Chemistry 122 [Tyvoll]
Spring 2008
Chapter 11 Homework Solutions for Moore
Problems # – 3, 5, 14, 16, 18, 20, 24, 33, 35, 40, 45, 90, 93 and 96
3. The energy required to expand a liquid’s surface is called the surface tension. It is higher for
liquids with strong intermolecular attractions. Water has a relatively high surface tension
compared to other liquids due to the stronger hydrogen bonding forces experienced among the
water molecules in the liquid. (for more info, see Section 11.1) Mercury is used as an example
in the text of a liquid that has extremely high surface tension. In mercury, the intermolecular
forces responsible for its high surface tension are metallic bonding.
5. The boiling point is when the vapor pressure of the liquid is equal to the atmospheric pressure
acting on the liquid. When the atmospheric pressure is 1 atm, the boiling point is called the
normal boiling point. (for more info, see Section 11.2)
14. Answer and Explanation: Some molecules have more kinetic energy than the potential energy
of the intermolecular attractive forces holding the liquid molecules together. If such a molecule
is at the surface of the liquid and moving in the right direction, it will leave the liquid phase
and enter the gaseous phase. This describes vaporization at the molecular level. Molecules in
the gas phase move randomly at various speeds and in every possible direction. That means
some of them will eventually impact the surface of the liquid. This can drive the molecule back
into the liquid phase where its high kinetic energy is absorbed by several of the molecules near
the surface where it hit. The loss of kinetic energy and attractive intermolecular forces causes
the molecule to reincorporate into the liquid state. This describes condensation at the molecular
level.
16. Answer/Explanation: The heat of vaporization is the heat that must be added to cause a liquid to
form a gas. The heat of condensation, which is the heat released by a gaseous substance when
forming a liquid, is the opposite in sign and identical in size to the heat of vaporization. The
molecules in the liquid state must gain sufficient energy to overcome the intermolecular
attractive forces holding the liquid molecules together in order to enter the gas phase; hence,
vaporization is endothermic.
18. Answer: see graphs below; molecules with hydrogen bonding, (HF, H2O, and NH3) all have
higher boiling points and larger heats of vaporization than those two that do not (Ne and CH4).
Strategy and Explanation:
Number of H atoms
0
1
2
3
4
Boiling Point (°C)
–246
19.7
100
–33.4
–161.5
ΔHvap (kJ/mol)
1.8
25.2
40.7
25.1
8.9
18. (continued) Now make graphs of the boiling point versus number of H atoms in one graph and
the heat of vaporization versus number of H atoms.
Trends for both graphs are the same, since the boiling points and heats of vaporization are
related to the strengths of the inter-particle forces that must be overcome to go from liquid to
gas. The molecules with hydrogen bonding, (HF, H2O, and NH3) all have higher boiling points
and larger heats of vaporization than those two that do not (Ne and CH4).
20. Strategy and Explanation: Given the molar enthalpy of vaporization of a compound, determine
the heat energy required to vaporize a given mass of a compound. Convert the mass to moles,
and use the molar enthalpy of vaporization to get total heat energy.
? kJ = (1.00 kg CCl3F)(1000 g/kg)(1 mol CCl3F/137.368 g CCl3F)(24.8 kJ/1 mol CCl3F)
? kJ = 181 kJ (3 significant figures)
24. Answer/Explanation: The larger boiling point of the alcohol is due to stronger attractive
intermolecular forces (hydrogen bonding). The ether is not able to undergo hydrogen bonding
interactions in the liquid state, since none of the H atoms are bonded to a highly electronegative
atom. The strongest interactive forces between the ether molecules are dipole-dipole forces,
which are weaker than hydrogen bonding.
33. Strategy and Explanation: Given the boiling point and the vapor pressure at a given
temperature, find the enthalpy of vaporization. See Moore, p.494 for details. R is 8.314 J/molK
T1 = 50.0°C + 273.15 = 323.2 K T2 = 78.3°C + 273.15 = 351.5 K
Use the two-set Clausius-Clapeyron equation: ln(P2/P1) = − (ΔH(vap)/R)(1/400 K − 1/360 K)
ln(P2/370 mmHg) = − (44.0 kJ/mol/8.314 J/molK)(1 kJ/103 J))(1/T2 − 1/T1)
ln(P2/370 mmHg) = 1.5
P2 = (370 mmHg)exp(1.5) = 1600 mmHg
35. Answer/Explanation: A low heat of fusion tells us that not much energy is required to melt a
solid. That is the case when the intermolecular interactions between the particles in the solid are
weak, such as in solids composed of small nonpolar molecules or atoms.
40. Strategy and Explanation: Given the molar enthalpies of fusion and vaporization of a
compound, determine the heat energy required to raise a given number of moles of solid to the
melting point, melt it, raise the liquid to the boiling point and boil it. Find the mass from the
moles. Use the molar enthalpies to get total heat energy for both of the phase transitions. Use
Equation 6.2 and Table 6.1 to get the heat energy needed to change the temperature of the solid
and liquid water (as described in Chapter 6, Section 6.3). Add together the heat energy for each
stage to get the total quantity of heat energy.
