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Intermolecular Forces:
Liquids, Solids, and Phase Changes
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Intermolecular Forces:
Liquids, Solids, and Phase Changes
12.1 An Overview of Physical States and Phase Changes
12.2 Quantitative Aspects of Phase Changes
12.3 Types of Intermolecular Forces
12.4 Properties of the Liquid State
12.5 The Uniqueness of Water
12.6 The Solid State: Structure, Properties, and Bonding
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12.1 An Overview of Physical States and
Phase Changes
ATTRACTIVE FORCES
electrostatic in nature
Intramolecular forces
bonding forces
These forces exist within each molecule.
They influence the chemical properties of the substance.
Intermolecular forces
nonbonding forces
These forces exist between molecules.
They influence the physical properties of the substance.
Phase Changes
3
exothermic
sublimination
vaporizing
melting
solid
liquid
freezing
gas
condensing
endothermic
4
Table 12.1
A Macroscopic Comparison of Gases, Liquids, and Solids
State
Shape and Volume
Compressibility Ability to Flow
Gas
Conforms to shape and volume
of container
high
high
Liquid
Conforms to shape of container;
volume limited by surface
very low
moderate
Solid
Maintains its own shape and
volume
almost none
almost none
5
Heats of vaporization and fusion for several common substances.
6
Phase changes and their enthalpy changes.
7
12.2 Quantitative Aspects of Phase Changes
A cooling curve for the conversion of gaseous water to ice.
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Quantitative Aspects of Phase Changes
Within a phase, a change in heat is accompanied by a
change in temperature which is associated with a change in
average Ek as the most probable speed of the molecules
changes.
q = (amount)(molar heat capacity)(∆T)
During a phase change, a change in heat occurs at a
constant temperature, which is associated with a change in
Ep, as the average distance between molecules changes.
q = (amount)(enthalpy of phase change)
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Sample Problem 12.1
Finding the Heat of a Phase Change Depicted by
Molecular Scenes
PROBLEM: These molecular scenes represent a phase change of water. Select
data from the previous text discussion to find the heat (in kJ) lost or
gained when 24.3 g of H2O undergoes this change.
PLAN: The scenes show a disorderly, condensed phase (liquid) changing to
separate molecules (gas) and represent the vaporization of water. Three
endothermic stages: (1) heating liquid 85.0 to 100.oC, (2) liquid to gas at
100.oC, and (3) heating gas 100. to 117oC.
SOLUTION:
mol H2O = 24.3 g H2O x
mol H2O
18.02 g H2O
= 1.35 mol H2O
q = n x Cwater(l) x ∆T = (1.35 mol)(75.4 J/mol·oC)(100. – 85.0oC) = 1527 J = 1.53 kJ
q = n(∆Hovap) = (1.35 mol)(40.7 kJ/mol) = 54.9 kJ
q = n x Cwater(g) x ∆T = (1.35 mol)(33.1 J/mol·oC)(117 – 100.oC) = 759.6 J = 0.760 kJ
qtotal = 1.53 + 54.9 + 0.760 kJ = 57.2 kJ
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Liquid-gas equilibrium.
11
The effect of temperature on the distribution
of molecular speed in a liquid.
12
Vapor pressure as a function
of temperature and
intermolecular forces.
A linear plot of the
relationship between vapor
pressure and temperature .
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The Clausius-Clapeyron Equation
ln P =
- ∆Hvap  1 
 +C
R T 
P2 - ∆Hvap  1 1 
 − 
ln =
P1
R  T2 T1 
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Sample Problem 12.2
The vapor pressure of ethanol is 115 torr at 34.9oC. If ∆Hvap of
ethanol is 40.5 kJ/mol, calculate the temperature (in oC) when
the vapor pressure is 760 torr.
PROBLEM:
PLAN:
Using the Clausius-Clapeyron Equation
We are given 4 of the 5 variables in the Clausius-Clapeyron
equation. Substitute and solve for T2.
SOLUTION:
ln
ln
P2 - ∆Hvap  1 1 
 − 
=
P1
R  T2 T1 
760 torr
115 torr
=
- 40.5 x103 J/mol
8.314 J/mol·K
34.9oC + 273.15 = 308.0 K
1
T2
−
1
308.0 K
T2 = 350. K – 273.15 = 77°C
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Iodine subliming.
test tube with ice
iodine solid
iodine vapor
iodine solid
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Phase diagrams for CO2 and H2O.
CO2
H2O
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12.3 Types of Intermolecular Forces
Covalent and van der Waals radii.
van der Waal’s distance
bond length
covalent radius
van der Waal’s radius
18
Periodic trends in covalent and van der Waals radii (in pm).
19
20
21
Polar molecules and dipole-dipole forces.
solid
liquid
22
Dipole moment and boiling point.
23
THE HYDROGEN BOND
a dipole-dipole intermolecular force
A hydrogen bond may occur when an H atom in a molecule,
bound to small highly electronegative atom with lone pairs of
electrons, is attracted to the lone pairs in another molecule.
