Ch10 Zumdahl Lecture Part21

1

Copyright © Houghton Mifflin Company. All rights reserved.Chapter 10 | Slide 1

Three CubicUnit Cellsand the

CorrespondingLattices

Figure 10.9

Simple Cubic

Cubic Unit Cells -Simple Cubic

• 8 particles, one at each corner ofa cube

• 1/8th of each particle lies in theunit cell– each particle part of 8 cells– 1 particle in each unit cell

• 8 corners x 1/8• edge of unit cell = twice the

radius• coordination number of 6

2r

Body-Centered Cubic

2

Cubic Unit Cells -Body-Centered Cubic

• 9 particles, one at each corner of a cube+ one in center

• 1/8th of each corner particle lies in the unitcell– 2 particles in each unit cell

• 8 corners x 1/8 + 1 center• edge of unit cell = (4/√ 3) times the radius

of the particle• coordination number of 8

!

4r

3

Face-Centered Cubic

Cubic Unit Cells -Face-Centered Cubic

• 14 particles, one at each corner of a cube +one in center of each face

• 1/8th of each corner particle + 1/2 of faceparticle lies in the unit cell– 4 particles in each unit cell

• 8 corners x 1/8 + 6 faces x 1/2• edge of unit cell = 2√ 2 times the radius of the

particle• coordination number of 12

22r


Concept Check

Determine the number of metal atoms in a unitcell if the packing is:

a) Simple cubicb) Body centered cubic

3


Concept Check

A metal crystallizes in a face-centered cubicstructure.

Determine the relationship between the radiusof the metal atom and the length of an edge ofthe unit cell.

Problem

Polonium crystallizes in a simple cubicarrangement, with the edge of a unit cellhaving a length d=334pm. What is theradius (pm) of a polonium atom?

fcc = 4 atoms/uc, Al = 26.982 g/mol, 1 mol = 6.022 x 1023 atoms

cm 1043.1m 10

cm 1

pm 1

m 10pm 431

8

2-

-12!

"=""

g 10792.1mol 1

g 982.26

atoms106.022

mol 1Al atoms 4 22

23

!"="

""

Calculate the density of Al if it crystallizes in a fcc andhas a radius of 143 pm

the accepted density of Al at 20°C is 2.71 g/cm3, so theanswer makes sense

face-centered cubic, r = 143 pm

density, g/cm3

Check:

Solution:

ConceptPlan:

Relation-ships:

Given:Find:

1 cm = 102 m, 1 pm = 10-12 m

lr Vmassfcc

dm, V# atoms x mass 1 atom

V = l3, l = 2r√ 2 , d = m/Vd = m/V

l = 2r√2 V = l3

face-centered cubic, r = 1.43 x 10-8 cm, m = 1.792 x10-22 g

density, g/cm3

cm 10544.0cm)(1.414) 1043.1(222 -88!=!==

"rl

( )323

383

cm 10618.6

cm 10045.4

!

!

"=

"== lV

3cm

g

323

22

71.2

cm 10618.6

g 10792.1

=

!

!==

"

"

V

md

Problem

Silver metal crystallizes in face centeredcubic arrangement with the edge of a unitcell having a length of d=407pm. What isthe radius (pm) of a silver atom?

What is the density of silver?


4

Crystal Structures

Another way to envision crystal structuresis to think of atoms stacking in layers.

Closest-Packed StructuresFirst Layer

• with spheres, it is more efficient to offseteach row in the gaps of the previousrow than to line-up rows and columns

Closest-Packed StructuresSecond Layer

• the second layer atoms can sit directly overthe atoms in the first – called an AA pattern

· or the second layer can sit over the holes in the first – called an AB pattern

Closest-Packed StructuresThird Layer – with Offset 2nd Layer

• the third layer atoms can align directly over theatoms in the first – called an ABA pattern

· or the third layer can sit over the uncovered holes in the first – called an ABC pattern

