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Chapter 12. Intermolecular Forces: Liquids, Solids and Phase Changes. Intermolecular Forces: Liquids, Solids and Phase Changes. 12.1 Overview of physical states and phase changes. 12.2 Quantitative aspects of phase changes. 12.3 Types of intermolecular forces.

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slide1

Chapter 12

Intermolecular Forces: Liquids, Solids and Phase Changes

slide2

Intermolecular Forces:

Liquids, Solids and Phase Changes

12.1 Overview of physical states and phase changes

12.2 Quantitative aspects of phase changes

12.3 Types of intermolecular forces

12.4 Properties of the liquid state

12.5 The uniqueness of water

slide3

Types of Molecular Forces

Intramolecular: bonding forces within a molecule; influence

chemical properties

Intermolecular: forces between molecules; influence

physical properties

Three states of water: water vapor, liquid water, ice

Phase change: liquid water ice

Liquid water and ice are examples of condensed phases.

slide4

Table 12.1

A Macroscopic Comparison of Gases, Liquids and Solids

state

shape and volume

compressibility

ability to flow

gas

conforms to shape and volume of container

high

high

liquid

conforms to shape of container; volume limited by surface

very low

moderate

solid

maintains its own shape and volume

almost none

almost none

Importance of interplay between potential and kinetic energies

slide5

Types of Phase Changes

condensation/vaporization

freezing/melting (or fusion)

Enthalpy changes accompany phase changes!

condensation and freezing: exothermic processes

vaporization and melting: endothermic processes

DHofus = heat of fusion (+)

DHovap = heat of vaporization (+)

DHosubl = heat of sublimation = DHofus + DHovap (Hess’s Law)

solid gas

gas solid (called deposition)

slide6

Heats of vaporization and fusion for some common substances

It takes more energy to vaporize than to melt!

Figure 12.1

slide8

Quantitative Aspects of Phase Changes

Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in the average Ekas the most probable speed of the molecules changes.

q = (amount in moles)(molar heat capacity)(DT)

During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes.

q = (amount in moles)(enthalpy of phase change)

slide9

A cooling curve for the conversion of gaseous water to ice

Five stages, two phase changes

Figure 12.3

slide10

Phase changes are reversible and reach equilibrium.

A. Liquid-Gas Equilibria

Pressure at equilibrium =

vapor pressure

At the BP, the rate of evaporation

equals the rate of condensation!

Figure 12.4

slide11

The effect of temperature on the distribution of molecular speed in a liquid

higher temperature = higher vapor pressure

Figure 12.5

slide12

Vapor pressure as a function of temperature and intermolecular forces

Weaker intermolecular forces

translate into higher vapor

pressure at a given temperature

Figure 12.6

slide13

A linear plot of the vapor pressure-temperature relationship

Plotting ln P against

1/T yields a straight

line with slope equal to

-DHvap/R (the Clausius-

Clapeyron equation)

Figure 12.7

slide14

Full form of the Clausius-Clapeyron equation:

Two-point version of the equation:

slide15

Utility of the Clausius-Clapeyron Equation

It provides a means to determine experimentally the heat of vaporization,

which is the energy required to vaporize 1 mole of molecules in the liquid state.

or

If DHvap is known, and vapor pressure at one T is known, then

vapor pressure at a new T can be calculated.

slide16

PROBLEM:

The vapor pressure of ethanol is 115 torr at 34.9 oC. If DHvap of ethanol is 40.5 kJ/mol, calculate the temperature (in oC) when the vapor pressure is 760 torr.

PLAN:

We are given four of the five variables in the Clausius-Clapeyron equation. Substitute and solve for T2.

1

760 torr

1

ln

-40.5 x103 J/mol

-

T2

115 torr

308 K

8.314 J/mol.K

SAMPLE PROBLEM 12.1

Using the Clausius-Clapeyron equation

SOLUTION:

T1 = 34.9 oC = 308.0 K

=

T2 = 350 K = 77 oC

slide17

Vapor Pressure and Boiling Point

If we assume an open container, then the boiling point (BP) is the temperature

at which the vapor pressure equals the external pressure (usually

atmospheric pressure, 760 mmHg).

Thus, the BP depends on

the applied pressure

(see Figure 12.6)

Water boils at 100 oC at sea level,

but at 72 oC on the peak of Mt.

Everest!

slide18

B. Liquid-Solid Equilibria

Characterized by a melting point (temperature at which the

rate of melting equals the rate of freezing)

The MP is not significantly affected by pressure (two condensed

phases are involved).

slide19

Iodine subliming

I2 vapor in contact

with a cold finger at atmospheric pressure

C. Solid-Gas Equilibria

Why??

External pressure and

intermolecular forces maintain the

liquid phase after melting. These

are too weak in some cases.

