chem chapter11 lec

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Chapter 11

Liquids,

Solids, andIntermolecular

Forces

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.

Nivaldo Tro

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA



Tro, Chemistry: A Molecular Approach 2

Comparisons of the

States of Matter• the solid and liquid states have a much higher density

than the gas state

therefore the molar volume of the solid and liquid states ismuch smaller than the gas state

• the solid and liquid states have similar densities

generally the solid state is a little denser

notable exception: ice is less dense than liquid water

• the molecules in the solid and liquid state are in closecontact with each other, while the molecules in a gasare far apart




Freedom of Motion

• the molecules in a gas have complete freedom ofmotion their kinetic energy overcomes the attractive forces between

the molecules

• the molecules in a solid are locked in place, they

cannot move around though they do vibrate, they don’t have enough kinetic

energy to overcome the attractive forces

• the molecules in a liquid have limited freedom – they

can move around a little within the structure of theliquid they have enough kinetic energy to overcome some of the

attractive forces, but not enough to escape each other




Properties of the 3 Phases of Matter

•Fixed = keeps shape when placed in a container

•Indefinite = takes the shape of the container

State Shape Volume Compressible Flow Strength of

Intermolecular

Attractions

Solid Fixed Fixed No No very strong

Liquid Indef. Fixed No Yes moderate

Gas Indef. Indef. Yes Yes very weak




Kinetic - Molecular Theory

• the properties of solids, liquids, and gases can beexplained based on the kinetic energy of themolecules and the attractive forces betweenmolecules

• kinetic energy tries to give molecules freedom ofmotion

degrees of freedom = translational, rotational,vibrational

• attractive forces try to keep the molecules together

• kinetic energy depends only on the temperature

KE = 1.5 kT




Gas Structure

Gas molecules are rapidly

moving in random straight

lines and free from sticking

to each other.




Explaining the Properties of Solids

• the particles in a solid are packed closetogether and are fixed in position

though they may vibrate

• the close packing of the particles results insolids being incompressible

• the inability of the particles to move around

results in solids retaining their shape andvolume when placed in a new container; and prevents the particles from flowing




Solids• some solids have their particles

arranged in an orderly geometric

pattern – we call these crystalline

solids salt and diamonds

• other solids have particles that do not

show a regular geometric pattern over

a long range – we call theseamorphous solids

plastic and glass




Explaining the Properties of Liquids

• they have higher densities than gases

because the molecules are in close contact

• they have an indefinite shape because the

limited freedom of the molecules allowsthem to move around enough to get to the

container walls

• but they have a definite volume becausethe limit on their freedom keeps them

from escaping the rest of the molecules




Compressibility




Phase Changes




Why are molecules attracted to

each other?• intermolecular attractions are due to attractive forces between

opposite charges

+ ion to - ion

+ end of polar molecule to - end of polar molecule

H-bonding especially strong

even nonpolar molecules will have temporary charges

• larger the charge = stronger attraction

• longer the distance = weaker attraction

• however, these attractive forces are small relative to the

bonding forces between atoms generally smaller charges

generally over much larger distances




Trends in the Strength of

Intermolecular Attraction?• the stronger the attractions between the atoms or

molecules, the more energy it will take to separate

them• boiling a liquid requires we add enough energy to

overcome the attractions between the molecules or

atoms

• the higher the normal boiling point of the liquid, the

stronger the intermolecular attractive forces




Attractive Forces

+ - + - + - + -

+

++

+

_

_ _ _

+ + + + + + +

- - - - - - -

++ + +

+

--

- -

-




Dispersion Forces

• fluctuations in the electron distribution in atoms andmolecules result in a temporary dipole

region with excess electron density has partial ( ─) charge

region with depleted electron density has partial (+) charge

• the attractive forces caused by these temporary dipolesare called dispersion forces

aka London Forces

• all molecules and atoms will have them• as a temporary dipole is established in one molecule, itinduces a dipole in all the surrounding molecules




Dispersion Force




Size of the Induced Dipole

• the magnitude of the induced dipole depends onseveral factors

• polarizability of the electrons

volume of the electron cloud

larger molar mass = more electrons = larger

electron cloud = increased polarizability =

stronger attractions

• shape of the molecule

more surface-to-surface contact = larger

induced dipole = stronger attraction




Effect of Molecular Size

on Size of Dispersion Force

Noble Gases are all

nonpolar atomic

elements.

As the molar mass

increases, the number

of electrons increase.

Therefore the strength

of the dispersion

forces increases.

The stronger the

attractive forces between the molecules,

the higher the boiling

point will be.




