Liquids, Solids, and Intermolecular Forces or

Liquids, Solids, and Intermolecular Forces or Why your Water Evaporates and Your Cheerios Don t Previous topics concerned Intramolecular forces & structure. These are the net forces of attraction that are responsible for the ionic and covalent bonds that bind atoms to form individual compounds. We now move to examining the various forces of attraction that cause individual molecules to be attracted to one another to form gases, liquids and solids. These are the intermolecular forces. 2 Nature Of Intermolecular Forces, Properties & Factors Affecting Behavior. 1

Why are molecules attracted to each other? Strengths of Intermolecular Forces VDW = Van der Waals 5 Intermolecular attractions determine how tightly liquids and solids pack The strength of intermolecular attractions determine many physical properties: b ili i t vapor pressure, boiling point viscosity surface tension density melting point, freezing point (range) 2

Freedom of Motion (aka degrees of freedom) the molecules in a gas have complete freedom of motion so kinetic energy overcomes the attractive forces. the molecules in a solid are locked in place, they cannot move around though they do vibrate. the molecules in a liquid have limited freedom they can move around a little within the structure - have enough kinetic energy to overcome some of the attractive forces, but not enough to escape each other. Properties of the 3 Phases of Matter State Shape Volume Compressible Flow Strength of Intermolecular Attractions Solid Fixed Fixed No No very strong Liquid Indef. Fixed No Yes moderate Gas Indef. Indef. Yes Yes very weak Fixed = keeps shape when placed in a container Indefinite = takes the shape of the container 3

Kinetic - Molecular Theory the properties explained based on the kinetic energy of the molecules and the attractive forces between them kinetic energy is proportional freedom of motion degrees of freedom = translational, rotational, vibrational attractive forces keep the molecules together kinetic energy depends only on the temperature KE = 1.5 kt Gas Structure Gas molecules generally don t stick to each other. Why not? Explaining the Properties of Solids solid particles packed close together and fixed in position retain their shape and volume when placed in a new container; and prevents the particles from flowing incompressible 4

crystalline solids salts and diamonds Solids partially ordered solids certain plastics (nylon, PE, PP) liquid crystals (2-dimensional) amorphous solids rubber, polystyrene and window glass Explaining the Properties of Liquids close contact higher densities than gases limited freedom of movement indefinite shape but definite volume (no escape) Compressibility 5

Phase Changes Why are molecules attracted to each other? intermolecular attractions are due to attractive forces between opposite charges + ion to - ion + end of polar molecule to - end of polar molecule -bonding especially strong even nonpolar molecules will have temporary charges larger the charge = stronger attraction longer the distance = weaker attraction 6

these attractive forces are small relative to the ionic, covalent and metallic bonding forces between atoms generally smaller charges generally over much larger distances Dipole-dipole attractions Polar molecules tend to align their partial charges The attractive force is about 1% of a covalent bond and drops off as 1/d 3 (d=distance between dipoles) ydrogen bonds (5-10% of covalent bond) Very strong dipole-dipole attraction when is covalently bonded to to a highly electronegative atom (F, O, or N) Typically about ten times stronger than other dipoledipole attractions Are responsible for the expansion of water as it freezes Partly responsible for twisting of proteins in a helix (DNA) 7

ydrogen bonds Van der Waals (including London or dispersion) forces The (very) weak attractions between nonpolar atoms and molecules Arise from the interactions of instantaneous dipoles on neighboring molecules London forces depend on the number of atoms in the molecule The increase in boiling point with chain length of hydrocarbons demonstrates this trend Formula C C 2 C 3 C 4 4 6 8 10 BP at 1atm ( -161.5-88.6-42.1-0.5 o C) Formula C 5 C 6 C 22 12 14 46 BP at 1atm ( 36.1 68.7 327 o C) 8

Trends in the Strength of Intermolecular Attraction the stronger the attractions between the atoms or molecules, the more energy it will take to separate them boiling a liquid requires we add enough energy to overcome the attractions between the molecules or atoms the higher the normal boiling point of the liquid, the stronger the intermolecular attractive forces Attractive Forces + - + - + - + - + + + + + + + + + + + - - - - - - - + + + + + - - - - - Dispersion Forces fluctuations in electron distribution in atoms and molecules 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 London or dispersion forces all molecules and atoms exert them as a temporary dipole is established in one molecule, it induces a dipole in all the surrounding molecules 9

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 orbitals more electrons larger electron cloud increased polarizability stronger attractions shape of the molecule more surface-to-surface contact larger induced dipole stronger attraction Effect of Molecular Size on Size of Dispersion Force Noble Gases are all nonpolar atomic elements. As the molar mass increases, the size of electron cloud increases. Therefore the strength of the dispersion forces increases. The stronger the attractive forces between the molecules, the higher the boiling point will be. 10