Hexagonal Closest-Packed

Cubic Closest-PackedFace-Centered Cubic

5


Hexagonal Closest Packing

Figure 10.14

Cubic Closest-Packed Structures

Figure 10.15


The Indicated Sphere Has 12 NearestNeighbors

• Each sphere inboth ccp and hcphas 12 equivalentnearest neighbors

Figure 10.16

Classifying Crystalline Solids

• classified by the kinds of units found• sub-classified by the kinds of attractive forces holding the

units together• molecular solids are solids whose composite units are

molecules• ionic solids are solids whose composite units are ions• atomic solids are solids whose composite units are

atoms– nonbonding atomic solids (Group 8A) are held together by

dispersion forces– metallic atomic solids are held together by metallic bonds– network covalent atomic solids are held together by covalent

bonds

6

Molecular Solids

• the lattice site are occupiedby molecules

• the molecules are heldtogether by intermolecularattractive forces– dispersion forces, dipole-dipole

attractions, and H-bonds (if it isable to)

• because the attractive forcesare weak, they tend to havelow melting point– generally < 300°C

Ionic Solids• held together by attractions between

opposite charges– nondirectional– therefore every cation attracts all anions

around it, and vice versa• the coordination number represents the

number of close cation-anion interactions inthe crystal

• the higher the coordination number, themore stable the solid– lowers the potential energy of the solid

Ionic Solids

CsClcoordination number = 8

Cs+ = 167 pmCl─ = 181 pm

NaClcoordination number = 6

Na+ = 97 pmCl─ = 181 pm

Lattice Holes

Simple CubicHole

OctahedralHole

TetrahedralHole

7

Lattice Holes• in hexagonal closest packed or cubic closest

packed lattices there are 8 tetrahedral holesand 4 octahedral holes per unit cell

• in simple cubic there is 1 hole per unit cell• number and type of holes occupied

determines formula (empirical) of salt

= Octahedral

= Tetrahedral

Cesium Chloride Structures

• coordination number = 8• ⅛ of each Cl─ inside the

unit cell• whole Cs+ inside the unit

cell– cubic hole = hole in simple

cubic arrangement of Cl─ions

• Cs:Cl = 1: (8 x ⅛), thereforethe formula is CsCl

Rock Salt Structures

• coordination number = 6• Cl─ ions in a face-centered cubic

arrangement– ⅛ of each corner Cl─ inside the unit cell– ½ of each face Cl─ inside the unit cell

• each Na+ in holes between Cl─– octahedral holes– 1 in center of unit cell– ¼ of each edge Na+ inside the unit cell

• Na:Cl = (¼ x 12) + 1: (⅛ x 8) + (½ x 6) = 4:4= 1:1,

• therefore the formula is NaCl

Zinc Blende Structures

• coordination number = 4• S2─ ions in a face-centered cubic

arrangement– ⅛ of each corner S2─ inside the unit cell– ½ of each face S2─ inside the unit cell

• each Zn2+ in holes between S2─

– tetrahedral holes– 1 whole in ½ the holes

• Zn:S = (4 x 1) : (⅛ x 8) + (½ x 6) = 4:4 =1:1,

• therefore the formula is ZnS

8

Fluorite Structures• coordination number = 4• Ca2+ ions in a face-centered cubic

arrangement– ⅛ of each corner Ca2+ inside the

unit cell– ½ of each face Ca2+ inside the

unit cell• each F─ in holes between Ca2+

– tetrahedral holes– 1 whole in all the holes

• Ca:F = (⅛ x 8) + (½ x 6): (8 x 1) =4:8 = 1:2,

• therefore the formula is CaF2

Atomic Solids

• atomic solids are solids that contain atomsat the lattice points– nonbonding atomic solids (Group 8A)

are held together by London dispersionforces

– metallic atomic solids are held togetherby metallic bonds

– network covalent atomic solids are heldtogether by covalent bonds

Nonbonding Atomic Solids (Group 8)

• noble gases (solid)• held together by weak dispersion

forces– very low melting

• tend to arrange atoms in closest-packed structure– either hexagonal cp or cubic cp– maximizes attractive forces and minimizes

energy

Metallic Atomic Solids

• solid held together by metallic bonds– strength varies with sizes and charges of cations

• coulombic attractions

• melting point varies• mostly closest packed arrangements of the

lattice points– cations

9

Metallic Bonding• metal atoms release their valence

electrons• metal cation “islands” fixed in a “sea” of

mobile electrons

e-

e- e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

+ + + + + + + + +

+ + + + + + + + +

+ + + + + + + + +

Network Covalent Solids

• atoms attached to its nearest neighbors bycovalent bonds

• because of the directionality of the covalentbonds, these do not tend to form closest-packed arrangements in the crystal

