Chapter 12
Intermolecular Forces and Liquids
Chapter 12
Intermolecular Forces and Liquids
INSTRUCTOR’S NOTES
We believe the material in this chapter has been imprudently neglected in introductory courses, particularly the
matter of intermolecular forces. Therefore, we set aside 3-4 lecture periods for this chapter.
These are some points to make in the coverage of this chapter:

Polarity of molecules, as introduced in the context of molecular geometry, is a central consideration when
studying intermolecular forces.

London forces may not seem to be important, but because even non-polar atoms and molecules do condense
into liquids the existence of these forces is proved.

Although London forces are generally considered to be the weakest of the intermolecular forces, for large
molecules the total energy involved may equal or exceed that of other forces.

Hydrogen bonding, though not truly a bond, is responsible for an incredible deviation from trends in
properties such as boiling points. The role of hydrogen bonding in biochemistry can also be noted.
SUGGESTED DEMONSTRATIONS
1.
Vapor Pressure

We have searched for good demonstrations of vapor pressure for some time. One of the demonstrations we
use is a so-called “Love Meter,” a toy that can be obtained in novelty and gift shops. The closed, thin walled
glass vessel contains methyl chloride (bp –23.7 °C). Warming the vessel in your hand raises the vapor
pressure of the liquid and forces the liquid to another compartment of the vessel.
3.
Colligative Properties

4.
Morse, J. G. “A Simple Demonstration Model of Osmosis,” Journal of Chemical Education 1999, 76, 64.
Intermolecular Forces

Kitson, T. M. “Purple or Colorless—Which Way Up? An Entertaining Solubility Demonstration,” Journal
of Chemical Education 2003, 80, 892.
5.
Hydrogen Bonding

Johnson, J. L. H.; Yalkowsky, S. H. “A Three-Dimensional Model for Water,” Journal of Chemical
Education 2002, 79, 1088.
6.
Boiling / Phase Change

McRae, R.; Rahn, J. A.; Beamer, T. W.; LeBret, N. “The Liquid Nitrogen Fountain,” Journal of Chemical
Education 2002, 79, 1220.
241
Chapter 12
Intermolecular Forces and Liquids
SOLUTIONS TO STUDY QUESTIONS
12.1
(a) hydrogen bonding
(b) induced dipole/induced dipole
(c) hydrogen bonding
12.2
solid iodine: induced dipole/induced dipole
CH3OH: hydrogen bonding
I2 and CH3OH: dipole/induced dipole
12.3
12.4
12.5
(a) induced dipole/induced dipole
(c) dipole-dipole
(b) induced dipole/induced dipole
(d) hydrogen bonding
(a) induced dipole/induced dipole
(c) dipole-dipole
(b) hydrogen bonding
(d) induced dipole/induced dipole
Based on increasing molar mass:
CH4 < Ne < CO < CCl4
—increasing intermolecular forces
However, boiling points are in the order Ne (–246 ºC) < CO (–192 ºC) < CH4 (–162 ºC) < CCl4 (77 ºC)
CH4, Ne, and CO are gases at 25 ºC and 1 atm
12.6
He < CH3CH2CH2CH3 < CH3OH
—increasing intermolecular forces
He and CH3CH2CH2CH3 are gases at 25 ºC and 1 atm
12.7
(c) HF, (d) CH3CO2H, and (f) CH3OH would be expected to form intermolecular hydrogen bonds in the
liquid state
12.8
HCO2H would be expected to form intermolecular hydrogen bonds in the liquid state.
12.9
(a) LiCl
Lithium ions are smaller than cesium ions
(b) Mg(NO3)2
Magnesium has a larger positive charge (+2) than sodium (+1) and it is a smaller ion
(c) NiCl2
Nickel has a larger positive charge (+2) than rubidium (+1) and it is a smaller ion
12.10
Mg2+ is most strongly hydrated because of its small size and large positive charge. Cs+ is least strongly
hydrated because of its large size and smaller positive charge.
0.7849 g 1 mol CH3CH2 OH 42.32 kJ
·
·
= 90.1 kJ
1 mL
46.069 g
1 mol
12.11
125 mL ·
12.12
0.500 mL ·
12.13
(a) Approximately 150 mm Hg. Appendix G shows a value of 149.4 mm Hg
13.6 g 1 mol Hg
59.11 kJ