Relationship between Induced Dipole and Molecular Size

-300

-250

-200

-150

-100

-50

0

50

100

150

200

250

1 2 3 4 5 6

Period

B o i l i n g P o i n

t , ° C

BP, Noble Gas

BP, Halogens

BP, XH4

P ti f St i ht Ch i Alk




Name Molar Mass BP, °C MP, °C Density, g/mL

Methane 16 -162 -183 0.47

Ethane 30 -89 -183 0.57

Propane 44 -42 -188 0.5

Butane 58 0 -138 0.58

Pentane 72 36 -130 0.56

Hexane 86 69 -95 0.66

Heptane 100 98 -91 0.68Octane 114 126 -57 0.7

Nonane 128 151 -54 0.72

Decane 142 174 -30 0.74

Undecane 156 196 -26 0.75

Dodecane 170 216 -10 0.76Tridecane 184 235 -5 0.76

Tetradecane 198 254 6 0.77

Pentadecane 212 271 10 0.79

Hexadecane 226 287 18 0.77

Properties of Straight Chain Alkanes

Non-Polar Molecules




Boiling Points of n-Alkanes




n-Alkane Boiling & Melting Points

-300

-200

-100

0

100

200

300

400

500

0 100 200 300 400 500Molar Mass

T e m p e r a t u r e ,

° C

BP, n-alkane

MP, n-alkane




Effect of Molecular Shape

on Size of Dispersion Force




Alkane Boiling Points

-20

0

20

40

60

80

100

120

140

58 72 86 100 114

Molar Mass

T e m p e r a t u r e ,

° C

n-alkanes

iso-alkanes

Alkane Boiling Points

• branched chains

have lower BPs

than straight

chains

• the straight chain

isomers have

more surface-to-surface contact




Practice – Choose the Substance in Each Pair with

the Highest Boiling Point

a) CH4 CH3CH2CH2CH3

b) CH3CH2CH=CHCH2CH3 cyclohexane

C

C

C

C

H

H H

HH

H H

H

H

HC

H

HHH

CC

CC

H

H

H

HH

H

CC

H

HH

H

HH

H

H H

H

H

HC

CH

H

HC

CH C

H

H

C






a) CH4 CH3CH2CH2CH3

b) CH3CH2CH=CHCH2CH3 cyclohexane

both molecules

are nonpolar

larger molar

mass

both molecules arenonpolar

flat molecule larger

surface-to-surface

contact

C

C

C

C

H

H H

HH

H H

H

H

HC

H

HHH

CC

CC

H

H

H

HH

H

CC

H

HH

H

HH

H

H H

H

H

HC

CH

H

HC

CH C

H

H

C




Dipole-Dipole Attractions

• polar molecules have a permanent dipole

because of bond polarity and shape

dipole moment

as well as the always present induced dipole

• the permanent dipole adds to the attractive forces between the molecules

raising the boiling and melting points relative to nonpolarmolecules of similar size and shape




Effect of Dipole-Dipole Attraction on

Boiling and Melting Points

M l B ili Di l



30

MolarMass

BoilingPoint

DipoleSize

CH3CH2CH3 44.09 -42°C 0.08 D

CH3-O-CH3 46.07 -24°C 1.30 D

CH3 - CH=O 44.05 20.2°C 2.69 D

CH3-C N 41.05 81.6°C 3.92 D




or



a) CH2FCH2F CH3CHF2

b)

C C

H

H

HHF

F

C C

H

H

HFH

F




or



more polar

polar nonpolar

a) CH2FCH2F CH3CHF2

b)

C C

H

H

HHF

F

C C

H

H

HFH

F




Attractive Forces and Solubility• Solubility depends on the attractive forces of solute

and solvent moleculesLike dissolves Like

miscible liquids will always dissolve in each other

• polar substance dissolve in polar solvents

hydrophilic groups = OH, CHO, C=O, COOH, NH2,Cl

• nonpolar molecules dissolve in nonpolar solvents

hydrophobic groups = C-H, C-C

• Many molecules have both hydrophilic andhydrophobic parts - solubility becomes competition between parts




Immiscible Liquids




Polar Solvents

Water

Dichloromethane

(methylene chloride)

Ethanol(ethyl alcohol)




Nonpolar Solvents

CH3

CH2

CH

2

CH2

CH

2

CH3

n-hexane

CH

CHCH

CH

CHCCH3

toluene

C

Cl

ClClCl

carbon tetrachloride




Hydrogen Bonding• When a very electronegative atom is bonded to

hydrogen, it strongly pulls the bonding electronstoward it

O-H, N-H, or F-H

• Since hydrogen has no other electrons, when itloses the electrons, the nucleus becomesdeshielded

exposing the H proton

• The exposed proton acts as a very strong centerof positive charge, attracting all the electronclouds from neighboring molecules




