General Chemistry 202 – CHM202
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General Chemistry 202 – CHM202
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1
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4
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7
Intermolecular Forces and
Liquids and Solids
Chapter 11
8
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A phase is a homogeneous part of the system in contact with
other parts of the system but separated from them by a welldefined boundary.
2 Phases
Solid phase - ice
Liquid phase - water
9
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
Hvap
Hfus
Hsub
10
Intermolecular Forces
Dipole-Dipole Forces
Attractive forces between polar molecules
Orientation of Polar Molecules in a Solid
11
Intermolecular Forces
Ion-Dipole Forces
Attractive forces between an ion and a polar molecule
Ion-Dipole Interaction
12
4
Interaction Between Water and Cations
in solution
13
Intermolecular Forces
Dispersion Forces
Attractive forces that arise as a result of temporary
dipoles induced in atoms or molecules
ion-induced dipole interaction
dipole-induced dipole interaction
14
Induced Dipoles Interacting With Each Other
15
5
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.
16
Example 11.1
What type(s) of intermolecular forces exist between the
following pairs?
(a) HBr and H2S
(b) Cl2 and CBr4
(c) I2 and
(d) NH3 and C6H6
Example 11.1
Strategy Classify the species into three categories: ionic, polar
(possessing a dipole moment), and nonpolar. Keep in mind that
dispersion forces exist between all species.
Solutions
(a) Both HBr and H2S are polar molecules. Therefore, the
intermolecular forces present are dipole-dipole forces, as
well as dispersion forces.
(b) Both Cl2 and CBr4 are nonpolar, so there are only dispersion
forces between these molecules.
6
Example 11.1
(c) I2 is a homonuclear diatomic molecule and therefore
nonpolar, so the forces between it and the ion,
, are
ion-induced dipole forces and dispersion forces.
(d) NH3 is polar, and C6H6 is nonpolar. The forces are dipoleinduced dipole forces and dispersion forces.
Intermolecular Forces
Hydrogen Bond
The hydrogen bond is a special dipole-dipole interaction
between the hydrogen atom in a polar N-H, O-H, or F-H bond
and an electronegative O, N, or F atom.
A
H…B
or
A
H…A
A & B are N, O, or F
20
Hydrogen Bond
HCOOH and water
21
7
Why is the hydrogen bond considered a “special”
dipole-dipole interaction?
Decreasing molar mass
Decreasing boiling point
22
Example 11.2
Which of the following can form hydrogen bonds with water?
a) CH3OCH3
b) CH4
c) F2
d) HCOOH
e) Na+
Example 11.2
Strategy A species can form hydrogen bonds with water if it
contains one of the three electronegative elements (F, O, or N)
or it has a H atom bonded to one of these three elements.
Solution There are no electronegative elements (F, O, or N) in
either CH4 or Na+. Therefore, only CH3OCH3, F2, and HCOOH
can form hydrogen bonds with water.
8
Example 11.2
Check Note that HCOOH (formic acid) can form hydrogen
bonds with water in two different ways.
HCOOH forms hydrogen bonds with two H2O molecules.
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
26
Properties of Liquids
Cohesion is the intermolecular attraction between like molecules
Adhesion is an attraction between unlike molecules
Adhesion
Cohesion
27
9
Properties of Liquids
Viscosity is a measure of a fluid’s resistance to flow.
Strong
intermolecular
forces
High
viscosity
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3-D Structure of Water
Water is a Unique Substance
Maximum Density
40C
Density of Water
Ice is less dense than water
29
A crystalline solid possesses rigid and long-range order. In a
crystalline solid, atoms, molecules or ions occupy specific
(predictable) positions.
An amorphous solid does not possess a well-defined
arrangement and long-range molecular order.
A unit cell is the basic repeating structural unit of a crystalline
solid.
lattice
point
Unit Cell
At lattice points:
Unit cells in 3 dimensions
•
Atoms
•
Molecules
•
Ions
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10
Seven Basic Unit Cells
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Three Types of Cubic Unit Cells
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Arrangement of Identical Spheres in a Simple Cubic Cell
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11
Arrangement of Identical Spheres in a Body-Centered
Cubic Cell
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A Corner Atom, an Edge-Centered Atom and
a Face-Centered Atom
Shared by 8
unit cells
Shared by 2
unit cells
Shared by 4
unit cells
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Number of Atoms Per Unit Cell
1 atom/unit cell
2 atoms/unit cell
4 atoms/unit cell
(8 x 1/8 = 1)
(8 x 1/8 + 1 = 2)
(8 x 1/8 + 6 x 1/2 = 4)
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12
Relation Between Edge Length and Atomic Radius
37
Closest Packing: Hexagonal and Cubic
hexagonal
cubic
38
Exploded Views
39
13
Example 11.3
Gold (Au) crystallizes in a cubic close-packed structure (the
face-centered cubic unit cell) and has a density of 19.3 g/cm3.
