Transcript Intermolecular forces

11
Intermolecular
Forces
1
INTERMOLECULAR FORCES
Hydrogen
bonds
Van der
Waals’ forces
Dipole-dipole forces
2
London
Dispersion forces
Johannes van der Waals
(1837−1923).
3
Fritz London
(1900−1954).
3 types of
dipoles
Permanent
dipole
4
Instantaneous
dipole
Induced
dipole
Permanent dipole
A permanent dipole exists in all polar
molecules as a result of the difference in the
electronegativity of bonded atoms.
5
Instantaneous dipole
An instantaneous dipole is a temporary dipole
that exists as a result of fluctuation in the
electron cloud.
6
Instantaneous dipole
An instantaneous dipole is a temporary dipole
that exists as a result of fluctuation in the
electron cloud.
7
Induced dipole
An induced dipole is a temporary dipole that
is created due to the influence of
neighbouring dipole (which may be a
permanent or an instantaneous dipole).
Permanent dipole
8
11.2
Van der Waals’
Forces
9
Van der Waals’ Forces
Van der Waals’
forces
DipoleDipole
Interaction
DipoleInduced
Dipole
Interaction
Instantaneous
DipoleInduced Dipole
Interaction
London dispersion forces
10
Dipole-dipole interactions
• Electrostatic interactions between polar
molecules
11
Dipole-dipole interactions
• In a sample containing many polar molecules
A balance of
attraction and
repulsion holding the
molecules together
12
Dipole-induced dipole interactions
• When a non-polar molecule approaches a polar
molecule (with a permanent dipole), a dipole will
be induced in the non-polar molecule.
Dispersion forces exist among all molecules and
contribute most to the overall van der Waals’ forces.
13
Polarization
Polarizability : - A measure of how easily
the electron cloud of an atom/molecule can
be distorted to induce a dipole
14
In general, size of electron cloud 




15
electron cloud is less controlled by
positive nuclei
extent of electron cloud distortion 
polarizability 
stronger dispersion forces
11.2 Van der Waals’ forces (SB p.277)
Instantaneous dipole-induced dipole
interactions
• The instantaneous dipole arises from
constant movement of electrons.
• Induces dipoles in neighbouring atoms or
molecules
16
17
Instantaneous dipole-induced dipole
interactions
18
19
Evidence for the presence of London
dispersion forces
1.
Condensation of noble gases at low
temperatures to form liquids and solids
 presence of attractive forces between
non-polar atoms
E.g. Xe(g)  Xe(s) Hsub = -14.9 kJ mol1
20
Evidence for the presence of London
dispersion forces
2. The non-ideal behaviour of gases

n
 P  a 
V 

2

V  bn  nRT

van der Waals’ equation
21
11.2 Van der Waals’ forces (SB p.279)
Strength of van der Waals’ forces
Much weaker than covalent bonds
Less than 10% the strength of covalent bonds
van der Waals’ radius > covalent radius
I2
22
Q.59
The electron clouds of adjacent iodine
molecules would repel each other strongly
until the equilibrium van der Waals’
distance is restored.
23
The strength of van der Waals’ forces can be
estimated by
melting point,
boiling point,
enthalpy change of fusion or
enthalpy change of vapourization.
Higher m.p./b.p./Hfusion/Hvap
 stronger van der Waals’ forces
24
Strength of van der Waals’ forces
Depends on three factors (in decreasing
order of importance) : 1. Size of molecule
2. Surface area of molecule
3. Polarity of molecule
25
1. Size of Molecule
Molecule
Helium
Neon
Argon
Fluorine
Chlorine
Bromine
Boiling point
(oC)
-269
-246
-186
-188
-34.7
58.8
Methane
Ethane
Propane
-162
-88.6
-42.2
26
Sometimes !
Rel.
Size
molecular
of molecule
mass 
Size of electron cloud 
Polarizability 
Dispersion forces 
2. Surface area of molecule
The van der Waals’ forces also increase
with the surface area of the molecule.
∵ van der Waals' forces are short-ranged forces
Atoms or molecules must come close together
for significant induction of dipoles.
27
2,2-dimethylpropane
(C5H12)
Pentane (C5H12)
Both are
non-polar
Same no. of
electrons
Boiling point: 36.1°C
28
Boiling point: 9.5°C
rod-shaped
pentane molecules
larger contact area
29
spherical in shape
2,2-dimethylpropane molecules
smaller contact area
Pentane (C5H12)
Boiling point
= 36.1C
Larger contact surface area
 Higher chance of forming induced dipoles
 stronger dispersion forces
30
2,2-dimethylpropane
(C5H12)
Boiling point
= 9.5C
Smaller contact surface area
 lower chance of forming induced dipoles
 weaker dispersion forces
31
3. Polarity of molecules
For molecules with comparable molecular sizes
and shapes, dispersion forces are approximately
equal.
