Nov 2015
Nov 2015
Nov 2015
B
B
C
C
B
11
12
13
14
15
D
C
D
B
D
21
22
23
24
25
B
D
B
A
D
31
32
33
34
35
D
A
C
A
B
6
7
8
9
10
D
B
C
A
B
16
17
18
19
20
D
A
C
B
A
26
27
28
29
30
D
D
C
B
A
36
37
38
39
40
A
C
C
A
B
Answer: B
Group IV elements have electronic configuration of ns2 np2. Hence the answer is 2s2 2p2.
2
Answer: B
Sigma bond is formed by a head-on overlap by either s or p orbitals. Pi bond is formed by
sideways overlap by p orbitals and not be s orbitals.
Answer: C
For OH- , number of electrons = 8 +1 + 1 = 10
Number of neutrons = 8
4
Answer: C
Calcium has a higher melting point than sodium as the melting point of metals is dependent on
the charge density of the cation (charge/ ionic radius). Ca2+ has a higher charge than Na+.
5
Answer: B
The amino acid has the bond angles 109.5o, 107o and 120o.
The zwitterion has the bond angles 109.5o and 120o.
Answer: D
The definition of lattice energy of sodium oxide is the enthalpy change when one mole of
solid sodium oxide is formed from the separate gaseous sodium ions and oxide ions
under standard conditions.
H2 Chemistry 9647 2015
Answer: B
NaOH + HCl NaCl + H2O
Number of moles of NaOH =
12.5
1000
25.0
Answer: C
10CH4 + 39/2O2 9CO2 + 1CO + 20H2O
Mol ratio of CH4 : O2 = 10 : 19.5
Volume ratio of CH4 : O2 = 1 : 1.95
Answer: A
The product CO2 in solution forms hydrogen bonds with water molecules. Bond formation
would result in H being exothermic (-ve).
There is lesser number of particles for products in comparison to the reactants, hence this
would result in more orderly arrangement S=-ve.
10
Answer: B
Students are expected to know the standard conditions for electrode potentials.
Standard conditions for concentration of solutions used are 1 mol dm-3 and for 1 mol dm-3 for
H2 Chemistry 9647 2015
[H+]
11
Answer: D
Ka = [CH3CO2-][H+]/[CH3COOH]
= (C)2 / (1 )C
= 2 / (1 )
12
Answer: C
Students are expected to do well in this question.
Applying Le Chateliers Principle, to increase the product methanol, the disturbance that would
shift POE to the right are to increase the pressure and to decrease the temperature.
Reason is there is lesser number of molecules of gases in comparison with reactant
13
Answer: D
H2O(g) + CO (g)
H2(g) + CO2(g)
Let the change in number of moles of H2O and CO be a mol and forming a mol of H2 and CO2
Total number moles of gas at eqm = 1-a + (1-a) + a +a = 2
Hence 33.3% of H2 in resulting mixture is a/2 = 33.3/100
a = 0.666
Kc = [H2] [ CO2] / [H2O] [CO]
= (a)(a )/ (1-a) (1-a)
= (0.666)2 / (1-0.666)2
= 3.97
14
Answer: B
Given that rate = k[H2][NO]2
H2 Chemistry 9647 2015
Using the values when [H2] kept constant and [NO] is doubled, rate of reaction would be 4 times,
hence x= 6/4 =1.5
Using the values when [NO] is kept constant and [H2] is doubled, rate of reaction would be 2 times,
hence y = x (2) = 1.5 x 2 = 3
Using the values when [H2] kept constant, rate of reaction y=3 compared to 0.75 is 4 times, hence
conclusion [NO] = 1.0 mol dm-3 is double that of value of z, hence z = 0.5 mol dm-3
15
Answer: D
A is incorrect as ionic radius of M2+ would increase down Group II, hence charge density would
decrease
B is incorrect as atomic radius of elements would increase down Group II, hence electronegativity
would decrease
C is incorrect as the number valence electrons in Group II is always 2
D is correct as the cationic size of M2+ would increase down Group II, hence its charge density and
polarizing power would decrease. Hence less able to distort the electron cloud of large nitrate
anion and weakening the covalent bonds between N and O atoms, hence thermal decomposition
would take place at a higher temperature, leading to thermal stability of the nitrate
16
Answer: D
A is not true as S also gives 2 acidic oxides SO2 and SO3
B is not as Ne has the highest 1st IE in period 3
C is not true there are chlorides of Mg, Si and S that gives acidic solutions
D is true as P forms P4 molecule with 4 atoms
17
Answer: A
Cu+ is colourless due to d10 electronic configuration.