= 2.00 kJ
1 mL
200.59 g
1 mol
(b) 93 ºC
(c) Ethanol (520 mm Hg) has a higher vapor pressure at 70 ºC than water (225 mm Hg)
242
Chapter 12
12.14
Intermolecular Forces and Liquids
(a) Approximately 400 mm Hg.
(b) diethyl ether < ethanol < water
(c) diethyl ether is a gas, and ethanol and water are in the liquid phase
12.15
The vapor pressure of diethyl ether at 30 ºC is approximately 590 mm Hg.


1 atm
 590 mm Hg ·
 (0.100 mL)
760
mm
Hg
PV 

n=
=
= 0.0031 mol
RT
(0.082057 L  atm/K  mol)(303 K)
0.0031 mol ·
74.1 g (C2 H5 )2 O
= 0.23 g diethyl ether needed to create pressure of 600 mm Hg in flask
1 mol
There is enough diethyl ether available (1.0 g) to create a pressure of 600 mm Hg. When the flask is
cooled, some ether will condense to a liquid.
12.16
(a) As the water cools, the vapor pressure of the water decreases, resulting in a decrease in the total
pressure inside the bottle. The bottle will collapse because the pressure inside is less than the external
pressure.
(b) The normal boiling point of diethyl ether is 34.6 °C so the liquid will evaporate completely at body
temperature (37 °C).
12.17
12.18
(a) O2 (higher molar mass)
(c) HF (hydrogen bonding)
(b) SO2 (polar molecule)
(d) GeH4 (higher molar mass)
Ne < C2H6 < NH3 < SCl2
—increasing boiling point
12.19
(a) CS2:
620 mm Hg
CH3NO2:
80 mm Hg
(b)
induced dipole/induced dipole
dipole-dipole
(c)
46 ºC
100 ºC
(d) 39 ºC
(e) 34 ºC
12.20
A<B<C
12.21
(a) 80.1 ºC
(b) T = 48 º C when P = 250 mm Hg and T = 75 ºC when = 650 mm Hg. This is actually best determined
from a plot of lnP vs. 1/T.
243
Chapter 12
Intermolecular Forces and Liquids
ln
ln(vapor pressure) vs. 1/temperature
7
6.5
6
5.5
5
4.5
4
3.5
3
0.0028
y = -4027.4x + 18.048
0.003
0.0032
0.0034
0.0036
1/T (1/K)
– vap H
 400. mm Hg 
1
 1

–
(c) ln 
=

299.3 K 
 100. mm Hg  0.0083145 kJ/K  mol  333.8 K
vapH = 33.4 kJ/mol
12.22
– vap H
 45.3 mm Hg 
1 
 1
ln 
=
 323 K – 298 K 
13.6
mm
Hg
0.0083145
kJ/K

mol




vapH = 38.5 kJ/mol
Use this value to calculate the temperature when vapor pressure = 760 mm Hg
 760 mm Hg 
–38.5 kJ/mol
1 
1
ln 
=
 T – 298 K 
13.6
mm
Hg
0.0083145
kJ/K