• because of the strength of the covalentbonds, these have very high melting points– generally > 1000°C

• dimensionality of the network affects otherphysical properties

The Diamond Structure:a 3-Dimensional Network

• the carbon atoms in a diamond each have 4covalent bonds to surrounding atoms– sp3

– tetrahedral geometry

• this effectively makes each crystal one giantmolecule held together by covalent bonds– you can follow a path of covalent bonds from any

atom to every other atom

Properties of Diamond• very high melting, ~3800°C

– need to overcome some covalent bonds• very rigid

– due to the directionality of the covalentbonds

• very hard– due to the strong covalent bonds holding

the atoms in position– used as abrasives

• electrical insulator• thermal conductor

– best known• chemically very nonreactive

Figure 10.12

10

The Graphite Structure:a 2-Dimensional Network

• in graphite, the carbon atoms in a sheet arecovalently bonded together– forming 6-member flat rings fused together

• similar to benzene• bond length = 142 pm

– sp2

• each C has 3 sigma and 1 pi bond– trigonal-planar geometry– each sheet a giant molecule

• the sheets are then stacked and held together bydispersion forces– sheets are 341 pm apart

Properties of Graphite• hexagonal crystals• high melting, ~3800°C

– need to overcome some covalent bonding• slippery feel

– because there are only dispersion forcesholding the sheets together, they can slide pasteach other

• glide planes– lubricants

• electrical conductor– parallel to sheets

• thermal insulator• chemically very nonreactive

Silicates

• Most common network covalent atomic solid• ~90% of earth’s crust• extended arrays of Si−O

– sometimes with Al substituted for Si –aluminosilicates

• glass is the amorphous form

Quartz -SiO2

• 3-dimensional array of Sicovalently bonded to 4 O– tetrahedral

• melts at ~1600°C• very hard

Figure 10.26

11


Classification of Solids- Summary

Table 10.3


Bonding Models for Metals

• Electron Sea Model• Band Model (MO Model)


The Electron Sea Model

• A regular array of cations in a “sea” of valence electrons.

Figure 10.18

Band Theory

• the structures of metals and covalentnetwork solids result in every atom’sorbitals being shared by the entirestructure

• for large numbers of atoms, this resultsin a large number of molecular orbitalsthat have approximately the sameenergy, we call this an energy band

12

Band Theory• when 2 atomic orbitals combine they produce both

a bonding and an antibonding molecular orbital• when many atomic orbitals combine they produce

a band of bonding molecular orbitals and a bandof antibonding molecular orbitals

• the band of bonding molecular orbitals is calledthe valence band

• the band of antibonding molecular orbitals iscalled the conduction band


The Band Model for Magnesium• Virtual continuum of levels, called bands.

Figure 10.20

Band Gap• at absolute zero, all the electrons will occupy

the valence band• as the temperature rises, some of the

electrons may acquire enough energy tojump to the conduction band

• the difference in energy between thevalence band and conduction band is calledthe band gap– the larger the band gap, the fewer electrons

there are with enough energy to make the jump

Band Gap and Conductivity

• the more electrons at any one time that a substance has inthe conduction band, the better conductor of electricity it is

• if the band gap is ~0, then the electrons will be almost aslikely to be in the conduction band as the valence bandand the material will be a conductor– metals– the conductivity of a metal decreases with temperature

• if the band gap is small, then a significant number of theelectrons will be in the conduction band at normaltemperatures and the material will be a semiconductor– graphite– the conductivity of a semiconductor increases with temperature

• if the band gap is large, then effectively no electrons willbe in the conduction band at normal temperatures and thematerial will be an insulator

13

Doping Semiconductors• doping is adding impurities to the semiconductor’s crystal

to increase its conductivity• goal is to increase the number of electrons in the

conduction band• n-type semiconductors do not have enough electrons

themselves to add to the conduction band, so they aredoped by adding electron rich impurities

• p-type semiconductors are doped with an electrondeficient impurity, resulting in electron “holes” in thevalence band. Electrons can jump between these holes inthe valence band, allowing conduction of electricity

