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Chapter 12 Intermolecular Forces: Liquids, Solids and Phase Changes
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Intermolecular Forces:
Liquids, Solids and Phase Changes 12.1 Overview of physical states and phase changes 12.2 Quantitative aspects of phase changes 12.3 Types of intermolecular forces 12.4 Properties of the liquid state 12.5 The uniqueness of water
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Types of Molecular Forces
Intramolecular: bonding forces within a molecule; influence chemical properties Intermolecular: forces between molecules; influence physical properties Three states of water: water vapor, liquid water, ice Phase change: liquid water ice Liquid water and ice are examples of condensed phases.
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A Macroscopic Comparison of Gases, Liquids and Solids
Table 12.1 A Macroscopic Comparison of Gases, Liquids and Solids state shape and volume compressibility ability to flow gas conforms to shape and volume of container high high liquid conforms to shape of container; volume limited by surface very low moderate solid maintains its own shape and volume almost none almost none Importance of interplay between potential and kinetic energies
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Types of Phase Changes condensation/vaporization
freezing/melting (or fusion) Enthalpy changes accompany phase changes! condensation and freezing: exothermic processes vaporization and melting: endothermic processes DHofus = heat of fusion (+) DHovap = heat of vaporization (+) DHosubl = heat of sublimation = DHofus + DHovap (Hess’s Law) solid gas gas solid (called deposition)
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Heats of vaporization and fusion for some common substances
It takes more energy to vaporize than to melt! Figure 12.1
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Phase changes and their associated enthalpy changes
Figure 12.2
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Quantitative Aspects of Phase Changes
Within a phase, a change in heat is accompanied by a change in temperature which is associated with a change in the average Ek as the most probable speed of the molecules changes. q = (amount in moles)(molar heat capacity)(DT) During a phase change, a change in heat occurs at a constant temperature, which is associated with a change in Ep, as the average distance between molecules changes. q = (amount in moles)(enthalpy of phase change)
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A cooling curve for the conversion of gaseous water to ice
Five stages, two phase changes Figure 12.3
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A. Liquid-Gas Equilibria
Phase changes are reversible and reach equilibrium. A. Liquid-Gas Equilibria Pressure at equilibrium = vapor pressure At the BP, the rate of evaporation equals the rate of condensation! Figure 12.4
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The effect of temperature on the distribution of molecular speed in a liquid
higher temperature = higher vapor pressure Figure 12.5
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Vapor pressure as a function of temperature and intermolecular forces
Weaker intermolecular forces translate into higher vapor pressure at a given temperature Figure 12.6
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A linear plot of the vapor pressure-temperature relationship
Plotting ln P against 1/T yields a straight line with slope equal to -DHvap/R (the Clausius- Clapeyron equation) Figure 12.7
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Full form of the Clausius-Clapeyron equation:
Two-point version of the equation:
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Utility of the Clausius-Clapeyron Equation
It provides a means to determine experimentally the heat of vaporization, which is the energy required to vaporize 1 mole of molecules in the liquid state. or If DHvap is known, and vapor pressure at one T is known, then vapor pressure at a new T can be calculated.
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SAMPLE PROBLEM 12.1 Using the Clausius-Clapeyron equation PROBLEM: The vapor pressure of ethanol is 115 torr at 34.9 oC. If DHvap of ethanol is 40.5 kJ/mol, calculate the temperature (in oC) when the vapor pressure is 760 torr. PLAN: We are given four of the five variables in the Clausius-Clapeyron equation. Substitute and solve for T2. SOLUTION: T1 = 34.9 oC = K 1 T2 308 K - ln 760 torr 115 torr -40.5 x103 J/mol 8.314 J/mol.K = T2 = 350 K = 77 oC
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Vapor Pressure and Boiling Point
If we assume an open container, then the boiling point (BP) is the temperature at which the vapor pressure equals the external pressure (usually atmospheric pressure, 760 mmHg). Thus, the BP depends on the applied pressure (see Figure 12.6) Water boils at 100 oC at sea level, but at 72 oC on the peak of Mt. Everest!
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B. Liquid-Solid Equilibria
Characterized by a melting point (temperature at which the rate of melting equals the rate of freezing) The MP is not significantly affected by pressure (two condensed phases are involved).
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C. Solid-Gas Equilibria
Iodine subliming I2 vapor in contact with a cold finger at atmospheric pressure C. Solid-Gas Equilibria Why?? External pressure and intermolecular forces maintain the liquid phase after melting. These are too weak in some cases. Figure 12.8
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Bringing It All Together: A Phase Diagram
A graph that describes phase changes of a substance under various combinations of temperature and pressure Key Characteristics Regions (bounded areas) (one phase) Interfaces (lines) between different regions (equilibria between two phases) Isolated Points (critical point, triple point) (unique T/P combinations)
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Phase diagrams for CO2 and H2O
(ice is less dense than liquid water) Figure 12.9
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Video: Phase Diagrams
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Defining/Quantifying Intermolecular Forces
Intramolecular (bonding) forces: strong, involve larger charges closer together Intermolecular forces: weak, involve smaller charges farther apart
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Covalent and van der Waals radii
solid Cl2 Figure 12.10
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Periodic trends in covalent and van der Waals radii (in pm)
blue: covalent radius black: van der Waals radius Figure 12.11
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dipole-induced dipole
Types of intermolecular (van der Waals) forces ion-dipole hydrogen bonding dipole-dipole ion-induced dipole dipole-induced dipole dispersion (London) decreasing strength
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Importance of polarizability!