mol
 2



T2 = 402 K (129 ºC)
The actual boiling point is 126 ºC
12.23
No. The critical temperature for carbon monoxide is below room temperature.
12.24
Yes. The critical temperature for propane is well above room temperature.
12.25
The layer of molecules on the surface of a liquid is harder to break through than the bulk liquid. The energy
needed to break through this skin is the surface tension. An application of the surface tension is that it is
possible to float a paperclip on the surface of water, but if the paperclip breaks through the surface, it will
sink. Surface tension results from intermolecular forces because molecules in the interior of a liquid
interact through intermolecular forces with molecules all around them, but molecules on the surface have
intermolecular forces only with molecules at or below the surface layer and thus feel a net inward force of
attraction.
12.26
Strength of intermolecular forces; stronger the intermolecular forces, the greater the viscosity. Also, the
shape of the molecules; long molecules that can become entagled similar to spaghetti have greater
viscosities.
244
Chapter 12
Intermolecular Forces and Liquids
Glycerol>ethylene glycol>ethanol>water; length of molecules from left to right allows for more
intermolecular interaction.
Yes. Increasing temperature provides more energy for intermolecular forces to be overcome.
12.27
Capillary action. The water is attracted to –OH groups in the paper. As a result of these adhesive forces,
some water molecules begin to move up the paper. Other water molecules remain in contact with these
water molecules by means of cohesive forces. A stream of water molecules thus moves up the paper.
12.28
Demonstration of adhesion vs. cohesion. Hg…Hg interactions are stronger then Hg…SiO2 interactions.
Interactions between glass surface and Hg are minimized to maximize Hg…Hg interactions. For water,
H2O…SiO2 interactions are stronger than H2O…H2O interactions.
Concave meniscus
12.29
Ar < CO2 < CH3OH
—increasing intermolecular forces
12.30
(a) induced dipole/induced dipole forces
(b) hydrogen bonding
12.31
Li2SO4 should have the more exothermic heat of hydration because the lithium ion is smaller than the
cesium ion.
12.32
(a) ICl (polar molecule)
(b) krypton (greater molar mass)
(c) CH3CH2OH (hydrogen bonding)
12.33
(a) 350 mm Hg
(b) ethanol (lower vapor pressure at every temperature)
(c) 84 ºC
12.34
(d) CS2: 46 ºC
C2H5OH: 78 ºC
C7H16: 99 ºC
(e) CS2: gas
C2H5OH: gas
C7H16: liquid
Compounds (a), (b), and (d) contain small, highly charged metal ions. These salts will most likely
experience strong ion-dipole forces between the metal cation and water molecules and exist as hydrated
solids. (d) Al(NO3)3 should have the most negative enthalpy of hydration.
12.35
C2H6 < HCl < CH3OH
—increasing enthalpy of vaporization
12.36
CO2 < CH3Cl < HCO2H
—increasing intermolecular forces
245
Chapter 12
12.37
Intermolecular Forces and Liquids
1 atm
0.00169 mm Hg ·
n
P
760 mm Hg
=
=
= 9.12  10–8 mol/L
V
RT
(0.082057 L  atm/K  mol)(297 K)
9.12  10–8 mol 6.022  1023 atoms
1L
·
· –3 3 = 5.49  1019 atoms/m3
1L
1 mol
10 m
12.38
(a)
ln(vapor pressure) vs. 1/temperature
7
6
5
ln
4
3
y = -5316.6x + 18.54
2
1
0
0.0022
0.0024
0.0026
0.0028
0.003
0.0032
0.0034
0.0036
1/T (1/K)
(b) Using the equation for the straight line, T = 135 ºC when P = 250 mm Hg and T = 168 ºC when
P = 650 mm Hg
(c) T = 173 ºC when P = 760 mm Hg
– vap H
 400. mm Hg 
1
 1

–
(d) ln 
=


 100. mm Hg  0.0083145 kJ/K  mol  424.6 K 381.5 K 
vapH = 43.3 kJ/mol
12.39
(a) 70.3 ºC
(b) Using the equation for the straight line, T = 39 ºC when P = 250 mm Hg and T = 66 ºC when
P = 650 mm Hg,
246
Chapter 12
Intermolecular Forces and Liquids
ln
ln(vapor pressure) vs. 1/temperature
7
6.5
6
5.5
5
y = -3885.2x + 17.949
4.5
4
3.5
3
0.0028
0.003
0.0032
0.0034
0.0036
0.0038
1/T (1/K)
– vap H
 650. mm Hg 
1
 1

–
(c) ln 
=


 250. mm Hg  0.0083145 kJ/K  mol  312.6 K 338.7 K 
vapH = 32.3 kJ/mol
12.40
The volatile liquid has a low boiling point. Warming the lower compartment in your hands increases the
number of molecules in the vapor phase, raising the vapor pressure of the liquid and forcing the liquid to the
upper chamber.
12.41
(a) When the water vapor in the can comes in contact with the cold water in the pan, the vapor condenses,
decreasing the vapor pressure inside the can. The pressure inside the can is now lower than the
atmospheric pressure outside the can, causing the can to collapse.
(b)
= water molecule
before heating
12.42
1.0  103 mL ·
after heating
1 cm3 0.785 g 1 mol C2 H5 OH
·
·
= 17 mol C2H5OH
1 mL
46.07 g
1 cm3
3
Vroom
 102 cm 
1 mL
1L
= 3.0 m · 2.5 m · 2.5 m · 
· 3
= 1.9  104 L
 ·
3
 1 m  1 cm 10 mL
247
Chapter 12
Intermolecular Forces and Liquids