H-Bonding

HF




H-Bonding in Water




Relationship between H-bonding

and Intermolecular Attraction

-200

-150

-100

-50

0

50

100

150

1 2 3 4 5

Period

B o i l i n P o i

n t , ° C

BP, HX

BP, H2X

BP, H3X

BP, XH4

CH4

NH3

HF

H2O

SiH4

GeH4

SnH4

H2S H2Se

H2Te




Practice – Choose the substance in each pair that

is a liquid at room temperature (the other is a gas)

a) CH3OH CH3CHF2

b) CH3-O-CH2CH3 CH3CH2CH2 NH2





is a liquid at room temperature (the other is a gas)

a) CH3OH CH3CHF2

b) CH3-O-CH2CH3 CH3CH2CH2 NH2

can H-bond

can H-bond





is more soluble in water

a) CH3OH CH3CHF2

b) CH3CH2CH2CH3 CH3Cl





is more soluble in water

a) CH3OH CH3CHF2

b) CH3CH2CH2CH3 CH3Cl

can H-bond with H2O

more polar




Ion-Dipole Attraction

• in a mixture, ions from an ionic compound areattracted to the dipole of polar molecules

• the strength of the ion-dipole attraction is one ofthe main factors that determines the solubility of

ionic compounds in water




Summary

• Dispersion forces are the weakest of theintermolecular attractions.

• Dispersion forces are present in all moleculesand atoms.

• The magnitude of the dispersion forcesincreases with molar mass

• Polar molecules also have dipole-dipoleattractive forces




Summary (cont’d)

• Hydrogen bonds are the strongest of the intermolecularattractive forces a pure substance can have

• Hydrogen bonds will be present when a molecule has

H directly bonded to either O , N, or F atomsonly example of H bonded to F is HF

• Ion-dipole attractions are present in mixtures of ioniccompounds with polar molecules.

• Ion-dipole attractions are the strongest intermolecularattraction

• Ion-dipole attractions are especially important inaqueous solutions of ionic compounds



Liquids

properties &

structure

S f T i



50

Surface Tension• surface tension is a property of liquids that results from

the tendency of liquids to minimize their surface area

• in order to minimize their surface area, liquids formdrops that are spherical as long as there is no gravity

• the layer of molecules on the surface behave differently

than the interior because the cohesive forces on the surface molecules have a

net pull into the liquid interior

• the surface layer acts like an elastic skin




Surface Tension

• because they have fewer neighborsto attract them, the surface

molecules are less stable than those

in the interior

have a higher potential energy

• the surface tension of a liquid is the

energy required to increase the

surface area a given amount at room temp, surface tension of H2O

= 72.8 mJ/m2




Factors Affecting Surface Tension

• the stronger the intermolecular attractive forces,

the higher the surface tension will be

• raising the temperature of a liquid reduces itssurface tension

raising the temperature of the liquid increases the

average kinetic energy of the moleculesthe increased molecular motion makes it easier to

stretch the surface




Surface Tension of Water vs. Temperature

50

55

60

65

70

75

80

-20 0 20 40 60 80 100 120

Temperature, °C

S u r

f a c e

T e n s i o

n ,

m J / m

2




Viscosity• viscosity is the resistance of a liquid to flow

1 poise = 1 P = 1 g/cm∙s

often given in centipoise, cP

• larger intermolecular attractions = larger viscosity

• higher temperature = lower viscosity




Viscosity of Water vs. Temperature

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100 120

Temperature, deg C

V i s c o s i t y

, c

P




Capillary Action

• the adhesive forces pull the surface liquid up theside of the tube, while the cohesive forces pullthe interior liquid with it

• the liquid rises up the tube until the force ofgravity counteracts the capillary action forces




Meniscus• the curving of the liquid surface in a thin

tube is due to the competition betweenadhesive and cohesive forces

• the meniscus of water is concave in aglass tube because its adhesion to theglass is stronger than its cohesion foritself

• the meniscus of mercury is convex in aglass tube because its cohesion for itselfis stronger than its adhesion for the glass

metallic bonds stronger than intermolecularattractions




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 toovercome the attractive forces

therefore – the larger the surface area,the faster the rate of evaporation

• this will allow them to escape theliquid and become a vapor




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

energy to escape

• but, as the temperature increases, the fraction of the moleculeswith “escape energy” increases

• the higher the temperature, the faster the rate of evaporation




Condensation

• some molecules of the vapor will lose energythrough molecular collisions

• the result will be that some of the molecules will

get captured back into the liquid when theycollide with it

• also some may stick and gather together to formdroplets of liquid

particularly on surrounding surfaces

• we call this process condensation




Evaporation vs. Condensation

• vaporization and condensation are opposite processes

• in an open container, the vapor molecules generallyspread out faster than they can condense

• the net result is that the rate of vaporization is greaterthan the rate of condensation, and there is a net loss ofliquid

• 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 timethe rates of vaporization and condensation will be equal

Effect of Intermolecular Attraction on




Effect of Intermolecular Attraction on

Evaporation and Condensation

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

• also, weaker attractive forces means that more energywill need to be removed from the vapor molecules

before they can condense

• the net result will be more molecules in the vapor phase, and a liquid that evaporates faster – the weakerthe attractive forces, the faster the rate ofevaporation

