CHEM 1342

• intermolecular attractions are due to attractive forces between opposite charges
– + ion to - ion
– + end of polar molecule to - end of polar molecule
• H-bonding especially strong
– even nonpolar molecules will have temporary charges
• larger the charge = stronger attraction
• longer the distance = weaker attraction
• however, these attractive forces are small relative to the bonding forces between atoms
– generally smaller charges
– generally over much larger distances

• the stronger the attractions between the atoms or molecules, the more energy it will take to separate them
• boiling a liquid requires we add enough energy to overcome the attractions between the molecules or atoms
• the higher the normal boiling point of the liquid, the stronger the intermolecular attractive forces

+ -
+ -
+ - + -
+
+
+
+
_
_
_
_
+ +
+ + + +
-
+
- -
- -
+
+ + +
+
-
-
-
-
-

• fluctuations in the electron distribution in atoms and molecules result in a temporary dipole
– region with excess electron density has partial (─) charge ( d -
)
– region with depleted electron density has partial (+) charge ( d
+ )
• the attractive forces caused by these temporary dipoles are called dispersion forces
– aka London Forces
• ALL molecules and atoms will have them!!

• the magnitude of the induced dipole depends on several factors
• polarizability of the electrons
– volume of the electron cloud
– larger molar mass = more electrons = larger electron cloud = increased polarizability = stronger attractions
• shape of the molecule
– more surface-to-surface contact = larger induced dipole = stronger attraction

Therefore the strength of the dispersion forces increases.
The stronger the attractive forces between the molecules, the higher the boiling point will be.

Practice – Choose the Substance in Each
Pair with the Highest Boiling Point a) CH
4
CH
3
CH
2
CH
2
CH
3
H H H H H
H
H
C
C C
H
C C H
H b) CH
3
CH
H H H
2
CH=CHCH
H
2
CH
3 cyclohexane
H
H H
H
C
H
C
H
C
C
H
H H
C
H
C
H
H H
H
C
H
H
C
C
H
H
H
H
C
C
H
C
H
H
H

• polar molecules have a permanent dipole
– because of bond polarity and shape
– dipole moment
– as well as the always present induced dipole
• the permanent dipole adds to the attractive forces between the molecules
– raising the boiling and melting points relative to nonpolar molecules of similar size and shape

Molar
Mass
Boiling
Point
Dipole
Size
CH
3
CH
2
CH
3
44.09 -42°C 0.08 D
CH
3
-O-CH
3
46.07 -24°C 1.30 D
CH
3
- CH=O 44.05 20.2°C 2.69 D
CH
3
-C

N 41.05 81.6°C 3.92 D
12

Practice – Choose the Substance in Each
Pair with the Highest Boiling Point a) CH
2
FCH
2
F
H
F
H
H
H
C C
F
CH
3
CHF
2
H
H
F
H
H
C C
F b) or

• Solubility depends on the attractive forces of solute and solvent molecules
– Like dissolves Like
– miscible liquids will always dissolve in each other
• polar substance dissolve in polar solvents
– hydrophilic groups = OH, CHO, C=O, COOH,
NH
2
, Cl
• nonpolar molecules dissolve in nonpolar solvents
– hydrophobic groups = C-H, C-C
• Many molecules have both hydrophilic and hydrophobic parts - solubility becomes competition between parts

Dichloromethane
(methylene chloride)
Ethanol Water
(ethyl alcohol)

H
3
C
H
2
C
C
H
2
H
2
C
C
H
2
CH
3 n -hexane
H C
CH
3
C
CH
H C
C
H
CH toluene
Cl
Cl
Cl
C
Cl carbon tetrachloride

• When a very electronegative atom is bonded to hydrogen, it strongly pulls the bonding electrons toward it
– O-H, N-H, or F-H
• Since hydrogen has no other electrons, when it loses the electrons, the nucleus becomes deshielded
– exposing the H proton
• The exposed proton acts as a very strong center of positive charge, attracting all the electron clouds from neighboring molecules

150
100
50
0
-50
1
-100
-150
-200
Relationship between H-bonding and Intermolecular Attraction
H
2
O
HF
NH
3
2
H
3
2
S
SiH
4
H
2
Se 4
GeH
4
CH
4
Period
H
2
Te
5
SnH
4
BP, HX
BP, H2X
BP, H3X
BP, XH4