The elements which are so electronegative are N, O, and F.
H
hydrogen bond
acceptor
..
O
..
O
..
..
..
..
F
..
hydrogen bond
donor
hydrogen bond
acceptor
hydrogen bond
acceptor
H
hydrogen bond
donor
..
..
N
..
F
..
H
..
N
hydrogen bond
donor
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Hydrogen bonding and boiling point.
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Sample Problem 12.3
PROBLEM:
Which of the following substances exhibits H bonding? For
those that do, draw two molecules of the substance with the H
bond(s) between them.
O
C2H6
(a)
PLAN:
(b) CH3OH
(c)
CH3C NH2
Find molecules in which H is bonded to N, O, or F. Draw H
bonds in the format —B: H—A—.
SOLUTION:
(b)
Drawing Hydrogen Bonds Between Molecules
of a Substance
(a) C2H6 has no H bonding sites.
H
H C O H
H
H
H O C H
H
H
(c)
H
O
H N
CH3O
H N CH3C CH3C
O
CH3C N H
H
O
N H
O
H
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Polarizability and Charged-Induced Dipole Forces
distortion of an electron cloud
•Polarizability increases down a group
size increases and the larger electron clouds are further
from the nucleus
•Polarizability decreases left to right across a period
increasing Zeff shrinks atomic size and holds the electrons
more tightly
•Cations are less polarizable than their parent atom
because they are smaller.
•Anions are more polarizable than their parent atom
because they are larger.
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Dispersion forces among nonpolar particles.
separated Ar
molecules
instantaneous
dipoles
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Molar mass and boiling point.
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Molecular shape and boiling point.
fewer points for
dispersion
forces to act
more points for
dispersion
forces to act
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Summary diagram for analyzing the intermolecular forces in a sample.
INTERACTING
INTERACTING PARTICLES
PARTICLES
(atoms,
(atoms, molecules,
molecules, ions)
ions)
ions present
ions
ions only
only
IONIC
IONIC BONDING
BONDING
(Section
(Section 9.2)
9.2)
ions not present
polar
polar molecules
molecules only
only
DIPOLE-DIPOLE
DIPOLE-DIPOLE
FORCES
FORCES
ion
ion ++ polar
polar molecule
molecule
ION-DIPOLE
ION-DIPOLE FORCES
FORCES
nonpolar
nonpolar
molecules
molecules only
only
DISPERSION
DISPERSION
FORCES
FORCES only
only
H bonded to
N, O, or F
HYDROGEN
HYDROGEN
BONDING
BONDING
polar
polar ++ nonpolar
nonpolar
molecules
molecules
DIPOLEDIPOLEINDUCED
INDUCED DIPOLE
DIPOLE
FORCES
FORCES
DISPERSION FORCES ALSO PRESENT
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Sample Problem 12.4
PROBLEM:
Predicting the Types of Intermolecular Force
For each pair of substances, identify the dominant
intermolecular force(s) in each substance, and select the
substance with the higher boiling point.
(a) MgCl2 or PCl3
(b) CH3NH2 or CH3F
(c) CH3OH or CH3CH2OH
CH3
(d) Hexane (CH3CH2CH2CH2CH2CH3)
CH3CCH2CH3
or 2,2-dimethylbutane
CH3
PLAN: Use the formula, structure, Table 12.2 and Figure 12.18.
• Bonding forces are stronger than nonbonding (intermolecular) forces.
• Hydrogen bonding is a strong type of dipole-dipole force.
• Dispersion forces are decisive when the difference is molar mass or
molecular shape.
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Sample Problem 12.4
Predicting the Types of Intermolecular Force
SOLUTION:
(a) Mg2+ and Cl− are held together by ionic bonds while PCl3 is covalently
bonded and the molecules are held together by dipole-dipole interactions. Ionic
bonds are stronger than dipole interactions and so MgCl2 has the higher boiling
point.
(b) CH3NH2 and CH3F are both covalent compounds and have bonds which are
polar. The dipole in CH3NH2 can H bond while that in CH3F cannot. Therefore
CH3NH2 has the stronger interactions and the higher boiling point.
(c) Both CH3OH and CH3CH2OH can H bond but CH3CH2OH has more CH for
more dispersion force interaction. Therefore CH3CH2OH has the higher boiling
point.
(d) Hexane and 2,2-dimethylbutane are both nonpolar with only dispersion
forces to hold the molecules together. Hexane has the larger surface area,
thereby the greater dispersion forces and the higher boiling point.
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12.4 Properties of the Liquid State
The molecular basis of surface tension.
hydrogen bonding
occurs across the surface
and below the surface
the net vector
for attractive
forces is downward
hydrogen bonding
occurs in three
dimensions
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Table 12.3 Surface Tension and Forces Between Particles
Surface Tension
Substance
Formula
(J/m2) at 200C
diethyl ether
CH3CH2OCH2CH3
1.7x10-2
dipole-dipole; dispersion
ethanol
CH3CH2OH
2.3x10-2
H bonding
butanol
CH3CH2CH2CH2OH
2.5x10-2
H bonding; dispersion
H2O
7.3x10-2
H bonding
Hg
48x10-2
metallic bonding
water
mercury
Major Force(s)
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Shape of water or mercury meniscus in glass.
