Chapter 12 INTERMOLECULAR FORCES. Covalent Radius and van der Waals Radius. Intraand. Intermolecular Forces. ½ the distance of non-bonded

Transcription

1 Chapter 2 INTERMOLECULAR FORCES Intraand Intermolecular Forces Covalent Radius and van der Waals Radius ½ the distance of bonded ½ the distance of non-bonded

2 Dipole Dipole Interactions Covalent and van der Waals Radii, cont d. Increase down a group Decrease across a period Cl 2 Dispersion Forces: (A) Isolated (B) Transient Instantaneous Dipoles (C) Bulk Sample View Dispersion Forces vs. Size (Atomic Mass)

3 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.

4 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

5 From the data below, calculate the total heat (in J) needed to convert 2.00 g of ice at o C to liquid water at o C. a) q=mc(t 2 -t ) b) q=h f (mass) c) q= mc(t 2 -t ) d) q=h v (mass) e) q = mc(t 2 -t ) Sum is total heat needed to vaporize solid mp at atm = 0.0 o C H o fus = 6.02 kj/mol c liquid = 4.2 J/g.o C c solid = 2.09 J/g.o C q= mcδt : q= nδh q t = q s + H fus + q liq =(2.00 g ) ( 2.09J/g o C)(5.00 o C) + (2.00 g )( mol/8 g )(6.02 x 0 3 J/mol) + (2.00 g )(4.2 J/g o C)(5.00 o C) = 25 J x 0 3 J J = 4.6 x 0 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

6 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 degree of gram of substance Normal Boiling Point: Vapor Pressure = atmospheric pressure (760 Torr) ( atm) (0 kpa)

7 A(s) + H sub A(g) : Vapor Pressure Temperature Relationship ) 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: ΔH ( sub ) P eq RT Peq e = Ce ΔH ( sub ) RT Peq = Pe e = Pe ΔH ΔH ( sub ) RT RT ΔH sub R T T ΔH RT let C = Pe Clausius-Clapeyron Equation Taking the natural log of both sides: -ΔHvap ln P = C R + T (y = m x + b) At two sets of conditions: P ln P 2 - ΔH = R vap T 2 T

8 Liquid ammonia boils at o C and has a heat of vaporization of 23.5 kj/mol. Calculate its vapor pressure at o C. Problem P2 - ΔH ln = P R - ΔH ln P2- ln P = - ΔH R vap ln P2 = + ln P R T2 T -ΔH vap R T2 T P2 = e + P P 2 = e -23,500 J/mol 8.34 J/K mol P2 = 36 Torr K vap K T vap 2 T T2 T Torr 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

9 Phase Diagrams Prob Prob. 2.2 Surface Tension

10 Capillary Action: concave meniscus and convex meniscus Page 439 Viscosity: Resistance to flow.

11 Ice forms tetrahedral centers Water and H-Bonds [4 Max] 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

12 Solid and liquid H 2 O densities Crystalline Solids Amorphous Solids: Glasses Unit Cells Unit Cells and Packing Efficiency

13 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)

14 Sphalerite (Zinc blende ZnS) Structure Based upon face-centered cubic lattice. Fluorite (CaF 2 ) and Antifluorite (Li 2 O) Structure (based upon FCC packing) S at the lattice points of fcc lattice. Fluorite (CaF 2 ) Antifluorite (Li 2 O) Zn in holes. Zn coordination # = S coordination # = # of Zn in unit cell = # of S in unit cell = Ca FCC lattice F in holes Ca C.N. = F C.N. = # Ca = # F = Adopted by MF 2 (M=Ca, Sr, Ba,Cd, Hg) O FCC lattice Li in holes O C.N. = Li C.N. = # O = # Li = Adopted by M 2 X (M=Li, Na, K; X=O,S,Se,Te) This structure is adopted by halides of large M +. Ex. CsCl, CsBr, CsI, TlX, NH 4 X Wurtzite Structure (based on HCP packing) Other Structures: Rutile (TiO 2 ) structure Ti coordination # = O coordination # = # of Ti in unit cell = # of O in unit cell = Tetrahedral holes. Packing of layers is in an ABABAB fashion. (Ex. CuBr, ZnS, NiAs, CdI and CdS)

15 Why do ions array in different ways? NaCl, CsCl, and CuCl all have different crystal packing. All the cations have + 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.

16 Fig Fig Atomic Solids Molecular Solids Fig. 2.3 Ionic Solid Fig Metallic Solids

17 Fig Network Solids Table 2.6 Carbon Allotropes Crystalline SiO 2 Amorphous Fig Molecular Orbital Band Theory of Solids Fig Electrical Conductivity in Solids