Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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10.7 Kinetic-Molecular Theory

Skill Review

73. Using the tenets of the kinetic-molecular theory, explain why

the absolute temperature is directly proportional to the pres-

sure of a trapped volume of gas.

74. Does the kinetic-molecular theory assist in the explanation

of Avogadro’s law? If so, how?

75. Two gases known for their bleaching power are ClO

and Cl

Assume you have separate 1-L containers containing 5 g of

each at 25°C.

a. Compare the average kinetic energy of the molecules of

each gas.

b. Compare the average speed of the molecules of each gas.

c. Which, if either, would be exerting more pressure?

d. If the samples were placed in the same container, which, if

either, would have the larger partial pressure?

76. Equal masses of two gases, B

and C

, are placed in sep-

arate containers at the same temperature.

a. Compare the average kinetic energy of the molecules of

each gas.

b. Compare the average speed of the molecules of each gas.

c. Which, if either, would exert a greater pressure?

d. If the samples were placed in the same container, which, if

either, would have the larger partial pressure?

77. Assuming that a single molecule of each of these gases is

moving at a speed of 1100 m/s, what would you calculate as

its kinetic energy?

a. CO

b. NH

c. Ne

78. Assuming that a single molecule of each of these gases is

moving at a speed of 1100 m/s, what would you calculate as

its kinetic energy?

a. H

b. C

c. Ar

79. Arrange these gases in order from lowest to highest average

molecular speed (assume that there are 1.00 mol of each at

the same temperature and pressure).

a. H

b. C

c. Ar

80. Arrange these gases in order from lowest to highest average

molecular speed (assume that there are 1.00 mol of each at

the same temperature and pressure).

a. CO

b. NH

c. Ne

81. At 25.0°C, what would be the root-mean-square speed of

propane (C

) used in bottle gas fuel systems?

ClO

438 Chapter 10 The Behavior and Applications of Gases

82. At extremely low temperatures, molecular motion also be-

comes relatively slow. What would the temperature be when

the root-mean-square speed of He was 100 m/s?

83. Calculate the rms speed of CO

and of H

O at 250.0 K.

84. What is the rms speed of a gas with a molar mass of

16.0 g/mol if the temperature is 45°C?

Chemical Applications and Practice

85. A device sometimes used in teaching the gas laws consists of

small marbles trapped inside a glass-walled container. The

container is open at one end, where it is ﬁtted with a move-

able piston. Using each of the six points of the kinetic-

molecular theory, discuss how this device compares to a gas

sample in a container. In addition to the marbles obviously

being larger than gas molecules, where does the analogy ﬁt

the kinetic-molecular theory and where does it not?

86. During exercise, our bodies heat up as a consequence of the

increase in metabolism. Using the kinetic-molecular theory,

explain how panting (breathing faster) can help reduce this

increase in temperature.

10.8 Effusion and Diffusion

Skill Review

87. Arrange these gases in order from lowest to highest rate of

effusion.

a. N

b. SO

c. CO

d. Xe

88. Arrange these gases in order from lowest to highest rate of

effusion.

a. O

b. Ar c. F

d. HF

89. A gas effuses at a rate of 0.300 m/s. A second gas has a molar

mass of 2.02 and effuses 6.00 times faster than the ﬁrst gas.

What is the molar mass of the ﬁrst gas?

90. Gas A diffuses 1.47 times as fast as gas B. If gas B has a molar

mass of 54.6 g/mol, what is the molar mass of gas A?

91. If a noble gas diffuses 0.317 times as fast as does helium at the

same pressure, which noble gas is it?

92. Whichwould takelonger to effuse, 1.00 mol Xegas or 2.00 mol

gas? Assume both are initially at the same temperature

and pressure. How much faster would 1.00 mol Xe gas effuse

if there were 10.0 mol CO

gas at the same temperature and

pressure?

93. Would it be possible to separate a mixture of propane (C

)

and carbon dioxide (CO

) using Graham’s law of effusion?

Explain your answer.