Copper (II) hydroxide is a blue precipitate.
18
Answer: C
Atomic radii decreases across the Period 3 due to increasing nuclear charge while shielding
effect remains effectively constant (because in the same period). Effective nuclear charge
increases and thus atomic radii of P < Si < Al < Mg
Across Period 3, melting point increases from Mg to Al (due to increasing metallic bond), with
Si having the highest melting point (due to strong covalent bonds between Si atoms), then
drop significantly for P4 (which exists as simple molecular structure with weak id-id between
H2 Chemistry 9647 2015
molecules)
19
Answer: B
Oxidation state of Cr changes from +6 to +3. Dichromate (VI) ions acts as an oxidizing agent for
Y. This reaction takes place in acidic medium. If it is in alkaline medium, Cr 2O3 maybe formed
instead.
Ketone, tertiary alcohol and carboxylic acids cannot be oxidised.
Y must be either primary or secondary alcohol or aldehyde.
20
Answer: A
PCl5 only eacts with either alcohol or carboxylic acid via nucleophilic substitution.
21
Answer: B
Alkynes have cabon-carbon triple bonds. There are two positional isomers and 1 branched
isomer, giving a total of 3 possible structural isomers.
CHC-CH2CH2CH3
H2 Chemistry 9647 2015
CH3CC-CH2CH3
CH3CH(CH3)CCH
22
Answer: D
sp2 carbons are those with one pi bond.
23
Answer: B
NaBH4 can reduce aldehyde to primary alcohol and ketone to secondary alcohol only.
24
Answer: A
H2 Chemistry 9647 2015
There are already 6 carbons in citric acid. Therefore, the reagent added has to be 3 carbon
only. There are 3 carboxylic acids functional groups and 1 alcohol functional group present in
citric acid. There is only an increase of one oxygen atom in the final organic product. Therefore,
the reagent added has to be a carboxylic acid, which reacts with 1 alcohol group in citric acid to
form an ester (with a removal of one water molecule).
25
Answer: D
Ease of C-Cl bond breaking
Ethanoyl chloride is an acid chloride and hence the bond breaks the fastest, giving rise to the
white ppt in the shortest amount of time.
26
Answer: D
As question did not provide students with the pKa values of the different groups, they need to
know that the generic pKa of the various groups.
pKa: -acid ~2-3, -amine ~9 R-acid ~4
Answer: D
Approach to this question is to do the conversion on the functional group and count the number
of atoms which have been lost or gained.
Conversion from aldehyde to ketone not possible so B is out.
28
Answer: C
Recall the relative acidities of phenol and alcohols and their reaction with NaOH.
Phenols will be deprotonated while alcohols will not.
29
Answer: B
Recognise that a multi-substitution on the ring has taken place due to the presence of the
phenol group and hence Br2(aq) is the reagent used.
30
Answer:A
Recall mechanism of electrophilic addition of Br2(aq)
Alkene is symmetrical hence need to take note of the following two points.
1) Br is inserted before OH
2) Generation of the carbocation intermediate
For each of the questions in this section, one or more of the three numbered statements 1 to 3 may
be correct.
Decide whether each of the statements is or is not correct (you may find it helpful to put a tick
against the statements that you consider to be correct).
The responses A to D should be selected on the basis of
A
1 only is correct
31
Answer: D
Draw out e configuration for ions
Mn2+: 3d5 no paired 3d
Fe2+: 3d6 has paired 3d
Co3+:3d6 has paired 3d
32
Answer: A
SiCl4 is covalent and replacing the two Cl with methyl groups would make it covalent still and
hence a low b.p. is expected.
Ge is in group 4 and just below Si and hence GeCl4 would be covalent and have a low b.p.
