation step.
(d)
Learn to draw skeletal structures when needed – it saves time.
(e)
(i)
CFCs are relatively volatile (weak van der waals’ forces holding the largely
non-polar molecules) thus it is useful to be used as refrigerant fluids (easily
liquefied under pressure) (evaporation requires heat to be absorbed by CFC
molecules, thus helps in cooling surroundings).
OR CFCs are relatively inert/non reactive (due to relatively strong C-F bond).
OR CFCs are non-toxic.
(ii)
CFCs are capable of travelling all the way to the stratosphere, where under
the action of uv light, Cl free radicals are generated (homolytic breaking of CCl bond) (NOT F radicals formed) and this would react with ozone molecules,
generating oxygen, thus ozone is being destroyed in a chain reaction.
(iii)
Alkanes are flammable.
2
(a)
(i)
For a weak acid dissolved in water, HA, HA(aq) ⇌ H+(aq) + A-(aq), the acid
dissociation constant, Ka, is defined as Ka =
[H  ][ A  ]
. pKa is defined as –lg Ka.
[HA ]
The smaller the value of pKa, the stronger is the acid.
(ii)
(explain to students why)
(b)
(i)
The tertiary structure of proteins is altered during coagulation. Tertiary (3o)
structure refers to the overall 3-dimenional shape of the protein involving
folding or coiling of the chains. It shows how protein molecules are arranged in
relation to each other due to the various R (residual) group interactions and
under external “stress”, example by adding weak acids, heating, enzymes etc.,
some of these R group interactions would be disrupted, and the protein will no
longer be able to fold or coil in its proper shape, and thus tertiary structure gets
destroyed, however, the exact amino acid sequence that makes up the protein
(primary structure) and secondary structure remains intact – this process is
called denaturation.
(ii)
Can disrupt the ionic interactions between –NH3+ and –CO2- groups by
instead forming ionic interactions with –CO2- groups. Form ion-dipole
interactions with –OH groups (or other suitable polar groups).
(iii)
Disrupt ionic interactions between –NH3+ and –CO2- groups, since addition of
weak acids converts –CO2- groups to carboxylic acids and –NH2 groups to –
NH3+, or disrupting ion-dipole interactions, example –CO2- groups with –OH
groups (may now form hydrogen bonding between –CO2H group formed with
–OH group instead).
(iv)
Kc =
[gluconic acid]
[GDL][H2O]
[GDL] initially =
(Question gives [H2O] thus must include in Kc)
1.00 1000
= 0.112 mol dm-3

178 50 .0
[GDL] at equilibrium = 0.112 – 0.0670 = 0.0450 mol dm-3
Therefore Kc =
0.0670
= 0.0268 mol-1 dm3
0.0450  55.5
(c)
(i)
(ii)
Diadzein with H2 and Ni: reduction of ketone to secondary alcohol and
addition to alkene to form corresponding alkane
1 mole of F reacts with 3 moles of Na: redox reaction involving 2 moles of
phenolic –OH group and 1 mole of secondary alcoholic –OH group
F dissolving in NaOH(aq): acid-base reaction between phenolic –OH group
and NaOH, sodium phenolate formed soluble in water due to formation of iondipole interactions
F reacting with K2Cr2O7/H+(aq): oxidation of secondary alcohol to ketone
G reacting with 2,4-DNPH: condensation reaction between ketone and 2,4DNPH to form corresponding hydrazone (orange ppt) (Ask students practise
drawing the products)
3
(a)
(i)
O2 + 4H+ + 4e-  2H2O
(ii)
O2 + 4H+ + 4e-  2H2O
x3
CH3OH + H2O  CO2 + 6H+ + 6ex2
2CH3OH + 3O2  2CO2 + 4H2O
Question ask for hence showing the working.
(iii)
Ecellº = Eredº - Eoxdº
(iv)
Methanol is a liquid and therefore easier to store or transport.
Less likely to be as explosive as hydrogen.