Then, strength of van der Waals’ forces depends
on the polarity of molecules involved
Polar/polar > polar/non-polar > non-polar/non-polar
32
H3C
+
C

O
H3C
Dipole-dipole
forces
H3C
+
+
Dispersion forces H3C
C

O
RMM = 58.0,
b.p. = 50C
H2
C
H3C
CH3
C
H2
RMM = 58.0,
b.p. = 0C
33
H2
C
H3C
Dispersion forces
only
CH3
C
H2
Other examples : 1. Graphite layers of large surface area
 strong van der Waals’ forces
2. Polyethene
(m.p. > 100C)
34
vs
ethene
(m.p. = 169C)
% contribution to the overall van
der Waals' forces
Molecule
35
DipoleInstantaneous
Dipoleinduced
dipoledipole
dipole
induced dipole
interaction
interaction
interaction
C4H10
0
0
100
HCl
15
4
81
Q.60(a)
CH3Cl
b.p./C
-24.2
<
CH3Br
3.56
<
CH3I
42.4
The strength of dispersion forces increases
with molecular size/mass.
Thus, b.p. increases with molecular size/mass
Although chloromethane is more polar,
the effect of dispersion forces outweights
that of dipole-dipole forces.
36
Q.60(b)
CH 3
C
H3C
CH 3
CH 3
9.5C
<
H2
C
CH3
H3C
CH
<
H3C
CH3
27.7C
Less spherical
Greater surface area
37
H2
C
C
H2
H2
C
CH3
36.1C
Q.60(c)
F2
F2
-188C
Cl2
<
ClF
<
-100C
ClF
CH2Cl2
Cl2
< CH2Cl2
-34.0C
39.6C
ClF > F2. It is because
1. ClF has a greater molecular size than F2 and
thus has stronger dispersion forces than F2
2. ClF is polar and its molecules are held by both
dipole-dipole forces and dispersion forces.
38
Q.60(c)
F2
-188C
<
ClF
-100C
<
Cl2
-34.0C
> CH2Cl2
39.6C
Cl2 > ClF. It is because
1. Cl2 has a greater molecular size than ClF and
thus has stronger dispersion forces than ClF.
2. Although ClF is polar, the effect of dispersion
forces outweights that of dipole-dipole forces.
39
Q.60(c)
F2
-188C
<
ClF
-100C
<
Cl2
-34.0C
> CH2Cl2
39.6C
CH2Cl2 > Cl2. It is because
1. CH2Cl2 has a greater molecular size than Cl2 and
thus has stronger dispersion forces than Cl2.
2. CH2Cl2 is polar and its molecules are held by
both dipole-dipole forces and dispersion forces.
40
Q.60(d)
NO
b.p./C
RMM
41
-151
28.0
<
C2H6
-89
28.0
1 pm = 0.001 nm
1 nm = 109 m
42
NO
b.p./C
RMM
-151
28.0
<
C2H6
-89
28.0
C2H6 > NO. It is because
1. C2H6 has a greater molecular size and contact
surface area than NO and thus has stronger
dispersion forces than NO.
2. Although NO is polar, the effect of dispersion
forces outweights that of dipole-dipole forces.
43
The melting of a solid involves the
separation of molecules from a regularly
packed molecular crystal.
Thus, m.p. of a solid depends on
1. The strength of van der Waals’ forces
2. Packing efficiency of molecules in the
crystal lattice
44
Symmetry of molecule 
 Packing efficiency 
 m.p. 
45
Q.61
CH 3
H2
C
CH3
H3C
CH
CH3
m.p. -160C
<
H3C
H2
C
C
H2
H2
C
CH3
<
C
H3C
-136C
Increasing symmetry
Increasing packing
efficiency
46
CH 3
CH 3
-20C
Q.61
H2
C
CH3
H3C
CH
CH3
47
<
H3C
H2
C
C
H2
H2
C
CH3
Greater surface area
Stronger van der Waals’
forces
11.4
48
Molecular
Crystals
Molecular crystals
A molecular crystal is a structure which
consists of individual molecules packed
together in a regular arrangement by weak
intermolecular forces.
49
Iodine
f.c.c. structure
A unit cell of iodine crystal showing the
orientation of I2 molecules
50
Dry ice
f.c.c. structure
A unit cell of dry ice (CO2)
51
Structure and bonding of
fullerenes
Fullerenes are molecules composed entirely
of carbon atoms, in the form of hollow
spheres or hollow tubes.
52
Buckminsterfullerene (or buckyball)
The first fullerene discovered was
buckminsterfullerene (C60).
Buckminsterfullerene.
53
A soccer ball.
R.F. Curl
H.W. Kroto
R.E. Smalley
Discovered C60 in 1985
Awarded Nobel prize for Chemistry in 1996
54
Buckminsterfullerene
C60
Cutting at
12 vertices
icosahedron
正二十面體
55
truncated
icosahedron
Buckminsterfullerene
12 pentagons by cutting at 12 vertices.
56
Buckminsterfullerene
20 hexagons by cutting 20 triangular faces.
