2. As gas P increases and/or T is lowered, intermolecular forces become significant, and deviations from ideal gas laws occur (van der Waal equation).

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1 A. Introduction. (Section 11.1) CHAPTER 11: STATES OF MATTER, LIQUIDS AND SOLIDS 1. Gases are easily treated mathematically because molecules behave independently. 2. As gas P increases and/or T is lowered, intermolecular forces become significant, and deviations from ideal gas laws occur (van der Waal equation). (P + n 2 a/v 2 )(V nb) = nrt 3. Eventually P can get sufficiently high or T low enough that intermolecular forces dominate and gas condenses to liquid. 4. Condensed phases (liquid and solid) have little empty space. 5. Random, chaotic motion of molecules remains in liquid phase. Molecules slide past each other and diffuse as in gases, but more slowly. 6. Liquids diffuse slowly into other liquids with which they are miscible. (example: food coloring and water) 7. Some liquid mixtures separate out due to immiscibility. (example: oil and water) 8. Additional cooling lowers KE of molecules until eventually they have insufficient energy to slide past one another. Solid forms. (crystallization) 9. Solids have ordered arrangement of particles with very restricted motion. Have definite shapes. A. Phase Transitions. (Figure 11.2) Part One: Changes of State (Sections ) Chapter 11 Page 1

2 1. Vaporization - escape of molecules from liquid surface into gas phase. a. Molecules of liquid have distribution of kinetic energies (KE). Figure 11.5 b. As T fraction with sufficient KE evaporation rate. c. Since only high energy molecules are escaping, the molecules left behind are the cooler ones. d. Your body is cooled by evaporation of perspiration. Chapter 11 Page 2

3 2. Vapor Pressure = (vp) pressure of gas molecules above surface of the liquid when equilibrium exists between gas and liquid. a. As T vapor pressure of liquid. b. Easily vaporized liquids are called volatile liquids, have high vapor pressures. Figure 11.7 c. Clausius-Clapeyron Equation. (Calculating vapor pressure vs T) ln P 2 = ΔH vap P 1 R 1 1 T 1 T 2 3. Condensation = reverse of evaporation, caused by gas molecules striking liquid surface (or other surface) and being captured there by cohesive forces. 4. Boiling Point. a. A liquid boils when its vapor pressure equals surrounding pressure. b. Boiling point = T at which vp = surrounding pressure. Chapter 11 Page 3

4 c. Normal boiling point = T at which vp = 1 atm pressure = boiling point at 1 atm. d. Two ways to boil a liquid: i) raise T until its vp = surrounding pressure ii) lower the surrounding pressure until vp = surrounding pressure 5. Molar heat of vaporization (ΔH vap ) = amount of heat needed to vaporize 1 mole of liquid at its boiling point. ΔH vap (H 2 O) = 40.7 kj/mol = 2260 J/gram 6. Specific heat capacity (C s ) = amount of heat needed to raise T of gram of substance 1 C. C s (for H 2 O liquid) = 1.0 calorie/g C (specific heat) 7. Molar heat capacity (C m ) = amount of heat needed to raise T of mole of substance 1 C. For water C m = 18.0 calorie/mol C = 75.3 J/mol C 1 cal = J 8. Freezing point (or melting point of the solid) = T at which solid and liquid phases coexist in equilibrium. liquid freezing solid melting 0 C for H 2 O at 1 atm 9. Molar heat of fusion (ΔH fus ) = amount of heat required to melt 1 mole of solid at its melting point. ΔH fus (H 2O) = 6.02 kj/mol C s (solid H 2 O) = 2.09 J/g C Chapter 11 Page 4

