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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 1
Exercise 1.1
An atom contains a dense nucleus surrounded by
shells of electrons. The nucleus contains the nucleons
(protons and neutrons). Protons are positively charged,
electrons have a negative charge and neutrons are
uncharged. The protons and neutrons have the same
relative mass. The mass of an electron is negligible.
d
The nucleus is extremely small so very few
particles get close enough to it to be repelled
completely.
e
The neutrons would go straight through the
foil because they are uncharged. They are not
deflected by the positive charge in the nucleus.
Exercise 1.2
f
Atoms of the isotope with more neutrons have
a greater mass. So since density = mass/volume,
the density of heavier isotopes is greater.
1 and C; 2 and B; 3 and D; 4 and A
Exercise 1.3
a
A 36, B 49, C 66, D 24, E 24, F 26
b
i
c
There is a mixture of isotopes present.
(Relative atomic mass is the weighted
mean of these isotopes.)
Exam-style questions
Question 1
a
ii
52
There are 3 protons, which are positively
charged.
iii
There are two more protons than
electrons in the strontium ion.
There are 4 neutrons, which have no
charge.
i
13 protons, 13 electrons, 14 neutrons
ii
55 protons, 54 electrons, 78 neutrons
There are 3 electrons, which are negatively
charged.
[1]
iii
8 protons, 10 electrons, 9 neutrons
Exercise 1.4
a
b
c
4 2+
2 He
Most of the atom is empty space because the
nucleus is extremely small. The electrons have
a much smaller mass than the alpha-particles
so do not cause a change in their momentum
if they do collide.
The positive charge on the nucleus (protons)
repels the positive charge on the alpha-particle.
b
[1]
Protons and neutron have a relative
mass of 1.
[1]
Electrons have hardly any mass / mass of
about 1/2000 that of a proton. [1]
The electron configuration is the same/
there is the same number of outer shell
electrons.
[1]
c
7 Li+ (1 mark for 7 and 3 in correct places,
3
1 mark for Li+)
[2]
d
It has 2 electrons and 3 protons / it has
one more proton than electrons.
1
[1]
[1]
[Total: 9]
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Question 2
c
a
Nickel, because it has one more proton than
cobalt / nickel has 28 protons and cobalt
has 27
[1]
b
Relative atomic mass depends on the
weighted mean
[1]
of the isotopes / idea of isotopes.
[1]
Cobalt has greater proportion of (isotopes)
with higher mass / nickel has greater
proportion (of isotopes) with lower mass. [1]
d
i
[1]
ii
Cobalt because the number of electrons
equals the number of protons.
[1]
iii
Co3+
[1]
The beam is deflected
[1]
towards the positive plate / away from the
negative plate.
[1]
Electrons are negatively charged / like
charges repel / unlike charges attract.
[1]
2
Cobalt-59 because it has 32 neutrons
whereas nickel-58 has 30 neutrons.
[Total: 10]
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may be different.
Workbook answers
Chapter 2
Exercise 2.1
a
A 2; B 1s; C 2; D 18; E 3; F 3s, 3p, 3d;
b
i
1s22s22p5
ii
Si
iii
1s22s22p63s23p64s13d5
iv
1s22s22p6
v
19
vi
Br−
f
H
g
Similar points further on in the curve have
lower values, e.g. I is lower than A, J is lower
than B, K is lower than C, etc.
Exercise 2.3
a
vii 1s22s22p63s23p63d2
c
i
Group 2
ii
Group 15
iii
Group 17
iv
Group 18
b
Exercise 2.2
a
b
c
1
Across a period, there is a general increase
in the value of IE1. This is because of the
increase in nuclear charge. Across a period
the electrons are added to the same principal
quantum shell so the attractive forces between
the nucleus and the outer electrons increases
gradually. So the first ionisation energy
increases gradually. Across a period there is
not much difference in shielding because there
are the same number of inner shell electrons.
i
There is a sudden jump in ionisation
energy when the second electron has been
removed (because that is in a shell closer
to the nucleus).
ii
Removal of electrons 10 and 11 takes
much more energy than the others
(because they are nearer the nucleus).
iii
There is a gradual change as electrons 2 to
9 are removed / there are no sudden jumps
in the ionisation energy on removal of
successive electrons from 2 to 9.
i
Ca 3+ ( g ) → Ca 4+ ( g ) + e −
ii
P + ( g ) → P2+ ( g ) + e −
Exercise 2.4
a
b
i
They are higher up on the diagram.
ii
The arrows are shown pointing in the
opposite direction.
i
ii 3p
iii 3p
3s
3s
A and I (ionisation energy increases across a
period to a maximum at Group 18)
2p
2p
2p
2s
2s
2s
i
A and B or I and J
1s
1s
1s
ii
F and G
iii
C and D or K and L
d
C and K
e
E
c
Electrons in the same orbital repel each other
more than electrons in separate orbitals. The
electron removed from P and S are from the
same 3p energy level. The 3 outer p electrons
in P are in separate 3p orbitals to minimise the
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repulsion between the electrons. There are 4
outer p electrons in S. So two of the electrons
have to go into the same orbital. The electron
removed from sulfur is from a 3p orbital,
which contains a pair of electrons in the same
orbital. The extra repulsion between this pair
of electrons makes the atom more unstable so
it is easier to remove an electron. This effect
is greater that the effect of the greater nuclear
charge of S compared with P.
This draws the outer electrons (in the ion)
nearer the nucleus.
[1]
d
i
Boron has lower first ionisation energy
than beryllium because the 5th electron
goes into a p subshell
[1]
which is further from the nucleus
[1]
so attractive force on nucleus on electron
is less.
[1]
a
The principal energy levels
Boron has lower first ionisation energy
than carbon because it has a lower
nuclear charge
[1]
b
The spacing gets smaller / the lines get closer
together.
so attractive forces between the nucleus
and outer electrons greater.
[1]
c
Electrons falling from higher to lower energy
levels
d
A
Similar amount of shielding since outer
electrons are in the same principal
quantum shell.
[1]
e
I
f
The energy levels get closer.
g
H
h
It takes more energy to remove the electron
from hydrogen as the electron is at a lower
energy level closer to the attractive force of the
nucleus / it takes less energy to remove the outer
electron from a lithium atom since the outer
electron is further from the attractive force of
the nucleus.
Exercise 2.5
ii
e
a
b
c
i
1s22s22p1
[1]
ii
1s22s22p63s23p63d104s24p1
[1]
iii
1s22s22p63s23p63d10
[1]
No / gallium ion is smaller (0 mark on own)
because the outer electrons in the atom are
further from the nucleus / the outer electrons
in the atom are not pulled in by the nuclear
charge as much.
[1]
Gallium. When a gallium ion is formed from
a gallium atom the outer quantum shell of
electrons is removed. [1]
[1]
gas state symbols
[1]
[1]
s is spherical shape
p is propeller / dumbbell shape (there
should be no electron density in the centre) [1]
[Total: 17]
Question 2
a
Exam-style questions
Question 1
B ( g ) → B+ ( g ) + e − Symbols
i
1s22s22p63s23p6
[1]
ii
1s22s22p63s23p64s13d5
[1]
iii
1s22s22p63s23p63d3
[1]
b
Argon
c
i
[1]
The electrons are being removed from
the same principal quantum shell.
[1]
There is a gradual increase in nuclear
charge
[1]
so attractive force between the nucleus
and electrons gradually increases.
[1]
ii
The 8th electron is in a quantum shell
nearer to the nucleus.
[1]
Attractive forces between the nucleus
and the outer electrons are higher.
[1]
Very little shielding.
[1]
So there is a greater force of attraction between
the nucleus and the outer electrons.
[1]
2
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iii
F2 + ( g ) → F3+ ( g ) + e − Symbols
[1]
gas state symbols
[1]
Correct levels and correct number of orbitals
filled / partly filled
[1]
[Total: 12]
Question 3
a
b
The energy needed to remove one electron
from each atom in one mole of an atoms
to form one mole of ions [1]
atoms and ions in gaseous state.
[1]
B and C
[1]
correct number of electrons in each
subshell [1]
correct spin
[1]
[Total: 18]
Question 4
a
Atomic radius decreases across a period.
[1]
The outer electrons are in the same principal
quantum shell but the nuclear charge
increases across a period.
[1]
Because there is a large jump in ionisation
energy between the first and second electrons
being removed
[1]
So there is a greater force of attraction
between outer electrons and the nucleus.
[1]
c
B
[1]
d
A
[1]
Decreases between groups 1 and 14 / decrease
across the group for positive ions.
[1]
Because there is a large jump in ionisation
energy between the third and fourth
electrons being removed
[1]
Large increase between groups 14 and 17 /
negative ions larger than positive ions in the
same period.
[1]
[1]
Decreases between groups 14 and 17 / decrease
across the group for negative ions.
[1]
So there are 3 electrons in the outer shell
Group 13 elements form stable ions by loss
of 3 electrons.
[1]
e
f
g
Accept values between 9000 and 11 000.
[1]
Element is in Group 2.
[1]
3rd, 4th and 5th ionisation energies are
increasing steadily (so must have at least
8 electrons in this shell).
[1]
i
1s 2s 2p 3s 3p 4s 3d [1]
ii
V + ( g ) → V 2 + ( g ) + e − Symbols
[1]
gas state symbols
[1]
3p
2
2
6
2
6
2
3
b
c
Down the group, there is one extra shell of
electrons from Cl to Br to I.
[1]
Greater number of inner shells means greater
amount of shielding.
[1]
This is greater than the effect of increased
nuclear charge.
[1]
d
i
ii
Atom (or molecule) having one (or
more) unpaired electrons
[1]
1s22s22p63s23p5
[1]
(same as a Cl atom)
[Total: 11]
3s
2p
2s
1s
3
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 3
Exercise 3.1
We compare the mass of atoms using the unified
atomic mass unit. This is defined as one twelfth of the
mass of an atom of carbon-12. Relative atomic mass is
the weighted average mass of atoms in a given sample.
The number of atoms in exactly 12 grams of the
isotope carbon-12 is called The Avogadro constant.
Its value is 6.02 × 1023 mol-1. So there are
6.02 × 1023 atoms of carbon in 12 g of the
carbon-12 isotope. A mole is the amount of
substance that has the same number of defined
particles as there are in exactly 6.02 × 1023 atoms
of carbon in 12 grams of the carbon-12 isotope.
Relative molecular mass is the weighted average
mass of a molecule to the unified atomic mass
unit. Relative molecular mass is found by adding
together the relative atomic masses of all the atoms
in the molecule. For ionic compounds we use the
term relative formula mass.
e
i
(NH4)2SO4
ii
Zn(NO3)2
iii
Ag3PO4
iv
Ca(OH)2
i
NiSO4•7H2O
ii
anhydrous
iii
water of crystallisation
iv
Heat the green nickel sulfate
Exercise 3.3
i
(3 × 207.2) + (4 × 16.0) = 685.6
ii
41.12
= 0.0600 mol (to 3 significant
685.6
figures)
Exercise 3.2
iii
0.0600 × 3 = 0.180 mol
a
iv
Mass = mol × molar mass = 0.180 × 207.2
= 37.3 g
35.61
= 0.300 mol
Mol Sn =
118.7
b
c
1
d
a
i
HCO3−
ii
OH−
iii
SO42−
iv
Ag+
i
ammonium
ii
phosphate
iii
nitrate
ii
Ratio of moles = Sn = 1, Cl = 2, SnCl4 = 1
iv
carbonate
iii
Equation: Sn + 2Cl2 → SnCl4
i
CO2
ii
MgO
iii
Ca3N2
iv
Al2S3
b
i
Mol Cl =
42.60
= 0.600 mol
71
Mol tin chloride =
78.21
= 0.300 mol
260.7
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Exercise 3.4
Exercise 3.6
a
a
i
Not the simplest ratio. The simplest ratio
is C3H5Cl.
ii
empirical formula mass of PNCl2 = 31 +
14 + 2 (35.5) = 116
=
mol
= 0.25
concentration 0.05
= 5 dm3 = 5000 cm3
5.4
U. 5.4 g NaOH =
mol = 0.135 mol.
40
0.135
Concentration =
= 0.90 mol dm−3
0.15
T.
multiply empirical formula mass by
3 = P3N3Cl6
i
Mass of oxygen = 19.78 − 14.98 = 4.80 g
ii
14.98
= 0.20 mol
74.9
4.8
= 0.30 mol
Mol O =
16
Simplest ratio = 2 As to 3 O so empirical
formula is As2O3
iii
c
Mol As =
b
i
ii
Molar mass
= 395.6 = 2.
empirical formula mass 197.8
iii
So molecular formula is As4O6
d
A. Moles of gas =
c
i
80
= 3.30 × 10−3
24000
B. Mass = mol × Mr = 3.3 × 10 × 34
= 0.112 g
mass
8
= 0.125 mol
=
Mr
64.1
E. Volume in dm3 = mol × 24 = 0.150 × 24
= 3.6 dm3 = 3600 cm3
D. Mol =
b
2
F.
Mass = mol × Mr = 0.150 × 32.0 = 4.80 g
i
20 cm3
ii
40 cm3 (because 60 cm3 total − 20 cm3
oxygen remaining)
iii
NO2 (because the stoichiometry is
2 volumes NO to 1 volume O2 to
2 volumes of NO2)
iv
2NO + O2 → 2NO2
35.4
= 7.08 × 10−3 mol
1000
7.08 × 10 −3
2
= 3.54 × 10−3 mol (from the stoichiometry)
Moles barium hydroxide =
Concentration =
3.54 × 10 −3 mol
0.02 dm3
20
× 1.5 = 0.03 mol
1000
from stoichiometry, mol Fe needed
= 0.03 / 2 = 0.015 mol
mol HCl =
actual number of mol of Fe = 0.9/55.8
= 0.016 (so Fe in excess)
−3
C. Volume in dm3 = mol × 24 = 0.125 × 24
= 3.0 dm3
Moles HCl = 0.2 ×
= 0.177 mol dm−3
Exercise 3.5
a
Volume (dm3) =
V. Mol NaCl = 0.20 × 2.0 = 0.40 mol so mass
= 0.04 × 58.5 = 23.4 g
Empirical formula mass = (2 × 74.9) +
(3 × 16.0) = 197.8
2 × 74.9
× 100 = 65.2%
(2 × 74.9) + (5 × 16.0 )
moles = 0.12
= 0.6 mol dm−3
0.2
volume
S. Mol = concentration × volume (dm3)
= 0.4 × 1.5 = 0.6 mol
ratio of molar mass to empirical formula
mass = 348/ 116 = 3
b
R. (200 cm3 = 0.2 dm3) concentration
ii
limiting reactant
Exercise 3.7
a
b
i
Ba(NO3)2(aq) + Na2SO4(aq) →
BaSO4(s) + 2NaNO3(aq)
ii
Ba2+(aq) + SO42−(aq) → BaSO4(s)
iii
Sodium ions and nitrate ions
Mole ratio is 2 moles of HCl reacting with 1
mole of iron oxide to make 1 mole of iron(II)
chloride
FeOx + 2HCl → FeCl2 + H2O
so x must be 1 for the equation to balance.
FeO(s) + 2HCl(aq) → FeCl2(aq) + H2O(l)
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c
Balanced equation: CaCO3(s) + 2HCl(aq) →
CaCl2(aq) + CO2(g) + H2O(l)
e
So ionic equation is: CaCO3(s) + 2H (aq) →
Ca2+(aq) + CO2(g) + H2O(l)
i
C2H535Cl = 64; C2H537Cl = 66
ii
The C2H535Cl is about three times higher
than the C2H537Cl peak because the ratio
of 35Cl : 37Cl is 3 : 1.
iii
Two:
+
Exercise 3.8
ClC2H537Cl = [M+2] peak
35
a
b
mass of isotope / fragment
charge on isotope / fragment
4
ClC2H535Cl = [M+2] peak (same mass as
ClC2H537Cl)
37
35
ClC2H537Cl = [M+4] peak
37
c
88
d
84
e
84 × 0.56 = 47.04
Sr
8771.02 ÷ 100 = 87.71
Sr
Exam-style questions
86 × 9.86 = 847.96
87 × 7.02 = 610.74
88 × 82.56 = 7265.28 = 72.7
100
Question 1
a
Exercise 3.9
a
b
c
Add suitable acid-base indicator to the acid in
the flask.
[1]
i
43
ii
77
iii
64 / 66
iv
17
v
45
i
[C3H7]+
ii
[CH2]+
iii
[OCH3]+ / [CH2OH]+
Add sodium hydroxide into the acid until the
indicator changes colour.
[1]
b
[C2H5] / [CHO]
v
[C6H5CH2]+
vi
[C2H4]+ / [CNH2]+
i
27 = [C2H3]+; 29 = [C2H5]+; 31 = [OCH3]+ /
[CH2OH]+; 45 = [C2H5O]+ /
[COOH]+; 46 = [C2H5OH]+
ii
+
Moles of NaOH = 2 × 2.687 × 10−2 mol
= 5.374 × 10−2 mol NaOH
Volume (dm3) =
d
= 4.30 × 10−1 dm3
[1]
Correct number of significant figures
[1]
c
Mol hydroxide =
n = 100 × 3.8 = 7 carbon atoms
1.10 4.93
49.3
0.0125 × 25
1000
[1]
Mol HCl = 0.05 × 12.5
1000
= 6.25 × 10−4 mol
3
5.374 × 10 −2
mol
=
0.125
concentration
= 3.125 × 10−4 mol
The relative molecular mass
[1]
Note that you do not have to use the 20 cm3
volume of water in the calculation.
[C2H5OH]+ is the molecular ion so 29
= [C2H5]+; 45 is [C2H5O]+
(not COOH) and 31 is [CH2OH]+ (not
OCH3)
iii
Repeat titration and take the best concurrent
readings (to within 0.1 cm3) for the
average titre.
[1]
3.60
= 2.687 ×
Moles of malic acid =
134
[1]
10−2 mol
From the stoichiometric equation,
iv
+
Idea of suitable apparatus, e.g. burette and
volumetric pipette.
[1]
[1]
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Stoichiometric ratio is 2 mol HCl to 1 mol
of metal hydroxide
[1]
X(OH)2 + 2HCl → XCl2 + 2H2O
[1]
Question 3
a
56 × 91.68 = 5134.08
[Total: 12]
57 × 2.17 = 123.69
Question 2
a
b
58 × 0.31 = 17.98
Number of specified particles (atoms, ions,
molecules or electrons) in a mole of
those particles
[1]
0.0011
= 2.50 × 10−5 mol
Mol CO2 =
44
Mol oxygen atoms = 2 × 2.50 × 10−5
= 5.00 × 10−5 mol
5591.11 ÷ 100 = 55.9
[1]
[1]
b
Number of atoms = 5.00 × 10−5 × 6.02 × 1023
= 3.01 × 1019
[1]
c
CxHy + O2 → CO2 + H2O
14 cm3 84 cm3
56 cm3
[1]
Mol ratio 1 mol 6 mol 4 mol
Mol of carbon per mol of butene =
4 so C4Hy
c
so 4H2O and formula for butene is C4H8.
i
d
Mol C = 37.25 = 3.1 mol
12
[1]
Correct answer to 3 significant figures
[1]
Mass of iron in 1 mole of limonite = 111.6 [1]
Molar mass of limonite = 177.6
[1]
111.6 × 100 = 62.8 %
177.6
[1]
i
[1]
Mol of iron = 2 × mol Fe2O3 = 10 mol
[1]
Mass of iron = 10 × 55.8 = 558 g
[1]
Moles CO2 = 3 × mol Fe2O3 = 15 mol
[1]
15 × 24 = 360 dm3
[1]
mol FeS2 =
mol Fe2O3 =
% yield =
60
= 0.5 mol
55.8 + 2 ( 32.1)
[1]
[1]
26.6
= 0.1667 mol [1]
2 (55.8 ) + 3 (16 )
0.1667
× 100 = 66.7 %
0.25
[1]
Dividing by smallest number (1.55)
gives 2 C, 5 H, 1 Cl.
[1]
Empirical formula is C2H5Cl.
[1]
Correct formulae
[1]
The relative molecular mass
[1]
Correct balance
[1]
e
f
[1]
[Total: 13]
3Fe + 4H2O → Fe3O4 + 4H2
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
Correct symbols
[1]
Correct balance and state symbols
[1]
4
Mol Fe2O3 = 798 = 5 mol
159.6
mol Fe2O3 expected if 100% conversion
= 0.5 / 2 = 0.25 mol The formula showing the number of each
type of atom in a compound.
Correct working of multiplications
[1]
Mol Cl = 55.0 = 1.55 mol (1 mark for
35.5
division by atomic masses)
[1]
e
[1]
ii
Mol H = 7.75 = 7.75 mol
1.0
ii
Multiplying isotopic masses by relative
abundance
[1]
Mol oxygen atoms remaining (for
combination with H in water) = 12 − 8 = 4 [1]
d
54 × 5.84 = 315.36
[Total: 19]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 4
Exercise 4.1
a
i
b
−
i
ii
F
2+
P
Cl
Cl
N
N
Cl
Mg
−
iv
iii
Br
H
F
S
Cl
Cl
ii
v
+
Li
c
O
O
H
C
2−
Cl
O
O
+
Na
b
C
O
Na
−
+
vi
An ionic bond is the strong force of
electrostatic attraction between positive and
negative ions in a crystal lattice. The charge on
the ions is spread out in all directions. It is not
like a covalent bond where the bonding is in
one direction between two particular atoms.
Metals conduct electricity because the
delocalised electrons are able to move
throughout the structure between the positive
metal ions. Many metals are strong and hard
because the metal ions are held together by
the strong electrostatic forces of attraction
between the ions and the delocalised electrons.
vii
H
H
N
viii
F
B
F
F
S
F
F
F
F
F
F
H
Exercise 4.3
a
b
i
R = 107°
ii
S = 109.5°
iii
T = 120°
iv
U = 109.5°, V = 104.5°
i
ii
O
H 104.5° H
+
O
H
Exercise 4.2
a
H
N
H
H
ammonia
H
H
H
H
C
C
H
H
ethanol
H
O
H
V - shaped
H
C
C
H
H
iii Cl
Be
iv
Cl
O
C
v
Linear
vi
P
109.5°
H
Cl Cl 107° Cl
Triangular pyramidal
O
180°
180°
ethene
H
107°
Triangular pyramidal
Linear
1
H
H
C
N
H
H 107° H
Tetrahedral around C atom.
Pyramidal around the N atom
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vii
H
viii
H
H
109.5°
+
H
107°
H
V - shaped
Triangular pyramidal
iii Cl
iv O C O
Be INTERNATIONAL
Cl
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180°
180°
Linear
Linear
v
vi
P
e
H
109.5°
C
H
N
vii
viii
H
C
H
ii
Pentane has a longer hydrocarbon chain
and more electrons than butane. So
pentane has more contact areas for id-id
forces and therefore has a higher boiling
point (high enough for it to be a liquid).
iii
CH3NH2 is hydrogen bonded but CH3Cl
has permanent dipole forces. Hydrogen
bonds are stronger than permanent dipole
forces so more energy is required to break
the bonds between CH3NH2 molecules
than between molecules of CH3Cl.
H
Tetrahedral around C atom.
Pyramidal around the N atom
Triangular pyramidal
C
Water is hydrogen bonded and pentane
has id-id forces between molecules.
Hydrogen bonding is stronger than id-id
forces so more energy is required to break
the bonds between water molecules than
between molecules of pentane.
H 107° H
Cl Cl 107° Cl
H
i
H
+
109.5°
N
120° (Allow 117–120°)
H
H
Planar with each H−C−H being
trigonal planar
H
H
Tetrahedral
ix
I
I
Ga
120°
I
Exercise 4.5
Trigonal planar
Exercise 4.4
a
a
fluorine > oxygen> nitrogen > chlorine >
hydrogen
b
i
H ± H
c
H + −
ii
Fluorine is more electronegative than
hydrogen, so it pulls the bonding electrons
closer towards it and this end of the
molecule is slightly more negative than the
hydrogen end
i
Permanent dipole-dipole forces
ii
Hydrogen bonding
iii
Instantaneous dipole-induced dipole forces
iv
Permanent dipole-dipole forces
v
Permanent dipole-dipole forces (not
hydrogen bonding because the Br is not
electronegative enough)
H
d
b
F
i
iii
ii
Cl
Cl+
δ
H
Br
None
−
δ
iv
δ−
δ+
N
C
Br
Cl
ii
The power of an atom to draw the pair of
electrons in a covalent bond toward itself
i
Electronegativity increases across a period.
The positive charge in the nucleus increases
across a period. So there is a greater
attraction for the electrons in the covalent
bond as you go across a period. The outer
electrons are in the same electron shell so
they are roughly the same distance from
the nucleus so this factor has less effect.
The amount of shielding is about the same.
ii
Chlorine is more electronegative than
bromine. The electrons in the bond are
attracted more to the Cl. A dipole is
formed with the Cl end of the molecule
slightly more negative than the Br end.
iii
Br─Cl is a polar molecule and so can
attack a centre of positive charge easily.
The difference between the electronegativity
of the Mg and Cl is large, so magnesium
chloride is ionic. The difference between the
electronegativity of the C and Cl is small, so
carbon tetrachloride is covalent.
d
i
The distance between the two nuclei of
the atoms which make up the bond.
ii
Bond lengths increase as the size of the
halogen atom increases (down the group).
Br
Br
−
H
A covalent bond in which both the
electrons in the bond come from the same
atom
c
Br
C
H
i
δ
+
δ
H
2
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As you go down the group the group the
outer electrons of the halogen atoms are
further from the nucleus and there is more
shielding. These two factors have a greater
effect than the effect of increasing nuclear
charge.
e
i
ii
b
A = sigma, B = pi, C = sp3, D = sp2
ii
The clouds of electrons have approximately
the same electron density close to each
other, so they repel each other equally.
The electrons arising from the p orbitals
are able to move around the rings.
iii
Benzene is a molecule, so although the
delocalised electrons can move within
the molecule, they cannot move between
molecules.
c
d
e
so the intermolecular forces are greater /
greater van der Waals’ forces with increasing
mass / size of molecules
[1]
so it is more difficult to break the weak forces
between the molecules.
[1]
b
The hydrogen bonding in ammonia is
weakest
Water has a higher boiling point than
hydrogen fluoride because it can form
(on average) two hydrogen bonds per
molecule.
[1]
Fluorine can only form (on average) one
hydrogen bond per molecule.
[1]
c
O
H
H
Bonding pair of electrons between each O
and H
[1]
Two lone pairs on the oxygen
d
i
[1]
Lone pair–lone pair repulsion is greater
than lone pair–bond pair repulsion. [1]
Idea of the lone pair–lone pair repulsion
pushing the OH bonds closer together /
into V-shaped position.
[1]
ii
3
[1]
because nitrogen is less electronegative than
either oxygen or fluorine.
[1]
The electron density of the pi bond in ethene
is more exposed to attacking reagents than the
sigma bonds
CH3CH2CH2Cl is a polar molecule because
the Cl is slightly electronegative. So this
molecule leaves the carbon slightly positive
and therefore open to attack by other reagents.
CH3CH2CH3 is non-polar so there are no
dipoles to allow attack by other molecules.
Methane has the lowest boiling point because
there are only instantaneous dipole-induced
dipole attractive forces between the molecules.
[1]
Ammonia, water and hydrogen fluoride all
have hydrogen bonding.
[1]
The rings of p electrons can combine so
that the electrons move over each layer of
the graphite.
The covalent bonds are weaker in oxygen
compared with nitrogen / the bond energy
of the double bond in oxygen is less than
the bond energy of the triple bond in
nitrogen. So the bonds in oxygen are more
easily broken.
The boiling points increase as the molecules
get larger / increase in molar mass.
[1]
The higher the molar mass, the more electrons
there are
[1]
The p orbitals join to form a ring above and
a ring below the plane of the carbon atoms.
ii
iv
a
They both vary considerably in strength.
(The bonding in sodium is not very strong
and neither is the covalent I─I bond. The
bonding in iron is very strong and so is
the bonding in H─F)
i
i
Question 1
B (id-id forces) → D (permanent dipole)
→ A (H-bonding) → C (ionic)
Exercise 4.6
a
Exam-style questions
104.5
[1]
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iii
Hydrogen attached to very electronegative
atom
[1]
c
H
[1]
(extreme) form of permanent-dipole
permanent dipole forces Two ammonia molecules with intermolecular
bond between N and H
[1]
[1]
Lone pairs shown on nitrogen which is
hydrogen bonded [1]
N----H−N in line (approximately 180°) [1]
δ+
C
d
H
i
H
[1]
The dipoles cancel each other / the centre of
positive and negative charge in the molecule
is the same.
[1]
ii
[Total: 22]
Question 2
2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
Correct formulae
[1]
correct balance
[1]
state symbols
[1]
Salts have ionic bonding.
[1]
Ionic bonds are strong / takes a high
temperature to break ionic bond
[1]
Ammonia is a molecule.
[1]
Weak forces of attraction between
molecules / hydrogen bonding between
molecules
[1]
2+
2−
Any five of:
Polymer C has the lowest value because the
side chains are non-polar.
e
Only temporary dipole-induced dipole forces
between the chains.
Mg
O
Polymer A has more bulky side chains so the
chains are further from each other than in
polymers B and C / there are fewer contact
points in A than in B and C.
Correct electron configuration [1]
Correct charges
[1]
Polymers A and B have polar side chains.
Nitrogen is more electronegative than Cl
b
N
H
H
dipole in correct direction (towards the Cl) [1]
a
H
bonded to lone pair on another very
electronegative atom Tetrahedral structure
f
N
[1]
H
Cl
H
such as N, O or F e
δ−
H
so greater forces of attraction between the
chains in polymer A than in polymer B.
[5]
−
C
N
[Total: 19]
Question 3
a
i
O
I
O
O
H
Correct electron configuration [1]
Structure completely correct
Correct charge shown
[1]
If 3 marks not scored allow correct
bonding around double bonded
oxygen atoms
[3]
ALLOW lone pair / correct structure
around the single bonded oxygen
4
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ii
iii
b
c
Lone pair–bond pair repulsion greater
than bond pair–bond pair repulsion
[1]
Idea of the lone pair–lone pair
repulsion pushing the O─I bonds
closer together [1]
ALLOW values between 98 and 104°
[1]
c
i
H C C H
[1]
ii
linear
[1]
180°
[1]
The triple bond is formed from two π
bonds / pi bonds
[1]
and one σ bond /sigma bond.
[1]
i
iodine has more electrons / iodine has
bigger molecule
[1]
so id-id forces are greater
[1]
Hydrogen fluoride is hydrogen bonded.
[1]
The two π bonds are at right angles to
each other / at right angles to the long
axis of the molecule.
[1]
[1]
The H─C bond is a σ bond /sigma
bond.
Hydrogen iodide has permanent dipole
bonds.
Hydrogen bonding is stronger than
permanent dipoles.
d
b
Sigma bonds formed by end-on overlap.
[1]
[1]
1-iodopropane has greater contact area /
2-iodopropane has less contact area
ii
[1]
[1]
sp carbon hybrid (with 1s of H)
[1]
[Total: 14]
because the chains can get closer in
1-iodopropane / chains cannot get as close
for 2-iodopropane / 2-iodopropane has side
chain sticking out.
[1]
Van der Waals’ forces / intermolecular forces
greater for 1-iodopropane / less for
2-iodopropane.
