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2
Chapter Goals
1. Kinetic-Molecular Description of Liquids and Solids
2. Intermolecular Attractions and Phase Changes
The Liquid State
3. Viscosity
4. Surface Tension
5. Capillary Action
6. Evaporation
7. Vapor Pressure
8. Boiling Points and Distillation
9. Heat Transfer Involving Liquids
3
Chapter Goals
The Solid State
10. Melting Point
11. Heat Transfer Involving Solids
12. Sublimation and the Vapor Pressure of Solids
13. Phase Diagrams (P versus T)
14. Amorphous Solids and Crystalline Solids
15. Structures of Crystals
16. Bonding in Solids
17. Band Theory of Metals
18. Synthesis Question
4
Kinetic-Molecular Description of Liquids and Solids
Solids and liquids are Solids and liquids are condensed states.condensed states. The atoms, ions, or molecules in solids and The atoms, ions, or molecules in solids and
liquids are much closer to one another than in liquids are much closer to one another than in gases.gases.
Solids and liquids are highly incompressible.Solids and liquids are highly incompressible.Liquids and gases are Liquids and gases are fluids.fluids.
They easily flow.They easily flow.The intermolecular attractions intermolecular attractions in liquids in liquids
and solids are strong.and solids are strong.
5
Kinetic-Molecular Description of Liquids and Solids
Schematic representation of the three common states of matter.
gas liquid solidcool cool
heatheat
6
Kinetic-Molecular Description of Liquids and Solids
If we compare the strengths of interactionsstrengths of interactions among particles and the degree of orderingdegree of ordering of particles, we see that
Gases< Liquids < SolidsMiscible liquidsMiscible liquids are soluble in each other.
Examples of miscible liquids:Water dissolves in alcohol.Gasoline dissolves in motor oil.
7
Kinetic-Molecular Description of Liquids and Solids
Immiscible liquidsImmiscible liquids are insoluble in each other. Two examples of immiscible liquids:
Water does not dissolve in oil. Water does not dissolve in cyclohexane.
8
Intermolecular Attractions and Phase Changes
There are four important intermolecular attractions. This list is from strongest attraction to the weakest
attraction.1. Ion-ion interactions
The force of attraction between two oppositely charged ions is governed by Coulomb’s law.
ions. ebetween th distance theis d
charges.ion theare q and q
d
qqF
-
2
-
9
Intermolecular Attractions and Phase Changes
Coulomb’s law determines: 1. The melting and boiling points of ionic
compounds.2. The solubility of ionic compounds.
Example 13-1: Arrange the following ionic compounds in the expected order of increasing melting and boiling points.
NaF, CaO, CaF2
You do it!You do it!
What important points must you consider?What important points must you consider?
11
Intermolecular Attractions and Phase Changes
2. Hydrogen bonding Consider H2O a very polar molecule.
12
Intermolecular Attractions and Phase Changes
2. Hydrogen bonding Consider H2O a very polar molecule.
One Water Molecule Water Dimer
Water Tetramer Ice Structure
14
Intermolecular Attractions and Phase Changes
3. Dipole-dipole interactions Consider BrF a polar molecule.
15
Intermolecular Attractions and Phase Changes
4. London Forces are very weak. They are the weakest of the intermolecular
forces. This is the only attractive force in nonpolar
molecules. Consider Ar as an isolated atom.
16
Intermolecular Attractions and Phase Changes
In a group of Ar atoms the temporary dipole in one atom induces other atomic dipoles.
17
Intermolecular Attractions and Phase Changes
Similar effects occur in a group of I2 molecules.The effect is shown in this movie.
18
The Liquid State
ViscosityViscosityViscosity is the resistance to flow.
For example, compare how water pours out of a glass compared to molasses, syrup or honey.
Oil for your car is bought based on this property. 10W30 or 5W30 describes the
viscosity of the oil at high and low temperatures.
20
The Liquid State
Surface TensionSurface Tension Surface tension is a measure of the
unequal attractions that occur at the surface of a liquid.
The molecules at the surface are attracted unevenly.
22
The Liquid State
Capillary ActionCapillary ActionCapillary action is the ability of a liquid to
rise (or fall) in a glass tube or other container
23
The Liquid State
Cohesive forces are the forces that hold liquids together.