• 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




Energetics of Vaporization

• when the high energy molecules are lost fromthe liquid, it lowers the average kinetic energy

• 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

• vaporization requires input of energy toovercome the attractions between molecules

Heat of Vaporization




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

of the liquid is called the Heat of Vaporization, Hvap

sometimes called the enthalpy of vaporization

• always endothermic, therefore DHvap is +

• somewhat temperature dependent

DHcondensation

= -DHvaporization

Example 11.3 – Calculate the mass of water that



can be vaporized with 155 kJ of heat at 100°C

since the given amount of heat is almost 4x the DHvap,

the amount of water makes sense

1 mol H2O = 40.7 kJ, 1 mol = 18.02 g

155 kJ

g H2O

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

kJ40.7

mol1

OHg8.66

mol1

g8.021

kJ40.7

OHmol1 kJ551

2

2

kJ mol H2O g H2O

mol1

g02.18




Dynamic Equilibrium

• in a closed container, once the rates of vaporization andcondensation are equal, the total amount of vapor andliquid will not change

• evaporation and condensation are still occurring, but

because they are opposite processes, there is no netgain or loss or either vapor or liquid

• when two opposite processes reach the same rate sothat there is no gain or loss of material, we call it a

dynamic equilibrium this does not mean there are equal amounts of vapor and

liquid – it means that they are changing by equal amounts




Dynamic Equilibrium




Vapor Pressure

• the pressure exerted by the vapor when it is in dynamic

equilibrium with its liquid is called the vapor pressure

remember using Dalton’s Law of Partial Pressures to accountfor the pressure of the water vapor when collecting gases by

water displacement?

• the weaker the attractive forces between the molecules,

the more molecules will be in the vapor

• therefore, the weaker the attractive forces, thehigher the vapor pressure

the higher the vapor pressure, the more volatile the liquid

Vapor Liquid Dynamic Equilibrium




Vapor-Liquid Dynamic Equilibrium• if the volume of the chamber is increased, that will decrease the

pressure of the vapor inside

at that point, there are fewer vapor molecules in a given volume, causingthe rate of condensation to slow

• eventually enough liquid evaporates so that the rates of thecondensation increases to the point where it is once again as fastas evaporation

equilibrium is reestablished

• at this point, the vapor pressure will be the same as it was before




Dynamic Equilibrium

• a system in dynamic equilibrium can respond to

changes in the conditions

• when conditions change, the system shifts itsposition to relieve or reduce the effects of the

change




Vapor Pressure vs. Temperature

• increasing the temperature increases the number

of molecules able to escape the liquid

•the net result is that

as the temperatureincreases, the vapor pressure increases

• small changes in temperature can make big

changes in vapor pressure

• the rate of growth depends on strength of the

intermolecular forces

Vapor Pressure Curves




Vapor Pressure Curves

Temperature vs Vapor Pressure

0

100

200

300

400500

600

700

800

9001000

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150Temperature, °C

V a p o r

P r e s s u r e ,

m m

H

w ater

TiCl4

chloroform

ether

ethanol

acetone

760 mmHg

normal BP

100°C

BP Ethanol at 500 mmHg

68.1°C




Boiling Point• when the temperature of a liquid reaches a point

where its vapor pressure is the same as the

external pressure, vapor bubbles can formanywhere in the liquid

not just on the surface

• this phenomenon is what is called boiling andthe temperature required to have the vapor pressure = external pressure is the boiling point




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

Heating C r e of a Liq id




Heating Curve of a Liquid• as you heat a liquid, its

temperature increaseslinearly until it reaches the boiling point q = mass x C s x DT

• once the temperature

reaches the boiling point, allthe added heat goes into

boiling the liquid – thetemperature stays constant

• once all the liquid has beenturned into gas, thetemperature can again startto rise

Cl i Cl E i




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

exponential growth curve

RT

H

vap

vap

P

D

e

• the logarithm of the vapor pressure vs.inverse absolute 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

RTH

vap

RT

H

vap

vap

vap

D

D

D

D

e

e

• the slope of the line x 8.314 J/mol∙K =D

Hvap in J/mol

Example 11.4 – Determine the DHvap of




p

dichloromethane given the vapor pressure vs.

temperature data

• enter the data into a spreadsheet and calculate theinverse of the absolute temperature and natural log ofthe vapor pressure

Temperature,

K Vapor Pressure,

torr Inverse Temperature,

K -1 ln(Vapor

Pressure) 200 0.8 0.00500 -0.2 220 4.5 0.00455 1.5 240 21 0.00417 3.0 260 71 0.00385 4.3 280 197 0.00357 5.3 300 391 0.00333 6.0




79

p


temperature data

• graph the inverse of the absolute temperature vs. thenatural log of the vapor pressure