[1]
e
Co-ordinate bond / dative covalent bond
[1]
Idea of completing octet of electrons around
the aluminium
[1]
[Total: 16]
Question 4
a
i
the average energy needed to break a
specific covalent bond [1]
averaged from a variety of molecules in
the gaseous state.
[1]
ii
The electron pairs in a double bond repel
each other. [1]
One of the electrons pairs forms a pi-bond
which is more easily broken (or has a
higher energy) / the electron pair in one of
the bonds have more energy so are more
easily separated.
[1]
iii
5
Allow values between 700 and 876 (actual
value 838)
[1]
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may be different.
Workbook answers
Chapter 5
Exercise 5.1
a
b
Substance
State at 25 °C and Proximity
1 atm pressure
Arrangement
Motion
bromine
liquid
close together /
not far apart
irregular / random
moving slowly / sliding
over each other
carbon
dioxide
gas
far apart
irregular / random
moving rapidly from
place to place
sulfur
solid
close together
regularly arranged
vibrating (around a
fixed position)
i
Giant; covalent; diamond; four;
tetrahedrally; high; atoms; strong; does
not conduct
ii
Carbon dioxide has a simple molecular
structure. It has strong covalent bonding
between each carbon and oxygen
atom. Carbon dioxide is a gas at room
temperature because the forces of
attraction between the molecules are weak.
Exercise 5.2
a
A is ionic, B is giant molecular / giant covalent, C is metal
b
Structure A
Structure B
Structure C
type of particles present positive and negative
in the diagram
ions / anions and cations
atoms of Si and O
metal ions and
delocalised electrons
melting point
high
high
(generally) high
electrical conductivity
of solid
does not conduct
does not conduct
conducts
conductivity when
molten
conducts
does not conduct
conducts
c
1 with D; 2 with C; 3 with B; 4 with E;
5 with A
d
Metal structure is formed by outer shell
electrons / valence electrons of metal atom
being lost forming delocalised electrons / sea
of electrons between the layers of metal ions.
Metals conduct electricity because the
delocalised electrons (are free to) move (when
voltage applied).
Metals are malleable because the attractive
forces between the metal ions and the
delocalised electrons are overcome (when a
force is applied)
causing the layers to slide over each other.
1
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When the force is removed, attractive forces
are reformed between the metal ions and
delocalised electrons.
b
i
Exercise 5.3
a
Diamond is a giant molecular / giant
covalent structure. The forces between all
the atoms are strong. So a lot of energy
is needed to overcome these attractive
forces.
pV = nRT so p = nRT/V
100 °C = 373 K, V in dm3 = 0.25 dm3
Substituting in the equation p = 0.2 × 8.31 ×
373 / 0.25 / 1000 = 2.48 kPa
b
i
The temperature of the vapour = 120 + 273
= 393 K
ii
Both C60 and benzene are non-polar. A
non-polar solvent generally dissolves a
non-polar solute / idea of like dissolves
like (in terms of polarity). Water is a polar
molecule so does not dissolve C60.
iii
Similarity: both contain rings of 6 carbon
atoms / arrangement of carbon atoms is
hexagonal.
The mass of L vaporised = 10.71 − 10.54
= 0.17 g
The volume of vapour in the gas syringe
in m3 = 69.0 − 4.0 = 65.0 cm3
= 6.5 × 10−5 m3
ii
iii
n = 1.0 × 10 × 6.5 × 10 = 2.0 × 10 −3 mol
8.31 × 393
M r = mass or 0.17 −3 = 85
moles 2.0 × 10
iv
v
5
−5
•Having 4 cm3 air - no effect because
the volume is being measured by the
difference at the same temperature and
pressure
•Loss of liquid - lower volume of gas
than expected so relative molecular
mass higher than expected (because
Mr is inversely proportional to V
mRT
))
(in M r =
pv
•Temperature too high - higher than
expected (1) because temperature is
proportional to the relative molecular
mRT
)
mass (in M r =
pv
Exercise 5.4
a
2
Difference: graphene is a single sheet /
layer but graphite has many layers
(attracted to each other by weak
intermolecular / van der Waals’ forces).
pv
RT
n=
Graphite is an allotrope of carbon. The
carbon atoms in graphite are arranged in
layers. The carbon atoms are arranged in
hexagons. Graphite has a high electrical
conductivity. This is because some of the
electrons are delocalised and are able to move
when a voltage is applied. Graphite has weak
van der Waals’ forces between the layers.
C60 is a simple molecule. There are weak
attractive forces / van der Waals’ forces
between the molecules. So not much
heat energy is needed to overcome these
attractive forces.
Exercise 5.5
a
Gas molecules are in continuous movement.
They collide with each other and with the walls
of the container. The pressure is due to the
force of the collisions over a particular area.
b
When the oxygen in the syringe is heated at
constant pressure its volume increases. This
is because the oxygen molecules move faster
at a higher temperature and collide with each
other with a greater frequency. They also hit
the walls of the container with greater force. If
the pressure is to remain constant, the greater
force of the molecules on the walls of the
syringe pushes the plunger out and the volume
of the gas increases.
c
In an ideal gas the particle volume is zero and
there are no intermolecular attractive forces.
d
B
The higher the pressure, the smaller the
volume. At higher pressure the molecules are
(on average) closer / distance between the
molecules is (on average) smaller. Idea of
concave downwards curve because pressure is
inversely proportional to volume.
e
The volume of the molecules is negligible / is
very small / can be ignored.
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The forces of attraction and repulsion
between the molecules can be ignored / are
negligible.
C is simple molecular [1] weak forces / van der
Waals’ forces
No (kinetic) energy is lost when molecules
collide / the collisions are elastic.
f
The kinetic theory assumes that the molecules
behave as hard spheres with no attractive
forces between them. At high pressures or low
temperatures the molecules are closer together.
So there are attractive forces between the
molecules which tend to reduce the volume
more than the ideal gas laws would suggest.
The kinetic theory also assumes that the
molecules do not have a volume, so this also
must be taken into account.
Exam-style questions
b
i
Different forms of the same element
ii
Regular arrangement of particles
(atoms/ions or molecules) in
three dimensions)
[Total: 18]
Question 2
a
Boron nitride rings are not planar / rings are
bent / boron atoms are not in
the same plane as the nitrogen atoms / in
graphene the rings are planar
[1]
b
i
It has a lot of strong (covalent) bonds /
all the bonds are strong.
[1]
ii
Some of the electrons are delocalised /
some of the electrons are mobile.
[1]
c
Weak forces between the layers / sheets
d
Any five of: (1 mark each)
•
C60 is a simple molecule
•
Weak bonds / van der Waals’ forces
between these molecules
•
Not much energy required to overcome
these forces
•
Polymer has a greater surface area /
greater contact area
Diamond has a giant molecular structure /
giant covalent structure
[1]
•
All the atoms are strongly bonded / it takes
a lot of energy to break the bonds
[1]
Total forces between polymer ‘tubes’
greater than for C60 molecules
•
More energy needed to overcome the
attractive forces between the polymer
chains
i
ii
[1]
Positive and negative ions / anions and
cations
[1]
When molten the ions can move.
[1]
Ions only vibrate in solid. ALLOW:
cannot move in solid.
[1]
A is ionic (structure and bonding).
[1]
B is metallic (structure and bonding).
[1]
[1]
Layers slide over each other [1] Must be
reference to layers or sheets not atoms.
[1]
Idea of alternating positive and negative
ions in every direction
[1]
3
[1]
NOTE: giant covalent = 2 marks
Strong forces of attraction between positive
ions and delocalised electrons / it takes a lot
of energy to overcome the forces of attraction
between the positive ions and delocalised
electrons
[1]
d
Covalent bonding within the molecules
[1]
Titanium has regular arrangement of positive
ions / cations / metal ions
[1]
Surrounded by delocalised electrons /
surrounded by mobile electrons
c
[1]
D is giant molecular / giant covalent structure
[1] bonding is covalent.
[1]
Question 1
a
between molecules.
[Total: 10]
Question 3
a
i
ii
Regular arrangement
[1]
in three dimensions
[1]
Solubility: insoluble
[1]
Electrical conductivity: does not
conduct
[1]
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b
c
The higher the temperature, the greater the
kinetic energy of the molecules
[1]
Molecules collide with (the walls of the
container with) more force (per second)
pv
n=
(or implication of this from
i
RT
working)
[1]
V = 0.400 × 8.31 ×5 313 = 5.20 × 10 −3 m 3 [1]
2.00 × 10
[1]
= 5.20 dm3
[Total: 14]
n = 1.1 × 10 × 8.5 × 10 = 3.1 × 10 −3 mol 8.31 × 363
[1]
M r = mass or 0.22 −3 (1) = 71 [1]
moles 3.1 × 10
5
4
nRT
(or implication of this from
P
working)
V=
[1]
Correct temperature (363) and volume
[1]
8.5 × 10−5 m3
ii
d
−5
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Workbook answers
Chapter 6
Exercise 6.1
a
b
Exercise 6.2
An enthalpy change is the exchange of heat
energy between a chemical reaction mixture
and its surroundings at constant pressure.
The symbol for enthalpy change is ∆H. If
heat is absorbed from the surroundings the
reaction is endothermic. If heat is released to
the surroundings the reaction is exothermic. In
comparing enthalpy changes we use standard
conditions. These are a pressure of 101 kilopascals,
a temperature of 298 kelvin with the reactants
and products in their normal physical state
under these conditions.
a
i
Standard enthalpy change of
neutralisation is the enthalpy change
when one mole of water is formed by the
reaction of an acid / alkali with an alkali /
acid under standard conditions. (last gas
dependent on what put in previous gap).
ii
Standard enthalpy change of combustion
is the enthalpy change when one mole of a
substance is burnt in excess oxygen under
standard conditions.
iii
Standard enthalpy change of reaction is
the enthalpy change when the amounts
shown in the equation react to give
products under standard conditions.
i
Enthalpy, H/kJ mol–1
Hreactants
Mg(s) + CuSO4(aq)
EA
b
1 with D; 2 with E; 3 with B; 4 with A; 5 with C
c
i
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
∆H⦵c[C3H8(g)]
ii
OH−(aq) + H+(aq) → H2O(l)∆H⦵n[OH−aq]
∆Hr
–531 kJ mol–1
Hproducts
MgSO4(aq) + Cu(s)
ii
c
Exothermic because the reactants have
more energy than the products / enthalpy
change negative
Enthalpy, H/kJ mol–1
2NaNO2(s) + O2(g)
∆Hr + 218 kJ mol–1
Hreactants
2NaNO3(s)
Reaction pathway
1
d
e
EA
Hproducts
MgCO3(s) → MgO(s) + CO2(g)
∆H⦵r [MgCO3(s)]
1
iv 2Na(s) + 2 O2(g) → Na2O(s) ∆H⦵f [Na2O(s)]
iii needs a constant input of energy.
iii
Reaction pathway
Standard conditions are a temperature of 298 K
and a pressure of 101 kPa.
Exercise 6.3
a
Put known volume / amount of water in the
can. Weigh the spirit burner with the fuel in it
(and cap on). Measure the temperature of the
water before lighting the spirit burner. Light
the spirit burner and let it heat up the water.
Keep the water stirred. After water has reached
a suitable temperature, blow out the spirit burner
and put the cap on it. Record the temperature of
the water. Reweigh the spirit burner.
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b
i
Mass of fuel burned = 92.33 − 92.19
= 0.14 g
ii
Temperature change = 35.2 − 20.5
= 14.7 °C
iii
mass of water × specific heat capacity
× temperature rise
80 × 4.18 × 14.7 = (−)4915.7 J
iv
C6H13OH = (6 × 12) + (14 × 1) + 16 = 102
102
(−)4915.7 × 0.14 = 3581438.5 J =
(−)3581 kJ mol−1
v
c
d
Exercise 6.5
a
i
b
2Fe(s) + 3CO2(g)
∆H2
∆H1
2Fe(s) + 3C(graphite) + 3O2(g)
ii
∆H⦵1 = ∆H⦵f[Fe2 O3(s)] + 3 × ∆H⦵f [CO(g)]
= (−824.2) + 3 × (−110.5)
= −1155.7 kJ
Heat losses from the walls of the can / from
the water surfaces / from the flame
C6H13OH(l) + 9O2(g) → 6CO2(g) + 7H2O(l)
∆H⦵2 = 3 × ∆H⦵f[CO2(g)] = 3 × (−393.5)
= −1180.5 kJ
∆H⦵c[C6H13OH(g)] = −3581 kJ mol−1
∆H⦵1 + ∆H⦵r = ∆H⦵2
So ∆H⦵r = ∆H⦵2 − ∆H⦵1
Exercise 6.4
a
∆Hr
Fe2O3(s) + 3CO(g)
The energy released in forming the bonds in the
products is greater than the energy absorbed
when the bonds in the reactants are broken.
i
= −1180.5 − (−1155.7) = −24.8 kJ mol−1
b
∆Hr
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(I)
Bonds broken
/ kJ
4 × (C−H) = 4 ×
410 = 1640 kJ
Bonds formed /
kJ
2 × (C═O) = 2 ×
805 = 1610 kJ
2 × (O═O) = 2 ×
496 = 992 kJ
4 × (O−H) = 4 ×
465 = 1860 kJ
∆H⦵1 = ∆H⦵f[C3H8(g)] = −104.5 kJ
Total = 2632 kJ
Total = 3470 kJ
∆H⦵2 = 3 × ∆H⦵f[CO2(g)] + 4 × ∆H⦵f [H2O(l)]
∆H2
∆H1
3C(graphite) + 4H2(g) + 5O2(g)
= (3 × −393.5) + (4 × −285.8) = −2323.7 kJ
ii
c
Enthalpy change = +2632 − 3470 =
−838 kJ mol−1
Bonds broken / kJ
1 × (C═C) = 1 × 612
= 612 kJ
Bonds formed / kJ
4 × (C═O) = 4 × 805
= 3220 kJ
4 × (C−H) = 4 × 410
= 1640 kJ
4 × (O−H) = 4 × 465
= 1860 kJ
∆H⦵1 + ∆H⦵r = ∆H⦵2
So ∆H⦵r = ∆H⦵2 − ∆H⦵1
= −2323.7 − (−104.5) = −2219.2 kJ mol−1
c
i
∆Hf
C4H10(g) (+6 12 O2)
4 × ∆H c [C(graphite)]
5 × ∆Hc [H2(g)]
3 × (O═O) = 3 × 496
= 1488 kJ
Total = 3740 kJ
4C(graphite) + 5H2(g) (+6 12 O2)
∆Hc [C4H10(g)]
4CO2(g) + 5H2O(I) (+6 12 O2)
Total = 5080 kJ
Enthalpy change = + 3740 − 5080
= −1340 kJ mol−1
ii
∆H⦵ left arrow = 4 × ∆H⦵c [C(graphite)] +
5 × ∆H⦵c[H2(g)]
= (4 × −393.5) + (5 × −285.8)
= −3003.0 kJ
∆H⦵ right arrow = ∆H⦵c[C4H10(g)]
= −2876.5 kJ
2
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4 × ∆H⦵c [C(graphite)] + 5 × ∆H⦵c[H2(g)] =
∆H⦵f [C4H10(g)] + ∆H⦵c [C4H10(g)]
∆H⦵r = ∆H⦵f [Ca(OH)2(s)] − (∆H⦵f [CaO(s)]
+ ∆H⦵f[H2O(l)])
So ∆H⦵f [C4H10(g)] = 4 × ∆H⦵c [C(graphite)]
+ 5 × ∆H⦵c [H2(g)] − ∆H⦵c [C4H10(g)]
or
= −3003.0 − (−2876.5) = −126.5 kJ mol
−1
iii
It is the reverse reaction, the products and
reactants being in the same state.
Exercise 6.6
a
Mg(NO3)2(aq) + Na2CO3(aq) →
MgCO3(s) + 2NaNO3(aq)
b
i
ii
d
e
∆H⦵r = −986.1 − (−635.1 − 285.8)
[1]
−65.2 kJ mol−1
[1]
Enthalpy change of solution of calcium
hydroxide
[1]
i
Correct enthalpy cycle with elements at
bottom
[1]
Volume of solution = 20 cm3 + 20 cm3
= 40 cm3
ii Enthalpy
∆Hr = –1081.1 kJ mol–1
2Fe(s) + 3CaO(s)
Reaction pathway
Reactants on left and products on right [1]
Reactants energy level above the
product
Question 1
[1]
i
It may exert a pressure on the inside of
the can
[1]
ii
There is a space above the calcium
oxide for it to expand
[1]
Indication of correct use of Hess’s Law, e.g.
∆H⦵f [CaO(s)] + ∆H⦵f [H2O(l)] + ∆H⦵r
= ∆H⦵f [Ca(OH)2(s)]
[1]
Enthalpy change shown with downward
pointing arrow and labelled
[1]
Activation energy shown and labelled [1]
The heat liberated is conducted through the
aluminium to heat the soup
[1]
c
[1]
Axes labelled correctly
Exam-style questions
b
EA
Ca(s) + Fe2O3(s)
The specific heat capacity of the solution
is the same as that of water.
Moles magnesium nitrate 20.0 × 1.0 =
1000
1
0.02 mol (−)718.96 ×
= 35 948 J =
0.02
(−)36.0 kJ mol−1
The reaction is exothermic
[1]
Arrows in correct directions
40 × 4.18 × 4.3 = (−)718.96 J
a
∆Hf [CaO(s)]
3Ca(s) + 2Fe(s) + 1 12 O2(g)
mass of solution × specific heat capacity ×
temperature rise
iv
2Fe(s) + 3CaO(s)
∆Hf [Fe2O3(s)]
Insulate the beaker / put a lid on the
beaker / stir the mixture with the
thermometer (so that there are no ‘hot
spots’ and the reaction takes place as
quickly as possible).
Temperature rise = 23.2 − 18.9 = 4.3 °C
iii
3Ca(s) + Fe2O3(s)
[Total: 15]
Question 2
a
i
Energy needed to break one mole of
bonds between two atoms
[1]
of a specified gaseous molecule
[1]
[1]
Correct rearrangement:
3
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ii
∆H⦵r = ∆H⦵at[C(graphite)] + (4 × ∆H⦵at
[ 1 Cl2(g)]) − ∆H⦵1
2
(If two marks not awarded, 1 mark for 4 ×
∆H⦵at[ 21 Cl2(g)])
Exact bond energies are for particular
bonds in ethanol / exact bond energy
depends on the ‘environment’ of the
bond
[1]
average bond energies are taken from the
same bonds in a variety of compounds /
do not take account of the ‘environment’
of the bonds [1]
iii
b
Br2(g) → 2Br(g) (1 mark for correct
formulae and balance, 1 mark for state
symbols)
[2]
Bond energy of reactants = +435.9 +
243.4 = 679.3 kJ
[1]
Bond energy of product = 2 × 432
= 864.0 kJ
[1]
Enthalpy change = +679.3 − 864.0
= −184.7 kJ
[1]
c
∆H⦵r = +716.7 + (4 × 121.7) − (−129.6)
= 1333.1 kJ
[1]
For 1 C─Cl bond, 1333.1 ÷ 4
= 333.3 kJ mol−1
[1]
[Total: 17]
Question 3
a
Enthalpy change when one mole of water [1]
is formed when an acid reacts with an
alkali under standard conditions
b
2H(g) + 2Cl(g)
[1]
Add measured volume of acid to measured
volume of alkali
[1]
Energy
Known concentrations of acid and alkali [1]
Carry out reaction in insulated vessel
H2(g) + Cl2(g)
∆Hr
Measure temperature of acid and alkali
before mixing and the maximum temperature
reached after mixing
[1]
2HCl(g)
Reaction pathway
Reactants on left and product on right with
product level below the reactant level
[1]
c
Arrows in correct directions for bond breaking
and bond making
[1]
e
Absorbed, because energy is released in the
reaction of chlorine with hydrogen / absorbed
because it is the reverse of the reaction when
chlorine and hydrogen combine
[1]
Use of Hess’s Law, e.g.
Stir reaction mixture
[1]
Volume of mixture is 75 cm3
[1]
Energy change = m × c × ∆T = 75 × 4.18
× 8.9 = 2790.15 J
[1]
50
Moles of NaOH = 1.0 × 1000 = 0.05 mol [1]
Arrow showing enthalpy change correct and
labelled
[1]
d
Enthalpy change per mole 2790.15 × 1.0
0.05
= 55.8 kJ mol−1
[1]
d
i
It is difficult to measure the temperature
of solids accurately during a reaction if
the reaction requires heating to start it.[1]
ii
Enthalpy change of formation of
copper(II) oxide
[1]
[1]
Cu(s) + 1 O2(g) → CuO(s)
2
[Total: 14]
∆H⦵1 + ∆H⦵r = ∆H⦵at [C(graphite)] + 4 × ∆H⦵at
4
1
[ 2 Cl2(g)]
[1]
[2]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 7
Exercise 7.1
Exercise 7.2
a
a
i
+3 to 0 reduction
ii
−1 to 0 oxidation
iii
−3 to +5 oxidation
The oxidation number of fluorine is −1.
iv
+1 to +2 oxidation
The oxidation number of oxygen is −2
except in peroxides where it is −1.
v
+3 to +5 oxidation
vi
+6 to −2 reduction
i
Reduction
ii
Oxidation
iii
Oxidation
Each O atom has an oxidation number of −2.
iv
Reduction
The oxidation number for 3 O atoms is −6.
v
Reduction
i
Ni → Ni2+ + 2e−
So each Fe atom has an oxidation number
of +3.
ii
HNO2 + H2O → NO3− + 3H+ + 2e−
iii
Te + 2H2O → TeO2 + 4H+ + 4e−
The sum of the oxidation numbers of all the
atoms in the NO3− ion is −1.
iv
Fe3+ + e− → Fe2+
v
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
i
The sum of the oxidation numbers in a
compound is zero.
ii
The sum of the oxidation numbers in an
ion is equal to the charge on the ion.
iii
iv
v
b
The total oxidation number of the sulfur
and the 4 oxygen atoms in SO42− is −2.
The sum of the oxidation numbers of all the
atoms in Fe2O3 is zero.
The oxidation number for 2 Fe atoms is +6.
c
Each O atom has an oxidation number of −2.
The oxidation number for 3 O atoms is −6.
The oxidation number for the N atom is +5
(N + (+6) = −1).
d
i
+3
ii
−1
iii
+6
iv
+5
v
+7
vi
+5
vii +4
1
b
c
Exercise 7.3
a
Oxidising agents gain electrons and get reduced.
b
A reducing agent decreases the oxidation
number of another atom by losing / donating
electrons. The reducing agent gets oxidised.
c
i
Bromine: it increases the oxidation
number of iodine from −1 to 0.
ii
Copper oxide: it increases the oxidation
number of the nitrogen from −3 to 0.
iii
Sulfuric acid: it increases the oxidation
number of I from −1 to 0.
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d
i
Iodide ion: It decreases the oxidation
number of oxygen from −1 to −2.
ii
Bromide ion: it decreases the oxidation
number of chlorine from 0 to −1.
iii
Hydrogen sulfide: it decreases the
oxidation number of iodine from 0 to −1.
Exercise 7.5
a
Exercise 7.4
a
b
c
d
A chemical reaction in which there is a
simultaneous oxidation and reduction
i
Ag+ + e− → Ag
ii
2
iii
Zn + 2Ag+ → Zn2+ + 2Ag
i
Cl2 + 2Fe2+ → 2Cl− + 2Fe3+
ii
6H+ + 2Al → 3H2 + 2Al3+
iii
iv
iron(II) carbonate
ii
manganese(IV) oxide
iii
iodine(V) chloride
iv
sodium bromate(V)
v
chromium(III) hydroxide
vi
potassium manganate(VI)
Deduce the formula for these compounds
to include the oxidation number as Roman
numerals.
i
KClO4
ii
AuCl3•2H2O
iii
NaIO3
iv
2IO3 + 10I + 12H → 6I2 + 6H2O
SnCl4
v
KClO
2Hg2+ + 2Cr2+ → Hg22+ + 2Cr3+
vi
NH4VO3
−
−
+
i MnO4− + Cr2+ + H+ → Mn2+ + Cr3+ + H2O
+7
ii
e
b
i
+2
+2
+3
Mn from +7 to +2 = −5
Exam-style questions
Cr from +2 to +3 = +1
Question 1
iii
MnO4 + 5Cr + H →
Mn2+ + 5Cr3+ + H2O
a
iv
MnO4− + 5Cr2+ + 8H+ →
Mn2+ + 5Cr3+ + H2O
v
MnO4− + 5Cr2+ + 8H+ →
Mn2+ + 5Cr3+ + 4H2O
i
2Cu → Cu + Cu
ii
iii
−
+
+
2+
b
2+
i
0
[1]
ii
+2
[1]
Ba + 2H2O → Ba(OH)2 + H2
Correct symbols
[1]
correct balance
[1]
i
Ba → Ba2+ + 2e−
2I− + 2Fe3+ → I2 + 2Fe2+
ii
2H2O + 2e → 2OH + H2
iii
v
Fe2O3 + 3CO → 2Fe + 3CO2
IO3− + 5Fe 2+ + 6H+ → 1 I2 + 5Fe3+ + 3H2O
2
3CuO + 2NH3 → 3Cu + N2 + 3H2O
Equation ii / 2H2O + 2e− → 2OH− + H2
(no mark alone) because electrons are
being gained.
[1]
vi
2Fe3+ + H2S → 2Fe2+ + 2H+ + S
i
Fe2+ because the oxidation number of
oxygen decreases from −1 to −2 / because
the oxidation number of Fe2+ is increasing
/because the Fe2+ has lost electrons
[1]
ii
2H+ + H2O2 + 2e− → 2H2O
iv
c
d
vii 2MnO42− + Cl2 → 2MnO4− + 2Cl−
viii 3MnO42− + 4H+ → 2MnO4− + MnO2 + 2H2O
f
2
i
A simultaneous reduction and oxidation
of the same species in a chemical reaction
ii
reactions e i and e viii
−
[1]
−
[1]
Correct formulae
[1]
correct balance with electrons
[1]
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e
i
+2 to +4 = +2
[1]
Question 3
ii
−1 to −2 = −1
[1]
a
iii
H2O2 + Mn + 2OH → MnO2 + 2H2O
2+
−
Correct formulae
[1]
correct balance
[1]
[Total: 14]
Question 2
a
i
+5
[1]
ii
+3
[1]
iii
N (in NaNO3) gains electron(s) and
oxygen (in NaNO3) loses electron(s) to
form O2
[1]
Oxidation number of nitrogen decreases
and oxidation number of (some of the)
oxygen increases
[1]
Idea of oxidation and reduction
happening simultaneously
iv
b
c
Sodium nitrate(III) (But with nitrates
and sulfates, nitrate and nitrite and
sulfate and sulfite are often preferred) [1]
i
−1 to 0 = +1
[1]
ii
+3 to +2 = −1
[1]
iii
The oxidation numbers are balanced /
+1 balances −1
[1]
iv
NO2− because it increases the oxidation
number of the iodine in iodide / it gets
reduced / it takes electrons from the
iodide ions
[1]
i
+7 to +2 = −5
[1]
ii
+3 to +5 = +2
[1]
iii
2MnO4− + 5NO2− + 6H+ → 2Mn2+ + 5NO2−
+ 3H2O
3
[1]
Correct formulae
[1]
correct balance
[1]
i
−2
[1]
ii
0
[1]
iii
Sulfur because it increases in oxidation
number
[1]
b
Hydrogen sulfide because it has decreased
the oxidation number of iodine atoms
[1]
c
i
Oxidation number change of 2 I atoms
is 2 × +5 , so −10 needed
[1]
ii
I2O5 + 5H2S → I2 + S + 5H2O
[1]
2I− → I2 + 2e− oxidation of iodide ions
[1]
d
Oxidation of iodide ions because the iodine
increases in oxidation number from −1 to 0
[1]
in I2
H2O2 + 2H+ + 2e− → 2H2O
[1]
Reduction of hydrogen peroxide because
the O in the hydrogen peroxide has decreased
in oxidation number from 0 to −2 (in O
in water)
[1]
[Total: 10]
[Total: 14]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 8
Exercise 8.1
a
1 with B; 2 with A; 3 with F; 4 with E; 5 with
C; 6 with D
iii
Position of equilibrium moves to the left
to reduce the increase in concentration.
b
i
Position of equilibrium moves to
the right to reduce the increase in
concentration.
iv
Position of equilibrium moves to the left
so concentration of reactant is increased.
i
ii
Position of equilibrium moves to the left
to reduce the increase in concentration.
Cl2 + ICl ⇌ ICl3
ii
The yellow solid changes to a brown
liquid as the chlorine escapes.
iii
Position of equilibrium moves to the left
in the direction of greater number of
molecules / moles.
iii
The equilibrium is shifted to the left to
increase the concentration of chlorine.
iv
Position of equilibrium moves to the left
to reduce the increase in temperature due
to exothermic reaction.
iv
The reaction is shifted to the right to
reduce the concentration of added chlorine.
c
1
v
Position of equilibrium moves to the
right (because ammonia is removed)
so concentration of ammonia to be
increased.
i
No effect because there is an equal
number of gaseous molecules / moles on
each side of the equation.
ii
No effect. A catalyst does not affect the
position of equilibrium, only the rate of
reaction.
d
e
When any of the conditions affecting the
position of equilibrium are changed, e.g.
pressure, concentration (or temperature) or
temperature (or concentration), the position
of equilibrium moves to oppose / counteract
the change.
Exercise 8.2
a
An equilibrium expression links the
concentration of reactants and products to the
stoichiometric equation. Under stated conditions
the value calculated from the equilibrium
expression is called the equilibrium constant.
b
C and D
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c
Chemical equation
Br2(g) + H2(g) ⇌ 2HBr(g)
Units
Kc =
[HBr ]
[Br2 ][H2 ]
none
Kc =
[ NH3 ]
3
[ N 2 ][H2 ]
dm6 mol−2
2
N2(g) + 3H2(g) ⇌ 2NH3(g)
2
CaCO3(s) ⇌ CaO(s) + CO2(g)
K c = [CO2 ]
2NO2(g) ⇌ 2NO(g) + O2(g)
mol dm−3
Kc =
[H2 ]4
[H2O]4
none
Kc =
[Cu 2+ ]
2
[ Ag ]
dm3 mol−1
Kc =
[Cr2O7 2− ] [ H2O]
[CrO42− ]2 [H+ ]2
dm6 mol−2
Equilibrium expression
Units
2
Cu(s) + 2Ag+(aq) ⇌ Cu2+(aq) + 2Ag(s)
2CrO42−(aq) + 2H+(aq) ⇌ Cr2O72−(aq) + H2O(aq)
Chemical equation
2NO2(g) ⇌ 2NO(g) + O2(g)
2SO2(g) + O2(g) ⇌ 2SO3(g)
2HI(g) ⇌ I2(g) + H2(g)
PCl5(g) ⇌ PCl3(g) + Cl2(g)
3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)
mol dm−3
[ NO] [O2 ]
Kc =
3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)
d
Equilibrium expression
[NO2 ]2
Kp =
2 × p
pNO
O2
2
pNO
2
Pa (or atm)
Kp =
2
pSO
3
2
pSO2 × pO2
Pa−1 (or atm−1)
Kp =
pI2 × pH2
2
pHI
no units
Kp =
pCl3 × pCl2
pCl5
Pa (or atm)
Kp =
pH42
pH4 2O
none
Exercise 8.3
a
b
2
i
nitric acid
ii
c
i
9–12
sulfuric acid
ii
0–1
iii
potassium hydroxide
iii
13–14
iv
ammonia
iv
4–6
i
CH3COOH
v
7
ii
HCl
i
2NaOH + H2SO4 → Na2SO4 + 2H2O
iii
NaOH
ii
HNO3 + NH3 → NH4NO3
d
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e
f
iii
3KOH + H3PO4 → K3PO4 + 3H2O
iii
Acid is NH4+, base is H2O
iv
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
iv
Base is NH2OH, acid is H2O
i
sodium sulfate
v
Acid is H2SO4, base is H2O
ii
ammonium nitrate
vi
Acid is H2SO4, base is HNO3
iii
potassium phosphate
c
HCl, H2SO4 and HNO3
iv
barium chloride
d
i
i
Add universal indicator to the solution.
