CHAPTER 11 Problems: 2, 14, 16, 18, 22, 28, 32, 36, 46, 56, 60, 64, 70, 75, 84, 103
2) Explain the term polarizability. What kind of molecules tend to have high polarizabilities? What is the relationship between polarizability and intermoleular forces?
Polarizability refers to the shift in electron density in an atom or molecule when it is brought close to a charge (positive attracts, negative repels).
Larger molecules and molecules where the outermost electrons are further away from the nucleus tend to be the ones that are most easily polarized. So, for example, Xe is more polarizable than He, and C2Cl6 is more polarizable than CCl4.
Polarizability is most important for molecules attracted to one another by dispersion forces. It is also important for “mixed” forces of attraction, such as ion-molecule or dipole-molecule attractive forces.
14) Arrange the following in order of increasing boiling point: RbF, CO2, CH3OH, CH3Br. Explain your reasoning.
Highest to lowestRbF > CH3OH > CH3Br > CO2
RbF is ionically bonded, and so should have a very high value for boiling point. CO2 is nonpolar (and so only experiences dispersion forces) and so should have a low boiling point (and in fact is a gas at room temperature). Both CH3OH and CH3Br are polar, but CH3OH can form hydrogen bonds between molecules, which is likely more important than the fact that CH3Br has the larger value for molecular mass.
16) Which member of each of the following pairs of substances would you expect to have a higher boiling point? Explain your answers.
a) O2 or Cl2Both are nonpolar, but Cl2 is larger and so should have stronger dispersion forces between molecules, and so a higher boiling point.
b) SO2 or CO2 SO2 is nonlinear (trigonal planar electron cloud geometry) and so polar, and it also has a larger molecular mass than CO2, which is linear and so nonpolar. Both of these factors favor SO2 as having a higher boiling point.
c) HF or HIHF molecules can hydrogen bond to one another, and so HF more likely has a higher boiling point than HI, even though HI is larger and so has stronger dispersion forces.
18) Explain the following in terms of intermolecular forces.
a) NH3 has a higher boiling point than CH4 The molecules are about the same size and molecular mass, but NH3 is polar, and NH3 molecules can hydrogen bond to one another, while CH4 is nonpolar.
b) KCl has a higher melting point than I2 The bonding in KCl is ionic, so KCl should have a high value for melting point. I2 is nonpolar, and the attractive forces between molecules are relatively weak dispersion forces, and so I2 should have a lower melting point.
22) Explain why liquids, unlike gases, are nearly incompressible.
The molecules in a liquid are in close contact with one another, so it takes a large pressure to force the molecules even a little bit closer together. Gases are mostly empty space, and so while increasing the pressure decreases the average distance between gas molecules, the molecules are still not generally in close contact.
28) What is viscosity? What is the relationship between intermolecular forces and viscosity?
Viscosity is the resistance of a liquid (or gas) to flow. Generally speaking, the stronger the intermolecular forces between molecules the larger the value for viscosity.
32) The vapor pressure of benzene (C6H6) is 40.1 mm Hg at 7.6 C. What is the vapor pressure at 60.6 C? The molar heat of vaporization for benzene is 31.0 kJ/mol.
We can use the Clausius-Clapeyron equation (eq 11.4 of Burdge)
ln(p2/p1) = - (Hvap/R) [ (1/T2) – (1/T1) ]
Let T1 = 7.6 C = 280.8 KT2 = 60.6 C = 333.8 K
p1 = 40.1 mm Hg
Then
ln(p2/p1) = - [(31000. J/mol.K)/(8.314 J/mol.K)] [ (1/333.8 K) – (1/280.8 K) ]
= 2.108
Taking the inverse ln of both sides of this equation
(p2/p1) = e2.108 = 8.235
p2 = p1 e2.108 = (40.1 mm Hg)(8.235) = 330. mm Hg = 330. torr
36) Vapor pressure measurements for mercury at several different temperatures are given below. Plot this data in an appropriate way, and determine the molar enthalpy of vaporization for mercury.
T (C)200.0250.0300.0320.0340.0
p (mm Hg) 17.3 74.4 246.8376.3557.9
We may use the form of the Clausius-Clapeyron equation in eq 11.1 of Burdge to analyze this data. We plot ln p vs (1/T). The slope of the line obtained from the plot is equal to - (Hvap/R), from which we can find Hvap.
T (C)p (torr)T (K)1/T (K-1)ln p
200.0 17.3473.20.0021132.8507
250.0 74.4523.20.0019114.3095
300.0246.8573.20.0017455.5086
320.0376.3593.20.0016865.9304
340.0557.9613.20.0016316.3242
The data are plotted below (using EXCEL, which was also used to find the parameters for the best fitting line).
