An Introduction to Chemistry

510 Chapter 13 Gases

Table 13.1 Percentage of total particles Partial pressure (kPa)

Gases in Dry Air at STP 78.081 79.116
20.948 21.226
Gas
0.934 0.995
nitrogen 0.035 0.035
oxygen 0.002 0.002
argon 0.0005 0.0005
carbon dioxide 0.0001 0.0001
neon 0.0001 0.0001
helium 0.00005 0.00005
methane 0.00005 0.00005
krypton 0.000007 0.000007
nitrous oxide 0.000009 0.000009
hydrogen
ozone
xenon

Objective 20 To get an understanding of the relationship between the total pressure of a mixture
of gases and the partial pressure of each gas in the mixture, let’s picture a luminous
tube being filled with neon and argon gases. Let’s say the neon gas is added to the tube
first, and let’s picture ourselves riding on one of the neon atoms. The particle is moving
rapidly, colliding constantly with other particles and with the walls of the container.
Each collision with a wall exerts a small force pushing out against the wall. The total
pressure (force per unit area) due to the collisions of all of the neon atoms with the
walls is determined by the rate of collision of the neon atoms with the walls and the
average force per collision. The only ways that the neon gas pressure can be changed are
either by a change in the rate of collisions with the walls or by a change in the average
force per collision.

Now consider the effect on gas pressure of the addition of argon gas—at the same
temperature—to the tube that already contains neon. Assuming that the mixture of
gases acts as an ideal gas, there are no significant attractions or repulsions between the
particles, and the volume occupied by the particles is very small. Except for collisions
between particles, each gas particle acts independently of all of the other particles in
the tube. Now that there are also a certain number of argon atoms in the tube, the neon
atoms, including the one we are riding on, have more collisions with other particles
and change their direction of motion more often, but they still collide with the walls
at the same rate as before. If the temperature stays the same, the average velocity of the
neon atoms and their average force per collision with the walls is the same.

If the neon atoms in the mixture are colliding with the walls of the tube at the same
rate and with the same average force per collision as they did when alone, they are
exerting the same pressure against the walls now as they did when alone. Therefore,
the partial pressure of neon in the argon-neon mixture is the same as the pressure that
the neon exerted when it was alone. Assuming ideal gas character, the partial pressure
of any gas in a mixture is the pressure that the gas would exert if it were alone in the
container.

13.4 Dalton’s Law of Partial Pressures 511

Because the argon atoms in the mixture are also acting independently, the partial
pressure of the argon gas is also the same as the pressure it would have if it were alone.
The total pressure in the luminous tube that now contains both neon gas and argon gas
is equal to the sum of their partial pressures (Figure 13.11).

Ptotal = P Ne + P Ar

Figure 13.11
Dalton’s Law of Partial
Pressures
The total pressure of a mixture of
ideal gases is equal to the sum of
the partial pressures of each gas.

1" 1#

1UPUBM 1" 1#

The situation is summarized in Dalton’s Law of Partial Pressures: the total pressure
of a mixture of gases is equal to the sum of the partial pressures of all the gases.

1UPUBM 31QBSUJBM Objective 21
Objective 22
Dalton’s law of partial pressures can be combined with the ideal gas equation to

solve many kinds of problems involving mixtures of gases. One way of combining

them begins by rearranging the ideal gas equation to express the partial pressures of

gases in a mixture—for example, the partial pressures of neon gas and argon gas in a

mixture of argon and neon gases:

PV = nRT or 1 O35
7
Therefore,

1 /F O /F35/F 1 "S O "S35"S
7 /F 7 "S

1UPUBM 1 /F 1 "S O / 7 F 3 / 5F / F

512 Chapter 13 Gases

The following sample study sheet describes calculations that can be done using
Dalton’s Law of Partial Pressures.

Sample Study Tip-off – The problem involves a mixture of gases and no chemical reaction. You
Sheet 13.4
Using Dalton’s are asked to calculate a value for one of the variables in the equations below, and you
Law of Partial
Pressures are given (directly or indirectly) values for the other variables.

Objective 21 General Steps - The following steps can be used to work these problems.
Objective 22
Step 1 Assign variables to the values that are given and the value that is unknown.

Step 2 From the following equations, choose the one that best fits the variables
assigned in Step 1.

1UPUBM 31QBSUJBM PS 1UPUBM 3O FBDI HBT
35
7

Step 3 Rearrange the equation to solve for your unknown.

Step 4 Plug in the values for the given properties.

Step 5 Make sure that the equation yields the correct units. Make any necessary unit
conversions.

Step 6 Calculate your answer and report it with the correct units and with the correct
significant figures.

Objective 21 Example 13.9 - Dalton’s Law of Partial Pressures

A typical 100-watt light bulb contains a mixture of argon gas and nitrogen gas. In
a light bulb with a total gas pressure of 111 kPa, enough argon is present to yield a
partial pressure of 102 kPa. What is the partial pressure of the nitrogen gas?

Solution

Step 1 Assign variables.

P T = 111 kPa P Ar = 102 kPa P N2 = ?

Step 2 Our variables fit the general equation

P total = S P partial or P total = P Ar + P N2

Steps 3-6 Solve the equation for the unknown variable, plug in the given values,
check the units, and calculate the answer.

P N2 = P total - P Ar = 111 kPa - 102 kPa = 9 kPa

13.4 Dalton’s Law of Partial Pressures 513

Example 13.10 - Dalton’s Law of Partial Pressures Objective 22

If an 85-mL light bulb contains 0.140 grams of argon and 0.011 grams of nitrogen at
20 °C, what is the total pressure of the mixture of gases?

Solution

Step 1 Although there are no mass variables in our partial pressure equations, we do
know that we can convert mass to moles using molar masses.

O"S NPM "S H "S NPM "S NPM "S
H "S

O/ NPM / H / NPM / NPM /
H /

V = 85 mL T = 20 °C + 273.15 = 293 K

Step 2 Our variables fit the following form of Dalton’s Law of Partial Pressures.

1UPUBM 3O FBDI HBT
35 1UPUBM O"S O /
35
7 7

Steps 3-6 We do not need to rearrange the equation algebraically, but when we plug
in our values with their units, we see that we need to convert 85 mL to liters
to get our units to cancel correctly.

1UPUBM NPM "S NPM /
-

514 Chapter 13 Gases

SPECIAL TOPIC 13.1 A Greener Way to Spray Paint

United States industries use an estimated 1.5 billion The critical temperature of CO2 is 31 °C. Above
liters of paints and other coatings per year, and much this temperature, carbon dioxide can be compressed
of this is applied by spraying. Each liter sprayed from to a very high pressure and relatively high density
a canister releases an average of 550 grams of volatile supercritical fluid. Like a liquid, the supercritical carbon
organic compounds (VOCs). Some of these VOCs are dioxide will mix with or dissolve the blend of coating
hazardous air pollutants. and low volatility solvent to form a product that is thin
enough to be sprayed easily, in very small droplets.
The mixture that comes out of the spray can has two The supercritical CO2 has a very high volatility, so it
kinds of components: (1) the solids being deposited evaporates from the droplets almost immediately after
on the surface as a coating and (2) a solvent blend that they leave the spray nozzle, leaving a mixture that is
allows the solids to be sprayed and to spread evenly. The thick enough not to run or sag when it hits the surface.
solvent blend must dissolve the coatings into a mixture The mixture is sprayed at temperatures of about 50 °C
that has a thin enough consistency to be easily sprayed. and a pressure of 100 atm (about 100 times normal
But a mixture that is thin enough to be easily sprayed room pressure).
will be too runny to remain in place when deposited
on a surface. Therefore, the solvent blend contains Because carbon dioxide is much less toxic than the
additional components that are so volatile they will VOCs it replaces and because it is nonflammable and
evaporate from the spray droplets between the time the relatively inert, it is much safer to use in the workplace.
spray leaves the spray nozzle and the time the spray hits It is also far less expensive. Moreover, because the
the surface. Still other, slower-evaporating components CO2 can be obtained from the production of other
do not evaporate until after the spray hits the surface. chemicals, the new process does not lead to an increase
They remain in the coating mixture long enough to in carbon dioxide in the atmosphere. In fact, since the
cause it to spread out evenly on the surface. Since the VOCs that it replaces can form CO2 in the atmosphere,
more volatile solvents have escaped, the mixture that hits the supercritical fluid spray process actually leads to a
the surface is thick enough not to run or sag. decrease in atmospheric CO2 levels compared to what
they would be with the traditional spray process.
The Clean Air Act has set strict limitations on the
emission of certain VOCs, so new safer solvents are Source: M. D. Donohue, J. L. Geiger, A. A. Kiamos, and K. A.
needed to replace them. One new spray system has Nielsen, “Reduction of VOC Emissions During Spray Painting.”
been developed that yields a high-quality coating while In Green Chemistry; ed. P. T. Anastas and T. C. Williams (American
emitting as much as 80% fewer VOCs of all types and Chemical Society: Washington DC, 1996), pp. 153-167.
none of the VOCs that are considered hazardous air
pollutants. This system is called the supercritical fluid By taking advantage of the properties of gases
spray process. The solvent mixture for this process still at high temperatures and pressures, scientists
contains some of the slow evaporating solvents that have invented a new, environmentally friendly
allow the coating to spread evenly but replaces the fast spray-painting.
evaporating solvents with high pressure CO2.

Some gases can be converted to liquids at room
temperature by being compressed into a smaller volume,
but for each gas, there is a temperature above which
the particles are moving too fast to allow a liquid to
form no matter how much the gas is compressed. The
temperature above which a gas cannot be liquefied is
called the critical temperature. The very high pressures
that are possible for a gas above its critical temperature
allow the formation of gas systems in which the density
approaches the densities of liquids. At this high pressure
and density, the substance takes on characteristics of
both gases and liquids and is called a supercritical fluid.