Compare the colour of the indicator to
the colour on the pH colour chart.
HSiO3− is conjugate with SiO32− and H2O
is conjugate with H3O+
ii
HCO2H is conjugate with HCO2− and
H2O is conjugate with H3O+
blue / purple • orange / yellow
iii
CH3NHCH2NH3+ is conjugate with
CH3NHCH2NH2 and H2O is conjugate
with H3O+
iv
NH2OH is conjugate with NH3OH+ and
H2O is conjugate with OH−
ii
Exercise 8.4
a
1 with C; 2 with G; 3 with E; 4 with F; 5 with
A; 6 with D; 7 with B
b
i
Acid is HCl, base is H2O
ii
Base is CH3NH2, acid is H2O
8.4f
e
Chemical equation
D
Equilibrium expression
C2H5CO2H(aq) ⇌ C2H5CO2 (aq) + H (aq)
+
−
N2H4(aq) + H2O(l) ⇌ N2H5+ (aq) + OH−(aq)
K=
[C 2H5CO2− ] [H+ ]
[C 2H5CO2H]
mol dm−3
K=
[ N 2H5+ ] [OH− ]
[ N 2 H 4 ] [H 2 O]
none
H2O2(aq) ⇌ HO2−(aq) + H+(aq)
[HO2− ] [H+ ]
K=
[ H 2 O2 ]
mol dm−3
Pb(OH)2(s) ⇌ PbOH+(aq) + OH−(aq)
K = [PbOH+] [OH−]
mol2 dm−6
HPO42− (aq) ⇌ PO43−(aq)+ H+(aq)
[ PO43− ] [H+ ]
K=
[HPO42− ]
mol dm−3
ii
Moles of ethanoic acid at equilibrium =
(1.00 × 10−3) − (7.84 × 10−4)
= 2.16 × 10−4 mol
iii
Concentration of pentene at equilibrium
= 7.025 mol dm−3
iv
The equilibrium concentrations on the top
and bottom of the equilibrium expression
cancel.
Concentration of ethanoic acid at
equilibrium = 2.70 mol dm−3
v
Kc =
Moles of pentene at equilibrium = (6.40 ×
10−3) − (7.84 × 10−4) = 5.62 × 10−3 mol
vi
(7.84 × 10 −4 )
= 646 dm3 mol −1
(5.62 × 10−3 ) × (2.16 × 10−4 )
Exercise 8.5
[HI]
[I2 ][H2 ]
2
a
i
Kc =
ii
(2.52 × 10−2 )
= 46.4(no units)
(1.14 × 10−2 )( 0.12 × 10−2 )
2
iii
b
3
i
Units
[CH3CO2C5H11 ]
[C5H10 ] [CH3CO2H]
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Exercise 8.6
a
Adding excess acid to this indicator shifts the
equilibrium to the left and the indicator turns
yellow. The colour of the indicator depends
on the relative concentrations of the ionised
and unionised forms.
number of moles of one particular gas in
mixture total number of moles of all gases in
mixture
OR
b
i
There is little change in pH (high pH /
pH10) at first / until about 17 cm3 of acid
has been added. The pH then decreases
more rapidly and at 20 cm3 there is a
sudden drop from pH 8 to pH 1.5. The
pH then decreases less rapidly as more
acid is added and remains steady at pH 1.
ii
20 cm3
i
Bromocresol green. The change in colour
corresponds to the sudden change in pH.
ii
Because the change in colour does not
correspond to the sudden change in pH.
It will not change colour fully until past
the equivalence point / point where equal
numbers of moles of acid and alkali have
reacted.
The molar proportion of one gas in a mixture
of gases
i
ii
iii
5.0 mol
1.0
N2 = 5.0 × 40 = 8 atm
3.5
H2 = 5.0 × 40 = 28 atm
c
Ar = 0.5 × 40 = 4 atm
5.0
c
i
mol He = 0.15, mol CH4 = 0.40,
mol O2 = 0.30
total number of moles = 0.85
mol fraction = 0.40/0.85 = 0.47
ii
Mol He = 0.15, mol CH4 = 0.40,
mol O2 = 0.30
Total number of moles = 0.85
0.40
Partial pressure of CH4 = 0.85 × 200
= 94 atm
d
i
d
i
14
12
pH of flask contents
b
Total number of moles = 1.02
0.96
Partial pressure of NO2 = 1.02 × 2 × 104
= 1.88 × 104 Pa
0
2
4
6
8 10 12 14 16
Volume of weak acid added to strong base/cm3
0.02
Partial pressure of O2 = 1.02 × 2 × 104
= 3.92 × 102 Pa
(1.88 × 10 4 )2
= 1.47 Pa −1
(7.84 × 10 2 )2 × (3.92 × 102 )
Exercise 8.7
+
−
yellow violet
4
14
12
An acid-base indicator changes colour over a
narrow pH range. These indicators are usually
weak acids in which the acid, HIn, and its
−
conjugate base, In , have different colours.
For example:
HIn ⇌ H + In
ii
pH of flask contents
a
Note that the equivalence point is 10 cm3
because there are twice as many moles
of ethanoic acid as there are moles
of potassium hydroxide. The starting
point is pH 13 because that is the pH of
0.1 mol dm−3 potassium hydroxide.
2
pNO
2
2
pNO × pO 2
iii
4
0
= 7.84 × 102 Pa
Kp =
8
7
6
2
0.04
Partial pressure of NO = 1.02 × 2 × 104
ii
10
10
8
7
6
4
2
0
0
2
4
6
8 10 12 14 16
Volume of acid added to strong base/cm3
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Note that the equivalence point is 10 cm3
because there are twice as many moles
of H+ ions which react in sulfuric acid
as there are moles of sodium hydroxide.
The starting point is pH is 12 because
that is the pH of 0.01 mol dm−3 potassium
hydroxide. The final pH is slightly above
pH 2 because the concentration of H+
ions in the excess acid is (slightly less
than) 0.02 mol dm−3.
e
Question 2
a
i
ii
Thymolphthalein
iii
Any two different ways (1 mark each), e.g.
increase concentration of hydrogen (or carbon
monoxide or remove methane) / increase the
pressure / decrease the temperature.
[2]
i
Products are being converted to reactants
at the same time as reactants are being
converted to products
[1]
The concentration of reactants and
products remains constant under specified
conditions / rate of forward and back
reaction is the same
[1]
ii
c
d
e
pCH3OH
pH22 × pCO
Kp =
ii
(9.92 × 101)/(6.67 × 104)2 × (3.33 × 104)
= 6.7 × 10−13 Pa−2
Correct numbers inserted, correct
answer, correct unit (1 mark each)
[3]
c
i
ii
[H2S ]2 [H ]2 [S2 ]
[2]
[1]
dm3 mol−1
[1]
( 0.442)
9.40 × 105 =
2
( 0.234) × [S2 ]
[S2] =
2
[1]
( 0.442 )2 × (9.40 × 105)
( 0.234)2
= 3.80 × 10−6 (dm3 mol−1) Correct
answer
[3]
If these not scored, 1 mark for correct
substitution in equation, 1 mark for correct
rearrangement of equation.
d
2
pSO
3
2
×
pSO
pO2
2
i
Kp =
ii
(80100)2
Kp = (10100)2 × (68800)
= 9.14 × 10−4 Pa−1 All correct
but if not scored allow 1 mark for correct
16.8
mole fraction, i.e. 7.2 + 16.8
[Total: 12]
5
K=
[1]
16.8
Partial pressure of hydrogen = 7.2 + 16.8 ×
Position of equilibrium moves to
the right
b
[1]
i
5.00 × 104 = 3.50 × 104 Pa Correct answer
[1]
Moves in the direction of fewer moles /
molecules to counteract the pressure
increase
[1]
No matter is lost to the surroundings or
gained from the surroundings
[1]
No effect / only increases the rate of
reaction
Position of equilibrium moves to the
right.
Because if there is a decrease in
temperature, the reaction goes in the
direction of more heat being produced.
ALLOW: because the reaction is exothermic
[1]
Question 1
b
[1]
To oppose the removal of sulfur so
that equilibrium is restored / idea of le
Chatelier’s principle
[1]
Exam-style questions
a
Position of equilibrium moves to
the left
[1]
[3]
If not then 1 mark for correct substitution
including the squares, 1 mark for correct
unit.
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e
i
ii
Position of equilibrium moves to the right
/ moves in direction of more products [1]
Question 4
Fewer moles / molecules of gaseous
products (in stoichiometric equation)
than reactants [1]
a
CH3COOH + OH− → CH3COO− + H2O b
i
High pressure dangerous / more corrosive
at higher pressures / higher pressures
more expensive [1]
Reaction is already far to the right iii
[1]
8
7
6
pH
5
4
[1]
450 °C (allow ±20 °C) [1]
2
200 atm pressure (allow ±20 atm / 2000 kPa) [1]
1
3
0
[Total: 24]
Volume of base
Axes labelled and one or more pH values
given
[1]
Ethanoic acid partially ionises /
dissociates in water
[1]
Nitric acid ionises / dissociates
completely in water
Curve typical ‘S’ shape with sudden rise
in pH
[1]
[1]
Midpoint of curve clearly below pH 7 /
curve levels of at pH 9 to 11 and starts at
pH 0 to 3
[1]
Kc =
i
[CH3CO2− ] [H+ ]
[CH3CO2H]
[CH3CO2− ] [H3O+ ]
ALLOW: CH CO H H O [ 3 2 ][ 2 ]
[1]
It is present in great excess
[1]
so its concentration is effectively
constant.
[1]
Nitric acid has a higher concentration of
hydrogen ions than ethanoic acid
[1]
ii
c
Its colour range does not coincide
with the sudden change of pH
Fe catalyst
Question 3
b
[1]
9
Higher temperature favours the
endothermic reaction (which is to the
left) so yield less
[1]
a
Phenolphthalein / other suitable
indicator
c
[1]
At lower temperatures rate of reaction
is too slow. [1]
f
ii
[1]
[Total: 6]
so more collisions per second compared
with ethanoic acid (or reverse argument). [1]
ALLOW: nitric acid is an oxidising acid for
1 mark
d
(0 to +2) = +2
[1]
(+5 to −3) = −8
[1]
H + OH → H2O
[1]
i
ii
e
+
6
−
[Total: 10]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 9
Exercise 9.1
i
B and C
e
ii
A: the molecules do not have enough
energy to react when they collide / the
molecules have less energy than the
activation energy.
Exercise 9.2
b
a
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) +
H2O(l)
The molecules have energy greater than or
equal to the activation energy and collide
with the correct orientation.
b
i
iv
effective collisions
i
Activation energy is the minimum energy
that colliding particles must have in order
to react. The symbol for activation energy
is EA.
ii
∆Hr
d
1
40
60
80
100 120
140
Mass 0.00 0.10 0.20 0.28 0.34 0.39 0.425 0.45
CO2
/g
Table continued with the other values.
ii
60
B
40
products
30
reaction pathway
c
20
50
EA
reactants
Time 0
/s
Mass of CO2 released / g
iii
Particles are closer together / more particles
per unit volume. Number of effective
collisions increases.
C: the molecules have not collided in the
correct orientation.
enthalpy H/kJ mol−1
a
20
iii
Exothermic because the reactants have
more energy than the products.
iv
The activation energy ‘hump’ is lower.
A reaction will speed up if the frequency of
effective collisions increases and the proportion
of particles with energy greater than the
activation energy increases.
The 0.8 mol dm−3 hydrochloric acid is more
concentrated. It has more hydrogen ions per
unit volume than the 0.4 mol dm−3 acid. So the
frequency of effective collisions is greater.
10
0
0
50
100
150
200
250
300
350
Time / s
iii
The one at 160 s
iv
1st 20 seconds:
rise = 0.1 g / 20 s = 5 × 10 − 3 gs −1
run
20−40 seconds:
rise = 0.2 − 0.1 g = 5 × 10 − 3 gs −1
run
40 − 20 s
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c
v
The graph begins to curve / the line is not
straight.
vi
Take a tangent to the curve.
vii rise = 0.60 − 0.20 g, run = 215−0 s = 215 s
gradient (rise / run) = 1.86 × 10−3 gs−1
i
0.53 g so moles = 0.53 = 0.012 mol
44
ii From the equation: 2 mol HCl form 1 mol
CO2. So moles HCl = 0.012 × 2
= 0.024 mol
Volume (dm3) =
mol
concentration (in mol dm − 3 )
ii
d
When the sample is heated, the mean energy
of the molecules increases. There is a wider
spread of energy values. A greater proportion
of the molecules exceed the activation energy
compared with the proportion at a lower
temperature.
iii
0.012 mol × 24 = 0.288 dm / 288 cm
iv
Yes: because mass of gas is very small / air
current can affect the balance / decimal
places
The activation energy for the catalysed
reaction is lower. So there is a greater
proportion of molecules with energy values
above the activation energy.
b
i
Enthalpy
E+P
2
EA for the
uncatalysed
reaction
EA for the catalysed
reaction
Exercise 9.3
b
∆Hr
E+S
See line B in answers to (b)(ii) above.
i
Reaction proceeds by a different route
with lower activation energy.
ii
No: because the graduations on a gas
syringe are thicker lines and only read
to nearest 2 cm3. Larger volumes being
read so less % error / syringe plunger
constantly moving so more difficult to
read accurately.
a
20°C: Ek = 3 × 8.31 × 293 = 3652.25 units
2
30°C: Ek = 3 × 8.31 × 303 = 3776.90 units
2
The % increase in kinetic energy is only
3.4% which is far less than the doubling
of rate every 10 °C.
a
3
OR
d
i
Exercise 9.4
= 0.024 = 0.6 dm 3 /60 cm 3
0.40
3
c
A few molecules have small amount of
energy, a large number of molecules have
energies near the average, a very small
number of molecules have very high energy.
ii
Activation energy. The minimum energy
required by molecules to react when they
collide.
iii
The proportion of molecules with energies
equal to and above the activation energy.
i
Average kinetic energy of the molecules
increases (slightly).
ii
Frequency of collisions increases.
iii
Reaction rate increases.
Progress of reaction
c
i
In homogeneous catalysis the reactants,
products and catalyst are all in the same
phase; in heterogeneous catalysis the
catalyst is in a different phase from the
reactants and products.
ii
Equation 1: heterogeneous
Equation 2: heterogeneous
Equation 3: homogeneous
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Exam-style questions
above products and to the left of
the products
[1]
Question 1
Correct activation energy for catalysed
reaction
[1]
Correct activation energy for uncatalysed
reaction
[1]
i
ii
Decreases rate (no mark on own but if
not correct max. 1 for question) Idea
that particles have less kinetic energy on
average
[1]
Question 2
Increases rate (no mark on own but if not
correct max. 1 for question)
a
Heterogeneous, because the catalyst is in
a different phase from the reactants and
products / because the catalysts are solid and
the reactants and products are not.
[1]
b
Catalyst lowers the activation energy.
[1]
Greater proportion of reactant molecules
has energy above the activation energy
[1]
for catalysed reaction.
[1]
Decrease in pressure causes molecules to
move further apart.
[1]
Frequency of effective collisions
increases
[1]
Increases rate (no mark on own but if not
correct max. 1 for question)
c
Adding a catalyst lowers the
activation energy
[1]
Greater proportion of particles has energy
above the activation energy
[1]
b
[Total: 17]
Lower proportion of reactant particles has
energy greater than activation energy [1]
Increasing the pressure forces particles
closer together
[1]
iii
Add known mass of catalyst to hydrogen
peroxide.
[1]
Record volume of gas in burette
[1]
at particular times.
[1]
Frequency of effective collisions is reduced. [1]
d
Enthalpy
a
C3H6 + H2
EA
C3H8
∆Hf
Repeat with same mass of different catalyst. [1]
Same volume and mass of hydrogen
peroxide.
[1]
All other conditions / named conditions kept
constant.
[1]
c
Enthalpy
EA uncatalysed
reaction
2H2O2
EA catalysed
reaction
2H2O + O2
Progress of reaction
Axes labelled
[1]
Reactants and products labelled
[1]
Reactants above and to the left of the
products
[1]
Correct activation energy
[1]
Arrow for enthalpy change in the upward
direction and labelled
[1]
[Total: 11]
Progress of reaction
3
Axes labelled
[1]
Reactants and products labelled
[1]
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Question 3
a
i
b
Greater frequency of effective
collisions.
ii
16 cm 3
(1) = 0.8 cm 3s −1 20 s
iii
[1]
Increases rate (no mark on own but if not
correct max. 1 for question).
Magnesium powder has a greater surface
area for the same mass
[1]
More magnesium ions / atoms exposed for
the hydrogen ions to collide with
[1]
iii
Greater proportion of reactant
particles has energy greater than
activation energy
4
iv
Increases rate (no mark on own but if not
correct max. 1 for question).
idea that particles have more kinetic
energy on average
[1]
Rate decreases
gradient decreases with time / gradient
gets shallower with time
[1]
16 cm 3
(1) = 0.8 cm 3s −1
[1]
20 s Increases rate (no mark on own but if not
correct max. 1 for question)
More particles per unit volume at higher
concentration / particles
closer together
[1]
ii
i
[1]
0.25 mol dm−3
[1]
The magnesium is in excess so the
acid is the limiting reagent.
[1]
The volume of hydrogen produced
is half that of B, so the moles
and (for the same volume) the
concentration is half that for B.
[1]
The increased gradient reflects the
increased rate of reaction / increase in
temperature increases reaction rate
[1]
The final volume is the same because
there are the same number of moles of
acid / the acid is the limiting reagent [1]
[1]
[Total: 15]
[1]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 10
Exercise 10.1
a
1 with C; 2 with F; 3 with A; 4 with D; 5 with
B; 6 with E
b
Sodium chloride dissolves in water to form
a neutral solution because the polar water
molecules surround the positive and negative
ions and separate them. Aluminium chloride is
hydrolysed by water and the solution becomes
acidic. Chlorides of silicon, phosphorus and
sulfur react with water. The gas, hydrogen
chloride, is released, some of which dissolves in
water and reacts to form an acidic solution.
c
i
PCl5 + 4H2O → H3PO4 + 5HCl
ii
SO3 + H2O → H2SO4
iii
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
iv
SiO2 + 2NaOH → Na2SiO3 + H2O
v
2Na + 2H2O → 2NaOH + H2
vi
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O
to break the strong metallic bonds between
the ions and delocalised electrons. S has no
delocalised electrons so does not conduct. Al
has delocalised electrons which move through
the whole structure between the ions when a
voltage is applied.
Exercise 10.2
a
i
There is a general increase in ionisation
energy across the period. This is because
successive atoms have one more proton
in their nucleus so there is an increased
nuclear charge. The outer electrons are
in the same principal quantum shell
and there is no significant difference in
shielding. So the attractive effect of the
increased nuclear charge on the electrons
makes it more difficult to remove an outer
electron from an atom across the period.
ii
The additional electron in aluminium
goes into a p orbital which is, on average,
slightly further from the nucleus and has
slightly more shielding. These effects
overcome the effect of the increased
nuclear charge. In sulfur, the added
electron goes into an orbital so that its
spin opposes that of the electron already
in the orbital. This gives the outer
electrons additional stability so an outer
electron is less easily removed.
iii
K: value should be slightly less than Na
(actual value = 419 kJ mol−1)
vii SiCl4 + 2H2O → SiO2 + 4HCl
viii 2P + 5Cl2 → 2PCl5
d
ix
4Al + 3O2 → 2Al2O3
i
sodium oxide pH 13–14 magnesium oxide
pH 9–11
ii
Sodium oxide reacts to form sodium
hydroxide which is very soluble so the
concentration of OH− ions is high.
Magnesium oxide reacts to form magnesium
hydroxide which is not very soluble so the
concentration of OH− ions is lower.
iii
e
1
Ca: value should be slightly less than Mg but
greater than K (actual value = 590 kJ mol−1)
MgO + H2O → Mg(OH)2
S is a simple molecule with weak
intermolecular forces. Only low temperature is
needed to separate the molecules. Al is a giant
metallic structure. High temperatures needed
b
Across a period, the number of protons
(positive charges) increases. So the nuclear
charge also increases. The number of electrons
(negative charges) also increases across a
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period. Each electron added to the atom
of successive elements goes into the same
principal quantum shell. So the shielding of
outer shell electrons by inner shell electrons
does not increase significantly. Across a
period, the greater attractive force between
the nucleus and the outer electrons pulls them
closer to the nucleus.
c
d
From sodium to silicon, the ionic radius
decreases for similar reasons detailed for the
atomic radius (part b). The outer electrons
are in the second quantum shell because the
ions are formed by loss of the outer electrons
and so the ionic radii are smaller than the
corresponding atomic radii. The values for
phosphide to chloride are much higher because
the outer electrons are in the next quantum
shell (the third). So the outer electrons are
further from the nucleus and the attraction to
the nucleus is much less. The values decrease
from phosphide to chloride for similar reasons
detailed for the atomic radius (part b).
i
Group 1
ii
Decreases from Group 1 to Group 13 or
14 and then increases to a maximum.
iii
The atomic radius decreases gradually
across a period but the atomic volume
decreases then increases.
iv
The atomic volume depends on how the
atoms are packed in the structure, whether
they are packed so that there is a lot of
space between some of the atoms or
whether they are packed more efficiently.
v
It increases (compare similar points on
the curve, e.g. Li with Na with K).
between the ions and the electrons so
making it more difficult to overcome
these.
iii
c
Ne: element before Na so about the same
level as argon. ALLOW: 5−100 K (actual
value 25 K)
K: next element after Ar value should be
nearer to that of Na but slightly lower.
ALLOW: 250−400 K (actual value 336 K)
Ca: next element after K so should be higher
than potassium. ALLOW: 900−1300 K (actual
value 1112 K)
d
e
f
i
They are metals so have delocalised
electrons which are responsible for the
transfer of charge throughout the whole
structure.
ii
Each aluminium atom can provide 3
electrons to the ‘sea’ of delocalised
electrons in the metallic structure. A
sodium atom can only provide 1 electron.
iii
It has a simple molecular structure so there
are no delocalised electrons (or mobile ions).
i
The ratio of chlorine to other element
increases to a maximum in Group 14 or
15 and then decreases.
ii
CCl4, SiCl4, NCl3, PCl5
iii
P = +5, S = +4
i
If the difference in electronegativity
between the ‘atoms’ is relatively large e.g.
1.0, the structure is ionic. If the difference
in electronegativity between the ‘atoms’
is relatively small e.g. 0.5, the structure
is covalent (although there is some ionic
character in PCl5).
ii
Magnesium and chloride ions are
stabilised by water molecules (ion-dipole
bonding) in solution and do not react. In
PCl5 the Cl is more electronegative than
the P. So P is δ+ and a highly polar water
molecule (O δ−)can attack the P) and
hydrolyse the molecule.
Exercise 10.3
a
It increases to a maximum (at Si) and then
decreases to very low values.
b
i
ii
2
They have molecular structures with only
weak forces of attraction (van der Waals’
forces) between the molecules.
Each aluminium atom can provide
3 electrons to the ‘sea’ of delocalised
electrons in the metallic structure. A
sodium atom can only provide 1 electron.
A greater number of delocalised electrons
and a higher ionic charge (3+ for Al) mean
that there are greater forces of attraction
Covalent giant structure (giant molecular
structure). Takes a very high temperature
/ a lot of energy to break all the bonds in
the lattice.
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Exercise 10.4
a
i
Element X is silicon and Y is germanium.
They have giant structures of covalent
bonds. It takes a lot of energy / high
temperature to break all these bonds.
iii
Element with atomic number 15
(phosphorus) is in group 15 and is a
simple molecule. There are only weak
forces of attraction between the molecules
so it does not take much energy / relatively
low temperature required to overcome
these forces.
v
3
High melting point → giant structure.
Alkaline oxide → group 1 or 2.
Neutral solution of chloride → group 1 or 2.
Chloride XCl2 → group 2.
3rd highest ionisation energy in Group →
period 4 (since 1st in group is period 2 and 1st
ionisation energies decrease down the group.
So X is Ca.
Exam-style questions
Question 1
a
i
ii
They are noble gases / Group 18. There are
only very weak forces of attraction between
their atoms (because the van der Waals’
forces are only between single atoms). So
not much energy / only low temperature is
required to overcome these forces.
ALLOW: between 500 and 1200 °C
(actual value is 931 °C)
ii
Conductor (the elements are more
metallic down the group)
iii
1.8
iv
(Weakly) acidic
v
The covalent bonding gets weaker as the
atoms get larger because there is less force
of attraction between the nucleus and the
bonding electrons.
They are metals and so do not form
covalent bonds.
i
Gets more negative
ii
It has a higher nuclear charge than the
atoms of the elements before it. So it is more
likely to attract an electron and fill the outer
p-electron shell to make it stable.
iii
The outer shell electrons are further from
the nucleus, so attractive forces between
the positive nuclear charge and an
incoming electron is less.
Mass of atoms
[1]
size of atoms
[1]
the way the atoms are packed together [1]
They rise from high values to very high
values then fall again.
i
vi
c
There is a regular pattern in the melting
points, e.g. in the d-block the melting
points rise to high values then fall again /
there is a peak for the Group 14 element.
ii
iv
b
d
b
c
i
The mass of the atoms increases as the
number of protons and neutrons
increases
[1]
the atomic radius / volume decreases
from sodium to aluminium
[1]
so ratio of mass / volume decreases.
[1]
Na2O(l)s) + H2O(l) → 2NaOH(aq)
Correct formulae
[1]
correct balance
[1]
correct state symbols
[1]
ii
Amphoteric
[1]
iii
Al2O3 + 2NaOH + 3H2O → 2NaAl(OH)4
Correct formulae
[1]
correct balance
[1]
Any six of:
Melting point of sodium, magnesium and
aluminium oxides are high because
they are giant ionic structures.
Increase in melting point from sodium to
aluminium reflects the increased electrostatic
forces between the ions because of the
increased charges on the ions.
Silicon dioxide has a high melting point
because it is a giant covalent structure.
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It takes a lot of energy / high temperatures
to break the electrostatic forces or covalent
bonds in a giant structure.
Phosphorus and sulfur do not conduct
because they are molecular and do
not have delocalised electrons.
[4]
ii
Phosphorus and sulfur oxides have low
melting points because they are simple
molecules.
b
i
With weak intermolecular forces / van der
Waals’ forces between the molecules
It does not take much energy / low
temperature to overcome these forces of
attraction.
d
i
iii
[6]
ii
In Period 2 there is a gradual increase
with increase in atomic number
up to Group 15 then decrease to
fluorine oxide.
[1]
Nitrogen: N2O5
[1]
Silicon: SiO2
[1]
Chlorine: Cl2O7
[1]
Maximum oxidation number increases
across a period [1]
i
so the shielding is approximately
the same
c
[1]
i
2Mg + O2 → 2MgO
[1]
ii
Electronegativity difference large
between Mg and O so ionic. [1]
Electronegativity difference small
between S and O so covalent. [1]
MgO reacts to form Mg(OH)2 [1]
iii
[Total: 26]
O2− ion behaves as a base / O2− ion accepts
H+ from water molecule. [1]
[1]
SO3 reacts to form H2SO4 Any four of:
so have delocalised electrons which flow
through the structure when a voltage is
applied.
Conductivity increases from sodium to
aluminium because there are more
electrons to conduct.
Silicon does not conduct very well / is a
semiconductor since it is a giant
covalent structure and does not have
delocalised electrons / a few electrons
can jump from place to place.
[1]
so there is the greatest force of attraction
between the nucleus and the outer
electrons.
[1]
In SO3 the O is more electronegative than
the S. So S is δ+ [1]
Sodium, magnesium and aluminium
are metals
4
[1]
Argon
The last (8th) electron added goes into the
same principal quantum shell as the
atoms of others elements in the same
period
[1]
Question 2
a
The energy needed to remove 1 mole of
electrons from 1 mole of atoms of an
element
[1]
Greatest nuclear charge in the period [1]
Number of outer shell electrons / valency
electrons forming bonds is the same as
the (maximum) oxidation number [1]
Example: Al3+ OxNo +3, SO3 OxNo
of S +6. [1]
in the gaseous state to form 1 mole of
gaseous ions.
[1]
In Period 3 there is a gradual increase
from Group 1 to 17.
[1]
ii
Any value between 10−11 and 10−22
(actual value is 10−17 S m−1)
d
Highly polar water molecule (O δ−)
can attack the S (and hydrolyse the
molecule). [1]
i
Covalent
[1]
ii
As4O10 ALLOW: As2O5
[1]
iii
Acid hydrolysis / forms an acid / forms
arsenic acid
[1]
[Total: 23]
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Question 3
a
i
c
Across a period the nuclear charge
increases.
Each electron added to the atom of
successive elements goes into the
same principal quantum shell
i
Melting points of sodium, magnesium
chlorides are high because they are
giant ionic structures.
[1]
It takes a high temperature / lot of energy
to break these strong attractive forces.
[1]
Decrease in melting point from sodium
to aluminium reflects the decreased
ionic bonding / greater proportion of
covalent bonding in aluminium chloride
/ aluminium chloride without water is
covalently bonded.
so the shielding of outer shell electrons by
inner shell electrons does
not increase significantly.
[1]
So across a period, the greater attractive
force between the nucleus and the
outer electrons pulls them closer to the
nucleus.
[1]
ii
Chlorides of silicon, phosphorus and
sulfur have low melting points because
they are simple molecules
The outer electrons in the magnesium ions
are in the second quantum shell
[1]
(because the two outer electrons of the
magnesium atom have been lost).
with weak intermolecular forces / van der
Waals’ forces between the molecules
It does not take much energy / low
temperature to overcome these forces
of attraction.
The sulfide ion has a greater ionic radius
because the outer electrons are
in the third quantum shell.
[1]
so the attraction to the nucleus is much
less than for magnesium.
[1]
b
i
ii
iii
Ease of hydrolysis increases across
Period 3.
SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(g)
Correct formulae
[1]
correct balance
[1]
correct state symbols
[1]
2Al + 3Cl2 → Al2Cl6 Correct formulae [1]
correct balance
5
[1]
[1]
Any five of:
[5]
ii
The melting point of aluminium chloride
is not measured at r.t.p. ALLOW:
phosphorus chloride sublimes so its
melting point cannot be measured.
[1]
iii
Sodium chloride dissolves in water
[1]
so its pH is the pH of water.