Adhesive forces are the forces between a liquid and another surface.Capillary rise implies that the:
Adhesive forces > cohesive forcesCapillary fall implies that the:
Cohesive forces > adhesive forces
24
The Liquid State
Water exhibits a capillary rise.
Water Mercury
Mercury exhibits a capillary fall.
26
The Liquid State
Evaporation Evaporation is the process in which
molecules escape from the surface of a liquid and become a gas.
Evaporation is temperature dependent.
28
The Liquid State
Vapor Pressure Vapor pressure is the pressure exerted by a liquid’s
vapor on its surface at equilibrium. Vapor Pressure (torr) and boiling point for three liquids
at different temperatures.0oC 20oC 30oC normal boiling point
diethyl ether 185 442 647 36oC
ethanol 12 44 74 78oC
water 5 18 32 100oC What are the intermolecular forces in each of these compounds?
You do it!You do it!
30
The Liquid State
Boiling Points and DistillationThe boiling pointboiling point is the temperature at
which the liquid’s vapor pressure is equal to the applied pressure.
The normalnormal boiling pointboiling point is the boiling point when the pressure is exactly 1 atm.
Distillation is a method we use to separate mixtures of liquids based on their differences in boiling points.
31
The Liquid State
DistillationDistillation is a process in which a
mixture or solution is separated into its components on the basis of the differences in boiling points of the components.
Distillation is another vapor pressure phenomenon.
32
The Liquid State
Heat Transfer Involving LiquidsFrom Chapter 1
q = m C T•Example 13-2: How much heat is released by 2.00 x 102 g of H2O as it cools from 85.0oC to 40.0oC? The specific heat of water is 4.184 J/goC.
You do it!You do it!
34
The Liquid State
Molar heat capacity is the amount of heat required to raise the temperature of one mole of a substance 1.00 oC.
Example 13-3: The molar heat capacity of ethyl alcohol, C2H5OH, is 113 J/moloC. How much heat is required to raise the T of 125 g of ethyl alcohol from 20.0oC to 30.0oC?
1 mol C2H5OH = 46.0 g
You do it!You do it!
35
The Liquid State
? mol = 125 g1 mol C H OH
46.0 g C H OH2.72 mol C H OH
? J = 2.72 mol J
mol C C kJ
2 5
2 52 5
oo
11330 0 20 0 307. . .
36
The Liquid State
The calculations we have done up to now tell us the energy changes as long as the substance remains in a single phase.
Next, we must address the energy associated with phase changes. For example, solid to liquid or liquid to gas and the reverse.
Heat of Vaporization is the amount of heat required to change 1.00 g of a liquid substance to a gas at constant temperature. Heat of vaporization has units of J/g.
Heat of Condensation is the reverse of heat of vaporization, phase change from gas to liquid.
J 2260-o
(g)2
J 2260o
)(2 C100.0at OH g 00.1C100.0at OH g 1.00
37
The Liquid State
Molar heat of vaporization or Hvap The Hvap is the amount of heat required to change 1.00
mole of a liquid to a gas at constant temperature.Hvap has units of J/mol.
Molar heat of condensation The reverse of molar heat of vaporization is the heat of
condensation.
kJ 40.7-o
(g)2
kJ 40.7o
)(2 C100.0at OH mol 00.1C100.0at OH mol 1.00
39
The Liquid State
Example 13-4: How many joules of energy must be absorbed by 5.00 x 102 g of H2O at 50.0oC to convert it to steam at 120oC? The molar heat of vaporization of water is 40.7 kJ/mol and the molar heat capacities of liquid water and steam are 75.3 J/mol oC and 36.4 J/mol oC, respectively.
You do it!You do it!
40
The Liquid State
? J = 27.8 mol J
mol J
40 7 101131 10
35.
.
? .
.. . .
mol = 500 g H O1 mol H O
g H O mol H O
1st let's calculate the heat required to warm water from 50 to 100 C
? J = 27.8 mol J
mol CC J
22
22
o
oo
1827 8
753100 0 50 0 105 105
Next, let’s calculate the energy required to boil the water.
Finally, let’s calculate the heat required to heat steam from 100 to 120oC.
? J = 27.8 mol J
mol C120.0 -100.0 C J
oo36 4
0 20 105..