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Table 12.4 Viscosity of Water at Several Temperatures
viscosity—resistance to flow
Temperature
(oC)
Viscosity
(N·s/m2)*
20
1.00x10−3
40
0.65x10−3
60
0.47x10−3
80
0.35x10−3
*The units of viscosity are Newton-seconds per square meter.
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12.5 The Uniqueness of Water
The H-bonding ability of the water molecule.
hydrogen bond donor
hydrogen bond acceptor
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The Unique Nature of Water
•great solvent properties due to polarity and
hydrogen bonding ability
•exceptional high specific heat capacity
•high surface tension and capillarity
•density differences of liquid and solid states
39
The hexagonal structure of ice.
40
The expansion and contraction of water
breaks rocks to sand and soil.
41
The macroscopic properties of water and their atomic
and molecular “roots.”
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12.6 The Solid State: Structure, Properties, and Bonding
The striking beauty of crystalline solids.
43
The crystal lattice and the unit cell.
lattice point
unit
cell
unit
cell
portion of 3-D lattice
44
The three cubic unit cells.
Simple Cubic
1/8 atom at
8 corners
Atoms/unit cell = 1/8 x 8 = 1
Coordination number = 6
45
The three cubic unit cells.
Body-centered
Cubic
1/8 atom at
8 corners
1 atom at
center
Atoms/unit cell = (1/8 x 8) + 1 = 2
Coordination number = 8
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The three cubic unit cells.
Face-centered
Cubic
1/8 atom at
8 corners
1/2 atom at
6 faces
Atoms/unit cell = (1/8 x 8) + (1/2 x 6) = 4
Coordination number = 12
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Packing identical spheres.
simple cubic
(52% packing efficiency)
body-centered cubic
(68% packing efficiency)
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layer a
layer b
hexagonal
closest
packing
layer a
cubic closest
packing
layer c
closest packing of first
and second layers
abab… (74%)
abcabc… (74%)
hexagonal
unit cell
expanded
side views
face-centered
unit cell 49
Edge length and atomic (ionic) radius in the
three cubic unit cells.
50
Sample Problem 12.5
PROBLEM:
PLAN:
Determining Atomic Radius from Crystal Structure
The crystal structure of copper adopts cubic closest packing
and the edge length of the unit cell is 361.5 pm What is the
atomic radius of copper?
Copper has a face-centered cubic unit cell with edge length A = 361.5
pm see Figure 12.29C. The diagonal of the cell’s face is 4r and the
Pythagorean theorem can be used to solve for r.
SOLUTION:
C= A +B
2
2
C = 2 A = 2(361.5 pm) = 511.2 pm
2
C = 4r
2
r = C/4 = 511.2 pm/4 = 127.8 pm
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Cubic closest packing for
frozen argon.
Cubic closest packing
of frozen methane.
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Table 12.5
Type
Atomic
Particle(s)
Atoms
Molecular Molecules
Ionic
Characteristics of the Major Types of Crystalline Solids
Interparticle
Forces
Dispersion
Dispersion,
dipole-dipole,
H bonds
Ion-ion
Positive &
negative ions
attraction
Physical
Properties
Soft, very low mp, poor
thermal & electrical
conductors
Fairly soft, low to moderate
mp, poor thermal &
electrical conductors
Examples (mp,oC)
Group 8A(18)
[Ne(-249) to Rn(-71)]
Nonpolar: O2[-219],
C4H10[-138], Cl2 [-101],
C6H14[-95], P4 [44.1]
Polar: SO2[-73],CHCl3
[-64], HNO3[-42], H2O
[0.0], CH3COOH[17]
Hard & brittle, high mp,
good thermal & electrical
conductors when molten
NaCl [801]
CaF2 [1423]
MgO [2852]
Soft to hard, low to very
high mp, excellent thermal
and electrical conductors,
malleable and ductile
Na [97.8]
Zn [420]
Fe [1535]
Metallic
Atoms
Metallic bond
Network
Covalent
Atoms
Covalent bond Very hard, very high mp,
usually poor thermal and
electrical conductors
SiO2 (quartz) [1610]
C (diamond) [~4000]
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The sodium chloride structure.
expanded view
space-filling
54
The zinc blende structure.
zinc sulfide
55
The fluorite (CaF2) structure.
56
Crystal structures of metals.
cubic closest packing
hexagonal closest packing
57
58
Crystalline and amorphous silicon dioxide.
Cristobalite (Silica)
Glass
59
Tools of the Laboratory
Bragg’s equation: 2dsinθ = nλ
Diffraction of x-rays by crystal planes.
60
Tools of the Laboratory
Monochromator
Formation of an x-ray diffraction pattern of the
protein hemoglobin.
61
Tools of the Laboratory
Scanning tunneling micrographs.
Gold surface
Cesium atoms on gallium
arsenide surface
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