q i = 0 aa + bb cc + dd q = si mi!t qsolid=>liquid = # moles!h fusion qliquid=>gas = # moles!h vaporization i=1

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1 Temperature C ere s the same curve now applying the conservation of energy (sum of the heats) q = s ice m ice Tq = m ice fus q = s 2 m 2 T eating solid ice to 0 C A Ice Phase transition Temperature does not change during a phase transition. B Ice + Water mix C Melting solid ice to 0 C water Water eating water to boiling 100 C D q = m 2 vap Phase transition Temperature does not change during a phase transition. Water + steam mix Boiling all water to steam 100 C q = s stm m stm T Steam eating steam past 100 C eat Added (kj/mole) E F Calculate the amount of heat required to convert 500 grams of ice at C to steam at 120. C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g C respectively, and the latent heat of fusion and vaporization,!f and!v, are 6.02 and 40.7 kj/mol respectively. n q i = 0 i=1 q = si mi!t sum the q s baby qsolid=>liquid = # moles! fusion qliquid=>gas = # moles! vaporization for non-phase transitions for phase transitions Calculate the amount of heat required to convert 500 grams of ice at -20 C to steam at 120 C. The specific heat capacities of water, ice and water vapor are 4.18, 2.06 and 1.84 J/g C respectively, and the latent heat of fusion and vaporization,!f and!v, are 6.02 and 40.7 kj/mol respectively. 1. eat ice from -20 C to ice at 0 C = 500. g x 2.06 J/g C x 20 C 2. Melt ice at 0 C to water at 0 C = 500. g/(18 g/mol) x 6.02 kj/mol 3. eat water from 0 C to water at 100 C= 500. g x 4.18 J/g C x 100 C 4. Evap water at 100 C to vap at 100 C = 500. g/(18 g/mol) x 40.7 kj/mol 5. eat vap from 100 C to vap at 120 C = 500. g x 1.84 J/g C x 20 C 1. = 20.6 kj 2. = kj 3. = kj 4. = kj 5. = 18.4 kj Total = kj In open containers molecules that have enough KE can overcome IMF s at the surface and evaporate into the atmosphere. Evaporation Liquid no equilibrium In closed containers, molecules, there is not evaporation. Instead molecule vaporize and condense until there is no further change in concentration. This forms an equilibrium vapor pressure over the liquid. Vapor Pressure Liquid equilibrium The equilibrium vapor pressure is the pressure exerted by a vapor over its liquid phase (measured under vacuum) when a dynamic equilibrium exists between condensation and evaporation. Dynamic chemical equilibrium is condition reached when the rate of evaporation and rate of condensation are equal (no net change). Equilibrium Dynamic equilibria also occurs with melting and sublimation and also in most chemical reactions. At the melting point a solid begins to change into a liquid as heat is added. As long no heat is added or removed melting (red arrows) and freezing (black arrows) occur at the same rate an the number of particles in the solid remains constant. Solid Liquid Time Molecules in liquid begin to vaporize Molecules vaporizing and condensing at such a rate that no net change in numbers occure aa + bb cc + dd Reactants Products Reaction Rate of the forward reaction = = Rate of Reverse reaction

2 As a liquid temperature increases, so does KE and so does the number of molecules with enough energy to overcome IMF s and escape to the vaporize. KE = Ek = 3 RT 2 Kinetic Energy T1 The boiling point of a pure liquid is the temperature at which the vapor pressure over its liquid phase is equal to the external pressure on the liquid. The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm. Temperature T2 Evaporation T1 Boiling Molecules with enough speed to vaporize T2 > T1 Speed KE needed to overcome IMF in liquid phase Vapor Pressure (torr) of Some Liquids The vapor pressure of a pure liquid depends on the IMF s between molecules. The stronger the attractive forces in the liquid phase the lower the vapor pressure--and the less volatile the liquid it is. Vapor Pressure of Some Liquids Note the shape of the curve is exponential eat of Vaporization, Boiling Pts of Liquids 2 atm.66 atm If we plot ln (vapor pressure) vs 1/temp we observe a linear relationship. Vapor pressure plotted as a function of temperature ln (vapor pressure) plotted as a function of 1/Temp Vapor pressure (torr) y = -!vap"$ 1 %' +C R #T & m slope =!vap/r x+ b R = 8.31 J/K mol ln P ln P By taking measurements at two temps, we get: ln 1/T Which is the least volatile at 1 atm? The Clausius-Claperyron equation relates the vapor pressure (P) of a pure liquid to the liquid s temperature (T) and the liquids molar heat of vaporization (!vap). ln P = Temperature C Which of the following has the highest vapor pressure at 1 atm? P2 P1 = -!vap# 1 1& " ( % R T $ 2 T1' 1/T