Practice – Choose the substance in each pair that is a liquid at room temperature (the other is a gas) a) CH
3
OH CH
3
CHF
2 b) CH
3
-O-CH
2
CH
3
CH
3
CH
2
CH
2
NH
2

Practice – Choose the substance in each pair that is more soluble in water a) CH
3
OH CH
3
CHF
2 b) CH
3
CH
2
CH
2
CH
3
CH
3
Cl

• in a mixture, ions from an ionic compound are attracted to the dipole of polar molecules
• the strength of the ion-dipole attraction is one of the main factors that determines the solubility of ionic compounds in water

• the pressure exerted by the vapor when it is in dynamic equilibrium with its liquid is called the vapor pressure
– remember using Dalton’s Law of Partial Pressures to account for the pressure of the water vapor when collecting gases by water displacement?
• the weaker the attractive forces between the molecules, the more molecules will be in the vapor
• therefore, the weaker the attractive forces, the higher the vapor pressure
– the higher the vapor pressure, the more volatile the liquid

• increasing the temperature increases the number of molecules able to escape the liquid
• the net result is that as the temperature increases, the vapor pressure increases
• small changes in temperature can make big changes in vapor pressure
• the rate of growth depends on strength of the intermolecular forces

normal BP
100°C
Temperature vs Vapor Pressure
1000
900
800
700
600
500
400
300
200
100
0
760 mmHg w ater
TiCl4 chloroform ether ethanol acetone
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Temperature, °C
BP Ethanol at 500 mmHg
68.1°C

• when the temperature of a liquid reaches a point where its vapor pressure is the same as the external pressure, vapor bubbles can form anywhere in the liquid
– not just on the surface
• this phenomenon is what is called boiling and the temperature required to have the vapor pressure = external pressure is the boiling point

• the normal boiling point is the temperature at which the vapor pressure of the liquid = 1 atm
• the lower the external pressure, the lower the boiling point of the liquid

solid
liquid melting or fusion vaporization
Section 11.4
gas sublimation
31
All of these processes require energy and are considered endothermic
- the energy associated with the process is given a positive sign

Conversion of
States of Matter Continued gas condensation liquid freezing
Section 11.4
solid deposition
32
All of these processes produce energy and are considered exothermic
- the energy associated with the process is given a negative sign

33
depict the relationship between the three states of matter and the conditions under which they exist
Pressure liquid solid gas
Temperature
Triple point –
All three states of matter exist at the same time

The point above which the vapor cannot be condensed to a liquid
34

Heating curves are diagrams which depict the energy associated with changes in the state of matter gas
100

C
Temperature
0

C liquid solid
38 Heat Added

The change in heat (also called the change in enthalpy) can be calculated by summing the heats associated with each process in the energy diagram.
39
Two types of heat to consider:
– Specific heat
– Process heat

40
Defined as the amount of heat required to raise the temperature of a substance
Specific heat (SH) = q m

T q = heat added m = mass of material

T = change in temperature

41
heat associated with a specific change in the state of matter
For example,

H (fusion)

H (vaporization) and so on…

Section 11.4
gas
100

C
Temperature
0

C
42 liquid solid
Heat Added
NOTE:
– All angled lines are associated with a specific heat change
– All horizontal lines are associated with a process change

How much heat has to be removed from
10.0 grams of Freon-11 gas, CCl
3
F, at
50 o C to convert it to a liquid at 0 o C?
43
CCl
3
F MW= 137.36 bp= 23.8

C sp heat(l)= 0.87 J/ g·K sp heat(g)=0.59 J/ g·K

H vap
= 24.75 kJ/mol

50 o C
23.8
o C
0 o C heat removed, J or kJ
Step 1: Heat removed (50 ° C to 23.8 ° C) – specific heat calculation
Step 2: Phase change from gas to liquid - process heat calculation
Step 3: Heat removed (23.8 ° C to 0 ° C) – specific heat calculation
44

q = (SH) m Δ T q = (0.59 J/ g·K) (10.0 g) (23.8 - 50.0 K) = -154.58 J
Δ H vap
= 10.0 g 1 mol
137.36 g
-24.75 kJ
1 mol
1 kJ
1000 J
Δ H vap
= -1801.3 J q = (SH) m Δ T q = (0.87 J/ g·K) (10.0 g) (0 - 23.8 K ) = -207.06 J
45
-2162.94 J (sum) or -2.2 kJ (Exothermic)