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1 Chapter 11. Liquid s, Solids, and Intermolecular Forces Student Objectives 11.1 Climbing Geckos and Intermolecular Forces Know that intermolecular forces are attractive forces between individual molecules. Know that the gecko s ability to climb smooth surfaces is due to intermolecular forces Solids, Liquids, and Gases: A Molecular Comparison Know the properties that differentia te 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 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 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. 154

2 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, H vap, 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 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 Heating Curve for Water Understand the different segments in the heating curve for H 2 O that ranges from below the melting point to above the boiling point. Calculate the energy changes associated with heating a substance (like H 2 O) through a series of temperature changes and phase changes 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 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 Crystalline Solids: Determining 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. 155

3 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 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 CaF 2. Know and identify atomic solid types nonbonding, metallic, and network covalent and some of their properties and examples Crystalline Solids: Band Theory Know that the organization of conduction and valence bands of molecular orbitals forms the basis for conductors, semiconductors, and insulators. 156

4 Section Summaries Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 157

5 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.1 Cli mbing Geckos and Intermolecular Forces Intermolecular forces defined Microhairs on a gecko s feet and intermolecular forces Intro figure: geckos and intermolecular forces unnumbered figure: electron micrograph of microhairs on a gecko s feet 11.2 Solids, Liquids, and Gases: A Molecular Comparison Phases of matter o gas o liquid o solid Phase properties o density, volume, mass o shape Phase changes o temperature o 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 158

6 Teaching Tips S uggestions and Examples Misconceptions and Pitfalls 11.1 Climbing Geckos and Intermolecular Forces The gecko s feet provide yet another illustration of practical chemistry 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 H 2 O, but they sometimes don t recognize for larger molecular compounds that boiling concerns only intermolecular forces rather than intramolecular. 159

7 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.3 Int ermolecular Forces: The Forces That Hold Condensed States Together Basics of intermolecular forces o Coulomb s law 1 q1q 2 E 4 o r Types of intermolecular forces o dispersion (London) forces exhibited by all molecular compounds mass shape, length o dipole dipole forces exhibited by polar compounds permanent dipoles dipole moments o hydrogen bonding H X, where X = N, O, F extreme case of dipole dipole o ion dipole solution and solvation Boiling point as a measure of intermolecular forces Hydrogen bonding in DNA o 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 T able 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 unnumbered table: formula, molar mass, structure, melting point, and boiling point of formaldehyde and ethane Figure 11.7 Dipole Dipole Interaction Figure 11.8 Dipole Moment and Boiling Point Figure 11.9 Polar and Nonpolar Compounds Example 11.1 Dipole Dipole Forces Figure Hydrogen Bonding in HF Figure Hydrogen Bonding in Ethanol unnumbered table: comparison of ethanol and dimethyl ether Figure Hydrogen Bonding in Water Figure Boiling Points of Group 4A and 6A Compounds Example 11.2 Hydrogen Bonding Figure Ion Dipole Forces Table 11.4 Types of Intermolecular Forces Chemistry and Medicine: Hydrogen Bonding in DNA Figure Nucleotides Figure Copying DNA 160

8 Teaching Tips S uggestions 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: o Table 11.3 Boiling Points of the Noble Gases o Figure 11.6 Boiling Points of the n Alkanes o Figure 11.8 Dipole Moment and Boiling Point o 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. 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. 161

9 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Ca pillary Action Surface tension o model o examples Viscosity o motor oil Capillary action o meniscus of water o meniscus of mercury 11.5 Vaporization and Vapor Pressure Vaporization and condensation o distribution of thermal energy o energetics, H vap o calculations using H vap as a conversion factor Vapor pressure and dynamic equilibrium o closed systems o piston: volume and temperature dependence Temperature dependence of vapor pressure o boiling point definition o Clausius Clapeyron equation H vap 1 ln P vap ln R T Critical point o critical temperature o critical pressure o supercritical fluid unnumbered figure: photo of trout fly and trout Figure The Origin of Surface Tension Figure Surface Tension in Action Figure Spherical Water Droplets Table 11.5 Viscosity of Several Hydrocarbons at 20 o C 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 Capillary Action Figure Meniscuses of Water and Mercury Figure Vaporization of Water Figure 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 Vaporization in a Sealed Flask Figure Dynamic Equilibrium Figure Dynamic Equilibrium in n Pentane Figure Vapor Pressure of Several Liquids at Different Temperatures Figure Boiling Table 11.8 Boiling Points of Water at Several Locations of Varied Altitudes Figure The Temperature during Boiling Figure A Clausius Clapeyron Plot for Diethyl Ether (CH 3 CH 2 OCH 2 CH 3 ) 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 Critical Point Transition 162