13.4 Dalton’s Law of Partial Pressures 515

SPECIAL TOPIC 13.2 Green Decaf Coffee

Caffeine, the primary stimulant found naturally in and pressure. It penetrates the beans more than a typical
coffee, tea, and chocolate, is artificially added to many liquid would, and then, like a liquid, dissolves the
other products, such as soft drinks, over the counter caffeine. When the supercritical carbon dioxide is later
stimulants, cold remedies, and pain relievers. It is the separated from the beans, it carries about 97% of the
world’s most widely used drug. Different types of coffee caffeine along with it. Any residue of carbon dioxide
contain different amounts of caffeine, and different remaining on the beans is quickly lost as gaseous CO2.
brewing techniques lead to further variations in content. The workers employed in this method of decaffeination
Depending on the type of coffee (an arabica coffee are safer without the contact with dichloromethane, and
contains about half the caffeine of a robusta) and the the coffee drinkers do not have to worry about the very
fineness of the grounds, the drip method of brewing slight residue of dichloromethane that remains when the
leads to 110 to 150 mg of caffeine per cup. Instant coffee is decaffeinated using the direct-contact method.
coffee has 40-108 mg of caffeine per cup.
The average cup of coffee contains from 40 All of these products contain caffeine.
to 150 mg of caffeine.

Some people like the taste of coffee and
the rituals associated with it (the customary
cup of coffee after dinner) but do not like
the effects of the caffeine. They may choose
to drink decaffeinated coffee, which has 2-4
mg of caffeine per cup. A coffee must have
at least 97% of its caffeine removed to be
considered decaffeinated coffee.

One of the commercial methods for
decaffeinating coffee is the direct-contact method. The
unroasted coffee beans are first softened with
steam and then brought in direct contact with
a decaffeinating agent, such as dichloromethane
(most often called methylene chloride in
this context), CH2Cl2. The caffeine dissolves
in the dichloromethane, after which the
dichloromethane/caffeine solution is removed
from the beans.

There is some evidence that dichloromethane is
carcinogenic. According to the United States Food
and Drug administration (FDA), most
decaffeinated coffee has less than 0.1
parts per million (ppm) of residual
dichloromethane, which is 100 times
less than the maximum level of 10 ppm
allowed by the FDA. Despite this low
residual concentration of dichloromethane,
there is still some concern about a possible health
hazard, so substitutes are being sought for many of its
applications.

One alternative to the current methods of
decaffeination uses supercritical carbon dioxide.
The unroasted beans are softened by steam and then
immersed in carbon dioxide at very high temperature

516 Chapter 13 Gases

Chapter Ideal gas model The model for gases that assumes (1) the particles are point-masses
Glossary (they have mass but no volume) and (2) there are no attractive or repulsive forces
between the particles.
Chapter
Objectives Ideal gas A gas for which the ideal gas model is a good description.
Pressure Force per unit area.
Boyle’s Law The pressure of a gas is inversely proportional to the volume it occupies

if the number of gas particles and the temperature are constant.
Gay-Lussac’s Law The pressure of an ideal gas is directly proportional to the Kelvin

temperature of the gas if the volume and the number of gas particles are constant.
Charles’ Law Volume and temperature are directly proportional if the number of gas

particles and pressure are constant.
Avogadro’s Law Volume and the number of gas particles are directly proportional if

the temperature and pressure are constant.
Universal gas constant, R The constant in the ideal gas equation.
Partial pressure The portion of the total pressure that one gas in a mixture of gases

contributes. Assuming ideal gas character, the partial pressure of any gas in a
mixture is the pressure that the gas would yield if it were alone in the container.
Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to
the sum of the partial pressures of each gas.

You can test yourself on the glossary terms at the textbook’s Web site.

The goal of this chapter is to teach you to do the following.
1. Define all of the terms in the Chapter Glossary.
Section 13.1 Gases and Their Properties
2. For a typical gas, state the percentage of space inside a gas-filled container that is

occupied by the gas particles themselves.
3. State the average distance traveled between collisions by oxygen molecules, O2, at

normal room temperature and pressure.
4. Write the key assumptions that distinguish an ideal gas from a real gas.
5. Describe the process that gives rise to gas pressure.
6. State the accepted SI unit for gas pressure.
7. Convert between the names and abbreviations for the following pressure units:

pascal (Pa), atmosphere (atm), millimeter of mercury (mmHg), and torr.
8. Convert a gas pressure described in pascals (Pa), atmospheres (atm), millimeters

of mercury (mmHg), or torr to any of the other units.
9. Convert between the names and variables used to describe pressure (P),

temperature, (T ), volume (V ), and moles of gas (n).
10. For each of the following pairs of gas properties, describe the relationship

between the properties, describe a simple system that could be used to
demonstrate the relationship, and explain the reason for the relationship:
(a) volume and pressure when the number of gas particles and temperature are
constant, (b) pressure and temperature when volume and the number of gas
particles are constant, (c) volume and temperature when pressure and the number
of gas particles are constant, (d) the number of gas particles and pressure when
volume and temperature are constant, and (e) the number of gas particles and
volume when pressure and temperature are constant.

Review Questions 517

11. Explain why decreased volume for the gasoline-air mixture in the cylinders of
a gasoline engine, increased number of gas particles, and increased temperature
lead to an increase in pressure in the cylinders.

12. Explain why air moves in and out of our lungs as we breathe.
13. Explain why balloons expand as they rise into the atmosphere.

Section 13.2 Ideal Gas Calculations
14. Write at least one value for the universal gas constant, R, including units.
15. Given values for three of the following four variables, calculate the fourth: P, V,

n, and T.
16. Given values for four out of the following five variables, calculate the fifth: P, V ,

g, M, and T.
17. Given the pressure and temperature of a gas, calculate its density.
18. Given (directly or indirectly) values for seven out of the following eight variables,

calculate the eighth: P 1, V 1, n 1, T1, P 2, V 2, n 2, and T2.
Section 13.3 Equation Stoichiometry and Ideal Gases
19. Convert between volume of gas and moles of gas in an equation stoichiometry

problem using either the molar volume at STP, the ideal gas equation, or R as a
conversion factor.

Section 13.4 Dalton’s Law of Partial Pressures
20. Explain why the total pressure of a mixture of ideal gases is equal to the sum of

the partial pressures of each gas.
21. Given all but one of the following properties for a mixture of gases, calculate the

one not given: total pressure of the mixture of gases and partial pressure of each
gas.
22. Given all but one of the following properties for a mixture of gases, calculate the
one not given: total pressure of the mixture of gases, mass or moles of each gas,
temperature, and volume of the container.

1. Describe the particle nature of gases. Review
2. What is 265.2 °C on the Kelvin scale? Convert 565.7 K into °C. Questions
3. About 55% of industrially produced sodium sulfate is used to make detergents. It

is made from the following reaction.

4NaCl + 2SO2 + 2H2O + O2 → 2Na2SO4 + 4HCl
a. What is the maximum mass of sodium sulfate that can be produced in the
reaction of 745 Mg of sodium chloride with excess SO2, H2O, and O2?
b. What is the maximum mass of sodium sulfate that can be produced in the
reaction of 745 Mg of sodium chloride with 150 Mg H2O and excess SO2
and O2?
c. If 868 Mg Na2SO4 are formed in the reaction of 745 Mg of sodium
chloride with 150 Mg H2O and excess SO2 and O2, what is the percent
yield?
d. What volume of 2.0 M Na2SO4 can be formed from 745 Mg of sodium
chloride with excess SO2, H2O, and O2?

518 Chapter 13 Gases

Key Ideas Complete the following statements by writing one of these words or phrases in each

blank.

0 °C increases
10-7 m Kelvin
0.1% Kelvin temperature
70% liters, L
99.9% molar mass
1 atm molarity
101.325 kPa moles, mol
273.15 K moles of gas
760 mmHg necessary unit conversions
algebra number of collisions
alone number of particles
attractions pascal, Pa
attractive or repulsive point-masses
Avogadro’s Law portion
collide pressure
continuous rapid
correct equation real
degrees Celsius, °C repulsions
direction of motion sum
directly temperature
empty space temperature and volume
force ten
force per unit area velocity
greater force volume

4. Under typical conditions, the average distance between gas particles is about
_____________ times their diameter.

5. Because of the large distances between gas particles, the volume occupied by the
particles themselves is very small compared to the volume of the _____________
around them.

6. For a gas at room temperature and pressure, the gas particles themselves occupy
about _____________ of the total volume. The other _____________ of the
total volume is empty space (whereas in liquids, about _____________ of the
volume is occupied by particles).

7. Because of the large distances between gas particles, there are very few
_____________ or _____________ between them.

8. The particles in a gas are in _____________ and _____________ motion.
9. As the temperature of a gas increases, the particles’ velocity _____________.
10. The particles in a gas are constantly colliding with the walls of the container

and with each other. Because of these collisions, the gas particles are constantly
changing their _____________ and their _____________.
11. Oxygen, O2, molecules at normal temperatures and pressures move an average of
_____________ between collisions.
12. The particles of an ideal gas are assumed to be _____________, that is, particles
that have a mass but occupy no volume.