94. Would it be possible to separate a mixture of diborane

) and acetylene (C

) using Graham’s law of effusion?

Explain your answer.

Chemical Applications and Practice

95. An interesting visual demonstration to show diffusion rates

of gases is to introduce some NH

gas at one end of a hollow

tube, while simultaneously placing some HCl at the other

end. Within a few minutes, the two gases will diffuse

through the air inside the tube and react to form a white

Cl precipitate when they meet. If the tube was 39 cm in

length, approximately where would the white precipitate

form inside the tube?

96. The demonstration described in Problem 95 was repeated,

this time using HF instead of HCl. Where would the precip-

itate form inside the tube?

97. During the Manhattan project, efforts were made to sepa-

rate

238

U from the ﬁssionable isotope

235

U. The uranium

sample was converted to UF

, which is a gas at low pressure.

How much faster does

235

effuse than

238

98. Atmospheric scientists have recently detected very low con-

centrations of a long-lived greenhouse gas containing sulfur

and ﬂuorine, SF

. What is the ratio of the rates of effu-

sion of this gas and of its suspected precursor SF

, a mater-

ial used in high-voltage insulators?

Comprehensive Problems

99. Examine Table 10.2 and answer these questions.

a. Assuming SATP conditions, how many liters of ammonia

gas were produced in 2004?

b. How many liters of oxygen gas at SATP were produced?

c. How many metric tons of hydrogen gas were produced at

SATP?

d. How many moles of N

gas were produced in the United

States in 2004 at SATP?

100. When you exhale, you are releasing carbon dioxide, all of

the nitrogen you inhaled, and unused oxygen gas. Under the

same conditions of temperature and pressure, rank these

three gases in order from

a. least dense to most dense.

b. fastest rate of diffusion to slowest rate of diffusion.

101. During exercise, we breathe faster and obtain more oxygen

for metabolism. If a handball player absorbs oxygen at a rate

of approximately 75 mL per kilogram of body mass per

minute, how many molecules of oxygen would a player with

a mass of 86.8 kg absorb during a 30.0-min match at SATP?

102. Although the ideal gas equation is sufﬁcient for most gas

law calculations, there are times when it needs to be modi-

ﬁed. Explain the speciﬁc physical purpose behind the van

der Waals constants a and b in the modiﬁed equation.

103. Hydrogen gas continues to gain attention as a possible fuel

for modiﬁed cars of the near future. Compare the pressure

Focus Your Learning

439

of H

in a tank with a volume of 25 L that contains 45 g of

at 25°C, using calculations from the ideal gas equation

and the van der Waals modiﬁcation to the gas equation.

104. In a quality test, one tennis ball is ﬁlled with N

gas. Another

tennis ball of the same volume is ﬁlled to the same pressure

as the ﬁrst with air. If the two tennis balls were at the same

temperature, what else would also be the same?

105. If you collected some oxygen in an experiment at 298 K,

how much of a change in the temperature (in kelvins), at

constant pressure, would be needed to quadruple the vol-

ume? If that same sample of oxygen had its temperature

lowered by 25°C, what volume change, at constant pressure,

would be expected?

106. If the CO

trapped above your favorite carbonated drink is

exerting a pressure of 7.2 atm, yet the total pressure above

the aqueous drink is 7.9 atm, what are the partial pressure

and percent of water in the trapped space? (Assume that no

other gas is present.)

107. The helium in a pressure tank at a carnival may be pressur-

ized to 21.0 atm. If the temperature of a 27.0-L tank is

29.5°C, how many grams of helium are in the tank? If it was

your job to inﬂate 1.50-L balloons with the helium, how

many could you inﬂate so that each of them would have

1.11 atm of pressure at 27.5°C?

