Chapter 11 Liquids, Solids, and Intermolecular Forces

Copyright 2011 Pearson Education, Inc.

Chapter 11Liquids,

Solids, and Intermolecular

Forces

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro


Climbing Geckos

• Geckos can adhere to almost any surface

• Recent studies indicate that this amazing ability is related to intermolecular attractive forces

• Geckos have millions of tiny hairs on their feet that branch out and flatten out on the endsetae = hairs, spatulae = flat ends

• This structure allows the gecko to have unusually close contact to the surface – allowing the intermolecular attractive forces to act strongly

2Tro: Chemistry: A Molecular Approach, 2/e


Properties of the three Phases of Matter

• Fixed = keeps shape when placed in a container • Indefinite = takes the shape of the container

State Shape Volume Density

Compressible?Will it Flow?

Strength of Intermolecular Attractions

Solid fixed fixed high No No very strongLiquid indefinite fixed high No Yes intermediate

Gas indefinite indefinite low Yes Yes weak



Notice that the densities of ice and liquid water are much larger than the density of steam

Notice that the densities and molar volumes of ice and liquid water are much closer to each other than to steam

Notice that the densities of ice is larger than the density of liquid water. This is not the norm, but is vital to the development of life as we know it.

Three Phases of Water



Degrees of Freedom

• Particles may have one or several types of freedom of motionand various degrees of each type

• Translational freedom is the ability to move from one position in space to another

• Rotational freedom is the ability to reorient the particles direction in space

• Vibrational freedom is the ability to oscillate about a particular point in space



Solids

• The particles in a solid are packed close together and are fixed in position though they may vibrate

• The close packing of the particles results in solids being incompressible

• The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container, and prevents the solid from flowing



Solids

• Some solids have their particles arranged in an orderly geometric pattern – we call these crystalline solidssalt and diamonds

• Other solids have particles that do not show a regular geometric pattern over a long range – we call these amorphous solidsplastic and glass


Copyright 2011 Pearson Education, Inc.8

Liquids

• The particles in a liquid are closely packed, but they have some ability to move around

• The close packing results in liquids being incompressible

• But the ability of the particles to move allows liquids to take the shape of their container and to flow – however, they don’t have enough freedom to escape or expand to fill the container

Tro: Chemistry: A Molecular Approach, 2/e


Gases• In the gas state, the particles have

complete freedom of motion and are not held together

• The particles are constantly flying around, bumping into each other and the container

• There is a large amount of space between the particlescompared to the size of the particlestherefore the molar volume of the

gas state of a material is much larger than the molar volume of the solid or liquid states



Gases• Because there is a lot of empty

space, the particles can be squeezed closer together – therefore gases are compressible

• Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow



Compressibility



Kinetic – Molecular Theory

• What state a material is in depends largely on two major factors

1. the amount of kinetic energy the particles possess

2. the strength of attraction between the particles

• These two factors are in competition with each other



States and Degrees of Freedom• The molecules in a gas have complete freedom of

motion 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 the liquid they have enough kinetic energy to overcome some of

the attractive forces, but not enough to escape each other



Kinetic Energy

• Increasing kinetic energy increases the motion energy of the particles

• The more motion energy the molecules have, the more freedom they can have

• The average kinetic energy is directly proportional to the temperatureKEavg = 1.5 kT



Attractive Forces• The particles are attracted to each other by

electrostatic forces• The strength of the attractive forces varies,

some are small and some are large• The strength of the attractive forces depends

on the kind(s) of particles• The stronger the attractive forces between the

particles, the more they resist movingthough no material completely lacks particle motion



Kinetic–Molecular Theoryof Gases

• When the kinetic energy is so large it overcomes the attractions between particles, the material will be a gas

• In an ideal gas, the particles have complete freedom of motion – especially translational

• This allows gas particles to expand to fill their containergases flow

• It also leads to there being large spaces between the particles therefore low density and compressibility



Gas Structure

Gas molecules are rapidly moving in random straight lines, and are free from sticking to each other.