Relationship between Induced Dipole and Molecular Size 250 200 150 100 BP, Noble Gas BP, alogens oint, C Boiling P 50 BP, X4 0 1 2 3 4 5 6-50 -100-150 -200-250 -300 Period Properties of Straight Chain Alkanes Non-Polar Molecules Name Molar Mass BP, C MP, C Density, g/ml Methane 16-162 -183 0.47 Ethane 30-89 -183 0.57 Propane 44-42 -188 0.5 Butane 58 0-138 0.58 Pentane 72 36-130 0.56 exane 86 69-95 0.66 eptane 100 98-91 0.68 Octane 114 126-57 0.7 Nonane 128 151-54 0.72 Decane 142 174-30 0.74 Undecane 156 196-26 0.75 Dodecane 170 216-10 0.76 Tridecane 184 235-5 0.76 Tetradecane 198 254 6 0.77 Pentadecane 212 271 10 0.79 exadecane 226 287 18 0.77 Boiling Points of n-alkanes 11

500 n-alkane Boiling & Melting Points 400 300 200 re, C 100 Temperatur 0-100 BP, n- alkane -200-300 0 50 100 150 200 250 300 350 400 450 500 Molar Mass Effect of Molecular Shape on Size of Dispersion Force branched chains (iso-)have lower BPs than straight chains the straight chain (n-) isomers have more surface-tosurface contact Alkane Boiling Points 12

Practice Choose the Substance in Each Pair with the igher Boiling Point a) C 4 C 3 C 2 C 2 C 3 C C C C C b) C 3 C 2 C=CC 2 C 3 cyclohexane C C C C C C C C C C C C Practice Choose the Substance in Each Pair with the ighest Boiling Point both molecules a) C 4 C 3 C 2 C 2 C 3 are nonpolar C C C C C larger molar mass b) C 3 C 2 C=CC 2 C 3 cyclohexane C C C C C C both molecules are nonpolar flat molecule larger surface-to-surface contact C C C C C C Dipole-Dipole Attractions polar molecules have a permanent dipole because of bond polarity and shape dipole moment as well as the always present induced dipole the permanent dipole adds to the attractive forces bt between the molecules l raising the boiling and melting points relative to nonpolar molecules of similar size and shape 13

Effect of Dipole-Dipole Attraction on Boiling and Melting Points Molar Boiling Dipole Mass Point Size C3C2C3 44.09-42 C 0.08 D C3-O-C3 46.07-24 C 1.30 D C3 - C=O 44.05 20.2 C 2.69 D C3-C N 41.05 81.6 C 3.92 D Practice Choose the Substance in Each Pair with the ighet Boiling Point a) C 2 FC 2 F C 3 CF 2 F C C F F C C F b) or 14

Practice Choose the Substance in Each Pair with the ighest Boiling Point a) C 2 FC 2 F C 3 CF 2 b) polar F C C F or F C C F nonpolar more polar Attractive Forces and Solubility Solubility depends on the attractive forces of solute and solvent molecules Like dissolves Like miscible liquids will always dissolve in each other polar substance dissolve in polar solvents hydrophilic groups = O, CO, C=O, COO, N 2, Cl nonpolar molecules dissolve in nonpolar solvents hydrophobic groups = C-, C-C Many molecules have both hydrophilic and hydrophobic parts - solubility becomes competition between parts Immiscible Liquids 15

Polar Solvents Dichloromethane (methylene chloride) Ethanol Water (ethyl alcohol) C 3 2 2 C C C 2 C 3 C 2 n-hexane Nonpolar Solvents C C C 3 C C C C toluene Cl Cl Cl C Cl carbon tetrachloride ydrogen Bonding When a very electronegative atom is bonded to hydrogen, it strongly pulls the bonding electrons toward it O-, N-, or F- Since hydrogen has no other electrons, when it loses the electrons, the nucleus becomes deshielded exposing the proton The exposed proton acts as a very strong center of positive charge, attracting all the electron clouds from neighboring molecules 16

-Bonding F -Bonding in Water Boilin Po oint, C 150 100 Relationship between -bonding and Intermolecular Attraction 2O 50 0 F 2Te -50 1 N 3 2 3 2Se 2S 4 5 Sn 4-100 Si 4 Ge 4-150 C 4-200 Period BP, X BP, 2X BP, 3X BP, X4 17

Practice Choose the substance in each pair that is a liquid at room temperature (the other is a gas) a) C 3 O C 3 CF 2 b) C 3 -O-C 2 C 3 C 3 C 2 C 2 N 2 Practice Choose the substance in each pair that is a liquid at room temperature (the other is a gas) a) C 3 O C 3 CF 2 can -bond b) C 3 -O-C 2 C 3 C 3 C 2 C 2 N 2 can -bond Practice Choose the substance in each pair that is more soluble in water a) C 3 O C 3 CF 2 b) C 3 C 2 C 2 C 3 C 3 Cl 18

Practice Choose the substance in each pair that is more soluble in water a) C 3 O C 3 CF 2 can -bond with 2 O b) C 3 C 2 C 2 C 3 C 3 Cl more polar 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 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 surface area (molar mass) Polar molecules also have dipole-dipole attractive forces 19

