Liquids, Solids, and
Phase Changes
Chapter 10
1
Gases, Liquids and Solids
• Gases have little or no interactions.
• Liquids and solids have significant
interactions.
2
Gases, Liquids and Solids
• Liquids and solids have well-defined
volume.
• Liquid molecules “flow,” while solids are
held “rigid.”
3
Intermolecular Forces
Intermolecular forces are attractive forces between molecules.
Intramolecular forces hold atoms together in a molecule.
Intermolecular vs Intramolecular
•
41 kJ to vaporize 1 mole of water (inter)
•
930 kJ to break all O-H bonds in 1 mole of water (intra)
“Measure” of intermolecular force
Generally,
intermolecular
forces are much
weaker than
intramolecular
forces.
boiling point
melting point
4
Intermolecular Forces
Dipole Moments
01
• Polar covalent bonds form between atoms of
different electronegativity. This is described
as a bond dipole.
5
Dipole Moments
02
• Dipole Moment (µ): The measure of net
molecular polarity or charge separation.
µ=Qr
r = distance between charges
+ = Q, – = –Q
6
Dipole Moments: µ = Q  r
• Q = Charge of electron: Q = 1.60 x 10-19 C,
• r = bond length, m
• μ , dipole moments are expressed in debyes (D) where
1 D = 3.336 x 10–30 C·m
What is the dipole moment if one proton separated
from one electron by a distance of 100 PM
μ = Q x r = (1.60 x 10-19 C) (100 x 10-12 m) (1D/3.336 x 10-30
C . m) = 4.80 D
7
Dipole Moments
03
• Polarity can be illustrated with an
electrostatic potential map. These show
electron-rich groups as red and electron-poor
groups as blue-green.
8
Dipole Moments
04
9
Intermolecular Forces
01
• Attractive forces between molecules and ions.
• Several types of forces:
–Dipole–dipole
–Instantaneous induced dipole (dispersion forces)
–Ion–dipole
–Hydrogen “bonds.”
11
Intermolecular Forces
Dipole-Dipole Forces
Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
12
Intermolecular Forces
Ion-Dipole Forces
Attractive forces between an ion and a polar molecule
Ion-Dipole Interaction
13
Intermolecular Forces
04
• London Dispersion Forces: Attraction is
due to instantaneous, temporary dipoles
formed due to electron motions.
14
Intermolecular Forces
Dispersion Forces Continued
Polarizability is the ease with which the electron distribution
in the atom or molecule can be distorted.
Polarizability increases with:
•
greater number of electrons
•
more diffuse electron cloud
Dispersion
forces usually
increase with
molar mass.
XXXXX
2
XXXX
15
What type(s) of intermolecular forces exist between
each of the following molecules?
HBr
HBr is a polar molecule: dipole-dipole forces. There are
also dispersion forces between HBr molecules.
CH4
CH4 is nonpolar: dispersion forces.
S
SO2
SO2 is a polar molecule: dipole-dipole forces. There are
also dispersion forces between SO2 molecules.
16
Hydrogen Bond
• Hydrogen Bond: Molecules containing N–
H, O–H, or F–H groups, and an
electronegative O, N, or F.
17
Why is the hydrogen bond considered a
“special” dipole-dipole interaction?
Decreasing molar mass
Decreasing boiling point
18
Which of the following molecules can
hydrogen bond with itself?
•
•
•
•
•
1, 2
2, 3
3, 4
1, 2, 3
1, 2, 3, 4
O
CH2F2
NH3
CH3-O-H
H3C C CH3
1
2
3
4
19
Which of the following molecules can
hydrogen bond with itself?
•
•
•
•
•
1, 2
2, 3
3, 4
1, 2, 3
1, 2, 3, 4
O
CH2F2
NH3
CH3-O-H
H3C C CH3
1
2
3
4
20
Of the following substances, predict
which has the lowest boiling point
based on London dispersion forces.
1.
2.
3.
4.
5.
He
Ne
Ar
Kr
Xe
21
Correct Answer:
1.
2.
3.
4.
5.
He
Ne
Ar
Kr
Xe
More massive species have more polarizability
and stronger London dispersion forces;
consequently, amongst the noble gases He has
the lowest boiling point.
22
Of the following substances, predict
which has the highest boiling point
based upon intermolecular forces?
1.
2.
3.
4.
5.
CH4
H2O
H2S
SiH4
H2Se
NH ……. O=C
23
Correct Answer:
1.
2.
3.
4.
5.
CH4
H2O
H2S
SiH4
H2Se
Of these, only H2O has any
hydrogen bonding.
Hydrogen bonding
substantially increases the
intermolecular forces, and
hence the boiling point.
24
© 2003 John Wiley and Sons Publishers
Courtesy Ken Karp
25
With nerves as steady as a chemical bond.
© 2003 John Wiley and Sons Publishers
Figure 13.1: “Floating” a tack on water.
26
© 2003 John Wiley and Sons Publishers
Courtesy Ken Karp
Figure 13.2: Place a tack on the surface of a glass of water.