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Orientation of polar molecules caused by dipole-dipole forces
More orderly in the solid phase than in the liquid phase Figure 12.12
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Dipole moment and boiling point
Higher dipole moment translates into higher BP. Figure 12.13
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Hydrogen bonding _B: H_A N-H O-H H-F General Model
Involves molecules that have an H atom bound at a small, highly electronegative atom with lone electron pairs N-H O-H H-F General Model _B: H_A electronegative atom bearing hydrogen (donor) H-bond electronegative atom with lone electron pair (acceptor)
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SAMPLE PROBLEM 12.2 Drawing hydrogen bonds between molecules of a substance PROBLEM: Which of the following substances exhibits hydrogen bonding? For those that do, draw two molecules of the substance with the H-bonds between them. (c) (a) (b) PLAN: Find molecules in which hydrogen is bonded to N, O or F. Draw H-bonds in the format, B: HA. SOLUTION: (a) C2H6 has no H-bonding sites (a non-polar molecule). (b) (c) Note: more than one H-bond per molecule is possible!
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Hydrogen bonding and boiling point ability to form H-bonds.
H2O, HF and NH3 exhibit aberrant behavior due to their ability to form H-bonds. binary hydrides of Groups 4-7 Figure 12.14
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Covalent and hydrogen bonding in the helical structure of DNA
A single H-bond is relatively weak, but the existence of many such bonds in a molecule can influence molecular structure significantly. strength in numbers! Figure 12.15
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Polarizability The ease with which a particle’s electron cloud can be distorted Pertinent to charge-induced dipole forces (ion-induced dipole and dipole-induced dipole) Increases down a group of atoms or ions (size) Decreases from left to right in a period (effective nuclear charge) Cations are less polarizable than their parent atoms; anions are more polarizable than their parent atoms.
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Dispersion forces among non-polar molecules
separated Cl2 molecules instantaneous dipoles Caused by momentary oscillations of electron charge More electrons, larger molecule, greater mass, greater dispersion forces Figure 12.16
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Molar mass and boiling point
Figure 12.17
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Molecular shape and boiling point
Figure 12.18
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SAMPLE PROBLEM 12.3 Predicting the Type and Relative Strength of Intermolecular Forces PROBLEM: For each pair of substances, identify the dominant intermolecular forces in each substance, and select the substance with the higher boiling point. (a) MgCl2 or PCl3 (b) CH3NH2 or CH3F (c) CH3OH or CH3CH2OH (d) hexane (CH3CH2CH2CH2CH2CH3) or 2,2-dimethylbutane PLAN: Bonding forces are stronger than nonbonding (intermolecular) forces. Hydrogen bonding is a strong dipole-dipole force. Dispersion forces are decisive when the difference is molar mass or molecular shape.
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SAMPLE PROBLEM 12.3 (continued) SOLUTION: (a) Mg2+ and Cl- are held together by ionic bonds (a salt) while PCl3 is covalently bonded and the molecules are held together by dipole-dipole interactions. Ionic attractions are much stronger than dipole interactions and so MgCl2 has the higher boiling point. (b) CH3NH2 and CH3F are both covalent compounds and have polar bonds. The dipole in CH3NH2 can H-bond while that in CH3F cannot. Therefore, CH3NH2 has the stronger interactions and the higher boiling point. (c) Both CH3OH and CH3CH2OH can H-bond but CH3CH2OH has more CH bonds for greater dispersion force interactions. Therefore, CH3CH2OH has the higher boiling point. (d) Hexane and 2,2-dimethylbutane are both non-polar with only dispersion forces to hold the molecules together. Hexane has a larger surface area, and therefore the greater dispersion forces and higher boiling point.