Based on the plot, I get
m = slope = - 7210. KHvap = - mR = - (- 7210. K) (8.314 J/mol.K)
= 59.9 kJ/mol
46) Metallic iron crystallizes in a cubic lattice. The unit cell edge is 287. pm. The density of iron is 7.87 g/cm3. How many iron atoms are present per unit cell?
The unit cell is a cube, and so if we call the length of the unit cell, the volume of the unit cell is
V = 3 = (287. pm)3 (1 cm/1010 pm)3 = 2.364 x 10-23 cm3
The number of unit cells contained in a volume of 1 cm3 is then
# unit cells = 1 cm3 = 4.230 x 1022 unit cells
(2.364 cm3/cell)
The mass of iron per unit cell is
mass Fe = 7.87 g 1 cm3 = 1.860 x 10-22 g
cm3 4.230 x 1022 unit cells
The mass of one iron atom is
m = 55.85 g 1 mol = 9.27 x 10-23 g/atom
mol 6.022 x 1023 atoms
And so (finally!) the number of iron atoms per unit cell is
# atoms = 1.860 x 10-22 g/unit cell = 2.006 = 2 atoms per unit cell
9.27 x 10-23 g/atom
56) Describe and give examples of each of the following types of crystals:
a) ionic crystals (examples: NaCl, KNO3, CuF2)
Held together by the attractive forces that act between ions of opposite charge. Compounds are hard and brittle, usually have a high melting and boiling point, do not conduct electricity in the solid phase, but do conduct electricity in the liquid phase or when dissolved in water (for those ionic compounds that are soluble in water).
b) covalent crystals (examples: the diamond form of C; the quartz form of SO2)
The entire crystal is held together by covalent bonds – there are no individual atoms or molecules present. Every atom in the solid is connected to every other atom by a network of covalent bonds (which is why these substances are sometimes called network covalent crystals). Compounds are extremely hard and have high melting and boiling points, and do not conduct electricity in the solid or liquid phase. The compounds are not soluble in water.
c) molecular crystals (examples: C10H8, H2O, CO2, C6H12O6)
Crystals consist of individual molecules held together by weak intermolecular forces (dipole-dipole and dispersion forces). Compounds are usually soft and have relatively low melting and boiling points (many are in fact liquids or gases at room temperature). Compounds do not conduct electricity in the solid or liquid state, and usually do not conduct electricity when dissolved in water.
d) metallic crystals (examples: Fe, Cu, Ag, Zn)
Crystals consist of individual metal atoms held together by “metallic bonding”, the bonding occurring due to the “electron sea” the atoms are immersed in. Metallic crystals can be soft to hard and can have melting and boiling points that range from low to high temperature. Compounds conduct electricity in the solid and liquid state, and do not dissolve in water.
60) A solid is very hard and has a high melting point. Neither the solid or its melt (liquid state) conducts electricity. Classify the solid as one of the four types from problem 56.
Based on problem 56 above (or the textbook) the solid is likely a covalent crystal.
64) What is an amorphous solid? How does it differ from a crystalline solid?
An amorphous solid is a solid that does not have a regular arrangement of the particles making up the solid, that is, there is no repeating crystal structure such as exists in crystalline solids.
70) How is the molar heat of sublimation related to the molar heat of fusion and the molar heat of vaporization? On what law is this relationshop based?
Since s g = s + g
We would expect, based on Hess’ law, that
Hsub = Hfus + Hvap
The above expression is only approximately correct if we use the usual values for H for each of the phase transitions, because the enthalpy changes for different phase transitions are usually measured at different temperatures and pressures (for example, the enthalpy of fusion for water is usually given at 1 atm and 0. C, while the enthalpy of vaporization is usually given at 1 atm and 100. C).
75) What is the critical temperature? What is the significance of the critical temperature in terms of the condensation of gases?
The critical temperature is defined as the temperature above which an isothermal (constant temperature) compression of a gas will never lead to a phase trase transition or formation of a liquid.
What this means is that if we have a gas at a temperature below the critical temperature we can liquefy the gas by increasing the applied pressure. We make use of this in refrigerators and air conditioners, which contain working fluids that can be liquified by compression.
84) Calculate the amount of heat (in kJ) required to convert 150.2 g of water into steam at T = 100. C. Note that for water at 100.0 C, Hvap = 40.7 kJ/mol.
M(H2O) = 18.01 g/mol, and so the moles of water is
n = 150.2 g 1 mol = 8.335 mol
18.02 g
Soq = (8.335 mol) (40.7 kJ/mol) = 339. kJ
103) Which of the following substances has the highest polarizability: CH4, H2, CCl4, SF6, H2S?
Generally speaking polarizability is higher for larger molecules, and molecules with valence electrons far away from the atomic nuclei. So I would expect
CCl4 > SF6 > H2S > CH4 > H2
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