108. In 2004, 17.7 billion m

of hydrogen gas was produced in

the United States, as shown in Table 10.2. Most of the gas is

delivered via trucks, either as a compressed gas or as a liquid

stored at very low temperature. If, for example, 50% of

the volume of gas produced were liqueﬁed and stored at

20 K, and each delivery truck held 6000 gal of liquid

hydrogen, how many truckloads of hydrogen would be

needed in one year? (Assume the density of liquid hydrogen

is 0.070 g/cm

Thinking Beyond the Calculation

109. In Problem 108 we noted that hydrogen gas can be trans-

ported via tanker trucks either as a compressed gas (at about

400 atm) or as a liquid at 20 K. If you are charged with

deciding whether to ship the hydrogen as a compressed gas

or as a liquid, what chemical, physical, demographic, eco-

nomic, and other considerations would factor into your

decision? Use the Internet and examine how some large gas

production companies make these decisions.

110. Under STP conditions, 100.0 mL of an unknown hydrocar-

bon gas was combusted in excess oxygen. The only products

of the combustion were 300.0 mL of carbon dioxide and

400.0 mL of water vapor.

a. What are the intermolecular forces common to hydro-

carbon gases?

b. Does the behavior of these gases tend to resemble ideal

gas behavior? How might these gases be expected to

deviate from this behavior?

c. How many moles of carbon dioxide are produced in the

combustion?

d. What is the formula of the hydrocarbon?

e. If the combustion were performed at 2.50 atm and

500°C, how many liters of water vapor would you expect

to produce?

and HCI are released at opposite ends of a glass tube.

The Chemistry

of Water and

the Nature

of Liquids

Water is a vital part of our lives. In the

developed world, most of us tend to take

clean water for granted. But many people,

like these girls in Nicaragua, are constantly

preoccupied with how they’ll obtain each

day’s supply of water.

440

Contents and Selected Applications

Chemical Encounters: Worldwide Water Use

11.1 The Structure of Water: An Introduction to Intermolecular

Forces

11.2 A Closer Look at Intermolecular Forces

11.3 Impact of Intermolecular Forces on the Physical Properties

of Water, I

11.4 Phase Diagrams

Chemical Encounters: CO

as a Dry Cleaning Solvent

11.5 Impact of Intermolecular Forces on the Physical Properties

of Water, II

11.6 Water: The Universal Solvent

11.7 Measures of Solution Concentration

Chemical Encounters: Composition of Seawater

11.8 The Effect of Temperature and Pressure on Solubility

Chemical Encounters: Impact of the Solubility of Oxygen in Fresh Water

11.9 Colligative Properties

Chemical Encounters: Meeting Municipal Water Needs

Go to college.hmco.com/pic/kelterMEE for online learning resources.

441

The Tampa Bay area has a severe water short-

age brought about by a ﬁvefold increase in popula-

tion since 1950. With over 2.5 million residents and

population growth estimated at over 50,000 per year, even

the abundant rainfall of Florida’s coastal areas cannot keep up

with the skyrocketing demand for water. Tampa Bay is not at all

unique in the world, or even in the United States, in its thirst for this pre-

cious natural resource. Figure 11.1 shows the expected trends in water use by

continent through the year 2025. Most of the world’s increase in water consump-

tion will be driven by population growth and rapid industrialization in Asian coun-

tries. Localized dramatic increases are also predicted in other areas with pockets of

rapid population growth, such as southern Florida and the southwestern states of

Nevada and Arizona. This will further stress the already limited water supplies.

What uses are there for water, and why is it so important in our lives? Domestic

use for washing, cooking, and drinking generates part of the demand, but agricul-

ture accounts for most of our consumption of water (Figure 11.2). Large amounts

of water are also used for industrial production, including the generation of electric-

ity. The patterns of water demand also reveal a great deal about the development of

a country’s economy. Examining these data across the countries of the world, as is

done in Figure 11.3, illustrates that water is used differently in developed countries

than in underdeveloped countries. For example, Figure 11.4 shows that the United

States earmarks most of its water supply for power generation and agriculture.

Application

HEMICAL

ENCOUNTERS:

Worldwide

Water Use

500

1000

1500

2000

2500

3000

1900 1920 1940 1960 1980 2000 2020 2040

Water consumption (km

/year)

Assessment Forecast

FIGURE 11.1

UNESCO data showing the trends

in water consumption.

