Liquids and Solids AP Chemistry Chapter 10
Liquids and Solids Gases are much easier to study because molecules move independent of each other. In liquids and solids forces between molecules become very important and they differ greatly from one substance to another substance.
10.1 Intermolecular Forces Intramolecular forces Intermolecular forces
Types of Intermolecular Forces: 1) London Dispersion Forces
Types of Intermolecular Forces: 2) Dipole-Dipole Forces
Types of Intermolecular Forces: 2) Dipole-Dipole Forces
Types of Intermolecular Forces: Hydrogen Bonding
Applications of Intermolecular Forces: Alcohols have boiling points compared to their alkane derivatives CH 3 OH CH 4 C 2 H 5 OH C 2 H 6
Applications of Intermolecular Forces: If comparing two polar substances, the one with typically has the higher b.p. Ex. HCl vs. HBr vs. HI
Applications of Intermolecular Forces Covalent Hydrides
IMF SUMMARY QUESTIONS: 1. Why is the boiling point of O 2 (-183 C) higher than N 2 (-196 C)? 2. Why is the boiling point of NO (-151 C) higher than both O 2 and N 2 even though it has approximately the same molar mass? 3. What has a higher boiling point Cl 2 or ICl? Why?
IMF SUMMARY QUESTIONS: 4. Which of the following would have hydrogen bonding? Ethyl alcohol dimethyl ether hydrazine, N 2 H 4 Acetic acid acetone
IMF SUMMARY QUESTIONS: 5. What types of intermolecular forces are present in a sample each of the following? a. H 2 b. CCl 4 c. OCS d. NH 3
10.3 An Introduction to Structure and Types of Solids amorphous solids crystalline solids lattice
10.3 An Introduction to Structure and Types of Solids x-ray diffraction used to determine crystalline structure diffraction when beams of light are scattered from a regular array of points in which spacing between the components are comparable with the wavelengths of light (due to constructive and destructive interference) See fig. 10.10,11 pg. 433 xy + yz = nλ xy + yz = 2d sin Ө nλ = 2d sin Ө Bragg Equation
10.3 An Introduction to Structure and Types of Solids xy + yz = nλ xy + yz = 2d sin Ө nλ = 2d sin Ө Bragg Equation
Types of Crystalline Solids: Ionic Compounds Examples - NaCl, Fe(NO 3 ) 3 - Salts - most metal compounds Properties - 1) Melting points <<not as high as network>> - Strong forces between oppositely charges ions. 2) Non-conductor as solid Conductor as liquid or dissolved 3) Often soluble in water
Types of Crystalline Solids: Ionic Compounds Ionic Bond Strength and Lattice Energy varies by Coulomb s Law. (Lattice Energy is H to form solid from gaseous ions) Coulomb s Law: E = k x Q 1 x Q 2 d Q 1 and Q 2 = ion charge d = inter nuclear distance Bond strength and lattice energies increase as ion charge and ion size.
Molecular Substance Properties ( bonded molecules) 1. Nonconductors of electricity when pure. 2. Usually insoluble in water but soluble in nonpolar solvents such as CCl 4 or benzene. 3. Volatile, with appreciable vapor pressures at room temperature. 4. Low melting and boiling points. * The stronger the intermolecular forces, the the boiling point.
Atomic Solids
Network Covalent Substances: Examples - Diamond, Graphite, Quartz, Mica, Asbestos, SiO 2 Properties - 1) melting points often above 1000 C - Covalent bonds must be broken 2) Non-conductors 3) Insoluble in water and all common solvents
Summary X X X Atomic X X X X X
10.2 The Liquid State Properties: low compressibility lack rigidity high density
10.2 The Liquid State Properties: low compressibility lack rigidity high density surface tension capillary action meniscus viscosity
Summary Types of Solids a. CO 2 b. SiO 2 c. Si d. CH 4 e. Ru f. I 2 g. KBr h. H 2 O i. NaOH j. U k. CaCO 3 l. PH 3
Summary Types of Solids
Summary Types of Solids
10.4 Structure and Bonding in Metals Properties 1. durable 2. b.p. ( of melting points) 3. thermal/ electrical conductor 4. malleable 5. ductile 6. Insoluble in water Bonding (see models below) (accounts for physical properties!)