Kinetic–Molecular Theory of Solids

• When the attractive forces are strong enough so the kinetic energy cannot overcome it at all, the material will be a solid

• In a solid, the particles are packed together without any translational or rotational motionthe only freedom they have is

vibrational motion



Kinetic–Molecular Theoryof Liquids

• When the attractive forces are strong enough so the kinetic energy can only partially overcome them, the material will be a liquid

• In a liquid, the particles are packed together with only very limited translational or rotational freedom



Explaining the Properties of Liquids

• Liquids have higher densities than gases and are incompressible because the particles are in contact

• They have an indefinite shape because the limited translational freedom of the particles allows them to move around enough to get to the container walls

• It also allows them to flow

• But they have a definite volume because the limit on their freedom keeps the particles from escaping each other



Phase Changes• Because the attractive forces between the molecules are

fixed, changing the material’s state requires changing the amount of kinetic energy the particles have, or limiting their freedom

• Solids melt when heated because the particles gain enough kinetic energy to partially overcome the attactive forces

• Liquids boil when heated because the particles gain enough kinetic energy to completely overcome the attractive forces the stronger the attractive forces, the higher you will need to raise

the temperature

• Gases can be condensed by decreasing the temperature and/or increasing the pressure pressure can be increased by decreasing the gas volume reducing the volume reduces the amount of translational freedom

the particles have



Phase Changes



Intermolecular Attractions

• The strength of the attractions between the particles of a substance determines its state

• At room temperature, moderate to strong attractive forces result in materials being solids or liquids

• The stronger the attractive forces are, the higher will be the boiling point of the liquid and the melting point of the solidother factors also influence the melting point



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 charge = stronger attraction• Longer 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 all the attractions between the particlesHowever, not breaking the covalent bonds

• The higher the normal boiling point of the liquid, the stronger the intermolecular attractive forces



Kinds of Attractive Forces

• Temporary polarity in the molecules due to unequal electron distribution leads to attractions called dispersion forces

• Permanent polarity in the molecules due to their structure leads to attractive forces called dipole–dipole attractions

• An especially strong dipole–dipole attraction results when H is attached to an extremely electronegative atom. These are called hydrogen bonds.



Dispersion Forces• Fluctuations in the electron distribution in atoms and

molecules 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 dipoles are called dispersion forcesaka London Forces

• All molecules and atoms will have them• As a temporary dipole is established in one molecule,

it induces a dipole in all the surrounding molecules



Dispersion Force



Size of the Induced Dipole• The magnitude of the induced dipole

depends on several 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 moleculemore surface-to-surface contact =

larger induced dipole = stronger attraction

+ + + + + + +

- - - - - - -

++ + +

+

--

- --

molecules that are flat have more surface

interaction than spherical ones

+ + + + + + +++

+ + + ++

− −− −− −− −− −−− −

larger molecules have more electrons, leading

to increased polarizability



The Noble gases are all nonpolar atomic elements

As the molar mass increases, the number of electrons increases. Therefore the strength of the dispersion forces increases.

Effect of Molecular Sizeon Size of Dispersion Force

The stronger the attractive forces between the molecules, the higher the boiling point will be.


Copyright 2011 Pearson Education, Inc.31Tro: Chemistry: A Molecular Approach, 2/e


Properties of Straight Chain AlkanesNonPolar Molecules



Boiling Points of n-Alkanes




Effect of Molecular Shapeon Size of Dispersion Force

35

the larger surface-to- surface contact between molecules in n-pentane

results in stronger dispersion force

attractions



Alkane Boiling Points

• Branched chains have lower BPs than straight chains

• The straight chain isomers have more surface-to-surface contact



a) CH4 C4H10

b) C6H12 C6H12

Practice – Choose the Substance in Each Pair with the Higher Boiling Point



Practice – Choose the Substance in Each Pair with the Higher Boiling Point

a) CH4 CH3CH2CH2CH3

b) CH3CH2CH=CHCH2CH3 cyclohexane

Both molecules are nonpolarlarger molar mass

Both molecules are nonpolar, but the flatter ring molecule has larger surface-to-surface contact



Dipole–Dipole Attractions• Polar molecules have a permanent dipole

because of bond polarity and shapedipole momentas 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

nonpolar molecules of similar size and shape



Effect of Dipole–Dipole Attraction on Boiling and Melting Points



replace with the figure 11.8



Example 11.1b: Determine if dipole–dipole attractions occur between CH2Cl2 molecules

molecules that have dipole–dipole attractions must be polar

CH2Cl2, EN C = 2.5, H = 2.1, Cl = 3.0

If there are dipole–dipole attractions

Solution:

Conceptual Plan:

Relationships:

Given:Find:

EN Difference Shape

Lewis Structure

BondPolarity

MoleculePolarity

Formula

Cl—C3.0−2.5 = 0.5polar

C—H2.5−2.1 = 0.4nonpolar

polar molecule; therefore dipole–dipole attractions

42

polar bonds andtetrahedral shape = polar molecule

4 bonding areasno lone pairs = tetrahedral shape



or

Practice – Choose the substance in each pair with the higher boiling point

a) CH2FCH2F CH3CHF2

b)



Practice – Choose the substance in each pair with the higher boiling point

more polara) CH2FCH2F CH3CHF2

b)

