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AQA A-Level Chemistry Year 2
Student Book Answers
CHAPTER 1
Assignment 1
A1.
a. ΔfHƟ[MgCl(s)] = 148 + 738 + 121 − 349 − 815 = −157 kJ mol−1
b. ΔfHƟ[MgCl3(s)] = 148 + 738 + 1451 + 7733 + 121 − 349 − 5540 = +4302 kJ mol−1
A2.
Its enthalpy of formation is highly endothermic.
A3.
Assignment 2
A1.
1000 g
A2.
98 g
A3.
98/18 moles
A4.
2.4 × 100 kJ
A5.
(2.4 × 100)/(98/18) = 44.1 kJ mol−1
A6.
373 K
A7.
(44.1 × 1000)/373 = 118 J mol−1 K−1
A8.
For example, reduce energy loss by improving the kettle’s insulation; make sure that
no water vapour condenses and returns to the kettle; try experiments with different
quantities of water and compare results.
Assignment 3
A1.
a.
ΔfHƟ [KCl(s)] = (+89) + (+112) + (+419) + (−348) + (−711)
= −439 kJ mol−1
b.
ΔfHƟ[KCl2(s)] = +(+89) + (2 × +122) + (+419) + (+3051) + (2 × −348) + (−2350)
= +757 kJ mol−1
Generally, a transfer of energy from the surroundings is needed to overcome attractive
forces, and energy is transferred to the surroundings (released) when particles are attracted
to each other and move towards each other.
Atomisation is an endothermic change. A transfer of energy from the surroundings is
needed.

In metals, a transfer of energy from the surroundings is needed for atoms to break free from the
lattice, where positive ions are held together by a sea of mobile, negative electrons.

In covalent molecules, a transfer of energy from the surroundings is needed to form separate
atoms by breaking covalent bonds.
The ionisation enthalpy is endothermic. The negative electrons are attracted by the positive
nucleus. A transfer of energy from the surroundings is needed to remove them.
The electron affinity is exothermic for the first electron and endothermic for any further
electrons. Electrons are attracted towards the positive nucleus, so energy is released. If
more electrons are introduced then the electron–electron repulsion means that a transfer of
energy from the surroundings is needed.
The enthalpy of lattice formation is exothermic. There are attractive forces between the
positive and negative ions, so energy is released. ΔfH for KCl is −429 kJ mol−1 (exothermic)
and ΔfH for KCl2 is +757 kJ mol−1 (endothermic); therefore, KCl is the more favoured in terms
of enthalpy changes.
A2.
ΔlattHƟ (NH4Cl) = 705 kJ mol−1
Therefore ΔsolHƟ = (+705) + (−307)+(−381)
= 17 kJ mol−1
A3.
For a reaction to occur spontaneously, the Gibbs free energy must be zero or
negative, and ΔG = ΔH − T ΔS. The process proceeds spontaneously because there
is an increase in entropy as the ordered crystal arrangement changes to a disordered
system in solution. Even though ΔH is positive, the increase in entropy (and,
combined with the temperature, −T ΔS) is sufficiently negative to make ΔG negative.
A4.
ΔH = mc ΔT = 200 g × 4.2 J g−1 K−1 × 16.5 K = 13 860 J = 13.86 kJ
A5.
The molar enthalpy change for ammonium nitrate is +26.5 kJ mol−1, so the number of
moles needed is
76
5 = 0.52 mol
26.5
Practice questions
1a
Mg2+(g) + 2e− + 2Cl(g) (Note: this is the only answer for the top line)
Mg2+(g) + 2e− + Cl2(g)
Mg+(g) + e− + Cl2(g)
Mg(g) + Cl2(g)
1b
642 + 150 + 736 + 2 × 121 = 2 × 364 + 2493 = = +1451 kJ mol−1
1c
ΔH = −ΔH(lattice formation) + ΣΔH(hydration)
= 2493 – 1920 – 2 × 364 = −155 kJ mol−1
1d(i)
Increase in disorder on dissolving or ΔS positive
ΔG is negative or TΔS > ΔH
1d(ii)
Moles of NH4Cl = 2/53.5 = 0.0374
Heat absorbed = 15 × 0.0374 = 0.561
Q = m c ΔT
ΔT = Q/mc = (0.561×1000)/(50×4.2) = 2.6°C
Final temperature = 20 – 2.6 = 17.4°C
2a
ΔHϴ = ΣΔH(formation products) - ΣΔH(formation reactants)
= 3 × −111 – (−1669) = +1336
ΔSϴ = ΣS(products) - ΣS(reactants)
= 2×28 + 3×198 – (51 + 3×6) = +581
ΔGϴ = ΔH – TΔS
= 1336 – (298×581)/1000 = 1163
ΔGϴ is positive (or free energy (G) increases)
2b
When ΔGϴ = 0 or T = ΔHϴ/ΔSϴ
= (1336×1000)/581 = 2299 K
2c
Enthalpy too high or reaction too slow
2d
Method: electrolysis
Conditions: molten or high T or 500–1500°C or dissolved Cryolite
3a
Particles are in the maximum state of order (entropy is zero at 0 K by definition)
1
3b
Ice melts (or water freezes)
3c
The increase in disorder is bigger (at T2)
3d(i)
Moles of water = 1.53/18 = 0.085
Heat change per mole = 3.49/0.085 = 41.1 kJ mol−1
3d(ii)
ΔG = ΔH – TΔS
3d(iii)
ΔH = TΔS or ΔS = ΔH/T
ΔS = 41.1/373 = 0.110 kJ K−1 (mol−1 (or 110 (J K−1 (mol−1))
4a
1s2 2s2 2p6 3s2 3p6
4b
S–(g)
4c
The negative S– ion repels the electron being added
4d(i)
Enthalpy of atomisation of sulfur
4d(ii)
Second ionisation enthalpy of calcium
4d(iii)
Second electron affinity of sulfur
4e
Electron more strongly attracted nearer to the nucleus or attracted by Ca+ ion
4f
+178 + 279 + 590 +1145 – 200 + ΔH –3013 + 482 = 0
ΔH = 539
5a
ΔS = 238 + 189 – 214 – 3 × 131 = –180 J K–1 mol–1
ΔG = ΔH – TΔS
= –49 – 523 × (–180)
1000
= +45.1 kJ mol–1
5b
When ΔG = 0, ΔH = TΔS
Therefore, T = ΔH/ΔS = –49 × 1000/–180 = 272 K
5c
Diagram of a molecule showing O–H bond and two lone pairs on each oxygen
Hydrogen bonding is a strong enough force to hold methanol molecules together in a liquid.
CHAPTER 2
Assignment 1
The temperature of the reaction mixture.
A1.
A pipette and safety filler. The size of the pipette depends on the volume required,
A2.
but 10.0 cm3 is a reasonable choice. Smaller pipettes will be slightly less accurate; larger
pipettes require a larger volume of reaction mixture to allow a sufficient sample to be taken.
Sodium hydroxide solution (concentration depends on initial concentration of ester).
A3.
Alkalis are corrosive. It is best to work with concentrations as low as the experiment
A4.
allows. Eye and hand protection should be used and a safety filler used to fill the pipette.
Should any alkali get onto a person’s skin, it should be washed off immediately with cold
running water.
A5.
A6.
a. At 0.200 mol dm−3, rate = 2.3 mol dm−3 s−1 (note: to calculate rate, time must be
converted from minutes to seconds)
At 0.150 mol dm−3, rate = 1.9 mol dm−3 s−1
At 0.100 mol dm−3, rate = 1.2 mol dm−3 s−1
At 0.050 mol dm−3, rate = 0.7 mol dm−3 s−1
Because tangents are difficult to draw with accuracy, the rates are given to just 2 significant
figures)
b.
Since the graph is a straight line, the order of the reaction with respect to methyl ethanoate
is 1.
c. The gradient of the rate against the ester concentration gives the rate constant k
for the hydrolysis of methyl ethanoate as 4 s−1.
Assignment 2
A light probe in, for example, a colorimeter can accurately record the moment at
A1.
which the dark blue colour appears. Linking the probe to an electronic timer allows the time
for the colour change to be measured accurately, avoiding human error when trying the
judge the endpoint.
Temperature
A2.
The final volume of reaction mixture in all experiments is 100 cm3, so the
A3.
concentration of sodium thiosulfate is 0.005 mol dm−3.
Reaction 1: 2.00 × 10−4 mol dm−3 s−1. Reaction 2: 1.61 × 10−4 mol dm−3 s−1. Reaction
3: 1.22 × 10−4 mol dm−3 s−1. Reaction 4: 0.81 × 10−4 mol dm−3 s−1. Reaction 5: 0.48 ×
10−4 mol dm−3 s−1.
Reaction 1: 0.05 mol dm−3. Reaction 2: 0.04 mol dm−3. Reaction 3: 0.03 mol dm−3.
A4.
Reaction 4: 0.02 mol dm−3. Reaction 5: 0.01 mol dm−3.
A5.
A6.
Straight-line graph, so first order with respect to hydrogen peroxide.
Assignment 3
A1.
a. Changing the concentration of OH− has no effect on rate, so zero order with
respect to OH−. Halving the concentration of C4H9Br halves the rate, so first order with
respect to C4H9Br.
b. Rate = k[C4H9Br]
c. From experiment 1, k = 0.001/2.5 = 0.0004 (or 4 × 10−4) s−1.
d. SN1 since the rate-determining step involves just one species (C4H9Br)
A2.
a.
b.
[OH−] (mol dm−3)
0.241
0.195
0.155
0.118
0.099
Rate (mol dm−3 s−1)
3.1
2.4
1.5
1
0.9
Note: Values for rate are not accurate because of the difficulty in drawing tangents.
c.
d. Since the graph in c is not a straight line, it appears that both 1-bromobutane and
sodium hydroxide affect the rate of reaction and so an SN2 mechanism is likely.
Required practical
P1.
a. Initial rate method
b. i. 10 cm3 pipette, ii. 25 cm3 pipette, iii. dropping pipette, iv. 15 cm3 pipette or,
since these are rare, a 50 cm3 burette.
c. It is simply an indicator to show the presence of iodine.
d. Instead of making a judgement by eye, a light meter can measure the precise
moment when a certain depth of darkness appears.
P2.
a. Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
b. Hydrogen – measure the volume produced.
c. Temperature.
d. Measure the volume of hydrogen produced, either using a gas syringe or by the
downward displacement of water.
e. Plot a graph of the volume of hydrogen produced against time.
If the graph is a straight line, the reaction is zero order and the rate constant is given by its
gradient.
If the graph is a curve, measure gradients at various points to calculate the rate of reaction
at different stages of the reaction. Plot rate against concentration, and if a straight-line graph
is produced the reaction is first order.
Practice questions
1a Initial rate of reaction is proportional to [(CH3)3Br]
1b Initial rate of reaction is not affected by changes in [OH–].