AlBr3 is expected to be covalent compound since Br is below Cl and AlCl3 is a covalent
compound.
33
Answer: C
Assumptions of the Kinetic Theory for Ideal Gas:
(i) The gas particles are in a constant state of random motion. (Statement 2)
(ii) The volume occupied by the gas particles is negligible as compared to the volume
occupied by the gas. (Statement 3)
(iii) There is negligible intermolecular forces of attraction between the gas particles.
34
Answer: A
[RECALL: Reduction reactions are favoured for positive Eored values, while oxidation reactions
are favoured for negative Eored values]
Statement (1): Based on Eored values, NH4+(aq) will undergo reduction, while Zn(s) will undergo
oxidation, therefore, Eocell = Eored (Reduction Half-Cell) - Eored (Oxidation Half-Cell)
= (+0.74) (-0.76) = +1.50 V
Statement (2): Overall cell reaction: 2NH4+(aq) + Zn(s) Zn2+(aq) + 2NH3(g) + H2(g)
Statement (3): Zinc casing will hence become thinner as Zn(s) is oxidized to form Zn2+(aq).
35
Answer: B
Answer: A
Statement (1): Methanal (HCHO) is a planar molecule. The shape about the central C-atom is
trigonal planar (i.e. 3 bond pairs & 0 lone pair).
37
Answer: C
[RECALL: Initiation Step of Free-Radical Substitution Homolytic fission of X-X bond]
Statement (1) & (2): Radicals have only one lone electron (or unpaired electron) and not a
lone pair of electrons.
Statement (3): Homolytic fission of a covalent bond is defined as the chemical dissociation/
cleaving of a bond in a molecules, whereby each of the atoms (involved in the bond) retains
one of the original bonded electrons.
38
Answer: C
Statement (1):
Statement (2):
atom Y is not H-atom.
Statement (3):
atom Y is not H-atom.
HINT: As long as atom Y is not H-atom, the compound will be a geometric isomer.
39
Answer: A
Thinking Point: Although the question did not specify the reaction conditions, students are
required to consider all positive reaction conditions (i.e. room temperature or heat etc.)
Statement (1): When heated, CH3-group will be oxidized with alkaline aqueous KMnO4 to form
CHO group.
Statement (2): At room temperature, phenol will undergo condensation with ethanoyl chloride
to form an ester.
Statement (3) At room temperature, phenol will undergo redox (acid-metal) reaction with
Na(s).
40
Answer: B
[RECALL: For a molecule to optically active, it should have a chiral centre (i.e. carbon bonded
to 4 different functional groups) with no plane of symmetry]
Molecule (3) is the only molecule with a plane of symmetry, hence it is not optically active.
PAPER 2
2015 A-Level Paper 2 Question 1 Planning
a)
Common Mistake: The equation to represent the reaction between propanoic acid and methanol
was generally well known. Some candidates had difficulty in expressing the ester as a displayed
formula. The use of the incorrect carboxylic acid and omission of water as a product were seen.
H2 Chemistry 9647 2015
Thinking Point: Student should be mindful of reversibility and displayed formula of ester.
Standardisation of HCl(aq)
Part 1: Prepare 250 cm3 Standard HCl(aq)
Procedure
1. Measure out 40.00 cm3 HCl(aq) from 50.00 cm3 burette into 250 cm3 standard flask.
2. Top up to 250 cm3 mark with DI water using dropper.
3. Stopper flask, and invert it a few times to obtain a homogeneous solution.
4. Pipette 25.0 cm3 from this 250 cm3 solution into 250 cm3 conical flask.
5. Add one drop of phenolphthalein into solution.
6. Titrate with 1.00M NaOH(aq) from 50.00 cm3 burette.
7. Add dropwise near endpoint.
8. Stop titration when solution changes from colourless to pink.
9. This is the titre volume.
10. Repeat Step 4 to 9 until at least two consistent results are obtained.
Note
Why measure out 40.00 cm3 HCl(aq)? Assume that it is 3.00 M HCl(aq), then the titre value
would be about 12.00 cm3. This allows 50.00 12.00 = 38.00 cm3 of NaOH(aq) to be used
for titrating equilibrium mixture later on. If 0.03 mole of CH3CH2COOH is used in Mixture 1
(shown later), then maximum another 30.00 cm3 of NaOH(aq) will be used. This is feasible.