44

or

polar nonpolar



Hydrogen Bonding• When a very electronegative atom is bonded to

hydrogen, it strongly pulls the bonding electrons toward itO─H, N─H, or F─H

• Because hydrogen has no other electrons, when its electron is pulled away, the nucleus becomes deshieldedexposing the H proton

• The exposed proton acts as a very strong center of positive charge, attracting all the electron clouds from neighboring molecules



H-Bonding

HF



H-Bonding in Water



H-Bonds

• Hydrogen bonds are very strong intermolecular attractive forcesstronger than dipole–dipole or dispersion forces

• Substances that can hydrogen bond will have higher boiling points and melting points than similar substances that cannot

• But hydrogen bonds are not nearly as strong as chemical bonds2 to 5% the strength of covalent bonds



Effect of H-Bonding on Boiling Point



For nonpolar molecules, such as the hydrides of Group 4, the intermolecular attractions are due to dispersion forces. Therefore they increase down the column, causing the boiling point to increase.

HF, H2O, and NH3 have unusually strong dipole-dipole attractions, called hydrogen bonds. Therefore they have higher boiling points than you would expect from the general trends.

Polar molecules, such as the hydrides of Groups 5–7, have both dispersion forces and dipole–dipole attractions. Therefore they have higher boiling points than the corresponding Group 4 molecules.



Step 3. Evaluate the strengths of the total intermolecular attractive forces. The substance with the strongest will be the liquid.Formaldehyde:

dispersion forces: MM 30.03, trigonal planardipole–dipole: very polar C=O bond uncancelledH-bonding: no O–H, N–H, or F–H therefore no H-bond

Fluoromethane: dispersion forces: MM 34.03, tetrahedraldipole–dipole: very polar C–F bond uncancelledH-bonding: no O–H, N–H, or F–H therefore no H-bond

Hydrogen peroxide: dispersion forces: MM 34.02, tetrahedral bentdipole–dipole: polar O–H bonds uncancelledH-bonding: O–H, therefore H-bond

Because the molar masses are similar, the size of the dispersion force attractions should be similar

Because only hydrogen peroxide has the additional very strong H-bond additional attractions, its intermolecular attractions will be the strongest. We therefore expect hydrogen peroxide to be the liquid.

Example 11.2: One of these compounds is a liquid at room temperature (the others are

gases). Which one and why?

51

Step 1. Determine the kinds of intermolecular attractive forces

MM = 30.03PolarNo H-Bonds

MM = 34.03PolarNo H-Bonds

MM = 34.02PolarH-Bonds

Step 2. Compare intermolecular attractive forces

Because all the molecules are polar, the size of the dipole–dipole attractions should be similar

Only the hydrogen peroxide also has additional hydrogen bond attractions



a) CH3OH CH3CHF2

b) CH3-O-CH2CH3 CH3CH2CH2NH2

Practice – Choose the substance in each pair that is a liquid at room temperature (the other is a gas)

can H-bond

can H-bond



Attractive Forces and Solubility• Solubility depends, in part, on the attractive forces

of the solute and solvent moleculeslike dissolves likemiscible liquids will always dissolve in each other

• Polar substances dissolve in polar solventshydrophilic groups = OH, CHO, C=O, COOH, NH2, Cl

• Nonpolar molecules dissolve in nonpolar solventshydrophobic groups = C-H, C-C

• Many molecules have both hydrophilic and hydrophobic parts – solubility in water becomes a competition between the attraction of the polar groups for the water and the attraction of the nonpolar groups for their own kind



Immiscible Liquids

54

Pentane, C5H12 is a nonpolar molecule.

Water is a polar molecule.

The attractive forces between the water molecules is much stronger than their attractions for the pentane molecules. The result is the liquids are immiscible.



Dichloromethane(methylene chloride)

Polar Solvents

55

Water

Ethanol(ethyl alcohol)



Nonpolar Solvents



Ion–Dipole Attraction• In a mixture, ions from an ionic compound are

attracted to the dipole of polar molecules

• The strength of the ion–dipole attraction is one of the main factors that determines the solubility of ionic compounds in water



a) CH3OH CH3CHF2

b) CH3CH2CH2CH2CH3 CH3Cl

Practice – Choose the substance in each pair that is more soluble in water

can H-bond with H2O

more polar



Summary

• Dispersion forces are the weakest of the intermolecular attractions

• Dispersion forces are present in all molecules and atoms

• The magnitude of the dispersion forces increases with molar mass

• Polar molecules also have dipole–dipole attractive forces



Summary (cont’d)• Hydrogen bonds are the strongest of the intermolecular

attractive forcesa 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 ionic compounds with polar molecules.