[1]
Magnesium chloride undergoes a small
amount of hydrolysis in water
[1]
So sufficient hydrogen ions formed to
lower the pH.
[1]
[Total: 23]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
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Workbook answers
Chapter 11
Exercise 11.1
a
b
c
Exercise 11.3
i
2Ca + O2 → 2CaO
ii
Mg + 2HCl → MgCl2 + H2
iii
Mg + H2SO4 → MgSO4 + H2
iv
Ba + 2H2O → Ba(OH)2 + H2
i
CaCO3
ii
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
iii
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
iv
heat
a
CaO + CO2
1
ALLOW 660−710 °C (actual = about
700 °C)
ii
ALLOW 4−6 g cm−3 (actual about
5.0 g cm−3)
iii
ALLOW 0.97 to 1.04 (actual 1.0)
iv
ALLOW 0.18 to 0.21 (actual = 0.197 nm)
b
Down the group density decreases to
minimum at calcium, then increases.
CaO + H2O → Ca(OH)2
c
Magnesium
i
MgO(s) + 2HCl(aq) →
MgCl2(aq) + H2O(l)
d
Down the group solubility of the hydroxides
increases.
ii
Ba(OH)2(aq) + H2SO4(aq) →
BaSO4(s) + 2H2O(l)
e
Strontium hydroxide is more soluble than
magnesium hydroxide.
So strontium hydroxide has higher
concentration of hydroxide ions in solution
than magnesium hydroxide.
The higher the concentration of hydroxide
ions, the higher the pH.
f
Down the group solubility of the sulfates
decreases.
Exercise 11.2
a
1 with D; 2 with G; 3 with A; 4 with F; 5 with
C; 6 with B; 7 with E
b
i
2Ca(NO3)2 → 2CaO + 4NO2 + O2
ii
BaCO3 + 2HNO3 → Ba(NO3)2 + CO2 +
H2O
c
i
iii
2Sr + O2 → 2SrO
iv
MgCO3
i
SrCO3(s)
ii
2Mg(NO3)2•6H2O(s) → 2MgO(s) +
4NO2(g) + O2(g) + 6H2O(l)
iii
SrCO3(s) + 2HCl(aq) → SrCl2(aq) +
CO2(g) + H2O(l)
iv
2Sr(s) + H2O(l) → Sr(OH)2(aq)
heat
heat
Exercise 11.4
a
MgO + CO2
SrO(s) + CO2(g)
i
Down the group, there is an increase in
reactivity.
ii
2Mg(s) + O2(g) → 2MgO(s)
iii
Litmus will turn blue due to alkaline pH.
The magnesium oxide reacts with water
to form magnesium hydroxide, which is
slightly alkaline.
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b
i
Constant heating is required (which
means energy is absorbed).
iii
2Ba + O2 → 2BaO
[1]
iv
Oxidation and reduction occur in the
same equation / together
[1]
ii
∆Hr
MgCO3(s)
Oxidation number of Ba increases from
[1]
0 in Ba to +2 in Ba2+ ion in BaO
MgO(s) + CO2(g)
∆Hf [MgO(s)]
∆Hf [MgCO3(s)]
Oxidation number of O decreases from 0
[1]
in O2 to −2 in O2− ion in BaO
∆Hf [CO2(g)]
1
Mg(s) + C(graphite) + 12 O2(g)
iii
iv
c
d
b
−1095.8 + ∆H⦵r = −601.7 − 393.5
So ∆H⦵r = −601.7 − 393.5 + 1095.8
= + 100.6 kJ mol−1
a
2Ca(NO3)2 → 2CaO + 4NO2 + O2
ii
Brown fumes of nitrogen dioxide
i
Group 2 nitrates are more difficult
to decompose / decompose at higher
temperatures as you go down the group.
Calcium hydroxide decomposes at a lower
temperature than barium hydroxide and
calcium is higher in the group, which is
the same pattern.
iii
Sr(OH)2
heat
ii
1s22s22p63s23p6
[1]
i
[1]
down the group
[1]
Calcium hydroxide is more soluble than
magnesium hydroxide
[1]
[1]
The higher the concentration of hydroxide
ions, the higher the pH.
[1]
b
The solubility of the hydroxides increases
down the group [1]
Molar mass of calcium hydroxide = 74.1
[1]
[1]
state symbols
[1]
Going down the group the ionisation
energies decrease
[1]
(in 100 g water)
[1]
In 500 cm3 water = 0.556 g
[1]
d
Solubility decreases down the group
[1]
e
i
Barium sulfate
[1]
because it is insoluble in water / its
solubility in water is very low
[1]
ii
because the outer electrons are in electron
shells further from the nucleus
[1]
[1]
and more shielding of outer electrons
by electrons in inner shells.
[1]
So the two outer electrons are more easily
transferred to the oxygen atom going
down the group.
[1]
2
Solubility of hydroxides increases
mass = moles × molar mass
= (1.50 × 10−3) × 74.1 = 0.111 g
Sr+(g) → Sr2+(g) + e−
Correct reactants and products
so there is less attraction between the
outer electrons and the nucleus [Total: 13]
so calcium hydroxide has higher
concentration of hydroxide ions in
solution than magnesium hydroxide.
c
SrO + H2O
Question 1
i
ii
ii
Exam-style questions
a
[1]
Question 2
i
ALLOW between 770 and 1300 °C.
Ionic radius increases
Strontium carbonate decomposes
less readily because, as you go down
the group, the temperature at which
decomposition occurs gets higher.
ii
i
f
i
Ba2+(aq) + SO42−(aq) → BaSO4(s)
Correct reactants and products
[1]
state symbols
[1]
BaCO3(s) + H2SO4(aq) → BaSO4(s) +
CO2(g) + H2O(l)
Correct reactants and products
[1]
state symbols
[1]
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ii
Barium sulfate is insoluble in water
[1]
so forms a layer round the outside of
the barium carbonate particles
[1]
c
(so the sulfuric acid cannot reach the
barium carbonate)
[Total: 13]
d
Question 3
a
O2− + H2O → 2OH−
b
Molar mass of strontium oxide = 103.6 g and
molar mass of strontium nitrate = 211.6 g [1]
41.4
[1]
Moles of SrO = 103.6 = 0.400 mol
Mass strontium nitrate = 0.400 × 211.6
= 84.6 g
[1]
2Sr(NO3)2(s) → 2SrO(s) + 4NO2(g) + O2(g)
Correct formulae
[1]
correct balance
[1]
correct state symbols
[1]
Oxidation and reduction occur in the same
equation / together.
[1]
Oxidation number of Sr increases from 0 in Sr
[1]
to +2 in Sr2+ ion in Sr3N2
[1]
Oxidation number of N decreases from 0 in
[1]
N2 to −3 in N3− ion in Sr3N2
e
i
Because the outer electrons in strontium
are in electron shells further from the
nucleus
[1]
ii
1s22s22p63s23p63d104s24p65s2
[1]
iii
Sr → Sr2+ + 2e−
[1]
3
[Total: 13]
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Workbook answers
Chapter 12
Exercise 12.1
a
1 with C; 2 with A; 3 with E; 4 with F; 5 with
D; 6 with B
b
hydrogen iodide < hydrogen bromide <
hydrogen chloride < hydrogen fluoride
c
2HI(g) ⇌ H2(g) + I2(g)
d
Exercise 12.2
a
b
c
1
The suspected halide is dissolved in dilute
nitric acid. A few drops of aqueous
silver nitrate are added. If chloride ions are
present a white coloured precipitate is formed
which goes grey / violet in the presence of
light. The precipitate dissolves in dilute
ammonia solution. If a bromide is present
a cream coloured precipitate is seen which
dissolves in concentrated ammonia solution.
i
AgNO3(aq) + KI(aq) →
AgI(s) + KNO3(aq)
ii
Ag+(aq) + I−(aq) → AgI(s)
i
Carry out the experiment in a fume
cupboard. Place sodium chloride in the
flask and sulfuric acid in the dropping
funnel. Remove the stopper from the
dropping funnel and open the tap so that
the sulfuric acid drips slowly onto the
sodium chloride. Close tap when sufficient
sulfuric acid has been added.
e
ii
Hydrogen chloride is denser than air.
iii
White fumes of hydrogen chloride would
be seen at the mouth of the measuring
cylinder (due to reaction with water
vapour in the air).
i
2HI + H2SO4 → I2 + SO2 + 2H2O
ii
6HI + H2SO4 → 3I2 + S + 4H2O
iii
8HI + H2SO4 → 4I2 + H2S + 4H2O
i
I from −1 to 0; S from +6 to +4
ii
I from −1 to 0; S from + 6 to 0
iii
I from −1 to 0; S from +6 to −2
f
Brown solution (aqueous iodine), purple
fumes (iodine vapour), black solid (iodine),
smell of rotten eggs (hydrogen sulfide)
g
Sulfuric acid reacts with NaCl to produce
NaHSO4 and HCl. The bond energy H─Cl
is relatively high so HCl can only undergo
thermal decomposition at very high
temperatures.
Sulfuric acid reacts with NaI to produce
NaHSO4 and HI. The bond energy H─I is
low high so HI readily undergoes thermal
decomposition to produce iodine and sulfur
dioxide. Iodide ions are good reducing agents
and so SO2 can be reduced to S and further
to H2S.
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Exercise 12.3
a
b
c
d
Halogen
Halide
Reaction or no
reaction
Colour of the aqueous
mixture after the
addition
Colour of the
hexane layer
Cl2(aq)
NaBr(aq)
reaction
orange
dark orange
I2(aq)
KCl(aq)
no reaction
brown (colour of I2(aq))
purple
Br2(aq)
KI(aq)
reaction
brown
purple
Cl2(aq)
LiBr(aq)
reaction
orange
dark orange
Br2(aq)
MgCl2(aq)
no reaction
orange (of bromine)
dark orange
Cl2(aq)
NaI(aq)
reaction
brown
purple
i
Chlorine because it is better at accepting
electrons from another reactant than the
others in the table.
ii
Iodide because it is better at donating
electrons to another reactant than the
others in the table.
i
Cl2(aq) + 2KI(aq) → I2(aq) + 2KCl(aq)
ii
Br2(aq) + 2NaAt(aq) → At2(aq) +
2NaBr(aq)
iii
Cl2(aq) + MgBr2(aq) → Br2(aq) +
MgCl2(aq)
i
Br2(aq) + 2KI(aq) → I2(aq) + 2KBr(aq)
ii
Cl2(aq) + 2NaBr(aq) → Br2(aq) +
2NaCl(aq)
e
f
i
Because they are measured at different
temperatures and density varies slightly
with temperature.
ii
The particles in a liquid are packed close
together and so the mass divided by the
volume of the atoms can still be compared.
The comparison with iodine may be less
valid because the packing will be different.
H2(g) + Cl2(g) → 2HCl(g)
Exercise 12.5
a
i
Mg(s) + Cl2(g) → MgCl2(s)
ii
Mg has been oxidised because its
oxidation number has increased from
0 to +2.
Chlorine has been reduced because its
oxidation number has decreased from
0 to −1.
Exercise 12.4
a
2
i
ALLOW: 200−450 °C (actual = 337 °C)
ii
ALLOW: 0.5−1.54 g cm−3
(actual = 1.54 g cm−3)
iii
Grey black / black
b
b
Colour gets darker going down the group.
c
Liquid, because −4 °C is above the melting
point but lower than the boiling point.
d
Less volatile down the group because the
boiling points are increasing down the group.
i
chlorine = 0; sodium chloride = −1;
sodium chlorate(I) = +1
ii
The chlorine has been reduced to chloride
ions (oxidation number change 0 to −1)
and it has also been oxidised to Cl in
chlorate (oxidation number change 0
to +1).
iii
Cl2 + 2OH− → Cl− + ClO− + H2O
1
−
−
−
2 Cl2 + 2OH → ClO + H2O + e
(oxidation because loss of electrons)
1
−
−
2 Cl2 + e → Cl (reduction because of
gain of electrons by Cl)
iv
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c
d
i
To kill bacteria / to kill microbes
ii
The chlorine has been reduced to chloride
ions (oxidation number change 0 to −1)
and it has also been oxidised to Cl in
chlorate (oxidation number change 0
to +1).
of nitrogen / has decreased its oxidation
number.
[1]
e
i
ii
H2(g) + Cl2(g) → 2HCl(g)
[1]
White precipitate formed
[1]
Precipitate dissolves
[1]
[Total: 19]
Exam-style questions
Question 2
Question 1
a
a
i
ii
iii
Gas because −32 °C is above the boiling
point of chlorine.
[1]
ALLOW: −250 to −150 °C (actual value
= −220 °C)
[1]
chlorination kills (harmful) bacteria
[1]
Cl2 + H2O ⇌ HClO + HCl
[1]
Hypochlorous acid and chlorine are strong
oxidising agents
[1]
b
i
Less volatile down the group / less volatile
with greater relative molecular mass
[1]
id-id forces. instantaneous dipole-induced
dipole forces
[1]
ii
These forces increase as the number of
electrons in the molecules increases
[1]
Number of electrons in the halogens
increases as the molar mass of the
molecules increases
[1]
b
1s22s22p63s23p63d104s24p5
c
Halogens are better oxidising agents, the
higher they are in the group / less good
oxidising agents down the group.
d
[1]
So electrons more easily gained
[1]
i
ii
iii
[1]
and the greater the pull of the nuclei on
electrons from other atoms.
c
i
ii
3NH3 + 3Cl2 → NCl3 + 3HCl
Correct formulae
[1]
correct balance
[1]
Chlorine is oxidising agent because it has
increased the oxidation number
[1]
Oxidation number of Cl in sodium
chlorate(V) = +5
[1]
Chlorine has been reduced to
chloride ions (oxidation number change
0 to −1)
[1]
It is disproportionation because the
chlorine has been both oxidised and
reduced at the same time.
[1]
Cl2(aq) + 6OH−(aq) → 5Cl−(aq) +
ClO3−(aq) + 3H2O(l)
[1]
Cl2(aq) + 2KI(aq) → I2(aq) + 2KCl(aq)
Correct formulae and balance
[1]
correct state symbols
[1]
Colourless (ALLOW: very light green)
to brown
[1]
Aqueous iodine is brown
iii
Oxidation number of nitrogen increases
from −3 to + 3.
[1]
Oxidation number of chlorine changes
from 0 to −1.
[1]
Oxidation number of Cl in sodium
chloride = −1
Chlorine has been oxidised to Cl in
chlorate(V) (oxidation number change
0 to +5)
[1]
[1]
The higher in the group, the smaller the bond
length / the smaller the atoms
[1]
3
Add nitric acid and aqueous silver
nitrate
d
i
[1]
Shake the mixture with hexane then allow
the layers to settle.
[1]
Colour in the hexane layer is purple.
[1]
Hydrogen iodide
[1]
Oxidation number of S decreases / goes
from +6 to 0 OR oxidation number
of I increases / goes from −1 to 0
[1]
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ii
+6 to −2
iii
1 mark each for any three of:
[1]
d
i
Cl2 + 2KBr → Br2 + 2KCl
[1]
ii
Add nitric acid and aqueous silver
nitrate
[1]
Cream precipitate formed
[1]
Precipitate dissolves
[1]
because a complex ion is formed
[1]
which is soluble in the solution.
[1]
Manganese(IV) oxide
[1]
Black solid / yellow solid / purple vapour /
brown solution / bad egg smell
[3]
[Total: 21]
iii
Question 3
a
i
Chlorine reacts faster than bromine with
hydrogen.
[1]
e
i
Hydrogen reacts explosively with chlorine
in sunlight.
[1]
b
Bromine reacts on heating.
[1]
i
No effect
[1]
ii
Position of equilibrium shifted to the left /
concentration of HI lowered / reaction
shifted in favour of the reactants
[1]
iii
c
ii
MnO2 + 4H+ + 2e− → Mn2+ + 2H2O
[1]
Reduction because gain of electrons by
Mn / decrease in oxidation number
of Mn from +4 to +2
[1]
Position of equilibrium shifted to the right /
concentration of HI increased / reaction
shifted in favour of the product
[1]
The atoms get larger as the atomic number is
increased / down the group.
[1]
Increases the oxidation number of iodine
from −1 to 0
[1]
1
I + e−
[1]
I− + e− →
2 2
Oxidation because of loss of electrons
from I− / increase in oxidation number
of I from −1 to 0
[1]
[Total: 21]
The attraction between the nuclei of the
atoms and the electrons in the covalent bond
decrease (as the atoms get bigger).
[1]
The energy needed to break the bond
decreases / bond energy gets smaller
(down the group).
4
[1]
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Workbook answers
Chapter 13
Exercise 13.1
a
b
Exercise 13.2
Nitrate fertilisers dissolve in rainwater and are
then leached into lakes and rivers. The nitrates
promote the excessive growth of water plants,
especially algae. The algae spread across the
surface of the water and block out the light so
that the water plants cannot grow. The plants
are decomposed by aerobic bacteria, which
multiply and use up the dissolved oxygen
in the water. Fish and other water creatures
cannot survive without oxygen and they die.
This process taking place in streams and rivers
leading to the death of plants and animals is
called eutrophication.
a
i
N
b
N
ii
It has a triple bond with very high bond
energy.
i
Lone pairs
ii
107° because lone pair─bond pair
repulsion is greater than bond pair─bond
pair repulsion. So the bond angle closes
up compared with 4 bonding pairs to
minimise repulsive forces of electron pairs.
iii
109.5° There are no lone pairs so there are
only bond pair─bond pair repulsions so
the bonds move to minimise these electron
pair repulsions.
i
Removal of crop plants depletes the
soil of nitrogen so there is not sufficient
nitrogen in the soil after several years for
plants to grow well. Plants need nitrogen
to grow / plants need nitrogen to make
proteins.
i
Ammonia is less dense than air.
ii
The products are ammonia which is a
weak base and HCl which is a strong
acid. So there are more hydrogen ions
than hydroxide ions (from ammonia) in
solution.
ii
Damp red litmus placed at the mouth of
the tube turns blue.
iii
To dry the ammonia (since water vapour
will also be formed).
iii
Lone pair of electrons on ammonia
accepts hydrogen ion from water.
Hydroxide ion formed.
iv
It will react with the ammonia to form
ammonium sulfate.
Displacement of ammonia
2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 +
2H2O
v
iv
vi
Neutralisation (of ammonia): the reaction
of an acid with an alkali.
v
Calcium carbonate
c
Displacement (of ammonia): The
replacement of one group by another.
In this case hydroxide ions in calcium
hydroxide have been displaced by
chloride ions.
1
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Thermal decomposition: The ammonium
chloride would break down when heated
to form ammonia and hydrogen chloride.
d
e
i
NH3 + H3PO4 → (NH4)3PO4
ii
Heat the solution to evaporate the water.
Melt the solid formed and then spray it in
a tower through which air is blown.
iii
To provide extra nutrients / extra nitrogen,
phosphorus and /or potassium to the soil,
to replace nutrients taken up by previous
crops. These elements increase the growth
of (crop) plants and therefore increase
food production.
g
h
i
i
By combination of nitrogen with oxygen
at the high temperature within the engine.
ii
4CO + 2NO2 → 4CO2 + N2
i
NO2 reacts with SO2 to form SO3 and NO.
The NO is then oxidised by atmospheric
oxygen to reform NO2. The NO2 acts as a
catalyst since it is reformed and can then
oxidise further molecules of SO2.
ii
NO2 is in the same phase (gas phase) as
the SO2 and the products.
i
Reduced growth (because of increased
leaching of nutrients from the soil);
increased likelihood of disease (due to
decomposition of the waxy covering on
the surface of leaves); leaf burn (especially
in conifers); reduction in ability to absorb
material through the roots, etc.
ii
Corrosion of metals; erosion of mortar;
erosion of carbonate building materials
(limestone / chalk, etc.).
Nitric acid
Exercise 13.3
a
In the mornings the concentration of nitrogen
dioxide increases. After midday there is a
slight decrease but there is a general increase
from mid-afternoon until early evening. There
is a decrease in concentration from the late
evening to early morning (except for day 3).
b
Increasing amount of traffic producing
nitrogen dioxide, e.g. diesel vehicles.
c
i
2NO + O2 → 2NO2
ii
NO2 + O2 → NO + O3
iii
The NO formed in the second reaction
can react with more oxygen in the air to
form more nitrogen dioxide so the reaction
continues in the presence of oxygen.
i
Nitrogen dioxide is brown. The
concentration of the nitrogen dioxide
increases during the day.
d
ii
e
f
2
Exam-style questions
Question 1
a
Nitrogen reacts with oxygen in the presence
of lightning. The lightning produces high
enough temperatures for the reaction to occur.
Nitrogen(II) oxide / nitric oxide is formed. This
reacts with oxygen to form nitrogen dioxide.
Line A because towns are more likely to
have more traffic.
[1]
Idea of nitrogen dioxide being produced
by car engines /high temperature furnaces
in industry.
[1]
The amount of ozone will gradually
increase during the day as the
concentration of nitrogen dioxide
increases (see reaction (c)(ii)).
Day 3. Temperatures are usually lower at
night so the amount of nitrogen dioxide in the
atmosphere during the night falls in the first
two days.
i
ii
Winds blow the nitrogen dioxide from
other areas where concentration is
higher
[1]
iii
−6 g(1) moles NO
400 ×× 10
10−6
g moles NO2 2
400
=
400 × 10 −6
[1]
or 8.7 × 10 −6 mol [1]
8.7 × 10 −6
mol
1000
= 8.7 × 10 −9 mol dm −3 [1]
46
In 1 dm 3 =
b
High temperature caused by (electric
discharge) of lightning
[1]
causes nitrogen to combine with oxygen to
form NO / nitric oxide.
[1]
Nitric oxide reacts with oxygen to form
nitrogen dioxide.
[1]
N2 + O2 → 2NO / 2NO + O2 → 2NO2
[1]
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c
d
4CO + 2NO2 → 4CO2 + N2
c
4NH3 + 5O2 → 4NO + 6H2O
[1]
Correct formulae
[1]
ii
for fertilisers
[1]
correct balance
[1]
iii
NO oxidised to NO2 in atmosphere
[1]
i
N in NO has oxidation number +2; N in
N2 has oxidation number +4 / oxidation
number change is −2.
[1]
NO2 reacts with rainwater to form nitric
acid
[1]
iv
C in CO has oxidation number +2; C in
CO2 has oxidation number +4 / oxidation
number change is +2.
[1]
Decrease in oxidation number is reduction
and increase in oxidation number is
oxidation.
[1]
ii
e
i
i
in oxidation of sulfur dioxide to sulfur
trioxide
[1]
d
in presence of high electric discharge from
lightening
[1]
e
free radical
[1]
The catalyst is a solid and the reactants
are gases / the catalyst is in a different
phase to the reactants.
[1]
f
i
sunlight / ultraviolet light
[1]
ii
peroxyacetyl nitrate / PAN
[1]
It has very strong bonds / its bond energy
is very high
[1]
N
a
i
N
[1]
the 2 lone pairs
[1]
Decrease in oxidation number is reduction
and increase in oxidation number is
oxidation.
[1]
[Total: 21]
3
[1]
H
H N H
(NH4)2SO4 + 2NaOH → 2NH3 + Na2SO4
+ 2H2O
Correct formulae
[1]
Correct balance
[1]
ii
Displacement
[1]
i
Incompletely ionised in solution /
incompletely dissociated in solution
[1]
Lone pair on ammonium ion accepts
proton / H+ ion
[1]
from water
[1]
Hydroxide ion / OH− formed which is
responsible for alkalinity
[1]
ii
+
ii
Question 2
b
N in NH4+ has OxNo −3; N in NO2− has
OxNo +3 / oxidation number change
is +6.
[1]
O in O2 has OxNo 0; O in H2O has OxNo
−2 / oxidation number change is −2.
[1]
3 pairs of bonding electrons
[Total: 18]
Question 3
ii
i
[1]
Reaction of nitrogen with oxygen
[1]
Heterogeneous catalysis
It needs a high temperature / lot of energy
to break these bonds in order to react [1]
a
[1]
acts as a catalyst
H
iii
iv
b
2NO2− + O2 → 2NO3− Correct species
[1]
balanced
[1]
Idea of run off / ions moving into deeper
soil / ions going to where plant roots
cannot reach them
[1]
Nitrogen has very strong bonds / its bond
energy is very high.
[1]
It needs a high temperature / lot of energy to
break these bonds in order to react.
[1]
c
Ammonia is a proton acceptor / ammonia is
[1]
taking H+ ions from the acid.
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d
e
4
2 × 14 × 100 = 35% [2]
80
(If 2 marks not scored, 1 mark for correct
molar mass of NH4NO3 (80))
i
Potassium nitrate is less soluble in water
so not so much is leached away.
[1]
ii
Ammonium nitrate is more soluble in
water so plants can absorb it more readily /
ammonium nitrate has a lower density so it
is more economical to spread it /
ammonium nitrate has greater % nitrogen.[1]
f
NH4+ + OH− → NH3 + H2O
[1]
[Total: 15]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
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Workbook answers
Chapter 14
Exercise 14.1
a
b
1 with C; 2 with A; 3 with G; 4 with B; 5 with
H; 6 with D; 7 with E; 8 with F
d
H C
C
C H
H
H
H C H
ii
Propanol
H
iii
But-1-ene
iv
Pentanal
v
1-bromopentane
vi
Propanone
iii
ix
butan-1-ol
x
Pent-1-ene
xi
Methanal
a
xiii pentylamine
2-methylbutane
T
Butan-2-ol
U
1,2-dibromoethene
V
Hex-2-ene
W 3-ethyl-3methylhexane
X
H
H
O
H
H C
C
C H
H
iv
O
H
H
H C
C
C O C
H
H
H
H
H
H
H
H
H C
C
C
H
H
H
O
C
H
Exercise 14.2
xii pentanenitrile
S
ii
O H
H
Hexane
viii Heptane
1
H
i
vii Propanoic acid
c
i
i
Addition
ii
Oxidation
iii
Hydrolysis
iv
Substitution
v
Elimination
vi
Reduction
vii Substitution
viii Condensation
b
i
In heterolytic fission a covalent bond
breaks so that two free radicals are
formed.
ii
In homolytic fission a covalent bond
breaks so that one of the atoms in the
bond accepts both the bonding electrons.
iii
A nucleophile is a species which donates
/ gives a pair of electrons to an electrondeficient atom.
iv
An electrophile is a species with a positive
or partially positive charge which accepts
a pair of electrons from another species.
2,2-dibromobutane
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c
d
v
A carbocation is an organic ion which has
a carbon atom which is positively charged.
vi
radical, unpaired
i
Termination
ii
Initiation
iii
Propagation
i
B
ii
D
e
R is electrophilic addition. Bromine is adding
across the double bond of ethene so it is an
addition reaction (no other product formed).
It is electrophilic because the electrophile is
the positively polarised end of the bromine
molecule which attacks the double bond of
the ethene. The bromine accepts accept a pair
of electrons from the double bond.
S is nucleophilic substitution. It is substitution
because the OH replaces the Br. It is
nucleophilic because the nucleophile is the OHion which donates donates a pair of electrons to
the positively charged area of the carbocation.
Exercise 14.3
a
Type of
formula
Compound
Butane
But-2-ene
Propan-2-ol
Propanone
displayed
formula
H H H H
H C C C C H
H H H H
H
H H
H C C C C H
H H
H
H H H
H C C CH
H
HO H
H OH
H CCCH
H H
condensed
structural
formula
CH3CH2CH2CH3
CH3CH=CHCH3
CH3CH(OH)CH3
CH3COCH3
molecular
formula
C4H10
C4H8
C3H8O
C3H6O
OH
O
C3H8O
C3H6O
skeletal
formula
empirical
formula
b
c
CH2
i
L = C2Cl4
M = C2H8N2
N = C4H6O5
O = C9H11NO2
ii
L = CCl2
M = CH4N
N = C4H6O5
O = C9H11NO2
Cl
Cl
Cl
Cl
Y
2
C2H5
OH
Cl
Z
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Exercise 14.4
Exercise 14.5
a
Structural isomers are compounds with
the same molecular formula but different
structural formulae.
a
b
i
H
H H
i
A group of organic compounds having the
same functional group, the same general
formula and similar chemical properties.
ii
An atom or group of atoms in an organic
molecule that determines the characteristic
reactions of a homologous series
i
CnH2n+2
ii
CnH2n+1OH / CnH2n+2O
C H
H C C
H O HH
ii
b
H
H H
H C C C
O
H H
iii
H
H
H
H C C
C
C H and H C
H C H
C
i
d
i
There is a general increase in density as
the number of carbon atoms increases.
ii
Ethanol has a density which is lower than
expected from the trend / methanol has a
higher density.
iii
allow values between 0.61 and 066
(inclusive)
iv
There is a general increase in boiling point
as the number of carbon atoms increases.
v
Butane ALLOW: values between 250 and
290 K (actual = 272.6 K) Pentan-1-ol
ALLOW: values between 400 and 420 K
(actual = 411.1 K).
vi
CH3CH2 is ethyl, CH3 is methyl
C H
H
H
H C H
H
H
c
C6H14O
H C H
H
H
H H
H H
c
H
H
H
H
H
H C H
C H
H
C
C H
Br
C
Br
C
C
Br
Br
C H
H H
cis-2,3-dibromobut-2-ene trans-2,3-dibromobut-2-ene
ii
d
Cis/trans isomerism / geometrical
isomerism
i
H
ii
Br
C3H7
C
C
CH3
Cl
C2H5
CH3
Exercise 14.6
a
i
H
HOOC
H
C
H
COOH HOOC
H
H
H
H
C
C
C
H
H
H
COOH
trans-cyclopropane
1,2-dicarboxylic acid
isomers of cyclopropane dicarboxylic acid
H
C C
H
methane
H
C H
C
cis-cyclopropane
1,2-dicarboxylic acid
H
C
C atom in centre of molecule (attached to
4 different groups)
e
3
The structural formula of methane is CH4.The
structural formula of ethane is CH3CH3.
ii
b
H
H
H
H
ethane
109.5°. There are only bonding pairs of
electrons so the repulsions between these
bonding pairs are the same, leading to the
minimum repulsion which corresponds to
the tetrahedral bond angle.
σ bonds are single bonds formed by the
hybridisation of the s orbitals of the hydrogen
atoms with the p orbitals of the carbon atom
forming sp2 hybrid orbitals. Then π bond
is formed by the sideways overlap of the
remaining p orbitals from the carbon atoms
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c
d
e
to form a bond which has its electron density
above and below the plane of the C and H
atoms and at right angles to this plane.
c
All the bonds in butane are single bonds.
There is free rotation of the atoms about
single bonds and so there are no isomers with
fixed positions of the atoms. But-2-ene has a
double bond. There is no free rotation around
a double bond, so two possible geometric
isomers are possible.
d
[1]
CH3CH2CH2CH2OH + 2[O] →
CH3CH2CH2COOH + H2O
Correct reactants and products
[1]
balance
[1]
H
HO
CH3
C2H5
Groups attached correctly to central c atom[1]
[1]
use of wedge and dashed line
f
i
ii
OH
[1]
Carbon atom attached to the OH [1]
This carbon has 4 different groups
bonded to it [1]
[Total: 16]
a
CH3CH2CH2Cl
[1]
C H O
47.4
10.5
42.1
12.0 1.00 16.0
(1 for correct division)
b
i
(Chlorine) free radical
[1]
ii
Homolytic
[1]
c
Carbocation
[1]
3.95
10.5
2.63
2.63
2.63
2.63
d
i
[1]
3
8
2 = C3H8O2
(1 for simplest ratio and formula)
ii
[3]
Empirical formula mass = (12 × 3) + (8 × 1) +
(16 × 2) = 76
This is also the molecular mass. Indication of
working required.