Need not to be kept under pressure, unlike hydrogen.
Will not accept answers for larger emf. In fact, the emf for (H2/O2) fuel cell is
larger.
1.18 = +1.23 - Eoxdº
Eº(CO2/CH3OH) = +0.05 V
(b)
(i)
CH3CH2OH + 3O2  2CO2 + 3H2O
(ii)
ΔGº = ΔHº - TΔSº = -1367 – 298(
(iii)
Compare equation in (b)(i) and (a)(ii), notice same amount of O2 consumed
and same amount of CO2 produced, this suggest same amount of electrons
involved in the redox reaction.
140
) = -1325 kJ mol-1
1000
Therefore z = 12
Therefore Ecellº =
(c)
(i)
( 1325  1000 )
12(9.65  10 4 )
= 1.14 V
bonds broken (BE)
5 C-H
5(+410)
1 C-O
1(+360)
3 O=O
3(+496)
1 C-C
1(+350)
bonds formed (BE)
4 C=O
4(-805)
5 O-H
5(-460)
ΔH = 5(410) + 1(+360) + 3(+496) + 1(+350) + 4(-805) + 5(-460)
= -1272 kJ mol-1
(ii)
Bond energies are average values, thus ΔH calculated in (i) is only an
approximation. OR bond energies are defined to be in gaseous state,
however ethanol is a liquid at standard conditions.
(d)
(i)
CxHyOH ≡
1
H2 and
2
CxHyOH ≡ xCO2
thus
1
H2 ≡ xCO2
2
Volume of H2 produced from 0.10 cm3 of J = 10.9 cm3
Volume of CO2 produced from 0.10 cm3 of J = 109 cm3
x=
1 109
(
)=5
2 10.9
CxHyOH + (x +
y 1
y 1
- )O2  xCO2 + (
)H2O
4 4
2
Reduction in gas volume (at 298 K) originally
= volume of O2 originally – volume of CO2 formed = 54.4 cm3
i.e. comparing reduction in volume originally and volume of CO2
y 1
y 1
20
- )O2 ≡ xCO2
i.e. ( - )O2 ≡ 5CO2 or O2 ≡
CO2
4 4
4 4
y 1
109
20
=
so y = 11
y  1 54.4
(
(ii)
You can solve part(ii) even if you do not know how to do part(i)!
(iii)
conc. H2SO4 and heat
(iv)
cannot show geometrical isomerism since there are no two different
substituent groups attached to each C in the C=C.
Incorrect to say that these identical groups are on the same side of the double
bond or there is a carbon atom having two identical groups.
4
(a)
(b)
Nucleon number is defined as the total number of protons and neutrons in an
atomic nucleus.
Electrons are deflected towards the positive plate, while protons to the
negative plate. Electrons are deflected at a much larger angle than protons.
(c)
(i)
(ii)
For NO2: bond angle is 136 º (accept any value between what is given for O3
and 170 º)
For O3: bond angle is 116 º (must be less than in NO2)(accept any value
between 110 – 120 º)
Both have 2 bond pairs and 1 lone pair of electrons on each central atom
(treat the lone electron in NO2 as a lone pair). However in NO2, since only a
lone electron is involved, the consequent electron repulsion (between lone
pair and bond pair) is not as pronounced as in O3, involving a real lone pair.
(iii)
Covalent bonding (comprising
double bond) between Cl and O
atoms, each double bond
comprises a σ and π bond
Hypothetical structure of FO2
Dative covalent bonding from F (donor)
to O atoms
Cl has available empty orbitals
to accept lone pair of electrons
from oxygen, can expand octet
F does not have empty orbitals to allow
expansion of octet
Even if considering dative covalent
bonding, F is more electronegative than
O, very unlikely F would donate its lone
pair of electrons to O
(d)
(i)
With a metal tong holding a small piece of calcium which is polished to
remove the oxide layer, immerse the calcium in a gas jar filled with
atmosphere of oxygen.