• Ion–dipole attractions are the strongest intermolecular attraction

• Ion–dipole attractions are especially important in aqueous solutions of ionic compounds




Liquids

Properties &

Structure



Surface Tension

• Surface tension is a property of liquids that results from the tendency of liquids to minimize their surface area

• To minimize their surface area, liquids form drops that are sphericalas long as there

is no gravity



Surface Tension• 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 allowing you to “float” a paper clip even though

steel is denser than water



Surface Tension

• Because they have fewer neighbors to attract them, the surface molecules are less stable than those in the interiorhave a higher potential energy

• The surface tension of a liquid is the energy required to increase the surface area a given amountsurface tension of H2O = 72.8 mJ/m2

at room temperature

surface tension of C6H6 = 28 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 its surface tensionraising the temperature of the liquid increases the

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

stretch the surface




Viscosity• Viscosity is the resistance of a liquid to flow

1 poise = 1 P = 1 g/cm∙soften given in centipoise, cP

H2O = 1 cP at room temperature

• Larger intermolecular attractions = larger viscosity



Factors Affecting Viscosity

• The stronger the intermolecular attractive forces, the higher the liquid’s viscosity will be

• The more spherical the molecular shape, the lower the viscosity will bemolecules roll more easily less surface-to-surface contact lowers attractions

• Raising the temperature of a liquid reduces its viscosity raising the temperature of the liquid increases the average kinetic

energy of the molecules the increased molecular motion makes it easier to overcome the

intermolecular attractions and flow



Insert Table 11.6



Capillary Action

• Capillary action is the ability of a liquid to flow up a thin tube against the influence of gravitythe narrower the tube, the higher the liquid rises

• Capillary action is the result of two forces working in conjunction, the cohesive and adhesive forces cohesive forces hold the liquid molecules togetheradhesive forces attract the outer liquid molecules to

the tube’s surface



Capillary Action• The adhesive forces pull the surface liquid up the side of

the tube, and the cohesive forces pull the interior liquid with it

• The liquid rises up the tube until the force of gravity counteracts the capillary action forces

• The narrower the tube diameter, the higher the liquid will rise up the tube



Meniscus• The curving of the liquid surface in a

thin tube is due to the competition between adhesive and cohesive forces

• The meniscus of water is concave in a glass tube because its adhesion to the glass is stronger than its cohesion for itself

• The meniscus of mercury is convex in a glass tube because its cohesion for itself is stronger than its adhesion for the glassmetallic bonds are stronger than

intermolecular attractions



The Molecular Dance

• Molecules in the liquid are constantly in motionvibrational, and limited rotational and

translational

• The average kinetic energy is proportional to the temperature

• However, some molecules have more kinetic energy than the average, and others have less



Vaporization• If these high energy molecules

are at the surface, they may have enough energy to overcome the attractive forcestherefore – the larger the surface

area, the faster the rate of evaporation

• This will allow them to escape the liquid 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 molecules with “escape energy” increases

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



Condensation

• Some molecules of the vapor will lose energy through molecular collisions

• The result will be that some of the molecules will get captured back into the liquid when they collide with it

• Also some may stick and gather together to form droplets of liquidparticularly on surrounding surfaces

• We call this process condensation



Evaporation vs. Condensation

• Vaporization and condensation are opposite processes

• 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



Effect of Intermolecular Attraction on Evaporation and Condensation

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

• Also, weaker attractive forces means that more energy will 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 weaker the attractive forces, the faster the rate of evaporation

• Liquids that evaporate easily are said to be volatilee.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 from the 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 processand condensation is an exothermic process

• Vaporization requires input of energy to overcome the attractions between molecules



Heat of Vaporization• The amount of heat energy required to vaporize

one mole of the liquid is called the heat of vaporization, Hvapsometimes called the enthalpy of vaporization

• Slways endothermic, therefore Hvap is +• Somewhat temperature dependent Hcondensation = −Hvaporization



Example 11.3: Calculate the mass of water that can be vaporized with 155 kJ of heat at 100 °C

82

because the given amount of heat is almost 4x the Hvap, the amount of water makes sense

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

155 kJg H2O

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:



Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point

(Hvap = 39.9 kJ/mol)



Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point

84

because the given amount of C3H7OH is more than 1 mole the amount of heat makes sense

1 mol C3H7OH = 39.9 kJ, 1 mol = 60.09 g

90.0 gkJ

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:



Dynamic Equilibrium• In a closed container, once the rates of vaporization

and condensation are equal, the total amount of vapor and liquid will not change

• Evaporation and condensation are still occurring, but because they are opposite processes, there is no net gain or loss of either vapor or liquid

• When two opposite processes reach the same rate so that 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 account

for 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, the higher the vapor pressure the higher the vapor pressure, the more volatile the liquid