Assessment Forecast

1900

400

800

1200

1600

2000

2400

2800

3200

1925 1950 1975 2000 2025

Global water consumption (km

)

FIGURE 11.2

Agriculture is a major consumer

of water around the world.

S. America

Africa

Australia

Europe

N. America

Asia

Wor ld

Agricultural

Domestic

Industrial

442 Chapter 11 The Chemistry of Water and the Nature of Liquids

FIGURE 11.3

The use of water can be broken down

into agricultural, industrial, and domestic

(personal) uses.

Water is so fundamental to life that the search for it has extended to the outer

reaches of the solar system. Detecting water elsewhere in the universe could provide

important information about the chemistry of the universe and, perhaps, the origin

of life.

What is it about the structure and properties of water that makes it so pervasive and so

vital to our world?

How can these properties help us to understand water’s many and

varied uses—and assist us in our efforts to provide clean water for places like

Tampa Bay? This chapter focuses on the nature of water. We will compare it to

other liquids in order to highlight its own special character. As you might sense,

water is truly unique, vital in so many ways, and it’s worth knowing why.

11.1 The Structure of Water: An Introduction to Intermolecular Forces 443

FIGURE 11.4

According to the U.S. Geological

Survey (USGS), more than half of

the water used in the United States

in the year 2000 was used for ther-

moelectric power generation and

agriculture.

11.1 The Structure of Water:

An Introduction to Intermolecular Forces

When we get thirsty, we can pour ourselves a glass of water. Huge numbers of

water molecules stream out of the faucet in the

liquid state. They fall into our glass

and, like the molecules of all liquids, conform to the shape of their container. We

can freeze water to make it a

solid, which has its own shape, or boil it, forming a

gas, which completely ﬁlls a container no matter how much is present. To make

our water hot, we pipe the water into a water heater, which may be operated using

another substance vital to our way of life,

natural gas (mostly methane, CH

+

–

+

–

444 Chapter 11 The Chemistry of Water and the Nature of Liquids

Both water and natural gas arrive at our homes via a system of underground

pipes. Even though they possess similar molar masses (18 g/mol for H

O and

16 g/mol for CH

), water is a liquid and methane is a gas. Why is this so? The

properties, chemical behavior, and day-to-day uses of these compounds are based

on their structures. As we’ve noted in previous discussions, the structure of a

molecule determines its properties.

Figure 11.5 provides several representations of a water molecule. Recall from

Chapter 7 that oxygen is substantially more electronegative than hydrogen,

leading to poles of charge on the individual O—H bonds in water. Using the

VSEPR model from Chapter 8, we saw that water is a bent molecule, with an

H—O—H bond angle of 104.5°. This led to our discovery of a net dipole mo-

ment for the polar molecule. Unlike water, methane is made up of atoms with

similar electronegativity values. Additionally, the VSEPR structure of CH

tetrahedral, and the molecule lacks a net dipole moment.

As we’ve seen in previous discussions, opposite charges attract. This is appar-

ent in the structure of table salt (NaCl), in which the positive charge on the

sodium ion is attracted to the negative charge on the chloride ion. But this at-

traction isn’t limited to complete charges. Even slight distortions of electron dis-

tribution give rise to this attraction. If we recall the concept of energy (Chapter 5)

as that which is needed to oppose a natural force, we can deduce that energy is re-

leased when opposite poles attract. This concept helps us to understand why water

is liquid under normal conditions: The hydrogen atoms at positive poles of the

dipole moment on water molecules can interact with the negative poles of oxygen

atoms on other molecules. The resulting release of energy is the key to the stabil-

ity of liquid water at room temperature. This interaction is an example of an

in-

termolecular force

. Unlike water, methane, does not have sufﬁciently strong inter-

molecular forces to exist as a liquid or solid at normal conditions, so it travels to

our homes as a gas.