40. (continued) g H2O = (0.50 mol H2O)(18.0152 g H2O/1 mol H2O) = 9.0 g H2O
Total heat energy is the sum of the heat energies to warm the ice to the melting point (0 °C), the
heat energy required to melt the ice, the heat energy to warm the water to the boiling point
(100°C), and the heat energy required to vaporize the water, so
qtotal = qheat ice + qmelt + qheat water + qboil
qtotal = (cice ⋅ m ⋅ ΔT) + (nΔHfusion) + (cliquid ⋅ m × ΔT) + (nΔHvap)
qtotal = (2.06 J/ g 0C)(1 kJ/1000 J)(9.0 g) × [0 °C – (–5 °C)] + (0.50 mol)(6.020 kJ/mol)
+ (4.184 J/ g 0C)(1 kJ/1000 J)(9.0 g)[0 °C – (–5 °C)] + (0.50 mol) (40.07 kJ/mol)
qtotal = 0.09 kJ + 3.0 kJ + 3.8 kJ + 20. kJ = 27. kJ
45. Strategy and Explanation: The highest melting point is a result of strongest inter-particle forces.
Ionic interactions are the strongest individual forces in solids. Looking at the two ionic
compounds LiBr and CaO, CaO has higher ionic charges (Ca2+ and O2– versus Li+ and Br–).
Coulomb’s law describes the attraction between charged particles; the ion-ion coulombic
interaction is stronger with higher ionic charges. Interionic attractions in the solid CaO are
stronger than those in solid LiBr, so (b) CaO has a higher melting point. The lowest melting
point is a result of the weakest intermolecular forces. We need to compare the intermolecular
forces in the two molecules, CO and CH3OH. The strongest forces that CO molecules can
experience are dipole-dipole forces. The strongest forces that CH3OH molecules experience are
hydrogen bonding. Dipole-dipole intermolecular forces are weaker than hydrogen bonding
interactions, so (c) CO has the lowest melting point.
Answer: Highest melting point is (b) CaO. Lowest melting point is (c) CO.
90. Answer/Explanation: Amorphous solids are compared to crystalline solids in Section 11.5 and
glasses are discussed in more detail in Section 11.11. The amorphous solids known as glasses
are different from NaCl, because they lack symmetry or long range order, whereas ionic solids
such as NaCl are extremely symmetrical. NaCl must be heated to melting temperatures, then
cooled very slowly, to make a glass.
93. Strategy and Explanation: Given the normal boiling point, molar heat of vaporization, and the
specific heat capacities of the gas and liquid states of a compound, determine the heat energy
evolved when a given mass of the substance is cooled from an initial to a final temperature.
Use the molar enthalpies to get total heat energy transferred during both of the phase transitions.
Use Equation 6.2 and Table 6.1 to get the heat energy for changing the temperature of the solid
and liquid water (as described in Chapter 6, Section 6.3). Add together the heat energy for each
stage to get the total quantity of heat energy.
93. (continued) Total heat energy evolved is the sum of the heat energy evolved when cooling the
gas to the boiling point (– 30 °C), the heat energy evolved when condensing the gas to a liquid,
and the heat energy evolved when cooling the liquid to the final temperature (–40 °C). A change
in temperature in Kelvin degrees is the same as the change in temperature in Celsius degrees.
So, use
cgas = 0.61 J g–1 °C–1 and cliquid = 0.97 J g–1 °C–1 (assume T is ± 1 °C)
qtotal = (cgas)mΔT) + (mΔHvap) + (cliquid)mΔT)
qtotal = (0.61 J/ g 0C)(10.0 g)[ –30 °C – 40 °C)] + (10.0 g)(165 J/g)
+ (0.097 J/ g 0C)(10.0 g)[ –40 °C – (–30 °C)] + (0.50 mol) (40.07 kJ/mol)
qtotal = [– 430 J + (– 1650 J) + (– 97 J)]  (1 kJ/1000 J) = – 2.18 kJ
A negative qtotal means that heat energy is evolved. The amount of heat energy evolved is 2.18
kJ.
96. Strategy and Explanation: (a) The molecular geometry of the SO2 molecule is determined in
the solution for SO2 (18 e–) The VSEPR formula is AX2E1, so the electron-pair geometry is
triangular planar, and the molecular geometry is angular (120°).
The asymmetric shape of the SO2 molecule means that it is polar. Therefore, the strongest
intermolecular forces between molecules in solid and liquid SO2 are dipole-dipole forces. All
molecules experience London forces, so that force is also part of the attractions in the liquid and
solid states of SO2 .
(b) The normal boiling point is smaller when the intermolecular attractions are weaker. The
normal boiling point is larger when the intermolecular attractions are stronger.
CH4 (–161.5 °C) < NH3 (– 33.4 °C) < SO2 (–10 °C) < H2O (100 °C)
Answer Check: Ordering the molecules by boiling points indicates that the
intermolecular attractions in SO2 are stronger than the attractions in NH3. This is probably a
result of variable size. The molar mass of SO2 is 64 g/mol, but the molar mass of NH3 is only 17
g/mol. SO2 has three atoms from period two and NH3 has only one atom from period two. So it
makes sense SO2 molecules may have significant enough London forces to be more attractive to
each other than are tiny, polar molecules of NH3 undergoing hydrogen bonding.