57
Named after the architect
Richard Buckminster Fuller
A geodesic dome
58
Each carbon atom is connected to three other
carbon atoms by one double covalent bond and
two single covalent bonds.
Buckminsterfullerene
59
Each pentagon is connected to five hexagons
Each hexagon is connected to three pentagons
and three hexagons alternately.
60
Graphite is planar because it is made of
hexagonal rings linked together.
61
In C60, pentagonal rings prevent the sheet from
being planar, making it spherical.
62
0.1434 nm
0.1386 nm
Why are there two types of bond in C60 ?
63
The surface of the sphere is NOT planar
 2pz orbitals are NOT parallel to one another
 Delocalization of  es is NOT favourable
64
Family of fullerenes
C28
C32
C50
Some of the more stable members of the
fullerene family. (a) C28 (b) C32 (c) C50 (d) C70
65
C70
Molecular structure
C60 molecules held by
dispersion forces
66
1. Melting point
Fullerene molecules are held together
by weak van der Waals’ forces.
Substance
Melting point (°C)
Graphite
Diamond
3730
3550
Buckminsterfullerene
1070
67
2. Solubility
Graphite
Diamond
Fullerenes
Molecular
structure
68
Giant covalent
structure
insoluble
in all liquid
solvents
dissolves
in benzene
3. Strength and hardness
Buckminsterfullerenes are relatively
strong and hard compared with most
other molecular solids.
The C60 molecules are packed
closely together in solid state.
buckminsterfullerene
molecule (C60)
69
4. Electrical conductivity
Pure buckminsterfullerene (C60) is an
electrical insulator.(no delocalized electrons)
buckminsterfullerene
potassium atom
70
The buckminsterfullerene with
potassium atoms filling the
spaces between its molecules
is a superconductor. Its
formula is K3C60.
Carbon nanotube (CNT) or buckytube
First discovered by Dr. Sumio Iijima in 1991
71
Carbon nanotube (CNT) or buckytube
It is formed by carbon atoms arranged in
a long cylindrical hollow tube.
72
Carbon nanotube (CNT) or buckytube
The diameter of a nanotube is in the
order of a few nanometres (109 m).
73
Carbon nanotube (CNT) or buckytube
74
Graphite sheet
75
Carbon nanotube (CNT) or buckytube
76
Properties of nanotubes
The tensile strength of carbon nanotubes is
exceptionally high due to the strong
covalent bonds holding the atoms together
The strongest materials on earth.
~100 times stronger than steel
Applications : clothes, sports equipments,
space elevators…
77
Properties of nanotubes
Carbon nanotube is an electrical
conductor because of the movement of
delocalized electrons along the graphite
sheets.
Depending on their structures, carbon
nanotubes can be semi-conducting or as
electrically conductive as metals.
78
11.5
79
Hydrogen
Bonding
Evidence of hydrogen bonding
Look at the boiling points of some simple
hydrides of Group IV to VII elements
(p.87).
80
B.p.  as molecular size 
Group 4 hydrides are non-polar, only dispersion forces exist
Dispersion forces  as molecular size .
81
All are polar
B.p.  as molecular size  (dispersion > dipole-dipole)
However, H2O, HF and NH3 have abnormally high b.p.
There exist unusually strong dipole-dipole forces (H-bond)
82
Formation of hydrogen bonding
When a hydrogen atom is directly
bonded to a highly electronegative atom
(e.g. fluorine, oxygen and nitrogen), a
highly polar bond is formed.
2.1
83
4.0
2.1
3.5
2.1
3.0
Electrostatic attractions exist between
this partial positive charge and the
lone pair electrons on a highly
electronegative atom (i.e. fluorine,
oxygen or nitrogen) of another molecule.
These attractions are called hydrogen
bonds
84
hydrogen bond
85
hydrogen bond
Formation of hydrogen bonds between H2O
molecules.
86
Reasons for abnormal strength of H-bond
1. the polarity of H–X bond is great when X is F ,
O , or N.
2. H atom does not have inner electrons.
 its nucleus (proton) is partially exposed due
to unequal sharing of electron.
 The partial positive charge on H is so
concentrated that it can come very close
to the lone pair of a small & highly
electronegative atom (F, O or N)
 Abnormally strong dipole-dipole forces
87
Two essential requirements for the
formation of a hydrogen bond:
• One molecule must contain at least one
H atom attached to a highly
electronegative atom (i.e. F, O or N).
• The other molecule must contain an F,
O or N atom that provides the lone
pair of electrons.
88
Identify the hydrogen atoms of the following
species that are capable of forming hydrogen
bonding with water molecules.