1c Rate = 𝑘 × [(CH3 )3 Br] so either by using a graph plot or by using values directly in
equation:
rate
6.0 × 10−3
𝑘=
=
= 3.0 s –1
[(CH3 )3 Br] 2.0 × 10−3
1d Rate = 𝑘 × [(CH3 )3 Br] = 3.0 × 4.0 × 10−3 = 12 mol dm−3 s−1
2a Table of data
Experiment
Initial [A] /mol
dm–3
Initial [B] /mol
dm–3
1
2
3
4
5
0.020
0.040
0.010
0.060
0.040
0.020
0.040
0.040
0.030
0.035
Initial rate of
reaction /mol dm–
3 s–1
1.2 × 10−4
𝟗. 𝟔 × 𝟏𝟎−𝟒
2.4 × 10−4
𝟖. 𝟏 × 𝟏𝟎−𝟒
7.2 × 10−4
2b 𝑘 =
rate
[A][B]2
=
1.2 × 10−4
0.020 × (0.020)2
= 15.0 mol−2 dm6 s –1
2c
Order with respect to A = second, order with respect to B = first
2d
Order with respect to A = zero, order with respect to B = zero
3a Table of data
Experiment
3.0
6.0
3.0
𝟏. 𝟓
1
2
3
4
3b 𝑘 =
Initial [X] /mol
dm–3
rate
[Y][X]2
×
×
×
×
10−2
10−2
10−2
𝟏𝟎−𝟐
Initial [Y] /mol
dm–3
4.0 × 10−2
4.0 × 10−2
𝟏𝟔. 𝟎 × 𝟏𝟎−𝟐
16.0 × 10−2
Initial rate of
reaction /mol dm–
3 s–1
1.6 × 10−5
𝟔. 𝟒 × 𝟏𝟎−𝟓
6.4 × 10−5
1.6 × 10−5
1.6 × 10−5
= 4.0 × 10−2 × (3.0 × 10−2 )2 = 0.444 mol−2 dm6 s–1
3c Rate constant increased as temperature is increased.
3d Rate constant unchanged as concentration of Y is increased at a fixed temperature.
4a(i)
Expt 2 2.68 ×10−4
Expt 3 10.7(2) ×10−4
Expt 4 2.08 ×10−3
4a(ii)
= 186 mol−1dm3s−1
4b
Increases (exponentially): straight line, not a curve
5a(i)
k = 0 65/(0 15)(0 24)2 = 75.23 to 74.7 mol−2dm6s−1
5a(ii)
0.081 (min sig. figs required)
5b(i)
2
5b(ii)
0
6a(i)
2
6a(ii)
1
6a(iii)
0
6b
k = rate/[D]2[E] or
8 36×10−4/(0.84)2(1.16)
= 1.0(2) × 10−3 to 1.05 × 10−3
mol−2dm6s−1
7a
k = rate/[CH3CH2COOCH3][H+]
or
= 1⋅15×10−4/(0.150)(0.555)
= 1.38×10−3 to 1.4×10−3
mol−1dm3s−1
7b
ans = rate constant × ( ½ × 0.150) × (½ × 0.555)
= rate constant × 0.0208
2.88 × 10−5
8a
Consider experiments 1 and 2:
[B constant], [A] increases × 3: rate increases by 32
Therefore, 2nd order with respect to A
Consider experiments 2 and 3:
[A] increases × 2: rate should increase × 22 but only increases × 2
Therefore, halving [B] halves rate and so 1st order with respect to B
Rate equation: rate = k[A]2[B]
8b
rate = k [C]2[D]; therefore, k = rate / [C]2[D]
k = 7.2 × 10−4/ ((1.9 × 10−2)2(3.5 × 10−2)) = 57.0
mol–2 dm6 s–1
8c
rate = 57.0 × (3.6 × 10–2)2 × 5.4 × 10–2 = 3.99 × 10–3 (mol dm–3 s–1)
8d
Reaction occurs when molecules have E≥Ea
Doubling T causes many more molecules to have this E
Whereas doubling [E] only doubles the number with this E
8e
Ea = RT(lnA – lnk)/1000
Ea = 8.31 × 300 (23.97 – (–5.03))/1000 = 72.3 (kJ mol–1)
CHAPTER 3
Assignment 1
A1.
Only a small amount of product is expected, so the reaction can be considered as
unlikely to occur.
A2.
At the higher temperature, the value of Kp will be higher and the yield of NO(g) will
increase.
A3.
Endothermic because Kp increases with an increase in temperature, where
Kp 
2
PNO
PN 2 PO2
so the yield of NO (right-hand side of equation) increases. Le Chatelier’s principle predicts
that if temperature is increased, a reaction will shift in the endothermic direction.
A4.
At midnight (the start of the graph), the concentration of NO is quite low. As traffic
starts to build up, the concentration of NO increases. Following the production of NO, it is
then converted to NO2 by oxidation by oxygen in the atmosphere. The NO is being removed
as it is formed, so its concentration reaches a maximum value. Also, as the light intensity
increases, the concentration of NO2 decreases as it undergoes a series of photochemical
reactions (one of the products is ‘ground-level ozone’). As the level of traffic decreases later
in the morning, the concentration of NO2 decreases. There is then a slight increase in the
level of NO from increased traffic in the afternoon (but this is removed as it is formed, by the
ground-level ozone present in the atmosphere from the previous reactions). From then the
levels of NO and NO2 decrease.
A5.
Arguments supporting opinion in favour of LEZs: Fossil-fuel-powered vehicles
produce pollution. The photograph shows a brown haze of nitrogen dioxide (which in turn
produces low-level ozone). This can cause breathing problems. Arguments against LEZs:
People may have to drive around cities to carry out their legitimate work, for example taxi
drivers and delivery drivers.
Practice questions
1a
Kp 
 H 2 3 C2 H 2 
CH 4 2
1b
𝐾p =
(C2 H2 )(H2 )3
(CH4 )2
𝐾p units =
kPa × kPa × kPa × kPa
= kPa2
kPa × kPa
Total number of moles = 0.44 + 0.28 + 0.12 = 0.84 moles
𝑃(CH4 ) = 200 ×
0.44
= 104.76 kPa
0.84
𝑃(C2 H2 ) = 200 ×
0.12
= 28.57 kPa
0.84
𝑃(H2 ) = 200 ×
𝐾p =
0.28
= 66.67 kPa
0.84
(28.57)(66.67)3
= 771.45 kPa2
(104.76)2
1c
To right or to product(s) or forwards; Increase
1d
To left or to reagent or backwards; no effect
2a(i)
Increase because higher P gives lower yield or moves to left, equilibrium shifts to reduce P
or equilibrium favours side with fewer moles
2a(ii)
Endothermic because increase T increases yield or moves to right, equilibrium shifts to
reduce T or equilibrium favours endothermic direction
2b(i)
Moles of iodine = 0.023
Moles of HI = 0.172
2b(ii)
2b(iii)
= 0.0179 or 1.79 × 10−2
3a(i)
Moles of C2F2 = 0.40
Moles of HCl = 0.80
3a(ii)
3a(iii)
mol dm−3
3b(i)
Increase
3b(ii)
Decrease
3c
Addition or radical
4a
mol Cl2 = 1.2
total mol = 3.8
4b
mol fraction PCl5 = 1.4/3 8 = 0.368
mol fraction Cl2 = 1.2/3 8 = 0.316
4c(i)
pp = mol fraction × total P
4c(ii)
min: pp PCl5 = 0.368 × 125 = 46(.0) 0.37 × 125 = 46.3
max: pp Cl2 = 0.316 × 125 = 39.47 0.32 × 125 = 40.0
4d
4e(i)
No effect
4e(ii)
Increase
4f
42 62 / 36.9 = 49.2
kPa
CHAPTER 4
Assignment 1
A1.
CH3CH(OH)COOH(aq) + H2O(l) ⇌ CH3CH(OH)COO−(aq) + H3O+(aq)
A2.
CH3CH(OH)COOH(aq) + OH−(aq) ⇌ CH3CH(OH)COO−(aq) + H2O(l)
A3.
Ka = 1.38 × 10−4 mol dm−3
A4.
Lactic acid in venous blood = 0.5 × 10−3 mol dm−3 to 2.2 × 10−3 mol dm−3
[H+] range =
1.38 104  0.5 103 to
1.38 104  2.2 103
2.63 × 10−4 mol dm−3 to 5.51 × 10−4 mol dm−3
The pH range would be from 3.58 to 3.26.
Lactic acid in arterial blood = 0.5 × 10−3 mol dm−3 to 1.6 × 10−3 mol dm−3
[H+] range =
1.38 104  0.5 103 to
1.38 104 1.6 103
2.63 × 10−4 mol dm−3 to 4.70 × 10−4 mol dm−3
The pH range would be from 3.58 to 3.23.
A5.
a. The pH values are much lower than blood pH.
b. Blood is buffered to resist changes in pH when acid is produced in blood.
Assignment 2
A1.
Ideas that may be included: CO2 causes warming by absorbing infrared. CO2 is
released into the atmosphere by burning, respiration and decomposition. The amount of CO2
released has increased rapidly in the last 200 years. CO2 dissolves in the oceans, sets up an
equilibrium and lowers the pH. CO2 reacts and forms HCO3− and CO32−, which react to form
shells of sea creatures and eventually rocks. Carbonates act as a buffer. The resulting
change in pH and the ability of oceans to buffer this must be fully understood. The impact of
climate change on weather and marine life is not fully understood.
Required practical
P1.
Apparatus and instruments for measurement have either a graduation mark or a
scale. Regular checks are needed and, if necessary, adjustments are made using standard
materials (ones whose properties are known to an appropriate level of accuracy). For a pH
meter, the standards used are buffer solutions of known pH. The checks and adjustments
are called calibration.
P2.
a. A pipette has a single graduation mark (though some graduated pipettes have
more than one mark). It is used to measure the required volume of the weak acid into a
beaker because this is a fixed volume that needs to be measured out accurately.
b. A burette is used to measure quantities of the strong base because the volume
needed to react completely with the acid is being measured. Since this is not known, a
burette has many graduation marks and, therefore, allows the volume of solution dispensed
from it to be measured accurately and to the appropriate precision.
P3.
A measured volume of the weak base is placed in a conical flask. A strong acid of
known concentration is added from a burette, with the pH being recorded. The endpoint is
identified from the change in the shape of the pH curve.
We need the equation for the reaction between the weak base and the strong acid, as this
gives the stoichiometry of the reaction.
The number of moles of acid added to reach the endpoint is calculated from the volume of
the acid and its concentration. The equation for the acid–base reaction enables the number
of moles of base to be calculated. Now, both the volume and the number of moles are
known and the concentration of the base can be calculated.