10.4 Structure and Bonding in Metals Structure - See fig. 10.14,15 pg. 437 (next slide) 1) hexagonal Closest packed (hcp) structure Ex. Mg, Zn (Ca) 2) cubic closest packed (ccp) structure Ex. Al, Fe, Cu, Co Ni
10.4 Structure and Bonding in Metals Structure - See fig. 10.13,14,15 pg. 437 1) hexagonal Closest packed (hcp) structure Ex. Mg, Zn (Ca) 74.04% packing efficiency aba arrangement; hexagonal prism 2) cubic closest packed (ccp) structure Ex. Al, Fe, Cu, Co Ni 74.04% packing efficiency abc arrangement; face-centered cubic
10.4 Structure and Bonding in Metals
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 simple cubic (Po) Seldom found in nature Only 52.3% packing efficiency
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 body-centered (alkali metals, U) 68.02% packing efficiency 20% metals have this packing Found for group IA metals and Barium
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 face-centered (Au) 73.04% packing efficiency (ccp) 40% metals have this packing Found in Calcium and Strontium
Unit Cell Types
Unit Cell Types
Counting Atoms in Unit Cells.
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 face-centered (Au) 73.04% packing efficiency (ccp) 40% metals have this packing Found in Calcium and Strontium
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 body-centered (alkali metals, U) 68.02% packing efficiency 20% metals have this packing Found for group IA metals and Barium
10.4 Structure and Bonding in Metals Unit Cells See fig 10.9 pg. 432 simple cubic (Po) Seldom found in nature Only 52.3% packing efficiency
Unit Cell Types Metal Radii S = 2r (4r) 2 = s 2 + s 2 4r = s 2 +s 2 + s 2 4r = s 2 Important to know the # atoms per cell & arrangement (ex. to calc. density) 4r = s 3
Calculations - density 1. Titanium has a body-centered cubic unit cell. The density of titanium is 4.50 g/cm 3. Calculate the edge length of the unit cell and a value for the atomic radius of titanium.
Calculations - density 2. The radius of gold is 144 pm, and the density if 19.32 g/cm 3. Does elemental gold have a face-centered cubic structure or a body-centered cubic structure.
Bonding Models for Metals electron sea model: band model/ molecular orbital (MO) model: electrons are assumed to travel around the metal crystal in molecular orbitals formed from the valence atomic orbitals of the metal atoms
Bonding Models for Metals Li 2 two widely spaced MO Energy levels result In a metal crystal the large # of resulting MO s become more closely spaced and form a virtual continuum of levels called bands See fig. 10.19
Bonding Models for Metals
Alloys Alloy a substance that contains a mixture of elements and has metallic properties Substitutional alloy Ex. Brass (Cu & 1/3 Zn), Sterling Silver (93% Ag & 7% Cu), pewter (85% Sn, 7% Cu, 6% Bi, 2% Sb) Interstitial alloy Ex. Steel (Fe w/c)
Conduction of Carbon and Silicon Fig 10.23 Diamond vs. typical metals MO E s Graphite has closely spaced π molecular orbitals Silicon
Semiconductors n-type A silicon semiconductor can be doped with Arsenic (each having one more valence e- than Si), the extra electrons become available for conduction. p-type p-n junction A silicon semiconductor can be doped with Boron (each having one less valence e- than Si), leaving vacancies/holes where e- would ve been. This leaves unpaired e- which can serve as conduction e-. involved contact of a p-type & an n-type semiconductor b) reverse bias no current flows; the junction resists the imposed current flow c) foward bias current flows; the movement of electrons and holes is in the favored direction.
Semiconductors See fig 10.31 electrons (-) (+) holes Negative terminal (-) (+) Positive terminal Positive terminal (+) (-) Negative terminal p n
10.7 Ionic Solids NOTE: Both and exist in a lattice structure. In all atoms/ions have the same radii. In there are different size ions.key: understand fundamental principles governing their structures (the rest is just details). 3 hole types: 1) trigonal 2) tetrahedral 3) octahedral trigonal < tetrahedral < octahedral
10.7 Ionic Solids
10.7 Ionic Solids * the type of hole used depends on the relative cation and anion sizes. Ex. ZnS 1. What type of packing do the gray atoms in (a) have? 2. How many tetrahedral holes, as shown in (b), are there in this unit cell? 3. How many Zinc ions in the unit cell in (c)? Sulfide ions? 4. If Cl - ions were used instead, how many of the tetrahedral holes would be used?