10 Teaching Tips S uggestions 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 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. 163

11 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 11.6 Sublimation and Fusion Sublimation and deposition o definitions o examples Fusion o melti ng poin t o energetics, Hfus o Hvap > H fus 11.7 Heating Curve for Water Calculations o T < m p o T = mp o b p > T > mp o T = bp o T > bp Figure The Sublimation of Ice unnumbered figure: photo of ice crystals on frozen peas unnumbered figure: photo of dry ice Figure Temperature during Melting Table 11.9 Heats of Fusion of Several Substances Figure Heat of Fusion and Heat of Vaporization Figure Heating Curve for Water 11.8 Phase Diagrams Major features o s, l, g regions o phase changes o triple point o normal mp, bp o critical point Navigation Comparison of water, iodine, carbon dioxide Figure Phase Diagram for Water Figure Navigation on the Phase Diagram for Water Figure 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 Phoenix Mars Lander Figure Boiling Points of Main Group Hydrides Figure Hydrogen Bonding in Water unnumbered figure: photo of frozen and fresh lettuce Chemistry in the Environment: Water Pollution 164

12 Teaching Tips S uggestions 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 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 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 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 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. 165

13 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Crystalline Solids: Determining Their Structure by X Ray Crystallography Review of constructive and destructive interference X ray diffraction o contact with atomic nuclei o interaction with multiple crystal layers: constructive interference Bragg s law o n = 2d sin unnumbered figure: photo of crystalline mineral unnumbered figure: photo of snowflake unnumbered figure: illustration of wave interference Figure Diffraction from a Crystal Figure X Ray Diffraction Analysis unnumbered figure: diffraction pattern of DNA Example 11.6 Using Bragg s Law Crystalline Solids: Unit Cells and Basic Structures Solids and unit cells o cubic structures equal edge lengths 90o angles Types of cubic lattice structures o simple o body centered o face centered Lattice properties o atoms per unit cell o unit cell composition o edge lengths o density and unit cell structure Closest packed structures o hexagonal o cubic o unit cells and layer composition unnumbered figure: illustration of simple cubic unit cell Figure The Cubic Crystalline Lattices Figure Simple Cubic Crystal Structure Figure Body Centered Cubic Crystal Structure unnumbered figure: illustration of bodycentered cubic unit cell Figure Face Centered Cubic Crystal Structure unnumbered figure: illustration of facecentered cubic unit cell Example 11.7 Relating Density to Crystal Structure unnumbered figures: illustrations of closest packed structures Figure Hexagonal Closest Packing Crystal Structure unnumbered figures: cubic closest packing structures Figure Cubic Closest Packing Crystal Structure 166

14 Teaching Tips S uggestions and Examples Misconceptions and Pitfalls C rystalline Solids: Determining Their Structure by X Ray Crystallography Layers of atoms or molecules in a crystalline solid diffract X rays in manners that cause constructive as well as destructive interference. The diffraction patterns can be interpreted to yield information about atom distances, radii, and bond lengths. Ultimately, the structures of molecules, including large, complex molecules such as proteins, can be determined C rystalline Solids: Unit Cells and Basic Structures Use models that can be moved and rotated to optimize students understanding of the cubic crystalline types. Collections of crystal structure models are viewable in CHIME and RASMOL. Cubic lattice structures have a common edge length, enabling calculations involving density, edge length, and mass. Closest packing models are well illustrated. In particular, the layers and how they stack are shown in detail for the hexagonal and cubic closest packed structures. Models are likely to be beneficial; for a large class, they can be placed below a document camera. Visualizing unit cells and the atoms per unit cell requires several perspectives. Closest packed geometries are determined by how one layer of atoms is oriented with respect to the layer above and below. 167