Practice questions
1a(i)
Kw = [H+][OH–]
1a(ii)
2.34 ×10−7
1a(iii)
2.34 ×10−7
1a(iv)
5.48 to 5.50 ×10−14
1b
[H+] = 10−14 / 0.136
= 7.35 ×10−14 OR pOH = 0.87 1
pH = 13.13
2a
Ka =
[H+] = √(1.35 ×10−5 × 0.169)
pH = 2.82
2b(i)
CH3CH2COOH + NaOH → CH3CH2COONa + H2O
or CH3CH2COOH + OH– → CH3CH2COO– + H2O
2b(ii)
mol propanoic acid = 0.250 – 0.015 = 0.235
mol propanoate ions = 0.190 + 0.015 = 0.205
2b(iii)
H  
 
K a   CH3CH 2COOH 
CH3CH 2COO  


(= 1.548 ×10–5)
4.81
3a Rearranging the dissociation constant equation gives
[H + ]2 = 𝐾a × [CH3 CH2 COOH] = 1.35 × 10−5 × 0.550 = 7.43 × 10−6 mol2 dm−6
So [H + ] = 0.002 72 mol dm−3 and pH =– log10[H + ] = 2.56
3b(i) Amount of propanoic acid = 0.0165 mol
3b(ii) Amount of sodium hydroxide = 0.002 30 mol
3b(iii) Amount of propanoic acid in buffer solution = 0.0165 – 0.00230 = 0.0142 mol
3b(iv) pH of buffer − log10[H + ] = − log10 √
4a(i)
Proton donor - alone
1.35 × 10−5 × 0.0142
(0.010 + 0.030)
= 2.66
4a(ii)
Completely dissociated
4b(i)
7.05 ×10−3 × 103/50 = 0.141
4b(ii)
−log [H+] or log 1/[H+]
4b(iii)
0.85
4b(iv)
pH = 1, [H+] = 0.10 (mol dm−3 )
(7.05 ×10−3)/0.10
vol = 7.05 ×10−2 dm3 or 70.5 cm3
4c(i)
4c(ii)
4d(i)
[H+] = 1.66 ×10−4
= 7.22 × 10−5
pKa = 4.14
4d(ii)
Effect = none/ negligible/v small decrease/v small change
Salt or Y− reacts with extra H+ or
equilibrium HY
H+  Y shifts to LHS or
H+ is removed as equilibrium shifts to LHS
Therefore, [H+] or ratio [HY]/[Y-] or ratio [Y-]/[HY] remains almost constant
5a
−log [H+]
4.57 × 10−3
5b(i)
[
5b(ii)
= 1.39 × 10−4
mol dm−3
5b(iii)
pKa = 3.86
5c(i)
1000
30/1000 × 0.480 = 0.0144 or 1.44 ×10−2
5c(ii)
18/1000 × 0.350 = 0.0063 or 6.3 × 10−3
5c(iii)
0.0144 × 2(0.0063) = 1.80 × 10−3
5c(iv)
1.80 × 10-3 × 1000/48 = 0.0375
5c(v)
10−14 / 0.0375 = 2.66 × 10−13
or pOH = 1.426
pH = 12.57
6a(i)
[H+][OH−]1
–log [H+]
6a(ii)
[H+] = [OH−]
6a(iii)
(2.0 × 10−3) × 0.5 = 1.0 × 10−3
6a(iv)
pH = 10.40
6b(i)
[H+] = √(1.35 × 10−5) × 0.125 ( = 1.30 × 10−3)
pH = 2.89
6c(i)
(50.0 × 10−3) × 0.125 = 6.25 × 10−3
6c(ii)
(6.25 × 10−3) – (1.0 × 10−3) = 5.25 × 10−3
6c(iii)
Mol salt formed = 1.0 × 10−3
CH COOH 
 H   Ka  CH
CH CH COO 



3
2
3
2


5.25  10 /V
 1.0  10 /V   7.088  10 
 1.35  105 
3
5
3
pH = 4.15
7a(i)
B
C
A
7a(ii)
Cresolphthalein or thymolphthalein
7b(i)
-log[H+]; 1
7b(ii)
[H+] = 1.259 × 10−12 or pOH = 14 – pH
or = 2.10
= 7.94 × 10−3
7c(i)
Ka = [H+]2/[CH3CH2COOH] or [H+]2/[HA] or [H+] = [A−]
[H+] = √1.35 × 10−5 × 0.117 = 1.257 × 10−3
pH = 2.90
7c(ii)
Ka = [H+] or pKa = pH
pH = 4.87
8a
pH = −log[H+]
[H+] = [A−]
[H+] = √ 1.74 × 10−5 × 0.15
pH = 2.79
8b(i)
Solution which resists change in pH /maintains pH despite the addition of (small amounts of)
acid/base (or dilution)
8b(ii)
CH3COO− + H+ → CH3COOH
8c(i)
= 2.61 × 10−5
pH = 4.58
8c(ii)
Moles H+ added = 10 × 10−3 × 1.0 = 0.01 1
Moles ethanoic acid after addition = 0.15 + 0.01 = 0.16 1
Moles ethanoate ions after addition = 0.10 – 0.01 = 0.09
pH = 4.51
9a
Concentration of acid: m1v1 = m2v2
Hence 25 × m1 = 18.2 × 0.150
or
moles NaOH = 2.73 ×10−3
m1 = 18.2 × 0.150 / 25 = 0.109
9b(i)
Ka = [H+][A−] / [HA]
9b(ii)
pKa = –log Ka
9b(iii)
[A−] = [HA]
Hence Ka = [H+][A−] / [HA] = [H+]
and –log Ka = –log [H+]
9c
Ratio [A−] : [HA] remains constant
Hence as [H+] = Ka [HA] / [A−], [H+] remains constant
10
[H+] = √Ka[HA] = √(1.15×10−4 × 0.5)
=7.58×10−3 mol dm−3
pH = -log10[H+] (or log or lg)
pH = 2.12
11a
Burette, because it can deliver variable volumes
11b
The change in pH is gradual / not rapid at the end point
An indicator would change colour over a range of volumes of sodium hydroxide
11c
[H+] = 10–pH = 1.58 × 10–12
–
Kw = [H+] [OH–]; therefore [OH ] = Kw / [H+]
Therefore, [OH–] = 1 × 10–14/1.58 × 10–12 = 6.33 × 10–3 (mol dm–3)
11d
At this point, [NH3] = [H+]
2
H 
Therefore, Ka   
 NH 4 


[H+] = 10–4.6 = 2.51 × 10–5
Ka = (2.51 × 10–5)2/2 = 3.15 × 10–10 (mol dm–3)
11e
When [NH3] = [NH4+], Ka = [H]
Therefore – log Ka = –log [H+]
Therefore pH = – log10(3.15 × 10–10) = 9.50
12a Ethanedioic acid dissociates more readily than ethanoic acid, partly because there are
two –COOH groups per molecule. Electronegativity measures the tendency of an atom to
attract a bonding pair of electrons. Oxygen atoms are more electronegative than carbon
atoms. –COOH groups tend to attract electrons onto the oxygen atoms, whereas –CH3
groups tend to push electrons away. Therefore ethanedioic acid has a greater tendency to
pull the electrons towards the oxygen atoms within the molecule, meaning that the first
proton more readily dissociates than with ethanoic acid, so ethanedioic acid is a stronger
acid than ethanoic acid.
12b
Moles of NaOH = Moles of HOOCCOO– formed = 6.00 × 10–2
Moles of HOOCCOOH remaining = 1.00 × 10–1 – 6.00 × 10–2
= 4.00 × 10–2
Ka = [H+][A–]/[HA]
[H+] = Ka × [HA]/[A–]
[H+] = 5.89 × 10–2 × (4.00 × 10–2/V)/(6.00 × 10–2/V) = 3.927 × 10–2
pH = –log10(3.927 ×10–2) = 1.406 = 1.41
CHAPTER 5
Assignment 1
A1.
A volumetric flask can measure a specified volume precisely.
A2.
The measured rotation of sample 4 is anomalous. The mass of sugar was not
weighed accurately and was too high, the volume of water was not measured
accurately and was too small, or the polarimeter was not calibrated properly.
A3.
0.5 g dm−3
A4.
Average measured rotation = 3.33°
Specific rotation =
10  3.3275
 66.6
(0.5 1)
A4.
D-(+)-sucrose
A5.
The measured rotations are small and close to the precision of the instrument, which
gives high percentage errors from the equipment. Larger values of measured
rotations would give lower percentage errors. Increasing the concentration of the
sugar solution by increasing the mass of sucrose or decreasing the volume of water
would increase the measured rotation.
A6.
Nine.
A7.
Specific rotation = 66.6° =
10 15
(c 1)
Concentration c = 2.25 g dm−3
Assignment 2
A1.
−102°
A2.
It contains a 50 : 50 ratio of D-(+)-limonene and L-(-)-limonene
A3.
Polarimeter
A4.
IR spectroscopy, 1H NMR, 13C NMR, mass spectroscopy, HPLC
A5.
The enantiomers have identical chemical and physical properties, including spectra.
Assignment 3
A1.
A2.
The type of isomerism is optical isomerism.
A3.
It is a 50 : 50 mixture of D- and L-enantiomers.
A4.
Both contain a chiral centre. For both, the D-isomer is biologically active. They both
have a benzene ring and a carboxylic acid functional group. Parts of the structures
are identical, as shown in the blue boxes.
A5.
In both drugs, the D-isomer is the active compound. In ibuprofen, the L-isomer is not
biologically active but it is not toxic, and so it can be sold as a racemic mixture of
isomers. For naproxen, because the L-isomer is toxic, it is important to only sell the
D-isomer. Naproxen will have to be prepared as a single enantiomer and so it will be
more expensive to prepare than the racemate of ibuprofen.
Practice questions
1a
4 (monochloro isomers)
1b
2
3
The major product exists as a pair of enantiomers
The third isomer is 1-bromobutane (minor product)
Because it is obtained via primary carbocation
4
Compounds/molecules with same structural formula but with bonds/atoms/groups arranged
differently in space or in 3D. Use plane-polarised light, which will be rotated by samples in
opposite directions.
5.
B. Four different atoms or groups attached to a carbon atom
6.
A. CH3CH2CH(OH)CH3
7.
C. Three
8.
B. Rotation of plane-polarised light
9.
B. One compound is optically inactive and the enantiomers of the other compound
are present in a 50 : 50 ratio.
CHAPTER 6
Assignment 1
A1.
Nucleophilic addition
A2.
A3.
A4.
No, it is a racemate because the addition of the cyanide anion can occur equally from
either side of the planar aldehyde to give a 50 : 50 mixture of enantiomers.
A5.
It will be faster if ionic KCN is used (a source of cyanide ion), as hydrogen cyanide is
a very weak acid. The degree of ionisation in solution, producing hydrogen ions (H+)
and the reactive nucleophilic cyanide ions (:CN−), is very small (Ka for HCN = 4.0 ×
10−10 mol dm−3).
A5.
Potassium cyanide is a highly toxic material that causes respiratory failure,
convulsions and death if inhaled or absorbed through the skin. Disposable gloves
should be used and all reactions undertaken in a fume cupboard, where vapours can
be removed. As the reaction involves the addition of acid in a second step, then any
unreacted potassium cyanide will react to give hydrogen cyanide gas, which is also
very toxic and might be accidently inhaled.
Assignment 2
A1.
A2.
This is a reduction or a nucleophilic addition.
A3.
A4.
A5.
Required Practical
P1.
Sample preparation
For each of the solids, add about 0.5 g to a clean test tube and dissolve it in approximately
5 cm3 of deionised water. If the sample does not fully dissolve, warm in a water bath until
dissolved. Label each test tube with a permanent marker in order not to confuse the
samples.
Fehling’s solution
For each sample to be tested, make up a fresh sample of Fehling’s solution by mixing about
1 cm3 of Fehling’s A and Fehling’s B solutions in a clean test tube with gentle swirling. When
transferring the solutions, use a clean pipette for each solution. Check the Fehling’s test
solution is blue in colour. Next add a few drops of the solution of one of the unknown
samples to be tested, swirling the tube gently. Warm the reaction mixture in a water bath (at
approximately 60 °C) for a few minutes. Note any colour change in your lab book. Repeat
the whole process (making fresh Fehling’s test solutions each time) for the other four
samples.
P2.
Compound tested
Outcome of test
No reaction
Solution remains blue
Oxidation state of
copper
Cu(II)
Reaction
Brick-red precipitate
formed
Cu(I)
No reaction
Solution remains blue
Cu(II)
P3.
Reaction
Brick-red precipitate
formed
Cu(I)
No reaction
Solution remains blue
Cu(II)
You can heat the sample by placing the test tube in a bath of hot water.