H+ is a catalyst. It is not used up in the reaction.
Assume it is 3.00 M HCl(aq).
Amount of H+(aq) in 40.00 cm3 = (0.040)(3) = 0.120 mol
6. Repeat Step 2 to Step 6 for another mixture of 2.96g CH3CH2COOH, 1.92g CH3OH and
50.00cm3 HCl(aq).
Note:
Given that the mixture is left for one week for completion. This necessitates the use of
stoppered flask. If not there will be problems of evaporation and oxidation of methanol by
O2 in air.
There is no need for water bath. Enthalpy change of esterification is about zero, based on
the estimation of bonds broken and bond formed. Hence there will be no large temperature
changes in mixture. Mixture is also expected to be at room temperature of 25C. One week
is given for completion. Enough time is given for mixture to equilibrate to the surrounding
room temperature
d) Titration Procedure on Equilibrium Mixture after ONE week
1. Using a thermometer, measure the temperature of mixture. It should be the same as
previous temperature.
2. Pipette out 25.0 cm3 mixture into another 250 cm3 conical flask.
3. Add one drop of thymol blue (pH indicator) into conical flask.
4. Add 1M NaOH(aq) from 50.00 cm3 burette into mixture.
5. Add dropwise near endpoint.
6. Stop when mixture changes from red to orange.
7. Record this endpoint volume. [Compare the endpoint in Part 1.]
8. Add one drop of phenolphthalein to mixture.
9. Continue adding 1M NaOH(aq) to mixture.
10. Add dropwise near endpoint.
11. Stop when mixture changes from yellow to orange.
12. Record this endpoint volume.
Question: Why use Thymol Blue, not Methyl Orange?
Answer: pH of solution at first endpoint is about 2.4. Working pH range of Thymol Blue from 1.2
(Blue) to 2.8 (Yellow), while that of methyl orange from 3.1 (Red) to 4.4 (Yellow).
mol
21
21
Change / mol (. )
(. )
Eqm / mol
21
1000
21
1000
0.01
CH3CH2COOCH3(aq) +
0
+(. )
(.
Ignoring the information provided when outlining experimental plans resulted in a lack of
appropriate experimental detail. Use of measuring cylinders to accurately measure volume
was common when a pipette or burette would have been more appropriate. Explanations of
how to process the results in order to solve the problem proved to be demanding. For
those candidates who included the processing of results, there was evidence of clear and
logical explanations. The relevance of water in the original amount of hydrochloric acid was
rarely recognised.
Thinking Point:
(a)
(i)
(i)
Answer: Using the data from the table, 0.0358 MJ of energy is released per dm 3
of methane at 25 oC and 101 kPa.
101 kPa - 0.0358 MJ
24.8 MPa 8.79 MJ
Thinking points: Remember to express answer in appropriate units.
Common mistakes made: Many students did not use the data in the table. They
used the Ideal Gas Equation.
H2 Chemistry 9647 2015
(ii)
(i)
Answer:
CH3OH (l)
-726 kJ mol-1
-283 kJ mol-1
Answer:
H2 Chemistry 9647 2015
G = H -TS
For a spontaneous reaction, G < 0.
H -TS < 0
+332
+129 T ( 1000 ) < 0
+332
-T < +129 (
)
1000
T > 388 K
The minimum temperature is 389 K.
Thinking points: When G = 0, the reaction is at equilibrium. Students need to
be mindful in solving equation involving inequality sign.
Common mistakes made: Many students did not convert to the correct units and
made rounding errors or did not express the answer in correct number of
significant figures.
(e)
(a)
(i)
Answer:
Enthalpy change = 244 + 436 2(431)
= -182 kJ mol-1
[1]
(ii)
Answer:
(i)
Answer:
Bright light provides sufficient energy to overcome the high activation energy
barrier to form chlorine radicals. [1]
Thinking points:
Must allude to the presence of activation energy barrier and not just bond
breaking of Cl2 as endothermic
Common mistakes made: Not including the fact that light provided energy for the
formation of chlorine radicals.