15 Lecture Outline T erms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Crystalline Solids: The Fundamental Types Solid types and properties o molecular o ionic CsCl NaCl ZnS CaF 2 o atomic nonbonding metallic network covalent Figure Types of Crystalline Solids Figure Cesium Chloride Unit Cell Figure Sodium Chloride Unit Cell Figure Zinc Sulfide (Zinc Blende) Unit Cell unnumbered figure: illustration of a tetrahedral hole Figure Calcium Fluoride Unit Cell Figure The Electron Sea Model Figure Closest Packed Crystal Structures in Metals Figure Network Covalent Atomic Solids Figure The Structure of Quartz Crystalline Solids: Band Theory Band theory Energy levels of lithium molecules Band gap o conductors: no band gap o semiconductors: small band gap o insulators: large band gap Figure Energy Levels of Molecular Orbitals in Lithium Molecules Figure Band Gap 168

16 Teaching Tips S uggestions and Examples Misconceptions and Pitfalls Crystalline Solids: The Fundamental Types The taxonomy of crystalline solid types includes many that are new to most students. Crystalline solids can be composed of molecules, ions, or atoms. Atomic solids can be nonbonding, metallic, or network covalent. Ionic solids and silicates are most commonly encountered previously by students Crystalline Solids: Band Theory The organization of MO energy levels is shown for lithium molecules. The behavior of atoms in a solid can be organized by the relationship between occupied and unoccupied orbitals. The band gap classifies the material as a conductor, semiconductor, or insulator. 169

17 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) CHF 3 b) OCS c) H 2 S 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. 170

18 Additional Problem for Using the Heat of Vaporization in Calculations (Example 11.3) Sort You are given a certain mass of water asked the energy necessary to vaporize it. 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 H vap as a conversion factor to obtain kj of energy. Calculate the energy in kj necessary to vaporize 100 g of water at its boiling point. Given 100 g water Find kj Conceptual Plan g mol kj 1 mol H O 40.7 kj 1 mol H O g H2O 2 Relationships Used H vap = 40.7 kj/mol (at 100 o C) 1 mol H 2 O = g H 2 O Solve Follow the conceptual plan to solve the problem. Check Solution 100 g H2O 1 mol H2O 40.7 kj = 226 kj g H2O 1 mol H2O The units of the answer are correct and the magnitude makes sense. The 100 g sample is slightly more than 5 moles. 171

19 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 o C and a heat of vaporization (H vap ) of 38.8 kj/mol. What is the vapor pressure of octane at 20 o C? Sort Given T 1 = o C 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. Strategize Use the Clausius-Clapeyron equation to find P 2. P 1 = 760 mmhg H vap = 38.8 kj/mol T 2 = 20 o C Find P 2 Conceptual Plan P Hvap 1 1 P1 R T1 T2 2 ln = Relationships Used Clausius-Clapeyron equation (two point form) T (K) = o C Solve Convert T 1 and T 2 from o C to K. Solution T 1 (K) = = K Substitute into the equation and solve for P 2. T 2 (K) P = = K 2 vap ln = P1 R T1 T2 P 760 torr 2 ln = H 1 1 J mol 1 J K mol K K = 4.21 P = 760 torr ( e ) = 760 torr (0.0148) = 11.3 torr 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. 172

20 Additional Problem for Relating Density to Crystal Structure (Example 11.7) Sort You are given the radius of an iron atom and its crystal structure. You are asked to find the density of solid iron. Strategize The conceptual plan is based on the definition of density. For a body-centered cubic unit cell, the atoms touch along the center diagonal. The cell length can be derived in terms of the atom radius. Solve First, find the length of the unit cell using the equations relating the length to the radius of an atom in the body-centered cell. Second, convert the length to a volume by cubing the value. Find the mass of the iron atoms in the cell. The body- centered cell contains two iron atoms. Use the molar mass and Avogadro s number to find the mass of each atom. The density is the mass divided by the volume. Compute the value. Check A form of iron crystallizes with a body-centered unit cell. The radius of iron is 124 pm. Calculate the density of this form of iron in g/cm 3. Given r = 124 pm, body-centered cubic Find d Conceptual Plan l = 4r / 3 V = l 3 [volume of a cube with a length of l] m = mass of unit cell = number of atoms in unit cell x mass of each atom d = m / V Solution l = 4r / 3 = x 10 8 cm / 1.73 = cm V = l 3 = cm g Fe m = 2 atoms Fe 1 mol Fe 22 = g m d = V g = cm 3 3 = 7.84 g/cm 1 mol Fe atoms Fe The units (g) are correct. The magnitude of the answer (7.84) seems to make physical sense. 173