Vapor–Liquid Dynamic Equilibrium• If the volume of the chamber is increased, it will decrease

the pressure of the vapor inside the chamber at that point, there are fewer vapor molecules in a given volume,

causing the rate of condensation to slow

• Therefore, for a period of time, the rate of vaporization will be faster than the rate of condensation, and the amount of vapor will increase

• Eventually enough vapor accumulates so that the rate of the condensation increases to the point where it is once again as fast as evaporation equilibrium is reestablished

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



Changing the Container’s Volume Disturbs the Equilibrium

Initially, the rate of vaporization and

condensation are equal and the system is in dynamic equilibrium

When the volume is increased, the rate of vaporization becomes faster than the rate of

condensation

When the volume is decreased, the rate of vaporization becomes slower than the rate of

condensation89Tro: Chemistry: A Molecular Approach, 2/e


Dynamic Equilibrium

• A system in dynamic equilibrium can respond to changes in the conditions

• When conditions change, the system shifts its position 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 temperature increases, the vapor pressure increases

• Small changes in temperature can make big changes in vapor pressurethe rate of growth depends on strength of the

intermolecular forces



Vapor Pressure Curves

760 mmHg

normal BP100 °C

BP Ethanol at 500 mmHg68.1°C



a) waterb) TiCl4c) etherd) ethanole) acetone

Practice – Which of the following is the most volatile?





Practice – Which of the following has the strongest Intermolecular attractions?




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 form anywhere in the liquidnot just on the surface

• This phenomenon is what is called boiling and the temperature at which 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




Practice – Which of the following has the highest normal boiling point?




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

temperature increases linearly until it reaches the boiling pointq = mass x Cs x T

• Once the temperature reaches the boiling point, all the added heat goes into boiling the liquid – the temperature stays constant

• Once all the liquid has been turned into gas, the temperature can again start to rise



• The logarithm of the vapor pressure vs. inverse absolute temperature is a linear function

• A graph of ln(Pvap) vs. 1/T is a straight line • The slope of the line x 8.314 J/mol∙K = Hvap

in J/mol

Clausius–Clapeyron Equation

• The graph of vapor pressure vs. temperature is an exponential growth curve



Example 11.4: Determine the Hvap of dichloromethane given the vapor pressure vs.

temperature data• Enter the data into a spreadsheet and calculate

the inverse of the absolute temperature and natural log of the vapor pressure




temperature data• Graph the inverse of the absolute temperature vs.

the natural log of the vapor pressure




temperature data• Add a trendline, making sure the display equation

on chart option is checked off




temperature data• Determine the slope of the line

−3776.7 ≈ −3800 K




temperature data• Use the slope of the line to determine the heat of

vaporizationslope ≈ −3800 K, R = 8.314 J/mol∙K



Clausius–Clapeyron Equation2-Point Form

• The equation below can be used with just two measurements of vapor pressure and temperature however, it generally gives less precise results

fewer data points will not give as precise an average because there is less averaging out of the errorso as with any other sets of measurements

• It can also be used to predict the vapor pressure if you know the heat of vaporization and the normal boiling point remember: the vapor pressure at the normal boiling point is 760 torr



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

P2, torr

T1 = BP = 337.8 K, P1 = 760 torr, Hvap = 35.2 kJ/mol, T2 = 285.2 K

P2, torr

Solution:

Example 11.5: Calculate the vapor pressure of methanol at 12.0 °C

106

the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures

Check:

Conceptual Plan:

Relationships:

Given:

Find:

T(K) = T(°C) + 273.15



Practice – Determine the vapor pressure of water at 25 C (normal BP = 100.0 C, Hvap= 40.7 kJ/mol)



T1 = BP = 100.0 °C, P1 = 760 torr, Hvap = 40.7 kJ/mol, T2 = 25.0 °C

P2, torr

T1 = BP = 373.2 K, P1 = 760 torr, Hvap = 40.7 kJ/mol, T2 = 298.2 K

P2, torr

Practice – Determine the vapor pressure of water at 25 C

108

the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

T(K) = T(°C) + 273.15



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 vapor disappears and the states commingle to form a supercritical fluid

• Supercritical fluids have properties of both gas and liquid states



The Critical Point

• The temperature required to produce a supercritical fluid is called the critical temperature

• The pressure at the critical temperature is called the critical pressure

• At the critical temperature or higher temperatures, the gas cannot be condensed to a liquid, no matter how high the pressure gets



Sublimation and Deposition• Molecules in the solid have thermal energy that

allows them to vibrate• 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 is called deposition

• The solid and vapor phases exist in dynamic equilibrium in a closed containerat temperatures below the melting point therefore, molecular solids have a vapor pressure

solid gassublimation

deposition



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• As you heat a solid, its

temperature increases linearly until it reaches the melting point q = mass x Cs x T