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Properties of the Liquid State
surface tension capillarity viscosity
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The molecular basis of surface tension
The energy required to increase surface area by a unit amount Surface molecules experience a net attraction downward. Stronger intermolecular forces translate into greater surface tension. Figure 12.19
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surface tension (J/m2) at 20 oC
Table 12.3 Surface Tension and Forces Between Particles surface tension (J/m2) at 20 oC substance formula major force(s) diethyl ether CH3CH2OCH2CH3 1.7 x 10-2 dipole-dipole; dispersion ethanol CH3CH2OH 2.3 x 10-2 H-bonding butanol CH3CH2CH2CH2OH 2.5 x 10-2 H-bonding; dispersion water H2O 7.3 x 10-2 H-bonding mercury Hg 48 x 10-2 metallic bonding
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Capillarity: the rising of a liquid through a narrow
space against the pull of gravity
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Shape of a water or mercury meniscus in glass
stronger cohesive forces strongeradhesive forces water-glass forces > water-water forces Hg-Hg forces > Hg-glass forces Figure 12.20
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Viscosity: a liquid’s resistance to flow
Affected by temperature (viscosity decreases at higher T) Affected by molecular shape (longer molecules exhibit higher viscosity)
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Table 12.4 Viscosity of Water at Several Temperatures
viscosity (N.s/m2)* temperature (oC) 20 1.00 x 10-3 40 0.65 x 10-3 60 0.47 x 10-3 80 0.35 x 10-3 *The units of viscosity are newton-seconds per square meter.
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Water solvent power high specific heat capacity
H-bonding ability solvent power high specific heat capacity high heat of vaporization high surface tension high capillarity density of liquid water vs ice
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The H-bonding ability of the water molecule
Four H-bonds per molecule in the solid state; fewer in the liquid state acceptor donor Figure 12.21
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The hexagonal structure of ice
Figure 12.22
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End of Assigned Material
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The macroscopic properties of water and their atomic and molecular “roots”
Figure 12.24
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The crystal lattice and the unit cell
Figure 12.26
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The three cubic unit cells
Figure 12.27 simple cubic 1/8 atom at 8 corners Atoms/unit cell = 1/8 x 8 = 1 coordination number = 6
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The three cubic unit cells
body-centered cubic coordination number = 8 1/8 atom at 8 corners 1 atom at center Atoms/unit cell = (1/8 x 8) + 1 = 2 Figure 12.27
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The three cubic unit cells
face-centered cubic coordination number = 12 1/8 atom at 8 corners 1/2 atom at 6 faces Atoms/unit cell = (1/8 x 8) + (1/2 x 6) = 4 Figure 12.27
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Packing of spheres simple cubic 52% packing efficiency
body-centered cubic 68% packing efficiency Figure 12.28
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hexagonal closest packing cubic closest packing layer a layer c
Figure 12.26 layer a layer b hexagonal closest packing cubic closest packing layer a layer c closest packing of first and second layers abab… (74%) hexagonal unit cell abcabc… (74%) expanded side views face-centered unit cell
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radius of a Ba atom SAMPLE PROBLEM 12.4
Determining atomic radius from crystal structure PROBLEM: Barium is the largest non-radioactive alkaline earth metal. It has a body-centered cubic unit cell and a density of 3.62 g/cm3. What is the atomic radius of barium? (volume of a sphere: V = 4/3pr3) PLAN: Use the density and molar mass to find the volume of 1 mol of Ba. Since 68% (for a body-centered cubic) of the unit cell contains atomic material, dividing by Avogadro’s number will give the volume of one atom of Ba. Using the volume of a sphere, the radius can be calculated. radius of a Ba atom density of Ba (g/cm3) V = 4/3pr3 reciprocal divided by M volume of 1 mol Ba metal volume of 1 Ba atom multiply by packing efficiency volume of 1 mol Ba atoms divide by Avogadro’s number
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SAMPLE PROBLEM 12.4 (continued) SOLUTION: 1 cm3 3.62 g x 137.3 g Ba mol Ba volume of Ba metal = = 37.9 cm3/mol Ba 37.9 cm3/mol Ba x 0.68 = 26 cm3/mol Ba atoms 26 cm3 mol Ba atoms x mol Ba atoms 6.022 x 1023 atoms = 4.3 x cm3/atom r3 = 3V/4p = 2.2 x 10-8 cm
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Cubic closest packing of frozen methane
Cubic closest packing for frozen argon Figure 12.29 Figure 12.30
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The sodium chloride structure
Figure 12.31
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The zinc blende structure
Figure 12.32
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The fluorite (CaF2) structure
Figure 12.33
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Crystal structures of metals
hexagonal closest packing cubic closest packing Figure 12.34
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The random coil shape of a polymer chain
Figure 12.47
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The semi-crystallinity of a polymer chain
Figure 12.48
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The viscosity of a polymer in solution
Figure 12.49
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tip of an atomic force microscope (AFM)
Manipulating atoms Figure 12.50 tip of an atomic force microscope (AFM)
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Manipulating atoms Figure 12.50 nanotube gear
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Tools of the Laboratory
Figure B12.1 Diffraction of x-rays by crystal planes
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Formation of an x-ray diffraction pattern of the protein hemoglobin
Figure B12.2 Tools of the Laboratory Formation of an x-ray diffraction pattern of the protein hemoglobin
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Tools of the Laboratory
Figure B12.3 Scanning tunneling micrographs gallium arsenide semiconductor metallic gold
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