Clausius-Clapeyron Plot for Dichloromethane

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

0.00000 0.00100 0.00200 0.00300 0.00400 0.00500 0.00600

Inv. Temperature K -1

l n ( v a p o r p r e s s u r e )




80

p


temperature data

• add a trendline, making sure the display equation onchart option is checked off

Clausius-Clapeyron Plot for Dichloromethane

y = -3776.7x + 18.719

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

0.00000 0.00100 0.00200 0.00300 0.00400 0.00500 0.00600


l n ( v a p o r p r e s s u r e

)




81

p


temperature data

• determine the slope of the line

-3776.7 ≈ 3800 K Clausius-Clapeyron Plot for Dichloromethane

y = -3776.7x + 18.719

-1.0

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

0.00000 0.00100 0.00200 0.00300 0.00400 0.00500 0.00600


l n ( v a p o r p r e s s u r e

)




82

p


temperature data

• use the slope of the line to determine the heat ofvaporization

-3776.7 ≈ 3800 K

mol

kJ

mol

J4vap

K mol

J

vap

vap

6.311016.3H

314.8

H-K 3800

R H- slope

D

D

D

Clausius-Clapeyron Equation




p y q

2-Point Form

• the equation below can be used with just two measurements ofvapor pressure and temperature

however, it generally gives less accurate results

fewer data points will not give as accurate an average because there is lessaveraging out of the errors

– as with any other sets of measurements

• can also be used to predict the vapor pressure if you know theheat of vaporization and the normal boiling point

remember: the vapor pressure at the normal boiling point is 760 torr

D

12

vap

1

2

T

1

T

1

R

H

P

Pln

Example 11.5 – Calculate the vapor pressure ofmethanol at 12 0°C



K 2.28515.2730.12T

K 8.33715.2736.64T

2

1

K 337.8

1

K 285.2

1

8.314

102.35

P

Pln

K mol

J

mol

J3

1

2

methanol at 12.0°C

the units are correct, the size makes sense since the

vapor pressure is lower at lower temperatures

T1 = BP = 64.6°C, P1 = 760 torr, DHvap = 35.2 kJ/mol,

T2 = 12.0°C

P2, torr

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

P1, T1, DHvap P2

D

12

vap

1

2

T

1

T

1

R

H

P

Pln

T(K) = T(°C) + 273.15

T1 = BP = 337.8 K, P1 = 760 torr, DHvap = 35.2 kJ/mol,

T2 = 285.2 K

P2, torr

D

12

vap

1

2

T

1

T

1

R

H

P

Pln

torr 4.75P

0993.0 torr 760

P

2

31.22

e31.2

P

Pln

1

2

S i i l Fl id




Supercritical Fluid• as a liquid is heated in a sealed container, more vapor collects

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 and vapordisappears and the states commingle to form a supercriticalfluid

• supercritical fluid have properties of both gas and liquid states




The Critical Point

• the temperature required to produce asupercritical fluid is called the critical

temperature

• the pressure at the critical temperature is calledthe critical pressure

• at the critical temperature or higher

temperatures, the gas cannot be condensed to aliquid, no matter how high the pressure gets

S bli ti d D iti




Sublimation and Deposition• molecules in the solid have thermal energy that allows

them to vibrate• surface molecules with sufficient energy may breakfree from the surface and become a gas – this process iscalled sublimation

• the capturing of vapor molecules into a solid is calleddeposition

• the solid and vapor phases exist in dynamic equilibriumin a closed container at temperatures below the melting point

therefore, molecular solids have a vapor pressure

solid gassublimation

deposition

Sublimation




Sublimation




Melting = Fusion

• as a solid is heated, its temperature rises and the

molecules vibrate more vigorously

• once the temperature reaches the melting point,the molecules have sufficient energy to

overcome some of the attractions that hold them

in position and the solid melts (or fuses)• the opposite of melting is freezing

Heating Curve of a Solid




Heating Curve of a Solid• as you heat a solid, its

temperature increases linearlyuntil it reaches the melting point

q = mass x C s x DT

• once the temperature reaches themelting point, all the added heat

goes into melting the solid – thetemperature stays constant

• once all the solid has been turnedinto liquid, the temperature canagain start to rise

ice/water will always have atemperature of 0°C

at 1 atm

E ti f M lti




Energetics of Melting

• when the high energy molecules are lost fromthe solid, it lowers the average kinetic energy

• if energy is not drawn back into the solid its

temperature will decrease – therefore, meltingis an endothermic process

and freezing is an exothermic process

• melting requires input of energy to overcomethe attractions between molecules

Heat of Fusion



92

• the amount of heat energy required to melt one mole of the solid iscalled the Heat of Fusion, Hfus

sometimes called the enthalpy of fusion• always endothermic, therefore DHfus is +

