CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 12 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 11 INTERMOLECULAR FORCES AND THE SOLID STATE
Intra- and Intermolecular Forces
Covalent Radius and van der Waals Radius ½ the distance of bonded ½ the distance of non-bonded
Covalent and van der Waals Radii, cont d. Increase down a group Decrease across a period
Dipole Dipole Interactions
Cl 2 Dispersion Forces: (A) Isolated (B) Transient Instantaneous Dipoles (C) Bulk Sample View
Dispersion Forces vs. Size (Atomic Mass)
Dispersion Forces vs. Surface Area
The temperature at which a phase change begins ( is constant throughout the transition) is called the melting point(or freezing point) for solids (or liquids) The temperature is the boiling point (or condensation point) for liquids (and gases)
Energy changes in State Melting and freezing occur with energy change Energy change depends upon type of IM attractions: Ionic substances have highest melting points Low molecular weight covalent is lowest Water is high because of additional H- bonds
Heat of Fusion Energy is needed to melt substances: this is an endothermic process. Energy is released upon freezing: this is an exothermic process The amount of heat transferred (cal or J) per unit mass (grams) is the heat of fusion for a substance.
Temperature Change in Phase Transitions Solids gain heat energy that increases the temperature of the substance until a phase change begins After the phase change begins, the temp remains constant Upon complete phase change, the temp begins increasing again.
Heat of Fusion and of Vaporization
Enthalpy Changes and Changes of State Endothermic --- Exothermic
A(s) + H sub A(g) : 1) A(s) + H fus A (l) 2) A (l) + H vap A (g) Since the sum of the energy must be the same For both pathways: H sub = H fus + H vap
a) q=mc(t 2 -t 1 ) b) q=h f (mass) c) q= mc(t 2 -t 1 ) d) q=h v (mass) e) q = mc(t 2 -t 1 ) Sum is total heat needed to vaporize solid
From the data below, calculate the total heat (in J) needed to convert 12.00 g of ice at -5.00 o C to liquid water at 5.00 o C. mp at 1 atm = 0.0 o C H o fus = 6.02 kj/mol c liquid = 4.21 J/g.o C c solid = 2.09 J/g.o C q mct : q nh
q t = q s + H fus + q liq =(12.00 g ) ( 2.09J/g o C)(5.00 o C) + (12.00 g )(1 mol/18 g )(6.02 x 10 3 J/mol) + (12.00 g )(4.21 J/g o C)(5.00 o C) = 125 J + 4.01 x 10 3 J + 25.3 J = 4.16 x 10 3 J
Equilibrium Vapor Pressure Condensation: change from gaseous to liquid state Vaporization: change from liquid to gaseous state Equilibrium : Rate of vaporization = rate of condensation Sublimation: solid to gas phase change
Dynamic Equilibrium
Fraction as gas is greater at higher temp
Heat of Vaporization The amount of heat (cal, J) required per gram to vaporize a liquid is the heat of vaporization ( or the heat removed to liquify a gas) The SPECIFIC HEAT is the heat required to change the temp by 1 degree of 1 gram of substance
Normal Boiling Point: Vapor Pressure = atmospheric pressure (760 Torr) (1 atm) (101 kpa)
Vapor Pressure Temperature Relationship
A(s) + H sub A(g) : 1) A(s) + H fus A (l) 2) A (l) + H vap A (g) Since the sum of the energy must be the same For both pathways: H sub = H fus + H vap
The extent that sublimation occurs at a Temperature is measured as the equilibrium Vapor pressure over the solid: P eq Peq e Ce H ( sub ) RT H ( sub ) RT RT1 P Pe e eq Pe 1 1 H H ( sub ) RT 1 1 R T T1 H sub let C Pe 1 H RT 1
Clausius-Clapeyron Equation Taking the natural log of both sides: ln P -H 1 vap R T (y m x b) C At two sets of conditions: ln P P 2 1 - H R vap 1 T 2 1 T 1
Problem Liquid ammonia boils at -33.4 o C and has a heat of vaporization of 23.5 kj/mol. Calculate its vapor pressure at -50.0 o C. 1 2 vap 1 2 T 1 T 1 R H - P P ln 1 2 vap 1 2 T 1 T 1 R H - - ln P ln P 1 1 2 vap 2 P ln T 1 T 1 R H - P ln 1 T 1 T 1 P2 P 1 2 R Hvap - e 316 Torr P 760 Torr P 2 239.75K 1 223.15K 1 2 mol 8.314J/K -23,500J/mol e
Molecular Polarity
Hydrogen Bonding
In each pair, identify all the intermolecular forces present for each substance, and select the substance with the higher boiling point: (a) CH 3 Br or CH 3 F (b) CH 3 CH 2 CH 2 OH or CH 3 CH 2 OCH 3 (c) C 2 H 6 or C 3 H 8
Phase Diagrams
Surface Tension
Capillary Action: concave meniscus and convex meniscus
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Viscosity: Resistance to flow.