1 atm
4
 59 mm Hg ·
 (1.9  10 L)
760
mm
Hg
PV

n=
= 
= 60. mol ethanol
RT
(0.082057 L  atm/K  mol)(298 K)
Less ethanol is available (17 mol) than would be required to completely fill the room with vapor
(60. mol), so all of the ethanol will evaporate.
12.43
Acetone readily absorbs water because hydrogen bonding occurs between the molecules.
O
O
H
H
C
H3C
12.44
CH3
Since cooking oil does not mix with water, the two substances have different types of intermolecular forces.
Water is a polar molecule, and strong hydrogen bonding occurs between water molecules. The molecules in
cooking oil are likely nonpolar.
12.45
The viscosity of ethylene glycol is greater than that of ethanol since ethylene glycol contains two O—H
groups per molecule while ethanol has only one.
12.46
The meniscus is concave since there are adhesive forces between the methanol and the silicate of the glass.
12.47
(a) Water has two O—H bonds and two lone pairs on oxygen, and ethanol has only one O—H bond.
This leads to less extensive hydrogen bonding in ethanol and a lower boiling point than water.
(b) The water molecules interact extensively with the ethanol molecules (through hydrogen bonding) and
can therefore occupy less than the anticipated 100 mL.
12.48
1-propanol has stronger intermolecular forces (hydrogen bonding) than methyl ethyl ether (dipole-dipole).
12.49
Two pieces of evidence for H2O() having considerable intermolecular attractive forces:
(1) Based on the trend for Group VIA hydrides, the boiling point of water should be approximately
–80 ºC. The actual boiling point (100 ºC) reflects the significant hydrogen bonding present.
(2) Liquid water has a heat capacity of 4.184 J/g·K, higher than almost all other liquids. This reflects the
fact that it takes a relatively large amount of energy to overcome intermolecular forces in order to raise
the temperature of a sample of water.
12.50
The frost–free refrigerator's freezer is warmed periodically to near the freezing point of water. This allows
the vapor pressure of the ice to increase so that sublimation readily occurs, keeping the frost from
accumulating in the freezer. It would therefore make the hailstones disappear over time. No such
mechanism exists in the older non–frost–free refrigerator.
12.51
(a) HI
(b) The large iodine atom in HI leads to a significant polarizability for the molecule and thus to a large
dispersion force.
248
Chapter 12
Intermolecular Forces and Liquids
(c) The dipole moment of HCl (1.07 D, Table 8.7) is larger than for HI (0.38 D).
(d) HI; see (b)
12.52

