Chapter 11 Liquids, Solids, and Intermolecular Forces

Chapter 11 Liquids, Solids, and Intermolecular Forces

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 end setae = 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

Properties of the three Phases of Matter
Fixed = keeps shape when placed in a container Indefinite = takes the shape of the container

Three Phases of Water 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.

Degrees of Freedom Particles may have one or several types of freedom of motion and 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 Particles packed close and are fixed Incompressible
though they may vibrate Incompressible Retention of shape and volume Do not flow Crystalline Solids - particles arranged in an orderly geometric pattern salt and diamonds Amorphous Solids – particles do not show a regular geometric pattern over a long range plastic and glass

Liquids Particles are closely packed, but they have some ability to move around Incompressible Take the shape of their container and to flow Do not have enough freedom to escape or expand to fill the container

Gases Particles have complete freedom of motion and are not held together Particles in constant motion, bumping into each other and the container There is a large amount of space between the particles compared to the size of the particles molar volume of the gas state of a material is much larger than the molar volume of the solid or liquid states Compressible 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 temperature KEavg = 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 moving though no material completely lacks particle motion

Kinetic–Molecular Theory of 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 container gases 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
Solids exhibit strong attractive forces limiting the kinetic energy of the particles Solid packing results in no translational or rotational motion the only freedom they have is vibrational motion

Kinetic–Molecular Theory of Liquids
Liquids exhibit attractive forces strong enough to partially overcome the kinetic energy Particles are packed together with only very limited translational or rotational freedom

Explaining the Properties of Liquids
Because the particles are in contact… Higher densities than gases Incompressible Because the particles have limited translational freedom… Indefinite shape allowing for flow Take the shape of the container Because the limit on their freedom keeps the particles from escaping each other Liquids have a definite volume

Phase Changes The attractive forces between the molecules are fixed changing the material’s state requires changing the amount of kinetic energy that is limiting the particles freedom Solids melt when heated because the particles gain enough kinetic energy to partially overcome the attractive 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 solid other 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 particles However, not breaking the covalent bonds The higher the normal boiling point of the liquid, the stronger the intermolecular attractive forces

Kinds of Attractive Forces
Dispersion Forces - Unequal electron distribution leads to attractions causing temporary polarity in the molecules Dipole–Dipole Attractions - Permanent polarity in the molecules due to their structure leads to attractive forces Hydrogen Bonds - An especially strong dipole–dipole attraction resulting when H is attached to an extremely electronegative atom.

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 forces aka 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 molecule more 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 + - + −

Effect of Molecular Size on Size of Dispersion Force
As the molar mass increases, the number of electrons increases. Therefore the strength of the dispersion forces increases. The Noble gases are all nonpolar atomic elements The stronger the attractive forces between the molecules, the higher the boiling point will be.

Properties of Straight Chain Alkanes NonPolar Molecules

Boiling Points of n-Alkanes

Effect of Molecular Shape on Size of Dispersion Force
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

Practice – Choose the Substance in Each Pair with the Higher Boiling Point
a) CH C4H10 b) C6H C6H12

a) CH4 CH3CH2CH2CH3 b) CH3CH2CH=CHCH2CH3 cyclohexane
Practice – Choose the Substance in Each Pair with the Higher Boiling Point a) CH4 CH3CH2CH2CH3 b) CH3CH2CH=CHCH2CH3 cyclohexane Both molecules are nonpolar larger 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 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 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
Given: Find: CH2Cl2, EN C = 2.5, H = 2.1, Cl = 3.0 If there are dipole–dipole attractions Conceptual Plan: Relationships: Lewis Structure Bond Polarity Molecule Formula EN Difference Shape molecules that have dipole–dipole attractions must be polar Cl—C 3.0−2.5 = 0.5 polar Solution: polar bonds and tetrahedral shape = polar molecule 4 bonding areas no lone pairs = tetrahedral shape C—H 2.5−2.1 = 0.4 nonpolar polar molecule; therefore dipole–dipole attractions

Practice – Choose the substance in each pair with the higher boiling point
a) CH2FCH2F CH3CHF2 b) or

Practice – Choose the substance in each pair with the higher boiling point
a) CH2FCH2F CH3CHF2 more polar or b) polar nonpolar

Hydrogen Bonding When a very electronegative atom is bonded to hydrogen, it strongly pulls the bonding electrons toward it O─H, N─H, or F─H Because hydrogen has no other electrons, when its electron is pulled away, the nucleus becomes deshielded exposing 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 forces stronger 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 bonds 2 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.