Summary (cont d) ydrogen bonds are the second strongest of the intermolecular attractive forces a pure substance can have ydrogen bonds will be present when a molecule has directly bonded to either O, N, or F atoms only example of f bonded dto Fis F 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 Force Strength (kj/mol) Distance (nm) Van der Waals 0.4-4.0 0.3-0.6 ydrogen Bonds 12-30 0.3 Ion - dipole 20 0.25 Covalent varies C-C 348 0.154 C=C 614 0.133 C=C 839 0.120 C-X 240-485 Ionic (coulombic lattice energy) NaCl 787 0.236 AlCl 3 5,492 0.20 Na 2 O 2,481 Al 2 O 3 15,916 59 20

Liquids properties & structure Surface Tension the tendency of liquids to minimize their surface area liquids minimize their surface area spherical as long as there is no gravity molecules of the surface behave differently molecules of interior because the cohesive forces on the surface molecules have a net pull into the liquid interior the surface layer acts like an elastic skin Surface Tension surface molecules have fewer neighbors to attract them the surface is less stable than the interior have a higher potential energy surface tension is the energy required to increase the surface area a given amount at room temp, surface tension of 2 O = 72.8 mj/m 2 surface tension of exane =18.4 mj/m 2 21

Factors Affecting Surface Tension intermolecular attractive forces, surface tension temperature, surface tension raising the temperature of the liquid increases the average kinetic energy of the molecules the increased molecular motion makes it easier to stretch the surface 80 Surface Tension of Water vs. Temperature 75 sion, mj/m 2 Surface Ten 70 65 60 55 50-20 0 20 40 60 80 100 120 Temperature, C Viscosity viscosity is the resistance of a liquid to flow, also called internal friction 1 poise = 1 P = 1 g/cm s often given in centipoise, cp intermolecular attractions = viscosity temperature t = viscosity it 22

Viscosity of Water vs. Temperature 1.2 1 0.8 ity, cp Viscosi 06 0.6 0.4 0.2 0 0 20 40 60 80 100 120 Temperature, deg C Capillary Action ability of a liquid to flow up a thin tube against the influence of gravity the narrower the tube, the higher the liquid rises results from two forces working in conjunction, ct o the cohesive and adhesive forces cohesive forces attract the molecules together adhesive forces attract the molecules on the edge to the tube s surface Capillary Action adhesive forces pull the surface liquid up the side of the tube, while 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 23

Meniscus meniscus 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 glass metallic bonds stronger than intermolecular attractions INTERMOLECULAR FORCES, RESULTING PROPERTIES & FACTORS AFFECTING BEAVIOR surface tension viscosity capillary action Van der Waals adhesion (sticky gecko feet) 72 24

Vaporization molecules are constantly in motion average KE α T some molecules have more kinetic energy than the average molecules at the surface may have enough energy to overcome the attractive forces to become a gas therefore the larger the surface area, the faster the rate of evaporation Distribution of Thermal Energy only a small fraction of the molecules in a liquid have enough energy to escape (recall kinetic molecular theory in Gases lecture) As temperature the fraction of the molecules with escape energy the higher the temperature, the faster the evaporation Condensation some molecules of the vapor will lose energy through molecular collisions some of the molecules will get captured back into the liquid when they collide with it some may stick and gather together to form droplets of liquid (e.g., rain, fog) particularly on surrounding surfaces Think of droplets on the inside surface of the vapor space in the water bottle you carry around. 25

Evaporation vs. Condensation 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 in a closed container at some time dynamic equilibrium is attaned when rate of vaporization = rate of condensation Equilibrium with George Pimmentel 1987.wmv Tro, Chemistry: A Molecular Approach 77 Dynamic Equilibrium* fish at start fish at equilibrium * Based on a lecture demonstration by the late Prof. George Pimentel at U.C., Berkeley 26

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 or 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 Tro, Chemistry: A Molecular Approach 81 27

Effect of Intermolecular Attraction on Evaporation and Condensation forces, energy needed to vaporize. also, attractive forces means that more energy must be removed from the vapor molecules before they can condense the 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, acetone (fingernail polish remover), perfume, sewage components liquids that do not evaporate easily are called nonvolatile e.g., motor oil, honey, most solids 84 28

Energetics of Vaporization Loss of higher energy molecules lowers the average kinetic energy of the liquid 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 eat of Vaporization the amount of heat energy required to vaporize one mole of the liquid is called the eat of Vaporization, vap sometimes called the enthalpy of vaporization always endothermic, therefore vap is +ive somewhat temperature dependent condensation = - vaporization Example 11.3 Calculate the mass of water that can be vaporized with 155 kj of heat at 100 C Given: Find: Concept Plan: Rlti Relationships: Solution: Check: q=155 kj, vap = 40.7 kj/mol g 2 O kj mol 2 O g 2 O 1mol 18.02 g 40.7 kj 1mol 1 mol 2 O = 40.7 kj, 1 mol = 18.02 g 1mol O 18.02 g 55 kj 40.7 kj 1mol 2 1 2 68.6 g O since the given amount of heat is almost 4x the vap, the amount of water makes sense 29