10.7 Ionic Solids Key: understand fundamental principles governing their structures (the rest is just details). 1. Ionic compounds have electrical neutrality (positive and negative must equal). 2. Typically the larger ions (usually the neg. ions) are packed in hcp or ccp, and the smaller cations fit into holes among the negative ions
10.7 Ionic Solids This unit cell uses octahedral holes in a cubic unit cell. Is this more likely sodium chloride or sodium oxide?
10.7 Ionic Solids 1. The structures of some common crystalline substances are shown below. Show that the net composition of each unit cell corresponds to the correct formula of each substance.
10.7 Ionic Solids 1. The structures of some common crystalline substances are shown below. Show that the net composition of each unit cell corresponds to the correct formula of each substance.
10.7 Ionic Solids
10.7 Ionic Solids * the type of hole used depends on the relative cation and anion sizes. Ex. ZnS NaCl <<see table 10.7 pg. 458>> Key: understand fundamental principles governing their structures (the rest is just details).
10.8 Vapor Pressure and Changes of State equilibrium vapor pressure: the stronger the IMF, the equilibrium vapor pressure the higher the temperature, the equilibrium vapor pressure <<see fig. 10.41 pg. 461 & table 10.8>>
10.8 Vapor Pressure and Changes of State P atm = P vap + P Hg column As long as both liquid and vapor are present, the pressure exerted by the vapor is independent of the volume of the container. This is different than a normal gas why???
10.8 Vapor Pressure and Changes of State PROBLEM: Given 1.00 g of H 2 O, at 35 C P vap = 42 mm Hg a) How much liquid water remains in a 1.00 L flask when equilibrium is established? b) How many grams of H 2 O (l) remain in a 5.0 L flask at equilibrium? c) How large a flask is needed to evaporate all the liquid water?
10.8 Vapor Pressure and Changes of State PROBLEM: Given 1.00 g of H 2 O, at 35 C P vap = 42 mm Hg a) How much liquid water remains in a 1.00 L flask when equilibrium is established?
10.8 Vapor Pressure and Changes of State PROBLEM: Given 1.00 g of H 2 O, at 35 C P vap = 42 mm Hg b) How many grams of H 2 O (l) remain in a 5.0 L flask at equilibrium?
10.8 Vapor Pressure and Changes of State PROBLEM: Given 1.00 g of H 2 O, at 35 C P vap = 42 mm Hg c) How large a flask is needed to evaporate all the liquid water?
10.8 Vapor Pressure and Changes of State The increase in equilibrium vapor pressure with temp is a logarithmic relationship between absolute temperature and the molar heat of vaporization of the liquid (ΔH vap ). ln P vap = -ΔHvap 1 + C R T Clausius-Clapeyron Equation: R is the ideal gas law constant 8.31 J/mol K
10.8 Vapor Pressure and Changes of State PROBLEMS: 1. Water has a H vap = 40.7 kj/mol and at 35.0 C the vapor pressure = 42.2 mm Hg. Find the vapor pressure at 55.0 C.
Heating Curve Boiling point a liquid will boil at a temp. when the vapor pressure of the liquid becomes equal to the pressure above its surface Normal boiling point b.p. when P atm = 1 atm Super cooling Super heating Melting point a substance will melt when the solid and liquid have the same equilibrium vapor pressure (see fig. 10.45 & cases 1-3)
Phase Diagrams - Water Critical temperature the highest temperature a substance can exist as a liquid (above this temperature it will only exist as a gas no matter how much pressure is applied) Critical pressure the pressure needed to condense a substance at its critical temp.
Phase Diagrams Carbon Dioxide Phase diagrams represent closed systems and cannot be used to explain thing occurring in nature.
Phase Diagram for Carbon