Answers
1 (a)
H2O2 + 2H+ + 2I-  2H2O + I2
[1]
Note: Must be IONIC equation.
(b)
The presence of ethanoic acid and sodium ethanoate forms a buffer solution to [1]
keep the pH relatively unchanged as the reaction progresses.
(c)
Apparatus:
250 cm3 graduated flask
Measuring cylinder (20 cm3)
25.0 cm3 pipette
Burettes
Stopwatch
White tile
Preparation of stock H2O2 solution
1. Using a burette, transfer 80.00 cm3 of 7.5 mol dm-3 H2O2 solution into a
250 cm3 graduated flask. Top up the graduated flask with deionised water,
and shake thoroughly.
[H2O2]stock = (7.5 x 80/1000)/(250/1000) = 2.4 mol dm-3
The resultant H2O2 solution in the graduated cylinder has a concentration of
2.4 mol dm-3. It is designated as Solution C.
Experiment 1
2. Fill up a burette with Solution C. Using the burette, transfer 25.00 cm3 of
Solution C to a dry conical flask.
3. Using a 25.0 cm3 pipette, add 25.0 cm3 of 1.0 mol dm-3 ethanoic acid
solution to the same conical flask. Mix the contents thoroughly by swirling
the flask. The resultant solution is designated as Solution D.
4. Fill up another burette with Solution A. Transfer 50.00 cm3 of Solution A
into another dry conical flask.
5. Place a white tile below the conical flask containing Solution A. Pour
Solution D rapidly into the conical flask containing Solution A. Start a
stopwatch at the same time when solutions A and D are mixed and
swirled.
6. Stop the stopwatch when the dark blue colour appears throughout the
mixture.
7. Note the time elapsed to the nearest 0.1s.
In experiment 1, the initial concentration of H2O2 in step 5 after mixing would be
(2.4 x 25/1000)/(100/1000) = 0.600 mol dm-3, which is significantly higher than
the concentration of H2O2 [(0.75 x 50/1000)/(100/1000) = 0.375 mol dm-3] in the
teacher’s demonstration. As all other reagents are present in the same
proportion as in the teacher’s demonstration, the mixture should turn dark blue
in significantly less time than 20 seconds.
Experiments 2-5
8. Perform four other experiments, repeating Steps 2 to 7, using different
volumes of Solution C and adding water using a measuring cylinder to ensure
the total volume is 100 cm3 after mixing in Step 5.
A suggested table of values for the volumes used, including the initial
concentration of H2O2:
Experiment
Volume
of Volume
Time/
Solution C of water s
[H2O2]
Rate
/mol dm-3
/s-1
/cm3
/cm3
1
25.00
0
2
20.00
5
0.480
3
15.00
10
0.360
4
10.00
15
0.240
5
5.00
20
0.120
<20
0.600
9. Calculate the rate of the reaction for each experiment based on the
formula:
Reaction rate = 1/time
Fill out the rate column in the table.
10. Plot a graph of reaction rate against [H2O2] on graph paper.
Results analysis
If the reaction rate does not change with a change in the concentration of
[H2O2], showing a horizontal straight line, the reaction is zero order with respect
to H2O2.
If the reaction rate increases proportionately with [H2O2] with the line passing
through the origin, then it is first order with respect to H2O2.
If the reaction rate increases exponentially with [H2O2] with the line passing
through the origin, then it is 2nd order with respect to H2O2.
Marks allocation:
Method, including appropriate apparatus used (max 3m):
1. Burette, graduated/standard/volumetric flask (preparation of stock
solution).
2. Burette, pipette, conical flask.
3. Mixes solutions & swirls.
4. Pour rapidly and start stopwatch when two solutions first mixed.
5. Stop when dark blue colour observed throughout mixture.
6. Repeat experiment with different volume of Solution C used, with
appropriate volume of water added to ensure total volume of mixture
kept constant.