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, that will decrease the pressure of the vapor inside at that point, there are fewer vapor molecules in a given volume, causing the rate of condensation to slow eventually enough liquid evaporates so that the rates 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 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 (Le Chatelier s Principle) 30

Vapor Pressure vs. Temperature temperature the number of molecules able to escape the liquid the net result is that as the temperature, the vapor pressure small changes in temperature can make large changes in vapor pressure the rate of change depends on strength of the intermolecular forces in a specific liquid. Boiling Point At a certain T in the liquid P vapor = P atm. this phenomenon is what is called boiling. T required for P vapor = P atm is the boiling point Boiling Point the normal boiling point is the temperature at which the P vapor = 1 atm the the external pressure, the the boiling point of the the external pressure, the the boiling point of the liquid 31

essure, mmg Vapor Pr Vapor Pressure Curves normal BP Temperature vs Vapor Pressure 100 C 1000 900 800 760 mmg 700 water 600 TiCl4 500 chloroform ether 400 ethanol 300 acetone 200 100 0 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 Temperature, C BP Ethanol at 500 mmg 68.1 C Boiling Water with Ice 95 32

Clausius-Clapeyron Equation the graph logarithm slope of of vapor the of the line pressure vapor x 8.314 pressure vs. J/mol K temperature vs. = vap is an exponential inverse J/mol absolute growth temperature curve is a linear function vap vap ln(p ) ln RT vap e P RT vap e vap ln(p ) ln( ) ln RT vap e vap ln(pvap) ln( ) RT vap 1 ln(pvap) ln( ) R T Example Determine the vap of dichloromethane given the vapor pressure vs. temperature data enter the P vapor = f(t) data into a spreadsheet, Calculate the inverse of the absolute temperature and natural log of the vapor pressure T P 1/T lnp K torr K-1 200 0.8 0.005-0.2 220 4.5 0.00455 1.5 240 21 0.00417 3 260 71 0.00385 4.3 C 2 Cl 2 280 197 0.00357 5.3 300 391 0.00333 6 33

Example Determine the vap of dichloromethane given the vapor pressure vs. temperature data add a trendline, making sure the display equation on chart option is checked off 7.0 Clausius-Clapeyron Plot for Dichloromethane y = -3776.7x + 18.719 6.0 5.0 ln(vapor pressure) 4.0 3.0 2.0 1.0 0.0 0.00000 0.00100 0.00200 0.00300 0.00400 0.00500 0.00600-1.0 Inv. Temperature K -1 Example 11.4 Determine the vap of dichloromethane given the vapor pressure vs. temperature data determine the slope of the line -3776.7 3800 K 7.0 Clausius-Clapeyron Plot for Dichloromethane y = -3776.7x + 18.719 6.0 5.0 ln(vapor pressure) 4.0 3.0 2.0 1.0 0.0 0.00000 0.00100 0.00200 0.00300 0.00400 0.00500 0.00600-1.0 Inv. Temperature K -1 Example Determine the vap of dichloromethane given the vapor pressure vs. temperature data use the slope of the line to determine the heat of vaporization -3776.7 3800 K - vap slope R - vap 3800 K J 8.314 vap 3.16 10 mol K 4 J mol 31.6 kj mol 34

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 accurate results fewer data points will not give as accurate an average because there is less averaging out of the errors as with any other sets of measurements 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 P2 ln P1 R vap 1 T2 1 T1 Example 11.5 Calculate the vapor pressure of methanol at 12.0 C Given: Find: Concept Plan: T 1 = BP = 64.6 C, 337.8 K, P 11 = 760 torr, vap vap = 35.2 kj/mol, T 2 = 12.0 C 285.2 K P 2, torr P 1, T 1, vap P 2 P 2 vap 1 1 ln P1 R T2 T1 Relationships: T(K) = T( C) + 273.15 Solution: 3 J T1 64.6 35.2 273.15 10 P 2 2.31 337.8 K 1 P P 2 mol vap 1 e1 ln J T 2 12.08.314 273.15 285.2 K 0.0993 2 ln 2.31 760 torr 337.8 K P 1 mol K P1 R P2 T75.4 2 torr T1 Check: the units are correct, the size makes sense since the vapor pressure is lower at lower temperatures Vapor Pressure of Solutions Containing Nonvolatile Solutes the vapor pressure of a solvent above a solution is lower than the vapor pressure of the pure solvent the solute particles replace some of the solvent molecules at the surface Addition Eventually, of a equilibrium nonvolatile is solute reestablished, The pure solvent establishes an reduces the but rate a of smaller vaporization, number of liquid vapor equilibrium decreasing vapor molecules the amount therefore of vapor the vapor pressure will be lower 35