Key Ideas 519
13. There are no _____________ forces at all between the particles of an ideal gas.
14. The ideal gas model is used to predict changes in four related gas properties:

_____________, _____________, _____________, and _____________.
15. Volumes of gases are usually described in _____________ or cubic meters, m3,

and numbers of particles are usually described in _____________.
16. Although gas temperatures are often measured with thermometers that report

temperatures in _____________ scientists generally use _____________
temperatures for calculations.
17. Each time a gas particle collides with and ricochets off one of the walls of its
container, it exerts a(n) _____________ against the wall. The sum of the forces
of these ongoing collisions of gas particles against all the container’s interior walls
creates a continuous pressure upon those walls.
18. The accepted SI unit for gas pressure is the _____________.
19. _____________ gases deviate somewhat from predicted behavior of ideal gases.
20. The observation that the pressure of an ideal gas is inversely proportional to the
volume it occupies if the _____________ and the temperature are constant is a
statement of Boyle’s Law.
21. When the volume of a chamber that contains a gas decreases but the number of
gas particles remains constant, there is an increase in the concentration (number
of gas particles per liter) of the gas. This leads to an increase in the number of
particles near any given area of the container walls at any time and to an increase
in the number of collisions against the walls per unit area in a given time. More
collisions mean an increase in the _____________, or pressure, of the gas.
22. The pressure of an ideal gas is directly proportional to the _____________ of the
gas if the volume and the number of gas particles is constant. This relationship is
sometimes called Gay-Lussac’s Law.
23. Increased temperature means increased motion of the particles of a gas. If the
particles are moving faster in the container, they will _____________ with the
walls more often and with _____________ per collision. This leads to a greater
overall force pushing on the walls and to a greater force per unit area or pressure.
24. For an ideal gas, volume and temperature described in kelvins are
_____________ proportional if the number of gas particles and pressure are
constant. This is a statement of Charles’ Law.
25. The increase in the number of gas particles in a constant-volume container leads
to an increase in the _____________ with the walls per unit time. This leads to
an increase in the force per unit area–that is, to an increase in gas pressure.
26. If the _____________ of an ideal gas are held constant, the number of particles
in a container and the gas pressure are directly proportional.
27. The relationship between the number of particles of an ideal gas and volume is
summarized by _____________, which states that the volume and the number
of gas particles are directly proportional if the temperature and pressure are
constant.
28. It is always a good idea to include the units in a solved equation as well as the
numbers. If the units cancel to yield a reasonable unit for the unknown property,
you can feel confident that you picked the _____________, that you did the
_____________ correctly to solve for your unknown, and that you have made
the _____________.

520 Chapter 13 Gases

29. STP stands for “standard temperature and pressure.” Standard temperature is
_____________, or _____________, and standard pressure is _____________,
or _____________, or _____________.

30. There are three different ways to convert between a measurable property and
moles in equation stoichiometry problems. For pure liquids and solids, we can
convert between mass and moles, using the _____________ as a conversion
factor. For solutions, _____________ provides a conversion factor that enables
us to convert between moles of solute and volume of solution.

31. The _____________ of the total pressure that one gas in a mixture of gases
contributes is called the partial pressure of the gas.

32. Assuming ideal gas character, the partial pressure of any gas in a mixture is the
pressure that the gas would exert if it were _____________ in the container.

33. Dalton’s law of partial pressures states that the total pressure of a mixture of gases
is equal to the _____________ of the partial pressures of gas all the gases.

Chapter Section 13.1 Gases and their Properties
Problems 34. For a gas under typical conditions, what approximate percentage of the volume is

Objective 2 occupied by the gas particles themselves?
35. One of the car maintenance tasks you might do at a service station is to fill your
Objective 3
Objective 4 tires with air. Why are the tires not really full when you get to the desired tire
Objective 5 pressure? Why can you still add more air to the tire?
36. Why is it harder to walk through water than to walk through air?
Objectives 6, & 7 37. What is the average distance traveled between collisions by molecules of oxygen
Objective 8 gas, O2, at room temperature and pressure?
Objective 8 38. What are the key assumptions that distinguish an ideal gas from a real gas?
39. Consider a child’s helium-filled balloon. Describe the process that creates the gas
Objective 10 pressure that keeps the balloon inflated.
40. A TV weather person predicts a storm for the next day and refers to the dropping
barometric pressure to support the prediction. What causes the pressure in the
air?
41. What is the accepted SI unit for gas pressure? What is its abbreviation?
42. The pressure at the center of the earth is 4 × 1011 pascals, Pa. What is this
pressure in kPa, atm, mmHg, and torr?
43. The highest sustained pressure achieved in the laboratory is 1.5 × 107 kilopascals,
kPa. What is this pressure in atm, mmHg, and torr?
44. What does inversely proportional mean?
45. For each of the following pairs of gas properties, describe the relationship
between the properties, describe a simple system that could be used to
demonstrate the relationship, and explain the reason for the relationship:
(a) volume and pressure when number of gas particles and temperature are
constant, (b) pressure and temperature when volume and the number of gas
particles are constant, (c) volume and temperature when pressure and the number
of gas particles are constant, (d) the number of gas particles and pressure when
volume and temperature are constant, and (e) the number of gas particles and
volume when pressure and temperature are constant.

Chapter Problems 521

46. Explain why decreased volume for the gasoline-air mixture in the cylinders of Objective 11
a gasoline engine, increased number of gas particles, and increased temperature
lead to an increase in pressure in the cylinders. Objective 12
Objective 13
47. Ammonia, NH3, is a gas that can be dissolved in water and used as a cleaner.
When an ammonia solution is used to clean the wax off a floor, ammonia gas Objective 15
escapes from the solution, mixes easily with the gas particles in the air, and Objective 15
spreads throughout the room. Very quickly, everyone in the room can smell the Objective 15
ammonia gas. Explain why gaseous ammonia mixes so easily with air. Objective 15

48. After gasoline vapor and air are mixed in the cylinders of a car, the piston in the
cylinder moves up into the cylinder and compresses the gaseous mixture. Explain
why these gases can be compressed.

49. Explain why air moves in and out of our lungs as we breathe.
50. Explain why balloons expand as they rise into the atmosphere.
51. With reference to the relationships between properties of gases, answer the

following questions.
a. In a common classroom demonstration, water is heated in a one gallon can,
covered with a tight lid, and allowed to cool. As the can cools, it collapses.
Why?
b. Dents in ping pong balls can often be removed by placing the balls in hot
water. Why?

52. Use the relationships between properties of gases to answer the following
questions.
a. When a balloon is placed in the freezer, it shrinks. Why?
b. Aerosol cans often explode in fires. Why?
c. Why do your ears pop when you drive up a mountain?

Section 13.2 Ideal Gas Calculations
53. Neon gas produced for use in luminous tubes comes packaged in 1.0-L

containers at a pressure of 1.00 atm and a temperature of 19 °C. How many
moles of Ne do these cylinders hold?
54. Fluorescent light bulbs contain a mixture of mercury vapor and a noble gas such
as argon. How many moles of argon are in a 121-mL fluorescent bulb with a
pressure of argon gas of 325 Pa at 40 °C?
55. A typical aerosol can is able to withstand 10-12 atm without exploding.

a. If a 375-mL aerosol can contains 0.062 moles of gas, at what temperature
would the gas pressure reach 12 atm of pressure?

b. Aerosol cans usually contain liquids as well as gas. If the 375-mL can
described in part (a) contained liquid along with the 0.062 moles of gas,
why would it explode at a lower temperature than if it contained only the
gas? (Hint: What happens to a liquid when it is heated?)

56. Ethylene oxide is produced from the reaction of ethylene and oxygen at
270-290 °C and 8-20 atm. In order to prevent potentially dangerous pressure
buildups, the container in which this reaction takes place has a safety valve set to
release gas when the pressure reaches 25 atm. If a 15-m3 reaction vessel contains
7.8 × 103 moles of gas, at what temperature will the pressure reach 25 atm?
(There are 103 L per m3.)

522 Chapter 13 Gases

Objective 15 57. Bromomethane, CH3Br, commonly called methyl bromide, is used as a soil
fumigant, but because it is a threat to the ozone layer, its use is being phased
Objective 15
Objective 15 out. This colorless gas is toxic by ingestion, inhalation, or skin absorption, so
Objective 15
Objective 16 the American Conference of Government Industrial Hygienists has assigned it
Objective 16
Objective 16 a threshold limit value, or TLV, a maximum concentration to which a worker

may be repeatedly exposed day after day without experiencing adverse effects.

(TLV standards are meant to serve as guides in control of health hazards, not to

provide definitive dividing lines between safe and dangerous concentrations.)

Bromomethane reaches its TLV when 0.028 moles escape into a room that
has a volume of 180 m3 and a temperature of 18 °C. What is the pressure in
atmospheres of the gas under those conditions? (There are 103 L per m3.)

58. Sulfur hexafluoride is a very stable substance that is used as a gaseous insulator for

electrical equipment. Although SF6 is considered relatively safe, it reaches its TLV
(defined in question 57) when 143 moles of it escape into a room with a volume
of 3500 m3 and a temperature of 21 °C. What is its pressure in pascals for these

conditions?
59. Gaseous chlorine, Cl2, used in water purification, is a dense, greenish-yellow

substance with an irritating odor. It is formed from the electrolysis of molten

sodium chloride. Chlorine reaches its TLV (defined in question 57) in a

laboratory when 0.017 moles of Cl2 are released yielding a pressure of
7.5 × 10-4 mmHg at a temperature of 19 °C. What is the volume in liters of the

room?

60. Hydrogen bromide gas, HBr, is a colorless gas used as a pharmaceutical

intermediate. Hydrogen bromide reaches its TLV (defined in question 57) in a

laboratory room when 0.33 moles of it are released yielding a pressure of 0.30 Pa
at a temperature of 22 °C. What is the volume in m3 of the room?
61. Hydrogen sulfide, H2S, used to precipitate sulfides of metals, is a colorless gas
that smells like rotten eggs. It can be detected by the nose at about 0.6 mg/m3. To
reach this level in a 295 m3 room, 0.177 g H2S must escape into the room. What
is the pressure in pascals of 0.177 g H2S that escapes into a 295-m3 laboratory
room at 291 K?