Example 11.2: Which of these compounds is a liquid at room temperature (the others are gases). Why?
MM = 30.03 Polar No H-Bonds MM = 34.03 Polar No H-Bonds MM = 34.02 Polar H-Bonds Step 2. Compare intermolecular attractive forces Step 3. Evaluate the strengths of the total intermolecular attractive forces. The substance with the strongest will be the liquid. Step 1. Determine the kinds of intermolecular attractive forces Because the molar masses are similar, the size of the dispersion force attractions should be similar Fluoromethane: dispersion forces: MM 34.03, tetrahedral dipole–dipole: very polar C–F bond uncancelled H-bonding: no O–H, N–H, or F–H therefore no H-bond Formaldehyde: dispersion forces: MM 30.03, trigonal planar dipole–dipole: very polar C=O bond uncancelled H-bonding: no O–H, N–H, or F–H therefore no H-bond Hydrogen peroxide: dispersion forces: MM 34.02, tetrahedral bent dipole–dipole: polar O–H bonds uncancelled H-bonding: O–H, therefore H-bond 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. 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) a) CH3OH CH3CHF2 b) CH3-O-CH2CH3 CH3CH2CH2NH2 can H-bond can H-bond

Attractive Forces and Solubility
Solubility depends, in part, on the attractive forces of the solute and solvent molecules like dissolves like miscible liquids will always dissolve in each other Polar substances 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 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 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.

Polar Solvents Dichloromethane (methylene chloride) Water Ethanol
(ethyl alcohol) Dichloromethane (methylene chloride) Water

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

Practice – Choose the substance in each pair that is more soluble in water
a) CH3OH CH3CHF2 b) CH3CH2CH2CH2CH3 CH3Cl can H-bond with H2O more polar

Summary Dispersion forces
The weakest of the intermolecular attractions Present in all molecules and atoms Magnitude of the dispersion forces increases with molar mass Polar molecules also have dipole–dipole attractive forces Hydrogen bonds the strongest of the intermolecular attractive forces a pure substance can have present when a molecule has H directly bonded to either O , N, or F atoms only example of H bonded to F is HF Ion–dipole attractions present in mixtures of ionic compounds with polar molecules. the strongest intermolecular attraction especially important in aqueous solutions of ionic compounds

Properties & Structure
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 spherical as long as there is no gravity

Surface Tension The layer of molecules on the surface behave differently than molecules in the interior 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 molecules have a higher potential energy and are less stable than those in the interior because they have fewer neighbors to attract them The surface tension of a liquid is the energy required to increase the surface area a given amount surface tension of H2O = 72.8 mJ/m2 at room temperature surface tension of C6H6 = 28 mJ/m2

Factors Affecting Surface Tension
Stronger intermolecular attractive forces  higher surface tension Raising the temperature of a liquid reduces its surface tension increases the average kinetic energy of the molecules the 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∙s often 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 The more spherical the molecular shape, the lower the viscosity will be molecules 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 gravity the 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 together adhesive 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 (dyed red here) 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 glass metallic bonds are stronger than intermolecular attractions

The Molecular Dance Molecules in the liquid are constantly in motion
vibrational, 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 forces therefore – 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 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 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 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 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 process and 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, DHvap 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
Given: Find: 155 kJ g H2O Conceptual Plan: Relationships: 1 mol H2O = 40.7 kJ, 1 mol = g kJ mol H2O g H2O Solution: Check: because the given amount of heat is almost 4x the DHvap, the amount of water makes sense

Practice – Calculate the amount of heat needed to vaporize 90
Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point (DHvap = 39.9 kJ/mol)

Practice – Calculate the amount of heat needed to vaporize 90
Practice – Calculate the amount of heat needed to vaporize 90.0 g of C3H7OH at its boiling point Given: Find: 90.0 g kJ Conceptual Plan: Relationships: g mol kJ 1 mol C3H7OH = 39.9 kJ, 1 mol = g Solution: Check: because the given amount of C3H7OH is more than 1 mole the amount of heat makes sense

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 condensation

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 pressure the rate of growth depends on strength of the intermolecular forces

Vapor Pressure Curves normal BP 100 °C BP Ethanol at 500 mmHg 68.1°C

Practice – Which of the following is the most volatile?
a) water b) TiCl4 c) ether d) ethanol e) acetone a) water b) TiCl4 c) ether d) ethanol e) acetone

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 liquid not 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 point q = mass x Cs x DT 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