The result of the attraction between opposite poles in a sample of water is the

interaction of 3 to 6 water molecules with each other at any one time, with an av-

erage of about 4.5. These molecules change partners constantly as they swirl

about in the sample exchanging intermolecular forces of attraction. Although the

number of interactions among the molecules of a substance does say something

about its properties, the relative strength of each individual intermolecular inter-

action is particularly important.

How do intermolecular (between molecules) forces compare with the

intramolecular (within molecules) forces that we call bonds? Consider what hap-

pens when we boil water. We produce water vapor (H

O(g)), but we do not pro-

duce H

and O

gas. In boiling water, the individual intramolecular forces

(bonds) within each water molecule have not been broken. Instead, the intermol-

ecular interactions between molecules have been disrupted.

Why is this so? About

44 kJ of energy is required to convert 1 mol of liquid water to the vapor, via

the breaking of intermolecular attractions. It takes about 940 kJ to break the

O—H bonds within a mole of water. This idea can be reinforced by examining

similar forces in a sample of methane. About 9 kJ of energy is needed to vaporize

H H

(a) (b) (c) (d)

FIGURE 11.5

There are several ways to represent

a water molecule: (a) ball-and-stick

model, (b) Lewis dot structure,

(d) electron dot surface model.

VSEPR structure of methane.

Dipoles attract each other.

Video Lesson: An Introduction

to Intermolecular Forces and

States of Matter

11.2 A Closer Look at Intermolecular Forces 445

Water boils and creates steam.

a mole of liquid methane, compared to 1650 kJ to break the four C—H bonds in

a mole of methane! We can extend this information to make the more general

statement that intermolecular interactions are much weaker than intramolecular

interactions. Less energy is required to boil a liquid than to break it into its

component elements.

11.2 A Closer Look at Intermolecular Forces

Can we use our understanding of intermolecular forces to explain the nature

of liquids other than water? In short, yes; the forces by which molecules come

together as liquids are based on the intermolecular interaction of oppositely

charged poles. They are collectively known as

van der Waals forces, after the Dutch

physicist Johannes Diderik van der Waals, who noted their existence in 1879 and

whose correction for the behavior of real gases we discussed in Chapter 10. van

der Waals forces can be quite weak, as reﬂected in the low boiling points of

methane, –164°C (109 K), and N

, –196°C (77 K), at 1 atm. van der Waals forces

connecting molecules, such as water, are different in nature and much stronger.

The stronger forces result in a much higher boiling point for water (boiling

point = 100°C at 1 atm). What are these forces and how do they vary among

types of molecules?

London Dispersion Forces—Induced Dipoles

In our comparison of water and methane, we noted that the polarity of water ex-

plains why it exists as a liquid at normal temperatures. Yet the ability of nonpolar

methane to liquefy at very low temperatures (boiling point =−164°C) suggests

that there is something even in nonpolar substances that can hold molecules

together. The molecule octane (C

) is nonpolar, yet it is a liquid at room

Five molecules of liquid water interact

with each other. Note the positions of

the hydrogen atoms. They are being

shared with neighboring oxygen atoms.

Video Lesson: Intermolecular

Forces

Octane

δ+δ –

FIGURE 11.6

In this representation, instantaneous induced dipoles

are shown on atoms.

temperature. It has a boiling point of 125.7°C, which is higher than that of our

small polar water molecule. How can we reconcile these two observations?

The physicist Fritz London, who deﬁned a concept we now call

London forces,

gave the explanation to this question in 1929. The key to his concept is a property

called

polarizability, the extent to which electrons can be shifted in their location

by an electric ﬁeld. Polarized electrons produce an

induced dipole, an uneven dis-

tribution of charge caused (or “induced”) by the electric ﬁeld. In the same way,

atoms or molecules can have temporary dipoles induced by instantaneous distor-

tions in neighboring electron positions, as shown in Figure 11.6. The larger the

number of electrons in the system, the larger the polarizability. The larger polar-

izability results in a larger induced dipole moment. In short, stronger attractive

forces between molecules are observed when the induced dipole moment is

larger.