Thirsty Solutions a concentrated solution will draw solvent molecules toward it due to the natural drive for materials in nature to mix similarly, a concentrated solution will draw pure solvent vapor into it due to this tendency to mix the result is reduction in vapor pressure Thirsty Solutions Beakers When equilibrium with equal is liquid established, levels the of pure liquid solvent level in and the solution a are beaker place is in higher a bell than jar. the Solvent solution molecules level in the pure evaporate solvent from beaker each the one thirsty and fill solution the bell jar, grabs establishing and holds an solvent vapor equilibrium more effectively with the liquids in the beakers. Raoult s Law the vapor pressure of a volatile solvent above a solution is equal to its mole fraction of its normal vapor pressure, P P solvent in solution = solvent P since the mole fraction is always less than 1, the vapor pressure of the solvent in solution will always be less than the vapor pressure of the pure solvent 36

Ex 12.5 Calculate the vapor pressure of water in a solution prepared by mixing 99.5 g of C 12 22 O 11 with 300.0 ml of 2 O Given: 99.5 g C 12 22 O 11, 300.0 ml 2 O Find: P 2O Concept Plan: g C 12 22 O 11 mola mol C 12 22 O 11 moltotal ml 2 O g 2 O mol 2 O Solve: P P P 2O Relationships: P 2O = 23.8 torr, 1 mol C 12 22 O 11 = 342.30 g, 1 mol 2 O = 18.02 g 1mol C1222 11 99.5 g C1222O11 16.65 mol 2O 0.2907 mol C1222O11 342.30 g C1222O 0.9828 0.2907 mol C 11 1222O11 16.65 mol 2O 1.00 g 1mol 2O 300.0 ml 2O 16.65 mol 2O P O O 1mL P O18.02 0.9828 g 2O 23.8 torr 2 2 2 P O 23.4 torr 2 Ionic Solutes and Vapor Pressure according to Raoult s Law, the effect of solute on the vapor pressure simply depends on the number of solute particles when ionic compounds dissolve in water, they dissociate so the number of solute particles is a multiple of the number of moles of formula units the effect of ionic compounds on the vapor pressure of water is magnified by the dissociation since NaCl dissociates into 2 ions, Na + and Cl, one mole of NaCl lowers the vapor pressure of water twice as much as 1 mole of C 12 22 O 11 molecules would Effect of Dissociation 37

Ex 12.6 What is the vapor pressure of 2 O when 0.102 mol Ca(NO 3 ) 2 is mixed with 0.927 mol 2 O @ 55 C? Given: Find: Concept Plan: 0.102 mol Ca(NO 3 ) 2, 0.927 mol 2 O, P = 118.1 torr P of 2 O, torr 2O, P 2O P P P 2O Relationships: P = P Solve: Ca(NO 3 ) 2 Ca 2+ + 2 NO 3 3(0.102 mol solute) 0.927 mol 2O O 3 0.102 P O O P O 0.7518 118.1torr 2 2 2 2 mol solute 0.927 mol 2O P O 88.8 torr 2 0.7518 2O Check: the unit is correct, the pressure lower than the normal vapor pressure makes sense Raoult s Law for Volatile Solute when both the solvent and the solute can evaporate, both molecules will be found in the vapor phase the total vapor pressure above the solution will be the sum of the vapor pressures of the solute and solvent for an ideal solution P total = P solute + P solvent the solvent decreases the solute vapor pressure in the same way the solute decreased the solvent s P solute = solute P solute and P solvent = solvent P solvent eating Curve of a Liquid heating a liquid causes its temperature to increase linearly until it reaches the boiling point 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 q = mass x Δ vap q = mass x Cp x ΔT 38

115 eating Curve of Water liquid + gas temperatu ure solid solid + liquid liquid energy eating Curve of Water liquid + gas: q=m Δ vap temperatu ure solid + liquid id liquid: q=mc p ΔT q=m Δ fus solid: q=mc p ΔT energy 39

Segment 1 heating 1.00 mole of ice at -25.0 C up to the melting point, 0.0 C q = mass x C s x T mass of 1.00 mole of ice = 18.0 g C s = 209J/ 2.09 J/mol C l C J.0 g 2.09 0.0 C 25.0 C q 18 g C q 941J 0.941kJ Segment 2 melting 1.00 mole of ice at the melting point, 0.0 C q = n fus n = 1.00 mole of ice fus = 6.02 kj/mol fus kj mol 6.02 q 1.00 mol q 6.02 kj Segment 3 heating 1.00 mole of water at 0.0 C up to the boiling point, 100.0 C q = mass x C s x T mass of 1.00 mole of water = 18.0 g C s = 209J/ 2.09 J/mol C l C q J 18.0 g 4.18 100.0 C 0.0 C g C 3 q 7.52 10 J 7.52 kj 40

