13 liquids and solids - dhspcoe.weebly.com/uploads/2/3/1/8/23185096/apchapter13.pdfevaporation 7....
TRANSCRIPT
13 Liquids and Solids
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 condensed states.
– The atoms, ions, or molecules in solids and
liquids are much closer to one another than in
gases.
– Solids and liquids are highly incompressible.
• Liquids and gases are fluids.
– They easily flow.
• The intermolecular attractions in liquids
and solids are strong.
5
Kinetic-Molecular Description of Liquids and Solids
• Schematic representation of the three
common states of matter.
6
Kinetic-Molecular Description of Liquids and Solids
• If we compare the strengths of interactions
among particles and the degree of ordering
of particles, we see that
Gases< Liquids < Solids
• Miscible 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 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.
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!
What important points must you consider?
10
Intermolecular Attractions and Phase Changes
Na F Ca F Ca O+ - 2+2
2+ 2-
11
Intermolecular Attractions and Phase Changes
2. Dipole-dipole interactions
– Consider BrF a polar molecule.
12
Intermolecular Attractions and Phase Changes
3. Hydrogen bonding
– Consider H2O a very polar molecule.
13
Intermolecular Attractions and Phase Changes
3. Hydrogen bonding
– Consider H2O a very polar molecule.
14
Intermolecular Attractions and Phase Changes
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.
18
The Liquid State
Viscosity
• Viscosity 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 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.
21
The Liquid State
• Floating paper clip demonstration of
surface tension.
22
The Liquid State
Capillary Action
• Capillary 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 forces
– Capillary fall implies that the:
• Cohesive forces > adhesive forces
24
The Liquid State
• Water exhibits a capillary rise.
Water Mercury
Mercury exhibits a capillary fall.
25
The Liquid State
• Capillary action also affects the meniscus of liquids.
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.
27
The Liquid State
29
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!
30
The Liquid State
Vapor Pressure as a function of temperature.
31
The Liquid State
Boiling Points and Distillation
• The boiling point is the temperature at which the liquid’s vapor pressure is equal to the applied pressure.
• The normal boiling 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.
32
The Liquid State
Distillation
• Distillation 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.
33
The Liquid State
Heat Transfer Involving Liquids
• From 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!
34
The Liquid State
kJ 37.6 J 1076.3J?
)C0.400.85)( g(4.184 102.00 J ?
4
o
Cg J2
o
35
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!
36
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 5
2 5
o
o
11330 0 20 0 307. . .
37
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
38
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
40
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!
41
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
2
2
o
o
o
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
o
o36 40 20 105..
42
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
43
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!
44
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
o
Cmol J
molkJo
Cmol J
o
o
45
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 T
2
1
vap
1 2
1 1
46
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
torr
760 torr
8.314 K T
T
2
1
vap
1 2
Jmol
JK mol 2
2
1 1
630 40 7 10 1
373
1
0 829 4895 0 0026811
3
ln.
ln . .
47
The Liquid State
0188
48950 002681
1
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
48
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
49
The Liquid State
Noble Gases
-300
-250
-200
-150
-100
-50
0
4 20 40 84 131 222
Molar Mass
Bo
ilin
g P
oin
t (C
)
50
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
Alkanes
-200
-150
-100
-50
0
50
16 30 44 58 72
Molar Mass
Bo
ilin
g P
oin
t (C
)
52
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
53
The Liquid State
Hydrogen Halides
-100
-80
-60
-40
-20
0
20
40
20 37 81 128
Molar Mass
Bo
ilin
g P
oin
t (C
)
54
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
55
The Liquid State
VIA Hydrides
-100
-50
0
50
100
150
18 34 81 130
Molar Mass
Bo
ilin
g P
oin
t (C
)
56
The Liquid State
• At the molecular level what happens when
a species boils?
57
The Liquid State
• Example 13-6: Arrange the following
substances in order of increasing boiling
points.
C2H6, NH3, Ar, NaCl, AsH3
You do it!
Ar < C2H6 < AsH3 < NH3 < NaCl
nonpolar nonpolar polar very polar ionic
London London dipole-dipole H-bonding ion-ion
58
The Solid State
Normal Melting Point
• The normal 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.
59
The Solid State
• Which requires more energy?
OHOH
or
NaClNaCl
2s2
s
What experimental proof do you have?
60
Heat Transfer Involving Solids
Heat of Fusion
• Heat 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 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
61
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
62
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
63
Heat Transfer Involving Solids
• Example 13-7: 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
64
Heat Transfer Involving Solids
? J = (150.0 g)(2.09 J
g C)(10 C) = 3.14 10 J
? J = (150.0 g)(334 J
g) = 5.01 10 J
? J = (150.0 g)(4.18 J
g C)(40 C) = 2.51 10 J
7.83 10 J
o
o 3
4
o
o 4
4
65
Sublimation and the Vapor Pressure of Solids
Sublimation
• In 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
66
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
Phase Diagrams (P versus T)
• Compare water’s phase diagram to
carbon dioxide’s phase diagram.
68
Amorphous Solids and Crystalline 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 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.
69
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.
70
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.
71
Structure of Crystals
• In a simple cubic unit
cell each atom, ion,
or molecule at a
corner is shared by 8
unit cells
– Thus 1 unit cell
contains 8(1/8) = 1
atom, ion, or
molecule.
72
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.
73
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.
74
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.
75
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 include:
– water, sugar, carbon dioxide, benzene
76
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
77
Bonding in Solids
• Ionic Solids have ions that occupy the positions in the unit cell.
• Examples of ionic solids include:
– CsCl, NaCl, ZnS
78
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, ……..
79
Bonding in Solids
• Variations in Melting Points for Molecular Solids
• What are the intermolecular forces in each solid?
• Compound Melting Point (oC)
• ice 0.0
• ammonia -77.7
• benzene, C6H6 5.5
• napthalene, C10H8 80.6
• benzoic acid, C6H5CO2H 122.4
80
Bonding in Solids
• Variations in Melting Points for Covalent Solids
• Substance Melting Point (oC)
• sand, SiO2 1713
• carborundum, SiC ~2700
• diamond >3550
• graphite 3652-3697
81
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
82
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
83
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?
84
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
85
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-
3
322-
86
Bonding in Solids
3. Determine the mass of one atom in the unit cell.
atomg
1044.3 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 207mole
atoms10022.6atom
g1044.3 2322
87
Bonding in Solids
• To determine an atomic radius requires
some geometry.
• For simple cubic unit cells:
– The edge length = 2 radii
88
Bonding in Solids
• For face-centered cubic unit cells:
– The face diagonal = 2 x edge length.
– The diagonal length = 4 radii.
89
Bonding in Solids
• For body-centered cubic unit cells:
– The body diagonal = 3 x edge length.
– The diagonal length = 4 radii.
90
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
.
. .
91
Band Theory of Metals
• Sodium’s 3s orbitals can interact to
produce overlapping orbitals
92
Band Theory of Metals
• The 3s orbitals can also overlap with unfilled 3p orbitals
93
Band Theory of Metals
• Insulators have a large gap between the s and p
bands.
– Gap is called the forbidden zone.
• Semiconductors have a small gap between the
bands.
94
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%.
95
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?
96
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
Synthesis Question
2
K molatm L
4
CO of mol 748,000n
K 3730.08206
L 107.633atm 300
RTPVn
nRTPV
98
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?
13 Liquids and Solids