7. Analyse results by plotting a suitable graph.
[2] Justification for concentration of standard solution used, including relevant
calculations
[1] Table of data to be tabulated
[1] Suitable range of concentrations shown in table
[1] Correct choice of graph to be plotted
[1] Interpretation of proposed graph
(d)
The 7.5 moldm-3 H2O2 solution is corrosive and produces fumes. The [1]
preparation of the stock solution should be conducted in a fume cupboard with
protective gloves and safety goggles.
2 (a)
(i)
[1]
Hydrocarbons
(ii)
[1]
NO is formed as the high temperatures in the internal combustion
engine enable the reaction between O2 and N2 molecules to form NO.
(iii)
[1]
Proteins
(some of the component amino acids can form disulphide bridges,
therefore they would have sulphur present; this is a cross-reference from
Organic Nitrogen Compounds)
Note: DNA is not acceptable, as they contain phosphorus, not sulphur
(b)
(i)
[1]
[1]
Apply formula
∆H1
= ∑∆Hf(products) – ∑∆Hf(reactants)
= -396 – (-297) = –99 kJ mol-1
(ii)
[1]
[1]
∆G
= ∆H - T∆S
= -99 x 1000 – 298(-93.5)
= –71137 J mol-1
(c)
(i)
[1]
Just apply the formula provided:
-71137 = -2.303 RT log Kc
log Kc = -71137 / (-2.303 x RT)
[1]
log Kc = 12.47
Kc = 2.97 x 1012 (no need to state units here)
(ii)
[1]
It implies that the equilibrium for oxidation lies to the right, favouring [1]
the formation of products in the reaction. The very large value suggests
that the reaction proceeds to completion.
(iii)
[1]
he reaction to form SO3 requires a high activation energy due to the
need to break more double O=O bonds.
(iv)
[1]
[1]
The presence of a catalyst lowers the activation energy of the reaction
and speeds up the reaction, enabling it to arrive at the equilibrium
faster.
3 (a)
(i)
[1]
[1]
Ksp = [Pb2+][Cl–]2
units: mol 3 dm-9
(ii)
[1]
Mr of PbCl 2 = 207 + 35.5 x 2 = 278
Amt of PbCl 2 in 4.7g = 4.7 / 278 = 1.6906 x 10-2 mol
[PbCl 2] = 1.6906 x 10-2 mol dm-3
PbCl 2  Pb2+ + 2Cl–
[1]
[1]
[Pb2+] = [PbCl 2] = 1.6906 x 10-2 mol dm-3
[Cl–] = 2 x [PbCl 2] =3.381 x 10-2 mol dm-3
Ksp
= [Pb2+][Cl–]2
[1]
= 1.6906 x 10-2 x (3.381 x 10-2)2
= 1.9 x 10-5 mol3 dm-9 (to 2 sf)
(b)
(i)
[1]
[1]
Immediately: white ppt
After one hour in sunlight - The precipitate darkens in colour as the
silver chloride decomposes in the presence of sunlight into its constituent
elements of silver metal and chlorine.
(ii)
[1]
[1]
[1]
Add dilute aqueous ammonia to the original precipitate.
The original precipitate will dissolve in aqueous ammonia.
AgCl(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Cl–(aq)
(iii)
[1]
[1]
PbCl 2 + 2Cl–  [PbCl4]2–
The precipitate dissolves due to the formation of the [PbCl 4]2– complex.
(c)
[1]
The addition of concentrated HCl results in the formation of the [CuCl4]2–
complex which is yellow in colour:
CuCl 2 + 2Cl–  [CuCl4]2–
[1]
[1]
Ligand exchange occurs. The addition of water replaces the chloride
ligands and forms a different copper complex which is blue in colour.
[CuCl4]2– + 6H2O  [Cu(H2O)6]2+ + 4Cl–
(d)
(i)
[1]
[1]
In acidic solution with high H+ ion concentration, the concentration of S2–
ions is low as the equilibrium shifts left to form more HS– ions.