Segment 4 boiling 1.00 mole of water at the boiling point, 100.0 C q = n vap n = 1.00 mole of ice fus = 40.7 kj/mol fus kj mol 40.7 q 1.00 mol q 40.7 kj Segment 5 heating 1.00 mole of steam at 100.0 C up to 125.0 C q = mass x C s x T mass of 1.00 mole of water = 18.0 g C s = 2.01 J/mol C s J.0 g 2.01 125.0 C 100.0 C q 18 g C q 904 J 0.904 kj eating Curve of Water liquid + gas temperatu ure solid solid + liquid liquid energy 41

Phase Diagrams describe the different states and state changes that occur at various temperature - pressure conditions areas 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 Pressure 1 atm Solid Sublimation Curve sublimation deposition melting freezing normal melting pt. Phase Diagrams triple point Fusion Curve Liquid vaporization condensation Temperature Gas critical point normal boiling pt. Vapor Pressure Curve 42

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 eat of Fusion the amount of heat energy required to melt one mole of the solid is called the eat of Fusion, fus sometimes called the enthalpy of fusion always endothermic, therefore fus is + somewhat temperature dependent crystallization = - fusion generally much less than vap sublimation = fusion + vaporization 43

ressure Pr 1 atm Ice Phase Diagram of Water normal melting pt. 0 C triple point 0.01 C 0.006 atm Water Steam Temperature critical point 374.1 C 217.7 atm normal boiling pt. 100 C Phase Diagram of CO 2 ressure Pr 1 atm Solid -56.7 C 5.1 atm normal sublimation pt. -78.5 C triple point Liquid Gas critical point 31.0 C 72.9 atm Temperature 44

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 solid sublimation deposition gas Sublimation 45

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 Supercritical CO 2 instead of Boiling point: 250 F (121.1 C) Formula: C 2 Cl 4 Molar mass: 165.83 g/mol Density: 1.62 g/cm³ IUPAC ID: Tetrachloroethene Melting point: -2.2 F (-19 C) Soluble in: Water 138 46

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 fluid have properties of both gas and liquid states Water An Extraordinary Substance water is a liquid at room temperature most molecular substances with small molar masses are gases at room temperature due to -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 l have solubility in water e.g., O 2, CO 2 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 47

Morphic Forms of Ice Solution homogeneous mixtures composition may vary from one sample to another appears to be one substance, though really contains multiple materials many homogeneous materials we encounter are actually solutions e.g., air and sea water nature has a tendency toward spontaneous mixing generally, uniform mixing is more energetically favorable 48

Solutions solute is the dissolved substance seems to disappear takes on the state of the solvent solvent is the substance solute dissolves in does not appear to change state when both solute and solvent have the same state, the solvent is the component present in the highest percentage solutions in which the solvent is water are called aqueous solutions Seawater drinking seawater will dehydrate you and give you diarrhea the cell wall acts as a barrier to solute moving the only way for the seawater and the cell solution to have uniform mixing is for water to flow out of the cells of your intestine and into your digestive tract Common Types of Solution Solution Phase Solute Phase Solvent Phase Example gaseous solutions gas gas air (mostly N 2 & O 2 ) liquid solutions gas liquid solid liquid liquid liquid soda (CO 2 in 2 O) vodka (C 2 5 O in 2 O) seawater (NaCl in 2 O) solid solutions solid solid brass (Zn in Cu) solutions that contain g and some other metal are called amalgams solutions that contain metal solutes and a metal solvent are called alloys 49

Brass Cu/Zn Solid Solutions (alloys) Solubility when one substance (solute) dissolves in another (solvent) it is said to be soluble or miscible salt is soluble in water bromine is soluble in methylene chloride when one substance does not dissolve in another it is said to be insoluble or immicible oil is insoluble in water the solubility of one substance in another depends on two factors nature s tendency towards mixing, types of intermolecular attractive forces 50

Spontaneous Mixing Solubility there is usually a limit to the solubility of one substance in another gases are always soluble in each other two liquids that are mutually soluble are said to be miscible alcohol and water are miscible oil and water are immiscible the maximum amount of solute that can be dissolved in a given amount of solvent is called the solubility the solubility of one substance in another varies with temperature and pressure Mixing and the Solution Process Entropy formation of a solution does not necessarily lower the potential energy of the system the difference in attractive forces between atoms of two separate ideal gases vs. two mixed ideal gases is negligible yet the gases mix spontaneously yet the gases mix spontaneously the gases mix because the energy of the system is lowered through the release of entropy entropy is the measure of energy dispersal throughout the system energy has a spontaneous drive to spread out over as large a volume as it is allowed 51

Intermolecular Forces and the Solution Process Enthalpy of Solution energy changes in the formation of most solutions also involve differences in attractive forces between particles must overcome solute-solute attractive forces endothermic must overcome some of the solvent-solvent attractive forces endothermic at least some of the energy to do this comes from making new solute-solvent attractions exothermic Intermolecular Attractions Relative Interactions and Solution Formation Solute-to-Solvent > Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent = Solute-to-Solute + Solvent-to-Solvent Solution Forms Solute-to-Solvent < Solute-to-Solute to + Solution May or Solvent-to-Solvent May Not Form when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy 52

