A Molecular Approach, 4e - Notes for Chapter (11).doc

Chapter 11. Liquids, Solids, and Intermolecular Forces Chapter 11. Liquids, Solids, and Intermolecular Forces Chapter 11. Liquids, Solids, and Intermolecular Forces Student Objectives 11.1 Water, No Gravity Know that intermolecular forces are attractive forces between individual molecules. The states of matter are due to intermolecular forces. 11.2 Solids, Liquids, and Gases: A Molecular Comparison Know the properties that differentiate the phases of matter: density, molar volume, molecular shape, and strength of intermolecular forces. Define crystalline and amorphous and recognize the difference in solids. Know that both temperature and pressure can affect phase changes. 11.3 Intermolecular Forces: The Forces That Hold Condensed States Together Know and understand that intermolecular forces originate from the interactions between charges, partial charges, and temporary charges on molecules, atoms, and ions. Know how Coulomb’s law describes the mathematical relationship between energy of attraction, magnitude of charge, and distance. Know and understand that dispersion (London) forces result from fluctuations of electron distribution within molecules and atoms. Identify and predict how the shape and sizes of molecules or atoms affects the magnitude of dispersion forces the particles exhibit as well as macroscopic physical properties like boiling point. Know and understand that polar molecules have permanent dipoles that attract each other through dipole-dipole interactions. Know and understand the phenomenon of hydrogen bonding. Predict the ability of molecules to exhibit hydrogen bonding. Recognize hydrogen bonding as the force that holds double-stranded DNA together. Rank a series of molecular compounds with respect to boiling point. Know and understand that the interaction of ions and dipoles leads to the dissolution and solvation of ions by water and other polar liquids. 11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action Know and understand that surface tension is due to intermolecular forces. Describe examples of surface tension. Know and understand that viscosity is due to intermolecular forces, mass, shape, and length. Know and understand that capillary action is the result of both cohesive and adhesive forces. 11.5 Vaporization and Vapor Pressure Understand the process of vaporization and how it changes with temperature, surface area, and the degree of intermolecular forces. Understand that molecules or atoms have a distribution of thermal energies that changes as a function of temperature. Know that the heat of vaporization, Hvap, is a quantitative measure for the process of vaporization. Calculate and interconvert mass, moles, and energy using the heat of vaporization. Know and understand how vapor pressure and dynamic equilibrium dictate vaporization and condensation. Know that the vapor pressure of a liquid depends on temperature and that the boiling point of a liquid depends on the external pressure. Use the Clausius-Clapeyron equation to relate temperature and vapor pressure. Define critical temperature and critical pressure. 11.6 Sublimation and Fusion Define and understand sublimation and deposition. Define and understand fusion in the context of phase changes. Use the heat of fusion, Hfus in calculations involving energy, masses, and moles. 11.7 Heating Curve for Water Understand the different segments in the heating curve for H2O that ranges from below the melting point to above the boiling point. Calculate the energy changes associated with heating a substance (like H2O) through a series of temperature changes and phase changes. 11.8 Phase Diagrams Know that a phase diagram relates the states of matter for a substance to temperature and pressure. Identify the main regions and significant points in a phase diagram. Understand the effect of changes in temperature and changes in pressure on the phase of a substance as shown by its phase diagram. 11.9 Water: An Extraordinary Substance Know that water has unique properties compared with similar molecules based on size, constituent atoms, and molar mass. Know that the unique properties of water are attributable to hydrogen bonding. 11.10 Crystalline Solids: Their Structure by X-Ray Crystallography Recall that waves can interfere constructively and destructively. Know that X-rays diffract when interacting with the atoms in crystalline solids, forming diffraction patterns. Know that diffraction patterns can be analyzed and used to identify the three-dimensional structure of the atoms or molecules in a crystalline solid. Use Bragg’s law to calculate the relationship between the distance between crystalline layers, the wavelength of electromagnetic radiation, and the angle of reflection. 11.11 Crystalline Solids: Unit Cells and Basic Structures Define and identify unit cells. Know and identify the cubic crystalline lattice types: simple, body-centered, and face-centered. Identify the kind of unit cell, the coordination number, and the edge length for the three cubic crystalline lattice types. Use the kind of unit cell and the radius of an atom to calculate the density of a metal. Identify the hexagonal and cubic closest-packing structures, and know their unit cells and component layers. 