Diodes

• when a p-type semiconductor adjoins an n-type semiconductor, the result is an p-njunction

• electricity can flow across the p-n junction inonly one direction – this is called a diode

• this also allows the accumulation of electricalenergy – called an amplifier


Metal Alloys

• Substitutional Alloy – some of thehost metal atoms are replaced byother metal atoms of similar size

• Interstitial Alloy – some of the holesin the closest packed metal structureare occupied by small atoms


Two Typesof Alloys

Figure 10.21

14

53

Vaporization• molecules in the liquid are constantly in motion• the average kinetic energy is proportional to the

temperature• however, some molecules have more kinetic

energy than the average• if these molecules are at the surface, they may

have enough energy to overcome the attractiveforces– the larger the surface area, the faster the

rate of evaporation• this will allow them to escape the liquid and

become a vapor

54

Distribution of Thermal Energy• only a small fraction of the molecules in a liquid have enough energy to

escape• as the temperature increases, the fraction of the molecules with “escape

energy” increases• the higher the temperature, the faster the rate of evaporation

Figure 10.41

55

Condensation

• some molecules of the vapor will loseenergy through molecular collisions

• the result will be that some of the moleculeswill get captured back into the liquid whenthey collide with it

• also some may stick and gather together toform droplets of liquid– particularly on surrounding surfaces

• we call this process condensation

56

Evaporation vs. Condensation• in an open container, the vapor molecules generally

spread out faster than they can condense• the net result is that the rate of vaporization is greater than

the rate of condensation, and there is a net loss of liquid• however, in a closed container, the vapor is not allowed to

spread out indefinitely• the net result in a closed container is that at some time the

rates of vaporization and condensation will be equal

15

57

Effect of Intermolecular Attraction onEvaporation and Condensation

• the weaker the attractive forces between molecules, theless energy they will need to vaporize

• also, weaker attractive forces means that more energy willneed to be removed from the vapor molecules before theycan condense

• the net result will be more molecules in the vapor phase,and a liquid that evaporates faster – the weaker theattractive forces, the faster the rate of evaporation

• liquids that evaporate easily are said to be volatile– e.g., gasoline, fingernail polish remover– liquids that do not evaporate easily are called

nonvolatile• e.g., motor oil

58

Energetics of Vaporization

• when the high energy molecules are lostfrom the liquid, it lowers the average kineticenergy

• if energy is not drawn back into the liquid,its temperature will decrease – therefore,vaporization is an endothermic process– and condensation is an exothermic process

59

Heat of Vaporization• the amount of heat energy required to vaporize one

mole of the liquid is called the Heat ofVaporization, ΔHvap– sometimes called the enthalpy of vaporization

• always endothermic, therefore ΔHvap is +• −ΔHcondensation = ΔHvaporization

• Examples;ΔHvap of water = 44.0kJ/mol

ΔHvap of isopropanol = 45.4kJ/mol ΔHvap of acetone = 29.1kJ/mol

Calculate the mass of water that can bevaporized with 155 kJ of heat at 100°C

since the given amount of heat is almost 4x the ΔHvap, the amount of watermakes sense

1 mol H2O = 40.7 kJ1 mol = 18.02 g

155 kJg H2O

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

kJ 40.7

mol 1

OH g 8.66 mol 1

g 8.021

kJ 40.7

OH mol 1 kJ 551

2

2 =!!

kJ mol H2O g H2Omol 1

g 02.18

16

61

Dynamic Equilibrium• once the rates of vaporization and condensation

are equal, the total amount of vapor and liquid willnot change

• evaporation and condensation are still occurring,but because they are opposite processes, there isno net gain or loss or either vapor or liquid

• when two opposite processes reach the same ratewe call it a dynamic equilibrium– this does not mean there are equal amounts of vapor

and liquid – it means that they are changing by equalamounts


The Rates of Condensation andEvaporation

Figure 10.39

63

Dynamic Equilibrium

Figure 10.38

64

Vapor Pressure• the pressure exerted by the vapor at equilibrium is called

the vapor pressure• the weaker the attractive forces between the molecules,

the more molecules will be in the vapor• therefore, the weaker the attractive forces, the higher

the vapor pressure– the higher the vapor pressure, the more volatile the

liquid

17


Vapor Pressure

Figure 10.40

66

Dynamic Equilibrium

• a system can respond to changes in theconditions while at equilibrium

• the system shifts its position torelieve or reduce the effects of thechange

67

Vapor Pressure vs. Temperature

• increasing the temperature increases thenumber of molecules able to escape theliquid