• Once the temperature reaches the melting point, all the added heat goes into melting the solid – the temperature stays constant

• Once all the solid has been turned into liquid, the temperature can again start to rise ice/water will always have a

temperature of 0 °C at 1 atm



Energetics of Melting

• When the high energy molecules are lost from the solid, it lowers the average kinetic energy

• If energy is not drawn back into the solid its temperature will decrease – therefore, melting is an endothermic processand freezing is an exothermic process

• Melting requires input of energy to overcome the attractions between molecules



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

solid is called the Heat of Fusion, Hfus sometimes called the enthalpy of fusion

• Always endothermic, therefore Hfus is +• Somewhat temperature dependent Hcrystallization = −Hfusion

Generally much less than Hvap

Hsublimation = Hfusion + Hvaporization



Heats of Fusion and Vaporization



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 Cs x Tmass of 1.00 mole of ice = 18.0 gCs = 2.09 J/mol∙°C



Segment 2

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

• q = n∙Hfus

n = 1.00 mole of iceHfus = 6.02 kJ/mol



Segment 3

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

• q = mass x Cs x Tmass of 1.00 mole of water = 18.0 gCs = 2.09 J/mol∙°C



Segment 4

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

• q = n∙Hvap

n = 1.00 mole of iceHfus = 40.7 kJ/mol



Segment 5

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

• q = mass x Cs x Tmass of 1.00 mole of water = 18.0 gCs = 2.01 J/mol∙°C



Practice – How much heat, in kJ, is needed to raise the temperature of a 12.0 g benzene

sample from −10.0 °C to 25.0 °C?



12.0 g benzene, seg 1 (T1 = −10.0 °C, T2 = 5.5 °C), seg 2 = melting, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ

12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = 1.51 kJ, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ

Practice – How much heat is needed to raise the temperature of a 12.0 g benzene sample from −10.0 °C to 25.0 °C?

125

Hfus 9.8 kJ/mol, 1 mol = 78.11 g, 1 kJ = 1000 J, q = m∙Cs∙TCs,sol = 1.25 J/g°C, Cs,liq = 1.70 J/g°C

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

g J kJSeg 1

12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = melting, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C)kJ

g J kJSeg 3 g mol kJSeg 2

12.0 g benzene, seg 1 = 0.2325 kJ,seg 2 = 1.51 kJ, seg 3 = 0.3978 kJkJ



Phase Diagrams• Phase diagrams describe the different states and

state changes that occur at various temperature/pressure conditions

• Regions represent states• Lines represent state changes

liquid/gas line is vapor pressure curveboth 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 DiagramsP

ress

ure

Temperature

vaporization

condensation

criticalpoint

triplepoint

Solid Liquid

Gas

1 atm

normalmelting pt.

normalboiling pt.

Fusion Curve

Vapor PressureCurve

SublimationCurve

melting

freezing

sublimation

deposition




Phase Diagram of Water

Temperature

Pre

ssur

e

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




Morphic Forms of Ice



Phase Diagram of CO2P

ress

ure

Temperature

criticalpoint

31.0 °C72.9 atm

triplepoint

Solid Liquid

Gas1 atm

-56.7 °C5.1 atm

normalsublimation pt.

-78.5 °C





• 20.0 °C, 72.9 atm

• −56.7 °C, 5.1 atm

• 10.0 °C, 1.0 atm

• −78.5 °C, 1.0 atm

• 50.0 °C, 80.0 atm

• 20.0 °C, 72.9 atm liquid

• −56.7 °C, 5.1 atm solid, liquid, gas

• 10.0 °C, 1.0 atm gas

• −78.5 °C, 1.0 atm solid, gas

• 50.0 °C, 80.0 atm scf

Practice – Consider the phase diagram of CO2

shown. What phase(s) is/are present at each of the following conditions?



Water – An Extraordinary Substance• Water is a liquid at room temperature

most molecular substances with similar molar masses are gases at room temperature

e.g. NH3, CH4

due to H-bonding between molecules

• Water is an excellent solvent – dissolving many ionic and polar molecular substances because of its large dipole moment even many small nonpolar molecules have some 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



Crystal Lattice

• When allowed to cool slowly, the particles in a liquid will arrange themselves to give the maximum attractive forcestherefore 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 of arrangement for all the particles is called the unit cell



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

unit cells throughout the lattice



7 Unit Cells

Cubica = b = c

all 90°

a

b

c

Tetragonala = c < b

all 90°

a

b

c

Orthorhombica b c

all 90°

a

b

c

Monoclinica b c

2 faces 90°

a b

c

Hexagonala = c < b

2 faces 90°1 face 120°

c

a

b

Rhombohedrala = b = cno 90°

ab

c

Triclinica b c

no 90°

a

b

c

140


Unit Cells

• The number of other particles each particle is in contact with 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 crystal together

