Chapter 11 Homework Solutions
12. a) the liquid state (phase)
b) In the solid phase, molecules are packed in a maximally efficient arrangement and are at or
very near the most stable arrangement for the material. On melting, energy is provided for
the molecules to slip past one another. For this to happen small gaps in the structure must
appear for the molecules to move. This leads to a swelling of the material and a
corresponding decrease in density.
16. a) hydrogen bonding (CH3OH has it, CH3SH doesn’t.)
b) London dispersion forces (Xenon is larger.)
c) London dispersion forces (Chlorine is larger.)
d) dipole-dipole forces (Acetone is polar, 2-methylpropane is not.)
17. a) Polarizability refers to the tendency of electrons in an atom or molecule to be displaced or
distorted by an external charge.
b) Antimony. It is the largest of the atoms.
c) CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4
d) CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4
20. a) Br2 b) CH3CH2CH2CH2CH2SH c) CH3CH2CH2Cl
24. a) HF has hydrogen bonds, while HCl has only dipole-dipole interactions.
b) CHBr3 has a larger electron cloud, yielding larger London dispersion forces.
c) ICl is polar, Br2 is not. dipole interactions + London dispersion forces > London
dispersion forces alone.
26. a) London dispersion forces: C8H18
b) dipole-dipole forces (CH3OCH3) and London dispersion forces (C3H8 and CH3OCH3
minor contributor): CH3OCH3
c) both have London dispersion forces and dipole-dipole forces, but hydrogen peroxide also
has hydrogen bonding: H2O2
d) hydrogen bonding, dipole-dipole, and London dispersion forces (minor contributor)
(NH2NH2), London dispersion forces (CH3CH3): NH2NH2
.. .. .. .. .. ..
32. a) N N O
H H H H H
hydrazine hydrogen peroxide water
b) All have O-H or N-H bonds. These bonds can lead to hydrogen bonding which would
increase surface tension.
34. a) condensation, exothermic c) vaporization, endothermic
b) sublimation, endothermic d) freezing, exothermic
40. The C2Cl3F3 will absorb heat in 3 distinct phases:
i) Heat the liquid (∆H1) (∆T = 36 K) iii) Heat the gas (∆H3) (∆T = 192 K)
ii) Boil the liquid (∆H2)
∆H1 = (37.6 K)(50.0 g)(0.91 J/g•K) = 1710 J
1 mol 27.49 kJ 1000 J
∆H2 = (50.0 g)
187.4 g mol kJ = 7330 J
∆H3 = (37.4 K)(50.0 g)(0.67 J/g•K) = 1250 J
∆HT = ∆H1 + ∆H2 + ∆H3 = 10290 J
= 10.29 kJ (after adjusting for significant figures and converting to kJ)
46. (b) is true. (a), (c), and (d) are false.
48. a) On a humid day there is more water vapor in the air than on a dry day. The greater
number of water molecules in the air means more will be striking the surface of the water
that is evaporating. Thus, since the return rate is faster, the net evaporation rate is slower.
b) At high altitude atmospheric pressure is lower. The boiling point of a liquid occurs when
the vapor pressure of the liquid equals the external pressure. Therefore, water boils at a
lower temperature because it must be heated less to reach the lower external pressure.
Lower temperatures increase the time necessary for cooking.
54. a) Solid CO2 sublimes to gaseous CO2 at ≈ -65 ºC.
b) Solid CO2 melts to liquid CO2 at ≈ - 50 ºC.
72. a) ionic c) ionic e) molecular
b) metallic d) molecular f) molecular
74. a) metallic c) ionic or network covalent e) ionic
b) molecular or metallic d) network covalent
78. a) HF – HF hydrogen bonds plus has dipole-dipole interactions, HCl has only the latter.
HCl isn’t so much larger than HF that London dispersion forces would make a significant
b) graphite – Graphite is a network covalent compound, whereas methane forms a molecular
c) KCl – Potassium chloride is an ionic compound, whereas bromine forms a molecular solid.
d) MgF2 – Both are ionic compounds with a common anion. The difference must lie in the
cation. The magnesium ion carries a +2 charge vs. the +1 charge on lithium. The higher
charge on magnesium results in a stronger attraction (see Sections 5.1 and 8.2) and,
hence, a higher melting point.
79. a) decrease c) increase e) increase g) increase
b) increase d) increase f) increase
82. The C-Br bond would be less polar than the C-Cl bond, while the C-F bond would be more
polar. Thus, CH2Br2 would be more polarizable than CH2Cl2, while CH2F2 would be less
polarizable. Bromine is a larger atom than chlorine, while fluorine is smaller. This means
that the contribution of dipole-dipole interactions would increase in CH2F2 and decrease in
CH2Br2. Thus, in CH2F2 the relative importance of dipole-dipole interactions would increase
relative to dispersive forces, while in CH2Br2 dispersive forces would increase at the expense
of dipole-dipole forces.
100. a) trigonal planar CH3
O C tetrahedral
c) London dispersion forces and dipole-dipole interactions
d) Propanol molecules hydrogen bond to each other, while the strongest interaction between
acetone molecules is dipole-dipole attraction.
102. a) Greater. Imagine filling the tank. We begin with an empty tank full of air. At 0 ºC, the
pressure is 1.00 atm (STP). When we add the liquid butane, its boiling point is close
enough to 0º to say the pressure is about 1 atm (just a little higher actually). When the
temperature is raised to 35º, some of the butane boils from the liquid phase which
increases the total pressure.
Answering this question requires two assumptions: (1) we put butane in a tank and seal it
and (2) not all of it vaporizes when the temperature increases to 35º. If this is the case,
then the volume does not affect the pressure.
b) It would cool. At the moment some of the pressure (mostly butane gas) is bled off, the
pressure drops and the equilibrium is disrupted. Now there is less in the gas phase then
there should be. Since it requires energy to vaporize a molecule, increasing the number
of molecules in the gas phase will require the input in energy (as heat). The loss of heat
from the liquid phase will result in a temperature drop.
1 mol 21.3 kJ
c) q = (250 g)
58.1 g mol = 92 kJ
1 mol 0.0821 L • atm
( 250 g)
58.1 g mol • K
V= = 110 L