1 chapter 11 liquids and solids. 2 chapter goals 1. kinetic-molecular description of liquids and...

1

CHAPTER 11

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 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


Schematic representation of the three common states of matter.

gas liquid solidcool cool

heatheat

6


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


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


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?

10


Na F Ca F Ca O+ - 2+2

2+ 2-

11


2. Hydrogen bonding Consider H2O a very polar molecule.

12


2. Hydrogen bonding Consider H2O a very polar molecule.

One Water Molecule Water Dimer

Water Tetramer Ice Structure

13


14


3. Dipole-dipole interactions Consider BrF a polar molecule.

15


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


In a group of Ar atoms the temporary dipole in one atom induces other atomic dipoles.

17


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.

19

The Liquid State

An example of viscosity of two liquids.

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.

21

The Liquid State

Floating paper clip demonstration of surface tension.

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.

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

This is an animation of evaporation

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?


29

The Liquid State

Vapor Pressure as a function of temperature.

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.


33

The Liquid State

kJ 37.6 J 1076.3J?

)C0.400.85)( g(4.184 102.00 J ?

4

oCg

J2o

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


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

38

The Liquid State

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.


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?


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

50

The Liquid State

Alkanes

-200

-150

-100

-50

0

50

16 30 44 58 72

Molar Mass

Boi

ling

Poi

nt (

C)

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

54

The Liquid State

VIA Hydrides

-100

-50

0

50

100

150

18 34 81 130

Molar Mass

Boi

ling

Poi

nt (

C)

55

The Liquid State

At the molecular level what happens when a species boils?

56

The Liquid State

Example 13-6: Arrange the following substances in order of increasing boiling points.

C2H6, NH3, Ar, NaCl, AsH3


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


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

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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


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


? 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.

66

Phase Diagrams (P versus T)

Compare water’s phase diagram to carbon dioxide’s phase diagram.

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


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


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


A face centered cubic (fcc) unit cell has a cubic unit cell structure with an extra atom, ion, or molecule in each face.

73


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

.

. .

90

Band Theory of Metals

Sodium’s 3s orbitals can interact to produce overlapping orbitals

91


The 3s orbitals can also overlap with unfilled 3p orbitals

92


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

96

Synthesis Question

2

K molatm L

4

CO of mol 748,000n

K 3730.08206

L 107.633atm 300RT

PVn

nRTPV

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?

98

End of Chapter 13

Our understanding of Band Theory was a major breakthrough in semiconductor knowledge.

Why computers work!

1 chapter 11 liquids and solids. 2 chapter goals 1. kinetic-molecular description of liquids and...

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