41
The Liquid State
The total amount of energy for this process is the sum of the 3 pieces we have calculated
105 10 1131 10 0 20 10
12 56 10
5 5 5
5
. . .
.
J J J
J or 1.26 10 kJ3
42
The Liquid State
Example 13-5: If 45.0 g of steam at 140oC is slowly bubbled into 450 g of water at 50.0oC in an insulated container, can all the steam be condensed?
You do it!You do it!
43
The Liquid State
condensed. becannot steam theof all Thus
kJ. 94.1 is absorbcan water liquid theheat that ofAmount
kJ. 105 is steam theof all condense heat to ofAmount
kJ 94.1 C)50.0 -(100.0 75.3 mol 25.0
water.liquid in the availableheat ofamount theCalculate (2)
kJ .105 40.7 mol 2.50 C100.0 -140.0 36.4mol 2.50
steam. thecondense torequiredheat ofamount theCalculate (1)
mol .025g 18
mol 1 waterg 450 mol 2.50
steam g 18
mol 1 steam g 0.45
oCmol
J
molkJo
Cmol J
o
o
44
The Liquid State
Clausius-Clapeyron equation determine vapor pressure of a liquid at a new T determine what T we must heat something to get a
specified vapor pressure way to determine Hvap if we know pressure at 2 T’s
lnP
P
H
R T T2
1
vap
1 2
1 1
45
The Liquid State
In Denver the normal atmospheric pressure is 630 torr. At what temperature does water boil in Denver?
lnP
P
H
R T T
torr760 torr
8.314 K T
T
2
1
vap
1 2
Jmol
JK mol 2
2
1 1
630 40 7 10 1373
1
0 829 4895 0 0026811
3
ln.
ln . .
46
The Liquid State
01884895
0 0026811
383 10 0 0026811
383 10 0 0026811
0 002721
368
5
5
..
. .
. .
.
T
T
T
T
T K or 95 C
2
2
2
2
2o
47
The Liquid State
Boiling Points of Various Kinds of Liquids
Gas MW BP(oC)He 4 -269
Ne 20 -246
Ar 40 -186
Kr 84 -153
Xe 131 -107
Rn 222 -62
48
The Liquid State
Noble Gases
-300
-250
-200
-150
-100
-50
04 20 40 84 131 222
Molar Mass
Boi
ling
Poi
nt (
C)
49
The Liquid State
Compound MW(amu) B.P.(oC)
CH4 16 -161
C2H6 30 -88
C3H8 44 -42
n-C4H10 58 -0.6
n-C5H12 72 +36
51
The Liquid State
Compound MW(amu) B.P.( C)
HF 20 19.5
HCl 37 - 85.0
HBr 81 - 67.0
HI 128 - 34.0
o
52
The Liquid State
Hydrogen Halides
-100
-80
-60
-40
-20
0
20
40
20 37 81 128
Molar Mass
Boi
ling
Poi
nt (
C)
53
The Liquid State
Compound MW(amu) B.P.( C)
H O 18 100
H S 34 - 61
H Se 81 - 42
H Te 130 - 2
o
2
2
2
2
56
The Liquid State
Example 13-6: Arrange the following substances in order of increasing boiling points.
C2H6, NH3, Ar, NaCl, AsH3
You do it!You do it!
Ar < C2H6 < AsH3 < NH3 < NaCl
nonpolar nonpolar polar very polar ionic
London London dipole-dipole H-bonding ion-ion
57
The Solid State
Normal Melting PointThe normal melting pointnormal melting point is the
temperature at which the solid melts (liquid and solid in equilibrium) at exactly 1.00 atm of pressure.
The melting point increases as the strength of the intermolecular attractions increase.
58
The Solid State
Which requires more energy?
OHOH
or
NaClNaCl
2s2
s
What experimental proof do you have?
59
Heat Transfer Involving Solids
Heat of FusionHeat of fusionHeat of fusion is the amount of heat
required to melt one gram of a solid at its melting point at constant temperature.
• Heat of crystallization Heat of crystallization is the reverse of the heat of fusion.