62. Nitrogen dioxide, NO2, is a reddish-brown gas above 21.1 °C and a brown
liquid below 21.1 °C. It is used to make nitric acid and as an oxidizer for rocket

fuels. Its TLV (defined in question 57) is reached when 1.35 g of NO2 escape
into a room that has a volume of 225 m3. What is the pressure in millimeters of
mercury of 1.35 g of NO2 at 302 K in a 225-m3 room?
63. Hydrogen chloride, HCl, is used to make vinyl chloride, which is then used to

make polyvinyl chloride (PVC) plastic. Hydrogen chloride can be purchased in

liquefied form in cylinders that contain 8 lb HCl. What volume in liters will

8.0 lb HCl occupy if the compound becomes a gas at 1.0 atm and 85 °C?

Chapter Problems 523

64. Sulfur dioxide, SO2, forms in the combustion of the sulfur found in fossil fuels Objective 16
such as coal. In the air, SO2 forms SO3, which dissolves in water to form sulfuric
acid. Thus SO2 is a major contributor to acid rain as well as a strong eye and lung Objective 16
irritant. The 1990 amendments to the Clean Air Act of 1967 call for a reduction Objective 16
in sulfur dioxide released from power plants to 10 million tons per year. This is Objective 16
about one-half the emission measured in 1990. The US atmospheric standard for Objective 16
SO2 is 0.35 mg/m3, which would yield a pressure due to sulfur dioxide of 0.013 Objective 16
Pa at 17 °C. To what volume in m3 must 1.0 × 107 tons of SO2 expand in order Objective 16
to yield a pressure of 0.013 Pa at 17 °C? (There are 2000 lb per ton.)

65. The bulbs used for fluorescent lights have a mercury gas pressure of 1.07 Pa at
40 °C. How many milligrams of liquid mercury must evaporate at 40 °C to yield
this pressure in a 1.39-L fluorescent bulb?

66. Low-pressure sodium lamps have a sodium vapor pressure of 0.7 Pa at 260 °C.
How many micrograms of Na must evaporate at 260 °C to yield a pressure of
0.7 Pa in a 27-mL bulb?

67. Flashtubes are light bulbs that produce a high-intensity, very short duration flash
of light. They are used in photography and for airport approach lighting. The
xenon gas, Xe, in flashtubes has large atoms that can quickly carry the heat away
from the filament to stop the radiation of light. At what temperature does
1.80 g Xe in a 0.625-L flashtube have a pressure of 75.0 kPa?

68. As part of the production of vinyl chloride—which is used to make polyvinyl
chloride (PVC) plastic—1,2-dichloroethane, ClCH2CH2Cl, is decomposed to
form vinyl chloride and hydrogen chloride. If 515 kg of 1,2-dichloroethane gas
are released into a 96 m3 container, at what temperature will the pressure become
3.4 atm?

69. When 0.690 g of an unknown gas are held in an otherwise-empty 285-mL
container, the pressure is 756.8 mmHg at 19 °C. What is the molecular mass of
the gas?

70. A 1.02 g sample of an unknown gas in a 537-mL container generates a pressure
of 101.7 kPa at 298 K. What is the molecular mass of the gas?

71. Consider two helium tanks used to fill children’s balloons. The tanks have equal
volume and are at the same temperature, but tank A is new and therefore has a
higher pressure of helium than tank B, which has been used to fill many balloons.
Which tank holds gas with a greater density, tank A or tank B? Support your
answer.

72. Two hot-air balloons have just been launched. They have equal volume and
pressure, but balloon A has a more efficient heating system, so the gas in balloon
A is at a higher temperature than the gas in balloon B. Which balloon holds
gas with a greater density, balloon A or balloon B? Support your answer. Which
balloon do you think will rise faster?

524 Chapter 13 Gases

Objective 17 73. Butadiene, CH2CHCHCH2, a suspected carcinogen, is used to make plastics
Objective 18 such as styrene-butadiene rubber (SBR). Because it is highly flammable and
Objective 18 forms explosive peroxides when exposed to air, it is often shipped in insulated
Objective 18 containers at about 2 °C. What is the density of butadiene when its pressure is
Objective 18 102 kPa and its temperature is 2 °C?

74. Picture a gas in the apparatus shown in Figure 13.2. If the temperature and the
number of gas particles remain constant, how would you change the volume to
increase the pressure of the gas?

75. In order to draw air into your lungs, your diaphragm and other muscles contract,
increasing the lung volume. This lowers the air pressure of the lungs to below
atmospheric pressure, and air flows in. When your diaphragm and other muscles
relax, the volume of the lungs decreases, and air is forced out.
a. If the total volume of your lungs at rest is 6.00 L and the initial pressure
is 759 mmHg, what is the new pressure if the lung volume is increased to
6.02 L?
b. If the total volume of your lungs at rest is 6.00 L and the initial pressure is
759 mmHg, at what volume will the pressure be 763 mmHg?

76. Dry air is 78% nitrogen gas, 21% oxygen gas, and 0.9% argon gas. These gases
can be separated by distillation after the air is first converted to a liquid. In one of
the steps in this process, filtered air is compressed enough to increase the pressure
to about 5.2 atm.
a. To what volume must 175 L of air at 1.0 atm be compressed to yield a
pressure of 5.2 atm?
b. The argon derived from the distillation of air can be used in fluorescent
light bulbs. When 225 L of argon gas at 0.997 atm are distilled from the air
and allowed to expand to a volume of 91.5 m3, what is the new pressure of
the gas in pascals?

77. A scuba diver has 4.5 L of air in his lungs 66 ft below the ocean surface, where
the pressure is 3.0 atm. What would the volume of this gas be at the surface,
where the pressure is 1.0 atm? If the diver’s lungs can hold no more than 7.0 L
without rupturing, is he in trouble if he doesn’t exhale as he rises to the surface?

78. Picture a gas in the apparatus shown in Figure 13.2. If the volume and
temperature remain constant, how would you change the number of gas particles
to increase the pressure of the gas?

79. Picture a gas in the apparatus shown in Figure 13.2. If the pressure and
temperature remain constant, how would you change the number of gas particles
to increase the volume of the gas?

80. A balloon containing 0.62 moles of gas has a volume of 15 L. Assuming constant
temperature and pressure, how many moles of gas does the balloon contain when
enough gas leaks out to decrease the volume to 11 L?

81. Picture a gas in the apparatus shown in Figure 13.2. If the volume and the
number of gas particles remain constant, how would the temperature have to
change to increase the pressure of the gas?

Chapter Problems 525

82. Consider a weather balloon with a volume on the ground of 113 m3 at 101 kPa Objective 18
and 14 °C. Objective 18
a. If the balloon rises to a height where the temperature is –50 °C and the Objective 18
pressure is 2.35 kPa, what will its volume be? (In 1958, a balloon reached
101,500 ft or 30.85 km, the height at which the temperature is about –50 Objective 19
°C.) Objective 19
b. When the balloon rises to 10 km, where the temperature is –40 °C, its
volume increases to 129 m3. If the pressures inside and outside the balloon
are equal, what is the atmospheric pressure at this height?
c. If the balloon has a volume of 206 m3 at 25 km, where the pressure is
45 kPa, what is the temperature at this height?

83. Ethylene oxide, which is used as a rocket propellant, is formed from the reaction
of ethylene gas with oxygen gas.
a. After 273 m3 of ethylene oxide are formed at 748 kPa and 525 K, the gas is
cooled at constant volume to 293 K. What will the new pressure be?
b. After 273 m3 of ethylene oxide at 748 kPa and 525 K is cooled to 293 K, it
is allowed to expand to 1.10 × 103 m3. What is the new pressure?
c. What volume of ethylene gas at 293 K and 102 kPa must be compressed to
yield 295 m3 of ethylene gas at 808 kPa and 585 K?

84. The hydrogen gas used to make ammonia can be made from small hydrocarbons
such as methane in the so-called steam-reforming process, run at high
temperature and pressure.
a. If 3.2 atm of methane at 21 °C are introduced into a container, to what
temperature must the gas be heated to increase the pressure to 12 atm?
b. If 4.0 × 103 L of methane gas at 21 °C is heated and allowed to expand
at a constant pressure, what will the volume become in m3 when the
temperature reaches 815 °C?
c. What volume of methane gas at 21 °C and 1.1 atm must be compressed to
yield 1.5 × 104 L of methane gas at 12 atm and 815 °C?

Section 13.3 Equation Stoichiometry and Ideal Gases

85. Chlorine gas is used in the production of many other chemicals, including
carbon tetrachloride, polyvinyl chloride plastic, and hydrochloric acid. It is
produced from the electrolysis of molten sodium chloride. What minimum mass
of sodium chloride in megagrams is necessary to make 2.7 × 105 L of chlorine gas
at standard temperature and pressure, STP?
/B$M M
& M F D U S P M Z T J T /B M
$M H


86. Sulfur is used in the production of vulcanized rubber, detergents, dyes,
pharmaceuticals, insecticides, and many other substances. The final step in the
Claus method for making sulfur is
SO2( g ) + 2H2S( g ) → 3S(s) + 2H2O(l )
a. What minimum volume of sulfur dioxide at STP would be necessary to
produce 82.5 kg of sulfur?
b. What maximum mass of sulfur in kilograms could be produced by the
reaction of 1.3 × 104 m3 of SO2( g ) with 2.5 × 104 m3 of H2S( g ), both at
STP?

526 Chapter 13 Gases

Objective 19 87. Air bags in cars inflate when sodium azide, NaN3, decomposes to generate
Objective 19 nitrogen gas. How many grams of NaN3 must react to generate 112 L N2 at
Objective 19 121 kPa and 305 K?
Objective 19
2NaN3(s) → 2Na(s) + 3N2( g )
Objective 19
88. When natural gas is heated to 370 °C, the sulfur it contains is converted to
Objective 19 hydrogen sulfide. The H2S gas can be removed by a reaction with sodium
Objective 19 hydroxide that forms solid sodium sulfide and water.

