Chapter 12 Intermolecular Forces
Chapter 12 Intermolecular Forces CHEM 132
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Date Created: 09/21/15
Chapter Notes winter 12 Intermolecular Forces IMFs Liquids gnd Solids Section 121 Kinetic Molecular Theory of Liquids and Solids In gases the distance between molecules is vast enough that at ordinary pressures and temperatures such as 1 atm and 25 C there are not any applicable interactions Use the word phase for changes of state that involve only one substance Use the word system for changes in phase that involve two or more substances Section 122 Intermolecular Forces IMFs IMFs are defined as the attractive forces that exist between molecules Intramolecular forces are the attractive forces that exist within the molecules themselves 0 As the temperature drops the average kinetic energy decreases 0 When the temperature gets low enough the gas undergoes condensation and condenses into a liquid Generally IMFs are much weaker than intramolecular forces 0 Much less energy is needed to evaporate a liquid than is needed to break the bonds in liquid molecules 0 Boiling point BP is directly related to the strength of the IMFs present Melting point MP is as well Van der Waals forces 0 Dipoledipole attractive forces between polar molecules 0 Dipoleinduced dipole O Dispersion 0 Hydrogen bonding strong dipoledipole force Iondipole forces an electrostatic force 0 Coulomb s law 0 Attracts cation andor anion and a polar molecule to each other 0 Strength depends on the charge size of the ion magnitude of dipole moment and the size of the molecule Attractive interactions occur in nonpolar molecules as well 0 If we place an ion or a polar molecule near an atom or a nonpolar molecule the electron distribution becomes distorted by the ion or the polar molecule this results in an induced dipole being formed 0 Induced Dipole a separation of positive and negative charges in the atom or nonpolar molecule which is caused by an ion or a polar molecule being in the proximity of another molecule Polarizability the ease with which the electron distribution in the atom or molecule can be distorted 0 Likelihood of a dipole moment being induced also depends on the charge on the ion and the strength of the dipole 0 Generally the larger the number of electrons and the more diffuse the electron cloud in the atom or molecule the greater its polarizability Dispersion forces attractive forces that arise as a result of temporary dipoles induced in atoms or molecules 0 Fritz London 1930 showed that the magnitude of this attractive interaction is directly proportional to the polarizability of the atom or molecule 0 Usually increase with molar mass because molecules with larger molar mass tend to have more electrons and dispersion forces increase in strength with the number of electrons Mel 39 Point 0 CH4 1825 CF4 150 CC14 23 CBI394 C14 171 Example 121 What types of IMFs exist between the following pairs a HBr and HZS b C12 and CBr4 c 12 and N03 d NH3 and C6H6 a Both HBr and HZS are polar molecules so dispersion and dipoledipole IMFs are present b Both C12 and CBr4 are nonpolar molecules so only dispersion IMFs are present c 12 is a diatomic molecule or molecule containing two atoms and is therefore nonpolar N0339 is an ion so there will be ioninduced dipole and dispersion forces present d NH3 is a polar molecule and C6H6 is nonpolar so dipoleinduced dipole and dispersion forces are present Hydrogen Bonding a special type of dipoledipole interaction that involves the hydrogen atom in a polar bond and an electronegative element NOF Hydrogen bonds occur only between pairs of atoms such as NH OH or FH Example 122 Which of the following can form hydrogen bonds with water CH30CH3 CH4 F HCOOH Na Solution Neither Na nor CH4 have electronegative elements F O and N so they will not form hydrogen bonds with water However CH3OCH3 F39 and HCOOH will form hydrogen bonds with water because HCOOH has a free oxygen that allows for the H in water to bond with it For CH30CH3 the oxygen is singlebound to two H3C atoms leaving it available to be bound to one of the hydrogen atoms in water For F39 the electronegative element is attracted to the hydrogen making it easier for it to form hydrogen bonds Although many compounds can form intermolecular hydrogen bonds the difference between H20 and other polar molecules such as NH3 and HF is that each oxygen atom can form two hydrogen bonds the same as the number of lone electron pairs on the oxygen atom Section 125 Bonding in Solids Crystals are classified according to the types of IMFs between the particles Table 124 Types of Crystals and General Properties Type of Forces Holding Units Together General Properties Examples Crystal Ionic Electrostatic attraction Hard brittle high MP poor conductor NaCl LiF MgO CaCO3 of heat and electricity Moleculafk Dispersion dipoledipole H Soft low MP poor conductor of heat Ar C02 12 H20 CIZHZZO11 bonds and electricity sucrose Covalent Covalent bonding Hard high MP poor conductor of heat C diamondf SiO2 quartz and electricity Metallic Metallic bonding Soft to hard low to high MP good All metallic elements conductor of heat and electricity Na Mg Fe Cu Included in this category are crystals made up of individual atoms Diamond is a good thermal conductor Diamond and graphite are two allotropes of carbon 126 Phase Changes Phase change transformations from one phase to another 0 Occur when energy usually in heat form is added or removed from a substance 