Solution Interactions Will It Dissolve? Chemist s Rule of Thumb Like Dissolves Like a solute will dissolve in a solvent if it has a similar structure to the solvent when the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other Classifying Solvents Solvent Class Structural Feature Water, 2 O polar O- Methyl Alcohol, C 3 O polar O- Ethyl Alcohol, C 2 5 O polar O- Acetone, C 3 6 O polar C=O Toluene, C 7 8 nonpolar C-C & C- exane, C 6 14 nonpolar C-C & C- Diethyl Ether, C 4 10 O nonpolar C-C, C- & C-O, (nonpolar > polar) Carbon Tetrachloride nonpolar C-Cl, but symmetrical 53

Example 12.1a predict whether the following vitamin is soluble in fat or water The 4 O groups make the molecule highly polar and it will also -bond to water. Vitamin C is water soluble Vitamin C Example predict whether the following vitamin is soluble in fat or water The 2 C=O groups are polar, but their geometric symmetry suggests their pulls will cancel and the molecule will be nonpolar. Vitamin K 3 is fat soluble C C C C C C O C C O C C Vitamin K 3 C 3 END 162 54

Solids properties & structure M.C. Escher Eight eads woodcut 1922 M.C. Escher Cubic Space-Division lithograph 1952 55

166 167 168 56

Icosahedral Symmetry in Viruses Icosahedral symmetry involves 6 five-fold rotation axes, 10 three-fold, and 15 two-fold. Robijn Bruinsma, PhD, Professor Physics and Astronomy, Virus Research Group at UCLA http://virus.chem.ucla.e du/node/8 169 Yum! 20,000 : 1 171 57

Example 11.6 An x-ray beam at =154 pm striking an iron crystal results in the angle of reflection = 32.6. Assuming n = 1, calculate the distance between layers Given: Find: Concept Plan: Relationships: Solution: Check: n = 1, = 32.6, = 154 pm d, pm n,, d n d 2 sin d n 2 sin 1 154 pm 2 sin 32.6 143pm the units are correct, the size makes sense since the iron atom has an atomic radius of 140 pm 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 58

Familiar Examples of 2-Dimensional Unit Cells 175 13-8 Crystal Structures 7 Unit Cells c c c c b b a b a a a b Cubic a = b = c all 90 Tetragonal a = c < b all 90 Orthorhombic a b c all 90 Monoclinic a b c 2f faces 90 c b a exagonal a = c < b 2 faces 90 1 face 120 c b a Rhombohedral a = b = c no 90 Triclinic a b c no 90 59

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 Volume of a Cube edge length 4 3 Volume of a Sphere π r 3 3 60

Counting Cell Occupancy Simple Cubic Cubic Unit Cells - Simple Cubic 8 particles, one at each corner of a cube 1/8 th of each particle lies in the unit cell each hparticle part of f8 cells 1 particle in each unit cell 8 corners x 1/8 edge of unit cell = twice the radius coordination number of 6 2r 61

Body-Centered Cubic Atomic Radii from Crystal Structures 62

Cubic Unit Cells - Body-Centered Cubic 9 particles, one at each corner of a cube + one in center 1/8 th of each corner particle lies in the unit cell 2 particles in each unit cell 8 corners x 1/8 + 1 center edge of unit cell = (4/ 3) times the radius of the particle coordination number of 8 4r 3 Face-Centered Cubic 63

Cubic Unit Cells - Face-Centered Cubic 14 particles, one at each corner of a cube + one in center of each face 1/8 th of each corner particle + 1/2 of face particle lies in the unit cell 4 particles in each unit cell 2r 2 8 corners x 1/8 + 6 faces x 1/2 edge of unit cell = 2 2 times the radius of the particle coordination number of 12 Closest-Packed Structures First Layer with spheres, it is more efficient to offset each row in the gaps of the previous row than to lineup rows and columns 64

Closest-Packed Structures Second Layer the second layer atoms can sit directly over the atoms in the first called an AA pattern or the second layer can sit over the holes in the first called an AB pattern Closest-Packed Structures Third Layer with Offset 2 nd Layer the third layer atoms can align directly over the atoms in the first called an ABA pattern or the third layer can sit over the uncovered holes in the first called an ABC pattern exagonal Cubic Closest-Packed Face-Centered Cubic exagonal Closest-Packed Structures 65

exagonal Close Packed (hcp) 66

Cubic Closest-Packed Structures 200 Classifying Crystalline Solids classified by the kinds of units found sub-classified by the kinds of attractive forces holding the units together molecular solids are solids whose composite units are molecules ionic solids are solids whose composite units are ions atomic solids are solids whose composite units are atoms nonbonding atomic solids 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 67

Molecular Solids the lattice site are occupied by molecules the molecules are held together by intermolecular attractive forces dispersion forces, dipole attractions, and -bonds because the attractive forces are weak, they tend to have low melting point generally < 300 C Ionic Solids Attractive Forces held together by attractions between opposite charges nondirectional therefore every cation attracts all anions around it, and vice versa the coordination number represents the number of close cationanion 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 generally, anions are larger than cations the number of anions that can surround the cation limited by the size of the cation the closer in size the ions are, the higher the coordination number is 68