H2S + 2NaOH → Na2S + 2H2O

What volume of H2S( g ) at 370 °C and 1.1 atm can be removed by 2.7 Mg
NaOH?
89. The hydrogen chloride gas used to make polyvinyl chloride (PVC) plastic is made
by reacting hydrogen gas with chlorine gas. What volume of HCl(g) at 105 kPa
and 296 K can be formed from 1150 m3 of H2( g ) at STP?

H2( g ) + Cl2( g ) → 2HCl( g )

90. The hydrogen gas needed to make ammonia, hydrogen chloride gas, and
methanol can be obtained from small hydrocarbons such as methane or propane
through the steam-reforming process, conducted at 41 atm and 760-980 °C.
If 2.7 × 107 L of C3H8(g) at 810 °C and 8.0 atm react in the first step of the
process, shown below, what is the maximum volume in liters of CO(g) at STP
that can form?

C3H8 + 3H2O → 3CO + 7H2

3CO + 3H2O → 3CO2 + 3H2

91. Ammonia and carbon dioxide combine to produce urea, NH2CONH2, and
water through the reaction shown below. About 90% of the urea is used to
make fertilizers, but some is also used to make animal feed, plastics, and
pharmaceuticals. To get the optimum yield of urea, the ratio of NH3 to CO2 is
carefully regulated at a 3:1 molar ratio of NH3 to CO2.

2NH3( g ) + CO2( g ) → NH2CONH2(s) + H2O(l )
a. What volume of NH3 at STP must be combined with 1.4 × 104 L CO2 at
STP to yield the 3:1 molar ratio?
b. The concentration of urea in a typical liquid fertilizer is about 1.0 M
NH2CONH2. What minimum volume of CO2 at 190 °C and 34.0 atm is
needed to make 3.5 × 104 L of 1.0 M NH2CONH2?

92. Alka-Seltzer contains the base sodium hydrogen carbonate, NaHCO3, which
reacts with hydrochloric acid in the stomach to yield sodium chloride, carbon
dioxide, and water. What volume of CO2 gas at 104.5 kPa and 311 K would be
formed when 10.0 mL of 1.0 M HCl is mixed with an excess of NaHCO3?

93. Some chlorofluorocarbons, CFCs, are formed from carbon tetrachloride, CCl4,
which can be made from the methane in natural gas, as shown below. What
minimum volume in cubic meters of methane gas at STP must be added to
completely react 1.9 × 106 L of Cl2( g ) at STP?

CH4( g ) + 4Cl2( g ) → CCl4(l ) + 4HCl( g )

Chapter Problems 527

94. The carbon tetrachloride made in the the process described in the previous Objective 19
problem can be used to make CFC-12, CF2Cl2, and CFC-11, CFCl3: Objective 19

3HF( g ) + 2CCl4( g ) → CF2Cl2( g ) + CFCl3( g ) + 3HCl( g ) Objective 19

a. What is the minimum volume of HF(g) in cubic meters at 1.2 atm and
19 °C that would be necessary to react 235 kg CCl4 in this reaction?

b. What is the maximum volume of CFC-12 at 764 mmHg and 18 °C that
can be formed from 724 m3 of hydrogen fluoride gas at 1397 mmHg and
27 °C?

95. Ethylene oxide, C2H4O, is a cyclic compound whose two carbon atoms and
oxygen atom form a 3-membered ring. The small ring is somewhat unstable,
so ethylene oxide is a very reactive substance. It is useful as a rocket propellant,
to sterilize medical plastic tubing, and to make polyesters and antifreeze. It is
formed from the reaction of ethylene, CH2CH2, with oxygen at 270-290 °C and
8-20 atm in the presence of a silver catalyst.

2CH2CH2( g ) + O2( g ) → 2C2H4O( g )

a. What minimum mass in megagrams of liquefied ethylene, CH2CH2, is
necessary to form 2.0 × 105 m3 of ethylene oxide, C2H4O, at 107 kPa and
298 K?

b. Ethylene can be purchased in cylinders containing 30 lb of liquefied
ethylene. Assuming complete conversion of ethylene to ethylene oxide,
how many of these cylinders would be necessary to make 2.0 × 105 m3 of
ethylene oxide at 107 kPa and 298 K?

c. How many liters of ethylene oxide gas at 2.04 atm and 67 °C can be made
from 1.4 × 104 L of CH2CH2 gas at 21 °C and 1.05 atm?

d. How many kilograms of liquefied ethylene oxide can be made from the
reaction of 1.4 × 104 L of CH2CH2 gas 1.05 atm and 21 °C and
8.9 × 103 L of oxygen gas at 0.998 atm and 19 °C?

96. Methanol, CH3OH, a common solvent, is also used to make formaldehyde,
acetic acid, and detergents. It is made from the methane in natural gas in two
steps, the second of which is run at 200-300 °C and 50-100 atm using metal
oxide catalysts:

3CH4( g ) + 2H2O( g ) + CO2( g ) → 4CO( g ) + 8H2( g )

CO( g ) + 2H2( g ) → CH3OH(l )

a. In the second reaction, how many megagrams of liquid methanol can be
produced from 187 m3 of carbon monoxide, CO, at 2.0 × 103 kPa and
500 K?

b. In the second reaction, what is the minimum volume of H2 gas at
4.0 × 103 kPa and 500 K necessary to convert 187 m3 of CO gas at
2.0 × 103 kPa and 500 K?

c. How many megagrams of liquid methanol can be made from the reaction
of 187 m3 of CO( g ) at 2.0 × 103 kPa and 500 K and 159 m3 of H2( g ) at
4.0 × 103 kPa and 500 K?

528 Chapter 13 Gases

Objective 19 97. Rutile ore is about 95% titanium(IV) oxide, TiO2, The TiO2 is purified in the
two reactions shown below. In the first reaction, what minimum volume in m3
Objective 19 of Cl2 at STP is necessary to convert all of the TiO2 in 1.7 × 104 metric tons of
rutile ore to TiCl4?
Objective 20 5J0 $ $M o $ 5J$M $0 $0
Objective 21 5J$M 0 o $ 5J0 $M
Objective 22
98. Acetic acid, CH3CO2H, can be made from methanol and carbon dioxide in a
Objective 21 reaction (shown below) that uses a rhodium/iodine catalyst. What is the percent
Objective 22 yield when 513 kg of liquid CH3CO2H form from the reaction of 309 kg
CH3OH with 543 m3 CO( g ) at 1 atm and 175 °C?
CH3OH + CO → CH3CO2H

Section 13.4 Dalton’s Law of Partial Pressures

99. Explain why the total pressure of a mixture of ideal gases is equal to the sum of
the partial pressures of each gas.

100. A typical 100-watt incandescent light bulb contains argon gas and nitrogen gas.
a. If enough argon and nitrogen are added to this light bulb to yield a partial
pressure of Ar of 98.5 kPa and a partial pressure of N2 of 10.9 kPa, what is
the total gas pressure in the bulb?
b. If a 125-mL light bulb contains 5.99 × 10-4 moles of N2 and 5.39 × 10-4
moles of Ar, what is the pressure in kPa in the bulb at a typical operating
temperature of 119 °C?

101. A typical fluorescent light bulb contains argon gas and mercury vapor.
a. If the total gas pressure in a fluorescent light bulb is 307.1 Pa and the
partial pressure of the argon gas is 306.0 Pa, what is the partial pressure of
the mercury vapor?
b. If a 935-mL fluorescent bulb contains 4108 mg of argon gas and 77 mg of
mercury vapor, what is the total gas pressure in the fluorescent bulb at
313 K?

102. If a 135-mL luminous tube contains 1.2 mg Ne and 1.2 mg Ar, at what
temperature will the total gas pressure be 13 mmHg?

103. The atmosphere of Venus contains carbon dioxide and nitrogen gases. At the
planet’s surface, the temperature is about 730 K, the total atmospheric pressure
is 98 atm, and the partial pressure of carbon dioxide is 94 atm. If scientists
wanted to collect 10.0 moles of gas from the surface of Venus, what volume of
gas should they collect?

104. Natural gas is a mixture of gaseous hydrocarbons and other gases in small
quantities. The percentage of each component varies depending on the source.
Consider a 21 m3 storage container that holds 11.8 kg methane gas, CH4,
2.3 kg ethane gas, C2H6, 1.1 kg propane gas, C3H8, and an unknown amount
of other gases.
a. If the total pressure in the container is 1.00 atm at 21 °C, how many moles
of gases other than CH4, C2H6, and C3H8 are in the container?
b. What percentage of the gas particles in the cylinder are methane molecules?