• The packing efficiency is the percentage of volume in the unit cell occupied by particles the higher the coordination number, the more efficiently

the particles are packing together



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 particles⅛ of each corner particle is within the cube½ of each particle on a face is within the cube¼ of each particle on an edge is within the cube




Cubic Unit Cells - Simple Cubic

• Eight particles, one at each corner of a cube

• 1/8th of each particle lies in the unit celleach particle part of eight cellstotal = one particle in each unit

cell8 corners x 1/8

• Edge of unit cell = twice the radius

• Coordination number of 6

2r



Simple Cubic



Cubic Unit Cells - Body-Centered Cubic

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

• 1/8th of each corner particle lies in the unit celltwo particles in each unit cell

8 corners x 1/8 + 1 center

• Edge of unit cell = (4/ 3) times the radius of the particle


146

4r

3



Body-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 face particle lies in the unit cell4 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


148

2r 2



Face-Centered Cubic



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

Example 11.6: Calculate the density of Al if it crystallizes in a fcc and has a radius of 143 pm

150

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:

Conceptual Plan:

Relationships:

Given:Find:

1 cm = 102 m, 1 pm = 10−12 m

lr Vmassfcc

dm, V

# atoms x mass of 1 atom

V = l 3, l = 2r√2, d = m/Vd = 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

face-centered cubic, V = 6.618 x 10−23 cm3, m =1.792 x 10−22g

density, g/cm3



Practice – Estimate the density of Rb if it crystallizes in a body-centered cubic unit cell and has an atomic

radius of 247.5 pm



Solution:bcc = 2 atoms/uc, Rb = 85.47 g/mol, 1 mol = 6.022 x 1023 atoms

Practice – Estimate the density of Rb if it crystallizes in a bcc and has a radius of 247.5 pm

the accepted density of Rb at 20°C is 1.53 g/cm3, so the answer makes sense

body-centered cubic, r = 247.5 pm

density, g/cm3

Check:

Conceptual Plan:

Relationships:

Given:Find:

1 cm = 102 m, 1 pm = 10−12 m

lr Vmassbcc

dm, V

# atoms x mass of 1 atom

V = l 3, l = 4r/√3, d = m/Vd = m/V

l = 4r/√3 V = l 3

body-centered cubic, r = 2.475 x 10−8 cm, m = 2.839 x 10−22 g

density, g/cm3

body-centered cubic, V= 1.868 x 10−22 cm3, m= 2.839 x 10−22 g

density, g/cm3



Closest-Packed StructuresFirst Layer

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



Closest-Packed StructuresSecond Layer

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

• Or the second layer can sit over the holes in the first layer – called an AB pattern



Closest-Packed StructuresThird Layer – with Offset 2nd Layer

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

• Or the third layer can sit over the uncovered holes in the first layer– called an ABC pattern

Hexagonal Closest-PackedCubic Closest-PackedFace-Centered Cubic



Hexagonal Closest-Packed Structures



Cubic Closest-Packed Structures



Classifying Crystalline Solids

• Crystalline solids are classified by the kinds of particles found

• Some of the categories are sub-classified by the kinds of attractive forces holding the particles together



Classifying Crystalline Solids• Molecular solids are solids whose composite

particles are molecules

• Ionic solids are solids whose composite particles are ions

• Atomic solids are solids whose composite particles are atomsnonbonding atomic solids are held together by

dispersion forcesmetallic atomic solids are held together by metallic bondsnetwork covalent atomic solids are held together by

covalent bonds




Molecular Solids

• The lattice sites are occupied by moleculesCO2, H2O, C12H22O11

• The molecules are held together by intermolecular attractive forcesdispersion forces, dipole–dipole attractions, and H-

bonds

• Because the attractive forces are weak, they tend to have low melting pointsgenerally < 300 °C



Ionic Solids• Lattice sites occupied by ions

• Held together by attractions between oppositely charged ions 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 the cations and anions that maintains charge balance generally, anions are larger than cations the number of anions that can surround the cation is limited by the

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



Ionic Crystals

CsClcoordination number = 8

Cs+ = 167 pmCl─ = 181 pm

NaClcoordination number = 6

Na+ = 97 pmCl─ = 181 pm



Lattice Holes

Simple CubicHole

OctahedralHole

TetrahedralHole



Lattice Holes• In hexagonal closest-packed or cubic closest- packed

lattices there are eight tetrahedral holes and four octahedral holes per unit cell