• the net result is that as the temperatureincreases, the vapor pressure increases

• small changes in temperature can make bigchanges in vapor pressure


VaporPressure vs.Temperature

Figure 10.42

18

69

Boiling Point

• when the temperature of a liquid reaches apoint where its vapor pressure is the sameas the external pressure then bubbles form(not just at the surface)

• called boiling and the temperaturerequired to have the vapor pressure =external pressure is the boiling point

70

Boiling Point• the normal boiling point is the temperature at which

the vapor pressure of the liquid = 1 atm• the lower the external pressure, the lower the boiling point

of the liquid


Concept Check

• What is the vapor pressure of water at100°C? How do you know?


Superheating

• Liquid can be heated to a temperaturehigher than its boiling point without actuallyboiling

• Surface tension suppresses the formationof bubbles

19

73

Clausius-Clapeyron Equation• the graph of vapor pressure vs. temperature is an

exponential growth curve

RT

H

vap

vap

P

!"

•= e#

Figure 10.4274

Clausius-Clapeyron Equation• the logarithm of the vapor pressure vs. inverseabsolute temperature is a linear function

)(lnT

1

R

H)ln(P

RT

H)(ln)ln(P

ln)(ln)ln(P

ln)ln(P

vap

vap

vap

vap

RT

H

vap

RT

H

vap

vap

vap

!

!

!

!

+""#

$%%&

' ()=

""#

$%%&

' ()+=

""

#

$

%%

&

'+=

""

#

$

%%

&

'•=

()

()

e

e

Figure 10.42

75

Clausius-Clapeyron Equation• if you know the heat of vaporization and the vapor

pressure at one temp, then we can calculate thevapor pressure at another temp– remember: the vapor pressure at the normal boiling

point is 760 torr

!!"

#$$%

&'

(=!

!"

#$$%

&

'(

='

(+==

(+

=

12

vap

2

1

12121

22

11

T

1

T

1

R

H

P

Pln

)11

()ln(P)ln(P

)ln(P)ln(P

Cconstant ln

TTR

H

RT

HC

RT

H

vap

vapvap

)

!

T1

= 64.6 + 273.15 = 337.8 K

T2

=12.0 + 273.15 = 285.2 K

!

lnP

1

P2

"

# $

%

& ' =

35,200J

mol

8.314J

mol•K

1

285.2 K(

1

337.8 K

"

# $

%

& '

Calculate the vapor pressure of methanolat 12.0°C

the units are correct, the size makes sense since the vapor pressure is lower atlower temperatures

T1 = BP = 64.6°C P1 = 760 torr ΔHvap = 35.2 kJ/mol T2 = 12.0°C

P2 ?

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

P1, T1, T2, ΔHvap P2

!

lnP1

P2

"

# $

%

& ' =

(Hvap

R

1

T2)1

T1

"

# $

%

& '

T(K) = T(°C) + 273.15

T1 = BP = 337.8K P1 = 760 torr ΔHvap = 35.2 kJ/mol T2 = 285.2K°C

P2, torr

!

lnP1

P2

"

# $

%

& ' = 2.31

!

lnP1

P2

"

# $

%

& ' =

(Hvap

R

1

T2)1

T1

"

# $

%

& '

!

760 torr

P2

= e2.31

=10.090

P2

= 75.3 torr

20

77

Sublimation and Deposition• Solids also have vapor pressure• surface molecules with sufficient energy may

break free from the surface and become a gas –this process is called sublimation

• the capturing of vapor molecules into a solid iscalled deposition

• the solid and vapor phases exist in dynamicequilibrium in a closed container

solid gassublimation

deposition

78

Melting = Fusion

• as a solid is heated, its temperature rises andthe molecules vibrate more vigorously

• once the temperature reaches the meltingpoint, the molecules have sufficient energy toovercome some of the attractions that holdthem in position and the solid melts (or fuses)

• melting is an endothermic process• the opposite of melting is freezing

79

Heat of Fusion• the amount of heat energy required to melt one

mole of the solid is called the Heat of Fusion, ΔHfus– sometimes called the enthalpy of fusion

• always endothermic, therefore ΔHfus is +• generally much less than ΔHvap• ΔHsublimation = ΔHfusion + ΔHvaporization


Concept Check

• Which would you predict should be larger fora given substance: ΔHvap or ΔHfus?