Soluble in water
adenine
89
glucose
An exceptional case : 
Cl
 Cl
C
Cl
+
H
H-bond
CH3
O
C
CH3

Due to the combined effect of the three
electronegative Cl atoms, the H atom becomes
sufficiently positive to form hydrogen bond
90
Relative strength of van der Waals’ forces,
hydrogen bond and covalent bond
Phenomenon
91
Energy
absorbed Forces overcome
(kJ mol-1)
He(s)  He(g)
0.11
Van der Waals’
forces
H2O(s)  H2O(g)
46.90
Hydrogen bonds
O2(g)  2O(g)
494.00
Covalent bonds
Q.65
Tendency of H-bond formation : EN
0.4
0.4
0.9
0.9
1.4
1.9
C – H < S – H < Cl – H < N – H < O – H < F – H
No lone
pair on C
N is smaller
than Cl
H can come
closer
92
Q.66
Substance
Relative
molecular mass
Boiling point
(°C)
NH3
17
-33.3
HF
20
19.5
H2O
18
100
HF > NH3 because
H – F bond is more polar than N – H bond
93
Q.66
Substance
Relative
molecular mass
Boiling point
(°C)
NH3
17
-33.3
HF
20
19.5
H2O
18
100
H2O > HF because
H2O can form H-bonds more extensively,
regardless of the fact that H-F bond is more
polar than H-O bond.
94
hydrogen bond
Each NH3 molecule has only ONE lone pair.

On the average, each NH3 molecule can form
only ONE hydrogen bond
95
Each HF molecule has only ONE hydrogen atom.

On the average, each HF molecule can form
only ONE hydrogen bond
96
hydrogen bond
Each H2O molecule has TWO hydrogen atoms and
TWO lone pairs.

97
On the average, each H2O molecule can form
TWO hydrogen bonds
Structure and bonding of ice
The lone pairs of
oxygen atom of each
water molecule forms
hydrogen bonds with
two hydrogen atoms of
nearby water molecules
a water
molecule
hydrogen
bond
hydrogen
bond
98
hydrogen atom
oxygen atom
The two hydrogen atoms of
each water molecule also
form hydrogen bonds with
the lone pairs of oxygen
atoms of nearby water
molecules.
hydrogen
bond
hydrogen
bond
99
hydrogen atom
oxygen atom
Each H2O molecule is bonded
tetrahedrally to four H2O molecules
1
4
2
3
100
In solid ice, the tetrahedral arrangement repeats
over and over again, resulting in an open and regular
network structure of water molecules.
Open : the maximum number
of hydrogen bonds can be
formed
Regular : all molecules are
held in positions by strong
hydrogen bonds
101
The oxygen atoms in
the structure of ice
are arranged in a
hexagonal shape.
102
The hexagonal symmetry
of a snowflake reflects
the structure of ice.
103
In liquid state, water molecules pack together more
closely and randomly.
Hydrogen bonds are
continuously formed and broken.
Liquid water takes the shapes
of the containers
104
Properties of ice
1. Density
Most substances have higher densities in
the solid state than in the liquid state.
Solid paraffin
is denser than
liquid paraffin.
liquid
paraffin
solid
paraffin
105
ice
water
Ice has a lower density than liquid water!
At 0°C, density of ice = 0.92 g cm−3
density of liquid water = 1.00 g cm−3
liquid
paraffin
solid
paraffin
106
ice
water
In cold weather, ice forms a layer on
the top of a pond.
Ice acts as an
insulator for the
water beneath.
Ice
107
This allows fish
and other aquatic
organisms to survive.
Explanation
In ice, water molecules are arranged in an
orderly manner in an open network structure
because of extensive formation of hydrogen
bonding.
108
In this open structure, water molecules are
further apart than they are in liquid water.
liquid water
ice
melts
open structure
collapses
water molecules tend to
pack more closely together
109
Energy is absorbed to break some of the
hydrogen bonds
Less H-bonds
 less stable
 Close & random
More H-bonds
 More stable
 Open & regular
110
Effect of hydrogen bonding on
properties of water
1. Melting point and boiling point
The melting point (0°C) and boiling point
(100°C) of water are much higher than
expected.
A lot of energy is required to overcome
the hydrogen bonds between water
molecules and separate them.
111
2. Surface tension
High surface tension of water allows water
striders to ‘walk’ on it.
112
2. Surface tension
Surface tension of molecular liquids arises
from intermolecular forces.
Stronger intermolecular forces leads to
higher surface tension
Hexane
Relative surface
tension at 25C
18.4
Methanol
22.6
Ethanol
Water
22.8
72.3
Liquid
113
2. Surface tension
Water molecules at the surface
are strongly attracted by
neighboring molecules on the
same surface.
The surface of water is like a
tightly-stretched skin such that
small insects can walk on it.
114
intermolecular
forces
2. Surface tension
Water forms droplets rather than
spreading out on leaf.
115
2. Surface tension
In a sample of water,
each water molecule is
attracted to neighboring
water molecules in all
directions and there is
a balance of force.
116
2. Surface tension
There is an imbalance of force for
the molecules at the surface.
The water molecules at the
surface tend to be pulled inwards
by other water molecules below
the surface.
As a result, water forms droplets
rather than spreading out on leaf.
In other words, water tends to
reduce its surface area by taking
the spherical shape.