Calculate the atomic radius of gold in picometers.
Example 11.3
Strategy We want to calculate the radius of a gold atom.
For a face-centered cubic unit cell, the relationship between
radius (r) and edge length (a), according to Figure 11.22,
is
.
Therefore, to determine r of a Au atom, we need to find a. The
volume of a cube is
. Thus, if we can
determine the volume of the unit cell, we can calculate a. We
are given the density in the problem.
Example 11.3
The sequence of steps is summarized as follows:
Solution
Step 1: We know the density, so in order to determine the
volume, we find the mass of the unit cell. Each unit cell
has eight corners and six faces. The total number of
atoms within such a cell, according to Figure 11.19, is
14
Example 11.3
The mass of a unit cell in grams is
From the definition of density (d = m/V), we calculate the
volume of the unit cell as follows:
Example 11.3
Step 2: Because volume is length cubed, we take the cubic
root of the volume of the unit cell to obtain the edge
length (a) of the cell
Step 3: From Figure 11.22 we see that the radius of an Au
sphere (r) is related to the edge length by
Example 11.3
Therefore,
= 144 pm
15
An Arrangement for Obtaining the X-ray Diffraction Pattern
of a Crystal.
46
Reflection of X-rays from Two Layers of Atoms
Extra distance = BC + CD = 2d sin = n
(Bragg Equation)
47
Example 11.4
X rays of wavelength 0.154 nm strike an aluminum crystal; the
rays are reflected at an angle of 19.3°.
Assuming that n = 1, calculate the spacing between the planes
of aluminum atoms (in pm) that is responsible for this angle of
reflection.
The conversion factor is obtained from 1 nm = 1000 pm.
16
Example 11.4
Strategy This is an application of Equation (11.1).
Solution Converting the wavelength to picometers and using
the angle of reflection (19.3°), we write
Types of Crystals
Ionic Crystals
• Lattice points occupied by cations and anions
• Held together by electrostatic attraction
• Hard, brittle, high melting point
• Poor conductor of heat and electricity
CsCl
ZnS
CaF2
50
Example 11.5
How many Na+ and Cl− ions are in each NaCl unit cell?
17
Example 11.5
Solution NaCl has a structure based on a face-centered cubic
lattice. One whole Na+ ion is at the center of the unit cell, and
there are twelve Na+ ions at the edges. Because each edge
Na+ ion is shared by four unit cells, the total number of Na+ ions
is 1 + (12 × ¼) = 4.
Similarly, there are six Cl− ions at the face centers and eight Cl−
ions at the corners. Each face-centered ion is shared by two
unit cells, and each corner ion is shared by eight unit cells, so
the total number of Cl− ions is (6 × ½) + (8 × 1/8) = 4. Thus,
there are four Na+ ions and four Cl− ions in each NaCl unit cell.
Check This result agrees with sodium chloride’s empirical
formula.
Example 11.6
The edge length of the NaCl unit cell is 564 pm. What is the
density of NaCl in g/cm3?
Example 11.6
Strategy
To calculate the density, we need to know the mass of the unit
cell. The volume can be calculated from the given edge length
because V = a3. How many Na+ and Cl− ions are in a unit cell?
What is the total mass in amu? What are the conversion factors
between amu and g and between pm and cm?
Solution
From Example 11.5 we see that there are four Na+ ions and
four Cl− ions in each unit cell. So the total mass (in amu) of a
unit cell is
mass = 4(22.99 amu + 35.45 amu) = 233.8 amu
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Example 11.6
Converting amu to grams, we write
The volume of the unit cell is V = a3 = (564 pm)3. Converting
pm3 to cm3, the volume is given by
Finally, from the definition of density
= 2.16 g/cm3
Types of Crystals
Covalent Crystals
• Lattice points occupied by atoms
• Held together by covalent bonds
• Hard, high melting point
• Poor conductor of heat and electricity
carbon
atoms
diamond
graphite
56
Types of Crystals
Molecular Crystals
• Lattice points occupied by molecules
•
Held together by intermolecular forces
•
Soft, low melting point
•
Poor conductor of heat and electricity
water
benzene
57
19
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-
58
Crystal Structures of Metals
59
Types of Crystals
60
20
An amorphous solid does not possess a well-defined
arrangement and long-range molecular order.