However, CuS has very low Ksp value and hence precipitates easily even
with very low concentrations of S2– ions.
(ii)
[1]
In an alkaline solution, the concentration of H+ ions is low, and hence the [1]
equilibrium will shift right, resulting in a high concentration of S2– ions.
NiS has the highest Ksp value for the 3 salts, but due to the high
concentration of S2– ions, the resulting ionic product will exceed the Ksp,
thus the precipitate will form.
(iii)
[1]
ZnS has a smaller Ksp value than NiS, therefore it will form a precipitate
in alkaline solution which has a high concentration of S2– ions. In neutral
solution, the concentration of S2– ions is higher than in an acidic solution, [1]
but lower than an alkaline solution.
Due to the lower Ksp of ZnS, the resulting ionic product of S2– ions and
Zn2+ ions still exceeds the Ksp, hence a precipitate is observed.
4 (a)
O
(i)
HO
OH
O
HO
O
OH
OH
OH
20 Alcohol
30 Alcohol
[1]
[1]
[1]
Carboxylic acid
[1]
Ester
[1]
Phenol
(ii)
4 chiral carbons are present
[1]
O
HO
OH
O
HO
O
OH
OH
OH
(b)
H
O
O
H
H
C
C
H
O
C
C
H
H
C
O
HH
C
O
H
H
C
H
H O
(c)
O
C
C
C
H
O
O
O
H
H
Non-aromatic compound
Aromatic compound R
[1]
[1]
O
Br
[2]
Br
Br
HO
Br
Br
OH
OH
H
H
(d)
O
(i)
[1]
+ -
Na O
OH
OH
(ii)
Only the carboxylic group is acidic enough to react with the [1]
hydrogencarbonate ion, due to the resonance stability of the conjugate
base. The phenol –OH groups are not acidic enough to lose their [1]
protons.
5 (a)
(i)
Reagent: HCN
[1]
Conditions: trace amount of NaOH(aq), 150C
[1]
Type of reaction: Nucleophilic addition
[1]
NaOH + HCN  Na+ + CN¯ + H2O
(ii)
[2]
OR NaCN  Na+ + CN–
(Nucleophile)
H3C
C
H

+
O + CN-
¯
sp2 hybridised
slow
rds
CH3
H
C
OCN
sp3 hybridised
tetrahedral
intermediate
CH3
H
C
OCN-
+
H
CN
CH3
fast
H
C
OH
CN
+
CNregenerated
cyanohydrin
(b)
(i)
[1]
[1]
Firstly, lactic acid contains a chiral carbon.
Secondly, due to its biological origin, the lactic acid in milk contains only
one optical isomer of the compound, therefore it can rotate plane
polarised light.
(ii)
[1]
During the synthesis of lactic acid, the mechanism involved the CN–
nucleophile attacking the electropositive carbon, which has a trigonal
planar orientation.
As a result, the nucleophile can attack from either the top or the
bottom of the plane with equal probability, thus leading to equal
amounts of the optical isomers of lactic acid formed.
[1]
(c)
Molecule
A
Explanation
COOH
COOH
C
C
OH
OH
H
[1]
H
Both chiral carbons and their groups are arranged in such a
way that its mirror image B is not superimposable on it.
B
COOH
COOH
C
C
OH
OH
H
[1]
H
B is the mirror image of A.
COOH
COOH
C
C
OH
OH
H
C
[1]
H
In C, there is an internal line of symmetry, so it has
superimposable mirror images, therefore the mirror images
are actually identical, so it has no optical activity.
Meso C
Mirror Image 1
Mirror Image 2
View 1
rotate mirror image 1 along a horizontal axis first… (focus on the red and
green spheres)
Then rotate it along a vertical axis
Which gives us mirror image 2. Therefore the mirror images are actually
identical, and superimposable