• In a simple cubic lattice there is one cubic hole per unit cell

• Number and type of holes occupied determines formula (empirical) of the salt

= Octahedral

= Tetrahedral



Cesium Chloride Structures

• Coordination number = 8

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

• Whole Cs+ (167 pm) inside the unit cellcubic hole = hole in simple

cubic arrangement of Cl─ ions

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



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

• Na+ (97 pm) in holes between Cl─

octahedral holes 1 in center of unit cell 1 whole particle in every octahedral hole 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 (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 particle 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 particle 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 structure when the cation:anion ratio is 2:1 the anions occupy the lattice sites and the

cations occupy the tetrahedral holes



Practice – Gallium arsenide crystallizes in a cubic closest-packed array of arsenide ions with gallium ions in ½ the tetrahedral holes. What is the ratio of gallium ions to arsenide ions in the structure and the empirical

formula of the compound?

As = cpp = 4 atoms per unit cell

Ga = ½ (8 tetrahedral holes per unit cell)Ga = 4 atoms per unit cell

Ga:As = 4 atoms :4 atoms per unit cell = 1:1The formula is GaAs



Nonbonding Atomic Solids

• Noble gases in solid form

• Solid held together by weak dispersion forcesvery low melting

• Tend to arrange atoms in closest-packed structureeither hexagonal cp or cubic cpmaximizes attractive forces and minimizes energy



Metallic Atomic Solids

• Solid held together by metallic bondsstrength varies with sizes and charges of cations

coulombic attractions

• Melting point varies

• Mostly closest-packed arrangements of the lattice pointscations



Metallic Structure



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 their nearest neighbors by covalent bonds

• Because of the directionality of the covalent bonds, these do not tend to form closest-packed arrangements in the crystal

• Because of the strength of the covalent bonds, these have 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 four covalent bonds to surrounding atomssp3

tetrahedral geometry

• This effectively makes each crystal one giant molecule held together by covalent bondsyou can follow a path of covalent bonds from any

atom to every other atom



Properties of Diamond • Very high melting point, ~3800 °C

need to overcome some covalent bonds

• Very rigid due to the directionality of the covalent

bonds

• 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



The Graphite Structure:a 2-Dimensional Network

• In graphite, the carbon atoms in a sheet are covalently bonded together forming six-member flat rings fused together

similar to benzenebond length = 142 pm

sp2 each C has three sigma and one pi bond

trigonal-planar geometryeach sheet a giant molecule

• The sheets are then stacked and held together by dispersion forcessheets are 341 pm apart



Properties of Graphite• Hexagonal crystals

• High melting point, ~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

• ~90% of Earth’s crust

• Extended arrays of SiOsometimes with Al substituted for Si – aluminosilicates

• Glass is the amorphous form



Quartz• SiO2 in pure form

impurities add color

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

• Melts at ~1600 °C• Very hard



Micas• There are various kinds of mica that have slightly different

compositions – but are all of the general form X2Y4-6Z8O20(OH,F)4 X is K, Na, or Ca or less commonly Ba, Rb, or Cs Y is Al, Mg, or Fe or less commonly Mn, Cr, Ti, Li, etc.Z is chiefly Si or Al but also may include Fe3+ or Ti

• Minerals that are mainly 2-dimensional arrays of Si bonded to Ohexagonal arrangement of atoms

• Sheets

• Chemically stable

• Thermal and electrical insulator



a) KCl ionic SCl2 molecular

b) C(s, graphite) cov. network S8 molecular

c) Kr atomic K metallic

d) SrCl2 ionic SiO2 (s, quartz) cov. network

Practice – Pick the solid in each pair with the highest melting point

a) KCl SCl2

b) C(s, graphite)

c) Kr K

d) SrCl2 SiO2 (s, quartz)



Band Theory

• The structures of metals and covalent network solids result in every atom’s orbitals being shared 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 two 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 band of antibonding molecular orbitals

• The band of bonding molecular orbitals is called the valence band

• The band of antibonding molecular orbitals is called the conduction band



Molecular Orbitals of Polylithium



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 to jump to the conduction band

• The difference in energy between the valence band and conduction band is called the band gapthe larger the band gap, the fewer electrons there

are with enough energy to make the jump



Types of Band Gaps andConductivity



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

• If the band gap is ~0, then the electrons will be almost as likely to be in the conduction band as the valence band and 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 the electrons will be in the conduction band at normal temperatures 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 will be in the conduction band at normal temperatures and the material will be an insulator



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 are doped by adding electron-rich impurities

• p-type semiconductors are doped with an electron-deficient impurity, resulting in electron “holes” in the valence band. Electrons can jump between these holes in 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 in only one direction – this is called a diode

• This also allows the accumulation of electrical energy – called an amplifier


Documents

Chapter 11 Liquids, Solids, and Intermolecular Forces