• somewhat temperature dependent

DHcrystallization = -DHfusion

generally much less than

DHvap

DHsublimation = DHfusion + DHvaporization

Heats of Fusion and Vaporization




Heats of Fusion and Vaporization

Heating Curve of Water



94

Heating Curve of Water




Segment 1

• heating 1.00 mole of ice at -25.0°C up to the melting point, 0.0°C

• q = mass x C s x DT

mass of 1.00 mole of ice = 18.0 g

C s = 2.09 J/mol∙°C

kJ0.941J941

C25.0C0.009.2g0.18Cg

J

q

q




Segment 2

• melting 1.00 mole of ice at the melting point, 0.0°C

• q = n∙DHfus

n = 1.00 mole of ice

DHfus = 6.02 kJ/mol

kJ6.02

02.6mol1.00mol

kJ

q

q




Segment 3

• heating 1.00 mole of water at 0.0°C up to the boiling point, 100.0°C


mass of 1.00 mole of water = 18.0 g

C s = 2.09 J/mol∙°C

kJ52.7J1052.7

C.00C100.018.4g0.18

3

Cg

J

q

q




Segment 4

• boiling 1.00 mole of water at the boiling point, 100.0°C

• q = n∙DHvap

n = 1.00 mole of ice

DHfus = 40.7 kJ/mol

kJ7.04

7.40mol1.00mol

kJ

q

q




Segment 5

• heating 1.00 mole of steam at 100.0°C up to 125.0°C


mass of 1.00 mole of water = 18.0 g

C s = 2.01 J/mol∙°C

kJ904.0J904

C.0100C125.001.2g0.18Cg

J

q

q

Phase Diagrams




• describe the different states and state changes thatoccur at various temperature - pressure conditions

• areas represent states

• lines represent state changesliquid/gas line is vapor pressure curve

both states exist simultaneouslycritical point is the furthest point on the vapor pressure

curve

• triple point is the temperature/pressure condition

where all three states exist simultaneously• for most substances, freezing point increases as

pressure increases

Phase Diagrams



101

P r e s s u r e

Temperature

vaporization

condensation

critical

point

triple

point

Solid Liquid

Gas

1 atmnormal

melting pt.normal boiling pt.

Fusion Curve

Vapor PressureCurve

Sublimation

Curve

melting

freezing

sublimation

deposition




Phase Diagram of Water

0

5000

10000

15000

20000

25000

-50 0 50 100 150 200 250 300 350 400 450

Temperature, (°C)

P r e s s u r e ,

( k P a )

Phase Diagram of Water



104

Temperature

P r

e s s u r e

critical

point

374.1°C

217.7 atm

triple

point

Ice Water

Steam

1 atm

normal

boiling pt.100°C

normal

melting pt.0°C

0.01°C

0.006 atm




Morphic Forms of Ice




Phase Diagram of CO2



107

P r

e s s u r e

Temperature

critical

point31.0°C

72.9 atm

triple

point

Solid Liquid

Gas1 atm

-56.7°C

5.1 atm

normalsublimation pt.

-78.5°C




Water – An Extraordinary Substance




Wate t ao d a y Substa ce• water is a liquid at room temperature

most molecular substances with small molar masses are gases at roomtemperature

due to H-bonding between molecules

• water is an excellent solvent – dissolving many ionic and polarmolecular substances because of its large dipole moment

even many small nonpolar molecules have solubility in water e.g., O2, CO2

• water has a very high specific heat for a molecular substance moderating effect on coastal climates

• water expands when it freezes

at a pressure of 1 atm about 9%

making ice less dense than liquid water



Solids

properties &structure




Determining Crystal Structure

• crystalline solids have a very regular geometricarrangement of their particles

• the arrangement of the particles and distances between

them is determined by x-ray diffraction• in this technique, a crystal is struck by beams of x-rays,

which then are reflected

• the wavelength is adjusted to result in an interference pattern – at which point the wavelength is an integralmultiple of the distances between the particles

X-ray Crystallography




y y g p y

Bragg’s Law




• when the interference between x-rays is constructive,the distance between the two paths (a ) is an integral

multiple of the wavelength

n =2a

• the angle of reflection is therefore related to thedistance (d ) between two layers of particles

sin q = a/d

• combining equations and rearranging we get anequation called Bragg’s Law

q

l

sin2

nd

Example 11.6 – An x-ray beam at l=154 pm striking an

iron crystal results in the angle of reflection q = 32.6°.



y g

Assuming n = 1, calculate the distance between layers

the units are correct, the size makes sense since the

iron atom has an atomic radius of 140 pm

n = 1, q = 32.6°, l = 154 pm

d , pm

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

n,q

,l

d

q

l

sin2

nd

pm1436.32sin2

pm1541sin2

q l nd

Crystal Lattice




Crystal Lattice

• when allowed to cool slowly, the particles in aliquid will arrange themselves to give themaximum attractive forces

therefore minimize the energy

• the result will generally be a crystalline solid• the arrangement of the particles in a crystalline

solid is called the crystal lattice

• the smallest unit that shows the pattern ofarrangement for all the particles is called theunit cell