117
The high surface tension
of water allows water to
be transported to the
top of trees by capillary
action.
The tallest tree on earth
115.56m
118
3. Viscosity
Viscosity
The resistance of
a liquid to flow.
The higher the viscosity of a liquid, the
more slowly it flows.
Viscosity arises from intermolecular
forces
119
Strong hydrogen bonds hold water
molecules together and do not allow them
to move past one another easily.
Liquid
Relative viscosity
Benzene
1
Water
15
Water has high melting and boiling points,
high surface tension and high viscosity.
Surface tension
120
Effect of hydrogen bonding on
properties of alcohols
Consider an ethanol molecule.
hydroxyl group
lone pairs of electrons
121
1. Boiling point
Ethanol molecules are held together by H-bonds.
 high boiling point
hydrogen bond
H-bond strength
122
Alcohols vs Thiols (p.90)
Alcohol CH3OH C2H5OH C3H7OH C4H9OH
b.p.(C)
Thiol
b.p.(C)
64.5
78
97
CH3SH C2H5SH C3H7SH C4H9SH
5.8
37
67
Dispersion forces : Thiol > alcohol
Boiling point : Alcohol > thiol
123
117
97
2. Solubility in water
hydrogen bonds
water
ethanol
Ethanol and water are completely miscible
124
3. Viscosity
Ethanol is viscous because of the
presence of extensive intermolecular
hydrogen bonds.
Ethanol is viscous, completely miscible
with water, and has a high boiling point.
125
propan-1-ol
propane-1,2-diol
propane-1,2,3-triol
Viscosity  as
no. of OH groups per molecule 
Viscosity
126
Explain the following.
Water is easily absorbed by tissue paper
rather than forming droplets on it.
Tissue paper is composed of cellulose which is a
natural polymer made of glucose molecules.
Thus, tissue paper can form extensive hydrogen
bonds with water molecules.
127
Carboxylic Acids
H
O
H3C
H-bonds
C
RMM = 60
O
O
H
C
CH 3
O
Ethanoic acid exists as dimers, (CH3COOH)2,
in vapour phase or in non-polar solvents
RMM = 260 = 120
128
Q.67
Ethanoic acid molecules form H-bonds with polar
solvent molecules rather than with other ethanoic
molecules.
H
H
H
O
O
H
C
O
H3C
H
O
129
Q.67
R
Ethanoic acid molecules form H-bonds with polar
solvent molecules rather than with other ethanoic
molecules.
C
H
O
H
O
H3C
O
H
O
R
130
Intramolecular Hydrogen Bonding
H
H
O
O
O
N
O
b.p. = 214C
N
O
O
b.p. = 279C
131
Formation of intramolecular hydrogen bonds
prevents the formation of intermolecular
hydrogen bonds  lower boiling point
H
H
O
O
O
N
O
b.p. = 214C
N
O
O
b.p. = 279C
132
Roles of Hydrogen Bonding in Biochemical Systems
Proteins : polymers of amino acids
Primary structure : sequence of amino acids
Peptide
linkage
133
Secondary structures : 1. -pleated sheet
2. -helix
134
1. -pleated sheet
Intermolecular H-bonds
formed between peptide
linkages of adjacent
protein chains
135
C
O
N
C
C
H
O
N
C
C
H
C
planar
Both N and C of the N – C bond are sp2 hybridized
to facilitate delocalization of  electrons
136
C
O
N
C
C
H
O
N
C
H
C
C
planar
N – C bond has double bond character
 free rotation w.r.t. the bond axis is restricted
 N – H and C = O groups are held in opposite
positions to facilitate the formation of
inter-chain H-bonds
137
138
2. -helical structure
C
Intramolecular
hydrogen bond
O
H
N
139
2. -helical structure
The molecular chains of
protein can be held in
position to give the
-helical structure by
forming intramolecular
hydrogen bonds.
140
Both - and - structures were first
suggested by Linus Pauling.
141
Tertiary structure
3-D arrangements of secondary structures
Myoglobin
142
Quaternary structure
3-D arrangements of tertiary structures
Haemoglobin
143
Hydrogen bonding in DNA
DNA (DeoxyribonNuclei Acid) carries genetic
information
hydrogen
bonds
144
Effect of hydrogen bonding on DNA
The presence of intermolecular H-bonds helps
maintain the double helical shape of DNA molecules.
hydrogen
bonds
145
Effect of hydrogen bonding on DNA
The double helical
structure is maintained by
intermolecular hydrogen
bonds formed between
specific base pairs
146
Effect of hydrogen bonding on DNA
Cytosine
147
Guanine
Effect of hydrogen bonding on DNA
Thymine
148
Adenine
Effect of hydrogen bonding on DNA
Sequence of bases = genetic code
CAGACTTGCAAT… Or
GTCTGAACGTTA…
149
Without hydrogen bond, life becomes impossible
hydrogen bonds
150
Without hydrogen bond, life becomes impossible
151
Without hydrogen bond, life becomes impossible
+
H
O
O
O

H
+
It allows oxygen to dissolve in water
152
Change of states and intermolecular
forces
• 3 different states: solid, liquid and gas
• Change of states involves breaking or
forming of intermolecular forces of the
molecular substances
153
Phase Diagram
A phase diagram is a graph summarizing the
conditions of pressure and temperature
under which the different phases of a
substance are stable.