3 Pressure (Atm) Vapor pressure of pure et is 115 torr at 34.9 C. If!vap = 40.5 kj/mol calculate the temperature when the vapor pressure of et is 760 torr. R is the gas constant given at J/mol K The following diagram shows a close-up view of part of the vapor-pressure curves for a solvent (red curve) and a solution of the solvent with a second liquid (green curve). Which solvent is more volatile?!vap =40.5 kj/mol P1=115 torr P2=760 torr T1= = 308.0K ln ln P 2 = -!vap # 1 " 1 & % ( P 1 R $ T 2 T 1 ' 760 torr 115 torr = x10 3 J/mol J/mol*K T 2 = 350K = 77 C 1 1 T K The more volatile solvent will have a higher vapor pressure (more gas molecules in the gas phase above its liquid). The green solvent has a higher vapor pressure at all temperatures. It has weaker IMF s. A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas phase of matter. 3 Freezing Point A Solid Melting Freezing Sublimation Deposition Liquid Evaporation Condensation Gas AB is the solid-liquid (sublimation) curve for ice; BD the vapor pressure curve for water-steam; BC the solid-liquid (melting point line); point B the triple point point D labels the critical point. D 374 Supercritical Fluid Critical Point Normal Boiling Point Triple Point A phase diagram summarizes the conditions at which a substance exists as a solid, liquid, or gas phase of matter. Line AB is the vapor-pressure curve for solid-vapor (sublimation curve); BD the vapor pressure curve for liquid-vapor water; BC the melting point line (notice how it has a negative slope = solid is lower density); point B the triple point: only point when three phases are in equilibrium; point D is the critical point--the temperature at which no pressure can liquefy a gas (new phase of matter). Terminology in a Phase Diagram The triple point is the single P,T point at which all three phases are in equilibrium with one another. Water is one of the few substances on earth for which the liquid phase is more dense than its solid phase. We see this in a phase diagram in the slope of the solidliquid line (higher pressure decreases the melting point) The critical temperature (T c ) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure. The critical pressure (P c ) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature. The normal boiling point is the temperature at which the vapor pressure of a liquid equals 1 atm. The normal freezing point is the temperature at which the liquid is in equilibrium with its solids phase at 1 atm. more pressure at a fixed temperature leads to a liquid not a solid Water more pressure at a fixed temperature leads to a solid not a liquid C2

4 Water Is A Unique Substance Ice is less dense than water Density of Water Generally speaking, which phase of a substance should have the highest density (mass/volume)? Name a compound that does not follow this trend? Maximum Density 4 0C great solvent due to high polarity and hydrogen bonding ability LIQUID GAS exceptional high specific heat capacity SLID high surface tension and capillarity large density differences of liquid and solid states Polar-covalent bonds result from differences in electronegativity between bonded atoms with geometry taken into consideration. When combined with a molecule s geometry they can produce a net dipole moment (dipole) in molecules. Po d lar on Dipole B B on lar d + o P Net Dipole Moment Polar Bond Electronegativity is an element s inherent property to draw electrons to itself when chemically bonded to another atom in a molecule. Lookout for bonds having: F,, Cl, N, Br, I F is the most electronegative element Think of the dipole moment as a molecule with separated charges + and - that can influence other charged molecules. Polar Bond No Net Dipole Moment No dipole (charge separation) Connecting Dots: large electronegativity differences in a bond => polar molecules => large dipole moments => gives rise to large IMF => decreased vapor pressure => increased boiling and melting points, higher viscosity, higher surface tension. Po lar nd o Dipole Forces + B B on lar + d London Dispersion o P Forces Net Dipole Moment Cs the least Identify polar bonds in a molecule, and then consider molecular geometry to gauge if there is a net dipole moment (dipole) in a molecule.!-!+ Polar with net dipole moment Polar Bond No Net Dipole Moment!+ net dipole!+!- non-polar with no net dipole moment moment!!non-polar with no net dipole moment No dipole forces London Dispersion Forces nly Polar with!+!polar Bond!-!- Polar with net dipole moment

5 Recall your molecular and electron geometry from Chem 7. Intermolecular forces are classified into four major types. 2 EG 3 EG 4 EG 1. Ion-dipole: occur between neighboring an ion solution and a polar molecule (dipole) also in solution. Na+ 2. Dipole-dipole: occur between a neutral polar molecules (they can be different or the same molecules). 5 EG 6 EG 3. Induced-dipoles: occur when a ion or a dipole induces a spontaneous dipole in a neutral polarizable molecule. Ion or Dipole Induced Dipole London Dispersion Forces are attractive IMF s that occur between all molecules. Spontaneous dipoles are formed randomly or induced by other charged species in neutral polarizable molecules. Comparing the strengths of Intermolecular Forces Flowchart for classifying intermolecular forces Force Strength Energy- Distance Interactions London Forces 1-10 kj/mole 1/r 6 All-molecules Dipole- Dipole ydrogen Bonding 3-40 kj/mole 1/r 3 Polar molecules kj/mole 1/r 3 - N- -F Ion-Dipole kj/mole 1/r 2 Ions-Polar molecules Ion-Ion >>200 kj/mole 1/r Cation-Anion As dipole moments increase in polar substances of the same mass, IMF s increase as do boiling points (lower vapor pressure) and melting points. Dipole-Dipole intermolecular forces are attractive forces that occur between polar molecules (same or not the same). Non-polar VS Polar Very polar Boiling point differences are explained by dipole-dipole interactions