Unit Cells




Unit Cells• unit cells are 3-dimensional,

usually containing 2 or 3 layers of particles• unit cells are repeated over and over to give the macroscopic

crystal structure of the solid

• starting anywhere within the crystal results in the same unit cell

• each particle in the unit cell is called a lattice point

• lattice planes are planes connecting equivalent points in unitcells throughout the lattice

7 Unit Cellsc c



118

Cubic

a = b = c

all 90°

a b

c

Tetragonal

a = c < b

all 90°

a

b

c

Orthorhombic

a b c

all 90°

a

b

c

Monoclinic

a b c

2 faces 90°

a b

c

Hexagonal

a = c < b

2 faces 90°

1 face 120°

c

a

b

Rhombohedral

a = b = c

no 90°

a b

c

Triclinic

a b c

no 90°

i C ll




Unit Cells

• the number of other particles each particle is in contactwith is called its coordination number for ions, it is the number of oppositely charged ions an ion is

in contact with

• higher coordination number means more interaction,therefore stronger attractive forces holding the crystaltogether

• the packing efficiency is the percentage of volume inthe unit cell occupied by particles

the higher the coordination number, the more efficiently the particles are packing together

Cubic Unit Cells




Cubic Unit Cells

• all 90° angles between corners of the unit cell• the length of all the edges are equal

• if the unit cell is made of spherical particles

⅛ of each corner particle is within the cube ½ of each particle on a face is within the cube

¼ of each particle on an edge is within the cube

3

3

π3

4 Sphereaof Volume

lengthedgeCubeaof Volume

r




Simple Cubic




Simple Cubic

Cubic Unit Cells -




Simple Cubic

• 8 particles, one at each cornerof a 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 theradius

• coordination number of 6

2r

Body-Centered Cubic




Cubic Unit Cells -

d d bi




Body-Centered Cubic

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

• 1/8th of each corner particle lies

in the unit cell

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


3

4r

Face-Centered Cubic




Cubic Unit Cells -

F C d C bi




Face-Centered Cubic

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

• 1/8th of each corner particle + 1/2

of face particle 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


22r

Example 11.7 – Calculate the density of Al if itcrystallizes in a fcc and has a radius of 143 pm



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

cm1043.1m10

cm1

pm1

m10 pm431

8

2-

-12

g10792.1mol1

g982.26

atoms106.022

mol1Alatoms4 22

23

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

answer makes sense

face-centered cubic, r = 143 pm

density, g/cm3

Check:

Solution:

Concept

Plan:

Relation-ships:

Given:

Find:

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

lr Vmass fcc

dm, V

# atoms x mass 1 atom

V = l 3, l = 2r √2, d = m/V d = m/V

l = 2r √2 V = l 3

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

density, g/cm3

cm10544.0cm)(1.414)1043.1(222 -88 r l

323

383

cm10618.6

cm10045.4

l V

3cm

g

323

22

71.2

cm10618.6g10792.1

V md

Closest-Packed Structures




First Layer

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

Closest-Packed StructuresSecond Layer




y

• the second layer atoms can sit directly over the

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




y y

• the third layer atoms can align directly over the

atoms 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-PackedCubic Closest-Packed

Face-Centered Cubic

Hexagonal Closest-Packed Structures




Hexagonal Closest Packed Structures

Cubic Closest-Packed Structures




Cubic Closest Packed Structures

Classifying Crystalline Solids




• classified by the kinds of units found

• sub-classified by the kinds of attractive forces holdingthe units together

• molecular solids are solids whose composite units aremolecules

• ionic solids are solids whose composite units are ions• atomic solids are solids whose composite units areatomsnonbonding atomic solids are held together by dispersion

forces

metallic atomic solids are held together by metallic bonds

network covalent atomic solids are held together bycovalent bonds



Molecular Solids




Molecular Solids

• the lattice site are occupied by molecules

• the molecules are held together by

intermolecular attractive forcesdispersion forces, dipole attractions, and H-bonds

• because the attractive forces are weak, they tend

to have low melting pointgenerally < 300°C

Ionic Solids

Att ti F




Attractive Forces

• 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 in the crystal

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

• the coordination number depends on the relative sizes of thecations and anions generally, anions are larger than cations

the number of anions that can surround the cation limited by the size ofthe cation

the closer in size the ions are, the higher the coordination number is

Ionic Crystals




CsCl

coordination number = 8

Cs+ = 167 pm

Cl ─ = 181 pm

NaCl

coordination number = 6

Na+ = 97 pm

Cl ─ = 181 pm

Lattice Holes




Simple Cubic

Hole

Octahedral

Hole

Tetrahedral

Hole

Lattice Holesi h l l t k d bi l t




• 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 occupieddetermines formula (empirical) of salt