5 10. Summary: Heating Curve for 1 gram of H 2 O (from -50 C to 150 C). a. heating ice from -50 C to 0 C: heat required = m C s ΔT = 1.0g x 2.09 J/g C x 50 C = J b. melting ice: heat required = n ΔH fus = (1.0g/18.0g) x 6.02 kj/mol = kj = 334 J c. heating liquid water from 0 to 100 : heat required = m C s ΔT = 1.0g x J/g C x 100 C = J d. boiling liquid to vapor: heat required = n ΔH fus = (1.0g/18.0g) x 40.7 kj/mol = 2.26 kj = 2260 J e. heating water vapor (steam) from 100 to 150 : heat required = m C s ΔT = 1.0g x 2.03 J/g C x 50 C = J TOTAL = 3,218 J = 769 calories Chapter 11 Page 5

6 11. Sublimation = process in which solids vaporize without first becoming a liquid. a. Examples: dry ice CO 2 (s) CO 2 (g) I 2 (s) I 2 (g) b. Opposite process is called deposition: gas solid B. Phase Diagrams. (Section 11.3) 1. Plot of P versus T (or other variables) showing conditions under which a system can exist in solid, liquid or gas phases. 2. Here, we show a pure substance in a closed vessel. 3. H 2 O P versus T diagram: Figure a. Line A-C is P, T combinations in which liquid and gas coexist in equilibrium. b. Line A-B is solid-liquid coexistence line. c. Line D-A is solid-gas coexistence line. d. Point A = triple point = coexistence of three phases. e. At pressures below the triple point, the liquid cannot exist. f. Point C = critical point. Chapter 11 Page 6

7 g. At temperatures above critical temperature, a gas cannot be made to liquefy. Figure supercritical CO 2 h. Critical pressure is the pressure exerted on a gas to get it to liquefy if T = critical temperature. i. Note that ice near the melting temperature can be melted by applying pressure. (This is because ice is less dense than liquid water. Applying pressure causes a stress which can be relieved by melting into a more compact liquid.) H 2 O is the only substance like this. Ice floats. All other solid substances sink in their own liquids! 4. CO 2 diagram: note the different slope Figure Chapter 11 Page 7

8 A. The Liquid State - Properties: Part Two: The Liquid State (Section ) 1. Viscosity - resistance to the flow of a liquid. a. Oil has high viscosity, gasoline low. b. Viscosity low when molecules can easily slide past one another (weak intermolecular forces). c. Intermolecular force strength viscosity. d. Measured by viscometer, which measures flow time through a capillary tube required for a specific volume of liquid. 2. Surface tension - forces that must be overcome to increase the surface area of a liquid. a. Due to imbalance of forces on molecules at liquid surface. 3. Capillary action: Figure A molecular-level view of the attractive forces experience by molecules at and below the surface of a liquid. a. Cohesive forces = all forces holding liquid together. b. Adhesive forces = forces attracting a liquid to another surface. c. Capillary action occurs when a liquid creeps up or down a capillary tube (caused by adhesion between liquid and the glass) until gravity forces eventually stop it. Chapter 11 Page 8

9 Figure B. Intermolecular Forces. (Section 11.5) 1. Responsible for formation of solids and liquids. (phase changes) 2. Weaker than covalent bonds. strong covalent force to complete octet weak interactions still remain even after octet rule is satisfied 3. Strength of these weak forces vary from species to species. Thus, boiling temperature (condensation temp.), melting temp., etc., are different for different substances. 4. Types of intermolecular forces: a. Dipole-Dipole Interactions: Two molecules with permanent dipole moment interact. -these are short-ranged forces compared to simple ion-ion force Chapter 11 Page 9

10 b. Hydrogen Bonding: A special case of very strong dipole-dipole interactions. In general: X = F, O, or N (highest Electronegativities) Most important examples: water alcohols (R = hydrocarbon group) ammonia C is not highly EN. methane (NOT!) Chapter 11 Page 10

11 Accounts for high boiling points of hydrogen bonded liquids. (called associated liquids ) Figure Importance in biology: Figure c. London Dispersion Forces: Very weak attractive forces and very short-ranged distance Accounts for liquification of substances with no dipole moments at very low T: He, H2, N2, O2, CO2, etc. Origin: Ne Ne interaction as example. Chapter 11 Page 11