• Explain why.

21


Phase Diagrams

• A convenient way of representing the phases of asubstance as a function of temperature andpressure:– Triple point (temperature/pressure condition where all

three states exist simultaneously– Critical point (furthest point on the vapor pressure

curve)

• for most substances, freezing point increasesas pressure increases

82

Phase Diagrams

Pres

sure

Temperature

vaporization

condensation

criticalpoint

triplepoint

Solid Liquid

Gas

1 atm

normalmelting pt.

normalboiling pt.

Fusion Curve

Vapor PressureCurve

SublimationCurve

melting

freezing

sublimation

deposition

83

Phase Diagram Navigation

Temperature

Pres

sure

Ice Water

Steam

1 atm

84

Phase Diagram of Water

Temperature

Pres

sure

criticalpoint

374.1°C217.7 atm

triplepoint

Ice Water

Steam

1 atm

normalboiling pt.

100°C

normalmelting pt.

0°C

0.01°C0.006 atm

22

85

Phase Diagram of CO2Pr

essu

re

Temperature

criticalpoint

31.0°C72.9 atm

triplepoint

Solid Liquid

Gas

1 atm

-56.7°C5.1 atm

normalsublimation pt.

-78.5°C

86

The Critical Point• the temperature required to produce a

supercritical fluid is called the criticaltemperature

• the pressure at the critical temperature iscalled the critical pressure

• at the critical temperature or highertemperatures, the gas cannot becondensed to a liquid, no matter how highthe pressure gets

87

Supercritical Fluid

• as a liquid is heated in a sealed container, more vaporcollects causing the pressure inside the container to rise– and the density of the vapor to increase– and the density of the liquid to decrease

• at some temperature, the meniscus between the liquid andvapor disappears and the states commingle to form asupercritical fluid

• supercritical fluid have properties of both gas and liquidstates

Considering the following phase diagram for UF6, which ofthe following conditions is true?

A. At 1 atm, UF6 boils.B. UF6 melts at 1 atm.C. At 1 atm, UF6 sublimes.D. UF6 does not boil until it

reaches the criticaltemperature.

Solid Liquid

Gas

Atmosphericpressure

Temp (°F)

P (atm)

23

Considering the following phase diagram for UF6, which ofthe following conditions is true?

A. At 1 atm, UF6 boils.B. UF6 melts at 1 atm.C. At 1 atm, UF6 sublimes.D. UF6 does not boil until it

reaches the criticaltemperature.

SolidLiquid

Gas

Atmosphericpressure

Temp (°F)

P (atm)


Concept Check

• As intermolecular forces increase, what happens toeach of the following? Why?– Boiling point– Viscosity– Surface tension– Enthalpy of fusion– Freezing point– Vapor pressure– Heat of vaporization


Oxygen has a Tt = 54.3K, Pt = 1.14mm Hg, Tc = 154.6K and Pc = 49.77atm. The densityof the liquid is 1.14/cm3, and the density of the solid is 1.33g/cm3. (small t stands for triplepoint and c for critical point)

A) Sketch a phase diagram (pressure (atm) versus temp (°C )) for oxygen, and label all thepoints of interest.

B) Using the phase diagram that you drew, tell what the physical state is under thefollowing conditions:

i) T = - 210°C, P = 1.5atm

ii) T = -100°C, P = 66atm

C) On your phase diagram above, trace a path starting from 0.0011atm and -225°C. First,increase P to 35atm while keeping the T constant. Next, increase T to -150°C whilekeeping P constant. Then, decrease P to 1.0atm while keeping T constant. Finally,decrease T to -215°C while keeping P constant. What is your starting phase and what isyour final phase?

Practice Problem

Ch10 Zumdahl Lecture Part21

Documents