Chapter Problems 529

Additional Problems
105. One balloon is filled with helium, and a second balloon is filled to the

same volume with xenon. Assuming that the two balloons are at the same
temperature and pressure and that the gases are ideal, which of the following
statements is true?
a. The helium balloon contains more atoms than the xenon balloon.
b. The helium balloon contains fewer atoms than the xenon balloon.
c. The two balloons contain the same number of atoms.
106. Although the temperature decreases with increasing altitude in the troposphere
(the lowest portion of the Earth’s atmosphere), it increases again in the upper
portion of the stratosphere. Thus, at about 50 km, the temperature is back to
about 20 °C. Compare the following properties of the air along the California
coastline at 20 °C and air at the same temperature but an altitude of 50 km.
Assume that the percent composition of the air in both places is essentially the
same.
a. Do the particles in the air along the California coastline and the air at

50 km have the same average kinetic energy? If not, which has particles
with the higher average kinetic energy? Explain your answer.
b. Do the particles in the air along the California coastline and the air at
50 km have the same velocity? If not, which has particles with the higher
average velocity? Explain your answer.
c. Does the air along the California coastline and the air at 50 km have the
same average distance between the particles? If not, which has the greater
average distance between the particles? Explain your answer.
d. Do the particles in the air along the California coastline and the air at 50
km have the same average frequency of particle collisions? If not, which has
the greater frequency of between particle collisions? Explain your answer.
e. Does the air along the California coastline and the air at 50 km have the
same density? If not, which has the higher density? Explain your answer.
f. Does the air along the California coastline and the air at 50 km have the
same gas pressure? If not, which has the higher gas pressure? Explain your
answer.
107. Compare the following properties of air on a sunny 28 °C day at a Florida
beach to the properties of air along the coast of Alaska, where the temperature
is –4 °C. Assume that the barometric pressure is the same in both places and
that the percent composition of the air is essentially the same.
a. Do the air in Florida at 28 °C and the air at –4 °C in Alaska have the same
density? If not, which has the higher density?
b. Do the air in Florida at 28 °C and the air at –4 °C in Alaska have the same
average distance between the particles? If not, which has the higher average
distance between the particles?
108. With reference to the relationships between properties of gases, answer the
following questions.
a. The pressure in the tires of a car sitting in a garage is higher at noon than at
midnight. Why?
b. The pressure in a car’s tires increases as it is driven from home to school.
Why?

530 Chapter 13 Gases

109. With reference to the relationships between properties of gases, answer the
following questions.
a. When the air is pumped out of a gallon can, the can collapses. Why?
b. A child’s helium balloon escapes and rises into the air where it expands and
pops. Why?

110. When people inhale helium gas and talk, their voices sound strange, as if it had
been recorded and played back at a faster rate. Why do you think that happens?

111. If the following reaction is run in a cylinder with a movable piston that allows
the pressure of gas in the cylinder to remain constant, and if the temperature is
constant, what will happen to the volume of gas as the reaction proceeds? Why?

N2( g ) + 3H2( g ) → 2NH3( g )

Discussion Problems

112. Consider two identical boxes, one containing helium gas and one containing
xenon gas, both at the same temperature. Each box has a small hole in the side
that allows atoms to escape. Which gas will escape faster? Why?

113. Is it easier to drink from a straw at the top of Mt Everest or at the seashore in
Monterey, California? What causes the liquid to move up through a straw?

114. Like any other object that has mass, gas particles are attracted to the Earth by
gravity. Why don’t gas particles in a balloon settle to the bottom of the balloon?

115. When you look at a beam of sunlight coming through the window, you can see
dancing particles of dust. Why are they moving? Why don’t they settle to the
floor?

116. Which of the following statements is/are true about a gas in a closed container?
a. When the particles collide, the temperature rises in the container.
b. When more gas is added to the container, the greater number of collisions
between particles leads to an increase in temperature.
c. If the gas is heated to a higher temperature, the number of collisions per
second between particles will increase.

117. Baking powder contains sodium hydrogen carbonate, NaHCO3, and a weak
acid. Why does the reaction between these two components cause a donut to
rise?

118. Use our model for gases to answer the following questions.
a. How does a vacuum cleaner work?
b. Why does a suction cup on a dart cause the dart to stick to a wall? Why
does it help to wet the suction cup first?
c. NASA is developing suction cup shoes for use in space stations. Why will
they work inside the space station but not outside the space station?

119. To ensure that tennis balls in Denver will have the same bounce as tennis balls
along the California coast, manufacturers put different gas pressures in them.
Would you expect the gas pressure to be greater in a tennis ball intended for
Denver or in one for San Francisco? Why?

Chapter Problems 531

120. About 90% of the atoms in the universe are thought to be hydrogen atoms.
Thus hydrogen accounts for about 75% of the mass of the universe. The Earth’s
crust, waters, and atmosphere contain about 0.9% hydrogen. Why is there so
much less hydrogen on the Earth than in the rest of the universe? The planet
Jupiter is thought to consist of about 92% hydrogen. Why is there so much
more hydrogen on Jupiter than on Earth?

121. When a scuba tank is filled from a storage tank, the scuba tank gets warmer
and the temperature of the storage tank goes down. Why?

122. Why does a helium filled balloon that escaped to the ceiling drop to the floor
overnight?

Chapter 14

Liquids: Condensation, Evaporation, and

Dynamic Equilibrium

ver the past weeks, you have seen numerous examples of how chemistry can 14.1 Changing from Gas
deepen your understanding of everyday phenomena. In this chapter, we revisit to Liquid and from
the topic of liquids and the changes they undergo, in order to explain some of Liquid to Gas—
the things you might experience on an unusually warm spring morning. An Introduction to
7:47 a.m. You get out of bed and take a quick shower. As you step out of the shower, Dynamic
you shiver with cold, even though the day is already warm. Equilibrium
7:51 a.m. You turn toward the bathroom mirror to comb your hair, but it’s so steamed
up you can hardly see yourself. What causes water to collect on the mirror’s surface? 14.2 Boiling Liquids
7:54 a.m. The water that dripped onto the floor has almost dried up now, but the water 14.3 Particle-Particle
that collects in the bottom of your toothbrush cup never seems to disappear. Why?
8:01 a.m. Drops of cologne land on the counter. Why do they evaporate so much more Attractions
quickly than water, and even more quickly in the heat from your blow dryer?
8:03 a.m. You notice a can of hair spray sitting on the window ledge in the hot sun. Why does water vapor
You’ve heard that if its temperature gets high enough, the can will explode. What condense on a bathroom
causes that to happen? mirror?
8:04 a.m. Heating the water for your morning tea, you wonder what causes bubbles
to form when the water boils—and why they don’t form until the water becomes
extremely hot. That reminds you—why did it take so long to boil the potatoes during
your backpacking trip to the high Sierras last week?
8:12 a.m. Whew! That’s a lot of wondering for the first few minutes of your day. Why
not relax, drink your tea, and settle down to read this chapter? All of your questions
are about to be answered.

Review Skills

The presentation of information in this chapter assumes that you can already perform
the tasks listed below. You can test your readiness to proceed by answering the Review
Questions at the end of the chapter. This might also be a good time to read the Chapter
Objectives, which precede the Review Questions.

Describe the relationship between the an ionic compound with polyatomic 533
temperature of a substance and the motion of ion(s), a binary covalent compound, a
its component particles. (Section 2.1.) binary acid, or an oxyacid. (Section 5.3.)
Compare the freedom of motion and the Convert between names and chemical
attractions between the component particles in formulas for binary ionic compounds,
solids, liquids, and gases. (Section 2.1.) ionic compounds with polyatomic ion(s),
Given a formula for a compound, classify it binary covalent compounds, binary acids,
as either a molecular compound or an ionic and oxyacids. (Section 5.3.)
compound. (Section 3.2.) Given the formula for a molecule or
Given the names or formulas for two elements, polyatomic ion, draw a reasonable Lewis
decide whether the bond that would form structure for it. (Section 12.2.)
between them would be covalent or ionic. Given the Lewis structure or enough
(Section 3.2.) information to produce it, draw the
Given a name or chemical formula, tell whether geometric sketch of a molecule, including
it represents a binary ionic compound, bond angles. (Section 12.4.)

534 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

14.1 Changing from Gas to Liquid and from Liquid to Gas—An Introduction to
Dynamic Equilibrium

Our discussion of liquids focuses on two opposing processes: condensation, in
which liquids are formed from gases, and evaporation, in which liquids return to
gases. An understanding of these processes will help us understand the dynamics of all
reversible physical and chemical changes.

Objective 2 The Process of Condensation
Objective 2
Objective 2 You have probably noticed that the steam from a hot shower deposits liquid droplets on
all the surfaces of your bathroom, but have you ever thought about what’s happening
on the submicroscopic level of atoms and molecules when this happens? How does
water vapor differ from liquid water, and why does water change from one state to the
other? (Note that we use the term vapor to describe the gaseous form of a substance
that is liquid at normal temperatures and pressures. We also use it to describe gas that
has recently come from a liquid.)

Picture yourself riding on a water molecule in a system in which water vapor is
present at a relatively high temperature. The first image in Figure 14.1 shows how
you might visualize this gas. The spheres in this figure represent the particles of any
substance; they could be atoms or molecules. For our discussion, each of the spheres
represents a water molecule. Initially, your molecule moves in a straight line, with no
significant interactions with other particles. Then, all of a sudden, it collides violently
with a slower-moving water molecule. Because of the collision, your particle changes
direction and slows down, the other molecule speeds up, and both molecules move
off along new, straight-line paths. The contact between the two particles was so brief
that at no time did you detect any attraction or repulsion between them. They simply
collided, bounced off each other, and moved apart.

Now picture the same system as we slowly decrease the vapor’s temperature, reducing
the average velocity of the molecules. The collisions between your molecule and others
decrease in violence, so much so that sometimes two colliding molecules stick together
for awhile, held by a mutual attraction. (One of the goals of this chapter is to describe this
attraction and other types of attractions between particles.) Initially, these attractions
do not last very long. A faster-moving water molecule slams into your pair, knocking
them apart, and all three particles continue on alone. As the temperature decreases
further, however, pairs of molecules are less likely to be knocked apart. Instead, they
begin to form trios and even larger clusters (Figure 14.1, middle image).

The slower the particles move, the harder it becomes for them to escape their mutual
attractions. The molecule clusters grow so large that they fall to the bottom of the
container (or cling to its sides), where they combine to form liquid water (Figure 14.1,
image on the right). This is the process of condensation.

In short, as the gaseous water molecules in the steam from a hot shower cool, they
cluster—or condense—into droplets of liquid water, forming a fog that fills the room.
These droplets then combine with other droplets and collect on every surface in the
bathroom, including your bathroom mirror.