P4.
a. You could use either Tollens’ reagent or Fehling’s solution. These distinguish
between aldehydes and ketones.
b. You would require two clean test tubes. For Fehling’s solution, you would require
Fehling’s A and B solutions, three clean pipettes and a hot water bath at 60 °C. For Tollens’
reagent, you would require Tollens’ reagent, two pipettes and possibly a water bath at 60 °C
(the reaction may go at room temperature).
c. You would add a few drops of each material to the test solution (Fehling’s or
Tollens) in a clean test tube. Then warm the tube and observe if any colour change occurs.
d. In the Tollens test, the aldehyde would be oxidised to a carboxylic acid and the
Ag(I) reduced to an Ag(0) mirror. This would be observed coating the test tube. The ketone
would not undergo any reaction and the colourless test solution would remain. In the
Fehling’s test, the aldehyde would be oxidised to the carboxylic acid and the blue Cu(II)
complex reduced to an insoluble Cu(I) brick-red precipitate. The ketone would not undergo
any reaction and the blue test solution would remain.
P5.
a. For every 1 mole of AgNO3 we require 2 moles of NH3.
5 cm3 of 1 M AgNO3 solution = 0.005 moles. We therefore require 0.01 moles of NH3.
This is 5 cm3 of a 2 M aqueous NH3 solution.
b. None. The Tollens reaction occurs only with aldehydes. Propanone is a ketone.
c. For every mole of aldehyde oxidised, 2 moles of Ag(0) are formed. Therefore we
would obtain 0.1 mol of Ag(0). The relative atomic mass of silver is 107.9 g per mole, so we
would expect 10.79 g of silver.
Practice questions
1a
CH3CH2COCH3 + 2[H] → CH3CH2CH(OH)CH3
1b
The carbonyl bond is planar in shape, so the [H] has an equal likelihood of attacking from
either side of the molecule. This means that when the product is chiral a racemic mixture will
be formed and so will not rotate the plane of polarised light.
2
b. 2-methylbutan-2-ol
3a
Nucleophilic addition
2-hydroxy-2-methylpentan(e)nitrile
3b
Product from Q is a racemic mixture/ equal amounts of enantiomers
Racemic mixture is inactive or inactive
Product from R is inactive (molecule) or has no chiral centre
3a
CH3CH2CHO + HCN → CH3CH2CH(OH)CN or C2H5CH(OH)CN
2-Hydroxybutanenitrile OR 2-hydroxybutanonitrile
3b
Nucleophilic addition
3c
(Propanone) slower or propanal faster
Inductive effects of alkyl groups
or C of C=O less δ+ in propanone
or alkyl groups in ketone hinder attack
or easier to attack at end of chain
4a
Nucleophilic addition
4b
Planar C=O (bond/group)
Attack (equally likely) from either side
Product is a racemic mixture formed, or a 50:50 mixture of each enantiomer equally likely
5
Nucleophilic addition
6a
Nucleophilic addition
6b(i)
6b(ii)
(Plane) polarized light
Rotated in opposite directions (equally) (only allow if M1 correct or close)
6c
2-Hydroxybutane(-1-)nitrile
6d
Weak acid / (acid) only slightly / partially dissociated/ionised
[CN–] very low
CHAPTER 7
Assignment A1
A1.
A2.
Olive oil contains two esters of oleic acid and one of palmitic acid, so we would need
a ratio of glycerol : oleic acid : palmitic
acid of 1 : 2 : 1.
A3.
If we mix the two carboxylic acids oleic acid and palmitic acid with glycerol,
statistically it is possible to make a range of triglycerides:
A4.
Animal fat contains the saturated fatty acid stearic acid, while olive oil contains the
unsaturated fatty acid oleic acid. Saturated fatty acids have higher melting points than
unsaturated fatty acids because they have better packing in the solid state and greater van
der Waals interactions. Consequently, animal fat has a higher melting point and is a solid at
room temperature.
A5.
Trimyristin is made up of saturated fatty acids so will be a solid at room temperature.
Assignment A2
A1.
A2.
Moles of methyl benzoate = 4/136 = 0.03 moles.
Moles of NaOH (20 cm3 of 2 mol dm−3) = 0.04 moles.
A3.
The condenser was added so that the reaction mixture could boil without the loss of
any of the reagents. Antibumping granules were added to ensure smooth boiling.
A4.
An excess of NaOH was used, so initially the reaction mixture will be basic and red
litmus would turn blue.
A5.
red:
In order to turn the carboxylate salt into the carboxylic acid. Blue litmus should turn
A6.
For efficient recrystallisation, a saturated solution of the compound in the hot solvent
is required. When cooled, the compound will start to crystallise from the saturated solution
as the solubility of the compound in the solvent decreases. If too much hot solvent is used
initially, the compound might not crystallise out of solution, even at room temperature.
A7.
Impure compounds have wide melting point ranges which are lower than that of the
pure compound. We can conclude that although the reaction worked, the recrystallisation did
not provide us with a very pure product.
Assignment A3
A1.
A2.
Second generation
A3.
The amount of carbon (usually carbon dioxide) released is equal to the amount of
carbon dioxide captured in its formation.
A4.
Particulates can cause asthma. Carbon monoxide is a poisonous gas which
interferes with oxygen transport in the blood. Nitrogen oxides can convert oxygen to groundlevel zone, which can affect breathing and the growth of plants. Greenhouse gases result in
warming of the atmosphere and cause climate change. Sulfur oxides are acidic and can
damage buildings and living organisms.
A5.
Land that could be used for growing food could be taken for biodiesel production,
thus increasing world hunger.
A6.
Saturated fatty acid esters will have a high melting point because of the van der
Waals interactions between the chains. Therefore the fuel could freeze in winter.
A7.
Assignment A4
A1.
Moles of salicylic acid used = 6.0/138 = 0.043 moles
Moles of aspirin produced = 6.1/164 = 0.037 moles
% yield = 0.037/0.043 × 100% = 86.0%
A2.
It is cheaper, reacts more slowly and does not evolve HCl.
A3.
% Atom economy = molecular mass of desired product/sum of molecular masses for
all reactants × 100%.
a. Molecule mass: ethanoyl chloride = 78.5, salicyclic acid = 138, aspirin = 164
Atom economy = 164/216.5 = 75.7%
b. Molecule mass: ethanoic anhydride = 102, salicyclic acid = 138, aspirin = 164
Atom economy = 164/240 = 68.3%
The use of ethanoyl chloride has greater atom economy.
A4.
Crush the sample into a powder using a spatula on a filter paper. Add a small amount
of the sample to a melting point capillary by pushing the open end of the capillary into the
powder. Tap the tube so that the powder settles to the bottom of the tube. Make sure that
about 3–4 mm of the tube is filled with the sample.
A5.
The ethanol solution contained both the aspirin and any unreacted impurities, such
as salicylic acid. Aspirin is not very soluble in water at room temperature, whereas salicylic
acid (being a carboxylic acid) is more soluble. Upon cooling in water, the aspirin
recrystallises and the salicylic acid remains in the water.
Required Practical 10b
P1.
So that any pressure build-up can escape to the atmosphere.
P2.
If the apparatus was disassembled or the water supply to the condenser was shut off
then the hot vapours in the flask would escape, which would be a safety hazard.
P3.
The reagents would evaporate and the reaction would fail.
P4.
Pour all the liquid back into the separating funnel, wait for the two layers to separate
and start again.
P5.
a. The liberation of carbon dioxide, as well as the exothermic nature of the acid–base
reaction causing some of the solvent to vaporise.
b. To remove the acidic ethanoic acid from the organic layer by neutralising it with a
base to provide the water-soluble sodium ethanoate salt.
c. The ethyl ethanoate will be in the top (organic) layer.
d. The anhydrous calcium chloride removes any residual water carried over from the
separation using the separating funnel.
P6.
Filtration
P7.
If the stopper is left in the top of the separating funnel and the tap opened, the liquid
does not drain out easily. This is because as the liquid leaves the funnel, a partial vacuum is
formed above the liquid, which stops the liquid draining.
Required Practical 10a
P1.
If the impurities are soluble, the solid compound should be highly soluble in the
solvent at high temperature but relatively insoluble at room temperature. The impurity should
remain soluble at room temperature.
If the impurities are insoluble, the solid compound should be highly soluble in the
solvent at high temperature but relatively insoluble at room temperature. The impurity should
remain insoluble at high temperature.
P2.
For efficient recrystallisation, a saturated solution of the compound in the hot solvent
is required. When cooled, the compound will start to crystallise from the saturated solution
as the solubility of the compound decreases. If too much hot solvent was used initially, the
compound might not crystallise out of solution, even at room temperature.
P3.
a. 10 g of compound A in mixture, solubility at 90 °C = 20 g in 100 cm3 in water. So
10 g will be soluble in 50 cm3.
b. At 20 °C the solubility of A is 0.2 g in 100 cm3, so in 50 cm3 only 0.1 g would
remain in solution.
c. The maximum recoverable amount is 9.9 g (or 99%). There is 1 g of B in the
mixture. The solubility of B is 2.5 g per 100 cm3 of water, so in 50 cm3 a maximum of
1.25 g would remain in solution. As only 1 g of B is present, it will all remain in
solution. Therefore recrystallised A will be pure.
P4.
To allow sufficient time for the heat to transfer from the heating solution through the
capillary to the sample. If the sample is heated too quickly, the heating block will be at a
higher temperature than the sample. In addition, slower heating allows a more accurate
melting range to be reported.
Practice questions
7
1(a)
1(b)
2a
(Nucleophilic) addition-elimination
2b(i)
To ensure the hot solution would be saturated / crystals would form on cooling
2b(ii)
Yield lower if warm / solubility higher if warm
2b(iii)
Air passes through the sample not just round it
2b(iv)
To wash away soluble impurities
2c
Water
Press the sample of crystals between filter papers
2d
Mr product = 135.0
3a
Propan(e)-1,2,3-triol or 1,2,3- propan(e)triol
3b
Soaps
3c(i)
Biodiesel
3c(ii)
3c(iii)
CH3(CH2)12COOCH3 + 21½ O2 → 15CO2 + 15 H2O
4
(Nucleophilic) addition-elimination
N-ethylpropanamide
5a
CH3CH2CH2COOH + CH3CH2OH (or C2H5OH) → CH3CH2CH2COOCH2CH3 + H2O
H2SO4 or HCl or H3PO4
5b
H3CH2CH2CH2OH + (CH3CO)2O → CH3COOCH2CH2CH2CH3 + CH3COOH
5c
(Nucleophilic) addition-elimination
5d
6
Nucleophilic) addition-elimination
N-propylethanamide
7a(i)
3CH3OH
HOCH2CH(OH)CH2OH
7a(ii)
C17H35COOCH3 + 27½ or 55/2 O2 → 19CO2 + 19H2O
8
Electron pair donor or lone pair donor
(Acid) anhydride
9a(i)
9a(ii)
(Nucleophilic) addition-elimination or (nucleophilic) addition followed by elimination
9a(iii)
Any two from: ethanoic anhydride is
• less corrosive
• less vulnerable to hydrolysis
• less dangerous to use,
• less violent/exothermic/vigorous reaction OR more controllable rxn
• does not produce toxic/corrosive/harmful fumes (of HCl) OR does not produce HCl
• less volatile
9b
CHAPTER 8
Assignment 1
A1.
Accept 4-nitromethylbenzene, 1-methyl-4-nitrobenzene or 4-nitrotoluene.
A2.
Electrophilic substitution by nitronium ion (NO2+).
A3.
A4.
Increases it.
A5.
Decreases it.
Assignment 2
A1.
A2.
The electrophile is NO2+:
H2SO4  HNO3
then H2NO3 

H2NO3 

 HSO4 
NO2   H2O
A3.
As the acids are corrosive, rubber gloves must be worn, as well as a lab coat and
safety spectacles. Because concentrated acids fume in air, the reaction should be carried
out in a fume cupboard.