Ionic Crystals CsCl coordination number = 8 Cs + = 167 pm Cl = 181 pm NaCl coordination number = 6 Na + = 97 pm Cl = 181 pm Sodium Chloride Coordination Number 69

Lattice oles in hexagonal closest packed or cubic closest packed lattices there are 8 tetrahedral holes and 4 octahedral holes per unit cell in simple cubic there is 1 hole per unit cell number and type of holes occupied determines formula (empirical) of salt = Octahedral = Tetrahedral Lattice oles Octahedral ole Tetrahedral ole Simple Cubic ole oles in Crystals 70

oles in Crystals Determining Crystal Structure crystalline solids have a very regular geometric arrangement of their particles the arrangement of the particles and distances between them is determined by x-ray diffraction in this technique, a crystal is struck by beams of x-rays, which then are reflected the wavelength is adjusted to result in an interference pattern at which point the wavelength is an integral multiple of the distances between the particles Bragg s Law when the interference between x-rays is constructive, the distance between the two paths (a) is an integral multiple of the wavelength n =2a the angle of reflection is therefore related to the distance (d) between two layers of particles sin = a/d combining equations and rearranging we get an equation called Bragg s Law d n 2 sin 71

X-ray Crystallography X-Ray Diffraction X-Ray Diffraction 72

Rock Salt Structures coordination number = 6 Cl ions (181 pm) in a face-centered cubic arrangement ⅛ of each corner Cl inside the unit cell ½ of each face Cl inside the unit cell each Na + (97 pm) in holes between Cl octahedral holes 1 in center of unit cell ¼ of each edge Na + inside the unit cell Na:Cl = (¼ x 12) + 1: (⅛ x 8) + (½ x 6) = 4:4 = 1:1, therefore the formula is NaCl Cesium Chloride Structures coordination number = 8 ⅛ of each Cl (184 pm) inside the unit cell whole hl Cs + (167 pm) inside id the unit cell cubic hole = hole in simple cubic arrangement of Cl ions Cs:Cl = 1: (8 x ⅛), therefore the formula is CsCl Cesium Chloride 73

Zinc Blende Structures coordination number = 4 S 2 ions (184 pm) in a face-centered cubic arrangement ⅛ of each corner S 2 inside the unit cell ½ of each face S 2 inside the unit cell each Zn 2+ (74 pm) in holes between S 2 tetrahedral holes 1 whole in ½ the holes Zn:S = (4 x 1) : (⅛ x 8) + (½ x 6) = 4:4 = 1:1, therefore the formula is ZnS Fluorite Structures coordination number = 4 Ca 2+ ions (99 pm) in a face-centered cubic arrangement ⅛ of each corner Ca 2+ inside the unit cell ½ of each face Ca 2+ inside the unit cell each F (133 pm) in holes between Ca 2+ tetrahedral holes 1 whole in all the holes Ca:F = (⅛ x 8) + (½ x 6): (8 x 1) = 4:8 = 1:2, therefore the formula is CaF 2 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 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 74

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 Crystal Structure of Metals at Room Temperature = body-centered cubic = hexagonal closest packed = other = cubic cp, face-centered = diamond 75

Network Covalent Solids atoms attached to its nearest neighbors by covalent bonds because of the directionality of the covalent bonds, these do not tend to form closest-packed 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 4 covalent bonds to surrounding atoms sp 3 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, ~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 76

The Graphite Structure: a 2-Dimensional Network in graphite, the carbon atoms in a sheet are covalently bonded together forming 6-member flat rings fused together similar to benzene bond length = 142 pm sp 2 each C has 3 sigma and 1 pi bond trigonal-planar geometry each sheet a giant molecule the sheets are then stacked and held together by dispersion forces sheets are 341 pm apart Properties of Graphite hexagonal crystals high melting, ~3800 C need to overcome some covalent bonding slippery feel because there are only dispersion 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 77

Quartz 3-dimensional array of Si covalently bonded to 4 O tetrahedral melts at ~1600 C very hard Micas minerals that are mainly 2-dimensional arrays of Si bonded to O hexagonal arrangement of atoms sheets chemically stable thermal and electrical insulator 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- + + + + + + + + + 78

Bonding in Metals Band theory. Extension of MO theory. N atoms give N orbitals that are closely spaced in energy. N/2 are filled. The valence band. N/2 are empty. The conduction band. 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 2 atomic orbitals combine they produce both a bonding and an antibonding molecular orbital when many atomic orbitals combine they produce a band of bonding molecular orbitals and a 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 79

2 ATOMIC ORBITALS 3 ATOMIC ORBITALS 10 ATOMIC ORBITALS Source: http://www.chemistry.uoguelph.ca/educmat/chm729/band/concept.htm 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 80

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 ifi 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 81

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 Photovoltaic Cells 82