[1]
b
Elimination / dehydration
C
1.50
3.99
1.00
(1 for dividing by smallest number)
ii
ii
Question 2
Question 1
i
[1]
Exam-style questions
a
Substitution
e
The repulsion between lone pairs of electrons
and bond pairs of electrons is greater than
the repulsion between two bonding pairs of
electrons. So the bond angle is less than the
tetrahedral bond angle to minimise repulsive
forces between the electron clouds.
There are 4 sp3 hybrid orbitals of the same
shape. The electron density in each of these is
exactly the same. The areas of electron density
arrange themselves so the electron repulsions
are minimised. This is the tetrahedral
structure. the bond angle is 109.5°.
i
e
Nucleophile
It donates a pair of electrons to a
positive / partially positive ion
[1]
It shows the direction of movement
of the electron pair
[1]
iii
Alcohol(s)
[1]
iv
2-methylpropan-2-ol
[1]
(Nucleophilic) substitution
[1]
[Total: 10]
Any 3 of: CH3CH2CH2CH2OH
CH3CH2CHOHCH3
(CH3)3COH
CH3CH(CH3)CH2OH ALLOW: displayed
formulae
[3]
4
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Question 3
a
b
i
with two hydrogen atoms and the other
carbon [1]
i and ii
O
O
C
C C C
(1 mark each)
remaining carbon electrons form pi-bond
(by sideways overlap of p orbitals) [1]
[2]
pi-bond electron density above and below
the plane of the C and H atoms [1]
[1]
Leads to planar arrangement of
ethene H
b
i
CnH2n+1CHO
ii
H H H
O
[1]
c
d
i
H
ii
Reduction [1]
Addition of hydrogen [1]
ii
Propan-1-ol [1]
i
Hydration [1]
ii
Propene [1]
iii
Alkene(s) [1]
c
i
5
electrophilic [1]
addition [1]
CH3
C
ii
i
Cl2 → 2Cl• [1]
ii
methyl free radical [1]
iii
propagation [1]
iv
CH3• + CH3• → C2H6 or CH3• + Cl• →
CH3Cl
[1]
iii
C
CH3
CH3
Cl
CH3
[1]
Cl Cl
C
[Total: 10]
Question 4
a
[1]
So H─C─H bond angle about 120° (allow
117-120°) to minimise repulsions [1]
H C C C C
H H H
Each C atom forms three sigma-bonds [1]
[1]
C
Cl
The C─C bond can rotate freely [1]
So there is no fixed positions for the
groups attached to the C atoms [1]
[Total: 16]
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Workbook answers
Chapter 15
Exercise 15.1
a
b
The top of the fractionating column is at a
lower temperature than the bottom. Petroleum
enters the column at the bottom in both the
gaseous state and liquid state. The more volatile
hydrocarbons with lower relative molecular
masses rise up the column further than the
less volatile hydrocarbons. The hydrocarbons
condense at different levels in the column as the
temperature falls and are collected as liquids.
The most volatile hydrocarbons are alkanes
with very short carbon chains (methane,
ethane, propane and butane). These leave the
top of the fractionating column as gases.
i
To provide more alkanes for which
demand is greater than supply / to produce
more petrol / to produce more diesel
ii
They are used to make other chemicals /
they are used to make plastics.
c
iii
They have a double bond, one of which
can break and add other atoms. The
activation energy for this is relatively
low compared with the activation energy
required to break a single bond in an
alkane.
iv
High temperature and zeolite catalyst
v
The alkane might undergo combustion.
vi
C─C bonds
i
C12H26 → C7H16 + C5H10
ii
C18H38 → C6H12 + C8H18 + C4H8
iii
C10H22 → C3H6 + C7H16
iv
C3H8 → C3H6 + H2
Exercise 15.2
Hydrocarbon
Displayed formula
Structural formula
butane
H H H H
H C C C C H
H H H H
CH3CH2CH2CH3
ethene
H
CH2=CH2
C2H4
H
but-2-ene
H
H
H C C C C H
H H H H
C4H10
C4H8
C=C
H
Skeletal formula
H
CH3CH=CHCH3
(Continued)
1
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Hydrocarbon
cyclopentane C5H10
Displayed formula
H
H
buta-1-3-diene C4H6
H
C
HH
C C
H
C C
H
H HH
H
H
C C C C
H
H H
H
i
ii
iii
iv
v
b
i
ii
iii
c
i
ii
iii
2
CH2 CH2
CH3=CHCH=CH2
The second step (propagation step) is attack of
the chlorine free radical on ethane.
C2H6 + Cl• → C2H5• + HCl
Hexane burns / blue flame / slightly smoky
flame. Combustion reaction. Alkanes
burn with blue flame in excess air or
smoky flame when air not in excess.
C2H5• can react further and regenerate a
chlorine free radical.
C2H5• + Cl2 → C2H5Cl + Cl•
No reaction. Ultraviolet light needed to
form chlorine free radicals with enough
energy to react with alkanes.
In excess chlorine this can continue until
all the hydrogen atoms are substituted by
chlorine.
Two layers, the top one of which is
(purple) coloured. No reaction but
aqueous solution is immiscible with water
so two layers formed. Iodine is more
soluble in organic solvent so the colour
migrates here.
C3H8 + 5O2 → 3CO2 + 4H2O
C10H22 + 15 1 O2 → 10CO2 + 11H2O (or
2
doubled)
C4H10 + 4 1 O2 → 4CO + 5H2O (or doubled)
2
C7H16 + 7 1 O2 → 7CO + 8H2O (or
2
doubled)
C12H26 + 12 1 O2 → 12CO + 13H2O (or
2
doubled)
The first step is homolytic fission.
Cl2 → 2Cl•
No reaction. Alkanes are unreactive apart
from combustion and reaction with free
radicals.
Green colour of chlorine disappears.
Ultraviolet light forms chlorine free
radicals which have enough energy to
react with alkanes.
C6H14 + 9 1 O2 → 6CO2 + 7H2O (or
2
doubled)
Skeletal formula
CH2
CH2 CH2
d
Exercise 15.3
a
Structural formula
The reaction can be terminated when free
radicals combine.
For example:
C2H5• + Cl• → C2H5Cl
C2H5• + C2H5• → C4H10
e
combustion/burning; ultraviolet
Exercise 15.4
a
1 with C; 2 with E; 3 with D; 4 with A; 5 with
F; 6 with B
b
i
Pent-2-ene
ii
Methylpropene
iii
Hept-2-ene
i
CH3CH2CH═CH2 + HBr →
CH3CH2CHBrCH2Br
ii
CH3CH═CHC2H5 + 4[O]
CH3COOH + C2H5COOH
hot
c
concentrated
KMnO4
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iii
CH2═CHCH2CH═CH2 + 2H2
So the charge density on the carbocation /
C+ ion is reduced. The greater the
reduction in charge on the C+ ion /
carbocation, the more stable the C+ ion /
carbocation.
Ni
CH3CH2CH2CH2CH3
iv
CH2═CH2 + H2O
v
CH3CH═CHCH3 + H2O + [O]
conc H3PO4
CH3CH2OH
The secondary carbocation is more stable
because it has more alkyl groups attached.
dilute cold
KMnO4
CH3CH(OH)CH(OH)CH3
d
Add aqueous bromine. Unsaturated
hydrocarbon: orange aqueous bromine turns
colourless. Saturated hydrocarbon: orange
aqueous bromine remains orange.
OR
Add aqueous acidified KMnO4. Unsaturated
hydrocarbon: purple aqueous KMnO4 turns
colourless. Saturated hydrocarbon: purple
aqueous KMnO4 remains purple.
Exercise 15.6
a
i
ii
Exercise 15.5
a
i
ii
The high electron density in the ethene
double bond repels the bonding electrons
in the bromine and pushes them towards
the further bromine atom making the Br
atom nearest the ethene δ+.
H
H
C
C
H
H
iii
v
H
H
C
C
H
Br
b
3
+
H n
C6H5 H
C
C
H
H n
F
F
C
C
F
F
n
i
The C─C bonds are not reactive.
There are few / no bacteria / organisms
which can break the C─C bonds in
hydrocarbons.
ii
Block watercourses / rivers / drains
(leading to flooding); litter / eyesore; gets
into animals’ or birds’ gullets and chokes
them; takes up space in landfill sites.
iii
Poisonous fumes released; carbon dioxide
(greenhouse gas) released which causes
enhanced global warming; acidic gases, e.g.
HCl, released which contribute to acid rain.
Br
Carbocation
H
c
δ−
iv
C
A is CH3CH2CH=CH2; B is CH3CH=CHCH3
Br
Heterolytic because one of the bromine
atoms takes both electrons of the electron
pair.
C
b
δ+
iii
CH3 H
H
Exercise 15.7
Br−
a
CH3CH2C+HCH3 ALLOW: displayed
formula
Disturbs the balance between respiration
and photosynthesis which are approximately
balanced. A small increase in concentration
of atmospheric carbon dioxide can cause a
significant increase in global warming.
b
Alkyl groups have a positive inductive
effect / alkyl groups push electrons away
from themselves.
Water vapour condenses easily to form liquid
which does not get high into the atmosphere.
c
Melting of polar ice caps; causing sea level rise;
desertification; more extreme weather patterns
vi
The carbocation. It is accepting a pair of
electrons.
i
1-chlorobutane, CH3CH2CH2CH2Cl.
ii
iii
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d
C5H12 + 8O2 → 5CO2 + 6H2O
e
Carcinogenic means cancer-causing.
Particulates, VOCs.
f
2NO + 2CO → N2 + 2CO2
g
We use the characteristic wavelength of
the infrared radiation absorbed by the
pollutant molecules to identify the molecule.
The intensity of the absorbed radiation at
characteristic wavelength gives us a measure
of the concentration of each pollutant.
h
b
i
breathing difficulties / asthma / heart
attacks irritates lungs / irritates throat /
irritates eyes / nose
ii
catalyst; in oxidation of sulfur dioxide to
sulfur trioxide
ii
b
ii
iii
iv
4
Br
H
The tertiary carbocation is more stable
because it has more alkyl groups
attached.
[1]
iii
CH3 H
C
C
CH3 H n
H H
H H
B
c
Basic structure without double bond
[1]
square brackets, n, and continuation
bonds
[1]
does not decompose / does not break down [1]
[3]
Hydrogen and high temperature
[1]
fills landfill / blocks drains / harms animals
when they ingest it [1]
nickel catalyst
[1]
Cold dilute acidified
[1]
Question 3
potassium manganate(VII)
[1]
a
Hot concentrated
[1]
potassium manganate(VII)
[1]
[Total: 13]
H
H
H
H H H
H
C
C C C
C H
H
H H H
H
H
[Total: 11]
Question 2
i
H
Steam, phosphoric acid, high
temperature / 330 °C (1 mark each)
a
C H
The greater the reduction in charge on
the C+ ion / carbocation, the more stable
the C+ ion / carbocation.
[1]
H C C O H H C C H (1 mark each)
i
C
Alkyl groups have a positive inductive
effect / alkyl groups push electrons away
from themselves.
[1]
Question 1
H H
A
H C
So the charge density on the carbocation /
C+ ion is reduced.
[1]
Exam-style questions
a
[1]
H C H
H
H
i
H H
H
b
H
H
C H
H
C
C
H
H
H
H
C
C
C
C
H
H
H
H C H
H
H
H C H
H
H
H
H
C
H
(1 mark each)
[3]
i
Free radical
[1]
substitution
[1]
(CH3)2CO
[1]
(C2H5)2CO
[1]
ii
Oxidation
[1]
ii
Cl2 → 2Cl•
[1]
iii
Ketones
[1]
iii
Excess chlorine (and continued UV
light)
[1]
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iv
c
Any two free radicals combining (1 mark
each), e.g.
Ethane has no effect / bromine remains
orange.
Cl• + Cl• → Cl2;
C5H11• + C5H11• → C10H22;
C5H10Cl• + Cl• → C5H10Cl2
OR
[2]
React with acidified KMnO4
[1]
i
C11H22 [1]
Ethene turns KMnO4 (from purple)
colourless
[1]
ii
heat
[1]
Ethane has no effect / KMnO4 remains
purple
[1]
aluminium oxide catalyst / zeolite
catalyst [1]
b
[Total: 12]
C
Question 4
a
[1]
C
H
H
C5H12 + 5 1 O2 → 5CO + 6H2O (or doubled)
2
Reactants and products
[1]
ethene
c
Sigma bonds between H and C and one of the
C─C bonds
[1]
balance
[1]
b
Carbon monoxide is poisonous
[1]
c
i
[1]
Pi bond formed by overlap of remaining p
orbital of C atoms
[1]
Under conditions of high temperature
and pressure
[1]
Is above and below the plane of the C and H
atoms
[1]
ii
d
i
Nitrogen combines with oxygen
ii
Reactants and products
[1]
balance
[1]
Carbon dioxide is a greenhouse gas
H
C
C
H
H
Br
δ+
δ−
So traps heat in the atmosphere
[1]
Correct polarisation of the bromine
molecule
[1]
[1]
curly arrow showing movement of
electron pair in bromine
[1]
curly arrow showing movement of
electron pair from ethene to δ+Br atom [1]
[2]
[Total: 13]
Question 5
a
H
[1]
Br
[1]
Any two effects (1 mark each), e.g.
melting of polar ice caps /
sea level rise / desertification / more
extreme weather patterns
i
[1]
Carbon dioxide absorbs infrared
radiation
Makes C and H in one plane
d
2NO2 + 4CO → N2 + 4CO2
So temperature of the atmosphere
increases
5
H
H
[1]
ii
Cl− ion from sodium chloride has lone
pair of electrons
[1]
Acts as a nucleophile / can attack the
positive carbon atom
[1]
[1]
Br− ion formed from attack of bromine
molecule on ethene
[1]
Ethene turns aqueous bromine (from orange)
colourless.
[1]
Also acts as a nucleophile / can attack
positive carbon atoms
[1]
React with aqueous bromine.
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e
6
HCl is polar molecule [1]
This stabilises the secondary carbocation [1]
HCl is an electrophile [1]
More than the primary carbocation electrons withdrawn from one of C=C
bonds by δ+ end of HCl [1]
So Cl− (from HCl) attracted to the secondary
carbocation [1]
CH3 group has positive inductive effect
[1]
[1]
[Total: 22]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 16
Exercise 16.1
a
Halogenoalkane
Structural formula
I-iodobutane
CH3CH2CH2CH2I
Displayed formula
Skeletal formula
HH H H
I
H C C C C I
HH H H
2-chloro-2methylpropane
(CH3)2CCICH3
H
Cl
H
H C
C
C H
H
3-bromopentane
CH3CH2CHBrCH2CH3
H C H
H
Cl
H
H H Br H H
H C C C C C H
Br
H H H H H
b
H
C
CH3
Cl
ii
H
C
CF3
CF3
Cl
HH
1 with D; 2 with C; 3 with A; 4 with E; 5
with B
d
When iodoethane is refluxed with aqueous
sodium hydroxide, the hydroxide ion acts as
a nucleophile and substitutes the iodine atom.
The reaction is also called a hydrolysis reaction
which means ‘breakdown’ by water. Hydrolysis
with sodium hydroxide is faster than hydrolysis
with water because the negatively charged
hydroxide ion is a more effective nucleophile.
The hydrolysis of chloroethane is slower than
the hydrolysis of iodoethane because the C−Cl
bond is stronger than the C−I bond.
1
Bromine is more electronegative than
carbon so it tends to draw the electrons in
the Br−C bond towards it.
iii
The ammonia molecule. Because it
donates an electron pair to a centre of
partial positive charge.
iv
An intermediate / transition state
v
The Br−C bond breaks and the NH3−C
bond forms.
vi
It has a positive charge on the nitrogen
which is not stabilised very much by the
alkyl group.
vii A hydrogen ion is lost from the NH3+
group.
viii S = substitiution, N = nucleophilic, 2 =
two molecules involved in the slow step.
Exercise 16.2
i
HH
δ+
Br C C H
CH3
c
a
δ−
H3N:
b
i
CH3
δ+
δ−
H3C C Cl
CH3
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ii
Carbocation
iii
The OH− ion because it donates a pair of
electrons to the + charged ion.
iv
S = substitiution, N = nucleophilic, 1 =
one molecule involved in the slow step
d
Add alcoholic silver nitrate solution to the
reaction mixture. White precipitate indicates
presence of chloride ions.
e
i
CH3CH2CH2CH2Br + NaOH →
CH3CH2CH=CH2 + NaBr + H2O
ii
CH3CH2CH2CH2Br + KCN →
CH3CH2CH2CH2CN + KBr
iii
CH3CHClCH3 + NaOH →
CH3CH(OH)CH3 + NaCl
iv
CH3CH2Cl + NH3 →
CH3CH2NH2 + HCl
Exercise 16.3
a
They are unreactive because of their strong
bonds which are not likely to be hydrolysed
under the aqueous conditions in the body
b
Aerosol propellants, solvents and refrigerants
c
Easily vaporised / has a low boiling point
Exercise 16.5
d
The Sun
a
e
By diffusion
f
Ultraviolet radiation causes fission of C─Cl
bonds to produce chlorine free radicals which
are very reactive and can react further by a
series of chain reactions.
g
h
i
Cl• + O3 → ClO• + O2
ii
ClO• + O3 → Cl• + 2O2
i
The propagation step involves the
formation of more chlorine free radicals
so that the process starts over again.
ii
a
b
R secondary, S primary, T secondary,
U tertiary
i
CH3CH2Br + NaOH → CH3CH2OH + NaBr
ii
CH3CH2I + NH3 → CH3CH2NH2 + HI
iii
CH3CH2Cl + NaOH → CH2=CH2 + NaCl
+ H2O
iv
CH3CHClCH3 + NaOH →
CH3CH(OH)CH3 + NaCl
v
CH3CH2CH2Cl + KCN →
CH3CH2 CH2CN + KCl
vi
C6H5CH2Br + KCN →
C6H5CH2CN + KBr
vii CH3CH2CH2Br + NaOH →
CH2=CHCH3 + NaBr + H2O
c
2
In b i nucleophilic substitution; in b iii
elimination
CH3CH2CH2OH + SOCl2 →
CH3CH2CH2Cl + SO2 + HCl
ii
3CH3CH2OH + PBr3 →
3CH3CH2Br + H3PO3
iii
CH3CH(OH)CH3 + HCl →
CH3CH(Cl)CH3 + H2O
iv
CH3CH2CH2OH + PCl5 →
CH3CH2CH2Cl + POCl3 + HCl
v
CH3CH2CH3 + Cl2
CH3CH2CH2Cl + HCl
uv light
b
Reaction v. It is a free radical reaction. More
chlorine free radicals can react to substitute
more hydrogen atoms. So a variety of
products can be formed.
c
Action of concentrated sulfuric or phosphoric
acids on potassium chloride.
d
i
Secondary.
ii
type: addition; mechanism: electrophilic
By free radicals combining (termination
step)
Exercise 16.4
i
Exam-style questions
Question 1
H
a
[1]
H C H
H
H
b
H C
C
C H
H
Br
H
i
The OH group replaces the Br atom.
[1]
ii
An electron pair is donated to a centre of
positive charge.
[1]
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c
d
Add nitric acid until the solution is neutralised
/ acidic.
[1]
Add aqueous silver nitrate.
[1]
Cream precipitate indicates bromide is
present.
[1]
e
Because the more alkyl groups that are
attached to a carbocation, the greater is the
positive inductive effect on reducing the
charge on the carbocation
[1]
Alkyl groups have a positive inductive effect /
alkyl groups
push electrons away from themselves.
The C─Hal bond is more easily broken
(without the help of a nucleophile)
[1]
So the nucleophile is needed to assist the
bond breaking
The greater the reduction in charge on the
C+ ion / carbocation, the more stable
the C+ ion / carbocation.
[1]
So the first stage in the reaction for tertiary
halogenoalkane is the self-ionisation
of the halogenoalkane.
[1]
a
b
a
[1]
The C─I bond in the halogenoalkane is
polar
[1]
because I is more electronegative than C
[1]
so the C has a partial positive charge.
[1]
OH− ion is a nucleophile.
[1]
CH3CH2I + CN− → CH3CH2CN + I−
ii
An additional carbon atom is being added
to the chain / the carbon chain is being
lengthened.
[1]
[1]
−
[1]
c
It has a pair of electrons which are donated to
an atom with a positive or partial
positive charge.
[1]
−
d
H H
δ+
H C C
δ−
I
H H
:CN−
I
H C C
H H
H H
CN
[1]
The C─Cl bond energy is greater than
the C─I bond energy.
[1]
It needs a higher activation energy to start the
reaction in the chloroalkane.
[1]
3
i
C N
CH3CH2CH2CH2I + NaOH →
CH3CH2CH2CH2OH + NaI
partially positive carbon atom.
d
[Total: 17]
b
A lone pair of electrons on the hydroxide ion
is donated to the
c
[1]
Question 3
[Total: 11]
Question 2
[1]
For a primary carbocation the C─Hal bond
bond is less easily broken / has a higher
activation energy
[1]
So the charge density on the carbocation /
[1]
C+ ion is reduced.
The tertiary carbocation is more stable
because it has more alkyl groups attached. [1]
tertiary carbocations have greater stability
than primary carbocations
[1]
i
CH3CH2CH2CH2I + OH− →
CH3CH2CH = CH2 + H2O + I− [1]
ii
Elimination
[1]
iii
Hydroxide ion is acting as a base
[1]
by accepting a proton from the
halogenoalkane.
[1]
e
Polarisation of C−I bond
[1]
arrow
[1]
intermediate
[1]
S = substitiution
[1]
N = nucleophilic
[1]
2 = two molecules involved in the
slow step.
[1]
[Total: 10]
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Workbook answers
Chapter 17
Exercise 17.1
A = primary; B = tertiary; C = secondary;
D = secondary; E = tertiary
iii
CH3CH(OH)CH3
b
A is propan-1-ol; B is 2-methypropan-2-ol;
C is butan-2-ol; D is pentan-2-ol; E is
2-methylbutan-2-ol
iv
2 3
CH3CH2CH2CH2CH2OH
CH3CH2CH2CH═CH2 + H2O
c
Primary alcohols: Potassium dichromate(VI)
turns from orange to green. The
product distilled off is an aldehyde. On
further oxidation a carboxylic acid is
formed. Secondary alcohols: Potassium
dichromate(VI) turns from orange to green.
The product is a ketone. On further oxidation
no reaction occurs.
Tertiary alcohols: Potassium dichromate(VI)
stays orange. No reaction takes place.
d
e
c
d
CH3CH═CH2 + H2O
Al O , heat
C3H7OH + 4 1 O2 → 3CO2 + 4H2O
2
ii C4H9OH + 4O2 → 4CO + 5H2O
iii C6H11OH + 8 1 O2 → 6CO2 + 6H2O
2
C because C contains the CH3CH(OH)─
group
i
Exercise 17.3
a
i
By reaction of sodium / potassium
bromide with concentrated sulfuric acid.
ii
1-bromopropane is insoluble in water /
immiscible with water and is denser than
water.
iii
The oxygen atom is more electronegative
than the carbon atoms. So the oxygen has
a slight negative charge and the carbon
atom has a slight positive charge.
i
acidified potassium manganate(VII)
ii
purple to colourless
i
CH3CH2CH2OH + [O] →
CH3CH2CHO + H2O
ii
CH3CH(OH)CH3 + [O] →
CH3COCH3 + H2O
iii
CH3CH2CHO + [O] → CH3CH2COOH
iv
iv
CH3CH(OH)CH2CH(OH)CH2CH3 + 2[O]
→ CH3COCH2COCH2CH3 + 2H2O
δ+ δ−
H—Br
v
The δ+ end of the hydrogen bromide
attacks the partially negative oxygen atom.
vi
CH3CH2CH2OH + HBr →
CH3CH2CH2Br + H2O
Exercise 17.2
a
1 with F; 2 with D; 3 with A; 4 with B;
5 with C; 6 with E
b
i
2C3H7OH + 2Li → 2C3H7O Li + H2
ii
2CH3OH + Ca → (CH3O )2 Ca + H2
−
−
+
b
C4H9OH + PCl5 → C4H9Cl + POCl3 + HCl
c
i
3C2H5OH + PI3 → 3C2H5I + H3PO3
ii
3C4H9OH + PBr3 → 3C4H9Br + H3PO3
2+
d
1
Al2O3, heat
a
CH3CH2CH2OH + SOCl2 →
CH3CH2CH2Cl + SO2 + HCl
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Exercise 17.4
Exercise 17.6
a
a
b
c
d
e
i
CH3CH2CH2CH2CH2COOH
i
Ethyl butanoate
ii
HCOOH
ii
Propyl hexanoate
iii
HOOCCOOH
iii
Methyl pentanoate
i
Concentrated acidified potassium
dichromate(VI), reflux
i
Butyl methanoate
ii
Pentyl ethanoate
ii
CH3CH2CH2CH2OH + 2[O] →
CH3CH2CH2COOH + H2O
iii
Propyl butanoate
iii
A bond in the carbon chain would have
to be broken to accommodate the extra
oxygen atom. This would lead to the chain
breaking.
i
It is a catalyst
ii
CH3COOC3H7 + H2O ⇌
CH3COOH + C3H7OH
i
HCOOC3H7 + NaOH →
HCOO−Na+ + C3H7OH
b
c
d
i
CnH2n+1CN
ii
The breakdown of a substance by water
(usually involving H+ or OH− as a catalyst).
ii
CH3COOCH3 + NaOH →
CH3COO−Na+ + CH3OH
i
CH3CN + HCl + 2H2O →
CH3COOH + NH4Cl
iii
CH3CH2COOC4H9 + NaOH →
CH3CH2COO−Na+ + C4H9OH
ii
CH3CH2CH2CH2CN + HCl + 2H2O →
CH3CH2CH2CH2COOH + NH4Cl
iii
CH3CH2CN + H+ + 2H2O →
CH3CH2COOH + NH4+
i
ethanenitrile
ii
pentanenitrile
Exercise 17.7
a
Exercise 17.5
a
b
c
d
2
i
CH2=CH2(g) + H2O(g) → CH3CH2OH
ii
CH3CH2CH2COOCH3 + NaOH →
CH3CH2CH2COO−Na+ + CH3OH
iii
CH3CH2Cl + NaOH →
CH3CH2OH + NaCl
iv
CH3CH2COOH + 2[H] →
CH3CH2CH2OH + H2O
v
CH3COCH3 + 2[H] → CH3CH(OH)CH3
vi
CH3CH2COOCH2CH3 + H2O →
CH3CH2CH2COOH + CH3OH
i
addition
ii
hydrolysis
iii
hydrolysis / substitution
i
It only partly dissociates in solution / it
ionises only partially in solution.
ii
CH3CH2COOH + H2O ⇌
CH3CH2COO− + H3O+
i
CH3COOH + KOH →
CH3COO−K+ + H2O
ii
2CH3CH2CH2COOH + Mg →
(CH3CH2CH2COO−)2Mg2+ + H2
iii
2C6H5COOH + Na2CO3 →
2C6H5COO− Na+ + CO2 + H2O
iv
reduction
v
reduction
i
Potassium ethanoate
vi
hydrolysis
ii
magnesium butanoate.
i
CH2=CH2 + [O] + H2O → CH2(OH)CH2(OH)
i
CH3COOH + 4[H] → CH3CH2OH + H2O
ii
CH3CH2CHO + 2[H] → CH3CH2CH2OH
ii
CH3CH2COOH + 4[H] →
CH3CH2CH2OH + H2O
iii
CH3COOCH2CH2CH3 + NaOH →
CH3COO−Na+ + CH3CH2CH2OH
iii
HOOCCH2CH2COOH + 8[H] →
HOCH2CH2CH2CH2OH + 2H2O
iv
CH3CH2COOH + 4[H] →
CH3CH2CH2OH + H2O
b
c
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Water has no alkyl group so negative
charge on oxygen not increased as much
(so it is more acidic / more likely to donate
a proton)
[1]
Exam-style questions
Question 1
a
b
(also the effect of the side chain on
hydration makes it harder for RO− to be
stabilised)
Heat with acidified potassium
dichromate(VI)
[1]
Propan-1-ol turns potassium dichromate
from orange to green
[1]
No reaction with 2-methylpropan-2-ol /
dichromate remains orange
[1]
Question 2
Acidified potassium dichromate
[1]
a
i
High concentration of dichromate and acid [1]
[1]
Reflux
c
H H
O
d
i
[1]
O H
[1]
e
Neutralise the acid with sodium
carbonate.
[1]
Distil and collect ester as distillate.
[1]
[1]
b
i
Sodium hydroxide reacts with ethanoic
acid
[1]
to form sodium ethanoate and water.
[1]
Ethanol does not react with sodium
hydroxide
[1]
2C2H5OH + 2Na → 2C2H5O−Na+ + H2
[1]
balance
[1]
Methyl propanoate
[1]
It is behaving as an acid because it is
donating a H+ ion to the NH2 group
2CH3COOH + 2Na →
2CH3COO−Na+ + H2
[1]
Correct formulae
[1]
balance
[1]
ii
Bubbles / effervescence
[1]
i
C2H5OH + PCl5 →
C2H5Cl + POCl3 + HCl
[1]
ii
HCl released as white fumes
[1]
iii
CH3COOH + PCl5 →
CH3COCl + POCl3 + HCl
[1]
CH3COOH + 4[H] → CH3CH2OH + H2O
[1]
i
The inductive effect is greater when more
alkyl groups are attached to the C next to
the O
[1]
When there are more electron donating
groups it increases the negative charge on
the O atom
[1]
c
More charge means less stability (compared
with the unionised molecule)
[1]
−
So the RO ion is less likely to form / ROH
form less likely to donate a proton
[1]
3
The ethanoate ion is relatively stable
in water.
[1]
iii
ii
[1]
Correct formulae
CH3CH2COOH + CH3OH ⇌
CH3CH2COOCH3 + H2O
ii
ii
Add methanol to mixture of propanoic
acid and sulfuric acid in flask.
[1]
Reflux.
Ethanoic acid donates a proton to
water.
Ethanol is hardly acidic (ALLOW: not
acidic) / the ethoxide ion is less stable
in water.
[1]
H C C C
H H
[Total: 19]
d
[Total: 15]
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Question 3
a
c
Step 1:Bubble ethene through concentrated
hydrochloric acid / React ethene
with hydrogen chloride
[1]
d
Phosphorus(V) chloride / phosphorus
pentachloride
Butan-1-ol absorbed into ceramic wool in
bottom of a test tube
[1]
Butan-1-ol is heated to form the vapour
At room temperature and
pressure
[1]
Butan-1-ol vapour passed over heated
aluminium oxide
Step 2: Add dilute sodium hydroxide
[1]
Step 3:Add concentrated acidic potassium
dichromate
[1]
[1]
Reflux
b
4
i
CH3CH2CH2OH + HCl →
CH3CH2CH2Cl + H2O
e
It is accepting a hydrogen ion from the
hydrogen chloride.