A4.
14 cm3 of liquids and 4.00 g of solid are reacted in the conical flask, which must be
stirred. As a conical flask should only be filled between a third and half full, either a 50 cm3
or 100 cm3 flask would be chosen.
A5.
Both the formation of the nitronium ion and the nitration reaction are exothermic.
Cooling the reagents ensures that the reaction mixture does not boil and the reaction does
not become too violent.
A6.
A Pasteur pipette.
A7.
a. It is important to prepare a hot saturated solution of the product to be
recrystallised. This ensures the maximum yield when the solution cools to room temperature
and the product crystallises out.
b. You can heat the ethanol using a water bath.
A8.
To assess the purity of the recrystallised product.
Assignment 3
A1.
A lone pair acceptor
A2.
A3.
A4.
Reduction reaction
A5.
Another reducing agent that can be used is nickel and hydrogen.
A6.
Number of moles of B = 1.34/134 = 0.1 mol. A 75% yield would give 0.075 mol of
product C.
0.075 mol C = 0.075 × 176 = 1.32 g
A7.
Ethanoyl chloride will react vigorously with water, producing ethanoic acid and
hydrochloric acid gas. Precautions must be taken so that moisture is not allowed into
the reaction flask. Clean, dry apparatus must be used.
A8.
It could be purified by distillation
Practice questions
1
2a
or C6H5NHCOCH3 + NO2+ → C6H4(NHCOCH3)NO2 + H+
2b
Electrophilic substitution
2c
Hydrolysis
2d
Sn/HCl
3a(i)
NO2+
HNO3 + 2H2SO4 →NO2+ + 2HSO4– + H3O+
or HNO3 + H2SO4 →NO2+ + HSO4– + H2O
3a(ii)
Electrophilic substitution
3b
H2/Ni or H2/Pt or Sn/HCl or Fe/HCl, dilute H2SO4
4a
CH3CH2COCl or CH3CH2CClO or propanoyl chloride or (CH3CH2CO)2O or propanoic
anhydride
AlCl3 or FeCl3
CH3CH2COCl + AlCl3 → CH3CH2CO+ + AlCl4–
4b
4c
Tollens or ammoniacal silver nitrate
5a
Benzene is more stable than cyclohexatriene
Expected ΔHϴ hydrogenation of C6H6 is 3(–120) = –360 kJ mol−1
Actual ΔHϴ hydrogenation of benzene is 152 kJ mol−1 (less exothermic) or 152 kJ mol−1
different from expected because of delocalisation or electrons spread out or resonance
5b
Conc HNO3, Conc H2SO4
2 H2SO4 + HNO3 → 2 HSO4– + NO2+ + H3O+
or H2SO4 + HNO3 → HSO4– + NO2+ + H2O
or via two equations:
H2SO4 + HNO3 → HSO4– + H2NO3+
H2NO3+ → NO2+ + H2O
6a(i)
Conc HNO3
Conc H2SO4
6a(ii)
2 H2SO4 + HNO3 → 2 HSO4– + NO2+ + H3O+
or H2SO4 + HNO3 → HSO4– + NO2+ + H2O
Via two equations:
H2SO4 + HNO3 → HSO4– + H2NO3+
H2NO3+ → NO2+ + H2O
6a(iii)
7a
Sn / HCl or Fe / HCl
Equation must use molecular formulae
C6H4N2O4 + 12 [H] → C6H8N2 + 4H2O
7b
H2 (Ni / Pt)
CH2
In benzene 120°
In cyclohexane 109° 28′ or 109½°
7c(i)
Nucleophilic addition
7c(ii)
Planar C=O (bond/group)
Attack (equally likely) from either side
About product: racemic mixture formed or 50:50 mixture or each enantiomer equally likely
CHAPTER 9
Assignment 1
A1.
Nucleophilic substitution
A2.
Reduction reaction
A3.
A4.
Ethanoic anhydride and a base, or ethanoyl chloride and a base.
A5.
The primary amine product is still nucleophilic and can react further, and mixtures of
secondary and tertiary amines and quaternary ammonium salts are inevitably
produced.
A6.
The overall yield of A → B → C → D is [0.91 × 0.87 × 0.96] × 100% = 76%.
The overall yield of E → C → D is [0.65 × 0.96] × 100% = 62.4%.
Despite the second route being shorter, the overall yield is lower so A → B → C → D
would be favoured.
This does not take into account the cost of the reagents and any cost of purification,
and the safety considerations of using toxic KCN. When this is taken into account, the
lower-yielding route might become more attractive.
A7.
Drug molecules must interact with their biological receptors. The binding sites have a
defined three-dimensional shape. H-bonding, ionic bonding or van der Waals
interactions between drugs and their receptors in the binding site are important for
binding. The drug molecule must fit into the binding site and its functional groups must
be displayed so that binding interactions are maximised. Molecule D is similar in size
and shape to benzedrex and has a group capable of H-bonding in a similar position.
Assignment 2
A1.
0–5 °C
A2
a. NaNO2 + HCl → HNO2 + NaCl
b. Nitrous acid and the diazonium salt are unstable and decompose above 5 °C.
c. 4-Methylazobenzene
A3.
The N=N double bond in the chromophore allows the delocalised electrons of one
benzene ring to be extended into the second ring. The delocalised electrons in the
chromophore can absorb certain wavelengths of visible light, and the light not absorbed
gives rise to the observed colour.
A4.
The NH2 group can hydrogen-bond with the surface molecules of the fabric.
A5.
The CO2− group can make an ionic bond with the surface molecules of the fabric.
Ionic bonds are normally much stronger than H-bonds.
A6.
Practice questions
1
3
2a
Dimethylamine
2b
Nucleophilic substitution
2c
Quaternary ammonium salt
(Cationic) surfactant / bactericide / detergent / fabric softener or conditioner / hair conditioner
2d
3a(i)
Concentrated HNO3, concentrated H2SO4
HNO3 + 2 H2SO4 → NO2+ + H3O+ + 2HSO4−
or HNO3 + H2SO4 → NO2+ + H2O + HSO4−
3a(ii)
Electrophilic substitution
3b
Sn or Fe / HCl or Ni / H2
3c(i)
NH3
Use an excess of ammonia
3c(ii)
Nucleophilic substitution
3d
Lone pair on N less available, delocalised into the ring (Q of L)
3e
3f
4a
(Nucleophilic) addition-elimination
N-ethylpropanamide
4b
CH3CN or ethan(e)nitrile or ethanonitrile
Step 1 Cl2
UV or above 300 °C
Step 2 KCN
aqueous and alcoholic (both needed)
Step 3 H2/Ni or LiAlH4 or Na/C2H5OH
5a
Proton acceptor
5b(i)
CH3CH2NH2 + H2O → CH3CH2NH3+ + OH–
5b(ii)
Reaction/equilibrium lies to left or low [OH–] or little OH– formed or little ethylamine has
reacted
5c
Ethylamine
Alkyl group is electron releasing/donating
or alkyl group has (positive) inductive effect that increases electron density on N(H2)
or increased availability of lp
or increases ability of lp (to accept H(+))
5d
CH3CH2NH3Cl
5e
Extra H+ reacts with ethylamine or OH–
or CH3CH2NH2 + H+ → CH3CH2NH3+
or H+ + OH– → H2O
Equilibrium shifts to RHS
or ratio [CH3CH2NH3+]/[ CH3CH2NH2] remains almost constant
6a
(Nucleophilic) addition-elimination
Propanamide
6b
Nucleophilic substitution
Propylamine or propan-1-amine or 1-aminopropane
6c
Electron-rich ring or benzene or pi cloud repels nucleophile/ammonia
7a
Nucleophilic) addition-elimination
N-propylethanamide
7b(i)
Primary
Secondary
Tertiary
7b(ii)
Absorption at 3300–3500 (cm−1) in spectrum
N―H (bond) (only) present in secondary amine or not present in tertiary amine
or
This peak or N―H absorption (only) present in spectrum of secondary amine or not present
in spectrum of tertiary amine
7c(i)
Route A: stage 1
KCN
Aqueous or ethanolic
Route A
Intermediate
CH3CH2CN or propanenitrile
Name alone must be exactly correct to gain
M1 but mark on if name close
correct formula gains M1 (ignore name if
close)
Route A: stage 2
contradiction of name and formula loses mark
H2
LiAlH4
H loses M4 but mark on
Ni or Pt or Pd
Route B
ether
NH3
Excess NH3
7c(ii)
Route A disadv
Toxic /poisonous KCN or
cyanide or CN− or HCN
Expensive LiAlH4
Ignore acidified
OR lower yield because 2
steps
Route B disadv
Further reaction/substitution likely
8
Lone pair on N labelled b more available / more able to be donated than lone pair on N
labelled a
lp or electrons or electron density on N labelled a: delocalized into (benzene) ring
lp or electrons or electron density on N labelled b: methyl/alkyl groups electron releasing or
donating or (positive) inductive effect or push electrons or electron density
9a
Quaternary (alkyl) ammonium salt / bromide
CH3Br or bromomethane
Excess (CH3Br or bromomethane)
9b
Nucleophilic substitution
10a(i)
Ammonia is a nucleophile
Benzene repels nucleophiles
10a(ii)
H2/Ni or H2/Pt or Sn/HCl or Fe/HCl
10a(iii)
Concentrated HNO3
Concentrated H2SO4
HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4–
or using two equations:
HNO3 + H2SO4 → H2NO3+ + HSO4–
H2NO3+ → H2O + NO2+
10a(iv)
Electrophilic substitution
CHAPTER 10
Assignment 1
A1.
a. Corrosive
b. Corrosive
c. Flammable, toxic
A2.
The reaction should be undertaken in a fume cupboard; lab coat, safety spectacles
and disposable gloves should be worn.
A3.
a. 2.2 g of 1,6-diaminohexane (molecular weight = 116) = 2.2/116 = 0.02 moles, in
50 cm3 = 0.02/50 × 1000 = 0.4 M.
b. 1.5 g of decandioyl dichloride (molecular weight = 239) = 1.5/239 = 0.006 moles in
50 cm3 = 0.006/50 × 1000 = 0.12 M.
A4.
Nylon 6,10
A5.
HCl
A6.
1:1 ratio
A7.
During the condensation polymerisation, HCl is produced. The excess 1,6diaminohexane reacts with the HCl, making an ammonium salt.
A8.
Unreacted monomers might be adhering to the fibre. As these are corrosive, they
must be washed off the nylon before it is used or disposed of.
Assignment 2
A1.
A2.
A3.
A4.
Hydrogen bonds
A5.
Both types of polyamides are capable of H-bonding between chains. But the more
rigid nature of the linking aromatic rings in Kevlar® and other aramids compared with the
flexible alkyl groups in nylons makes aramids stronger and tougher.
A6.
Kevlar® because the 1,4 monomers cause the polymer chain to be less zigzag
shaped, so the molecules lie closer together and the intermolecular forces can be more
effective.