Figure 12.8

slide20

Bringing It All Together: A Phase Diagram

A graph that describes phase changes of a substance under

various combinations of temperature and pressure

Key Characteristics

Regions (bounded areas) (one phase)

Interfaces (lines) between different regions (equilibria between two phases)

Isolated Points (critical point, triple point) (unique T/P combinations)

slide21

Phase diagrams for CO2 and H2O

CO2

H2O

(ice is less dense than

liquid water)

Figure 12.9

slide23

Defining/Quantifying Intermolecular Forces

Intramolecular (bonding) forces: strong,

involve larger charges closer together

Intermolecular forces: weak, involve

smaller charges farther apart

slide25

Periodic trends in covalent and van der Waals radii (in pm)

blue: covalent radius

black: van der Waals radius

Figure 12.11

slide26

Types of intermolecular (van der Waals) forces

ion-dipole

hydrogen bonding

dipole-dipole

ion-induced dipole

dipole-induced dipole

dispersion (London)

decreasing

strength

slide29

Orientation of polar molecules caused by dipole-dipole forces

More orderly in the solid phase than in the liquid phase

Figure12.12

slide30

Dipole moment and boiling point

Higher dipole moment translates into higher BP.

Figure 12.13

slide31

Hydrogen bonding

Involves molecules that have an H atom bound at a small,

highly electronegative atom with lone electron pairs

N-H O-H H-F

General Model

_B:H_A

electronegative atom bearing

hydrogen (donor)

H-bond

electronegative atom with

lone electron pair (acceptor)

slide32

PROBLEM:

Which of the following substances exhibits hydrogen bonding? For those that do, draw two molecules of the substance with the H-bonds between them.

(b)

(a)

(c)

(b)

SAMPLE PROBLEM 12.2

Drawing hydrogen bonds between molecules of a substance

Find molecules in which hydrogen is bonded to N, O or F.

Draw H-bonds in the format, B: ----- HA.

PLAN:

SOLUTION:

(a) C2H6 has no H-bonding sites (a non-polar molecule).

(c)

Note: more than one H-bond per molecule is possible!

slide33

Hydrogen bonding and boiling point

H2O, HF and NH3

exhibit aberrant

behavior due to their

ability to form H-bonds.

binary hydrides of Groups 4-7

Figure 12.14

slide34

Covalent and hydrogen bonding in the helical structure of DNA

A single H-bond is relatively weak,

but the existence of many such bonds

in a molecule can influence molecular

structure significantly.

strength in

numbers!

Figure 12.15

slide35

Polarizability

The ease with which a particle’s electron cloud can be distorted

Pertinent to charge-induced dipole forces

(ion-induced dipole and dipole-induced dipole)

Increases down a group of atoms or ions (size)

Decreases from left to right in a period (effective nuclear charge)

Cations are less polarizable than their parent atoms; anions are

more polarizable than their parent atoms.

slide36

Dispersion forces among non-polar molecules

separated Cl2 molecules

instantaneous dipoles

Caused by momentary

oscillations of electron

charge

More electrons, larger

molecule, greater mass,

greater dispersion forces

Figure 12.16

slide39

(d) hexane (CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane

SAMPLE PROBLEM 12.3

Predicting the Type and Relative Strength of Intermolecular Forces

PROBLEM:

For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point.

(a) MgCl2 or PCl3

(b) CH3NH2 or CH3F

(c) CH3OH or CH3CH2OH

PLAN:

  • Bonding forces are stronger than nonbonding (intermolecular) forces.
  • Hydrogen bonding is a strong dipole-dipole force.
  • Dispersion forces are decisive when the difference is molar mass or molecular shape.
slide40

SAMPLE PROBLEM 12.3

(continued)

SOLUTION:

(a) Mg2+ and Cl- are held together by ionic bonds (a salt) while PCl3 is covalentlybonded and the molecules are held together by dipole-dipole interactions. Ionic attractions are much stronger than dipole interactions and so MgCl2 has the higher boiling point.

(b) CH3NH2 and CH3F are both covalent compounds and have polar bonds. The dipole in CH3NH2can H-bond while that in CH3F cannot. Therefore, CH3NH2 has the stronger interactions and the higher boiling point.

(c) Both CH3OH and CH3CH2OH can H-bond but CH3CH2OH has more CH bonds for greater dispersion force interactions. Therefore, CH3CH2OH has the higher boiling point.

(d) Hexane and 2,2-dimethylbutane are both non-polar with only dispersion forces to hold the molecules together. Hexane has a larger surface area, and therefore the greater dispersion forces and higher boiling point.

slide41

Properties of the Liquid State

  • surface tension
  • capillarity
  • viscosity
slide42

The molecular basis of surface tension

The energy required to increase

surface area by a unit amount

Surface molecules experience

a net attraction downward.

Stronger intermolecular forces translate into greater surface tension.