= Octahedral

= Tetrahedral

Cesium Chloride Structures




Cesium Chloride Structures

• coordination number = 8

• ⅛ of each Cl ─ (184 pm) insidethe unit cell

• whole Cs+ (167 pm) inside theunit cell

cubic hole = hole in simple cubicarrangement of Cl ─ ions

• Cs:Cl = 1: (8 x ⅛), therefore theformula is CsCl

Rock Salt Structures




Rock Salt Structures

• coordination number = 6• Cl ─ ions (181 pm) 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+ (97 pm) 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




Zinc Blende Structures

• coordination number = 4• S2 ─ ions (184 pm) 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+ (74 pm) 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

Fluorite Structures• coordination number = 4




• Ca2+ ions (99 pm) 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 ─ (133 pm) 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 fluorite structure common for 1:2 ratio

• usually get the antifluorite structurewhen the cation:anion ratio is 2:1 the anions occupy the lattice sites and the

cations occupy the tetrahedral holes

Nonbonding Atomic Solids




Nonbonding Atomic Solids

• noble gases in solid form

• solid held together by weak dispersion forces

very low melting• tend to arrange atoms in closest-packed

structure

either hexagonal cp or cubic cpmaximizes attractive forces and minimizes energy



Metallic Structure




Metallic Bonding




• metal atoms release their valence electrons

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

e-

e-e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

e-

+ + + + + + + + +

+ + + + + + + + +

+ + + + + + + + +

Crystal Structure of Metals at Room Temperature

b d t d bi other




= body-centered cubic

= hexagonal closestpacked

= other

= cubic cp, face-centered

= diamond

Network Covalent Solids




Network Covalent Solids

• atoms attached to its nearest neighbors by covalent bonds

• because of the directionality of the covalent bonds,these do not tend to form closest-packed arrangementsin the crystal

• because of the strength of the covalent bonds, thesehave very high melting pointsgenerally > 1000°C

• dimensionality of the network affects other physical properties

The Diamond Structure:




a 3-Dimensional Network

• the carbon atoms in a diamond each have 4

covalent bonds to surrounding atoms

sp3

tetrahedral geometry

• this effectively makes each crystal one giant

molecule held together by covalent bonds

you can follow a path of covalent bonds from any

atom to every other atom



The Graphite Structure:

a 2 Dimensional Network




a 2-Dimensional Network

• in graphite, the carbon atoms in a sheet are covalently 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




Properties of Graphite• hexagonal crystals

• high melting, ~3800°C need to overcome some covalent bonding

• slippery feel

because there are only dispersion forces holding

the sheets together, they can slide past each other glide planes

lubricants

• electrical conductor

parallel to sheets

• thermal insulator

• chemically very nonreactive

Silicates




Silicates

• ~90% of earth’s crust

• extended arrays of SiO

sometimes with Al substituted for Si – aluminosilicates• glass is the amorphous form

Quartzdi i l f i l l b d d




• 3-dimensional array of Si covalently bonded to 4 O

tetrahedral

• melts at ~1600°C

• very hard

Micas




Micas

• minerals that are mainly 2-dimensional arrays of

Si bonded to O

hexagonal arrangement of atoms

• sheets

• chemically stable

• thermal and electrical insulator

Band Theory




Band Theory

• the structures of metals and covalent network

solids result in every atom’s orbitals beingshared by the entire structure

• for large numbers of atoms, this results in a

large number of molecular orbitals that have

approximately the same energy, we call this an

energy band

Band Theory




• when 2 atomic orbitals combine they produce

both a bonding and an antibonding molecularorbital

• when many atomic orbitals combine they

produce a band of bonding molecular orbitalsand a band of antibonding molecular orbitals

• the band of bonding molecular orbitals is calledthe valence band

• the band of antibonding molecular orbitals iscalled the conduction band

Molecular orbitals of polylithium




Band Gap




• at absolute zero, all the electrons will occupy thevalence band

• as the temperature rises, some of the electrons

may acquire enough energy to jump to theconduction band

• the difference in energy between the valence

band and conduction band is called the band gap

the larger the band gap, the fewer electrons there are

with enough energy to make the jump

Types of Band Gaps and

C d ti it




Conductivity

Band Gap and Conductivity• the more electrons at any one time that a substance has in the

conduction band the better conductor of electricity it is




conduction band, the better conductor of electricity it is

• if the band gap is ~0, then the electrons will be almost as likelyto be in the conduction band as the valence band and thematerial 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 normal temperaturesand 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 will be inthe conduction band at normal temperatures and the materialwill be an insulator

Doping Semiconductorsi ddi i i i h i d




• 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 holesin the valence band, allowing conduction of electricity

Diodes



• when a p-type semiconductor adjoins an n-type

semiconductor, the result is an p-n junction

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

• this also allows the accumulation of electrical

energy – called an amplifier

chem chapter11 lec

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