Practice questions
1
Can be hydrolysed or can be reacted with/attacked by acid/base/nucleophiles/H2O/OH−
2a
Addition
2b
2c
Q is biodegradable
Polar C=O group or δ+ C in Q (but not in P)
Therefore, can be attacked by nucleophiles (leading to breakdown)
3
Kevlar is biodegradeable but polymerised alkenes are not
Kevlar has polar bonds / is a (poly) amide / has peptide link
Can be hydrolysed/attacked by nucleophiles/acids/bases/enzymes
Polyalkenes are non polar /have non-polar bonds
4
Advantages:




reduces landfill
saves raw materials
lower cost for recycling than making from scratch
reduces CO2 emissions by not being incinerated
Disadvantages:
 difficulty/cost of collecting/sorting/processing
 product not suitable for original purpose, easily contaminated
5a(i)
5a(ii)
In polyamides – H bonding
In polyalkenes – van der Waals forces
Stronger forces (of attraction) in polyamides or H bonding is stronger
6a(i)
As a soap
6a(ii)
Biodiesel or biofuel or fuel for cars/lorries
6a(iii)
Cationic surfactant /detergent /fabric softener /germicide / shampoos /(hair) conditioners
/spermicidal jelly
6b(i)
(Poly)ester
Terylene or PET
6b(ii)
(Poly)amide
Kevlar or nylons
6b(iii)
Hydrogen bonding in b(ii)
Imfs in (b)(ii) are stronger
or
H bonding stronger than dipole-dipole/van der Waals/ dispersion/London forces in b(i)
7a
Methyl propanoate
7b(i)
Pentane-1,5-diol
7b(ii)
7b(iii)
(Base or alkaline) hydrolysis
δ+ C in polyester reacts with OH– or hydroxide ion
8a(i)
Condensation
8a(ii)
Propane-1,3-diol
8b(i)
Addition
8b(ii)
9
Hydrogen bonding
10
d. NH2CH2COCl
11
b. Amide
12
c. There are no gas emissions from a landfill site.
13
a. Poly(ethene)
CHAPTER 11
Assignment 1
A1.
a. Negative electrode
b. Positive electrode
A2.
No movement.
A3.
The structures are:
Alanine: HOOCCH(CH3)NH2
Aspartic acid: HOOCCH(CH2COOH)NH2
Lysine: HOOCCH(CH2CH2CH2CH2 NH2)NH2
A4.
They are chemically and physically the same except that they cause plane polarised light to
rotate in opposite directions.
A5.
Lysine has an extra basic NH2 group, causing it to be the most alkaline. Aspartic acid
has an extra acid COOH group, causing it to be the most acidic.
A6.
a
b
c
A7.
Perform an electrophoresis experiment at pH 6.1. Use a buffer to fix the pH.
At pH 6.1, glycine will be in its zwitterionic form and will not move in an electric field.
At pH 6.1 (above the isoelectric point), cysteine will be protonated on the nitrogen
and move to the negative terminal.
At pH 6.1 (below the isoelectric point), aspartic acid will be deprotonated at the
carboxylic acid group and move to the positive terminal.
Assignment 2
A1.
5 mol dm−3 hydrochloric acid for 24 hours.
A2.
Glycine – no carbon atom is bonded to four different groups.
A3.
Dipeptides = A and C. Tripeptide = B.
A4.
A5.
Assignment 3
A1.
The sequence of α-amino acids linked together in the protein chain
A2.
Yes. The α-helix is elastic and can be stretched but will hold its shape owing to the
H-bonds. If it is stretched too far then the H-bonds will break.
A3.
α-Helix and β-sheet
A4.
The cysteine residues could covalently link adjacent α-helices of α-keratin together.
A5.
When hair is wetted the H-bonding of the α-helices is disrupted, meaning that the
hair can be stretched.
Assignment 4
A1.
A2.
a.
b.
A3.
H-bond
A4.
The active sites of enzymes are stereospecific and the two enantiomers interact with
different enzymes, thereby producing different biological effects.
A5.
The structures of thalidomide and pomalidomide are very similar (differing only in one
amine group). They both might be able fit into the active site of the target enzyme and so
both inhibit it.
A6.
By forming a salt with a carboxylic acid active group in an enzyme
A7.
Owing to the similarity to thalidomide, the patient must not be pregnant and must
agree to use contraception during the treatment.
Assignment 5
A1.
AZT has an N3 group at the 3′-position on the sugar, while 2′-deoxythymidine-5′monophosphate has a hydroxyl group.
A2.
It gets incorporated into the growing DNA strand as it binds to the complementary nucleotide
during DNA replication.
A3.
The N3 group at the end of the growing chain cannot bond with another nucleotide,
so the chain stops growing and the DNA strand cannot finish replicating.
A4.
In high doses AZT can stop human DNA replication as it can also be incorporated
into human DNA, so it will be toxic. However, it is 100 times more toxic to the HIV virus than
to human cells.
Practice questions
1a
Secondary
1b
Nitrogen and oxygen are very electronegative
Therefore, C=O and N–H are polar, which results in the formation of a hydrogen bond
between O and H in which a lone pair of electrons on an oxygen atom is strongly attracted to
the δ+H
2a
A nucleotide is the monomer that makes up DNA. Each nucleotide consists of three
components linked together, a phosphate, a sugar and a base.
2b
2-deoxyribose does not have an alcohol group at the 2-position whereas ribose does.
2c
H-bonds between bases are being broken. A high temperature is required as there are a
large number of H-bonds to break in the DNA strand.
2d
TCTCATGCAA
3a
2-Deoxyribose
3b
Base A:
Top N–H forms hydrogen bonds to lone pair on O of guanine
The lone pair of electrons on N bonds to H–N of guanine
A lone pair of electrons on O bonds to lower H–N of guanine
3c
Either of the nitrogen atoms with a lone pair NOT involved in bonding to cytosine
3d
Use in very small amounts / target the application to the tumour
4a(i)
4a(ii)
4b
4c
5a
5b
Ionic bonding in aminoethanoic acid
Stronger attractions than hydrogen bonding in hydroxyethanoic acid
6
7
(a) Glycine does not contain a chiral C atom i.e 1 with 4 different groups bonded to it and
alanine and phenylalanine do.
(b)(i)
(ii)
(c)(i)
(ii)
(d)
8
Facts (b) and (d) are correct
9
(a) A cation
10
Facts (b) and (d) are correct
CHAPTER 12
Assignment 1
A1.
A2.
Concentrated nitric and concentrated sulfuric acid. Electrophilic substitution
(nitration).
A3.
Phenylamine
A4.
N-methylphenylamine
A5.
CH3Cl in ethanol solvent; heat
A6.
Ethanoic anhydride. It is cheaper, reacts slowly and does not evolve HCl.
A7.
Only ibuprofen contains a carbon atom attached to four different groups and so
exhibits optical isomerism.
A8.
A9.
A10.
Aluminium trichloride and 2-methylpropanoyl chloride.
The electrophile is NO2+.
Practice questions
1a
Diethylamine or ethyl ethanamine or ethyl aminoethane
1b
Three valid routes: A, B, C
Route A
Route B
Route C
F
CH3CH2Br or CH3CH2Cl
C2H6
CH3CH2OH
G
CH3CH2NH2 ethylamine OR
ethanamine OR aminoethane
CH3CH2Br OR
CH3CH2Cl
CH3CH2Br OR
CH3CH2Cl
1c
Route A
Step 1
Step 2
Step 3
Route B
Route C
Reagent(s)
HBr OR HCl
H2/ Ni (Not NaBH4)
H2O & H3PO4 OR H2O &
H2SO4
Mechanism
Electrophilic
addition
addition (allow
electrophilic OR catalytic
but not nucleophilic)
ignore hydrogenation
Electrophilic addition
Reagent(s)
NH3
Cl2 OR Br2
HBr OR KBr & H2SO4 OR
Pcl3 OR Pcl5 OR SOcl2
Mechanism
Nucleophilic
substitution
(free) radical substitution
Nucleophilic substitution
Reagent(s)
CH3CH2Br OR
CH3CH2Cl
CH3CH2NH2 OR NH3 but
penalise excess ammonia
here
CH3CH2NH2 OR NH3 but
penalise excess ammonia
here
Mechanism
Nucleophilic
substitution
Nucleophilic substitution
Nucleophilic substitution
1d
Tertiary amine or triethylamine or (CH3CH2)3N
Quaternary ammonium salt or tetraethylammonium bromide or chloride or ion or
(CH3CH2)4N+ (Br– or Cl–)
Further substitution will take place or diethylamine is a better nucleophile than ethylamine
2a [First part of question 2]
Nucleophilic addition
2b
2-bromobutanenitrile
2c
Reagent: ammonia, NH3
Conditions: excess ammonia, heat
Mechanism: nucleophilic substitution
2d(i)
2d(ii)
The electrostatic attraction between the oppositely charged parts of the ion accounts for the
relatively high melting-point values of amino acids.
2 [Second part of question 2]
Compound L
Compound M
Step 1 NaBH4 or LiAlH4, (nucleophilic) addition
Step 2 conc H2SO4 or conc H3PO4 or Al2O3, elimination
Step 3 HBr, electrophilic addition
3a(i)
3a(ii)
3a(iii)
3b
CHAPTER 13
Assignment 1
A1.
The sample contains paracetamol but it is impure; the impurity is a minor component
of the mixture.
A2.
Silica
A3.
a. Compound B (there are four peaks between 100 and 160 ppm)
b. C=O
c. Compound B because it has an aromatic group, whereas A does not.
A4.
4
A5.
Z. This is the only structure that has two CH3 groups, which will give singlets. The
triplet is from the CH2 next to the carbonyl group; it is coupled to the neighbouring
CH2 group, which is at higher ppm (being next to the electronegative O atom).
Required practical
P1.
0.25
P2.
Ibuprofen
P3.
D contains both paracetamol and ibuprofen. E contains a third painkiller that is not
paracetamol or ibuprofen.
P4.
Paracetamol
P5.
I hour: mainly alcohol, a little aldehyde formed. 2 hours: mainly product aldehyde, a
little alcohol remains.
P6.
After three hours, as pure aldehyde is produced.
P7.
Another product is forming, probably the carboxylic acid.
Practice questions
1a
5
1b
a, singlet QWC
b, triplet QWC
2a
or Si(CH3)4
Inert/non -toxic/volatile or low bp
2b
2
2c(i)
a = quartet or 4
b = triplet or 3
2c(ii)
3230 – 3550 (cm−1)
3a(i)
3a(ii)
3a(iii)
CH2CH2 or two adjacent methylene groups
3a(iv)
or CH3COCH2CH2OCH3
3b(i)
OH in acids or (carboxylic) acid present
3b(ii)
4
IR:
Absorption at 3360 cm–1 shows OH alcohol present
NMR:
There are 4 peaks which indicates 4 different environments of hydrogen
The integration ratio = 1.6 : 0.4 : 1.2 : 2.4
The simplest whole number ratio is 4 : 1 : 3 : 6
The singlet (integ 1) must be caused by H in OH alcohol
The singlet (integ 3) must be due to a CH3 group with no adjacent H
Quartet + triplet suggest CH2CH3 group
Integration 4 and integration 6 indicates two equivalent CH2CH3 groups
5a
Chromatography: GLC, TLC, GC, HPLC
5b
5
320.0 or 322.0
5c
Use of excess air/oxygen or high temperature (over 800 °C) or remove chlorine-containing
compounds before incineration
5d(i)
Si(CH3)4
5d(ii)
3
6a
6b
6c
E
F
CH3CH2COOCH3
CH3COOCH2CH3
6d
6e
7a
J acid amide
K secondary amine or amino
7b
δ = 3.1–3.9
Doublet or duplet
7c(i)
Solvent must be proton-free or CHCl3 has protons or has H or gives a peak
7c(ii)
CDCl3 is polar or CCl4 is non-polar
7d
11
7e(i)
Si(CH3)4 or SiC4H12
7e(ii)
A single number or range within 21-25
7e(iii)
CHAPTER 14
Assignment 1
A1.