(ii)
Free radical Substitution [1]
Cl2 2Cl
[1]
Thinking points:
Requires some inference and transfer of learning skills from organic
(iii)
Answer:
The brief exposure to light generates the radicals required for the propagation
step which no longer requires the light energy. [1]
Moreover, since the initiation stage is the energy requiring step whilst the
propagation step is nett exothermic resulting in an increase in temperature, the
rate of the reaction increases. [1]
Thinking points:
Two terms to be explained explicitly as they are in bold print.
Common mistakes made: Not realising that the exothermic nature of the
reaction and thus the increase in temperature resulted in an increase in rate.
(c)
Answer:
Although the BE of Br2 (+193 kJ mol-1) is less than that of Cl2
(+244 kJ mol-1), less energy required for bond breaking for Br 2, the BE of HBr (366 kJ mol-1) is also relatively lower than HCl (-431kJ mol-1), less energy given
out.
This results in the reaction having an enthalpy change of -103 kJ mol-1, which is
less exothermic than the reaction between Cl2 and H2 (-182 kJ mol-1). Thus Br2
reacts with hydrogen less reactively.
For 3 mks,
Thinking points:
What does the term less reactive really allude to?
Spontaneity?
Kinetics?
Common mistakes made: Unnecessary answers on electrode potential data or
discussions as to why H-Br bond is weaker.
(d)
(i)
Answer:
NaCl (s) + H2SO4 (l) NaHSO4 (s) + HCl (g)
[1]
(ii)
Answer:
2HBr (g) + H2SO4 (l) Br2 (g) + 2H2O (l) + SO2(g) [1]
HBr is a stronger reducing agent as compared to HCl and hence is able to
reduce H2SO4 to SO2.
[1]
Common mistakes made: Need to be clear on which is the reducing agent.
(e)
Answer:
Add Cl2 (aq) to separate samples of HX (aq).
[1]
For HBr (aq), reddish-brown solution of aqueous Br2 will be formed. For HCl
(aq), the solution remains colourless.
[1]
Cl2 + 2Br Br2 + 2Cl
[1]
Thinking points:
Use of oxidising agent such as acidified hydrogen peroxide or acidified
chlorate(I) or (III)
Observations to be made
3 mks what could be the last point?
Common mistakes made: Concentrated sulfuric acid is not suitable because
starting point is an aqueous solution.
a)
i)
[1] amine attack + C with lone pair of e, bond breaking of C-I bond.
[1] pentavalent transition state with correct charge separation
[1] formation of positively charge species with abstraction of proton
[1] other products (dependent on base used to abstract proton)
Common mistakes made: Not draw the positively charged intermediate along
with the relevant bond breaking to produce the products.
ii)
Methyl iodide is a primary alkyl halide with little steric hindrance [1] due to only H
attached to the C [1].
Or
Methyl iodide will form an unstable CH3+ carbocation [1] due to lack of e
donating groups attached if it were to undergo SN1 reaction [1].
Reason plus explanation to get full 2 marks
c
i)
i)
(a)
This results indicates that the compounds are aryl chlorides [1] as they did
not undergo a nucleophilic substitution reaction as expected with alkyl
chlorides. Aryl chlorides do not undergo nucleophilic substitution because the
lone pair of electrons on Cl can delocalise into the aromatic benzene ring and
this leads to a partial C Cl double bond character which strengthens the
bond,[1] and leads to a lower reactivity towards nucleophilic substitution.
Common mistake: to focus on what the compounds identity could not be
rather than what it actually was.
(b)
(i)
element
% by mass
Ar
% mass
Ar
ratio
C
49.0
12.0
49.0
= 4.08
12.0
4.08
=3
1.36
H
2.7
1.0
2.7
= 2.70
1.0
2.70
2
1.36
Cl
48.3
16.0
48.3
= 1.36
35.5
1.36
=1
1.36
m
Mr
mRT
pV
0.3448.31(273+181)
(101 000) (87.4 10-6 )
= 147
[1]
[1]
[1]
[1]
(c)
PAPER 3
1
(a)
(i)
ROH + Na RONa + H2
2ROH H2
Amount of menthol = 2 x 1.32 x 102 = 2.64 x 102 mol
Mass of menthol = 2.64 x 102 x 156 = 4.118 g
4.12
% by mass of menthol in peppermint =
x 100 % = 41.2 %
10
Note:
Students need to recognise mole ratio = 1 menthol : H2 and not 1:1.