Do London forces explain the differences among boiling points of nonpolar

substances? Table 11.1 shows that the boiling points of substances are quite reli-

ably related to their size, such that, for example, ﬂuorine is a gas at room temper-

ature, whereas bromine is a liquid and iodine is a solid. Similarly, ethane (C

)

is a gas, whereas hexane (C

) is a liquid and tetracosane (C

) is a solid.

However, the structure of a molecule does have some effect on its polarizabil-

ity. From Table 11.1 we see that the boiling point of hexane is 69°C. However,

2,2-dimethylbutane, an isomer of hexane (that is, it has the same chemical

formula but a different structure) shown in Figure 11.7, boils at 50°C, and

2,3-dimethylbutane boils at 58°C.

Why is this so? Polarizability is highly related to

distance. Within a molecule of hexane, electrons are relatively close to those on

446 Chapter 11 The Chemistry of Water and the Nature of Liquids

Hexane

bp 68C

2,2-Dimethylbutane

bp 50C

2,3-Dimethylbutane

bp 58C

CH CH

FIGURE 11.7

The “linear” molecule hexane has a higher boiling point than the two

isomers in which some polarizable electrons are “hidden.”

Boiling Points of Compounds at 1 atm

Molar Mass Boiling Point

Compound Name (g/mol) (°C)

Fluorine 38 –188.1

Chlorine 71 –34.6

Bromine 160 58.8

Iodine 254 184.4

Methane 16 –164

Ethane 30 –88.6

Butane 58 –0.5

Hexane 86 69

2,2-Dimethylbutane 86 50

2,3-Dimethylbutane 86 58

Decane 142 174.1

Tetracosane 339 391

TABLE 11.1

Visualization: Intermolecular

Forces: London Dispersion

Forces

other hexane molecules. On the other hand, some of the carbon atoms in the iso-

mers of hexane are “hidden,” farther away from other neighboring molecules, so

the polarizability of compact molecules is not as great. Moreover, within a series

of molecules of approximately the same molecular mass, those molecules that

can interact more with their neighbors have a higher boiling point. Branched

molecules lack some of this interaction; linear molecules have more such interac-

tion but, because of their ﬂexibility, do not interact perfectly; cyclic molecules

have the most interaction with their neighbors and can stack together like dinner

plates. This greater intermolecular interaction leads to higher boiling points.

The data in Table 11.1 also show us that the molar mass (related to the num-

ber of electrons in a molecule) of nonpolar substances is a major factor in their

boiling points. The impact of molar mass on the boiling point is much greater

than the effect observed in the isomers of hexanes. By comparing the boiling

points of the straight-chain molecules in the table (such as ethane, –88.6°C;

butane, –0.5°C; hexane, 69°C; decane, 174.1°C), we can see that as the molar mass

gets larger, the boiling point increases. In fact, with a sufﬁciently large molar

mass, it is possible for a nonpolar molecule to have a boiling point higher than

that of water. Why is it that water itself has such a high boiling point? Water must

have additional intermolecular forces of attraction that the nonpolar molecules

do not possess.

EXERCISE 11.1 Molar Mass, Structure, London Forces, and Boiling Point

One of the following nonpolar substances is a gas at 200°C. All the others are liquids.

Which one has the lowest boiling point at 1 atm, and why?

First Thoughts

The two most important criteria in determining the relative boiling points of non-

polar substances are their molar masses and their structure. When we evaluate the

structures to see which substance has the lowest boiling point, what we are looking

for is an especially low molar mass, or perhaps a nonlinear structure.

Dodecane

2,2,4,6,6-Pentamethylheptane

CH C

Pentadecane

11.2 A Closer Look at Intermolecular Forces 447

Cyclic molecules

Branched molecules

Linear molecules

Cyclic, linear, and branched molecules

interact through London forces differ-

ently. Those molecules that can get

closer to each other have greater inter-

molecular forces of attraction and

higher boiling points.