Phase  state
E.g. C in the same state may have different phases
Graphite, diamond, C60
154
A phase is any homogeneous and physically
distinct part of a system which is
separated from other parts of the system
by a definite physical boundary known as
the phase boundary.
155
A system having two phases in the same
liquid state
oil
Phase boundary
water
156
A system having three phases in two states
oil
Phase boundaries
water
157
glass
A system having four phases in three states
air
oil
Phase boundaries
water
158
glass
Phase Diagram of Carbon Dioxide
Three regions in each of which only one phase is stable
P / atm
Solid
Liquid
Vapour
T / C
159
Phase Diagram of Carbon Dioxide
The three regions meet at three lines, along which
two phases coexist in equilibrium.
P / atm
Solid
Liquid
Vapour
T / C
160
Phase Diagram of Carbon Dioxide
AT is the sublimation curve
P / atm
Liquid
Solid
T
Vapour
A
161
T / C
Phase Diagram of Carbon Dioxide
CO2(s)
sublimation
CO2(g)
P / atm
Liquid
Solid
T
Vapour
A
162
T / C
Phase Diagram of Carbon Dioxide
AT shows the variation of sublimation temperature of
carbon dioxide with external pressure
P / atm
Liquid
Solid
T
Vapour
A
163
T / C
Phase Diagram of Carbon Dioxide
TB is the melting curve
B
P / atm
positive(most common)
slope
Liquid
Solid
T
Vapour
A
164
T / C
Phase Diagram of Carbon Dioxide
melting
CO2(s)
CO2(l)
B
P / atm
Liquid
Solid
T
Vapour
A
165
T / C
Phase Diagram of Carbon Dioxide
TB shows the variation of melting temperature of
carbon dioxide with external pressure
B
P / atm
Liquid
Solid
T
Vapour
A
166
T / C
Phase Diagram of Carbon Dioxide
TC is the boiling curve
B
P / atm
C
Liquid
Solid
T
Vapour
A
167
T / C
Phase Diagram of Carbon Dioxide
boiling
CO2(l)
CO2(g)
B
P / atm
C
Liquid
Solid
T
Vapour
A
168
T / C
Phase Diagram of Carbon Dioxide
TC shows the variation of boiling temperature of
carbon dioxide with external pressure
B
P / atm
C
Liquid
Solid
T
Vapour
A
169
T / C
Q.62
(a) Condensation by  T
P / atm
Liquid
Solid
T
Vapour
A
170
T / C
Q.62
(a) Condensation by  P
P / atm
Liquid
Solid
T
Vapour
A
171
T / C
Q.62
(b) Boiling by  T
P / atm
Liquid
Solid
T
Vapour
A
172
T / C
Q.62
(b) Boiling by  P
P / atm
Liquid
Solid
T
Vapour
A
173
T / C
Q.62
(c) Freezing by  T
P / atm
Liquid
Solid
T
Vapour
A
174
T / C
Q.62
(c) Freezing by  P
P / atm
Liquid
Solid
T
Vapour
A
175
T / C
Q.62
(d) Melting by  T
P / atm
Liquid
Solid
T
Vapour
A
176
T / C
Q.62
(d) Melting by  P
P / atm
Liquid
Solid
T
Vapour
A
177
T / C
Q.62
(e) Sublimation by  T
P / atm
Liquid
Solid
T
Vapour
A
178
T / C
Q.62
(e) Sublimation by  P
P / atm
Liquid
Solid
T
Vapour
A
179
T / C
T is the triple point where all three phases
coexist in equilibrium.