(ii)
Chiral centres: 3
Number of optical isomers = 23 = 8
Note:
General formula for number of stereoisomers = 2n where n is the number of
chiral centre or double bonds with different R groups. Common mistakes are
to create six isomers (3 x 2) or nine isomers (32).
(b)
(i)
(ii)
Nickel catalysed
Energy / kJ mol1
Ea = +42 kJ
mol1
Platinum catalysed
Menthone +
Hydrogen
H = 80 kJ mol1
Menthol
Reaction pathway
Note:
The Maxwell Boltzmann distribution diagram is different from the energy
profile diagram.
Reaction is exothermic (H = - 80 kJmol1)
Reagents and products are to be labelled on diagram
As the mechanism is not given, the reaction pathway will be assumed to be
a single step. Hence there will be one hump.
Common mistake: to draw an endothermic reaction or draw a Boltzmann
distribution curve or not name the reactants and products or mix up the reactant
and product.
(iii)
1. Diffusion of reactants towards catalyst
Hydrogen gas and methone diffuse towards the nickel catalyst
2. Adsorption
Adsorption of hydrogen and menthone onto the catalyst surface
weakens the HH bond in H2 and the C = O bond in menthone
3. Reaction
CH and OH bonds form while HH bond in H2 and C = O bond
breaks.
4. Desorption
Menthol diffuses away from the nickel surface.
Note:
Question requires a description of how a heterogeneous catalyst works.
Question is not about how the role of a catalyst increases the rate of
reaction by reducing Ea.
(iv)
Palladium
Palladium is lower in the group than Ni and it has a similar electronic
configuration as Ni and Pt. So it is expected to behave in a similar mechanism
as Ni and Pt.
Note:
Students need to explain their choice of element.
(c)
A:
B:
C:
D:
E:
F:
G:
Observations
All 3 isomers react with Br2 (aq)
Inferences
Alkene functional group present
in all 3 isomers
Electrophilic addition
Ketone functional group present
in all 3 isomers
Condensation reaction
Reduction of carbonyl and
alkene functional groups.
Oxidative cleavage of alkene
functional groups
Common mistake:
Not comment on the use of hot KMnO4 or to think that the decolourisation of
bromine imply the presence of benzene.
(a)
Answer: All three exist as non-polar molecules; all possess instantaneous-dipole
induced-dipole (idid) interactions. As the size of the polarisable electron cloud
increases down Group VII, the idid gets stronger. More energy is required to
vaporise it hence the volatility decreases down Group VII.
Thinking points: What influences the volatility of a substance?
Common mistakes made: Students cannot link volatility to intermolecular forces of
attraction. Some students link the trend to bond energies or to relative molecular
mass.
(b)
Answer:
6NaOH (aq) + 3Cl2 (aq) NaClO3 (aq) + 3H2O (l) + 5NaCl (aq) [1]
The oxidation state of chlorine started off as zero and ended up as +5 in NaClO3 and
-1 in NaCl. [1]
Thinking points:
Keyword is hot, so the chlorine-containing product is chlorate(V) rather than
chlorate(I).
Best to include state symbols because question itself included it for NaOH.
Common mistake: unbalanced equation or did not mention changes in oxidation
number of chlorine.
(c)
(i)
Answer:
The bond energy of the H-F bond (+ 562 kJ mol-1) is the strongest out of the
four H-X bonds (+ 431, + 366, + 299 kJ mol-1). As a result while HCl, HBr and
HI dissociate completely, HF only dissociates partially.
Common mistakes made:
Failing to quote the bond energy values from the Data Booklet.
Some overzealous students might even write about extent of orbital overlap as
we go down Group VII. Not needed based on the context!
