AP CHEMISTRY NOTES 15-1 INTERMOLECULAR FORCES
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1 AP CHEMISTRY NOTES 15-1 INTERMOLECULAR FORCES INTERMOLECULAR FORCES In addition to the covalent bonds that exist between atoms in a molecule (H2O for instance), there are also weak attractions between the molecules themselves. They are responsible (among other things) for whether a compound is a liquid, solid, or gas at room temperature. 1. London Dispersion Forces (AKA Induced dipole-induced dipole, van der Waal s Forces) Caused by the attraction of the positively charged nucleus of one atom for the electron cloud of an atom in a nearby molecule, inducing a temporary dipole in neighboring atoms/molecules. *The weakest of the intermolecular attractive forces *All covalently bonded substaces are attracted by London dispersion forces *Bigger atoms / molecules have larger London dispersion forces and are therefore attracted more strongly (their large electron clouds are not attracted as strongly to their own nuclei, so they are more easily polarizable For example: The bigger the molecule, the higher the boiling point (note the halogens) 2. Dipole-Dipole Interactions - Caused by the attraction of polar molecules to each other 1 *Similar to ionic bonds, but much weaker since only partial charges are involved *Occur only in polar molecules
2 3. Dipole-Induced Dipole Interactions Present between a polar and nonpolar molecule. The strength of these forces increases with the magnitude of the dipole of the polar molecule and with the polarizability of the nonpolar molecule 4. Hydrogen Bonding Occurs when a hydrogen atom that is covalently bonded to a very electronegative element (F, O, N) is attracted to a lone electron pair on another atom in a nearby molecule *The strongest of the intermolecular attractive forces (~5% the strength of a covalent bond) *Explains many of the odd characteristics of water 5. Ion-Dipole Interactions The largest of the intermolecular forces of attraction; it exists between a fully-charged positive or negative ion and a polar molecule PROPERTIES OF LIQUIDS Vapor Pressure (also known as equilibrium vapor pressure and often referred to as volatility ) *The tendency of the molecules of a liquid to escape the liquid phase and enter the vapor phase *Vapor pressure increases as temperature increases *The higher the vapor pressure, the more volatile the liquid 2 Boiling Point the temperature at which the vapor pressure of a liquid is equal to the external (atmospheric) pressure on the system Normal Boiling Point the boiling point at one atmosphere of pressure
3 The Clausius-Clapeyron Equation relates vapor pressure and liquid temperature R = J/K. mol EXAMPLE: Estimate the vapor pressure of water at 85 o C. Note that the normal boiling point of water is 100 o C and that its heat of vaporization is 40.7 kj/mol. In addition, recall that the normal boiling point of a liquid is its boiling point at 1 atm (760 mmhg). KNOW IT CONCEPTUALLY PROPERTIES OF SOLIDS Network Solid a lattice structure of atoms all covalently bonded to each other 3
4 AP CHEMISTRY NOTES 15-2 PHASE DIAGRAMS Triple Point the point on a phase diagram representing the temperature and pressure at which the solid, liquid, and gaseous forms of a substance coexist in equilibrium Critical Temperature above this temperature, a conventional liquid does not exist no matter how high the pressure Matter above this temperature is sometimes referred to as a supercritical fluid. It has unexpected properties in this state (ie the ability to dissolve normally insoluble materials; companies are already using supercritical carbon dioxide to extract caffeine from coffee) Critical Pressure the vapor pressure at the critical temperature Critical Point the point at which the critical temperature and critical pressure meet 4
5 P For the diagram above, what is the triple point? the critical point? What phase change occurs when: the temperature at 1 atm moves from -10 o C to 50 o C? the pressure at 10 o C moves from mmhg to 10.0 mmhg? Comparative Densities of Solids and Liquids: *A positive slope for the solid/liquid interface = solid state more dense than liquid state (common) *A negative slope for the solid/liquid interface = liquid state more dense than solid state (common) 5
6 AP CHEMISTRY NOTES 15-3 COLLIGATIVE PROPERTIES SOLUTIONS Molality (m) the number of moles of solute per kilogram solvent EXAMPLE: How many grams of potassium nitrate (KNO3) must be dissolved in 550 g of water to produce a m solution? Colligative Properties properties which are based on the number of solute particles or ions dissolved in a given amount of solvent; these properties are based on molality 1. Lowering of Vapor Pressure lower tendency to evaporate *With nonpolar solutes: *With polar solutes: Solute particles at the surface of the liquid inhibit evaporation Ion dipole attraction also creates a greater attractive force, inhibiting evaporation Raoult s Law: P solvent = X solvent P o solvent 6 Where Psolvent = vapor pressure of solvent above the solution Xsolvent = mole fraction of the solvent P o solvent = vapor pressure of the pure liquid
7 EXAMPLE: Nitroglycerin is used in the manufacture of dynamite and in the treatment of certain coronary problems. If you add g of the compound to 100. g of benzene at 25 o C, the vapor pressure of the solution is mmhg. What is the molar mass of nitroglycerin? (the vapor pressure of pure benzene at the same temperature is mmhg.) 2. Elevation of the Boiling Point (same reasoning as for vapor pressure) To determine the boiling point of a solution: a. Use the following equation to determine the change in the boiling point: T b = k b im b. Calculate the new boiling point: Where Tb kb i m = change in boiling point = boiling point constant (0.512 o C/m) = van t Hoff factor = molality of solution New BP = Normal BP + T b EXAMPLE: What is the boiling point of a solution that contains 65.0 grams of NaCl in grams of water? 7
8 3. Depression of the Freezing Point solvation interferes with crystal formation To determine the freezing point of a solution: a. Use the following equation to determine the change in the freezing point: T f = k f im Where Tf kf i m = change in boiling point = freezing point constant (1.86 o C/m) = van t Hoff factor = molality of solution b. Calculate the new boiling point: New FP = Normal FP - T f EXAMPLE: What is the freezing point of a solution that contains 123 g of Ba(NO3)2 dissolved in g of water? EXAMPLE: Either camphor (C10H16O, molecular weight = 152 g/mol) or naphthalene (C10H8, molecular weight = 128 g/mol) can be used to make mothballs. A 5.2-g sample of mothballs was dissolved in 100. grams of ethyl alcohol, and the resulting solution had a boiling point of o C. Were the mothballs made of camphor or naphthalene? Pure ethyl alcohol has a boiling point of o C and its kb = 1.22 o C/m. 8
9 4. Elevation of Osmotic Pressure when solvent molecules move through a semi-permeable membrane from a region of lower solution concentration to a region of higher solute concentration A higher concentration of solute results in a higher osmotic pressure Osmotic pressure can be found using the following equation: π = M R T Where π = osmotic pressure (atm) M = solution concentration (mol/l) R = Gas Law Constant ( L. atm/mol. K) T = temperature (K) EXAMPLE: Beta-carotene is the most important of the A vitamins. Its molar mass can be determined by measuring the osmotic pressure generated by a given mass of carotene dissolved in the solvent chloroform. Calculate the molar mass of carotene if 7.68 mg dissolved in 10.0 ml of chloroform give an osmotic pressure of mmhg at 25 o C. 9
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