Clausius–Clapeyron Equation
A graph of ln(Pvap) vs. 1/T is a straight line The slope of the line x J/mol∙K = DHvap in J/mol The graph of vapor pressure vs. temperature is an exponential growth curve The logarithm of the vapor pressure vs. inverse absolute temperature is a linear function

Example 11.4: Determine the DHvap 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

Graph the inverse of the absolute temperature vs. the natural log of the vapor pressure

Add a trendline, making sure the display equation on chart option is checked off

Determine the slope of the line − ≈ −3800 K

Use the slope of the line to determine the heat of vaporization slope ≈ −3800 K, R = J/mol∙K

Clausius–Clapeyron Equation 2-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 errors 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

Example 11.5: Calculate the vapor pressure of methanol at 12.0 °C
Given: Find: T1 = BP = 64.6 °C, P1 = 760 torr, DHvap = 35.2 kJ/mol, T2 = 12.0 °C P2, torr T1 = BP = K, P1 = 760 torr, DHvap = 35.2 kJ/mol, T2 = K P2, torr Conceptual Plan: Relationships: P1, T1, DHvap P2 T(K) = T(°C) Solution: Check: the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures

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

Practice – Determine the vapor pressure of water at 25 C
Given: Find: T1 = BP = K, P1 = 760 torr, DHvap = 40.7 kJ/mol, T2 = K P2, torr T1 = BP = °C, P1 = 760 torr, DHvap = 40.7 kJ/mol, T2 = 25.0 °C P2, torr Conceptual Plan: Relationships: P1, T1, DHvap P2 T(K) = T(°C) Solution: Check: the units are correct, the size makes sense because the vapor pressure is lower at lower temperatures

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 container at temperatures below the melting point therefore, molecular solids have a vapor pressure sublimation solid gas deposition

Sublimation & Deposition

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 DT 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 process and 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, DHfus 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

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 DT mass of 1.00 mole of ice = 18.0 g Cs = 2.09 J/mol∙°C

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

Segment 3 Heating 1.00 mole of water at 0.0 °C up to the boiling point, °C q = mass x Cs x DT mass of 1.00 mole of water = 18.0 g Cs = 2.09 J/mol∙°C

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

Segment 5 Heating 1.00 mole of steam at 100.0 °C up to 125.0 °C
q = mass x Cs x DT mass of 1.00 mole of water = 18.0 g Cs = 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?

Practice – How much heat is needed to raise the temperature of a 12
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? Given: Find: 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 = kJ, seg 2 = melting, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C) kJ 12.0 g benzene, seg 1 = kJ, seg 2 = 1.51 kJ, seg 3 = kJ kJ 12.0 g benzene, seg 1 = kJ, seg 2 = 1.51 kJ, seg 3 (T1 = 5.5 °C, T2 = 25.0 °C) kJ Conceptual Plan: Relationships: g g g J mol J kJ kJ kJ Seg 2 Seg 1 Seg 3 DHfus 9.8 kJ/mol, 1 mol = g, 1 kJ = 1000 J, q = m∙Cs∙DT Cs,sol = 1.25 J/g°C, Cs,liq = 1.70 J/g°C Solution:

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 curve both states exist simultaneously critical 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 Liquid Pressure Solid Gas Temperature melting freezing
Fusion Curve critical point melting freezing Liquid Solid Pressure normal melting pt. normal boiling pt. 1 atm Sublimation Curve triple point Vapor Pressure Curve vaporization condensation sublimation deposition Gas Temperature

Phase Diagram of Water Pressure Water Ice Steam Temperature 1 atm
critical point 374.1 °C 217.7 atm triple Ice Water Steam 1 atm normal boiling pt. 100 °C melting pt. 0 °C 0.01 °C 0.006 atm

Morphic Forms of Ice

Phase Diagram of CO2 Liquid Pressure Solid Gas Temperature 1 atm
critical point 31.0 °C 72.9 atm triple Solid Liquid Gas 1 atm -56.7 °C 5.1 atm normal sublimation pt. -78.5 °C

Practice – Consider the phase diagram of CO2 shown
Practice – Consider the phase diagram of CO2 shown. What phase(s) is/are present at each of the following conditions? 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 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