[1]
iii
Suitable metal chloride, e.g. sodium
chloride
[1]
Concentrated sulfuric acid
[1]
[1]
i
CH3CH(OH)CH2CH3
[1]
ii
yellow precipitate
[1]
the isomer contains the
CH3CH(OH)─ group
[1]
[1]
ii
[1]
Butene collected (in inverted test tube) over
water
[1]
[1]
Reflux / heat
[1]
[Total: 18]
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Workbook answers
Chapter 18
Exercise 18.1
a
b
c
d
e
i
Methanal
ii
Hexan-3-one
iii
Butanal
i
CH3CH2COCH3
ii
CH3CH2CH2CH2CHO
iii
CH3COCH2COCH3
i
Acidified potassium dichromate(VI)
ii
Cr2O72− and Cr3+
iii
It distils off before the alcohol
iv
If refluxed it would be oxidised further to
propanoic acid
v
vi
i
gently with Fehling’s solution, the colour
changes from blue to orange (precipitate).
Propanal is oxidised to propanoic acid.
Copper(II) ions are reduced to copper(I) ions
When a ketone is heated with Tollen’s reagent
there is no reaction / no silver mirror
b
i
Propanone
c
Ketones cannot be further oxidised to
carboxylic acids / very strong
oxidation would break the carbon chain.
Two (larger) molecules join together and a
small molecule is eliminated
d
i
There are two compounds with a melting
point of 126 °C so X is either butanal or
propanone.
ii
Heat with Tollen’s reagent. Butanal
will react to form a silver mirror but
propanone will not. OR, heat with
Fehling’s solution. Butanal will react
to form an orange-red precipitate but
propanone will not react.
CH3COCH2CH3 + 2[H] →
CH3CH(OH)CH2CH3
ii
CH3CH2CHO + 2[H] →
CH3CH2CH2OH
iii
CH3COCH2COCH2CH3 + 4 [H] →
CH3CH(OH)CH2CH(OH)CH2CH3
i
Butan-2-ol
ii
Propan-1-ol
iii
Hexan-2,4-diol
Exercise 18.3
a
i
Oxygen is more electronegative than
carbon. So the electron density in the
bond is attracted to the oxygen more than
the carbon.
ii
C2H5
Exercise 18.2
a
1
A solution of 2,4-DNPH is added to a
carbonyl compound. An orange coloured
precipitate of a dinitrophenylhydrazone is
formed. The precipitate is purified by
recrystallisation and the melting point is
measured. Each dinitrophenylhydrazone
derivative of an aldehyde or ketone has a
characteristic melting point which can be
compared with known data book values.
When propanal is warmed gently with Tollen’s
reagent the colour changes from
colourless to a silver mirror. Propanal is
oxidised to propanoic acid. Silver ions are
reduced to silver. When propanal is warmed
δ+
δ−
C=O
H
:CN−
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b
iii
CN− ion because it donates a lone pair of
electrons to an electron-deficient area of
a molecule (which has a partial positive
charge).
iv
An intermediate
v
From ionisation of hydrogen cyanide /
from the acid used to make the hydrogen
cyanide (by reaction with sodium cyanide)
vi
The pair of electrons on the oxygen forms
a bond with the hydrogen ion.
i
b
ii
OH
CH3 C C N
CH3
H
CH3CH2CH2
Ethanol: Broad strong peak at 3200−3600
indicates hydrogen bonding of O─H in an
alcohol. Peak near 1300−1400 may indicate
C─O bond in alcohol.
Ethanoic acid: Broad less strong peak between
2500−3200 indicates hydrogen bonding
of O─H in carboxylic acid. Strong peak near
1600 may indicate C═O bond.
OH
CH3 C CN
iii
pattern of their infrared spectrum.
The spectrum shows the percentage absorbance
(vertical axis) and wavenumber (horizontal axis).
Wavenumber is the reciprocal of the wavelength.
Exam-style questions
OH
Question 1
C CN
a
i
CH3CH2CH2CH2OH + [O] →
CH3CH2CH2CHO + H2O [1]
ii
Butanal
[1]
iii
From orange
[1]
to green
[1]
CH3
Exercise 18.4
a
b
c
i
butanone;
ii
ethanal;
iii
pentane-2,4-dione;
v
propan-2-ol
i
Substitution
ii
RCOCI3
iii
Hydrolysis
iv
Triiodomethane
v
Sodium propanoate
b
2
To prevent further oxidation / to prevent
butanoic acid from forming
[1]
i
Add Tollen’s reagent to the carbonyl
compound in a test tube and warm
gently.
[1]
If a silver mirror forms on the side of
the tube, the unknown is an aldehyde. [1]
If no silver mirror forms the unknown
is a ketone.
[1]
ii
The alkaline iodine oxidises a CH3C(OH)
group to a CH3CO group which does give a
positive test.
Exercise 18.5
a
iv
The bonds in organic compounds vibrate by
stretching, bending and twisting. They have
a natural frequency at which they vibrate.
When molecules absorb infrared radiation
that corresponds to these natural frequencies,
it stimulates larger vibrations and energy
is absorbed. This frequency is called the
resonance frequency. Each type of bond absorbs
infrared radiation at a characteristic range of
frequencies. We can identify different functional
groups from the absorbance
c
(Aqueous) silver nitrate and (excess)
ammonia
[1]
iii
Ag+ + e− → Ag
[1]
i
CH3CH2COCH3 + 2[H] →
CH3CH2CH(OH)CH3
[1]
ii
Butan-2-ol
[1]
iii
yellow precipitate is triiodomethane
[1]
CHI3
[1]
other product is sodium propanoate
−
+
CH3CH2CO2 Na [1]
[1]
(allow error carried forward from part ii)
[Total: 16]
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Question 2
Warm the mixture
[1]
a
Yellow precipitate with pentan-2-one
[1]
Sodium cyanide / potassium cyanide /
other suitable cyanide of a reactive
metal
[1]
ii
The cyanide ion has a pair of electrons
which it donates to an electron-deficient
area on the organic molecule.
[1]
iii
CH3
δ+
No precipitate / no reaction with pentan3-one
[1]
b
δ−
C=O
CH3
:CN−
c
•Stating that propanone is a polar
molecule
d
•Correct polarisation of C═O bond
•Electron pair on cyanide ion
•Curly arrow going from CN (electron
pair) to carbonyl carbon
•Curly arrow going from C═O bond to
C═O oxygen
•
iv
v
precipitate
[1]
Recrystallise to get pure sample
[1]
Take the melting point
[1]
Compare the melting point with table of data
for dinitrophenylhydrazones
[1]
[5]
Any five of:
Add 2,4-DNPH to the carbonyl compound
and filter the
Two species add together
[1]
And a small molecule is eliminated
[1]
[1]
from acid added to the cyanide
[1]
i
OH
CH3 C CN
[1]
CH3
b
1000
Correct labelling of x-axis
[1]
Correct labelling of y-axis
[1]
Sharp and deep peak at 3000 cm−1
[1]
Sharp and weak peak at 1600 cm−1
[1]
[1]
[1]
C═O
[1]
Peak at 1600 cm−1 is due to a C═C
bond
in aldehyde or ketone
[1]
Peak at 2900 cm−1 is due to a C─H bond [1]
ii
In an alkene
[Total: 13]
Question 3
3
3000
2000
Wave number/cm−1
About 1700
a
50
100
4000
Bond forming from CN to C of C═O
From dissociation of HCN
0
i
Absorbance/%
i
i
CH3CH2CH2COCH3
[1]
ii
Add alkaline solution of iodine
[1]
[1]
[Total: 18]
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may be different.
Workbook answers
Chapter 19
Exercise 19.1
a
1 with E; 2 with C; 3 with A; 4 with F; 5 with
D; 6 with G; 7 with B
b
Lattice energy is the enthalpy change when one
mole of an ionic compound is formed from its
gaseous ions under standard conditions.
b
2Na+(g) + S2−(g)
2Na+(g) + S(g) + 2e−
∆Hat
First electron affinity is the enthalpy change
when one mole of electrons is added to one
mole of gaseous atoms to form one mole
of gaseous ions with a charge of 1− under
standard conditions.
EA1
EA2
2Na+(g) + S−(g) + e−
2Na+(g) + S(s) + 2e−
∆Hlatt
2IE1
2Na(g) + S(s)
2∆Hat
2Na(s) + S(s)
∆Hf
Na2S(s)
c
mole; gaseous; element; standard
d
i
negative; less; first
ii
attraction; positively; charge; electron;
energy; force; radius; decreases
i
Al+(g) → Al2+(g) + e−
ii
N2−(g) + e− → N3−(g)
∆H⦵x = 2(107.3) + 2(496) + 278.5 + −200.4 + 640
= +1924.7 kJ mol−1
iii
Mg(s) + S(s) + 2O2(g) → MgSO4(s)
∆H⦵latt + ∆H⦵x = ∆H⦵f
iv
2K+(g) + O2−(g) → K2O(s)
∆H⦵latt = ∆H⦵f − ∆H⦵x
e
c
∆H⦵x = 2∆H⦵ at[Na] + 2IE1⦵[Na] + ∆H⦵ at[S] +
EA1[S] + EA2⦵[S]
Exercise 19.2
= −364.8 − 1924.7 = −2289.5 kJ mol−1
a
Ca2+(g) + 2Br(g) + 2e−
2∆Hat
∆H⦵latt + sum of the other values, ∆H⦵x
= ∆H⦵f [Na2 S]
Ca2+(g) + Br2(I) + 2e−
2EA1 Ca2+(g) + 2Br−(g)
IE2
d
As the size of the cation increases from Li
to Rb, the lattice energy decreases (gets less
exothermic).
As the size of the anion increases from oxide
to sulfide, the lattice energy decreases (gets less
exothermic).
Ca+(g) + Br2(I) + e−
IE1
∆Hat
∆H Latt
Ca(g) + Br2(I)
Ca(s) + Br2(I)
∆Hf
1
CaBr2(s)
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Exercise 19.3
Exercise 19.4
a
a
Mg2+ because it has the highest charge density
(charge to radius ratio).
b
S2− because the polarisability of an anion
increases with increased size and increased
charge.
c
The magnesium ion is small and has a
relatively high charge.
absorbed; one; solid; dilute, mole; gas(eous);
dilute
b
∆Hlatt
Mg2+(g) + 2I−(g)
MgI2(s)
∆Hhyd[Mg2+]
∆Hsol
+
2∆Hhyd[I−]
Therefore, its (positive) charge density is high.
The nitrate ion is a large ion and its outer
electrons are attracted to the small highly
charged magnesium ion. This distorts the
shape of the nitrate ion.
−
Mg (aq) + 2I (aq)
2+
c
Mg2+(g) + 2I–(g)
d
∆HLatt [MgI2(s)]
∆H [Mg2+(g)]
hyd
+
2∆H [I–(g)]
hyd
MgI2(s)
∆H
d
i
Magnesium
ii
Magnesium
iii
Magnesium nitrate
iv
Nitrogen dioxide and oxygen
Exam-style questions
MgI2(aq)
The larger the cation, the smaller the
value of ΔH⦵hyd (the sulfate ion is the
same in both).
ii
It decreases as the size of the cation
decreases.
iii
Lattice energy is inversely proportional
to the size of the ions. The sulfate ion
contributes a larger part to the lattice
energy because it is much larger than the
cation. Therefore, the decrease in lattice
energy is small. The hydration energy is
more dependent on the size of the cation
because the hydration energy of sulfate is
smaller because it is a larger ion.
iv
2
sol
i
The more exothermic or less endothermic
the value of ∆H⦵r, the more likely is the
process to take place. The ∆H⦵sol value
for calcium sulfate is less endothermic
than the ∆H⦵sol for strontium sulfate, so
calcium sulfate is more likely to have a
higher solubility.
Question 1
a
The enthalpy change when one mole of an
ionic compound is formed from its gaseous
ions under standard conditions.
Idea of one mole of compound being formed
from its ions
[1]
(Ions) are gaseous and standard
conditions apply
b
[1]
As the size of the cation increases, the lattice
energy decreases (gets less exothermic)
[1]
As the size increases, the charge density of
the cation decreases. This results in weaker
electrostatic forces between the ions.
[1]
As the charge on the cation increases,
the lattice energy increases (gets more
exothermic).
[1]
As the charge increases, the charge density
also increases.
This results in stronger electrostatic forces
between the ions.
[1]
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c
K+(g) + I−(g)
∆Hlatt
Question 2
KI(s)
a
Mg(NO3)2(s)
[K(s)]
at
∆H
+
1
∆H [ I2(s)]
at
2
+
∆H
ion
[K(g)]
∆H [KI(S)]
f
2 marks for all boxes correct, but 1 mark if
two correct
[2]
2 marks for all boxes correct but 1 mark if two
correct
[2]
[1]
Arrows in correct direction
b
[1]
Arrows in correct direction
∆H⦵f[Mg(NO3)2 ] + ∆H⦵r = ∆H⦵ f[MgO]
+ 2∆H⦵f [NO2]
−790.7
+ ∆H⦵r = −601.7
+ 2(33.20)
∆H⦵latt + sum of the other values, ∆H⦵x
= ∆H⦵f [KI]
∆H⦵x = ∆H⦵ at[K] + IE1⦵ [K] + ∆H⦵ at[ 1 I2 ]
2
+ EA1⦵
[1]
= −535.3
∆H⦵r = ∆H⦵f [MgO] + 2∆H⦵f[NO2]
− ∆H⦵f[Mg(NO3)2]
∆H⦵x = (89.20) + (419.0) + 106.8 + −295.4 =
+319.6 kJ mol−1
= −535.3
− (−790.7) = +255.4 kJ mol−1
Correct value of ∆H⦵f[MgO] + 2∆H⦵f[NO2]
(or indication of this)
[1]
∆H⦵latt + ∆H⦵x = ∆H⦵f
∆H⦵latt = ∆H⦵f − ∆H⦵x
Idea of ∆H⦵f[Mg(NO3)2] + ∆H⦵r
= ∆H⦵f[MgO] + 2∆H⦵f[NO2]
[1]
Correct value of ∆H x (or indication that all
these values are added)
[1]
Correct answer from calculation done
correctly
[1]
Idea of ∆H⦵latt + ∆H⦵x = ∆H⦵f or ∆H⦵latt =
∆H⦵f − ∆H⦵x
[1]
Correct number of significant figures +
units
[1]
Correct answer from the calculation done
correctly
[1]
Correct number of significant figures +
units
[1]
An electron is being added to a negative
ion.
[1]
= −327.9 − 319.6 = −647.5 kJ mol−1
⦵
An input of energy is needed to overcome
the repulsive forces of two negative / like
charges
[1]
3
(Sum of) ∆Hf
Mg(s) + N2(g) + 3O2(g)
1
K(s) + 2 I2(s)
EA1[I(g)]
e
MgO(s) + 2NO2(g) + 12 O2(g)
∆Hf
+
d
∆Hr
c
The ease of decomposition decreases down
the group / increases up the group
[1]
Because the values get more endothermic
down the group
d
i
[1]
The distortion of the electron cloud of a
(large) ion
[1]
by a smaller / more highly charged ion [1]
ii
[Total: 15]
The size of the cation
[1]
The charge on the cation
[1]
iii
The larger the cation, the less easy it is
to decompose the nitrate
[1]
[Total: 14]
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Question 3
a
ii
V shaped water molecule with bond angle of
104.5 (see answer to Exercise 4.3(b)(i))
[1]
Arrow drawn in the direction relatively
smaller between the hydrogen atoms with
the head towards the oxygen /δ− on O atoms
and δ+ between the H atoms
[1]
b
i
H H
H
H
O
H O
Na+ O
H
O
H
O
H
H
H
ii
c
[1]
Ion-dipole bonding
When forces are formed between the water
and the ions, energy is given out.
[1]
When forces between ions in the ionic giant
structure are broken energy is absorbed. [1]
Energy change is exothermic / not too
endothermic.
d
[1]
i
Na (g) + Br−(g)
+
∆Hlatt
∆Hhyd
−390 − 337
[1]
= −742
+ ∆H⦵sol[NaBr]
∆H sol[NaBr] = (−390 − 337) + 742
= +15 kJ mol−1
iii
[1]
The enthalpy change when one mole of a
solid dissolves in a solvent
[1]
to form a very dilute / infinitely dilute
solution under standard conditions. [1]
e
Note: The oxygen of the water molecules
must be pointing towards the ion.
[1]
⦵
[1]
∆H⦵hyd [Na+] + ∆H⦵hyd [Br−] = ∆H⦵latt
[NaBr] + ∆H⦵sol[NaBr]
So both the lattice energy and hydration
energy of Mg(OH)2 are greater than those
of Ba(OH)2 because Mg2+ ion smaller than
Ba2+ ion
[1]
The difference between the lattice energies of
Mg(OH)2 and Ba(OH)2 is relatively larger / the
difference between the hydration energies of
Mg(OH)2 and Ba(OH)2 is relatively smaller [1]
enthalpy change of solution = enthalpy
change of hydration − lattice energy
[1]
So the enthalpy change of solution of
Ba(OH)2 is less positive (or more negative)
than the enthalpy change of solution of
Mg(OH)2
[1]
Less positive / more negative enthalpy change
of solution means greater solubility
[1]
(of Ba(OH)2) NaBr(s)
∆Hsol
[Total: 20]
Na+(aq) + Br–(aq)
2 marks for all boxes correct but 1
mark if two correct
[2]
Arrows in correct direction and enthalpy
changes correctly labelled
[1]
4
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may be different.
Workbook answers
Chapter 20
Exercise 20.1
a
b
i
The right hand electrode because it is
negative.
ii
Anions are negatively charged. The anode
is positively charged. Opposite charges
attract.
iii
The ions are not free to move from place
to place / the ions only vibrate / the ions
are fixed in a lattice.
iv
They conduct electricity; they do not react
with the electrolyte.
v
From the anode to the positive pole of the
cell from the negative pole of the cell to
the cathode.
vi
The outer shell electrons from the metal
atoms are free to move throughout the
whole structure / the delocalised electrons
move throughout the structure.
i
Mg + 2e → Mg (reduction)
2+
ii
Cathode: hydrogen because it is low in
the discharge series / H+ accepts electrons
more readily.
iii
iv
v
−
Al3+ + 3e− → Al (reduction)
vi
2O2− → O2 + 4e− (oxidation)
iv
Li+ + e− → Li (reduction)
i
Anode: chlorine because it is below OH−
in the discharge series.
Cathode: hydrogen because it is below
sodium in the discharge series / H+ accepts
electrons more readily.
1
Exercise 20.2
a
2I− → I2 + 2e− (oxidation)
c
Anode: oxygen because water / OH− from
water is being electrolysed.
Cathode: silver because it is low / below
H+ in the discharge / reactivity series.
Ca2+ + 2e− → Ca (reduction)
2Br − → Br2 + 2e− (oxidation)
v
Anode: chlorine because it is below OH−
in the discharge series.
Cathode: hydrogen because H+ is the only
positive ion.
2O → O2 + 4e (oxidation)
iii
Anode: oxygen because water / OH− from
water is being electrolysed / OH− below
SO42− in discharge series.
Cathode: copper because it is low / below
H+ in the discharge series.
−
Zn2+ + 2e− → Zn (reduction)
2−
Anode: oxygen because it is below SO42−
in the discharge series / OH− from water is
being electrolysed.
Cathode: hydrogen because H+ is the only
positive ion.
2Cl− → Cl2 + 2e− (oxidation)
ii
Anode: oxygen because the solution is
very dilute so the water is electrolysed / a
mixture of chlorine and oxygen because
Cl− is not very far below OH− in the
discharge series.
i
0.200 × 96 500 = 19 300 C
ii
3 × 96 500 = 289 500 C to deposit 1 mol
of Al
So 5 × 289 500 = 1 447 500 C to deposit
5 mol
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iii
b
i
2 × 96 500 = 193 000 C to deposit 1 mol
of Pb
Since the Ag+ ion has a single charge, the
charge on a mole of electrons (F) is
So 0.400 × 193 000 = 77 200 C to deposit
0.4 mol
410.4
−1
4.2632 × 10 −3 = 96 265.7 C mol
23
L = 96265.7
1.6 × 10 −19 = 6.009 × 10
Step 1: charge transferred = 3 × 10 × 60
= 1800 C
Step 2: number of coulombs to deposit
1 mol Cu = 2 × 96 500
= 193 000 C
step1
Step 3: moles Cu deposited =
step2
= 9.326 × 10−3 mol
d
Step 4: mass of copper deposited
= 9.326 × 10−3 × 63.5 = 0.59 g
ii
Charge transferred = 0.90 × 10 × 60
= 540 C
Number of coulombs to deposit 1 mol Ag
= 1 × 96 500 = 96 500 C
Moles Ag deposited = 540
96500
= 5.596 × 10−3 mol
Mass of Ag deposited
= 5.596 × 10−3 × 107.9 = 0.60 g
iii
iv
v
Charge transferred = 0.15 × 20 × 60
= 180 C
(Multiply by 4 because each O atom has a
charge of −2)
Moles O2 = 180 = 4.663 × 10−4 mol
386000
1 mol O2 occupies 24.0 dm3 at r.t.p.
So volume of O2 = 4.663 × 10−4 × 24 dm3
= 0.011 dm3 / 11 cm3
2
a
Charge transferred = 0.07600 × 90 × 60 =
410.4 C
Moles Ag = 0.4600 = 4.2632 × 10−3 mol
107.9
i
Co(s) → Co2+(aq) + 2e−
Pb2+(aq) + 2e− → Pb(s)
ii
Zn(s) → Zn2+(aq) + 2e−
Cu2+(aq) + 2e− → Cu(s)
Number of coulombs to deposit 1 mol Pb
= 2 × 96 500 = 193 000 C
Moles Pb deposited = 900
193000
= 4.663 × 10−3 mol
Number of coulombs to release 1 mol O2
= 4 × 96 500 = 386 000 C
c
Exercise 20.3
Charge transferred = 0.50 × 30 × 60
= 900 C
Mass of lead deposited
= 4.663 × 10−3 × 207.2 = 0.97 g
Electrolyse solution of copper sulfate between
copper electrodes for a known amount of
time (40 min) keeping current constant.
Determine the increase in mass of the cathode
(or decrease in mass of anode). Calculate
the number of moles of copper deposited.
Determine the increase in mass of the cathode
(or decrease in mass of anode). Calculate
quantity of electricity passed in coulombs by
Q = It (t in seconds). Calculate the quantity of
electricity passed to deposit 1 mole of copper.
Divide this by 2 to get the value of F (because
for every mole of copper deposited, 2 moles of
electrons are required (Cu2+ + 2e− → Cu).
iii
Al(s) → Al3+(aq) + 3e−
Ag+(aq) + e− → Ag(s)
iv
Mg(s) → Mg2+(aq) + 2e−
Sn2+(aq) + 2e− → Sn(s)
b
i
Zn(s) → Zn2+(aq) + 2e−
Cu2+(aq) + 2e− → Cu(s)
ii
Zn(s) → Zn2+(aq) + 2e− is oxidation (loss
of e−)
Cu2+(aq) + 2e− → Cu(s) is reduction (gain
of e−)
iii
Reduction always takes place at a cathode.
iv
The electrons move from the electrode
where the electron density is higher (the
zinc) to where it is lower (the copper).
v
To maintain electrical contact between
the solutions so a complete circuit is
obtained.
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c
i
The voltage increases because silver is
lower in the reactivity series than copper.
ii
The voltage increases because magnesium
is higher in the reactivity series than zinc.
iii
The voltage decreases because tin is lower
in the reactivity series than zinc.
iv
The voltage decreases because iron is
higher in the reactivity series than copper.
v
Exercise 20.5
a
b
i
H+ ions are easier to reduce than Zn2+ ions
as they have a more positive value of E⦵.
ii
H+ ions
iii
2H+(aq) + Zn(s) ⇌ H2(g) + Zn2+(aq)
i
Cl2(aq)
Cl2(aq) + 2Fe2+(aq) ⇌ 2Cl−(aq) + 2Fe3+(aq)
There will be no voltage because the
electrodes are the same.
ii
MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) ⇌
Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)
Exercise 20.4
a
i
iii
Hydrogen gas at a pressure of
approximately 101 kPa
ii
H+ ion concentration of 1.00 mol dm−3
iii
Electrode is Pt
iv
Temperature is 298 K
c
d
e
iv
Br2(aq)
Br2(aq) + 2I−(aq) ⇌ 2Br−(aq) + I2(aq)
c
The voltage of a standard hydrogen
electrode is 0 V
H+(aq) + e− ⇌ 1 H2
2
The more negative (or less positive) the
electrode potential, the more difficult it is
to reduce the ions on the left hand side of
the equation. So the metal on the right is a
relatively good reducing agent. The ions on
the left hand side are relatively good oxidising
agents.
i
1.36 − 0.8 = 0.56 V (Ag electrode is
negative pole)
ii
−0.76 − (−0.13) = −0.63 V (Zn electrode is
negative pole)
iii
0.8 − 0.54 = 0.26 V (Iodine electrode is
negative pole)
iv
−0.25 − 1.47 = −1.72 V (Ni electrode is
negative pole)
d
i
Potassium manganate(VII) to Mn2+ ions,
E⦵ = +1.52 V Fluorine to fluoride ions,
E⦵ = +2.87 V
i
Zinc
ii
Zinc
iii
Copper
iv
Copper
i
V2+(aq)
So Ni to Ni2+, E⦵ = + 0.25 V
ii
Cl2(aq)
Fe3+ to Fe2+, E⦵ = +0.77 V
iii
Chlorine is better at releasing electrons
than iodine / iodide is better at accepting
electrons than bromide.
Fe3+ to Fe, E⦵ = −0.04 V + 0.25 + 0.77 =
+1.02 V (reaction feasible)
iv
3
Pb2+(aq)
Pb2+(aq) + 2Cr2+(aq) ⇌ Pb(s) + 2Cr3+(aq)
v
b
MnO4−(aq)
So Fluoride ions to fluorine E⦵
= −2.87 V + 1.52 − 2.87 = −1.35 V
(reaction not feasible)
ii
Bromine is better at releasing electrons
than iodine / iodide is better at accepting
electrons than bromide so the reaction
will be the reaction of bromine with
iodide ions.
Ni2+ to Ni, E⦵ = −0.25 V
+0.25 − 0.04 = + 0.21 V (reaction feasible)
iii
Mn2+ to Mn, E⦵ = −1.18 V
Iodine to iodide, E⦵ = +0.54 V
So iodide to iodine, E⦵ = −0.54 V −1.18 +
(−0.54) = 1.72 V (reaction not feasible)
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Exercise 20.6
a
b
i
Shifts the equilibrium to the right, so the
value of E⦵ is less negative than −0.76 V.
ii
Shifts the equilibrium to the left, so the
value of E⦵ is more negative than −0.76 V.
iii
Greater concentration of Cr2+ ions shifts
the equilibrium to the left, so the value of
E⦵ is more negative than −0.41 V.
iv
No effect, because the concentrations of
the ions are the same / the changes cancel
each other out.
R is the gas constant = 8.314 J K−1 mol−1
E is the electrode potential under non-standard
conditions
Exam-style questions
Question 1
a
b
c
log10 is the logarithm to the base 10
i
ii
The reduced form is the metal whose
concentration, being a solid, does not
change.
E = −0.76 + 0.059 log10(2.0)
2
d
e
Rest of calculation correct
[1]
Charge = 2.6 × 10 × 60 = 1560 C
[1]
H2(g) → 2H+(aq) + 2e−
[1]
i
[1]
ii
4H (aq) + O2(g) + 4e → 2H2O(l)
iii
2H2(g) + O2(g) → 2H2O(l)
ii
−
The electron flow is from the negative
(hydrogen electrode) to the positive
(oxygen) electrode. The E⦵ value for the
hydrogen electrode is more negative.
v
Zero because it is acting as a standard
hydrogen electrode.
i
NiO2 + 2H2O + 2e− → Ni(OH)2 + 2OH−
This has a more positive E⦵ value so
proceeds in the forward direction.
4
[1]
Hydrogen
Hydrogen is more readily discharged
than zinc /
Oxygen comes from OH− ions in water [1]
Cl− ions are close to OH− ions in the
discharge series / reactivity series so
some Cl2 formed
The only product is water which is not a
pollutant.
b
[1]
hydrogen accepts electrons more readily
than zinc
[1]
i
iv
[1]
The ions cannot move from place to place /
the ions only vibrate / the ions are fixed in a
lattice
Exercise 20.7
+
[1]
[1]
= −0.75 V (to 2 significant figures)
a
[1]
Multiplying by 4 for 4e− required for each
O2 molecule
Charge required for 1 mole = 3 × 96 500
= 289 500 C
Moles Al = 1560 = 5.3886 × 10−3 mol
289500
Mass Al = 5.3886 × 10−3 × 27 = 0.15 g
F is the Faraday constant in coulombs per mole
(96 500 C mol−1)
c
O2− → 1 O2 + 2e− (or doubled)
2
56 cm3 O2 = 56 = 2.333 × 10−3 mol
24000
Charge = 4 × 96 500 × 2.333 × 10−3 mol
= 900.5 C (2)
[1]
[Total: 13]
Question 2
a
b
The standard electrode potential of a
half cell [1]
when measured against a standard
hydrogen electrode
[1]
To maintain electrical connection between the
half-cells / so that there is a complete electrical
circuit.
[1]
ii
1.30 V
Filter paper / inert support / named inert
support
[1]
iii
Cd + NiO2 + 2H2O → Ni(OH)2 + Cd(OH)2
Soaked in saturated potassium nitrate
[1]
iv
Ni(OH)2 + Cd(OH)2 → Cd + NiO2 + 2H2O
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c
d
e
f
1.00 mol dm−3 zinc ions
[1]
298°C and 101 kPa
[1]
b
[1]
H2 gas passed over Pt at 105 Pa
[1]
Zn(s) + 2H+(aq) → H2(g) + Zn2+(aq)
[1]
Electrons more easily released at the negative
electrode (or by copper) / copper more easily
oxidised than iron(II)
[1]
H+ ions at concentration of 1.00 mol dm−3 [1]
Pt coated with Pt black
Electrons flow from − to + electrode in
the wires.
c
0.77 − 0.34 = 0.43 V
d
Cu + 2Fe3+ → Cu2+ + 2Fe2+
[1]
Correct formulae
[1]
Correct formulae
[1]
balance and state symbols
[1]
balance
[1]
i
1.52 − (−0.76) = 2.28 V
[1]
e
[2]
ii
Electrode potential of Zn / Zn ion cell will
get less negative
[1]
E = 0.34 + 0.059 log10 (0.15) = 0.32 V 2
1 mark for error carried forward using
incorrect number of electrons
f
i
The likelihood of a reaction occurring
when two or more substances are added
together. [1]
ii
PbO2 is a better oxidising agent since E⦵
more positive / PbO2 more likely to accept
electrons since E⦵ more positive
[1]
since the equilibrium is pushed to the
right / in favour of the reductant.
[1]
So difference between the electrode
potential of the half-cells decreases.