B
P / atm
C
Liquid
Solid
5.1
atm
T
Vapour
A
56.4C
180
T / C
Vapour
Triple
point
Solid
Liquid
vapour pressure above solid = vapour pressure above liquid
181
Dry ice sublimes when heated at 1atm
B
P / atm
C
Liquid
Solid
5.1
atm
1 atm
T
Vapour
A
56.4C
182
T / C
Dry ice is so called because it never melts (goes wet)
at normal pressure
B
P / atm
C
Liquid
Solid
5.1
atm
1 atm
T
Vapour
A
56.4C
183
T / C
At P > 5.1 atm, dry ice melts to give liquid CO2
when heated
B
P / atm
C
10 atm
Liquid
Solid
5.1
atm
T
Vapour
A
56.4C
184
Triple point
video
T / C
TC curve terminates at C beyond which the
boundary between liquid and vapour disappears
B
P / atm
C
Pc= 73atm
Critical point
Liquid
Solid
T
Vapour
A
Tc= 31C
185
T / C
Above Tc, the vapour cannot be condensed no
matter how high the external pressure is
B
P / atm
C
Pc= 73atm
Liquid
Solid
T
Gas
Vapour
A
Tc= 31C
186
T / C
As dense as a liquid
As mobile as a gas
Supercritical
fluid
B
P / atm
C
Pc= 73atm
Liquid
Solid
Gas
T
Vapour
A
Tc= 31C
187
T / C
Decaffeination using
supercritical CO2
Supercritical
fluid
B
P / atm
C
Pc= 73atm
Liquid
Solid
Gas
T
Vapour
A
Tc= 31C
188
T / C
Q.63
In winter, T < Tc
Supercritical
fluid
CO2 is in liquid phase
B
P / atm
C
Pc= 73atm
Liquid
Solid
Gas
T
Vapour
A
Tc= 31C
189
T / C
Q.63
In summer, T > Tc
Supercritical
fluid
CO2 is in gas phase
B
P / atm
C
Pc= 73atm
Liquid
Solid
Gas
T
Vapour
A
Tc= 31C
190
T / C
H>0
Q.64 As P , CO2(l)  CO2(g)
So T  and CO2(g)  CO2(s)
B
P / atm
Supercritical
fluid
C
Pc= 73atm
Liquid
Solid
Gas
T
1 atm
Vapour
A
Tc= 31C
191
T / C
Phase Diagram of Water
P / atm
B
Negative
slope
(very rare)
Supercritical
fluid
C
Liquid
Solid
Gas
T
A
Vapour
m.p.  when external P 
192
T / C
Phase Diagram of Water
P / atm
H2O(s)
Supercritical
fluid
B
P 
C
H2O(l)
Liquid
Solid
Gas
T
A
Vapour
T / C
193
Phase Diagram of Water
P / atm
H2O(s)
Supercritical
fluid
B
P 
C
H2O(l)
Liquid
Solid
Gas
T
A
Vapour
T / C
194
H2O(s)
P 
H2O(l)
Ice melts below 0C when an extremely high
pressure is applied to it.
This results in a decrease in friction between
contact surfaces and makes possible
ice-skating and
the movement of tremendously massive glaciers.
195
Phase Diagram of Water
P / atm
Supercritical
fluid
B
C
Liquid
1 atm
Solid
0.006
atm
Gas
T
A
Vapour
0C 0.01C
196
100C
T / C
Phase Diagram of Water
P / atm
Supercritical
fluid
B
C
Pc=217.2atm
Liquid
Solid
Gas
T
A
Vapour
Tc=374C
197
T / C
Critical temperature and the strength of
intermolecular forces
Gas
He
H2
Ne
N2
O2
CO2 NH3 H2O
Tc(K) 4.2 33.3 44.5 126
154
304 405 647
non-polar
Higher Tc
 stronger intermolecular forces
Polar
with H
bond
 greater deviation from ideal gas behaviour
198
Q.68
Cu(NH3)4SO4NH3 does not exist.
Reason : Unlike H2O, each NH3 has one lone pair and
three H atoms.
Thus, NH3 cannot form hydrogen bonds in the
same way as H2O in the crystal lattice.
199
The END
200
11.2 Van der Waals’ forces (SB p.280)
Back
How is the enthalpy of vaporization related to
intermolecular forces of a simple molecular substance
like neon?
Answer
The enthalpy of vaporization of a substance is the energy needed
to vaporize one mole of the substance at its boiling point. Consider
a substance like neon, which consists of single atoms, Neon
liquefies when the temperature is lowered to –246 oC at 1 atm.
The enthalpy of vaporization of the liquid at this temperature is
1.77 kJ mol-1. Some of this energy is needed to push back the
atmosphere when the vapour forms. The remaining energy must
be supplied to overcome the intermolecular attractions. Because
each molecule in a liquid is surrounded by several neighbouring
molecules, this remaining energy is some multiple of a single
molecule-molecule interaction. Typically, this multiple is about 5.
201
11.2 Van der Waals’ forces (SB p.280)
(a) Comment on the relative strength of van der Waals’ forces
in solid, liquid and gaseous bromine.
Answer
(a) The relative strength of van der Waals’ forces decreases in the
order:
Solid bromine > liquid bromine > gaseous bromine
The van der Waals’ forces are highly dependent on the distance
between adjacent molecules. It decreases exponentially with the
separation between the molecules. Going from solid to liquid and
then to gaseous state, the separation between molecules
increases, so the van der Waals’ forces become weaker and
weaker.
202
11.2 Van der Waals’ forces (SB p.280)
(b) Plastics are substances which have very strong van der
Waals’ forces. Explain why the van der Waals’ forces are
so strong in plastics.
Answer
(b) A large size of a molecule of plastics indicates that it has a large
electron cloud which is more easily polarized. Therefore, the
molecule of plastics is more likely induced to form an
instantaneous dipole. Moreover, the molecule of plastics has an
extensive surface area. These make plastics have very strong
van der Waals’ forces between the molecules.