Calcium metal is resistant to burning at first due to its protective oxide layer,
however after a short while it bursts into flames, burning with a red or brickred flame, and leaves a white residue of calcium oxide.
(ii)
There is increase in solubility of oxides of Group II from magnesium to barium
(energetically more exothermic) and hence the [OH-] concentration increases,
leading to increase in pH.
(e)
Ba(O3)2 + H2O 
(f)
(i)
I2 ≡ 2Na2S2O3
Amount of I2 =
(ii)
15 .0
1
 0.100  = 7.50  10-4 moles
1000
2
O3 ≡ I2
Amount of O3 = 7.50  10-4 moles
Volume of O3 = 7.50  10-4  22 400 = 16.8 cm3
% of O3 =
(g)
(i)
5
O2 + Ba(OH)2
2
16.8
 100 = 3.36 %
500
To get the aldehyde, the reaction mixture has to be warmed and immediately
distilled and the distillate collected will be largely the aldehyde. To get the
carboxylic acid, the reaction mixture has to be refluxed.
(ii)
5
(a)
Anode: impure Cu
Cathode: pure Cu.
Electrolyte: Aqueous CuSO4
At the anode, Cu dissolves:
Cu(s) → Cu2+(aq) + 2e- (oxidation)
At the cathode, pure Cu is deposited. Cu2+(aq) + 2e- → Cu(s) (reduction)
State symbols required.
Pure Cu cathode grows in size as Cu2+ ions are discharged.
o
Cu2+ + 2e- ⇌ Cu
E = +0. 34V
Zn2+ + 2e- ⇌Zn
E = − 0.76V
Ag+ + e- ⇌ Ag
E = +0.80V
o
o
Impurities such as zinc is oxidised to Zn2+ at the anode and dissolves into the
solution as well.
However, zinc’s electrode potential is more negative than copper, thus Zn2+ is
not reduced at the cathode and remained in the electrolyte as ions.
While impurities such as silver, with an electrode potential more positive than
copper will not be oxidised, and remained as elemental silver and settle as
‘anode sludge’ at the bottom of the electrolyte.
(b)
When only a few drops of aqueous NH3 added to solution of Cu2+, a blue
precipitate of Cu(OH)2 is formed initially. Further addition of aqueous NH3
results in the blue precipitate dissolving and a deep blue solution of
[Cu(NH3)4]2+ is formed.
(c)
CuO + 2NH3  Cu(NH2)2 + H2O (clue in question: O2- is behaving as a strong
Bronsted base, will grab H+ from NH3)
Colour of solution: deep blue (due to [Cu(NH3)4]2+ formed)
(d)
[Cu(H2O)6]2+(aq) + 4 Cl-(aq) ⇌ [CuCl4]2-(aq) + 6H2O(l)
(I)
pale blue
pale yellow-green
(note: analogous to [Cu(NH3)4]2+)
When concentrated HCl added to solution of Cu2+(aq), ligand exchange
reaction occurs when Cl- displaces the H2O ligands.
On adding concentrated H2SO4 to solution of Cu2+(aq), no ligand exchange
reaction occurs as SO42- is a very poor ligand.
On dilution of yellow-green solution, concentrations of both [Cu(H2O)6]2+ and
Cl- decreases, position of eqm (I) shifts to left, and pale blue colour of
[Cu(H2O)6]2+ is restored.
(e)
(i)
2Cu2+ + 4I-  2CuI + I2
(ii)
Ecellº = +0.15 – (+0.54) < 0, reaction is not expected to be spontaneous.
(iii)
CuI is insoluble in water, and since Cu+(aq) are involved to form the
precipitate, [Cu+] is lower than expected, hence the E(Cu2+/Cu+) would be
higher, and can result in Ecell > 0.
(f)
(i)
The test is used for aldehydes. A positive result will give rise to a brick red
precipitate of Cu2O formed.
(ii)