Intermolecular Forces, Liquids, and Solids

Intermolecular Forces, Liquids, and Solids Key Questions 1. Is the average kinetic energy of molecules greater or lesser than the energy of intermolecular forces of attraction in (a) solids, (b) liquids, and (c) gases? (a) In solids, kinetic energy is less than intermolecular energy. (b) In liquids, kinetic energy is less than intermolecular energy, but the disparity is less than in solids. (c) In gases, the kinetic energy is much greater than the energy of intermolecular forces of attraction. 2. Why does increasing the temperature cause a substance to change in succession from a solid to a liquid to a gas? Kinetic energy increases with temperature. As the kinetic energy rises with temperature, the intermolecular forces of attraction are overcome by increasing molecular motion. 3. Why do substances with high surface tension also tend to have high viscosities? Liquids with stronger intermolecular forces of attraction hold the molecules closer together, which causes stronger surface tension and greater resistance to flowing (viscosity). 4. Why do surface tension and viscosity decrease with increasing temperature? The increased kinetic energy with rising temperature overcomes the cohesive intermolecular forces of attraction. 5. Name the kind or kinds of intermolecular forces that must be overcome to convert the following from liquid or solid to gas: (a) Br2, (c) CH3OH, (d) CO2, (e) HCN, (f) NH3 (a) (b) (c) (d) (e) 6. Br2 – London dispersion CH3OH – London dispersion, dipole-dipole, hydrogen bonding CO2 – London dispersion (CO2 is linear and therefore nonpolar.) HCN – London dispersion, dipole-dipole (HCN is linear but polar.) NH3 – London dispersion, dipole-dipole, hydrogen bonding Normal alkanes are hydrocarbons with unbranched carbon chains, having a general formula CnH2n+2. At room temperature, ethane, C2H6, is a gas; hexane, C6H14, is a liquid; and octadecane, C118H38, is a solid. Describe the intermolecular forces present in each substance and explain the differences in their room-temperature phases. All are nonpolar and therefore only have London dispersion forces. However, London dispersion forces rise with molecular weight, as the numbers of electrons increase, which in turn cause the polarizabilities to increase. Thus, the order of increasing intermolecular forces is C2H6 < C6H14 < C118H38. As the London dispersion forces increase the tendency to be in a condensed phase increases. 7. Arrange the following in order of increasing boiling point: The compounds that have -OH groups will have hydrogen bonding and therefore much higher boiling points than those that do not. Within the two groups (hydrogen bonding or no hydrogen bonding), the ordering follows increasing London dispersion forces (increasing molecular weight) and additional hydrogen bonding. Note that the last compound, ethylene glycol, has twice as many sites for hydrogen bonding per molecule as the two alcohols, CH3OH and C2H5OH. The observed boiling point (oC) is shown for each compound. 8. How much heat is required to heat 10.0 g of ice at -5.00 oC to become liquid water at +7.00 o C? In this temperature range, the heat capacity of H2O(s) is 37.7 J/molAK, and the heat capacity of H2O(l) is 75.8 J/molAK. The molar heat of fusion of ice is 6.01 kJ/mol. Calculate the heat to warm the ice, then the heat to melt the ice, and finally the heat to warm the liquid water. The sum is the total heat required. mol H2O = 10.0 g/18.02 gAmol-1 = 0.555 mol Heat ice to 0 oC: q1 = CpΔT = (37.7 J/molAK)(0.555 mol)(5.00 K) = 105 J Melt ice at 0 EC: q2 = ΔHmelt × moles = (6.01 kJ/mol)(0.555 mol) = 3.34 kJ = 3340 J Heat liquid to 7 oC: q3 = CpΔT = (75.8 J/molAK)(0.555 mol)(7.00 K) = 294 J Total heat: 9. qt = q1 + q2 + q3 = 105 J + 3340 J + 294 J = 3739 J = 3.74 kJ Explain how each of the following affects the vapor pressure of a liquid: (a) the volume of the liquid, (b) the volume of the container, (c) the surface area of the liquid, (d) the temperature, (e) intermolecular forces of attraction, (f) the density of the liquid. (a) (b) (c) (d) (e) (f) No effect No effect No effect Higher vapor pressure at higher temperature Lower vapor pressure with stronger intermolecular forces of attraction Density is mass per unit volume. It tends to increase as molecular weight increases. London dispersion forces also increase with molecular weight, and this would cause a decrease in volatility, resulting in lower vapor pressure. Thus, denser liquids tend to have lower vapor pressures. 10. Describe the phases and/or phase transitions experienced by CO2 under the following conditions: (a) Heating from –100 oC to 30 oC at 1.0 atm Solid to gas, with a normal sublimation point at –78 oC. (b) Heating from –100 oC to 50 oC at 70 atm Solid to liquid to gas (c) A sample at 35 oC and 100 atm A point above the critical point, so a supercritical fluid (d) A sample at –50 oC and 6.0 atm A point in the liquid range, just above the triple point 11. Does carbon dioxide have a normal boiling point? Explain. The normal boiling point is the temperature at which the vapor pressure above a liquid reaches exactly 1 atm. At 1 atm there is no liquid-vapor equilibrium for CO2, so it does not have a normal boiling point. 12. Describe the conditions under which liquid carbon dioxide boils. At a pressure and temperature just above the triple point (–57 oC, 5.1 atm), liquid CO2 would exist. This could be boiled by raising the temperature to achieve a point on the liquid-vapor line.