J 334-o
)(2
J 334 o
(s)2 C0at OH g 1.00 C0at OH g 1.00
60
Heat Transfer Involving Solids
Molar heat of fusion or Hfusion
The molar heat of fusion is the amount of heat required to melt a mole of a substance at its melting point.
The molar heat of crystallization is the reverse of molar heat of fusion
J 6012-o
)(2
J 6012o
(s)2 C0at OH mole 1.00 C0at OH mole 1.00
61
Heat Transfer Involving Solids
Here is a summary of the heats of transformation for water.
J 6012-o
)(2
J 6012o
(s)2 C0at OH mole 1.00 C0at OH mole 1.00
kJ 40.7-o
(g)2
kJ 40.7o
)(2 C100.0at OH mol 00.1C100.0at OH mol 1.00
62
Heat Transfer Involving Solids
Example 11-7: Calculate the amount of heat required to convert 150.0 g of ice at -10.0oC to water at 40.0oC.specific heat of ice is 2.09 J/goC
you do it
63
Heat Transfer Involving Solids
? J = (150.0 g)(2.09 J
g C)(10 C) = 3.14 10 J
? J = (150.0 g)(334 Jg
) = 5.01 10 J
? J = (150.0 g)(4.18 J
g C)(40 C) = 2.51 10 J
7.83 10 J
oo 3
4
oo 4
4
64
Sublimation and the Vapor Pressure of Solids
SublimationIn the sublimation process the solid transforms
directly to the vapor phase without passing through the liquid phase.
Solid CO2 or “dry” ice does this well.
oncondensati
nsublimatio
gas solid
65
Phase Diagrams (P versus T)
Phase diagrams are a convenient way to display all of the different phase transitions of a substance.
This is the phase diagram for water.
67
Amorphous Solids andCrystalline Solids
Amorphous solids do not have a well ordered molecular structure. Examples of amorphous solids include waxes,
glasses, asphalt. Crystalline solids have well defined structures
that consist of extended array of repeating units called unit cellsunit cells. Crystalline solids display X-ray diffraction patterns
which reflect the molecular structure. The Bragg equation, detailed in the textbook,
describes how an X-ray diffraction pattern can be used to determine the interatomic distances in crystals.
68
Structure of Crystals Unit cells are the smallest repeating unit of a
crystal. As an analogy, bricks are repeating units for
buildings. There are seven basic crystal systems.
69
Structure of Crystals
We shall look at the three variations of the cubic crystal system.
Simple cubic unit cells.The balls represent the positions of atoms,
ions, or molecules in a simple cubic unit cell.
70
Structure of Crystals
In a simple cubic unit cell each atom, ion, or molecule at a corner is shared by 8 unit cellsThus 1 unit cell contains 8(1/8) = 1 atom,
ion, or molecule.
71
Structure of Crystals
Body centered cubic (bcc) has an additional atom, ion, or molecule in the center of the unit cell.
On a body centered cubic unit cell there are 8 corners + 1 particle in center of cell. 1 bcc unit cell
contains 8(1/8) + 1 = 2 particles.
72
Structure of Crystals
A face centered cubic (fcc) unit cell has a cubic unit cell structure with an extra atom, ion, or molecule in each face.
73
Structure of Crystals
A face centered cubic unit cell has 8 corners and 6 faces.1 fcc unit cell contains
8(1/8) + 6(1/2) = 4 particles.
74
Bonding in Solids
Molecular Solids have molecules in each of the positions of the unit cell.Molecular solids have low melting points,
are volatile, and are electrical insulators.Examples of molecular solids inlude:
water, sugar, carbon dioxide, benzene
75
Bonding in Solids
Covalent Solids have atoms that are covalently bonded to one another
Some examples of covalent solids are:Diamond, graphite, SiO2 (sand), SiC
76
Bonding in Solids
Ionic Solids have ions that occupy the positions in the unit cell.
Examples of ionic solids include:CsCl, NaCl, ZnS
77
Bonding in Solids
Metallic Solids may be thought of as positively charged nuclei surrounded by a sea of electrons.
The positive ions occupy the crystal lattice positions.
Examples of metallic solids include:Na, Li, Au, Ag, ……..
