1.
Molecules are at the lattice points of the crystalline solid
2.
Ice = H
2
O at each lattice point a.
It takes 6 kJ/mol of energy to melt ice (470 kJ/mol to break O—H bond) b.
Weak, intermolecular H-bonds hold ice together
3.
CO
2
= no dipole moment; London forces hold solid together (gas at 25 o C)
4.
P
4
, S
8
= strong covalent bonds within the unit, but weak intermolecular forces between molecules = larger molecules, more London force, solids at higher T
1.
Ions at lattice points of the crystalline solid
2.
Strong (Coulombic) inter-ionic forces: stable, high melting point solids
3.
Packing a.
Larger anions are usually in one of the closest packing forms b.
Smaller cations fit into the holes among the anions to maximize the attraction and minimized repulsions

c.
Types of holes for the smaller ion i.
Trigonal holes = between 3 spheres in the same layer ii. Tetrahedral holes = between 3 spheres in one layer and 1 in another iii. Octahedral holes = between 3 spheres in one layer and 3 in another d.
Hole size: trigonal < tetrahedral < octahedral
Trigonal too small to be occupied in most ionic solids e.
There are more holes than cations needed, usually i.
ZnS Example: twice the number of tetrahedral holes as cations needed ii. Ccp structure: 4 S 2in the unit cell, 8 tetrahedral holes, 4 Zn 2+ needed iii. NaCl Example: ccp Cl-: each octahedral hole surrounded by 6 Cliv. One hole per each Cl-, one Na+ needed for each Cl-

1.
Vapor Pressure = pressure of the vapor over a liquid at equilibrium
2.
Equilibrium = molecules evaporating at the same rate as condensing a.
Dynamic = millions of individual molecules are changing state b.
Net change in gaseous molecules is Zero
3. Vapor Pressure depends on
Kinetic Energy = Temperature

4. Vapor pressure depends on Intermolecular Forces a.
Volatile Liquid = high vapor pressure i.
Small molecular weight ii. Small intermolecular forces: no dipole or H-bonding b.
Non-volatile Liquid = low vapor pressure i.
Large molecular weight (high London Forces): oils ii. Large intermolecular forces (H-bonding): water

ln( P
5.
Quantitative description of Vapor Pressure vap
) a.
Heat of Vaporization =
D
H vap b.
Relationship of
DH vap
, P
= heat added to free molecules to gas vap
, and T (in Kelvin)
 
D
H vap
R
1
T

C
b
c.
Plots as a straight line: y = mx + b

d.
The Clausius-Clapeyron Equation: We can calculate P vap at a new T ln( P vap , T
1
)

D
H
RT
1

C
 ln( P vap , T
2
)

D
H
RT
2 ln( P vap , T
1
)
 ln( P vap , T
2
)

D
H vap
R


1
T
2

1
T
1

 ln
P vap
P vap
, T
2
, T
1

 D
H vap
R


1
T
2

1
T
1

 e.
Example : P vap
(H
2
O) = 23.8 torr and
D
H vap
What is P vap at T = 50 o C?
= 43.9 kJ/mol at T = 25 o C. ln
P vap , T
2
P vap , T
1

 ln


P vap , T
2
23 .
8 torr



D
H vap
8
R
.



1
T
2
3145 J /

1
T
1
43 , 900 J
K
/


 mol mol
1
323 K

1
298 K ln


P vap , T
2
23 .
8 torr



1 .
37

P vap , T
2
23 .
8 torr

3 .
94

P vap , T
2

93 .
7 torr

1.
States of matter can be reversibly changed without changing the substance
2.09 J/g o C
ice
D
q = n
D
H fus
4.18 J/g o C
liq
D
q = n
D
H vap
2.01 J/g o C
gas
D
2.
Changing states a.
Melting (Fusion) = Solid to Liquid (opposite is Freezing) [
D
H fus
] b.
Vaporization = Liquid to Gas (opposite is Condensation) [
D
H vap
] c.
Sublimation = Solid to Gas (opposite is Vapor Deposition)

2.09 J/g o C
ice
D
q = n
D
H fus
4.18 J/g o C
liq
D
q = n
D
H vap
2.01 J/g o C
gas
D

2
o
o
1.
(500g)(2.09J/g o C)(20 o C) = 20,900 J
2.
(500g)(1mol/18g)(6,020J/mol) = 16,700 J
3.
(500g)(4.18J/g o C)(100 o C) = 209,000 J
4.
(500g)(1mol/18g)(40,700J/mol) = 1,130,000 J
5.
(500g)(2.01J/g o C)(150 o C) = 151,000 J
______________________________________________________________
6. Total = 20,900 + 16,700 + 209,000 + 1,130,000 + 151,000 = 1,527,600 J

1.
Phase diagrams show us the temperatures and pressures of state changes a.
Critical Temperature = temp. above which substance is always a gas b.
Critical Pressure = pressure needed to make liquid at Critical Temp c.
Critical Point = intersection of critical temp. and critical pressure d.
Triple Point = single point where all three phases can coexist

2. Navigating in Phase
Diagrams
3. Phase diagrams for other compounds: Iodine
Water is unusual

2. Applications of the Water Phase Diagram a.
Solid/Liquid Boundary has a negative slope: density of ice is less than liquid water (this is why ice floats on water) b.
Ice Skating: narrow skate blade applies large pressure + friction = ice melts to liquid. This lubricates the blade and allows smooth skating.
c.
Elevation Effects: Lower pressure = lower boiling point

3.
The CO
2
Phase Diagram a.
Solid/Liquid boundary has positive slope (normal) b.
Solid is more dense than liquid c.
Triple point = 5.1 atm, -56.6 o C d.
Critical point = 72.8 atm, 31 o C e.
Solid sublimes directly to gas at 1 atm and 25 o C = Dry Ice f.
Fire Extinguisher = liquid CO
2
; high pressure makes it liquid
As it is opened to atmosphere, gaseous CO
2 puts out fire (fog = water)