(ii)
Answer:
pH of HCl = - log (0.50) = 0.0301
pH of HF = - log (0.50 * 5.6 * 10^-4) = 2 log (2.8) = 1.776
Common mistakes made:
Failing to realise that HF is a weak acid, as implied by the Ka value provided.
H2 Chemistry 9647 2015
(d)
Answer:
On the addition of aqueous silver nitrate, a white precipitate (AgCl) and a pale yellow
precipitate (AgI) is formed, in the case of chloride and iodide ions respectively. On
adding excess aqueous ammonia, the precipitate dissolves to give a colorless
solution in the case of chloride but not iodide. The compound that caused AgCl to
dissolve in excess ammonia is Ag(NH3)2Cl.
Common mistakes made:
Since the question mentioned products, it is safer to write Ag(NH3)2Cl than just
write the formula of the silver diammine complex ion.
Students did not identify the ppt or gave the wrong colour or did not identify or
mention the silver complex.
(e)
(i)
Answer:
The value of PV is fixed at all temperatures.
(ii)
Answer:
V = 0.4 x 8.31 x 300 / 12 x 105 = 8.31 x 10-4 m3= 831 cm3
Thinking points:
Rather straight-forward calculation. No unit conversion required.
(iii)
Answer:
pV = 9.26 x10^5.
pV = 8.88 x 10^5.
Thinking points:
Also does not involve unit conversion but not as obvious as (ii). Look carefully
at the units in the third column.
Answer:
Assuming that there is a linear relationship and interpolating from the given
values,
PV = (8.88 x105) + (0.6 x 0.38 x 105) = 9.108 x 105
V = (9.108 x105) / (12 x 105) = 0.759 dm3
(iv)
Answer:
The volume in (iii) is less than the one in (ii) because HCl is a polar molecule
with relatively strong permanent dipole permanent dipole interactions between
molecules. [1] These interactions cause the actual volume of the gas (iii) to be
smaller than the ideal volume (ii). [1]
Thinking points:
Your answer must include the comparison between your answers in (ii) and (iii)
and the link with the appropriate property of HCl.
(f)
(i)
Answer:
S = + 16.8 / 188 = + 0.0893 kJ mol-1 K-1 [1]
The positive entropy change is due to an increase in the number of
gaseous molecules on vapourisation and there are more ways to arrange
the particles hence greater disorder. [1]
Thinking points:
It is expected of you to know that is G is zero for phase changes (the
examiner is not obliged to tell you so).
No need for unit conversion because the temperature is already in Kelvins and
the energy units of H and S are compatible.
Common mistakes:
Failing to insert the positive sign. Mix up the J and KJ units.
(ii)
Answer:
G = + 16.8 298(0.0893)= - 9.811 kJ mol-1
The negative sign of G indicates that the reaction is spontaneous.
Thinking points:
No need for unit conversion because the temperature is already in Kelvins and
the energy units of H and S are compatible.
There would be ECF, based on your answer to (i).
(a)
(i)
Magnesium burns readily with dazzling white flame and greyish white solid of
MgO is formed.
Mg(s) + 1/2O2(g) MgO(s)
Ca(s) + 1/2O2(g) CaO(s)
Calcium burns reluctantly with intense white flame with red tinge/brick red flame
H2 Chemistry 9647 2015
Thinking points: Students have not observed how calcium burns in air in the lab.
Common mistakes made: Not many knew the colour of the flame is brick red.
Descriptions with equations must be included. Solid is referred as precipitate.
(ii)
MgO is insoluble in water.
CaO dissolves quickly in water with a crackle sound to form a colourless
solution. Large amount of heat is released.
CaO (s) + H2O (l)
Ca(OH)2 (s)
Ca2+ (aq) + 2OH (aq)
Common mistakes made: Many students did not know the observations.
Thinking points: Students need to observe the reactions in the lab.
b
(i)
Ksp = [Mg2+][OH-]2
= 1.6 x 10-4 x (2 x 1.6 x 10-4)2 = 1.64 x 10-11 mol3 dm-9
Common mistakes made: Some omit the negative and positive signs on the
ions. Some did not use the correct [OH-]
(iv)
As the total volume is doubled, the concentration of barium ions and magnesium
ions would be halved. However the concentration of hydroxide would need to
consider the total concentration of hydroxide from both Mg(OH) 2 and Ba(OH)2.