6 ydrogen bonding is a special case of dipole-dipole intermolecular force that occurs between a hydrogen atom and an unshared pair of electrons in a polar N-, -, or -F bond. The boiling points of covalent binary hydrides increase with increasing molecular mass down a Group but the hydrides of N3, 2 and F have abnormally high BP because of hydrogen bonding (. 100? Increased London Forces C3C2 C3C3 Temperature C ethanol 78.3 C dimethyl ether C Period Ion-dipole forces are attractive forces between an ion in solution and a neighboring polar molecule. London Dispersion Forces are attractive intermolecular forces that occur when temporary dipoles are formed due to random electron motions in all polarizable molecules. - end of dipole attracted to + ions Molecules with a dipole moment in solution 2-neutral non-polar polarizable molecules end of dipole attracted to -ions spontaneous movement of electrons forming an instantaneous dipole moment An anion in solution A cation in solution Induced-Dipole Moment and IMF between molecules Polarizability is the ease with which an electron distribution (cloud) in the atom or molecule can be distorted by an outside ion or dipole. non-polar molecule (electron cloud) spontaneous induced dipole that depends on polarizability of substance London Dispersion Force induced dipole dispersion All molecules have at least this IMF Polarizability and London dispersion trends mirror atomic size trends in the periodic table. Size and polarizability increases down a group Polarizability increases right to left (bigger size more distortable. PLARIZABILITY Polarizability increases with molar mass (# e-) of a molecule Cations are less polarizable than their parent atom because they are smaller and more compact. Anions are more polarizable than their parent atom because they are larger.

7 Examples of increasing London dispersion forces with increase in size and polarizability. Increasing Size/ Polarizability Larger unbranched molecules are more polarizable than compact branched molecules. Increasing Size/ Polarizability neopentane bp=10 o C tetrahedral normal pentane bp=36 o C extended structure Larger polarizability in unbranched molecules explains boiling and melting points trends in isomers. Some Generalizations About IMF s 1. Ion-Ion > Ion-Dipole > Dipole-Dipole > Dispersion Which of the following substances exhibits bonding? For those that do, draw two molecules of the substance with the bonds between them. 2. For polar molecules of approximately the same mass and shape and volume (i.e. polarizability the same), dipole-dipole forces dictate the difference in physical properties. (a) C 2 6 (b) C 3 (c) C 3 C N 2 3. ydrogen bonding occurs with polar bonds in particular - F, -, -N and an unshared pair of electrons on a nearby electrognegative atom usually F,, or N. PLAN: Find molecules in which is bonded to N, or F. Draw bonds in the format -B: -A-. 4. For non-polar molecules of the same molecular mass, longer less compact molecules are generally more polarizable and have greater dispersion forces and show higher boiling and melting points. 5. For non-polar molecules of widely varying molecular mass, those with more mass are typically more polarizable and experience greater London dispersion forces and exhibit higher boiling points and melting points. SLUTIN: (b) (a) C 2 6 has no bonding sites. C (c) C 3 C N C N C 3 C C 3 C N N C 3 C Which substances experience dipole-dipole intermolecular forces? SiF4,CBr3, C2, S2! Arrange the following substances in order of increasing boiling points. SiF4, geometry tetrahedral, Si-F bonds are polar, but no molecular dipole; bond dipoles cancel. No dipole-dipole interactions. C 2 6, N 3, Ar, NaCl, As 3 C2, linear geometry, C- bonds are polar but symmetry cancels net diple, but no molecular dipole; bond dipoles cancel and no dipole IMF s S2, bent geometry, S- bonds are polar and do not cancel. Sulfur lone pair dipole only partially offsets net bond dipole. as dipoledipole forces CBr3, tetrahedral geometry, C- and C-Cl bonds are polar and do not cancel S2 and CCl3 experience dipole-dipole intermolecular forces.