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

Properties & Structure
Solids Properties & Structure

Crystal Lattice When allowed to cool slowly, the particles in a liquid will arrange themselves to give the maximum 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 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 c c c c b b b b a a a a Cubic a = b = c all 90°
Tetragonal a = c < b all 90° Orthorhombic a ¹ b ¹ c all 90° Monoclinic a ¹ b ¹ c 2 faces 90° Hexagonal a = c < b 2 faces 90° 1 face 120° c a b c c b b a a Rhombohedral a = b = c no 90° Triclinic a ¹ b ¹ c no 90°

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 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

Cubic Unit Cells - Simple Cubic
Eight particles, one at each corner of a cube 1/8th of each particle lies in the unit cell each particle part of eight cells total = one particle in each unit cell 8 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 cell two particles in each unit cell 8 corners x 1/8 + 1 center Edge of unit cell = (4/Ö 3) times the radius of the particle Coordination number of 8

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 cell 4 particles in each unit cell 8 corners x 1/8 + 6 faces x 1/2 Edge of unit cell = times the radius of the particle Coordination number of 12

Face-Centered Cubic

Example 11.6: Calculate the density of Al if it crystallizes in a fcc and has a radius of 143 pm
Given: Find: face-centered cubic, V = x 10−23 cm3, m =1.792 x 10−22g density, g/cm3 face-centered cubic, r = 143 pm density, g/cm3 face-centered cubic, r = 1.43 x 10−8 cm, m = x 10−22 g density, g/cm3 Conceptual Plan: Relationships: mass fcc l r V # atoms x mass of 1 atom l = 2r√2 V = l 3 d m, V d = m/V 1 cm = 102 m, 1 pm = 10−12 m V = l 3, l = 2r√2, d = m/V fcc = 4 atoms/uc, Al = g/mol, 1 mol = x 1023 atoms Solution: Check: the accepted density of Al at 20°C is 2.71 g/cm3, so the answer makes sense

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

Practice – Estimate the density of Rb if it crystallizes in a bcc and has a radius of 247.5 pm
Given: Find: body-centered cubic, V= x 10−22 cm3, m= x 10−22 g density, g/cm3 body-centered cubic, r = x 10−8 cm, m = x 10−22 g density, g/cm3 body-centered cubic, r = pm density, g/cm3 Conceptual Plan: Relationships: mass bcc l r V # atoms x mass of 1 atom l = 4r/√3 V = l 3 d m, V d = m/V 1 cm = 102 m, 1 pm = 10−12 m V = l 3, l = 4r/√3, d = m/V bcc = 2 atoms/uc, Rb = g/mol, 1 mol = x 1023 atoms Solution: Check: the accepted density of Rb at 20°C is 1.53 g/cm3, so the answer makes sense 147

Closest-Packed Structures First 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 Structures Second 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 Structures Third 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 Cubic Closest-Packed Face-Centered Cubic Hexagonal Closest-Packed

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 153

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

Molecular Solids The lattice sites are occupied by molecules
CO2, H2O, C12H22O11 The molecules are held together by intermolecular attractive forces dispersion forces, dipole–dipole attractions, and H-bonds Because the attractive forces are weak, they tend to have low melting points generally < 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 CsCl coordination number = 8 Cs+ = 167 pm Cl─ = 181 pm
NaCl coordination number = 6 Na+ = 97 pm Cl─ = 181 pm

Lattice Holes Tetrahedral Hole Octahedral Hole Simple Cubic Hole

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 cell cubic 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

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

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 cp maximizes attractive forces and minimizes energy

Metallic Atomic Solids
Solid held together by metallic bonds strength varies with sizes and charges of cations coulombic attractions Melting point varies Mostly closest-packed arrangements of the lattice points cations

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 points generally > 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 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

Properties of Diamond Very high melting point, ~3800 °C Very rigid
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 benzene bond length = 142 pm sp2 each C has three sigma and one pi bond trigonal-planar geometry each sheet a giant molecule The sheets are then stacked and held together by dispersion forces sheets 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 SiO
sometimes 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 O hexagonal arrangement of atoms Sheets Chemically stable Thermal and electrical insulator

Practice – Pick the solid in each pair with the highest melting point
KCl SCl2 C(s, graphite) Kr K SrCl2 SiO2 (s, quartz) KCl ionic SCl2 molecular C(s, graphite) cov. network S8 molecular Kr atomic K metallic SrCl2 ionic SiO2 (s, quartz) cov. network

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 gap the larger the band gap, the fewer electrons there are with enough energy to make the jump

Types of Band Gaps and 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 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

Chapter 11 Liquids, Solids, and Intermolecular Forces

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Chapter 11 Liquids, Solids, and Intermolecular Forces

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