Chem 112 Dr. Kevin Moore
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1 Chem 112 Dr. Kevin Moore
2 Gas Liquid Solid
3 Polar Covalent Bond Partial Separation of Charge Electronegativity: H 2.1 Cl 3.0 H Cl δ + δ - Dipole Moment measure of the net polarity in a molecule Q Q magnitude of charge r distance between charges r
4 Consider H 2 O O H H Dipole Moment is the net of all bond dipoles N H H H
5 Non-polar molecules have no net dipole Even if individual bonds have dipoles H C H H H No Net Dipole!
6 Measured in Debye Coulomb-meters Q electron = 1.60 x Coulomb = -1 For an electron at r=100 pm C m 1 Debye Cm 480. D
7 Compound Dipole Moment (D) NaCl 9.00 CH3Cl 1.87 H2O 1.85 NH CO2 0 CCl4 0
8 Determined Experimentally Use to determine ionic character of bond % ionic Observed Ionic 100%
9 Calculate the % Ionic Character of H-Cl if the bond length is 127 pm and µ=1.03 D. e 127 pm 480. D pm D % Ionic D D 100% %
10 Calculate the % Ionic Character of HF if µ=1.82 D and the bond length is 92 pm. Draw the Lewis Structure and Dipole Moment. e 92 pm 480. D pm D % ionic 182. D 44. D 100% 414. % H-F
11 Draw Dipole Moments along the bond Show 3D structure Draw the Lewis Structure and Dipole Moment for Methanol (CH 3 OH) H H C O H H
12 Solid and Liquid Phase implies a binding force Often called van der Waals forces Electrical in nature Ion-Dipole Interaction between a molecule and a charged ion O H H O H H
13 Strength of interaction Size Charge
14 Weak Force of interaction between dipoles Only relevant when molecules are very close together O S O O S O O S O
15 Orientation of Polar Molecules in a Solid
16 Strength of interaction is a function of dipole moment Substance Molar Mass Dipole Moment (D) Boiling Point (K) CH3CH2CH CH3OCH CH3Cl CH3CN
17 Benzene (µ=0) Boiling Point = 80.1 C Melting Point = 5.5 C H C H C C H H C C C H H
18 1000 Instantaneous asymmetric distribution of charge 800 Causes a distortion in a molecule closeby ALL 600 substances experience LDF Boiling Pt 400 Melting Pt Substance Melting Point (K) Boiling Point (K) 200 F Cl Br I
19
20 Non-polar compounds interact through random induced dipoles Electron Polarizability
21 Generally become stronger as molar mass and surface area increase n-pentane vs. neo-pentane
22 Interaction between a hydrogen bonded to a high EN atom (N, O & F) and unshared electrons (lone pairs) on another high EN atom. N O H H H H H O H H N H H H
23 Stronger interaction than dipole-dipole Through space interaction
24 Decreasing Molar Mass Decreasing Boiling Point
25 Summary Force Strength Effect Example Dispersion (LDF) Very Weak Depends on MW CCl4, C6H6, All Substances Dipole-Dipole Weak Solubility, BP, MP SO2, CH3COCH3 H-bonding Moderate BP, MP, 3D Structure H2O, NH3, Organic Acids, Organic Bases Ion-Dipole Strong Solubility
26 Draw Lewis Structure No Dipole? Yes LDF No Hydrogen Bonding? Yes Dipole-Dipole LDF H-Bonding LDF
27 Hydrocarbons are non-polar C 3 H 8 Symmetric molecules with no lone pairs on the central atom are non-polar Tetrahedral (CCl 4 ) Trigonal Planar (SO 3 ) Linear (CO 2 ) Different atoms connected to central atom is always polar CH 2 F 2 CH 3 COCH 3
28 Viscosity Measure of a liquid s resistance to flow Dependent on intermolecular forces
29 Resistance of a liquid to spreading out Least Surface Area Sphere
30 Capillary Action Interaction between a liquid and a porous solid in a narrow passage
31 Symbol Name Beginning Phase Ending Phase ΔH fus Fusion (Melting) Solid Liquid -ΔH fus Freezing Liquid Solid ΔH vap Vaporization Liquid Gas -ΔH vap Condensation Gas Liquid ΔH sub Sublimation Solid Gas -ΔH sub Deposition Gas Solid
32 Measure of disorder in a system Increases solid --> liquid --> gas 2 nd Law of Thermodynamics A closed system will undergo changes which increase the entropy of the system ΔS
33 Gibbs Free Energy G H T S ΔG<0 spontaneous ΔG>0 non-spontaneous ΔG=0 equilibrium Phase changes are equilibrium conditions
34 Process Enthalpy (ΔH) Entropy (ΔS) Free Energy (ΔG=0) Fusion >0 >0 Melting Point Vaporization >0 >0 Boiling Point Sublimation >0 >0 Situational Freezing <0 <0 Melting Point Condensation <0 <0 Boiling Point Deposition <0 <0 Situational
35 Strong intermolecular forces High Heat of vaporization