11.12 Crystalline Solids: The Fundamental Types Know the organization of crystalline solids—molecular, ionic, and atomic—including basic properties and examples. Know and identify constituent atoms, lattice types, and unit cells for some common ionic solids: CsCl, NaCl, ZnS, and CaF2. Know and identify atomic solid types—nonbonding, metallic, and network covalent—and some of their properties and examples. 11.13 Crystalline Solids: Band Theory Know that the organization of conduction and valence bands of molecular orbitals forms the basis for conductors, semiconductors, and insulators. Section Summaries Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Teaching Tips Suggestions and Examples Misconceptions and Pitfalls Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.1 Water, No Gravity Intermolecular forces defined Intro figure: water on a space station 11.2 Solids, Liquids, and Gases: A Molecular Comparison Phases of matter gas liquid solid Phase properties density, volume, mass shape Phase changes temperature pressure Table 11.1 The Three States of Water Table 11.2 Properties of the States of Matter Figure 11.1 Liquids Assume the Shapes of Their Containers Figure 11.2 Gases Are Compressible Figure 11.3 Crystalline and Amorphous Solids unnumbered figure: molecular representation of phase changes unnumbered figure: molecular illustration of propane Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 11.1 Water, No Gravity Water forms a sphere in the absence of gravity. 11.2 Solids, Liquids, and Gases: A Molecular Comparison The phases of matter provide a meaningful connection between the microscopic level and the macroscopic level. Conceptual Connection 11.1 State Changes Students easily grasp that steam is still H2O, but they sometimes don’t recognize for larger molecular compounds that boiling concerns only intermolecular forces rather than intramolecular. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.3 Intermolecular Forces: The Forces That Hold Condensed States Together Basics of intermolecular forces Coulomb’s law Types of intermolecular forces dispersion (London) forces exhibited by all molecular compounds mass shape, length dipole-dipole forces exhibited by polar compounds permanent dipoles dipole moments hydrogen bonding H–X, where X = N, O, F extreme case of dipole–dipole ion–dipole solution and solvation Boiling point as a measure of intermolecular forces Hydrogen bonding in DNA base pairing A-T G-C unnumbered figure: distances of intramolecular and intermolecular forces unnumbered figure: polarization of He atom Figure 11.4 Dispersion Interactions Table 11.3 Boiling Points of the Noble Gases unnumbered figure: molecular models of n-pentane and neopentane Figure 11.5 Dispersion Force and Molecular Shape Figure 11.6 Boiling Points of the n-Alkanes Figure 11.7 Dipole–Dipole Interaction unnumbered table: formula, molar mass, structure, melting point, and boiling point of formaldehyde and ethane Figure 11.8 Dipole Moment and Boiling Point Figure 11.9 Polar and Nonpolar Compounds Example 11.1 Dipole–Dipole Forces Figure 11.10 Hydrogen Bonding in HF unnumbered table: comparison of ethanol and dimethyl ether Figure 11.11 Hydrogen Bonding in Ethanol Figure 11.12 Hydrogen Bonding in Water Figure 11.13 Boiling Points of Group 4A and 6A Compounds Example 11.2 Hydrogen Bonding Figure 11.14 Ion–Dipole Forces Table 11.4 Types of Intermolecular Forces Chemistry and Medicine: Hydrogen Bonding in DNA Figure 11.15 Nucleotides Figure 11.17 Copying DNA Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 11.3 Intermolecular Forces: The Forces That Hold Condensed States Together Intermolecular forces are based on the distribution of charge, ranging from fluctuations in electron distribution to polar bonds and permanent dipoles to ions. Conceptual Connection 11.2 Dispersion Forces Several tables and figures warrant exploration: Table 11.3 Boiling Points of the Noble Gases Figure 11.6 Boiling Points of the n-Alkanes Figure 11.8 Dipole Moment and Boiling Point unnumbered table: comparison of ethanol and dimethyl ether Conceptual Connection 11.3 Dipole–Dipole Interaction Hydrogen bonding accounts for many of the properties of water and aqueous solutions as well as important behaviors in organic and biochemical systems. Conceptual Connection 11.4 Intermolecular Forces and Boiling Point All intermolecular forces are due to the distribution of charge. Even in the case of dispersion forces, Coulomb’s law is applicable. Hydrogen bonding is an example of a relatively strong intermolecular force, but it still is a weak interaction when compared with covalent bond energies. Specific conditions must be met for hydrogen bonding. Not every compound containing hydrogen nor every compound containing hydrogen and oxygen/nitrogen/fluorine exhibits this intermolecular force. The two molecules involved in a hydrogen bonding interaction can be identical or different. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action Surface tension model examples Viscosity motor oil Capillary action meniscus of water meniscus of mercury unnumbered figure: photo of trout fly and trout Figure 11.18 The Origin of Surface Tension Figure 11.19 Surface Tension in Action Figure 11.20 Spherical Water Droplets Table 11.5 Viscosity of Several Hydrocarbons at 20 oC Table 11.6 Viscosity of Liquid Water at Several Temperatures Chemistry in Your Day: Viscosity and Motor Oil unnumbered figure: photo of drawing blood using capillary action Figure 11.21 Capillary Action Figure 11.22 Meniscuses of Water and Mercury 11.5 Vaporization and Vapor Pressure Vaporization and condensation distribution of thermal energy energetics, Hvap calculations using Hvap as a conversion factor Vapor pressure and dynamic equilibrium closed systems piston: volume and temperature dependence Temperature dependence of vapor pressure boiling point definition Clausius-Clapeyron equation Critical point critical temperature critical pressure supercritical fluid Figure 11.23 Vaporization of Water Figure 11.24 Distribution of Thermal Energy Table 11.7 Heats of Vaporization of Several Liquids at Their Boiling Points and at 25 oC Example 11.3 Using the Heat of Vaporization in Calculations Figure 11.25 Vaporization in a Sealed Flask Figure 11.26 Dynamic Equilibrium Figure 11.27 Dynamic Equilibrium in n-Pentane Figure 11.28 Vapor Pressure of Several Liquids at Different Temperatures Figure 11.29 Boiling Table 11.8 Boiling Points of Water at Several Locations of Varied Altitudes Figure 11.30 The Temperature during Boiling Figure 11.31 A Clausius-Clapeyron Plot for Diethyl Ether (CH3CH2OCH2CH3) Example 11.4 Using the Clausius-Clapeyron Equation to Determine Heat of Vaporization from Experimental Measurements of Vapor Pressure Example 11.5 Using the Two-Point Form of the Clausius-Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature Figure 11.32 Critical Point Transition Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action Students can provide other examples of surface tension (undissolved powder on a water surface, water bugs that walk on the surface of a pond); viscosity (maple syrup, corn syrup); capillary action (the wick of a candle, a sponge or paper towel to soak up a spill). Surface tension, viscosity, and capillary action are macroscopic properties attributable to intermolecular forces. 11.5 Vaporization and Vapor Pressure Vaporization occurs at the surface interface between a liquid and the atmosphere in part because it is easier for molecules to break free. Like molecules in a gas, molecules in a liquid are characterized by a distribution of energies that changes with temperature. Evaporation has a cooling effect because the highest energy molecules escape, leaving behind a collection of molecules with a lower average kinetic energy. Vaporization and dynamic equilibrium can be demonstrated with sealed containers. Students experience this with water bottles. Ask for other examples. The supercritical fluid is a unique state of matter, combining aspects of liquid and gas phases. Supercritical carbon dioxide has a low critical temperature and can provide an exciting demonstration. Supercritical carbon dioxide is now used in many commercial applications: as a solvent for reactions, extraction of food components (e.g. caffeine), and as an alternative to halogenated organic solvents in dry cleaning. Conceptual Connection 11.5 Vapor Pressure The difference between vaporization and boiling is that vaporization is a phase transition that occurs only at the surface, whereas boiling is a phase transition that can occur at any point within the liquid. Vaporization is affected by temperature and pressure; the mathematical dependence on temperature is given by the Clausius-Clapeyron equation. The rate of vaporization increases with surface area, but vapor pressure is a constant for a given liquid at a given temperature, regardless of the container size or liquid volume (greater than 0). The two-point form of the Clausius-Clapeyron presents some challenges especially for the student with weak algebra skills. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.6 Sublimation and Fusion Sublimation and deposition definitions examples Fusion melting point energetics, Hfus Hvap > Hfus Figure 11.33 The Sublimation of Ice unnumbered figure: photo of dry ice Figure 11.34 Temperature during Melting Table 11.9 Heats of Fusion of Several Substances Figure 11.35 Heat of Fusion and Heat of Vaporization 11.7 Heating Curve for Water Calculations T < mp T = mp bp > T > mp T = bp T > bp Figure 11.36 Heating Curve for Water 11.8 Phase Diagrams Major features s, l, g regions phase changes triple point normal mp, bp critical point Navigation Comparison of water, iodine, carbon dioxide Figure 11.37 Phase Diagram for Water Figure 11.38 Navigation on the Phase Diagram for Water Figure 11.39 Phase Diagrams for Other Substances 11.9 Water: An Extraordinary Substance Boiling point comparison of main group