Water and H-Bonds [4 Max]
Ice forms tetrahedral centers
Solid water expands with the tetrahedral shape.
Water is more dense as a liquid: As the temp is decreased, the volume of the phase increases
Solid and liquid H 2 O densities
Crystalline Solids Amorphous Solids: Glasses
Unit Cells
Unit Cells and Packing Efficiency
Important Ionic Structures Rock-salt Caesium chloride Flourite Rutile Zinc blende (Cubic ZnS) Wurtzite (hexagonal ZnS)
The Cesium Chloride Structure (based upon a simple cubic lattice) Cl at the lattice points. Cs in cubic holes. Cs coordination # = Cl coordination # = # of Cs in unit cell = # of Cl in unit cell = This structure is adopted by halides of large M +. Ex. CsCl, CsBr, CsI, TlX, NH 4 X
Rock-salt (NaCl) Structure Based upon face-centered cubic lattice. Cl at the lattice points of fcc lattice. Na in holes. Na coordination # = Cl coordination # = # of Na in unit cell = # of Cl in unit cell = This structure is adopted by many MX compounds. Halides M + X - (M + = Li, Na, K, Rb, Ag) M 2+ X 2- (M 2+ = Mg, Ca, Sr; X 2- = O,S,Se)
Sphalerite (Zinc blende ZnS) Structure Based upon face-centered cubic lattice. S at the lattice points of fcc lattice. Zn in holes. Zn coordination # = S coordination # = # of Zn in unit cell = # of S in unit cell = This structure is adopted by halides of large M +. Ex. CsCl, CsBr, CsI, TlX, NH 4 X
Fluorite (CaF 2 ) and Antifluorite (Li 2 O) Structure (based upon FCC packing) Fluorite (CaF 2 ) Antifluorite (Li 2 O) Ca FCC lattice F in holes O FCC lattice Li in holes Ca C.N. = F C.N. = O C.N. = Li C.N. = # Ca = # F = # O = # Li = Adopted by MF 2 (M=Ca, Sr, Ba,Cd, Hg) Adopted by M 2 X (M=Li, Na, K; X=O,S,Se,Te)
Wurtzite Structure (based on HCP packing) Tetrahedral holes. Packing of layers is in an ABABAB fashion. (Ex. CuBr, ZnS, NiAs, CdI and CdS)
Other Structures: Rutile (TiO 2 ) structure Ti coordination # = O coordination # = # of Ti in unit cell = # of O in unit cell =
Why do ions array in different ways? NaCl, CsCl, and CuCl all have different crystal packing. All the cations have +1 charge, all the anions are the same Why? Start by calculating the enthalpy (or energy) for forming a solid ionic compound from a collection of gaseous atoms (NOTE: this is another model) X-ray data tells us that atoms are arranged in an alternating layered fashion with each Na + surrounded by 6 Cl - ions.
(r - / r + ) = 1.37 (r - / r + ) = 2.41 (r - / r + ) = 4.44
How are ionic radii determined? From X-ray crystallography we can measure the distance between two ions, d. d = r + + r - d We can determine radii from comparing numerous co We assume that (r k+ + r Na+ )is constant. r k+ + r Na+ = d KF - d NaF = 0.35Å = d KCl - d NaCl = 0.33Å = d KBr - d NaBr = 0.32Å = d KI - d NaI = 0.30Å
Fig. 12.29 Atomic Solids
Fig. 12.30 Molecular Solids
Fig. 12.31 Ionic Solid
Fig. 12.34 Metallic Solids
Fig. 12.35 Network Solids Crystalline SiO 2 Amorphous
Table 12.6 Carbon Allotropes
Fig. 12.36 Molecular Orbital Band Theory of Solids
Fig. 12.37 Electrical Conductivity in Solids