12 Figure instantaneous, temporary dipole-dipole interaction creates weak attraction -strength depends on polarizability ( distortability ) of the electron cloud around a species Polarizability as number of electrons as MolarMass A. Types of Solids. (Section 11.6) Part Three: The Solid State (Section ) Chapter 11 Page 12

13 1. Four categories based on bonding involved. metallic solids ionic solids molecular solids covalent (network) solids 2. Metallic Solids: a. Exist as positive metal ions embedded in a sea of delocalized electrons. b. Nearly all metals crystallize as: bcc (body-centered cubic) fcc (face-centered cubic - also called cubic close-packed) hcp (hexagonal close-packed) (fcc and hexagonal called close-packed because atoms are packed as tight and efficiently as possible - atoms occupy the highest percent of space available.) c. fcc (cubic close-packed) versus hexagonal close-packed. (See Figures and 11.39) imagine stacking layers of marbles: Either way, coordination # = % of volume is filled with atoms, only 26% is empty space. Chapter 11 Page 13

14 Figure Ionic Solids. a. Examples are ionic compounds (i.e., salts): NaCl, ZnS. b. Cations and anions occupy the sites of the unit cell. 4. Molecular Solids. Figure a. Sites of unit cell are occupied by neutral molecules. b. Examples: H 2 O (ice), CO 2, NH 3, etc. c. Held together by Hydrogen bonds in case of H 2 O and NH 3 ; by very weak London forces in case of CO 2, O 2, CH 4, and noble gas solids. 5. Covalent (Network) Solids: a. Atoms held together by covalent bonds. Chapter 11 Page 14

15 b. Very strong binding. c. Examples: diamond, quartz, graphite. Figure C. Summary of Properties: Structure of Crystalline Solids. (Section 11.7) 1. Chapter 11 Solids come as either crystalline or amorphous. The latter has a disordered structure (glass is an example). Page 15

16 2. Crystals contain orderly repeating arrays of atoms, molecules, or ions, called a crystal lattice. 3. The smallest repeat unit is called a unit cell. 4. See Figure for the 7 different crystal system. We will focus on the cubic crystal system for time s sake. 5. Different unit cells in the cubic system: simple cubic body-centered cubic (bcc) face-centered cubic (fcc); also called cubic close packed Figure In simple cubic: a. Every site is at corner. Include Figure b. Each site is shared by 8 other unit cells, so only 1/8 of that atom belongs to a given unit cell: 8 sites/cell x (1/8 occupation) = 1 atom per unit cell Chapter 11 Page 16

17 7. In bcc: a. Eight corner sites and one center site. b. 8 corner sites x 1/8 occup = 1 atom per unit cell from corner sites 1 central site x 1/1 occup = 1 atom per unit cell from center site = 2 atoms per unit cell c. Examples: Fe and Cr. 8. In fcc: a. Eight corner sites and 6 face-centered sites. b. 8 corner sites x 1/8 occup = 1 atom per unit cell from corner sites 6 face sites x 1/2 occup = 3 atom per unit cell from faces = 4 atoms per unit cell c. Examples: Ca and Ag metal. 9. Counting Nearest Neighbors. (Coordination Number) a. Simple cubic - each atom has 6 nearest neighbors. (inefficient packing) b. bcc - each atom has 8 nearest neighbors. c. fcc - each atom has 12 nearest neighbors. (packing at maximum density) 10. Calculations involving Unit Cells a. Problem: Titanium crystallizes in a body centered cubic unit cell with an edge length of Å. The density is g/ml. Use these data to calculate Avogadro s number. Chapter 11 Page 17

18 b. Problem: LiBr is in a face-centered cubic arrangement with a unit cell length of Å. Because Li + is so small, the Br - ions are in contact with each other through the face centered Br - site. Determine the ionic radius of the Br - ion. Chapter 11 Page 18

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