78
Bonding in Solids
Variations in Melting Points for Molecular Solids What are the intermolecular forces in each solid? Compound Melting Point (oC)ice 0.0ammonia -77.7benzene, C6H6 5.5
napthalene, C10H8 80.6
benzoic acid, C6H5CO2H 122.4
79
Bonding in Solids
Variations in Melting Points for Covalent SolidsSubstance Melting Point (oC)sand, SiO2 1713
carborundum, SiC ~2700diamond >3550graphite 3652-3697
80
Bonding in Solids
• Variations in Melting Points for Ionic Solids• Compound Melting Point (oC) LiF 842 LiCl 614 LiBr 547 LiI 450 CaF2 1360
CaCl2 772
CaBr2 730
CaI2 740
81
Bonding in Solids
Variations in Melting Points for Metallic Solids
Metal Melting Point (oC) Na 98 Pb 328 Al 660 Cu 1083 Fe 1535 W 3410
82
Bonding in Solids
Example 13-8. A group IVA element with a density of 11.35 g/cm3 crystallizes in a face-centered cubic lattice whose unit cell edge length is 4.95 Å. Calculate the element’s atomic weight. What is the atomic radius of this element?
83
Bonding in Solids
Face centered cubic unit cells have 4 atoms, ions, or molecules per unit cell.
Problem solution pathway:1. Determine the volume of a single unit cell.2. Use the density to determine the mass of a single
unit cell. 3. Determine the mass of one atom in a unit cell.4. Determine the mass of 1 mole of these atoms
84
Bonding in Solids
1. Determine the volume of a single unit cell.
322-38-
3
8-0
8-0
cm 101.21 cm 104.95
V so cubic are cellsunit cubic centered Face
cm 104.95 A 4.95 thuscm 10 A 1
2. Use the density to determine the mass of a unit cell.
cellunit one
g 101.38
cm
g 35.11 cm 101.21 21-
3322-
85
Bonding in Solids
3. Determine the mass of one atom in the unit cell.
atomg1044.3
4
101.38
fashion. in this determined becan atom one of mass the
cellunit per atoms 4 has cubic centered face Because
22
cellunit atoms
cellunit g21-
4. Determine the mass of one mole of these atoms.
g/mole 207moleatoms10022.6atom
g1044.3 2322
86
Bonding in Solids
To determine an atomic radius requires some geometry.
For simple cubic unit cells:The edge length = 2 radii
87
Bonding in Solids
For face-centered cubic unit cells:The face diagonal = 2 x edge length.The diagonal length = 4 radii.
88
Bonding in Solids
For body-centered cubic unit cells:The body diagonal = 3 x edge length.The diagonal length = 4 radii.
89
Bonding in Solids
Determine the diagonal length then divide by 4 to get the atomic radius.
diagonal = 2 4.95 10 cm
10 cm
radius = 10 cm4 10 cm
-8
-8
-8 -8
7 00
7 00 175
.
. .
92
Band Theory of Metals
Insulators have a large gap between the s and p bands. Gap is called the forbidden zoneforbidden zone.
Semiconductors have a small gap between the bands.
93
Synthesis Question
Maxwell House Coffee Company decaffeinates its coffee beans using an extractor that is 7.0 feet in diameter and 70.0 feet long. Supercritical carbon dioxide at a pressure of 300.0 atm and temperature of 100.0oC is passed through the stainless steel extractor. The extraction vessel contains 100,000 pounds of coffee beans soaked in water until they have a water content of 50%.
94
Synthesis Question
This process removes 90% of the caffeine in a single pass of the beans through the extractor. Carbon dioxide that has passed over the coffee is then directed into a water column that washes the caffeine from the supercritical CO2. How many moles of carbon dioxide are present in the extractor?
95
Synthesis Question
L 107.633
mL) L/1000 )(1mL/cm )(1cm10(7.633
(2134cm)06.7cm)(3.1416)(1hr vesselof Volume
cm 2134 cm/ft) ft)(30.48 (70.0 vesselofLength
cm 106.7 cm/2 213.4 vesselof Radius
cm 213.4cm/ft) ft)(30.48 (7.0 vesselofDiameter
4
337
22
97
Group Question
How many CO2 molecules are there in 1.0 cm3 of the Maxwell House Coffee Company extractor? How many more CO2 molecules are there in a cm3 of the supercritical fluid in the Maxwell House extractor than in a mole of CO2 at STP?