Based on calculation, magnesium hydroxide would be precipitated while barium
hydroxide would not precipitate.
The bonding is hydrogen bonds between the N-H bond of 1 amino acid and the
C=O bond of the 4th amino acid per turn.
Common mistakes made: Students need to indicate where the hydrogen bond is
formed.
Thinking point: Given is 2m, students should be more precise in their answer.
(iii)
6M acid solution and heat for several hours OR moderately concentrated NaOH
solution and heat for several hours.
Common mistakes made: The word dilute, aq or solution are omitted
Thinking point: Peptide bond is strong and hence needs to use appropriate
solutions of suitable concentration.
(iv)
(a)
Answer:
(l) +
15
O2(g)
2
6CO2(g) + 4H2O(l)
Hc
15
= 3BE(CC) + 8BE(CH) + 2BE(C=C) + 2BE(CO) + 2 BE(O=O) 12BE(C=O)
8BE(OH)
15
= 3(350) + 8(410) + 2(610) + 2(360) + 2 (496) 12(805) 8(460)
= -3350 kJ mol-1 (correct to 3 s.f.)
Common mistake: unbalanced equation or omission of some relevant bond energies
in their calculations or careless mistakes.
(c)
Answer:
q = 200 g 4.18 JK-1g-1 32K = 26752 J
Heat released by burning 1.00 g of DMF = 26752 J
100
80
= 33440 J
(d)
Common mistake: using the 80% and calculating the Mr was not easy for some
students.
(i)
Answer:
Common mistake: SN 2 reacting with Cl- in the rate determining step or ionizing
the alcohol to RO- in the first step
(ii)
Answer: K2Cr2O7, dilute H2SO4, heat under reflux
(Do not accept KMnO4, dilute H2SO4, heat under reflux because it will cleave
C=C bonds)
(e)
(i)
Answer: Ester
(ii)
Answer: CH2(OH)CH2(OH)
(Accept condensed formula, structural formula, or skeletal formula)
(iii)
Answer: Concentrated H2SO4, heat under reflux.
Common mistake: to write conc H2SO4 (aq) which contradicts
(f)
Answer:
Initiation:
Cl
Propagation:
R-CH3 + Cl
R-CH2 + Cl2
Cl
UV light
2Cl
R-CH2 + HCl
R-CH2Cl + Cl
Termination:
Cl + Cl
ClCl
R-CH2 + Cl
RCH2Cl
R-CH2 + CH2-R
R-CH2-CH2-R
Common mistake: to produce H in the propagation steps
(a)
(i)
Let % of 6Li be x
x
(100 x)
6.015
7.016 6.942
100
100
1.001x 694.2 701.6
7.4
x
7.39
1.001
%6Li = 7.39%
%7Li = 92.61%
[1] for each relative percentage abundance
Take note that answer is to be given to 2 d.p. as stated in question and not 3
sig figures.
(iii)
X: 3He
Y: 7Li
Students should show the nucleon number in their answer.
Should include the charge on 7Li to account for the conservation of charge.
Both charge and mass of compounds on both sides of equation must be
balanced.
Must identify the element as asked in the question and not leave answer in
terms of X and Y.
(i)
Metallic bond. Both lithium and carbon atoms are held together by a sea of
delocalised electrons.
Must make reference to the delocalised e in graphite.
(ii)
Before: +4
After: +3
(iii)
Making use of dot and cross diagram to show the bonding pairs.
4 bond pair of bonding e, no lone pairs of e hence tetrahedral shape.
(iv)
(i)
Li+ has a small ionic radius and hence its charge density is high enough to
distort the electron cloud of CO32, weakening the CO bonds and
allowing it to decompose thermally.
NOTE: Need to be clear that it was the C-O bond in the carbonate ion that
was polarised and that the cation has high charge density.
d
(i)
(ii)
Common mistake: not relating the position of the OH group to the number of
carbon atoms in each of the starting compounds and using RLi as one of the
reagents instead of RBr.
e