27
Surface Tension
Water strider walks on a pond without penetrating the surface
28
Surface Tension
29
Properties of Liquids
Surface tension is the amount of energy required to stretch
or increase the surface of a liquid by a unit area.
Strong
intermolecular
forces
High
surface
tension
30
Viscosity
Viscosity is the measure of a liquid’s resistance to flow
and is related to the ease with which molecules move
around, and thus to the intermolecular forces.
31
Surface Tension
Tensiometer
32
Intermolecular Forces
09
• Viscosity is the measure of a liquid’s
resistance to flow and is related to the ease
with which molecules move around, and thus
to the intermolecular forces.
Another unit of viscosity is kg/m.s
33
Phase Changes 01
34
Phase Changes
Phase Change (State Change): A change in physical
form but not the chemical identity of a substance.
Fusion (melting): solid to liquid
Freezing: liquid to solid
Vaporization: liquid to gas
Condensation: gas to liquid
Sublimation: solid to gas
Deposition: gas to solid
Phase Changes 02
(∆Hfus)
kJ/mol
(∆Hvap )
36
(∆Hfus) = 6.01 KJ/mol
(∆Hvap ) = 40.67 KJ/mol
Phase Changes
• Vapor Pressure: The pressure exerted by
gaseous molecules above a liquid.
37
Molar heat of vaporization (DHvap) is the energy required to
vaporize 1 mole of a liquid.
Clausius-Clapeyron Equation
DHvap
ln P = +C
RT
P = (equilibrium) vapor pressure
T = temperature (K)
R = gas constant (8.314 J/K•mol)
38
Vapor Pressure
• The boiling point of a
liquid is the temperature
at which its vapor
pressure equals
atmospheric pressure.
• The normal boiling
point is the temperature
at which its vapor
pressure is 760 torr.
39
Evaporation, Vapor Pressure, and
Boiling Point
ln Pvap = y
=
DHvap
1
R
T
m
x + b
+C
Evaporation, Vapor Pressure, and
Boiling Point
Vapor Pressure
42
Clausius-Clapeyron Equation
ln P = -
DHvap
RT
+C
• By taking measurements at two temps, we
get:
P1 DHvap  1 1 
ln 
  
P2
R  T2 T1 
43
The normal boiling point of benzene is 80.1 °C, and
ΔHvap = 30.8 kJ/mol, what is boiling point of benzene on top
of Mount Everest, where P = 260 mm Hg
• P1 = 760 mm Hg; P2 = 260 mm Hg; t1 = 80.1oC, T2 = ?
• ΔHvap = 30.8 kJ/mol
P1 DHvap  1 1 

• ln
  
P2
R  T2 T1 
, R = 8.3145 J / K . mol
• Solve for T2 (the boiling point for benzene at 260 mm Hg).
T2 = 320 K ; t = 47oC
(boiling point is lower at lower pressure)
44
•
•
•
•
•
Boiling point ~120°C
Boiling point ~95°C
Boiling point ~75°C
Melting point ~95°C
Melting point ~75°C
Vapor Pressure (mm Hg)
Which
statement is
true?
800
600
400
200
0
0
25
50
75
100
Temperature (°C)
45
•
•
•
•
•
Boiling point ~120°C
Boiling point ~95°C
Boiling point ~75°C
Melting point ~95°C
Melting point ~75°C
Vapor Pressure (mm Hg)
Which
statement is
true?
800
600
400
200
0
0
25
50
75
100
Temperature (°C)
46
Solids
An amorphous solid does not possess a well-defined
arrangement and long-range molecular order they do not have
a fixed sharp melting point. .
A glass is an optically transparent fusion product of inorganic
materials that has cooled to a rigid state without crystallizing
•Structure of a crystalline solid is based on the unit cell, a basic
• repeating structural unit.
Crystalline
quartz (SiO2)
Non-crystalline
quartz glass
47
Kinds of Solids
Amorphous Solids: Particles are randomly arranged and
have no ordered long-range structure. Example - rubber.
Crystalline Solids: Particles have an ordered
arrangement extending over a long range.
• ionic solids
• molecular solids
• covalent network solids
• metallic solids
Kinds of Solids
Ionic Solids: Particles are ions ordered in a regular
three-dimensional arrangement and held together
by ionic bonds. Example - sodium chloride.
lattice
point
At lattice points:
Unit Cell
Unit cells in 3 dimensions
•
Atoms
•
Molecules
•
Ions
50
Simple Cubic
Packing
Body-Centered Cubic
Packing
Crystal Structure
Simple Cube
Body-Centered Cube:
52
Crystal Structure
04
Face-Centered Cube:
53
Crystalline Solids
We can determine the
empirical formula of
an ionic solid by
determining how
many ions of each
element fall within the
unit cell.
54
Hexagonal Closest Pack
A-B-A-B-
Space used 74%
Cubic Closes Pack
• Space used 74%
• A-B-C-A-B-C
56
57
Unit Cells and the Packing of
Spheres in Crystalline Solids
Unit Cell: A small repeating unit that makes up a
crystal.