14.1 Changing from Gas to Liquid and from Liquid to Gas—An Introduction to Dynamic Equilibrium 535

Objective 4







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Figure 14.1
Objective 2 The Change from Gas to Liquid

Why does rainwater on a flat street evaporate so much faster than the same amount of Objective 3
water in a deep puddle? Why does nail polish remover evaporate so much more quickly
than water, and why do all liquids evaporate more quickly when they are hot? Many Figure 14.2
factors affect the rate of evaporation, the amount of liquid changing to gas per second. Evaporation
We begin our exploration of them by reviewing some information first presented in
Section 2.1.

To get an idea of the changes taking place when a liquid evaporates, picture yourself
riding on a molecule in a liquid held inside a closed container (Figure 14.2). In a
typical liquid, about 70% of the volume is occupied by particles, so there is not much
empty space for you and your molecule to move into. Your movement is further
hindered by the attractions between your molecule and the other molecules around
it. Nevertheless, you bump and jostle to all parts of the container, colliding with other
particles, changing velocity and direction, breaking old attractions, and making new
ones.

Now picture your particle moving toward the surface of the liquid. If it continues
along the same trajectory when it gets to the surface, its momentum (mass times

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536 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 4 velocity) will take you out of the liquid, into the space above—until the attractions
that hold the particles together in the liquid pull you back. If your particle is moving
fast enough, though, its momentum can take it so far beyond the surface that the
attractions will be broken. In that case, the particle escapes the liquid and joins the
vapor above the liquid (Figure 14.2 on the previous page).

For a particle to escape from the surface of a liquid, it must meet the following
criteria:

The particle must be at the liquid’s surface.
Its direction of motion must take it beyond the liquid’s surface.
Its momentum must be great enough to overcome the backward pull of the
other particles at the surface.

How Evaporation Causes Cooling Objective 5

The fastest water Particles must have a relatively high velocity in order to break the attractions holding
molecules are them in the liquid. Particles with relatively low velocity do not have enough momentum
escaping from the to escape. Thus fast moving particles escape, and slow moving particles do not. After the
water on his skin. The more rapidly moving particles have escaped, the particles left in the liquid have a lower
average velocity, and average velocity, and lower average velocity means lower temperature. In other words,
thus the temperature, evaporation lowers a liquid’s temperature. We experience this change in temperature
of the remaining water when we step out of a shower or a swimming pool.
molecules is lower.
Rate of Evaporation
Objective 6
The rate of evaporation—the number of particles moving from liquid to gas per
Objectives 7 & 8 second—depends on three factors:

Water evaporates much Surface area of the liquid
more quickly from a flat Strength of attractions between the particles in the liquid
street than from a deep Temperature of the liquid
pothole. Greater surface area means more particles at the surface of a liquid, which leads to a
greater rate of evaporation. Picture two identical glasses of water. One glass is left on
a table, and the second glass is emptied onto the floor. At the same temperature, the
percentage of particles with the momentum needed to escape from the surface is the
same for both samples of water, but the only particles that get an opportunity to escape
are particles at or near the surface. Because there are a lot more particles at the surface
of the water on the floor, the number of particles that escape from that water into
the gas phase is much greater than for the water in the glass. This explains why rain
evaporates so much faster from a flat street than from a deep puddle.
Weaker attractions between particles also lead to a higher rate of evaporation. Consider
the differences between liquid acetone and liquid water. Acetone, CH3COCH3, is a
common laboratory solvent and the main ingredient in many nail polish removers. If
you have ever used it, you probably noticed that it evaporates much more rapidly than
water. The attractions between acetone molecules are weaker than those between water
molecules, so it is easier for particles to escape from the surface of liquid acetone than
from liquid water. At the same temperature, more molecules will escape per second
from the acetone than from the water (Figure 14.3).

14.1 Changing from Gas to Liquid and from Liquid to Gas—An Introduction to Dynamic Equilibrium 537

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rapidly when the sun comes out.

Objective 9 Figure 14.4
Temperature Influences

Rate of Evaporation

Dynamic Equilibrium Between Liquid and Vapor

In a closed container, evaporation and condensation can take place at the same time. Objective 10
Our model helps us visualize this situation. Objective 10

Picture a container that is partly filled with liquid and then closed tightly so that
nothing can escape. As soon as the liquid is poured in, particles begin to evaporate
from it into the space above at a rate that is dependent on the surface area of the liquid,
the strengths of attractions between the liquid particles, and the temperature. If these
three factors remain constant, the rate of evaporation will be constant.

Imagine yourself riding on one of the particles in the vapor phase above the liquid.
Your rapidly moving particle collides with other particles, with the container walls,
and with the surface of the liquid. When it collides with the surface of the liquid
and its momentum carries it into the liquid, your particle returns to the liquid state.
The number of gas particles that return to liquid per second is called the rate of
condensation.

Now let’s go back to the instant the liquid is poured into the container. If we assume
that the container initially holds no vapor particles, there is no condensation occurring
when the liquid is first added. As the liquid evaporates, however, particles of vapor
gradually collect in the space above the liquid, and the condensation process slowly
begins. As long as the rate of evaporation of the liquid is greater than the rate of
condensation of the vapor, the concentration of vapor particles above the liquid will
increase. However, as the concentration of vapor particles increases, the rate of collisions
of vapor particles with the liquid increases, boosting the rate of condensation.

If there is enough liquid in the container, not all of it will evaporate. Instead, the
rising rate of condensation will eventually become equal to the rate of evaporation.

538 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 11 At this point, for every particle that leaves the liquid, a particle somewhere else in
the container returns to the liquid. Thus there is no net change in the amount of
substance in the liquid state or the amount of substance in the vapor state (Figures
14.5 and 14.6). There is no change in volume either (because our system is enclosed in
a container), so there is no change in the concentration of vapor above the liquid and
no change in the rate of collision with the surface of the liquid. Therefore, the rate of
condensation stays constant.

When two opposing rates of change are equal—such as the rate of evaporation
and the rate of condensation in our closed system—we say the system has reached
a dynamic equilibrium. Dynamic equilibrium is found in many of the systems that
you will encounter in chemistry, so it is important to have a general understanding
of the conditions necessary to create it. First, the system must exhibit two ongoing,
opposing changes, from state A to state B and from state B to state A. In our example,
state A is the liquid, state B is the vapor, and the two opposing changes are evaporation
and condensation. For a dynamic equilibrium to exist, the rates of the two opposing
changes must be equal, so that there are constant changes between state A and state
B but no net change in the components of the system. In the dynamic equilibrium of
our liquid-vapor system, the liquid is constantly changing to vapor, and the vapor is
constantly changing to liquid, but no net change in the amounts of liquid and vapor
is observed (Figure 14.6).

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14.1 Changing from Gas to Liquid and from Liquid to Gas—An Introduction to Dynamic Equilibrium 539

An analogy might help with this visualization of dynamic equilibrium. It is common Objective 12
these days for children’s indoor play areas to have a large bin filled with plastic balls. Objective 12
The bin is like a wading pool where the kids do not get wet. Picture two kids in a bin
throwing balls at one another. Most of the balls they throw hit the nets surrounding the
bin, but a few escape through the openings and fall onto the floor. The escape of the
balls can be likened to particles evaporating from the surface of a liquid.

The escaping balls attract the attention of an attendant, who rushes over to return
them to the bin. At first, there are only a few balls on the floor, so the attendant has to
move a relatively long way to get from one ball to another and throw them back in the
bin. This makes the rate of return rather low, and the kids are able to throw the balls
out faster than he can return them, resulting in a steady increase in the number of balls
on the floor. Now there is less distance between the balls, and the attendant is able to
return them to the bin at a faster and faster rate. This situation is like the increasing
rate of condensation that results from an increase in the
concentration of gas above a liquid.

Eventually the attendant is reaching the balls quickly
enough to return them to the bin just as quickly as the
kids can throw them out. At this point, the process
continues at a frantic pace but with no net change in the
numbers of balls in the bin and on the floor. This is like
the dynamic equilibrium reached between the rates of
evaporation and condensation of liquid and vapor in a
closed container. When the rates are equal, there is no net
change in the amount of liquid or vapor in the system.

The dynamic equilibrium lasts until something
happens to disrupt the system in a way that changes one
or both of the rates. For example, the liquid vapor system
would be disrupted if the top were removed from the closed container, allowing vapor
to escape and decreasing the rate of condensation. The plastic ball equilibrium could
be disrupted when a mother comes over to rescue the attendant by convincing the kids
to stop throwing balls and go get ice cream.

Equilibrium Vapor Pressure

When a liquid is placed in a closed container and begins to evaporate, the vapor
exerts a pressure against the container’s walls (and against the surface of the liquid).
When this system reaches a dynamic equilibrium between the rates of evaporation and
condensation, the amount of vapor above the liquid remains constant. According to
the ideal gas equation, the pressure of the vapor (or gas) is dependent on the vapor’s
concentration (moles divided by volume or n/V ) and its temperature:

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If the temperature of the system, as well as the concentration of the vapor, remains
constant, the pressure of the vapor stays constant as well.

The space above a liquid in a closed container usually contains substances other than
the vapor of the evaporating liquid. The pressure exerted by the vapor that has escaped
the liquid is therefore the partial pressure of that substance in the total amount of gas

540 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 13 inside the container. For example, if the container held both air and liquid at the time
we closed it, any vapor that escapes the liquid will mix with the oxygen, nitrogen,
and other gases in the air and contribute only a portion of the total pressure. The
partial pressure of vapor above a liquid in a closed system with a dynamic equilibrium
between the rate of evaporation and the rate of condensation is called the equilibrium
vapor pressure, Pvap.