36 To break apart substances Requires Energy Increases Disorder Favorable when temp allows ΔS to dominate ΔG To organize substances Releases Energy Decreases Disorder Favorable when temp allows ΔH to dominate ΔG G H T S
37 Calculate the boiling point of water if the ΔH vap =40.67 kj/mol and ΔS vap =109 J/K-mol. Boiling is a phase change T H S J / mol 109 J / K mol 373 K T 100 C
38 Molar Heat Capacity amount of heat required to heat 1 mole of a substance by 1 C Specific Heat Capacity amount of heat required to heat 1 g of a substance by 1 C
39 Heat from -10 C to 110 C Temperature ( C) 100 C ice = 36.5 Jmol -1 C -1 C H2O = 75.3 Jmol -1 C -1 C steam = 33.6 Jmol -1 C -1 ΔH vap = kjmol -1 ΔH fus = 6.01 kjmol Heat Added (kj)
40 #1 Warming the ice q 365. J ( 1 mol)( C) 365 J molc #2 melting the ice kj q ( 1 mol) 6010 J #3 warming the water fus mol q J ( 1 mol)( 100. C) 7530 J molc #4 vaporizing the water kj q ( 1 mol) J vap mol
41 #5 Heating of steam Total Heat J q 336. ( 1 mol)( C) 336 J steam molc q q q q q q tot ice fus water vap steam q 365 J 6010 J 7530 J J 336 J tot q J kj tot
42 Pressure of a liquid on a closed container Depends on Temperature All Liquids and many solids # of molecules Kinetic Energy ΔH vap
43 Passing of a liquid to a vapor without being at the boiling point Liquid cools as evaporation occurs
44 Vapor Pressure of a liquid can be used to derive the relationship Clapeyron Relationship dp dt PH RT vap 2
45 Boiling Point Temperature at which the vapor pressure is equal to atmospheric Pressure Integrating Liquid Vapor Relationship Claussius-Clapeyron Equation ln P vap H vap R 1 T C
46 P(mmHg) vs. T(K) H 2 O Pvap T 1/T Ln P vap
47 Plot ln P vap vs. 1/T 7 ln P vs. 1/T
48 Using equation ln ln ln P P vap 1 P H R vap 1 T Hvap Hvap ln P R 1 T R T2 Hvap 1 Hvap 1 ln P R T R T 1 2 ln P P 1 C 2 1 H 1 1 vap R T T 2 2 1
49 Calculate the ΔH vap of ethanol, if the boiling point is 78.4 C and it has a P vap =100. mmhg at 34.7 C. T 1 =78.4 = K T 2 =34.7 = K P 1 =760. mmhg P 2 =100. mmhg ln 760 H vap J mol K H vap J mol K H J kj vap mol 418. mol
50 The normal boiling point of water is C. What is the vapor pressure at 58.5 C if ΔH vap =40.67 kj/mol? ln ln J mol x J mol K ln x ln x x 147 mmhg
51 Hydrogen Bonding Density of Water Specific Heat Density
52 Crystalline Solids Amorphous
53 Crystalline Solids form repeating cells
54 d represents the edge of a unit cell Packing Methods Coordination # Lattice Points Simple Cubic
55 Body Centered Cubic Packing Face-Centered Cubic Packing Hexagonal Closest Packing
56 Closet Packing: Hexagonal and Cubic hexagonal cubic
57 Shared by 8 unit cells Shared by 4 unit cells Shared by 2 unit cells
58 Simple Cubic Atomic radius d r 2r d 2
59 Body-Centered r 3 d 4
60 Face-Centered r d 8
61 Silver crystallizes in a face-centered (FCC) unit cell (cubic closest packing). If the edge length is 407 pm, what is the radius of Ag? r 407 pm pm
62 Nickel has an edge length of pm. If it is FCC, what is its density? d D cm V d ( cm) cm m V FCC 4 atoms 4 atoms mol atoms g 1 mol g D 22 ( g) cm g cm 3
63 Titanium crystallizes in a BCC format. If the radius of a Ti atom is pm, what is its density? 8 4( cm) d 3 D m V BCC 2 2 atoms cm V d ( cm) cm atoms mol atoms g 1 mol g D 22 ( g) cm g cm 3
64 Cannot see atoms X-Rays can be used to determine crystal structure William Bragg Bragg Equation n=integer λ=wavelength Θ=angle of X-ray to crystal Used to determine atomic distances Usually measure in pm (10-12 m) d n 2sin
65 Energy required to break ionic solid into gas ions Solubility Melting Point
66 Intermolecular forces Ionic Crystals Ionic compounds High Lattice Energy
67 NaCl crystallizes in a face centered unit cell with an edge length of 564 pm. What is the density? Cell Contains: 4 Na + & 4 Cl - ions 4 Na 1 mol atoms g 1 mol g 4 Cl 1 mol atoms g 1 mol Total Mass: x g V d ( ) cm g D D g cm 22 g cm 22 3
68 Allotropes Different forms of substance in identical physical state Phosphorus Carbon Silicon
69 Plot of phases for Pressure vs. Temperature SuperCritical Fluid Solid Liquid Gas
70 Triple Point Critical Temperature Critical Pressure Slope of solid/liquid line
71
72
Honors Chemistry Dr. Kevin D. Moore
Honors Chemistry Dr. Kevin D. Moore Key Properties: Solid is less dense than liquid Water reaches maximum density at 4 C Very high specific heat Dissolves many substances Normal Boiling Point: 100 C Normal
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