Types of Crystals
What are the empirical formulas for these compounds?
(a) Orange:chlorine; Gray:cesium
(b) Blue:sulfur;
Gray: zinc
(c) Green:fluorine,
Gray: calcium
(a)
(b)
(c)
59
CsCl
ZnS
CaF2
Types of Crystal
04
• Carbon:
60
Carbon Allotropes
Types of Crystals
Covalent Crystals
• Lattice points occupied by atoms
• Held together by covalent bonds
• Hard, high melting point
lattice points:
•
Atoms
•
Molecules
•
Ions
carbon
atoms
diamond
graphite
62
Types of Crystals
Molecular Crystals
• Lattice points occupied by molecules
• Held together by intermolecular forces
• Soft, low melting point
63
11.6
Types of Crystals
Metallic Crystals
• Lattice points occupied by metal atoms
• Held together by metallic bonds
• Soft to hard, low to high melting point
• Good conductors of heat and electricity
Cross Section of a Metallic Crystal
nucleus &
inner shell emobile “sea”
of e-
64
•
•
•
•
Ionic Crystals
Covalent Crystals
Molecular Crystals
Metallic Crystals
65
Simple cubic cell
Body-centered cubic
66
Face-centered cubic
When silver crystallizes, it forms face-centered cubic
cells. The unit cell edge length is 409 pm. Calculate
the density of silver.
d=
m
V
V = a3 = (409 pm)3 = 6.83 x 10-23 cm3
4 atoms/unit cell in a face-centered cubic cell
1 mole Ag
107.9 g
-22 g
x
m = 4 Ag atoms x
=
7.17
x
10
mole Ag 6.022 x 1023 atoms
7.17 x 10-22 g
m
3
=
=
10.5
g/cm
d=
V
6.83 x 10-23 cm3
67
X-Ray Crystallography
• Diffraction is the scattering of radiation by an object
containing regularly spaced lines, with a spacing
that is equivalent to the wavelength of radiation.
• Diffraction is due to interference between two
waves passing through the same region of space at
the same time.
68
69
Extra distance = BC + CD = 2d sinq = nl (Bragg Equation)
70
X rays of wavelength 0.154 nm are diffracted from a
crystal at an angle of 14.170. Assuming that n = 1,
what is the distance (in pm) between layers in the
crystal?
nl = 2d sin q
n=1
q = 14.170 l = 0.154 nm = 154 pm
nl
1 x 154 pm
=
= 315 pm
d=
2 x sin14.17
2sinq
71
1 mol
6.022 x 1023 atoms
Titanium metal has a density of 4.54 g/cm3, and an atomic
radius of 144.8 pm. What is the structure of cubic unit cell?
•
mass of one Ti atom = 7.951 x 10-23 g/atom (problem 10.78)
• r = 144.8 pm = 144.8 x 10-12 m
• r = 144.8 x 10-12 m = 1.448 x 10-8 cm
• Calculate the volume and then the density for Ti assuming
it is Simple (primitive) cubic, body-centered cubic, and
face-centered cubic. Compare the calculated density with
the actual density to identify the unit cell.
• For primitive cubic:
• a = 2r; volume = a3 = [2(1.448 x 10-8 cm)]3 = 2.429 x 10-23
cm3
• density = m/v = 3.273 g/cm3
72
• For face-centered cubic:
• a = √8. r; volume = a3 = [2√2 . (1.448 x 10-8
cm)]3 = 6.870 x 10-23 cm3
• density = = 4.630 g/cm3
• For body-centered cubic:
• a = 4r/√3; volume = a3 = 3.739 x 1023 cm3
• density = = 4.253 g/cm3
• The calculated density for a face-centered cube
(4.630 g/cm3) is closest to the actual density of
4.54 g/cm3. Ti crystallizes in the face-centered
cubic unit cell.
73
Phase Diagrams
01
74
Water
Phase Diagrams
Normal BP: Occurs at 1 atm.
Critical Point: A combination of temperature and pressure
beyond which a gas cannot be liquefied.
• Critical Temperature: The temperature beyond which
a gas cannot be liquefied regardless of the pressure.
• Critical Pressure: The pressure beyond which a liquid
cannot be vaporized regardless of the temperature.
Supercritical Fluid: A state of matter beyond the critical point
that is neither liquid nor gas.
Triple Point: A point at which three phases coexist in
equilibrium.
Supercritical CO2
Caffeine extraction from green coffee with supercritical CO2
Application of Supercritical CO2 in dry cleaning
77
Phase Diagrams
Carbon Dioxide
Phase Diagrams
05
•
Approximately, what is the normal boiling point
and what is the normal melting point of the
substance?
•
What is the
physical state
when:
i. T = 150 K, P = 0.5 atm
ii. T = 325 K, P = 0.9 atm
iii. T = 450 K, P = 265 atm
79
The End
80
Coordination Numbers
81
82