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CHAPTER 11: STATES OF MATTER, LIQUIDS AND SOLIDS A. Introduction. (Section 11.1)

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 + n2a/V2)(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.

Part One: Changes of State (Sections 11.2-3) A. Phase Transitions. (Figure 11.2)

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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.

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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 P2P1

=ΔHvap

R1T1−1T2

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.

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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 (ΔHvap) = amount of heat needed to vaporize 1 mole of

liquid at its boiling point. ΔHvap (H2O) = 40.7 kJ/mol = 2260 J/gram

6. Specific heat capacity (Cs) = amount of heat needed to raise T of gram of substance

1°C. Cs (for H2O liquid) = 1.0 calorie/g °C (specific heat) 7. Molar heat capacity (Cm) = amount of heat needed to raise T of mole of substance

1°C. For water Cm = 18.0 calorie/mol °C = 75.3 J/mol °C 1 cal = 4.184 J 8. Freezing point (or melting point of the solid) = T at which solid and liquid phases co-

exist in equilibrium.

liquid melting

freezing⇔ solid 0°C for H2O at 1 atm

9. Molar heat of fusion (ΔHfus) = amount of heat required to melt 1 mole of solid at its

melting point. ΔHfus (H2O) = 6.02 kJ/mol Cs (solid H2O) = 2.09 J/g °C

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10. Summary: Heating Curve for 1 gram of H2O (from -50°C to 150°C).

a. heating ice from -50°C to 0°C:

heat required = m Cs ΔT = 1.0g x 2.09 J/g °C x 50°C = 104.5 J

b. melting ice:

heat required = n ΔHfus = (1.0g/18.0g) x 6.02 kJ/mol = 0.334 kJ = 334 J

c. heating liquid water from 0° to 100°:

heat required = m Cs ΔT = 1.0g x 4.184 J/g °C x 100°C = 418.4 J

d. boiling liquid to vapor:

heat required = n ΔHfus = (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 Cs ΔT = 1.0g x 2.03 J/g °C x 50°C = 101.5 J TOTAL = 3,218 J = 769 calories

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11. Sublimation = process in which solids vaporize without first becoming a liquid. a. Examples: dry ice CO2 (s) → CO2 (g) I2 (s) → I2 (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. H2O P versus T diagram:

Figure 11.11

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.

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g. At temperatures above critical temperature, a gas cannot be made to liquefy.

Figure 11.13 - supercritical CO2

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.) H2O is the only substance like this. Ice floats. All other solid substances sink in

their own liquids!

4. CO2 diagram: note the different slope

Figure 11.12

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Part Two: The Liquid State (Section 11.4-5) A. The Liquid State - Properties:

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.

Figure 11.16 - A molecular-level view of the attractive forces experience by molecules at and below the surface of a liquid.

3. Capillary action:

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.

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Figure 11.19

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

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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

methane (NOT!) C is not highly EN.

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Accounts for high boiling points of hydrogen bonded liquids. (called “associated liquids”)

Figure 11.24

Importance in biology:

Figure 11.26

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.

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Figure 11.22

-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 ↑

Part Three: The Solid State (Section 11.6-10)

A. Types of Solids. (Section 11.6)

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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 11.38 and

11.39) imagine stacking layers of marbles:

Either way, coordination # = 12. 74% of volume is filled with atoms, only 26% is empty space.

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Figure 11.40

3. Ionic Solids.

a. Examples are ionic compounds (i.e., salts): NaCl, ZnS. b. Cations and anions occupy the sites of the unit cell.

Figure 11.42

4. Molecular Solids.

a. Sites of unit cell are occupied by neutral molecules. b. Examples: H2O (ice), CO2, NH3, etc. c. Held together by Hydrogen bonds in case of H2O and NH3; by very weak London

forces in case of CO2, O2, CH4, and noble gas solids. 5. Covalent (Network) Solids:

a. Atoms held together by covalent bonds.

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b. Very strong binding. c. Examples: diamond, quartz, graphite.

Figure 11.28

6. Summary of Properties:

C. Structure of Crystalline Solids. (Section 11.7) 1. Solids come as either crystalline or amorphous. The latter has a disordered structure

(glass is an example).

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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 11.32 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 11.33

Include Figure 11.34

6. In simple cubic:

a. Every site is at corner. 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

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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 3.306 Å. The density is 4.401 g/mL. Use these data to calculate Avogadro’s number.

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b. Problem: LiBr is in a face-centered cubic arrangement with a unit cell length of 5.501 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.