17.03 g 
qNH3 = 1.00 mol NH3 ·
 (4.70 J/g  K)(229.9 K – 239.9 K)
1 mol NH3 

qNH3 = –800. J
18.02 g 

qH2O = 1.00 mol H2 O ·
 (4.18 J/g  K)(–10.0 K)
1 mol H2 O 

qH2O = –753 J
Cooling liquid ammonia evolves more heat.
12.53
A gas can be liquefied at or below its critical temperature. The critical temperature for CF4 (–45.7 ºC) is
below room temperature (25 ºC), so it cannot be liquefied at room temperature.
12.54
(a) –27 ºC
(b) 6.5 atm
(c) The flow is rapid at first because the liquid has vaporized to form the maximum amount of vapor
possible at that temperature, with a pressure of approximately 6.5 atm. As the gas leaves the cylinder,
additional liquid vaporizes in an attempt to reach liquid-vapor equilibrium. The flow becomes much
slower because the pressure in the cylinder has dropped and cannot increase unless the cylinder is
closed or enough heat is supplied to vaporize the remaining liquid. This vaporization requires energy,
so the temperature of the cylinder drops and ice forms on the outside of the cylinder.
(d) (2) Cool the cylinder to –78 ºC and open the valve. Cooling the cylinder will condense the vapor,
allowing it to be easily removed as a liquid.
12.55
Hydrogen bonding is most likely at the –OH oxygen, the –C=O oxygen, and the –NH group.
12.56
The nitrogen and oxygen atoms and the hydrogen atoms attached to nitrogen can be involved in hydrogen
bonding.
H
N
N
O
H
H
N
NH
N
HN
N
O
H
N
H
cytosine
guanine
12.57
Boiling point, enthalpy of vaporization, viscosity, surface tension
12.58
K+ < Na+ < Ca2+ < Mg2+
—increasing enthalpy of hydration
249
Chapter 12
Intermolecular Forces and Liquids
The smallest ion with the greatest charge (Mg2+) has the most negative enthalpy of hydration. The largest
ion with the lowest charge (K+) has the least negative enthalpy of hydration.
12.59
The more branching in the hydrocarbons, the lower the boiling point (weaker dispersion forces). As
branching increases, the shape of the molecule becomes more compact, decreasing the molecule’s surface
area. As surface area decreases, intermolecular contact decreases, decreasing the strength of the dispersion
forces.
12.60
The amount of bromine that will vaporize at its boiling point is
n=
PV
(1.0 atm)(1.0 L)
= 0.037 mol Br2
=
RT
(0.082057 L  atm/K  mol)(332.0 K)
The amount of Br2 available is
8.82 g ·
1 mol Br2
= 0.055 mol Br2
159.8 g
More Br2 is available than can vaporize at the boiling point, so the flask will contain both liquid bromine
and bromine vapor in equilibrium.
12.61
F2 < Cl2 < Br2 < I2
He < Ne < Ar < Kr < Xe
Molar mass or boiling points could be used to determine the rankings.
12.62
(d) and (e)
12.63
 29.70 psi 
–40.7 kJ/mol
1 
1
ln 
=
 T – 373 K 
14.70
psi
0.0083145
kJ/K

mol
 2



T2 = 394 K = 121 ºC
250
Chapter 12
Intermolecular Forces and Liquids
12.64
ln(vapor pressure) vs. 1/temperature
4
3
y = -2808.1x + 11.704
ln
2
1
0
-1
-2
-3
0.003
0.0035
0.004
0.0045
0.005
1/T (1/K)
slope = –2808.1 K–1 =
– vap H
–  vap H
=
R
0.0083145 kJ/K·mol
vapH = 23.3 kJ/mol
12.65
Using the equation for the straight line, P = 8.3 atm when T = 20 ºC
12.66
(a) y = –5141.8 x + 21.283
(b) The slope of the line is equal to
– vap H
. Set up an equation and solve for vapH.
R
(c) Using the equation for the straight line, P = 11.6 mm Hg when T = 0.00 ºC and P = 1804 mm Hg when
T = 100 ºC.
12.67
(a) Liquid water in tube with some vapor.
(b) Vapor pressure of water at 100 oC is 1.000 atm.
(c) Assume all water is liquid water.
(10.0 g)(1 mL/0.958 g) = 10.4 mL
(d) Volume of gas = 50.0 mL – 10.4 mL = 39.6 mL = 0.0396 L
PV=nRT
n = PV/RT = (1.000 atm)(0.0396 L)/(0.082057 L . atm/mol . K)(373 K) = 1.29 x 10-3 mol
(1.29 x 10-3 mol)(18.02 g/mol) = 0.0233 g H2O
251
Chapter 12
Intermolecular Forces and Liquids
SOLUTIONS TO APPLYING CHEMICAL PRINCIPLES: CHROMATOGRAPHY
1.
(a) 1,5-pentanediol
Most important is H-bonding but dipole-dipole and dispersion forces are also present.
(b) propyl ethyl ether
Dispersion and dipole-induced dipole forces
(c) First 1,5-pentanediol, then 1-pentanol, then propyl ethyl ether
2.
(First to last) pentane, hexane, heptanes, octane
Greater dispersion forces occur for larger molecules.
252