Figure 12.19

slide43

Table 12.3

Surface Tension and Forces Between Particles

surface tension (J/m2) at 20 oC

substance

formula

major force(s)

diethyl ether

CH3CH2OCH2CH3

dipole-dipole; dispersion

1.7 x 10-2

ethanol

CH3CH2OH

2.3 x 10-2

H-bonding

butanol

CH3CH2CH2CH2OH

2.5 x 10-2

H-bonding; dispersion

water

H2O

7.3 x 10-2

H-bonding

mercury

Hg

48 x 10-2

metallic bonding

slide45

stronger cohesive forces

strongeradhesive forces

Shape of a water or mercury meniscus in glass

water-glass forces >

water-water forces

Hg-Hg forces >

Hg-glass forces

Figure 12.20

slide46

Viscosity: a liquid’s resistance to flow

  • Affected by temperature (viscosity decreases at higher T)
  • Affected by molecular shape (longer molecules exhibit higher viscosity)
slide47

Table 12.4 Viscosity of Water at Several Temperatures

viscosity (N.s/m2)*

temperature (oC)

20

1.00 x 10-3

40

0.65 x 10-3

0.47 x 10-3

60

80

0.35 x 10-3

*The units of viscosity are newton-seconds per square meter.

slide48

Water

  • H-bonding ability
  • solvent power
  • high specific heat capacity
  • high heat of vaporization
  • high surface tension
  • high capillarity
  • density of liquid water vs ice
slide49

The H-bonding ability of the water molecule

Four H-bonds per

molecule in the solid

state; fewer in the liquid state

acceptor

donor

Figure 12.21

slide54

1/8 atom at 8 corners

The three cubic unit cells

Figure 12.27

simplecubic

Atoms/unit cell = 1/8 x 8 = 1

coordination number = 6

slide55

1/8 atom at 8 corners

1 atom at center

coordination number = 8

The three cubic unit cells

body-centered cubic

Atoms/unit cell = (1/8 x 8) + 1 = 2

Figure 12.27

slide56

1/8 atom at 8 corners

1/2 atom at 6 faces

coordination number = 12

The three cubic unit cells

face-centered cubic

Atoms/unit cell = (1/8 x 8) + (1/2 x 6) = 4

Figure 12.27

slide57

Packing of spheres

simple cubic

52% packing efficiency

body-centered cubic

68% packing efficiency

Figure 12.28

slide58

layer a

layer b

hexagonal closest packing

cubic closest packing

layer c

layer a

hexagonal unit cell

face-centered unit cell

expanded side views

Figure 12.26

closest packing of first and second layers

abab… (74%)

abcabc… (74%)

slide59

PROBLEM:

Barium is the largest non-radioactive alkaline earth metal. It has a body-centered cubic unit cell and a density of 3.62 g/cm3. What is the atomic radius of barium? (volume of a sphere: V = 4/3pr3)

reciprocal divided by M

V = 4/3pr3

multiply by packing efficiency

divide by Avogadro’s number

volume of 1 mol Ba atoms

SAMPLE PROBLEM 12.4

Determining atomic radius from crystal structure

PLAN:

Use the density and molar mass to find the volume of 1 mol of Ba. Since 68% (for a body-centered cubic) of the unit cell contains atomic material, dividing by Avogadro’s number will give the volume of one atom of Ba. Using the volume of a sphere, the radius can be calculated.

radius of a Ba atom

density of Ba (g/cm3)

volume of 1 mol Ba metal

volume of 1 Ba atom

slide60

137.3 g Ba

1 cm3

x

mol Ba

3.62 g

26 cm3

mol Ba atoms

x

mol Ba atoms

6.022 x 1023atoms

SAMPLE PROBLEM 12.4

(continued)

SOLUTION:

volume of Ba metal =

= 37.9 cm3/mol Ba

37.9 cm3/mol Ba

x 0.68

= 26 cm3/mol Ba atoms

= 4.3 x 10-23 cm3/atom

r3 = 3V/4p

= 2.2 x 10-8cm

slide61

Cubic closest packing of frozen methane

Cubic closest packing for frozen argon

Figure 12.29

Figure 12.30

slide66

hexagonal closest packing

Crystal structures of metals

cubic closest packing

Figure 12.34

slide72

Manipulating atoms

Figure 12.50

tip of an atomic force microscope (AFM)

slide73

Manipulating atoms

Figure 12.50

nanotube gear

slide74

Figure B12.1

Tools of the Laboratory

Diffraction of x-rays by crystal planes

slide75

Figure B12.2

Tools of the Laboratory

Formation of an x-ray diffraction pattern

of the protein hemoglobin

slide76

Tools of the Laboratory

Figure B12.3

Scanning tunneling micrographs

gallium arsenide semiconductor

metallic gold