[1]
[Total: 16]
Question 3
I− better reducing agent since E⦵ less
positive / I− better reducing agent since
more likely to release electrons [1]
a
voltmeter
copper
rod
I2 weaker oxidising agent and PbO2
weaker reducing agent so reaction not
feasible [1]
salt bridge
E⦵cell suggested reaction is +0.54 −
(1.47) = −0.93 V Negative value of E⦵cell suggested reaction
not feasible [1]
platinum
Cu2+, 1.00 mol dm−3
298K
Fe3+, 1.00 mol dm−3
Fe2+, 1.00 mol dm−3
[1]
[Total: 19]
Two half-cells connected to a (high resistance)
voltmeter labelled V or voltmeter
[1]
Salt bridge labelled
[1]
Negative and positive electrodes labelled
correctly / Cu negative and Pt positive
[1]
Cu dipping into solution of Cu2+ ions or
suitable soluble Cu salt
[1]
Pt electrode dipping into solution of Fe3+ and
Fe2+ ions or suitable soluble
iron(II) and iron(III) salts
[1]
All solutions 1.00 mol dm−3
5
[1]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 21
Exercise 21.1
Exercise 21.2
a
a
1 with C; 2 with A; 3 with F; 4 with E; 5 with B;
6 with D
b
i
The ionic product for water
ii
The dissociation constant for a weak acid
iii
Solubility product
iv
Partition coefficient
c
i
1.7
d
i
6.31 × 10−11 mol dm−3
ii
3.98 × 10−7 mol dm−3
iii
0.0126 mol dm−3
i
Kw = [H+(aq)][OH−(aq)]
b
c
d
i
A proton donor / H+ ion donor
ii
ionised / dissociated; solution
iii
weak; dissociated / ionised; strong; it is
completely dissociated / ionised in water
Ethanoic acid is a weak acid and hydrochloric
acid is a strong acid; there is a lower
concentration of H+ ions in ethanoic
acid/ higher concentration of H+ ions in
hydrochloric acid; the collision frequency of
the H+ ions with magnesium in ethanoic acid
is lower / the collision frequency of the H+ ions
with magnesium in hydrochloric acid is higher.
i
acid HNO3; base H2O
ii
acid H2O; base NH3
iii
acid CH3OH; base NH2−
iv
acid H2O; base NH2OH
v
acid H2SO4; base H2O
vi
acid H2SO4; base HIO3
i
NH3 base; NH4+ conjugate acid
e
iii
iv
v
4.2
= 1.56 × 10−12 mol dm−3
So pH = 11.8
ii
Kw = [H+(aq)][OH−(aq)]
so [H+(aq)] =
H2O base; H3O+ conjugate acid
CH3COOH acid; CH3COO− conjugate base
−14
Kw
= 1.00 × 10 −4
[OH − (aq)]
3.00 × 10
H2O base; H3O+ conjugate acid
= 3.33 × 10−11 mol dm−3
CH3NHCH2NH3+ acid;
CH3NHCH2NH2 conjugate base
So pH = 10.5
H2O base; H3O+ conjugate acid
iii
[H+(aq)] (from pH)
= 3.16 × 10−13 mol dm−3
HSiO3− acid; SiO32− conjugate base
Kw = [H+(aq)][OH−(aq)]
H2O base; H3O+ conjugate acid
so [OH−(aq)] =
NH2OH base; NH3OH+ conjugate acid
H2O acid; OH− conjugate base
1
iii
−14
Kw
= 1.00 × 10
[OH − (aq)] 6.40 × 10 −3
HCO2H acid; HCO2− conjugate base
vi
0.9
so [H+(aq)] =
H2O acid; OH− conjugate base
ii
ii
−14
Kw
= 1.00 × 10 −13
−
[O H (aq)] 3.16 × 10
= 0.032 mol dm−3
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Exercise 21.3
a
b
Chemical equation
Equilibrium expression
Units
Fe2+(aq) + 2OH−(aq) ⇌ Fe(OH)2(s)
Ksp = [Fe2+(aq)] [OH−(aq)]2
mol3 dm−9
Sn2+(aq) + CO32−(aq) ⇌ SnCO3(s)
Ksp = [Sn2+(aq)] [CO32−(aq)]
mol2 dm−6
2Ag+(aq) + CrO42−(aq) ⇌ Ag2CrO4(s)
Ksp = [Ag+(aq)]2 [CrO42−(aq)]
mol3 dm−9
3Ag+(aq) + PO43−(aq) ⇌ Ag3PO4(s)
Ksp = [Ag+(aq)]3 [PO43−(aq)]
mol4 dm−12
Cr3+(aq) + 3OH−(aq) ⇌ Cr(OH)3(s)
Ksp = [Cr3+(aq)] [OH−(aq)]3
mol4 dm−12
2Ag+(aq) + S2−(aq) ⇌ Ag2S(s)
Ksp = [Ag+(aq)]2 [S2−(aq)]
mol3 dm−9
i
Ksp = [Ag+(aq)]2[S2−(aq)]
[Sr2+(aq)] × [CO32−(aq)] = (0.50 × 10−4) ×
(4.0 × 10−6) = 2.0 ×10−10 mol2 dm−6
= (2 × 5.25 × 10−17)2 × (5.25 × 10−17)
Remember: there are 2 Ag+ ions per
mole Ag2S
= 5.79 × 10−49 mol3 dm−9
ii
iii
This value is greater than the solubility
product, so strontium carbonate will
precipitate.
e
Ksp = [Pb2+(aq)][SO42−(aq)]
= (1.48 × 10−6) × (1.48 × 10−6)
Concentrations are divided by two because
each dilutes the other twofold.
= 2.18 × 10−12 mol2 dm−6
[Sr2+(aq)] = 2.5 × 10−7 mol dm−3
Ksp = [Ba (aq)][BrO3 (aq)]
−
2+
2
[SO42−(aq)] = 0.025 mol dm−3 (ignoring the
sulfate ions from the SrSO4)
= (9.86 × 10 ) × (2 × 9.86 × 10 )
Remember: there are 2BrO3−
ions per mole Ba(BrO3)2
−5
−5 2
[Sr2+(aq)] × [SO42−(aq)] = (2.5 × 10−7) × (0.025)
= 6.25 × 10−9 mol2 dm−6
= 3.83 × 10−12 mol3 dm−9
c
i
This value is greater than the solubility
product, so strontium carbonate will
precipitate.
[Ba (aq)] = [SO4 (aq)]
2+
2−
So Ksp = [Ba2+(aq)]2
−10
So [Ba2+(aq)] = K
Kspsp == 1.0
1.0 ××10
10 −10
= 1.0 × 10−5 mol dm−3
ii
[Cd2+(aq)] = [S2−(aq)]
So Ksp = [Cd2+(aq)]2
ksp =
−27
So [Cd2+(aq)] = K
sp = 8.0
× 10
1.0 ×
10 −10
= 8.9 × 10−14 mol dm−3
d
Equation: Sr2+(aq) + SO42−(aq) ⇌ SrSO4(s)
Exercise 21.4
a
i
pH = −log[H+(aq)]. So [H+(aq)]
= 1.26 × 10−3 mol dm−3
+
2
−3 2
Ka = [H (aq)] = (1.26 × 10 )
0.1
[HA(aq)]
= 1.59 × 10−5 mol dm−3
Equation: Sr (aq) + CO3 (aq) ⇌ SrCO3(s)
pH = −log[H+(aq)]. So [H+(aq)]
= 5.13 × 10−8 mol dm−3
Concentrations are divided by two because
each dilutes the other twofold.
Ka =
2+
2−
[Sr2+(aq)] = 0.50 × 10−4 mol dm−3
ii
[H + (aq)]2 (5.13 × 10 −8 ) 2
=
0.002
[HA(aq)]
= 1.32 × 10−12 mol dm−3
[CO32−(aq)] = 4.0 × 10−6 mol dm−3
2
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CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
iii
pH = −log[H+(aq)]. So [H+(aq)]
= 1.78 × 10−5 mol dm−3
Ka =
c
[H + (aq)] 2 (1.78 × 10 −5 ) 2
=
0.005
[HA(aq)]
The concentration of hydrogen ions does not
fall significantly because more butanoic acid
ionises to maintain the equilibrium.
= 6.34 × 10−8 mol dm−3
b
[H + (aq)]2
[HA(aq)]
The concentration of C3H7COOH / butanoic
acid does not fall significantly because it is
present in relatively high concentration.
i
Ka =
ii
[H+(aq)]2 = Ka × [HA(aq)]
iii
[H+(aq)]2 = (1.5 × 10−5) × 0.2 = 3.0 × 10−6
So [H+(aq)] =
The concentration ratio [C3H7COO−] to
[C3H7COOH] does not change much so the
pH does not change significantly.
( 3.0 × 10 −6 )
= 1.73 × 10−3 mol dm−3
c
iv
−log (1.73 × 10 ) = pH 2.76 / pH 2.8
i
[H+(aq)]2 = Ka × [HA(aq)]
−3
d
(1.3 × 10 −6 )
−log (1.14 × 10−3) = pH 2.94 / pH 2.9
[H+(aq)] = 1.69 × 10−5 mol.
So pH = 4.77 / 4.8
e
[H+(aq)]2 = Ka × [HA(aq)]
Convert pH to [H+(aq)] = 6.31 × 10−6 mol dm−3
Ka =
[H+(aq)]2 = (4.7 × 10−4) × 0.15 = 7.05 × 10−5
So [H+(aq)] =
[H + (aq)] [A − (aq)]
[HA(aq)]
Rearranging: [A−(aq)]
( 7.05 × 10 −5 )
=
= 8.40 × 10−3 mol dm−3
−log (8.40 × 10−3) = pH 2.07 / pH 2.1
Exercise 21.5
a
A solution which contains a conjugate acidbase system that minimises change in pH on
addition of acid or alkali.
b
In this buffer solution the conjugate base is
C3H7COO− / butanoate ion
K a × [HA(aq)] (1.35 × 10 −5 ) × 1.00
=
[H + (aq)]
(6.31 × 10 −6 )
[A−(aq)] = 2.14 mol (since in 1 dm3)
f
Addition of acid shifts the equilibrium to
the left because hydrogen ions from the acid
combine with C3H7COO− / butanoate ions
from the buffer solution. The concentration
of C3H7COO− / butanoate ions does not
fall significantly and the concentration of
C3H7COOH / butanoic acid does not rise
significantly because the acid and base (salt)
are both in relatively high concentrations.
The concentration ratio [C3H7COO−] to
[C3H7COOH] does not change much so the
pH does not change much / does not change
significantly.
3
[H + (aq)] [A − (aq)]
[HA(aq)]
K a × [HA(aq)] 1.35 × 10−5 × 0.5
=
0.4
[A − (aq)]
= 1.14 × 10−3 mol dm−3
ii
Ka =
Rearranging: [H+(aq)] =
[H+(aq)]2 = (1.3 × 10−5) × 0.10 = 1.3 × 10−6
So [H+(aq)] =
Addition of alkali shifts the equilibrium to
the right, because hydrogen ions from the acid
combine with OH− ions from the alkali.
Concentrations have to be calculated because
each solution dilutes the other.
The total volume is 400 cm3.
[ethanoic acid] = 0.50 × 300
400
= 0.375 mol dm−3
100
400
= 0.20 mol dm−3
[sodium ethanoate] = 0.80 ×
Ka =
[H + (aq)] [A − (aq)]
[HA(aq)]
Rearranging: [H+(aq)]
=
K a × [HA(aq)] (1.70 × 10 −5 ) × 0.375
=
[A − (aq)]
(0.20)
[H+(aq)] = 3.19 × 10−5 mol. So pH = 4.5
g
dissolved; hydrogencarbonate; hydrogen;
direction; excess; hydrogencarbonate;
concentrations; pH
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Exercise 21.6
a
b
c
more soluble in C6H6
iii
more soluble in water
[H+(aq)] =
i
Titrate with strong acid (hydrochloric or
sulfuric).
pH = 4.2
ii
The gradient of the graph gives the value.
e.g 1.57 − 0.63 ; Kpc = 23.5
0.05 − 0.01
The value of the partition coefficient
(much greater than 1) shows that iodine is
much more soluble in CCl4 than in water.
Iodine is a non-polar molecule. It dissolves
well in CCl4 because this molecule has no
overall polarity. New instantaneous dipoleinduced dipole attractions can be formed
between iodine and CCl4.This is because
the strength of the instantaneous dipoleinduced dipole forces in both molecules
are similar. Water is a polar molecule
which is hydrogen-bonded. The strength of
hydrogen bonds is much greater than the
strength of instantaneous dipole-induced
dipole forces. So the iodine molecules
cannot penetrate the structure of the water
very easily (the reaction would be highly
exothermic).
ii
The concentration of H2CO3 is the same
as the equilibrium concentration.
[1]
[1]
Justified by assumption that the ionisation
of water is so small compared with the
ionisation of the acid.
[1]
[Total: 13]
Question 2
a
b
The equilibrium shifts slightly to the right
[1]
because OH− ions remove H+ ions.
[1]
Butanoic acid ionises to maintain
equilibrium.
[1]
The concentrations of butanoic acid and
butanoate ions do not change much.
[1]
So pH does not increase much / remains
(fairly) constant.
[1]
[butanoic acid] = 0.2 × 50 = 0.05 mol dm−3
200
[sodium butanoate] = 0.4 × 150
200
= 0.30 mol dm−3
(1 mark for both correct)
It donates hydrogen ions
[1]
to water / the base
[1]
H2CO3 (acid) and HCO3− (base)
[1]
H2O (base) and H3O+ (acid)
[1]
[H + (aq)] 2
[H + (aq)] [A − (aq)]
or Ka =
[HA(aq)]
[HA(aq)]
[1]
Ka =
[1]
[1]
The concentration of HCO3− equals
the concentration of H+ / water does
not ionise.
Question 1
i
( 4.5 × 10 −9 ) = 6.71 × 10−5
Justified by low value of Ka / % of
hydrogen which ionises is extremely low.
[1]
Exam-style questions
4
[1]
ii
[H+(aq)]2 = (4.5 × 10−7) × 0.01
= 4.5 × 10−9
(The graph does not go through the origin
because some of the ammonia combines with
the copper(II) sulfate as a complex.)
b
[1]
More soluble in CCl4
ii
a
Rearranging: [H+(aq)]2 =
Ka × [HA(aq)]
i
Using methyl orange indicator / methyl
red / bromophenol blue / indicator that
changes colour in acidic region.
c
i
Rearranging: [H+(aq)] =
[K a × [HA(aq)]
[A − (aq)]
(1.50 × 10 −5 ) × 0.05
0.3
= 2.5 × 10−6 mol dm−3
[1]
[1]
[H+(aq)] =
c
[1]
pH = 5.6
[1]
Hydrogen ions produced could lower
the pH of the blood [1]
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which could disrupt enyme function / body
function
[1]
HCO3− ions combine with H+ ions to remove
them.
[1]
CO2 + H2O ⇌ HCO3− + H+
Question 4
a
i
ii
[Total: 13]
Question 3
CH3COOH + OH− → CH3COO− + H2O
[1]
b
pH 4.8 from graph
[1]
[H+(aq)] = 1.58 × 10−5 mol dm−3
[1]
Kw
[OH− (aq)]
[1]
c
[H+(aq)] =
(1.00 × 10 −14 )
= 5 × 10−13 mol dm−3
0.02
i
Phenolphthalein / other suitable
indicator
=
d
ii
e
b
ii
c
i
Its colour range does not coincide with
the sudden change of pH.
[1]
Use a pH meter to find the pH of both
solutions
mol dm [1]
Ksp = [Mg2+(aq)]2
[1]
[Mg2+(aq)] = k sp = 1.0 × 10 −5 [1]
3.2 × 10−3 mol dm−3
[1]
Ksp = [Ag+(aq)]2[CO32−(aq)]
[1]
mol3 dm9
[1]
[Ag+(aq)] = 2 × 1.2 × 10−5
= 2.4 × 10−5 mol dm−3
[1]
Ksp = (2.4 × 10−5)2 × 1.2 × 10−5
[1]
Ksp = 6.9 × 10−15
[1]
mol3 dm−9
[1]
Reduction in the solubility of a
dissolved salt
[1]
by adding a solution containing an ion in
common
[1]
Make a concentrated solution of sodium
ethanoate by dissolving in water / make a
1 mol dm−3 (or greater) solution of sodium
ethanoate by dissolving in water [1]
Make a concentrated solution of ethanoic
acid by dissolving in water / make a 1 mol dm−3
(or greater) solution of ethanoic acid by
dissolving in water.
[1]
ii
Some of the precipitate may dissolve
in water.
[1]
Washing with sulfuric acid reduces the
possibility of the precipitate dissolving /
sulfate both in sulfuric acid and barium
sulfate causes common ion effect
[1]
d
[1]
Concentrations are divided by two because
each dilutes the other twofold.
[Ca2+(aq)] = 0.005 mol dm−3
Slowly add a solution of one to a fixed volume
of the other (with stirring) until the required
pH is reached
[1]
i
[1]
[1]
[1]
−6
2
[1]
a
Ksp = [Mg2+(aq)] [CO32−(aq)]
[Total: 11]
[SO42−(aq)] = 0.01 mol dm−3 (1 mark if
both correct)
[1]
[Ca2+(aq)] × [SO42−(aq)] = 0.005 × 0.01 =
5.0 × 10−5 mol2 dm−6
[1]
This value is greater than the solubility
product, so calcium sulfate will precipitate. [1]
5
[Total: 18]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 22
Exercise 22.1
a
i
ii
iii
0.004 × 1000 = 0.02 mol dm−3
200
0.02 = 1.1 × 10−5 mol dm3 s−1
30 × 60
80 × 1000 = 6.67 × 10−2 mol dm−3
24000
50
d
6.67 × 10 −2 = 5.6 × 10−4 mol dm3 s−1
2 × 60
3 g propanol = 3 = 0.0517 mol in
58
250 cm3 = 0.207 mol dm−3
i
The power to which the concentration of
a reactant is raised in the rate equation.
e.g. [X]2 is second order
ii
Graph should show approximately
proportional relationship.
iii
First order. Idea of some variability in the
data so more tangents should be taken to
get wider spread of data.
iv
Deduce the time when the concentration
drops to half its original value, then to a
quarter of its original value, then to oneeighth.
v
When [H2O2] = 7.75 time = 800 s
(approximately) When [H2O2] = 3.875 time
= 16500 s (approximately) (2nd half-life
= 850 s). So the half-life is more or less
constant. So the reaction is first order
vi
8.66 / 8.7 × 10−4 (s−1)
i
An equation showing the relationship
between the rate constant and the
concentration of those reactants that
affect the rate reaction.
ii
rate = k[H2O2]
0.207 = 2.3 × 10−4 mol dm3 s−1
15 × 60
b
i
Measure increase in electrical conductivity
/ take samples of solution for analysis of
H+ ion concentration by titrating with
sodium hydroxide
ii
Measure increase in pressure (in closed
reaction vessel) due to increasing moles
of gas.
Colorimetry: measure colour change
at appropriate wavelength either of
decrease in dichromate orange or increase
in copper or chromium ion colours /
decrease in electrical conductivity (H+ ion
more conductive than others).
e
iv
Increase in volume of oxygen produced
using gas syringe.
Exercise 22.2
v
Take samples of solution for analysis of OH−
ion concentration by titrating with acid.
Intial rate 15.5 = 0.0148 mol dm−3 s−1;
1050
500 s 13.0 = 0.00765 mol dm−3 s−1;
1700
iii
c
i
ii
1
1000 s 12.0 = 0.00462 mol dm−3 s−1
2600
The rate decreases as the concentration
decreases.
a
order; rate constant; rate; concentrations;
concentration
b
i
Left hand graph:
y-axis − concentration of reactant
(in mol dm−3)
x-axis − time (in s)
Right hand graph:
y-axis − rate of reaction (in mol dm−3 s−1)
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x-axis − concentration of reactant
(in mol dm−3)
ii
Zero order = A and D
Exercise 22.3
a
Rate equation
First order = B and F
Second order = C and E
c
Details of reaction
Rate
equation
Overall
order of
reaction
rate proportional to
rate
2nd
concentration of I2 and = k[H2][I2] order
concentration of H2
d
rate proportional to
the square of the
concentration of NO2
rate
2nd
= k[NO2]2 order
rate proportional to
concentration of I2
and the square of the
concentration of O2
rate =
k[I2][O2]2
rate proportional to
concentration of HI
and concentration of
H2O2
rate =
2nd
k[HI][H2O2] order
rate independent on
the concentration of
any of the reactants
rate = k
i
b
Rate =
2nd
k[HCOOCH3][H+]
dm3 mol−1 s−1
Rate =
k[H2O2]
1st
s−1
Rate =
k[NH3]0
0
mol dm−3 s−1
Rate =
k[N2O]
1st
s−1
Rate =
4th
k[BrO3−][Br −][H+]2
dm9 mol3 s−1
Rate =
k[NO2]2
dm3 mol−1 s−1
i
Hydrogen: (compare the 2nd and 4th lines
of data to keep NO constant). Doubling
concentration doubles rate. So first order
with respect to hydrogen.
NO: (compare the first three lines of
data to keep H2 constant). Doubling
concentration increases rate × 4. So
second order with respect to NO.
0 order
Zero order reaction: Successive half-lives
decrease with time. First order reaction:
Successive half-lives are constant. Second
order reaction: Successive half-lives
increase with time.
2nd
ii
Rate = k[H2][NO]2
iii
2.4 × 10−6 = k (1.0 × 10−6) × (1.25 × 10−2)2
k = 1.92 × 104 dm6 mol2 s−1
i
First order because rate is proportional to
concentration.
For A, the first half-life is 10 s (6.4 to 3.2),
the second half-life is 10 s (3.2 to 1.6).
Half-life is constant so it is first order. For
B, the first half-life is somewhere between
20–30 s (6.4 to 3.2), the second half-life is
somewhere between 30–40 s (3.2 to 1.6).
Half-life decreases so it is zero order.
ii
Take any rate and corresponding
concentration, e.g
iii
Rate = k[H2O2][catalyst]
i
A temperature rise of 10°C approximately
doubles the rate of reaction.
iv
ii
As the temperature increases the rate
constant also increases as the rate of
reaction increases.
Graph of concentration against time
shows a gentle curve downwards as in
line B in figure in Exercise 22.1 part (a).
i
Graph should show a gentle downward curve
as in line B in figure in Exercise 22.1 part (a).
ii
First half-life is from 100% to 50% =
54 × 103 s
ii
e
3rd
order
Overall order Units of k
of reaction
c
0.2 mol dm−3 and 0.0075 mol dm−3 s−1
0.0075 = k (0.2) So k = 0.0375 s−1
d
Second half-life is from 50% to 25% =
(108 − 54) × 103 s = 54 × 103 s Half-life is
constant so order of reaction is first order.
2
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b
iii
0.693 = 1.28 × 10−5 s−1
54 × 10 3
iv
Because the % cyclopropane is
proportional to concentration.
Exercise 22.4
a
i
Rate = k[C2H5Br][OH ]
ii
CH3CH2(OH)Br−
iii
The slow step has two species, C2H5Br and
OH− colliding.
−
ii
The Fe2+ ions are being reformed.
They are the same at the end as at the
beginning.
iii
Homogeneous
iv
The iodide ions get oxidised first and the
Fe3+ ions get reduced to Fe2+. The Fe2+
ions then react with the S2O82−(aq) to form
sulfate ion and Fe3+ ions.
Nucleophile. It is negatively charged and
attacks the C atom attached to the Br
atom. This C atom has a δ+ charge.
i
The sodium hydroxide
ii
Rate = k[CH3COCH3][OH−]
uncatalysed
reaction
Enthalpy change
b
S2O82−(aq) + 2Fe2+(aq) + 2I−(aq) →
2SO42− (aq) + 2Fe3+ (aq) + I2(aq)
v
These two species appear in the rate equation.
iv
i
S2O82− + 2I−
Fe3+ → Fe2+
reaction 1
Fe2+ → Fe3+
reaction 2
(CH3COCH3 and OH− both appear in the
rate determining step)
iii
c
Electrophile. It attacks an area of negative
charge.
i
Rate = k[RSO4−]
ii
Water is in excess so its concentration is
effectively constant.
iii
Since the concentration of water is
constant, only the sulfate is involved in
the slow (rate determining) step and this
appears alone in the rate equation.
3
i
Mixture of nitrogen oxides / NO + NO2
ii
CO is poisonous and NO2 causes acid rain
/ catalyst for oxidation of sulfur dioxide /
NOx associated with photochemical smog /
ALLOW: its a greenhouse gas / potentiates
asthma / increase risk of heart attacks
iii
D, F, E, A, C, G, B
iv
The molecules of NOx and CO would not
be able to move away from the surface of
the catalyst / the molecules would be too
strongly bonded to allow bonds to form
between NOx and CO.
v
The molecules of NOx and CO would
diffuse away from the surface before they
could react.
vi
Desorption
2SO42– + I2
Progress of reaction
Exam-style questions
Question 1
a
Exercise 22.5
a
catalysed
reaction
Measure the volume of gas / carbon dioxide
given of using a gas syringe.
[1]
Volume of gas increases with time.
[1]
Some carbon dioxide may dissolve in the
reaction mixture. [1]
ALLOW: Measure volume of gas in upturned
measuring cylinder full of water in trough full
of water for the first marking point.
Measure the electrical conductivity (no marks)
The conductivity decreases with time.
[1]
Other substances in the mixture, e.g. + ions
have not been taken into account.
[1]
b
i
C2O42− Second order
[1]
(Compare the first two lines of data.
Doubling concentration increases rate by 4.)
HgCl2 first order
[1]
(Compare the lines of data 1 and 3.
Doubling concentration doubles the rate.)
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ii
rate = k[C2O42−]2[ HgCl2]
[1]
iii
Correct substitution, e.g. 4.3 × 10−4 =
k(0.11)2 × (0.0418)
[1]
Answer: 0.85
[1]
Unit: dm6 mol−2 s−1
[1]
Titrate with standard acid using an
indicator.
ALLOW: back titration for marks 2 and
3, i.e.
–
Question 2
Catalyst
[1]
b
Reaction rate changes when temperature
changes.
[1]
Points correct
[1]
Curved line joining all points [1]
c
d
[1]
iii
(CH3)3CBr spontaneously ionises [1]
to form (CH3)3C+
[1]
The concentration of water is not known
first order in rate equation.
[1]
OH− ion reacts with (CH3)3C+
[1]
This is the fast step.
[1]
[Total: 17]
Question 4
a
[1]
uncatalysed
reaction
Enthalpy change
First order
[Total: 10]
Question 3
C2H4 + H2
catalysed
reaction
C2H6
i
2nd order
[1]
ii
4th order
[1]
b
dm9 mol−3 s−1
[1]
c
Slow step involves HBr and HBrO3
[1]
[HBr] depends on the concentration of
Br− and H+
[1]
[HBrO3] depends on the concentration
of H+ and BrO3−
Energy hump of uncatalysed reaction higher
than catalysed, and labelled.
[1]
[1]
Energy hump of catalysed reaction lower than
uncatalysed, and labelled.
[1]
−
Two moles of H involved.
i
[1]
[1]
+
d
Progress of reaction
Axes labelled and reactants on left and
product on right with product level
below reactant level.
−
One mole of each of Br and BrO3
involved.
Take small portions of the reaction
mixture at various time intervals.
Quench them [1]; ALLOW: cool them
immediately.
4
[1]
The experiment has not been done varying the
concentration of HCl
[1]
a
Rate = k[(CH3)3CBr]
Rates: 0 = 0.046; 0.036; 0.027; 0.017
The rates are (roughly) proportional to the
concentrations
[1]
f
ii
This is the slow step corresponding to
(CH3)3CBr being
(2 marks if all within ± 0.005; 1 mark if 2 are
within ± 0.005)
e
add excess acid of known concentration;
–titrate with standard sodium hydroxide
using indicator.
[Total: 11]
a
[1]
[1]
b
[1]
Lower the activation energy.
[1]
Mechanism is different / provides an
alternative route.
[1]
(Second mark dependent on the first mark
being correct)
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c
C2H4 and H2 diffuse to the catalyst surface. [1]
C2H4 and H2 adsorbed on the catalyst
surface / C2H4 and H2 form bonds with
atoms on the catalyst surface.
[1]
Adsorbed molecules of C2H4 and H2 which are
close together start to form bonds.
[1]
Idea of desorption / bonds between atoms
on surface and ethane broken and ethene
diffuses from the surface.
[1]
d
i
Nitrogen dioxide is reformed / nitrogen
dioxide is not used up.
[1]
ii
Homogeneous
Bonds between C2H6 and the surface atoms
are weakened.
[1]
The catalyst and the reactants are both
in the same phase.
[1]
5
[1]
[Total: 13]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 23
Exercise 23.1
a
b
A spontaneous change is a change that, once
started, will carry on until it is finished.
Examples are diffusion, or the reaction of
sodium with water. Spontaneous reactions
happen because statistics tell us that there is
a greater likelihood of the particles having
more ways of arranging their energy. In a
spontaneous reaction, the entropy increases.
Entropy is a measure of the randomness or
disorder. The greater the randomness, the
greater is the entropy. The total entropy takes
into account the system (the reactants and
products) and the surroundings (everything
else around the reactants and products).
The system is more stable when it is more
disordered. The entropy is also greater when
there are more ways of arranging the energy.
is possible. So HBr has a higher entropy. The
Cl in HCl is also more electronegative than
Br, so the dipole– dipole forces are stronger,
leading to a more ordered structure.
CH4 (methane) and C3H8 (propane) are both
gases but methane has fewer atoms and
electrons and so has more order than propane,
which has a higher entropy.
Exercise 23.2
a
System: magnesium and hydrochloric acid.
Surroundings: test tube, thermometer,
air, water in the hydrochloric acid (rest of
solution).
c
1 with E; 2 with D; 3 with A; 4 with B; 5 with C
d
H2O (water) and C2H5OH (ethanol) are both
liquids but water has a greater amount of
hydrogen bonding which means more order.
So it has a lower entropy. In addition, ethanol
has a greater variety of atoms and more
atoms, both of which contribute to a larger
entropy.
NaCl and NaClO3 are both solids. NaClO3 has
more atoms and a greater variety of atoms. So
it has a higher entropy. Br2 liquid has a lower
entropy than the vapour because there is more
order in a liquid and there are fewer ways of
arranging the energy in a liquid than a vapour
where there are more quanta of energy.
HCl and HBr are both gases but HBr has
more electrons, so a greater degree of disorder
1
b
i
The left. It has two types of molecule and
seems to have more disorder.
ii
The left (reactant) square
iii
Higher, because it is a more complex
molecule.
iv
The reactants have greater entropy
because there are more molecules and two
types of molecule.
v
The entropy will decrease.
vi
The reaction is not likely to be spontaneous
(feasible) at room temperature considering
the system alone because the entropy
change of the system is negative.
i
The reactants are likely to have higher
entropy because chlorine is a gas and has
a high entropy. NaCl and Na are solids
and have low entropy. There are also two
different substances on the left.
ii
Greater entropy on the left because the
iron(II) chloride and sodium hydroxide
are both aqueous and the ions are free to
move. The iron(II) hydroxide on the right
is solid so it has low entropy. This makes
it likely that the products have lower
entropy.
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iii
c
v
Oxygen gas is released which has high
entropy and there are three molecules
on the right and only two on the left. It
is likely that the products have greater
entropy.
iv
There are two gas molecules on the left
and only one on the right, so it is likely
that there is greater entropy on the left.
v
There is one molecule of solid on the left
but three on the right, two of which are
gases with high entropy. So it is likely that
there is greater entropy on the right.