A glass is an optically transparent fusion product of inorganic
materials that has cooled to a rigid state without crystallizing
Crystalline
quartz (SiO2)
Non-crystalline
quartz glass
61
Phase Changes
Least
Order
Greatest
Order
62
Effect of Temperature on Kinetic Energy
T2 > T1
63
21
The equilibrium vapor pressure is the vapor pressure
measured when a dynamic equilibrium exists between
condensation and evaporation
H2O (l)
H2O (g)
Dynamic Equilibrium
Rate of
Rate of
= evaporation
condensation
64
Measurement of Vapor Pressure
At
Equilibrium
Before
Evaporation
65
Molar heat of vaporization (Hvap) is the energy required to
vaporize 1 mole of a liquid at its boiling point.
Clausius-Clapeyron Equation
ln P = -
Hvap
RT
+C
P = (equilibrium) vapor pressure
T = temperature (K)
R = gas constant (8.314 J/K•mol)
Vapor Pressure Versus Temperature
66
22
Alternate Forms of the Clausius-Clapeyron Equation
At two temperatures
or
67
Example 11.7
Diethyl ether is a volatile, highly flammable organic liquid that is
used mainly as a solvent.
The vapor pressure of diethyl ether is 401 mmHg at 18°C.
Calculate its vapor pressure at 32°C.
Example 11.7
Strategy We are given the vapor pressure of diethyl ether at
one temperature and asked to find the pressure at another
temperature. Therefore, we need Equation (11.5).
Solution Table 11.6 tells us that Hvap = 26.0 kJ/mol. The data
are
From Equation (11.5) we have
23
Example 11.7
Taking the antilog of both sides (see Appendix 4), we obtain
Hence
P2 = 656 mmHg
Check We expect the vapor pressure to be greater at the
higher temperature. Therefore, the answer is reasonable.
The boiling point is the temperature at which the
(equilibrium) vapor pressure of a liquid is equal to the
external pressure.
The normal boiling point is the temperature at which a liquid
boils when the external pressure is 1 atm.
71
The critical temperature (Tc) is the temperature above which
the gas cannot be made to liquefy, no matter how great the
applied pressure.
The critical pressure
(Pc) is the minimum
pressure that must be
applied to bring about
liquefaction at the
critical temperature.
72
24
The Critical Phenomenon of SF6
T < Tc
T > Tc
T ~ Tc
T < Tc
73
Solid-Liquid Equilibrium
H2O (s)
H2O (l)
The melting point of a solid
or the freezing point of a
liquid is the temperature at
which the solid and liquid
phases coexist in
equilibrium.
74
Molar heat of fusion (Hfus) is the energy required to melt
1 mole of a solid substance at its freezing point.
75
25
Heating Curve
76
Solid-Gas Equilibrium
H2O (s)
H2O (g)
Molar heat of sublimation
(Hsub) is the energy required
to sublime 1 mole of a solid.
Hsub = Hfus + Hvap
( Hess’s Law)
77
Example 11.8
Calculate the amount of energy (in kilojoules) needed to heat
346 g of liquid water from 0°C to 182°C.
Assume that the specific heat of water is 4.184 J/g · °C over the
entire liquid range and that the specific heat of steam is 1.99
J/g · °C.
26
Example 11.8
Strategy The heat change (q) at each stage is given by q =
mst (see p. 247), where m is the mass of water, s is the
specific heat, and t is the temperature change.
If there is a phase change, such as vaporization, then q is given
by nHvap, where n is the number of moles of water.
Solution The calculation can be broken down in three steps.
Step 1: Heating water from 0°C to 100°C
Using Equation (6.12) we write
Example 11.8
Step 2: Evaporating 346 g of water at 100°C (a phase change)
In Table 11.6 we see Hvap = 40.79 kJ/mol for water, so
Step 3: Heating steam from 100°C to 182°C
Example 11.8
The overall energy required is given by
Check All the qs have a positive sign, which is consistent with
the fact that heat is absorbed to raise the temperature from 0°C
to 182°C. Also, as expected, much more heat is absorbed
during the phase transition.
27
A phase diagram summarizes the conditions at which a
substance exists as a solid, liquid, or gas.
Phase Diagram of Water
82
Phase Diagram of Carbon
Dioxide
At 1 atm
CO2 (s)
CO2 (g)
83
Effect of Increase in Pressure on the Melting Point
of Ice and the Boiling Point of Water
84
28