8 ! Arrange the following substances in order of increasing boiling points. C 2 6, N 3, Ar, NaCl Ar < C 2 6 < N 3 < NaCl nonpolar nonpolar polar ionic London London dipole-dipole -bonding ion-ion Arrange the following non-polar molecules in order of increasing melting point. SiF4, CS2, CI4, GeCl4 Solution. None of these molecules poses a net dipole moment. nly dispersion forces exist and these are expected to increase with increasing molecular mass (more polarizable as a molecule gets larger). The molar masses of these substances follow: Substance Molar Mass CS SiF CI GeCl The intermolecular forces, and the melting points, should increase in the following order: CS2 < SiF4 < GeCl4 < CI4 The experimentally determined melting points are , -90, -49.5, and 171 o C, respectively. For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point. SLUTIN: (a) Mg 2+ and Cl - are held together by ionic bonds while PCl 3 is covalently bonded and the molecules are held together by dipole-dipole interactions. Ionic bonds are stronger than dipole interactions and so MgCl 2 has the higher boiling point. (b) C 3 N 2 and C 3 F are both covalent compounds and have bonds which are polar. The dipole in C 3 N 2 can bond while that in C 3 F cannot. Therefore C 3 N 2 has the stronger interactions and the higher boiling point. (c) Both C 3 and C 3 C 2 can bond but C 3 C 2 has more C for more dispersion force interaction. Therefore C 3 C 2 has the higher boiling point. (d) exane and 2,2-dimethylbutane are both nonpolar with only dispersion forces to hold the molecules together. exane has the larger surface area, thereby the greater dispersion forces and the higher boiling point. What type Describe of intermolecular the intermolecular forces forces can you that recognize exist between in the following each of ionic the following and covalent molecules? compounds. Br(g) Br is a polar molecule: dipole-dipole forces. There are also dispersion forces between Br molecules. C 4 S 2 C 4 is non-polar: London dispersion forces. S S 2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between S 2 molecules. In which substances would hydrogen bonding forces occur between molecules? C26, CCl3, C3C2, N3, P3 Solution. ydrogen is bonded to one of the very electronegative atoms in C3C2 and N3. ydrogen bonding should occur in both of these substances. Which of the following statements is incorrect? (A) C3C2 should have a viscosity and boiling point greater than C2C2C2 (B) The large specific heat capacity and the large heat of vaporization, vap, of water are both a result of strong hydrogen-bonding forces. (C) The surface tension of a liquid is a result of strong and imbalanced intermolecular attractive forces acting at the surface of the liquid relative to its interior. (D) The observation that water forms a U-shaped meniscus climbing up the side of glassware is a result of cohesive forces being greater than adhesive forces. (E) Polar molecules is a result of moderate electronegativity differences between bonded atoms in molecules, but molecular geometry must be considered to determine if a molecule is polar overall.

9 Water Is Wild and Unique and Taken For Granted --igh -bonding = high cohesive forces responsible for the transport of water in roots and xylem of trees and plants. --Base-pairing in double-stranded DNA --eat capacity, high heat of vaporization, inverted density of water and ice. Three common physical properties of liquids that depend on the magnitude of IMF s include surface tension, capillarity and viscosity. surface tension Viscosity capillarity Surface tension is the amount of energy required to stretch or increase the surface area of a liquid. molecules at the surface feel a net force downward Surface Tension of Some Liquids Substance Formula Surface Tension (J/m 2 ) at 20 0 C diethyl ether C 3 C 2 C 2 C 3 1.7x10-2 Major IMF s dipole-dipole; dispersion molecules in the interior experience equal force in 3-D ethanol butanol C 3 C 2 C 3 C 2 C 2 C 2 2.3x x10-2 bonding bonding; dispersion Scientists say that the leaf has a low surface energy as water will not spread out (it takes energy to spread the water out) water mercury 2 g 7.3x x10-2 bonding metallic bonding bserved Physical Properties and IMF s The surface tension of a liquid is a function of temperature. Why might this be? (what forces compete?) ot water works better than cold water in cleaning your clothes or hands as it can more effectively wet dirt (water gets into pores and not get stuck ).

10 Capillary effect occurs when the adhesive IMFʼs between a liquid and a substance are stronger than the cohesive IMFʼs inside the liquid. Water will rise to different heights depending on the diameter of the capillary. Why? Cohesion is the intermolecular attraction between like molecules. Adhesion is an attraction between unlike molecules Cohesion Adhesion Capillary rise implies that the: Adhesive forces > cohesive forces Capillary fall implies that the: Cohesive forces > adhesive forces Viscosity is the measure of a liquid s resistance to flow relative to one another, and is thus related to the intermolecular forces. Figure Copyright The McGraw-ill Companies, Inc. Permission required for reproduction or display. The viscosity of a polymer in solution.! il for your car is bought based on this property: 10W30 or 5W30 describes the viscosity of the oil at high and low temperatures (W means winter dudes higher the number the more viscous). In general, viscosity decreases as temperature increases and visa versa. 12-