The first fuel cells powered a vehicle for only 30 seconds, but now with technological
advances more than 300 miles is possible. The voltage of the fuel cell and the chemical
reactions – which depend upon the scientific principles – are just the same.
A2.
No pollutants of carbon dioxide and nitrogen compounds are produced while the bus
is using its fuel cell. Fuel cells are quiet.
A3.
Air is a mixture of nitrogen and oxygen; when the fuel burns at high temperatures,
these two combine to form nitrogen oxides.
A4.
Fuel cells are nonpolluting when they are being used, and they have no moving
parts, so they are reliable. But there needs to be a store of hydrogen, which is flammable
and has to be pressurised.
Assignment 2
A1.
a. Zn2+(aq) + 2e− ⇌ Zn(s)
b. Mn4+(aq) + 2e− ⇌ Mn2+(aq)
c. ½O2(g) + H+ + 2e− ⇌ OH−(aq)
Reduction is more likely because the equilibrium goes from left to right. Therefore,
the cell potential will be less negative.
A2.
a.
i. Zn(s) | Zn2+(aq) ||
ii. Li+(aq) | Li(s) ||
iii. Pt(s) | H2(g) | H+(aq) ||
b. Under standard conditions this is the standard hydrogen electrode. Its potential is
assigned a value of zero and all other standard electrode potentials are measured relative to
this.
c. Hydrogen pressure = 1 bar, temperature = 298 K, [H+] = 1 mol dm−3.
A3.
a. Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
b. Pt(s) |H2(g) | H+(aq) || Zn2+(aq) | Zn(s)
A4.
a. The zinc/silver cell gives a larger e.m.f. (1.56 V) than the zinc/copper cell (1.10 V).
b.
c. −0.76 + (+0.80) V = 1.56 V
d. It has a higher cell potential. The cell reactants have a lower density.
A5.
a. Silver will be displaced. For zinc to displace silver from aqueous silver solution, the
following two half-reactions must occur:
Zn(s) ⇌ Zn2+(aq) + 2e−, EƟ = −(−0.76) V
Ag+(aq) + e− ⇌ Ag(s), EƟ = +0.80 V
The sum of these two half-reactions is
Zn(s) + Ag+(aq) ⇌ Zn2+(aq) + Ag(s), EƟ = +1.56 V
For this reaction to be spontaneous, EƟ must be positive. Therefore zinc does displace silver
from solution.
b. No reaction. This is the reverse of the reaction in part a. EƟ is negative. For this
reaction to be spontaneous EƟ must be positive, so no reaction occurs.
c. No reaction. For iodine to displace bromine from bromide solution, the following
two half-reactions must occur:
I2(s) + 2e− ⇌ 2l−(aq), EƟ = +0.54 V
2Br−(aq) ⇌ Br2(l) + 2e−, EƟ = −1.07 V
The sum of these two half-reactions is I2(s) + 2Br−(aq) ⇌ 2l−(aq) + Br2(l).
For this reaction to be spontaneous, EƟ must be positive. But (−1.07 V) + (+0.54 V) =
−0.53 V. Therefore, iodine does not displace bromine from solution.
d. This is the reverse of the reaction in part c. EƟ is positive. +0.54 V − (−1.07 V) =
+0.53 V. Therefore, bromine does displace iodine from iodide solution.
Required practical 8
P1.
The metals are cleaned with abrasive paper to remove the layer of metal oxide and
then with propanone to remove any grease coming from handing them
P2.
Stay away from flames and sources of heat, use protective glasses and gloves,
perform the operation in a fume hood
P3.
The salt bridge maintains the charge neutrality in the solutions
P4.
Temperature needs to be 25 degrees Celsius, pressure 1 Bar (100 KPa)
P5.
Take a number of readings at different temperatures e.g. every 10 degrees from 10
to 90 and plot them against the e.m.f.
P6.
Assemble an electrochemical cell consisting of two different metals, A and B, as
described in Required Practical 8. Test each metal of the four in turn against the others in
order to determine the cell with the highest e.m.f. Alternatively, test each metal against a
standard electrochemical cell, in order to determine their standard electrochemical potential.
The two metals with the highest electrochemical potentials will form the cell with the highest
e.m.f.
Practice questions
1a(i)
None or no reaction
E(Zn2+/Zn) more negative than E(Fe2+/Fe)
1a(ii)
Fe2+
Cr3+
E(Fe3+/Fe2+) more positive than E(Cr3+/Cr2+)
1b
E.m.f. = -0.41 – (-0.76) = 0.35
Zn + 2Cr3+ → Zn2+ + 2Cr2+
2a(i) Iron(II), Fe2+
2a(ii) Oxygen difluoride, F2O
2a(iii) Fe2+, Cl–
2b(i) E.m.f. = E (right) – E (left) = 1.52 – 0.77 = 0.75 V
2b(ii) Fe2+  Fe3+ + e–
2b(iii) Decrease Fe3+ concentration to increase e.m.f. of cell. Equilibrium shifts in favour of
more Fe3+, so electrode potential for Fe3+/Fe2+ becomes less positive.
3a
Most powerful reducing agent: Zn
3b(i)
Reducing species: Fe2+
3b(ii)
Oxidising species: Cl2
3c(i)
Standard electrode potential 1.25 V
3c(ii)
Equation: Tl3+ + 2 Fe2+ → 2Fe3+ + Tl+
4a(i)
0.60 V
4a(ii)
H2O + H2SO3 → SO42− + 4H+ + 2e−
4b(i)
2IO3− + 2H+ + 5H2O2 → 5O2 + I2 + 6H2O
4b(ii)
The concentration of the ions change or are no longer standard or the e.m.f. is determined
when no current flows
4b(iii)
Unchanged
4b(iv)
Increased
Equilibrium IO3− /I2 displaced to the right
Electrons more readily accepted or more reduction occurs or electrode becomes more
positive
CHAPTER 15
Assignment 1
A1.
[Cu(H2O)6]2+ ⇌ [Cu(H2O)5(OH)]+ + H+;
[Fe(H2O)6]3+ + H2O ⇌ [Fe(H2O)5(OH)]2+ + H3O+
A2.
These are hydrolysis reactions because the ion reacts with water and releases
hydrogen ions, giving an acidic solution.
A3.
The Fe3+ ion has a higher ionic charge than Cu2+, so the attraction for the lone pairs
on the water molecule is greater and this weakens the O–H bond more.
A4.
The equilibrium for Fe3+ lies further to the right, the solution is more acidic and
therefore the pH is lower for Fe3+ than for Cu2+.
A5.
Any test from the following.
Reaction with aqueous sodium hydroxide gives a blue precipitate of copper
hydroxide:
[Cu(H2O)6]2+ + 2OH− → [Cu(H2O)4(OH)2] + 2H2O
Reaction with aqueous ammonia gives a blue precipitate of copper hydroxide:
[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+
A6.
Fe2+ ions. The green precipitate is caused by the reaction
[Fe(H2O)6]2+ + 2OH− → [Fe(H2O)4(OH)2] + 2H2O
The iron(II) slowly oxidises in air to give a brown precipitate of iron(III) hydroxide.
Assignment 2
A1.
The water is acidic and contains copper ions. In acid conditions the equilibrium
favours [Cu(H2O)6]2+, which is the blue ion, and does not favour the insoluble hydroxo
complex, which would make the water cloudy.
A2.
a. A green–blue precipitate of CuCO3
b. [Cu(H2O)4(OH)2]
c. Experiment 1:
[Cu(H2O)6]2+ + CO32− → CuCO3 + 6H2O
Experiment 2:
[Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2] + 2NH4+
which gives a blue precipitate, and then reaction with more aqueous ammonia to give
[Cu(NH3)4(H2O)2]
d. In low concentrations it is an irritant, and at higher concentrations it is corrosive
and toxic. Use with gloves in a fume cupboard.
A3.
There may be hidden pollution problems, and companies may be liable for
compensation claims in the future.
A4.
Plants growing in different areas will be exposed to different levels of copper ions in
the soil.
A5.
Plant A, because the average amount of copper extracted from the soil is greater.
With Plant A, the average amount of copper is 498 mg per kg of leaves, while the
value for plant B is 67.4 mg per kg of leaves.
A6.
The hydrolysis reactions of transition metal aqua complexes make the soil acidic.
Different levels of metals will give rise to different levels of pH.
A7.
If the post was made from iron, then the hexaaquairon(III) ions in solution in the soil
could undergo hydrolysis,
[Fe(H2O)6]3+ + H2O → [Fe(H2O)5(OH)]2+ + H3O+
so as to acidify the soil. This would produce blue flowers.
A8.
Hydrolysis of the aluminium aqua ion in soft water lowers the pH of the water. Copper
is more soluble at low pH values: the equilibrium [Cu(H2O)6]2+ ⇌ [Cu(H2O)5(OH)]+ +
H+ lies further to the left. In tap water, the copper comes from the copper pipes.
Required practical
P1.
At low concentration it is an irritant; at a higher concentration (>2 M) it is corrosive.
P2.
Tap water will contain dissolved metal ions, which can affect the analysis.
P3.
A = FeSO4
[Fe(H2O)6]2+ + 2OH− → [Fe(H2O)4(OH)2] + 2H2O
B = CuCl2
[Cu(H2O)6]2+ + 2OH− → [Cu(H2O)4(OH)2] + 2H2O
C = Fe(NO3)3
[Fe(H2O)6]3+ + 3OH− → [Fe(H2O)3(OH)3] + 3H2O
P4.
Experiment a suggests the salt contains sulfate ions:
Ba2+(aq) + SO42−(aq) → BaSO4(s)
Experiment b suggests the salt contains Cr(III) ions as CO2 is produced. Metal(III)
complexes are more acidic than metal(II) complexes and enough acid is liberated to
turn carbonate into CO2 and water:
2H3O+ + CO32− → CO2 + 3H2O
The precipitate is [Cr(H2O)3(OH)3]:
2[Cr(H2O)6]3+ + 3CO32− → 2[Cr(H2O)3(OH)3] + 3CO2 + 3H2O
Therefore the compound is Cr2(SO4)3.
Practice questions
1a(i)
MgO: ionic
P4O10: covalent
1a(ii)
Electronegativity difference small or electronegativities similar or big difference in
electronegativity leads to ionic bonding
1b
Na2O + H2O → 2Na+ + 2OH− (or 2NaOH)
SO2 + H2O → H2SO3
1c
MgO + 2HCl → MgCl2 + H2O (or MgO + 2H+ → Mg2+ + H2O)
1d
P4O10 + 12NaOH → 4Na3PO4 +6H2O (or P4O10 + 12OH− → 4PO43− +6H2O)
2
Na2O: vigorous or violent or exothermic reaction; or forms a colourless solution
pH of solution formed = 13 or 14
Na2O + H2O → 2NaOH
P4O10 or P2O5: vigorous or violent or exothermic reaction; or forms a
Colourless solution; 1
pH of solution formed = 0 or 1
P4O10 + 6H2O → 4H3PO4
3 From sodium on the left-hand side of the Periodic Table, to sulfur on the right, the trend of
the pH in solutions of the oxides is to decrease. Oxides on the left of the Period are alkaline,
oxides on the right are acidic.
Sodium oxide is ionic and contains the oxide ion, O2–, that readily combines with hydrogen
ions.
Na2O + H2O  2NaOH
Magnesium oxide is ionic but reacts less strongly with water than sodium oxide.
Aluminium oxide is amphoteric, so can react as either an acid or a base. It is ionic but the
oxide ions are too strongly held in the lattice to react with water.
Silicon dioxide has a giant covalent structure which is difficult to break apart, and so does
not react with water.