0 These are physical changes that are characterized by changes in molecular order LiquidVapor Equilibrium o EvaporationVaporization at any given temperature a certain number of the molecules in a liquid possess a sufficient kinetic energy to escape from the surface 0 When a liquid evaporates its gaseous molecules exert a vapor pressure State of dynamic equilibrium the rate of a forward process is exactly balanced by the rate of the reverse process This is reached when the rate of condensation and the rate of evaporation become equal Equilibrium vapor pressure the vapor pressure of condensation and evaporation measured at dynamic equilibrium 0 Maximum vapor pressure a liquid exerts at a given temperature 0 Constant at constant temperature 0 We know that the number of molecules with higher kinetic energies is greater at higher temperatures and therefore the evaporation rate is also higher For this reason the vapor pressure VP of a liquid ALWAYS increases with temperature ie VP of water at 20 C is 175 mmHg while at 100 C it is 760 mmHg t of Vaporization AHEn and Boiling Point BP Molar heat of vaporization a measure of how strongly molecules are held in a liquid 0 Defined as the energy usually in kiloj oules kJ required to vaporize one mole of a liquid 0 Directly related to the strength of IMFs that exist in liquid I If IMF attraction is strong it takes a lot of energy to free the molecules from the liquid phase Consequently the liquid has a relatively low VP and a high AHvap Quantitative relationship between the equilibrium vapor pressure P of a liquid and the absolute temperature T the ClausiusClapeyron equation AHvap 0 In P RT C P l AHvap T l T 2 0 For multiple pressures and temperatures In E R T1 T 2 ble 125 M0131 Heats of Vgporization for Selected Liquids Substance Boiling Point C AHvap in kJmol Argon Ar 186 63 Benzene C6H6 801 31 Diethyl ether C2H50C2H5 346 26 Ethanol CZHSOH 783 393 Mercury Hg 357 59 Methane CH4 164 92 Water H20 100 4079 Example 125 Diethyl ether is a volatile highly ammable organic liquid that is used mainly as a solvent The vapor pressure of diethyl ether is 401 mmHg at 18 C Calculate its vapor pressure at 29 C Using table 125 we see that AHvap for diethyl ether is 26 kJmol which is 26000 Jmol Given P1 401 mmHg T1 18 C 291 K T2 29 C 302 K R 8314 JKmol we use the ClausiusClapeyron equation to find the second VP 401mmHg 26000 Jmol 291 K 302 K In P2 8314JKm0l 291K302K 1 Simplify the fractions and take equot of both sides to get 401P2 e A 0391 And solving for P2 we get 593 mmHg Boiling Point BP the temperature at which the vapor pressure of a liquid is equal to the external pressure The normal boiling point is the boiling point when the external pressure is 1 atm Hydrostatic pressure pressure caused by the presence of a liquid Pressure inside a bubble is due solely to a liquid s VP 0 If the VP in the bubble were lower than the external pressure the bubble would collapse before it could rise We can thus conclude that the BP of a liquid depends on the external pressure 0 Ultimately both the BP and the AHvalp are determined by the strength of the IMFs Critical Temperature T9 The highest temperature at which a substance can exist in liquid form Above this temperature the substance s gas form cannot be made to liquefy no matter how great the applied pressure Critical Pressure PS The minimum pressure needed to bring about liquefaction at the critical temperature 0 The melting point of a solid or the freezing point of a liquid is the temperature at which solid and liquid phases coexist in equilibrium ble 126 Critical Temperatures gnd Criticgl Pressures of Selected Substances Substance Tc C Pc atm Ammonia NH3 1324 1 1 15 Argon Ar 186 63 Benzene C6H6 2889 479 Carbon Dioxide C02 31 73 Diethyl ether C2H50C2H5 1926 356 Ethanol C2H50H 243 63 Mercury Hg 1462 1036 Methane CH4 83 456 Molecular Hydrogen H2 2399 128 Molecular Nitrogen N2 l47l 335 Molecular Oxygen 02 1 188 497 Sulfur Hexa uoride SF6 455 376 Water H20 3744 2195 ble 127 Molgr Hth of Fusion AHf for Selected Substances Substance Melting Point C AHfus in kJmol Argon Ar 190 13 Benzene C6H6 55 109 Diethyl ether CZHSOCZHS 1162 69 Ethanol CZHSOH 1173 761 Mercury Hg 39 234 Methane CH4 183 084 Water H20 0 601 Example Diethyl ether has a AHvalp of 26 kJmol and a vapor pressure of 0703 atm at 25 C Calculate its vapor pressure at 85 C We are given P1 0703 atm T1 25 C 298 K T2 85 C 358 K Need to find P2 Again we use the ClausiusClapeyron equation to find P2 0703atm 26000Jmol 298 358 111 P2 8314JKm0l 298358 1 111 P2 31273gt 106684 0703atm ln P2 31273562410394 Take the equot of both sides to get 0703 atmPz e139759 and solve for P2 to get 409 atm Molar heat of fusion AHEE The energy usually in kJ required to melt one mole of a solid 0 A comparison of the data in the two tables shows that for each substance AHfus lt AHvap Molar heat of sublimation AHsub The energy usually in kJ required to sublime one mole of a solid 127 Phase Diagrams A phase diagram summarizes the conditions under which a substance exists as a solid liquid or gas The triple point is the only location at which the temperature and pressure support all three phases at equilibrium with one another Phase diagrams enable us to predict changes in the melting point and boiling point of a substance as a result of changes in the external pressure We can also anticipate directions of phase transitions brought about by changes in temperature and pressure
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