203
11.2 Van der Waals’ forces (SB p.280)
Back
(c) Arrange the following substances in an increasing order
of boiling point:
Answer
(i) N2, O2, Cl2, Ne
(ii) H2, Br2, He
(c) (i) Ne < N2 < O2 < Cl2
(ii) He < H2 < Br2
204
11.3 Van der Waals’ radii (SB p.284)
Back
What is the consequence of two molecules approaching each
other at a distance less than the sum of their van der Waals’
radii?
Answer
The electron clouds of the two molecules will repel
each other, and the distance between the two
molecules will increase until the repulsion is just
balanced by the attraction.
205
11.5 Hydrogen bonding (SB p.291)
The relative molecular masses and boiling points of five
compounds are given below:
Compound
Relative
Boiling point
molecular mass
(oC)
Ammonia (NH3)
17
-33.4
Ethanol (C2H5OH)
46
78
Hydrogen fluoride (HF)
20
19.5
Methanol (CH3OH)
32
66
Water (H2O)
18
100
206
11.5 Hydrogen bonding (SB p.291)
(a) Ammonia, hydrogen fluoride and water have similar
relative molecular masses, yet their boiling points are
different. Explain why.
Answer
(a) H2O can form 2 hydrogen bonds per molecule while NH3
and HF can only form 1 hydrogen bond per molecule.
Thus, the boiling point of water is higher than those of
NH3 and HF. Besides, as F is more electronegative than N,
the intermolecular hydrogen bond formed between HF
molecules is stronger than that between NH3 molecules.
207
11.5 Hydrogen bonding (SB p.291)
Back
(b) Ethanol and methanol have similar structures, yet their
boiling points are different. Explain why.
Answer
(b) For molecules with similar structures, their boiling points
depend on their relative molecular masses. As the relative
molecular mass of ethanol is greater than that of
methanol, the boiling point of ethanol is higher.
208
11.5 Hydrogen bonding (SB p.293)
Why it takes much longer time to boil an egg on a
mountain peak?
Answer
The boiling point of water decreases with decreasing pressure.
Although water boils easily at mountain peak, the cooking of an
egg takes longer time. It is because the amount of heat delivered
to the egg is proportional to the temperature of water.
Back
209
11.5 Hydrogen bonding (SB p.296)
(a) The formation of a hydrogen bond between two
molecules RAH and R’B may be represented as:
R  A  H · · · · · · · B  R’
(i) Suggest possible elements for A and B. What are their
common features?
(ii)In which of the following ranges would you expect
the strength of hydrogen bonds to lie?
0.1 – 10 kJ mol-1
10 – 50 kJ mol-1
100 – 400 kJ mol-1
210
Answer
11.5 Hydrogen bonding (SB p.296)
(a) (i) A and B can be nitrogen, oxygen or fluorine. All of them are
highly electronegative atoms, thus they form highly polar
molecules, resulting in the formation of hydrogen bonds.
(ii) 10 – 50 kJ mol-1
211
11.5 Hydrogen bonding (SB p.296)
(b) Benzoic acid has an apparent relative molecular mass of
244 in hexane, but only 122 in aqueous solution. With the
aid of diagrams, explain this phenomenon.
Answer
212
11.5 Hydrogen bonding (SB p.296)
(b) The relative molecular mass of benzoic acid (C6H5COOH) is 122.
In hexane, benzoic acid molecules form dimers with hydrogen
bondings between the molecules.
However, in water, the benzoic acid molecules form hydrogen
bonds with the water molecules.
213
11.5 Hydrogen bonding (SB p.296)
(c) Cyclohexane (C6H12) is insoluble in water whereas glucose
(C6H12O6) is miscible with water in all proportions.
Answer
214
11.5 Hydrogen bonding (SB p.296)
Back
(c) Cyclohexane is non-polar, and there are only weak van der Waals’
forces holding the molecules together. Thus, cyclohexane
molecules do not form hydrogen bonds with water. On the other
hand, glucose can form hydrogen bonds with water molecules via
its OH groups. Therefore, glucose is soluble in water but
cyclohexane is not.
Cyclohexane
215
Glucose
11.5 Hydrogen bonding (SB p.297)
Name the types of bonding or intermolecular forces that are
broken and formed in the following processes.
•
H2O(s)  H2O(g)
•
2Mg(s) + O2(g)  2MgO(s)
•
H2(g) + F2(g)  2HF(g)
•
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
•
CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
216
Answer
11.5 Hydrogen bonding (SB p.297)
Back
(a) Bond broken: hydrogen bond
(b) Bonds broken: metallic bond and covalent bond
Bond formed: ionic bond
(c) Bond broken: covalent bond
Bonds formed: covalent bond and hydrogen bond
(d) Bonds broken: covalent bond, metallic bond and hydrogen bond
Bonds formed: ionic bond and covalent bond
(e) Bonds broken: covalent bond and hydrogen bond
Bonds formed: covalent bond and hydrogen bond
217