hydrides Hydrogen bonding Environmental solvent Freezing water and biological systems unnumbered figure: photo of Mars Curiosity Rover Figure 11.40 Boiling Points of Main Group Hydrides Figure 11.41 Hydrogen Bonding in Water unnumbered figure: photo of frozen and fresh lettuce Chemistry in the Environment: Water Pollution Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 11.6 Sublimation and Fusion Provide some examples of sublimation and deposition: dry ice, the freezer burn example of frozen peas, formation of frost in winter, freeze-dried foods, etc. Sublimation and deposition are phase transitions between gas and solid without involvement of the liquid phase. Students often are surprised to learn that freezing is an exothermic process, likely because their experience with freezing has mostly to do with water. Ask them what happens when molten iron turns to solid. Students sometimes interpret fusion as going from liquid to solid rather than the reverse. 11.7 Heating Curve for Water The heating curve for water plots temperature as a function of added heat. The largest requirement comes from the heat of vaporization. Figure 11.36 shows not only the plot but also microscopic views of each segment. Conceptual Connection 11.6 Cooling of Water with Ice The heat of fusion and heat of vaporization are much larger than the molar heat capacity of any of the phases. Students often are surprised to learn that melting/freezing, sublimation/deposition, and boiling/condensation are constant temperature processes. 11.8 Phase Diagrams Compare the phase diagrams of water, iodine, and carbon dioxide. Iodine’s diagram is similar to that of many substances, but water's and carbon dioxide’s are relatively unique. For water, the slope of the fusion curve is opposite that of most substances, and for carbon dioxide, the liquid exists only at relatively high pressures. Conceptual Connection 11.7 Phase Diagrams These are relatively simple phase diagrams for pure substances. A more detailed phase diagram for water shows the different phases of ice. Mixtures also have phase diagrams. 11.9 Water: An Extraordinary Substance The unique properties of water are attributable mostly to hydrogen bonding. Water is the solvent for the environment. Changes and negative impacts accompany the positive aspects of its role. Ask students to suggest advantages and disadvantages of the lower density of solid water versus liquid water. Additional Problem for Dipole–Dipole Forces (Example 11.1) A polar molecule will exhibit dipole–dipole forces. Which molecules will show dipole–dipole forces? a) CHF3 b) OCS c) H2S Solution 1) Electronegativity values from Figure 9.10: H = 2.1 C = 2.5 F = 4.0 2) There are three polar C–F bonds. The net dipole moment points down along the H–C bond. Solution 1) Electronegativity values from Figure 9.10: O = 3.5 C = 2.5 S = 2.5 2) The C=S bond is not polar; the C=O bond is polar. The net dipole moment points along the C=O bond. Solution 1) Electronegativity values from Figure 9.10: S = 2.5 H = 2.1 2) Each H–S bond is only weakly polar. A very small net dipole exists and points toward S bisecting the H–S–H angle. Additional Problem for Using the Heat of Vaporization in Calculations (Example 11.3) Calculate the energy in kJ necessary to vaporize 100 g of water at its boiling point. Sort You are given a certain mass of water asked the energy necessary to vaporize it. Given 100 g water Find kJ Strategize The heat of vaporization gives the relationship between heat absorbed and moles of water vaporized. Begin with the grams of water and convert to moles using the molar mass. Then use Hvap as a conversion factor to obtain kJ of energy. Conceptual Plan g mol kJ Relationships Used Hvap = 40.7 kJ/mol (at 100 oC) 1 mol H2O = 18.02 g H2O Solve Follow the conceptual plan to solve the problem. Solution Check The units of the answer are correct and the magnitude makes sense. The 100 g sample is slightly more than 5 moles. Additional Problem for Using the Two Point Form of the Clausius-Clapeyron Equation to Predict the Vapor Pressure at a Given Temperature (Example 11.5) Octane has a normal boiling point of 125.5 oC and a heat of vaporization (Hvap) of 38.8 kJ/mol. What is the vapor pressure of octane at 20 oC? Sort The problem gives you the normal boiling point of octane (the temperature at which the vapor pressure is 760 mmHg) and the heat of vaporization. You are asked to find the vapor pressure at a different temperature. Given T1 = 125.5 oC P1 = 760 mmHg Hvap = 38.8 kJ/mol T2 = 20 oC Find P2 Strategize Use the Clausius-Clapeyron equation to find P2. Conceptual Plan Relationships Used Clausius-Clapeyron equation (two point form) T (K) = oC + 273.15 Solve Convert T1 and T2 from oC to K. Substitute into the equation and solve for P2. Solution T1 (K) = 125.5 + 273.15 = 398.65 K T2 (K) = 20 + 273.15 = 293.15 K Check The units of the answer (torr) are correct. The magnitude of the answer (11.3) makes physical sense since octane is not a very volatile molecule. 156 Copyright © 2017 by Education, Inc. 155 Copyright © 2017 by Education, Inc.