Ice has very low entropy because the particles
are in fixed positions and only vibrate.
The particles are highly ordered. As the
temperature of the solid is increased there is a
small increase in entropy because the particles
vibrate more. The entropy increases to a
greater extent on melting because the water
molecules begin to move randomly.
+ 1 O2(g)
2
164.0 26.9 + (2 × 240.0) + 1 (205)
2
∆S = +445.4 J K−1 mol−1
Exercise 23.3
a
b
i
ii
i
The equilibrium will be over to the left.
ii
There is exact equilibrium.
iii
The reaction is unlikely to occur.
iv
The equilibrium is to the right.
i
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆Hr = −1367.3 kJ mol−1
160.7 (3 × 205) (2 × 213.6) (3 × 69.9)
∆Ssystem = −138.8 J K−1 mol−1
ΔG = ΔH⦵ − TΔSsystem = −1 367 300 −
(298 × −138.8) = −1326 kJ mol−1
There is still some hydrogen bonding in the
liquid so the particles do not move completely
randomly. The entropy of the liquid water
increases slowly as temperature increases as
more and more hydrogen bonds are broken
and the particles move faster and faster. The
energy is more spread out with increase in
temperature so the entropy increases. At the
boiling point, the entropy increase is very
large because the attractive forces between the
water molecules are broken and a vapour is
formed where the molecules move freely and
randomly (very high entropy).
d
ii
ΔSsystem = + 170.9 J K−1 mol−1
ΔG = ΔH⦵ − TΔSsystem = +1418600 −
(298 × +170.9) = +1367.7 kJ mol−1
iii
(3 × 32.7)
(3 × 26.9) (2 × 27.3)
87.4
2SO2(g) + O2(g) → 2SO3(g)
ΔSsystem = −50.2 J K mol−1
(2 × 248.1) + 205 (2 × 95.6)
∆S = −510 J K−1 mol−1
ΔG = ΔH⦵ − TΔSsystem = −980 900 −
(298 × −50.2) = −965.9 kJ mol−1
H2O(g) + C(graphite) → H2(g) + CO(g)
c
i
Fe2O3(s) + 3CO(g)
SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
∆Gr
2Fe(s) + 3CO2(g)
∆G1
B2O3(s) + 3Mg(s) → 2B(s) + 3MgO(s)
∆G2
2Fe(s) + 3C(s) + 3O2(g)
ii
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
−742.2
41.8 + (4 × 173.7) 282.4 + (2 × 69.9)
∆S = −314.4 JK−1 mol−1
(3 × − 137.2)
0
(3 × −394.4)
ΔG = −29.4 kJ mol−1
iii
2
3Mg(s) + Fe2O3(s) → 3MgO(s) + 2Fe(s)
ΔHr = −980.9 kJ mol−1
−1
54.0 + (3 × 32.7) (2 × 5.9) + (3 × 26.9)
∆S = −59.6 J K−1 mol−1
iv
SrCO3(s) → SrO(s) + CO2(g)
ΔHr = + 1418.6 kJ mol−1
97.1 54.4 213.6
188.7 + 5.7 130.6 + 197.6
∆S = +133.8 J K−1 mol−1
iii
Mg(NO3)2(s) → MgO(s) + 2NO2(g)
It is spontaneous (feasible) since ΔG is
negative.
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d
i
SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)
for the extraction of aluminium by
carbon. A lower temperature is required
for the zinc extraction so less energy is
required (in the extraction of zinc by
reduction with carbon, carbon burns
to carbon monoxide which is a better
reducing agent)
−856.7 (4 × −273.2) − 1572.7 (2 × −69.9)
ΔG = +237 kJ mol−1
ii
SrCO3(s) → SrO(s) + CO2(g)
−1140.4
−561.9 −394.4
e
ΔG = +184.1 kJ mol−1
i
Exercise 23.4
a
iΔS ⦵system = ΣS ⦵products − ΣS ⦵reactants =
ΔG ⦵ = ΔH ⦵ − TΔS ⦵, the large positive
value of TΔS ⦵ (large negative value
of −TΔS ⦵) may be greater than the small
positive value of ΔH ⦵ so ΔG ⦵ is negative
and the reaction is feasible.
239.2 − 49.3 = +189.9 J mol−1 K−1
ΔH ⦵ = −110.5 − (−348.3) =
+ 237.8 kJ mol−1 = + 237 800 J mol−1
ΔG ⦵ = + 237 800 − (298 × 189.9) =
+ 181 209.8 J mol−1 = + 181.2 kJ mol−1
ii
ΔS ⦵system = + 21.8 J mol−1 K−1
ΔH⦵ = +318 700 J mol−1
ΔG ⦵ = +312.2 kJ mol−1
iii
ΔS ⦵system = + 160.4 J mol−1 K−1
ΔH ⦵ = +178 300 J mol−1
ΔG ⦵ = +103.5 kJ mol−1
iv
ii
The value of ΔH ⦵ is large and negative.
The entropy change is small and negative.
In the equation ΔG ⦵ = ΔH ⦵ −TΔS ⦵, the
large negative value of ΔH ⦵ is much
greater than the small value of TΔS ⦵
so ΔG ⦵ s negative and the reaction is
feasible.
iii
The value of ΔH ⦵ is large and positive.
The entropy change is large and positive
since a solid (low entropy) is being
converted to a gas and liquid (high
entropy). In the equation
ΔS ⦵system = −407.5 J mol−1 K−1
ΔH ⦵ = − 583 200 J mol−1
ΔG ⦵ = −461.8 kJ mol−1
v
ΔG ⦵ = ΔH ⦵ −TΔS ⦵, the large positive
value of TΔS ⦵ is unlikely compensate
large positive value of ΔH ⦵ because the
relative value of ΔH ⦵ (kJ mol−1) is larger
than the relative value of ΔS ⦵ (only in
J mol−1 K−1) so ΔG ⦵ positive and the
reaction is not feasible.
ΔS ⦵system = + 581.4 J mol−1 K−1
ΔH ⦵ = +1344 200 J mol−1
ΔG ⦵ = +1287.2 kJ mol−1
b
reaction iv (sulfur dioxide + oxygen forming
sulfur trioxide) because the value of ΔG ⦵ is
negative.
c
i
Rearranging the equation to find T with
ΔG ⦵ at 0 for the reaction to just become
feasible.
237 800
ΔH ⦵
= 1252 K
= T
189.9
ΔS ⦵system
ii
d
3
ΔH ⦵
1344 200
= 2312 K
= T
581.4
ΔS ⦵system
i
the values assume that the enthalpy
change and the entropy change do not
vary with temperature (in fact they do).
ii
It would take too much energy to
maintain the high temperatures required
The value of ΔH ⦵ is small and positive.
The entropy change is large and positive
since a solid (low entropy) is being
converted to a gas and liquid (high
entropy). In the equation
f
i
E ⦵cell = + 0.15 V ΔG ⦵ = −nFE ⦵cell
= −2 × 96 500 × 0.15 = −28 950 J mol−1
(−29.0 kJ mol−1). ΔG ⦵ is negative so
reaction is feasible.
ii
E ⦵cell = −0.43 V ΔG ⦵ = −nFE ⦵cell
= −2 × (96 500 × −0.43)
= + 82 990 J mol−1 (+ 83 kJ mol−1).
ΔG ⦵ is positive so reaction is not feasible.
iii
E ⦵cell = + 0.26 V ΔG ⦵ = −nFE ⦵cell
= −2 × 96 500 × 0.26 = −50 180 J mol−1
(−50.2 kJ mol−1). ΔG ⦵ is negative so
reaction is feasible.
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Exam-style questions
Question 1
a
b
ii
The number of possible arrangements of the
particles and their energy in a given system.[1]
Hydrogen has high entropy because it is a gas.[1]
c
i
b
ii
iii
Use of Hess’s Law ∆H ⦵r
= (−393.5) − (−110.5 + −157.3) = −125.7 kJ mol−1 [1]
[1]
Use of Gibbs equation: ΔG ⦵
= ΔH ⦵ −TΔS ⦵
[1]
conversion of kJ to joules for ΔH ⦵
and 200 °C to 473K for T [1]
v
Reaction feasible because value of
ΔG ⦵ is negative [1]
The values of ΔH ⦵ or S ⦵ does not
change with temperature [1]
[Total: 15]
c
i
ii
TiO2(s) + 2C(g)
∆G1
= −0.09 V
E⦵
cell
[1]
Use of ΔG ⦵ = −nFE ⦵
cell
[1]
Position of equilibrium so far over to the
left that the reaction is not feasible [1]
value of E ⦵
negative (so not feasible) [1]
cell
[Total: 15]
Question 3
a
b
2Rb(s) + S(s) → Rb2S(s)
∆H ⦵r = + 360.7 kJ mol−1 (at 298 K)
(2 × 76.8) + 32.6 → 134.0
[1]
∆Ssystem = −52.2 J K−1 mol−1
[1]
i
Use of ΣSproducts − ΣSreactants
= (121.5 + 56.5) − (95.9)
∆G2
Ti(s) + 2C(s) + O2(g)
4
Ti(s) + 2CO(g)
[1]
Value of ΔG ⦵ positive (so not feasible) [1]
i
∆Gr
No it is not spontaneous because ∆G ⦵r is
positive.
[1]
−2 × 96 500 × (−0.09) = + 17 370 J mol−1
(+ 17.4 kJ mol−1)
[1]
Question 2
a
[2]
Use of Gibbs equation:
ΔG ⦵ = ΔH ⦵ −TΔS ⦵ ΔG ⦵ = −125 700 − 473 (+6.6)
= −128 821.8 J mol−1 / −128.8 kJ mol−1 [1]
iv
[1]
−1
At higher temperatures the increased value
of T mean that the value of −TΔS ⦵ is large
enough to overcome the positive enthalpy
change of the reaction so the value of ΔG ⦵
negative so reaction is feasible.
[1]
Use of ΣSproducts − ΣSreactants
[1]
[1]
∆G ⦵r = 2 × ∆G ⦵f[CO(g)] − ∆G ⦵f[TiO2 (s)]
At lower temperatures value of −TΔS ⦵ is not
large enough to overcome the positive enthalpy
change of the reaction (and the value of ΔG ⦵
still positive so the reaction is not feasible) [1]
[1]
= (33.2 + 213.6) − (42.6 + 197.6)
= + 6.6 J mol−1 K−1
[1]
∆G r = +610.1 kJ mol The products have higher entropy than the
reactants.
[1]
The higher the entropy, the greater the
stability of the system.
Rest of cycle correct with arrows in
correct direction
⦵
iii
[1]
[1]
= (2 × −137.2) − (−884.5)
Sodium has low entropy because it is a solid. [1]
Aqueous sodium hydroxide has a higher
entropy than water because there are
more particles.
Top line of Hess cycle correct
[1]
= + 82.1 J mol−1 K−1
ii
[1]
⦵
Use of Gibbs equation: ΔG = ΔH
−TΔS ⦵
⦵
[1]
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conversion of kJ to joules for ΔH ⦵ and
25 °C to 298 K for T
[1]
The entropy increase on going into
solution is large and positive/TΔS ⦵ is
large and positive.
ΔG ⦵ = + 9 500 − 298 (+ 82.1) = −14 965.8 J
mol−1/−14.97 kJ mol−1
[1]
iii
5
This value is greater than the value of
[1]
ΔH ⦵ so ΔG ⦵ is negative.
Solid rubidium chloride has a very low
entropy because there are few possible
arrangements of the particles/energy. [1]
When in solution the particles move about
randomly and there are many more ways
of arranging the particles so entropy
higher.
[1]
[1]
Negative value of ΔG ⦵ mean reaction is
feasible.
[1]
[Total: 12]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 24
Exercise 24.1
a
variable oxidation state; catalytic activity; have
coloured ions; form complex ions
iii
There is a greater stability in a half full d
level than in a full 4s level.
b
Melting point: iron = 1808 K; calcium = 1112 K
(transition elements have higher melting points)
iv
Sc only forms Sc3+ ions, so there are no d
electrons present. Zn only forms Zn2+ ions,
so Zn2+ ions have a full 3d shell (not a d
shell deficient in (electrons)).
i
1s22s22p63s23p63d6
Metallic radius: calcium = 0.197 nm; iron
= 0.126 nm (transition elements have lower
radius for the same Period)
ii
1s22s22p63s23p63d3
iii
1s22s22p63s23p63d7
Boiling point: calcium 1380 K; iron = 3023 K
(transition elements have higher boiling
points)
iv
1s22s22p63s23p63d10
i
1 with H; 2 with F; 3 with B; 4 with A; 5
with G; 6 with C; 7 with D; 8 with E
ii
Colour in transition elements complexes
is due to d-orbital splitting in the presence
of ligands. The chiral centre must have an
unpaired d-electron for this to occur readily.
The colour is due to the energy absorbed
when a d-electron is excited from the lower
to the higher split d-orbitals. A scandium
ion, Sc3+, has no d-orbitals so there is no
colour. A zinc ion, Zn2+, has a complete
d-subshell and so an excited electron will
have to go into an energy sublevel which is
higher. This requires too much energy. So
zinc compounds are colourless.
Density: iron = 7.86 g cm−3; calcium
1.54 g cm−3 (transition elements have higher
densities)
d
e
Ionic radius (X2+): iron = 0.061 nm; calcium =
0.100 nm (transition elements have lower ionic
radius for the same Period)
First ionisation energy: iron 759 kJ mol−1;
calcium = 590 kJ mol−1 (in the same period
transition elements have higher first ionisation
energy)
Electrical conductivity: calcium = 2.82 ×
107 S m−1; iron = 9.04 × 106 S m−1 (many
transition elements have lower conductivity)
c
i
Co = 1s22s22p63s23p63d74s2
Ni = 1s22s22p63s23p63d84s2
Exercise 24.2
Cu = 1s22s22p63s23p63d104s1
ii
i
3Cr3+ + 3e− → 3Cr2+
ii
VO2+ + 2H+ + e− → V3+ + H2O
4s
iii
2Ni → 2Ni2+ + 4e−
3p
iv
Fe2+ → Fe3+ + e−
v
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
vi
CrO42− + 4H2O + 3e− → Cr(OH)3 + 5OH−
a
3d
3s
As shown in diagram or with 4s above 3d
1
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b
c
i
ii
2VO2+ + 4H+ + Ni → 2V3+ + 2H2O + Ni2+
iii
2Cr2+ + Ni2+ → 2Cr3+ + Ni
iv
Cu2+ + 2I− → I2 + Cu
v
2MnO4− + 5C2O42− + 16H+ →
2Mn2+ + 10CO2 + 8H2O
i
The more positive the value of E ⦵, the
easier it is for the oxidised form to accept
electrons. The Cr2O72− ions have a more
positive value of E ⦵, so are better at
accepting electrons than Fe3+. Fe3+ ions are
better at releasing electrons.
ii
d
MnO4− + 8H+ + 5Fe2+ →
Mn2+ + 4H2O + 5Fe3+
d
NC
b
c
NC
2
2−
+
e
i
iii
Cr +6 to +3, Fe +2 to +3
iv
If dichromate is used, both the colours
of the dichromate and the Cr3+ ions are
intense so it is difficult to see a distinct
end point (the end would be a ‘muddy’
colour which is not clear). Potassium
manganate is purple but Mn2+ ions are
very pale pink (essentially colourless) and
iron(III) ions are very light yellow, so the
end point is more easily seen.
CN
ii
complex; transition; ligands
iii
dative covalent; co-ordination
i
A = nitrile / cyanide B = chloride
ii
A = 4; B = 4
iii
A = square planar; B = tetrahedral
iv
A = +2; B = +2
v
A = [Ni(CN)4]2−; B = [CoCl4]2−
vi
A
90° B
109.5°
i
+2
ii
+1
+3
iv +6
v
Pt
Cl
CN
ii
en
ii
Cl
iii
Cl
Cr
Cl
en
Cl
en
Cr
en
Cl
f
Any E ⦵ values less positive than 1.52 V will be
oxidised (Mn2+, I− and Cu).
ion; lone; ligand
Cl
Bidentate, because it has a lone pair of
electrons on each N atom which are able
to form dative covalent bonds with the
transition element ion.
iii
i
Cl
H3N Ag NH3
i
Cr2O72−(aq) + 14H+(aq) + 6Fe2+(aq) →
2Cr3+(aq) + 6Fe3+(aq) + 7H2O(l)
iii
4−
CN
Fe
i
Bidentate, because it has lone pairs of
electrons on each O atom which are able
to form dative covalent bonds with the
transition element ion.
ii
Bidentate, because it has lone pairs of
electrons on the OH oxygen atom and the
C─O− oxygen atom which are able to form
dative covalent bonds with the transition
element ion.
iii
Hexadentate because it has lone pairs
of electrons on each C─O− oxygen atom
and on each N which are able to form
dative covalent bonds with the transition
element ion.
Exercise 24.3
a
CN
g
F more electronegative than O so there is a
dipole with the δ− end towards the 3 F atoms
and the δ+ end towards the H2O
+3
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Exercise 24.4
a
i
ii
c
[[Co(NH3 )6 (aq)]2 + ][(H2O)(1) ]6
[[Co(H2O)6 (aq)]2 + ][NH3 (aq)]6
no units
Co-ordinate / dative covalent
[1]
ii
Lone pairs of electrons
[1]
iii
90°
[1]
The structure shown is planar so this
minimises electron-pair–electron-pair
repulsion.
[1]
NH3
[1]
[[CoCl 4 )2 − (aq)][(H2O)(l)]6
[[Co(H2O)6 ]2 + (aq)][Cl − (aq)]4
−6
mol dm
2
iii
i
[[Cu(EDTA)]2 − (aq)][H2 O(l)]6
[[Cu(H2 O)6 ]2 + (aq)][(EDTA)4− ]
Pt
mol5 dm−15
b
i
Adding ammonia: goes from green to
light violet because the stability constant
of the ammonia complex is greater so the
equilibrium goes to the right.
Adding EDTA: goes from light violet
to blue because the stability constant of
the EDTA complex is greater than the
ammonia complex so the equilibrium goes
to the right.
ii
[Ni(H2O)6]2+(aq) + 6NH3(aq) ⇌
[Ni(NH3)6]2+(aq) + 6H2O(aq)
iii
ALLOW: any value between 9 and 18.
a
The bonds are co-ordinate / dative covalent.[1]
b
ii
[Cu(H2O)6]2+(aq) + 2(en) ⇌
[Cu(en)2(H2O)2]2+(aq) + 4H2O(l)
[Cu(en)2(H2O)2]2+(aq) correct
[1]
Rest of equation correct
[1]
[[Cu(en)2 (H2O)2 ]2 + (aq)][H2 O(l)]4 [[Cu(H2O)6 ]2 + (aq)][(en)]2
[1]
mol2 dm−6
[1]
en because it has a higher stability
constant
[1]
[Total: 8]
Question 3
a
higher boiling or melting point
smaller atomic or ionic radius
Zn better at releasing electrons than Fe2+
so Zn reduces Fe3+ / so Fe3+ oxidises Zn
higher 1st ionisation energy
lower electrical conductivity The more positive the value of E ⦵, the easier it
is for the oxidised form to accept electrons. [1]
The Fe3+ ions have a more positive value than
Zn2+ ions so more readily oxidised
[1]
higher density
[3]
ALLOW: reverse argument for Ba
3
i
Pt has:
b
[Total: 9]
Ligand: molecule / ion with (one or more) lone
pair of electrons forming bond with transition
element ion.
[1]
iii
Any three of:
Cl
Bidentate: (ligand) forming two bonds with a
transition element ion.
[1]
Exam-style questions
Question 1
H3N
Question 2
Adding water: no effect because the edta
complex has such a high Kstab value that
even if excess water is added, it will not
shift the equilibrium significantly.
a
Cl
iv
The ions have electron configuration with an
incomplete d electron shell
[1]
b
[1]
i
+3
[1]
ii
octahedral
[1]
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c
i
ii
2Fe3+ + 2I− → 2Fe2+ + I2
iv
Correct formulae
[1]
correct balance
[1]
A complex is formed with the
F / FeF63− formed
[1]
Adding ammonia has no / little effect [1]
because the edta complex has such a high
Kstab value that a small amount added will
not shift the equilibrium significantly. [1]
c
2−
Cl
The stability constant of the complex /
F is greater than that with water
[1]
all the iron(III) ions are involved in
complex formation so cannot react
with iodine
d
i
ii
[1]
[1]
[1]
Rest of structure drawn correctly (wedges
and dashed lines)
[1]
concentration of ethanedioic acid
1000
= 20 × 9.30 × 10−4 = 0.0465 mol dm−3 [1]
[Total: 15]
a
b
i
1s 2s 2p 3s 3p 3d 4s [1]
ii
1s22s22p63s23p63d6
[1]
i
Adding ammonia: colour goes from pink
to green
[1]
2
6
2
6
7
2
because the stability constant of the
ammonia complex is greater so the
equilibrium shifted to the right.
ii
d
i
atoms / orbitals are at the same energy
level [1]
ii
see diagram TG 4d ii iii
tetrahedral: 3 orbitals higher energy, 2
orbitals lower energy
[1]
iv
4
(Ligand) forming 6 bonds with a
transition element ion
[1]
The bonds are co-ordinate / dative
covalent
[1]
[1]
wavelength / frequency of absorbed light
depends on energy difference between
non-degenerate sub-levels, ∆E
[1]
light of complementary wavelength /
frequency transmitted / reflected
Correct structure of [Co(NH3)6]2+(aq) [1]
iii
electrons in d energy sub-level absorb light
energy of particular wavelength
[1]
and move to a higher d energy nondegenerate sub-level
[Co(H2O)6]2+(aq) + 6NH3(aq) ⇌
[Co(NH3)6]2+(aq) + 6H2O(l)
[1]
[1]
octahedral: 2 orbitals higher energy, 3
orbitals lower energy
[1]
[1]
rest of equation correct
[1]
109.5°
Question 4
2
Cl
Tetrahedral structure drawn with Co in
centre
[1]
moles ethanedioic acid =
2.5 × 3.72 × 10−4 mol = 9.30 × 10−4 mol [1]
Cl
KMnO4 is pink / magenta and ethanedioic
acid is colourless [1]
at the end point there is a slight
permanent pink colour
18.6
× 0.02
moles of MnO4− =
1000
−4
= 3.72 × 10 mol
Co
Cl
[1]
∆E depends on types of ligand attached to
central (transition element) ion / atom [1]
[Total: 22]
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Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 25
Exercise 25.1
a
b
Each carbon atom in the benzene ring forms
sp2 hybrid orbitals, sharing one pair of
electrons with each of the two neighbouring
carbon atoms and one pair with a hydrogen
atom. These are sigma bonds. The remaining p
electron from each carbon atom contributes to
a pi bond by sideways overlap of p type atomic
orbitals. These form two rings of delocalised
electrons above and below the benzene ring.
To allow maximum overlap of electrons, the
benzene ring must be planar. The bond angles
around each carbon atom are 120°.
Exercise 25.2
a
ii
ii
OH
stage B
CH2Cl
CH3
Br
i
H
H
C
C
H
H
C
C
C
C
H
1
C
iii
H
C
H
H
H
C
C
C
C
H
C
N
H
H
H
stage C
Product
Aluminium chloride. AlCl3 is a polar
molecule with δ+ on the Al and δ− on the
Cl. The polar AlCl3 with draws electrons
from the bond in the Cl2 molecule so that
Cl+ and AlCl4− are formed. The Cl+ acts
as an electrophile and is attracted to the pi
electrons in the benzene ring.
i
Sulfuric acid. It donates a hydrogen ion to
the nitric acid.
ii
NO2+ is deficient in an electron pair
and receives an electron pair from the
electrons in the benzene ring.
iii
H
C
H
C
H
O
O
O
H
b
C2H5
H
H
+ HBr
+
vii 2-chloromethylbenzene and
4-chloromethylbenzene
Br
ii
H
Br
vi
Br
Br
NO2
d
vi
OH
Br
The attacking reagent is deficient in an
electron pair and receives an electron pair
from the electrons in the benzene ring. Br+
is the electrophile.
NO2
v
[FeBr4]−
v
CH3
iv
Br+ +
FeBr3
Iron is electropositive and bromine in
iron(III) bromide is electronegative. The
positive charge on the iron attracts the
electrons in the Br−Br bond towards it.
Br+
iii
CH3
δ−
Br
H
The lengths of each carbon–carbon bond in
benzene are the same, the bond length being
intermediate between that of a C─C and C═C
bond in alkanes and alkenes.
i
δ+
Br
iii and iv
Benzene does not react with bromine water, so
it does not have a C═C double bond.
c
i
H
C
C
C
C
H
H
C
H
C
C
H
H
NO2
+
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iv
In the intermediate there are 4 pi bonding
electrons and the positive charge on the
ring is spread over 5 atoms. The C─H
bond breaks heterolytically and both
electrons from this ond go to complete pi
bonded system. The H+ ion is released.
from the electrons in the benzene ring.
+
CH2CH3 is the electrophile.
b
Exercise 25.3
a
i and ii
H3C
Stage A
H
δ+
C
c
δ−
Cl
+
CH3CH2 + [AlCl4]−
AlCl3
H
H
+
CH2CH3
Stage B
CH2CH3
+
Stage C
+ [AlCl4]−
CH2CH3
+ HCl + AlCl3
Products
d
iii
Ethylbenzene
iv
HCl and AlCl3
v
The attacking reagent is deficient in an
electron pair and receives an electron pair
vi
C3H7COCl
i
Type: substitution; mechanism: free radical
ii
Initiation by light splitting the chlorine
molecule by homolytic fission to form two
Cl• free radicals.
i
C6H5−CH3 + 3[O] → C6H5 −COOH + H2O
ii
Benzoic acid
iii
Alkaline solution of potassium
manganate(VII) then acidify
iv
Reflux + alkaline conditions
i
C6H5−CH3 + 3H2 → C6H11−CH3
ii
Methylcyclohexane
iii
Catalyst
iv
150 °C / high temperature
Exercise 25.4
a
b
c
2
either add water to benzendiazonium chloride, warm / heat sodium hydroxide with chlorobenzene to
form sodium phenoxide then add acid
Reaction
Reaction conditions Reaction conditions
benzene
with phenol
Comparison of extent of
reaction
bromination
liquid bromine
halogen carrier e.g.
iron(III) bromide /
Fe + Bromine
bromine water;
room temperature
benzene: need excess Br2 to get
2 Br atoms substituted phenol: 3
Br atoms easily substituted
nitration
concentrated nitric
and sulfuric acid;
reflux at 55 °C
dilute nitric acid;
room temperature
benzene: second substitution
quite difficult phenol: second
substitution relatively easy; third
substitution needs concentrated
acid
substitution of
SO3H group
fuming sulfuric acid
reflux for 8 hrs / a
long time
heat with H2SO4 at
100 °C
benzene: second substitution
difficult phenol: quite difficult to
substitute a second SO3H group
substitution of
NO group
does not work
because if heated
electrophile
decomposes
NaNO2, H2SO4,
(electrophile only
stable at low
temperatures)
The −OH group in phenol is an activating
group. This means that the reaction with
electrophiles is much more rapid than with
benzene. The oxygen atom in the −OH group
in phenol has an electron pair that can be
partly donated to the carbon atom next to it.
The intermediate formed when an electrophile
is substituted has partial positive charge in
various parts of the ring. The reduction in the
positive charge by the donation of electrons
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means that the electron density in various
parts of the ring increases. This makes the aryl
ring more open to attack by electron-deficient
electrophiles.
d
e
2C6H5OH + 2Na → 2C6H5ONa + H2
a
NO2+
[1]
ii
C6H5OH + KOH → C6H5OK + H2O
b
Substitution
[1]
iii
C6H5O−Na+ + C10H7N≡N+ Cl− →
C10H7N=NC6H4OH + NaCl
Electrophilic
[1]
i
stronger, higher
The C−H bond breaks
[1]
ii
p, oxygen, reduces, negative, more
heterolytically.
[1]
iii
delocalisation, ions
iv
weaker, sodium carbonate
Both electrons from this bond go to
complete pi bonded system / go into the
benzene ring.
[1]
H+ ion is released
[1]
It makes the ring more reactive
[1]
to electrophiles
[1]
There is a higher electron density in the
benzene ring
[1]
i
[1]
c
d
i
ii
NH2
iii
CH3
COOH
NO2
e
Cl
Br
NO2
iv
Cl
Br
ii
f
or
vi
OH
Br
vii
CN
Cl
Cl
Br
i
ii
c
3
CH2CH3
Br
Cl
[1]
Free radical
Methyl benzene: add concentrated nitric acid
and concentrated sulfuric acid
[1]
[1]
reflux
NO2
v
Light
Methylbenzene should be boiling / hot [1]
Cl
NO2
b
Question 1
i
Exercise 25.5
a
Exam-style questions
Cl
Oxygen is more electronegative than S.
So the molecule is dipolar with S being
δ+. The δ+ end of the molecule is electron
deficient so it can act as an electrophile.
An electron pair for the benzene ring is
donated to form a bond between a carbon
atom and the S atom. A further movement
of an electron pair from one of the S═O
bonds causes the O to ionise: C─SO2─O−
iii
In the 3- (and 5-) positions.
i
chlorine, uv light, heat
ii
free radical (substitution)
2-nitromethylbenzene / 2-nitromethylbenzene
[1]
Phenol: add dilute nitric acid
[1]
room temperature
[1]
2-nitromethylbenzene / 2-nitromethylbenzene
[1]
OR
Phenol: add concentrated nitric acid
[1]
room temperature
[1]
2, 4, 6-trinitromethylbenzene
[1]
[Total: 19]
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY: WORKBOOK
Question 2
a
i
d
C6H5OH + H2O ⇌ C6H5O− + H3O+
[2]
(1 mark if → instead of ⇌ / C6H5OH ⇌
C6H5O− + H+)
ii
N=N
2C6H5OH + 2Na ⇌ 2C6H5O− + Na+ + H2
a
correct balance
[1]
i
2,4,6-tribromophenol
[1]
ii
The −OH group is an activating group. [1]
So pair of electrons donated to
electrophile more readily. [1]
Br2 liquid
[1]
Fe / FeBr3 / any suitable named halogen
carrier
[1]
Br+
[1]
Electrophile is deficient in electron
pair.
[1]
OH
[1]
[Total: 19]
Question 3
[1]
ii
[1]
No stabilisation in ethoxide ion / C2H5O
ion (so H+ ion easily recombines)
[1]
Correct formulae
i
low temperature / ice bath / 5-10oC The substitution
[1]
of an alkyl group into a compound.
[1]
Substitution
[1]
Electrophilic
[1]
Dry AlCl3
[1]
2-chlorobutane
[1]
Reflux (at 50 °C)
[1]
d
C10H14
[1]
e
i
C2H5COCl
[1]
ii
ketone
[1]
i
Ni catalyst
[1]
150 °C / high temperature
[1]
Ethylcyclohexane
[1]
b
c
It causes the benzene ring in phenol to
have a greater electron density
(in certain places).
[1]
c
[1]
It reduces the negative charge density on
[1]
the O−
−
b
alkaline solution / aqueous sodium
hydroxide
ii
so hydrogen ion is not so readily accepted.
[1]
iii
i
f
ii
[Total: 13]
Electron pair moves from benzene ring to
form bond with Br.
[1]
4
Cambridge International AS & A Level Chemistry © Cambridge University Press 2020