Oxides of phosphorus are covalent and exist in different forms, they form weak acids in
solution. For example
P4O10 + 6H2O  4H3PO4
Oxides of sulfur are covalent and exist in different forms: sulfur dioxide and sulfur trioxide.
Sulfur dioxide forms a weak acid but sulfur trioxide forms a strong acid
SO3 + H2O  H2SO4
4a(i) Na2O + H2O  2NaOH
pH 14
4a(ii) SO2 + H2O  H2SO3
But the main species present in solution is hydrated sulfur dioxide SO2.xH2O
pH approximately 1
4b Generally, ionic bonded oxides are basic, so pH is high. Covalent bonded oxides are
acidic, so pH is low.
5a
MgO (is a white solid that) forms a suspension (or slightly soluble)
MgO + H2O → Mg(OH)2 (or → Mg2+ + 2OH−)
pH is 8 to 10
SO2 dissolves (or forms (colourless) solution)
SO2 + H2O → H2SO3 (or → H+ + HSO3−) (or → 2H+ + SO32−)
pH is 1 to 4
5b
Al(OH)3 + OH− → Al(OH)4 (or forms Al(OH)63−)
Al(OH)3 + 3H+ + 3H2O → Al(H2O)63+ (or forms [Al(H2O)5(OH)]2+, Al3+, AlCl3
6(i)
Green solution (not blue-green or grey-green)
[Cr(OH)6]3− (or Cr(H2O)(OH)5]2− or Cr(H2O)2(OH)4]−)
6(ii)
Green precipitate
Bubbles (or gas or fizzing or effervescence)
Cr(H2O)3(OH)3 (or Cr(OH)3)
7a
The number of protons increases (across the period) / nuclear charge increases
Therefore, the attraction between the nucleus and electrons increases
7b
S8 molecules are bigger than P4 molecules
Therefore, van der Waals / dispersion / London forces between molecules are stronger in
sulfur
7c
Sodium oxide contains O2– ions
These O2– ions react with water forming OH– ions
7d
P4O10 + 12OH– → 4PO43– + 6H2O
8a
An electron pair on the ligand
Is donated from the ligand to the central metal ion
8b
Blue precipitate
Dissolves to give a dark blue solution
[Cu(H2O)6]2+ + 2NH3 → Cu(H2O)4(OH)2 + 2NH4+
Cu(H2O)4(OH)2 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2OH– + 2H2O
9a
MgO is ionic
Melt it
Molten oxide conducts electricity
9b
Macromolecular
Covalent bonding
Water cannot (supply enough energy to) break the covalent bonds / lattice
9c
(Phosphorus pentoxide’s melting point is) lower
Molecular with covalent bonding
Weak / easily broken / not much energy to break intermolecular forces or weak vdW / dipoledipole forces of attraction between molecules
9d
Reagent (water or acid)
Equation e.g. MgO + 2HCl → MgCl2 + H2O
9e
12NaOH + P4O10  4Na3PO4 + 6H2O
10a
Yellow / purple (solution)
Brown precipitate / solid
[Fe(H2O)6]3+ + 3OH– → Fe(H2O)3(OH)3 + 3H2O
10b
Blue (solution)
Dark / deep blue solution
[Cu(H2O)6]2+ + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 4H2O
10c
Colourless (solution)
White precipitate / solid
Bubbles / effervescence / gas evolved / given off
2[Al(H2O)6]3+ + 3CO32– → 2Al(H2O)3(OH)3 + 3CO2 + 3H2O
11
Moles NaOH = 0.0212 × 0.5 = 0.0106
Moles of H3PO4 = 1/3 moles of NaOH (= 0.00353)
Moles of P in 25000 l = 0.00353 × 106 = 3.53× 103
Moles of P4O10 = 3.53 × 103/4
Mass of P4O10 = 3.53 × 103/4 × 284 = 0.251 × 106 g = 251 kg
CHAPTER 16
Assignment 1
A1.
Since the concentration is only given to one significant figure, precise measurements
are not needed. Calculate (a) the mass of 1 mol of CuCl2.2H2O(s) and (b) the number of
moles to be dissolved in 10 cm3 to make 0.1 mol dm−3 of solution. Weigh out the calculated
quantity and dissolve it in 10 cm3 of deionised water. Note: this is the preparation of a
reagent solution, not a standard solution for volumetric analysis.
A2.
Both are corrosive and can cause burns. In both case the bottles should be stoppered when
not in use. They should be stored in separate places.
A3.
Octahedral aqua complex
A4.
a. Water ligands are displaced by chloride ion ligands. The green colour is caused by a
mixture of blue [Cu(H2O)6]2+ and yellow CuCl42− complex ions, and when there is little or no
[Cu(H2O)6]2+ remaining the solution is yellow.
b. Similar to a. except that [Co(H2O)6]2+ is pink and CoCl42− is blue.
c. Successive replacement of water ligands in [Cu(H2O)6]2+ by ammonia to give a mixture of
ammine complex ions; the most abundant one in concentrated ammonia solution is [Cu(NH 3)5H2O]2+.
d. Successive replacement of water ligands in [Co(H2O)6]2+ by ammonia to give [Co(NH3)6]2+.
Assignment 2
A1.
60 s, c = 0.049 mol dm−3; 120 s, c = 0.101 mol dm−3; 180 s, c = 0.169 mol dm−3;
240 s, c = 0.223 mol dm−3; 300 s, c = 0.261 mol dm−3
A2.
A3.
This is given by the slope of the graph, 9.1 × 10−4 mol dm−3 s−1.
A4.
0.22 = 9.1 × 10−4 t; t = 242 s
A5.
kr = 0.44/9.1 × 10−4 = 483 s−1
Assignment 3
A1.
MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
A2.
a. Tablet 1b titre 17.1 cm3, therefore number of moles of MnO4− = (0.02 × 17.1)/1000
= 3.42 × 10−4
b. 0.00171 mol of Fe2+
c. Mass = 56 × 0.00171 = 95.76 mg
d. 95.76/845 = 0.113, or 11.3% w/w
A3.
The expression is percentage of iron = titre (cm3) × 0.56/mass of the tablet (g).
Tablet 1a: 17.4 × 0.56/0.850 = 11.5%
Tablet 2a: 16.5 × 0.56/0.845 = 10.9%
Tablet 2b: 16.9 × 0.56/0.855 = 11.1%
A4.
Only batch 1
A5.
a. ±0.135%
b. The same apparatus, equipment and method are used for all tablets.
A6.
By taking more sample tablets from a batch and calculating the mean.
Assignment 4
A1.
The unused feedstocks can be recycled to save costs and are cracked because the
amounts in the feedstocks will not necessarily match the demand for the products.
A2.
A build-up of carbon will reduce the surface area of the catalyst available for reaction.
Burning off the carbon will regenerate the catalyst.
A3.
To expose the maximum amount of catalyst to the reactants.
A4.
Catalytic converters remove hydrocarbons, carbon monoxide, sulfur oxides and
nitrogen oxides. They do not remove carbon dioxide, a greenhouse gas.
A5.
The use of nonrenewable resources, dust hazards, health risks to miners and
damage to the landscape due to mining.
A6.
A catalyst in a different phase from that of the reactants is considered
heterogeneous.
A7.
To provide a greater surface area for the reactants.
A8.
The presence of double bonds increases the distance between adjacent molecules
and this reduces the van der Waals forces of attraction between the molecules, reducing the
melting point.
A9.
Animal fats contain a greater proportion of saturated fats, which, it is claimed,
contribute to heart disease.
Practice questions
1a(i)
SO2 + V2O5 → SO3 + V2O4
V2O4 + 1/2O2 → V2O5
V(IV) or 4 and V(V) or 5
1a(ii)
MnO4– + 8H+ + 4Mn2+ → 5Mn3+ + 4H2O
2Mn3+ + C2O42– → 2Mn2+ + 2CO2
Mn(III) or 3 and Mn(II) or 2
1b
[Co(NH3)6]2+ formed
Complex easy to oxidise
H2O2 (or air or oxygen)
1c
Moles of dichromate = (29.2/1000)×0.04 = 0.001168
Moles of Q2+ = (25/1000)×0.140 = 0.00350
Each mole of dichromate needs 6 electrons or half equation with 6 e−
Moles of electrons = 6×0.001168 = 0.007008 or moles Q2+:moles dichromate = 3:1
Moles of electrons per mole of Q = 0.007008/0.0035 = 2.002 = 2
Q(IV) or Q4+
2a(i)
Fe + 2HCl → FeCl2 + H2
or Fe + 2H+ → Fe2+ + H2
2a(ii)
PV = nRT, n = PV/RT
= 4.53 × 10−3 mol
2a(iii)
Moles of iron = 4.53 × 10−3 mol (or = 4.25 × 10−3)
Mass of iron = 4.53 × 10−3 × 55.8 = 0.253 g
2a(iv)
0.253 × 100/0.263 = 96.1%
2b(i)
Fe2+ → Fe3+ + e−
Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O
Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 7H2O + 6Fe3+
2b(ii)
Moles of dichromate = moles Fe2+/6
= 4.53 × 10−3/6 = 7.55 × 10−4
Volume of dichromate = moles/concentration
= (7.55 × 10–4 × 1000)/0.0200
V = 37.75 cm3
2b(iii)
KMnO4 will also oxidise (or react with) Cl– (or chloride or HCl)
3a
Initially slow because reaction is between two negative ions (or between two negative
reactants or two negative species) which repel each other.
Then Mn2+ (or Mn3+) ions are formed acting as an autocatalyst (or Mn2+ ions formed in the
reaction act as a catalyst)
2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 8H2O + 4CO2
MnO4− + 4Mn2+ + 8H+ → 5Mn3+ + 4H2O
C2O42− + 2Mn3+ → 2Mn2+ + 2CO2
3b
Active sites are where reactants are adsorbed onto a catalyst surface (or bind or react on a
catalyst surface).
Number of active sites increases if the surface area is increased (or catalyst is spread thinly
or on honeycomb or powdered or decreased particle size.
Active sites blocked by another species or poison (or species adsorbed more strongly or
species adsorbed irreversibly or species not desorbed).
4a
Effect on reaction rate: catalyst provides an alternative reaction route with a lower Ea, more
molecules able to react or rate increased
Equilibrium: forward and backward rates changes by the same amount ; hence
concentration of reactants and products constant or yield unchanged
4b
Heterogeneous: catalyst in a different phase or state to that of the reactants
Active site: place where reactants adsorbed or attached or bond; where reaction occurs
Reasons: large surface area; reduce cost or amount of catalyst
Catalyst poison: lead adsorbed; lead not desorbed or site blocked
4c
Reaction slow as both ions negatively charged or ions repel
2Fe2+ + S2O82− → 2Fe3+ + 2SO42–
2Fe3+ + 2I− → 2Fe2+ + I2
5a
3d7
5b
[Co(H2O)6]2+
Pink
5c(i)
[Co(NH3)6]2+
Pale brown or straw
5c(ii)
[Co(H2O)6]2+ + 6NH3 → [Co(NH3)6]2+ + 6H2O
5d
[Co(NH3)6]3+
An oxidising agent
6a
Iron
Heterogeneous; catalyst in a different phase from that of the reactants
Poison; a sulphur compound
Poison strongly adsorbed onto active sites / blocked
Poison not desorbed or reactants not adsorbed or catalyst surface area reduced
6b
Pale green solution
Green precipitate formed
Insoluble in excess ammonia
Equation:
e.g. [Fe(H2O)6]2+ + 2NH3 → [Fe(H2O)4(OH)2] + 2NH4+