Different substances have different equilibrium vapor pressures at the same
temperature. For example, the vapor pressure above liquid acetone in a closed container
is higher than for water at the same temperature. Because the attractions between
acetone molecules are weaker than the attractions between water molecules, it is easier
for an acetone molecule to break them and move into the vapor phase. Therefore,
the rate of evaporation from liquid acetone is greater than for water at the same
temperature. When the dynamic equilibrium between evaporation and condensation
for each liquid is reached, the rate of condensation for the acetone is higher than for
water. The rate of condensation is higher because the concentration of vapor above
the liquid is higher. The higher concentration of acetone particles creates a higher
equilibrium vapor pressure. This logic is summarized in Figure 14.7. The weaker the
attractions between particles of a substance, the higher the equilibrium vapor pressure
for that substance at a given temperature.

Figure 14.7 Objective 13
Relative Equilibrium Vapor Pressures

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Objective 14 The equilibrium vapor pressure of a liquid increases with increased temperature.
At a higher temperature, the faster-moving particles can escape from the liquid more
easily, so the rate of evaporation increases, leading to a higher rate of condensation
at equilibrium. To reach this higher rate of condensation, the concentration of vapor
above the liquid must rise to yield a higher rate of collisions with the surface of the
liquid. A higher concentration of vapor leads to a higher vapor pressure (Figure 14.8).
Figure 14.9 shows how the equilibrium vapor pressure varies with temperature for
acetone and water.

Figure 14.8 Objective 14
Effect of Temperature on Equilibrium Vapor Pressure

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14.1 Changing from Gas to Liquid and from Liquid to Gas—An Introduction to Dynamic Equilibrium 541

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can cause an aerosol can to explode. If you leave 5FNQFSBUVSF ¡$

an aerosol can in the hot sun, the temperature
of the substances in the can will go up, and the Figure 14.9
rates of evaporation of the liquids that it contains Equilibrium Vapor Pressure, Pvap, and Temperature
will increase. The concentration of vapor in the
can will rise accordingly, and the vapor pressure
will increase. If the pressure becomes too high,
the can will explode. A typical aerosol can is
built to withstand pressures up to about 10 to
12 times normal room pressure.

SPECIAL TOPIC 14.1 Chemistry Gets the Bad Guys

We’ve all seen it on TV and in the movies… with a puzzling problem. An 8-year-old was
the detective arrives at the crime scene and stalks kidnapped in Knoxville, Tennessee, but,
into the room wearing a trench coat. The area is fortunately, she was able to escape and identify
taped off, and the experts from the crime lab are the car used in the kidnapping. When the car was
already “dusting for fingerprints.” In the movies, checked for fingerprints, the investigators found
this usually means spreading black powder that many latent prints of the car’s owner, but they
will stick to the fingerprints, but in the real could not find any fingerprints of the kidnapped
world, forensic scientists have developed new and better girl. Knowing that this would raise doubts in the minds
ways to make fingerprints visible and help identify the of the jury, they began to study the differences between
“perp.” the fingerprints of children and adults in hopes of finding
an explanation. They found that an adult’s fingerprints
There are three types of fingerprints. If the people contain a relatively high concentration of molecules called
investigating a crime are lucky, the fingerprints will be esters produced by the oil glands on a grown person’s face
either plastic, impressions left in soft material like wax and then transferred to the fingers when they touch the
or paint, or visible, imprints made by blood, dirt, ink, face. These esters are not produced until after puberty,
or grease. However, usually the fingerprints are latent, so they are not found in a child’s fingerprints. The ester
invisible patterns that must be made visible before molecules are very large with strong attractions that lead
they can be of use. Latent fingerprints are composed of to low rates of evaporation. This discovery explained why
perspiration, oils, amino acids, proteins, inorganic salts the car owner’s fingerprints could be found even after the
(such as sodium and potassium chloride), and many other child’s prints had evaporated and helped convince the
substances. In order to make latent fingerprints visible, jury that the car’s owner was guilty.
the investigator covers the area with a substance that Source: Doris R. Kimbrough and Ronald DeLorenzo,
either sticks to them more than to the surfaces they’re “Solving the Mystery of Fading Fingerprints with London
sitting on or, in some cases, reacts chemically with the Dispersion Forces.” Journal of Chemical Education, 75, 10
fingerprint components. (October 1998), pp. 1300-1301.

A 1993 kidnapping case provided forensic chemists

542 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

14.2 Boiling Liquids

Why do bubbles form in very hot water but not in water that is cold or even warm?
Why does it take longer to boil your potatoes on a backpacking trip to the high Sierras?
This section presents a model for visualizing the process of boiling and then uses it to
answer questions such as these.

Objective 15 How Do Bubbles Form in Liquids?

To understand why bubbles form in a boiling liquid, picture yourself riding on particle
1 in the liquid shown in Figure 14.10. Your particle is moving constantly, sometimes
at very high velocity. When it collides with particles 2 and 3, it pushes them aside and
moves in between them, leaving a small space in your wake. If at the same time particle
4 moves across this space and collides with particles 5 and 6, the space will grow in
volume. Collisions such as these are continuously causing spaces to form within the
liquid. These spaces can be thought of as tiny bubbles (Figure 14.10).

Objective 15

Figure 14.10
Spaces in Liquids

Each of these tiny bubbles is enclosed in a spherical shell of liquid particles. The
shell acts as a surface that, except for its shape, is the same as the surface at the top of
the liquid. Particles can escape from the surface (evaporate) into a vapor phase inside
the bubble, and when particles in that vapor phase collide with the surface of the
bubble, they return to the liquid state (condense). A dynamic equilibrium can form in
the bubble between the rate of evaporation and the rate of condensation, just like the
liquid-vapor equilibrium above the liquid in the closed container (Figure 14.11).

14.2 Boiling Liquids 543

Figure 14.11 Objective 15
Bubble in a Liquid Objective 16

Each time a particle moves across a bubble’s interior and collides with the bubble’s
surface, it exerts a tiny force pushing the wall of the bubble outward. All such collisions
with the shell of the bubble combine to produce a gas pressure inside the bubble. This
pressure is the same as the equilibrium vapor pressure for the vapor above the liquid in
a closed container.

There is an opposing pressure pushing inward on the bubble equal to the sum of the
gas pressure on the upper surface of the liquid and the weight of the liquid above the
bubble. You can view this external pressure as competing with the equilibrium vapor
pressure inside the bubble. If the external pressure pushing in on the bubble is greater
than the bubble’s vapor pressure, the liquid particles of the bubble surface are pushed
closer together, and the bubble collapses. If the vapor pressure of the bubble is greater
than the external pressure, the bubble will grow. If the two pressures are equal, the
bubble maintains its volume (Figure 14.12).

Figure 14.12 Objective 16
Bubble formation

544 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 16 Therefore, if the vapor pressure of the bubble is greater than or equal to the external
pressure acting on it, the bubble continues to exist. Because the vapor in bubbles is
less dense than the liquid, the bubble will rise through the liquid and escape from its
upper surface. When this happens, the liquid is boiling. Boiling can be defined as the
conversion of liquid into vapor anywhere in the liquid rather than just at the top surface.
The temperature at which a liquid boils is called the boiling-point temperature, or
often just boiling point.

Remember that an increase in temperature increases the equilibrium vapor pressure,
which is also the pressure of bubbles as they form in the liquid. Below the boiling-
point temperature, the vapor pressure is too low to counteract the greater external
pressure, so our tiny bubbles that are always forming will quickly collapse. They do
not last long enough to move to the surface, so the liquid does not boil. As the liquid
is heated, the vapor pressure of the liquid increases until the temperature gets high
enough to make the pressure within the bubble equal to the external pressure. At
this temperature, the bubbles can maintain their volume, and the liquid boils. The
boiling-point temperature can therefore be defined as the temperature at which the
equilibrium vapor pressure of the liquid becomes equal to the external pressure acting
on the liquid.

Objective 17 Response of Boiling-Point Temperature to External Pressure Changes

If the external pressure acting on a liquid changes, then the internal vapor pressure
needed to preserve a bubble changes, and therefore the boiling-point temperature
changes. For example, if water is placed in a sealed container, and if the pressure above
the water is decreased by pumping some of the gas out of the space above the water,
the water can be made to boil at room temperature (Figure 14.13). If the external
gas pressure is lowered sufficiently, the particles have enough energy even at room
temperature to escape into bubbles and maintain sufficient pressure inside them to
keep them from collapsing (remember, the external pressure pushing in on them is
now lower).

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Objective 18 This helps us to explain why potatoes take longer to cook in boiling water in the
mountains than at the seashore, and why they cook much faster in a pressure cooker.
When pure water begins to boil at a constant external pressure, its temperature levels off
and stays constant at the boiling temperature as long as there is liquid in the container.
(The heat being added to keep the liquid boiling works to increase the potential energy
of the gas compared to that of the liquid, instead of increasing the kinetic energy of the
liquid. If the average kinetic energy of the liquid is constant, the temperature remains
constant.)

14.2 Boiling Liquids 545

At 8840 meters above sea level, on the top of Mount Objective 18
Everest, the atmospheric pressure is about 34 kPa (0.34 atm).
At this pressure, water boils at about 72 °C, and as we have Figure 14.14
just learned, the water remains at that temperature until all Variation in Boiling-Point
of it has boiled away. The potatoes will take much longer Temperature with Changing
to cook at this temperature, which is 28 °C lower than the External Pressure
normal boiling temperature. If we could heat the water in a
pressure cooker, however, the pressure above the water could Objective 18
increase to higher than the atmospheric pressure, causing the
boiling point of the water to increase. For example, water
boils at about 120 °C when the pressure above the water is
202 kPa (2 atm). Potatoes cook much faster at this temperature
(Figure 14.14).

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Because the external pressure determines the temperature at which a liquid boils,
we need to specify the external pressure associated with a reported boiling point. If the
pressure is not specified, it is understood to be 1 atmosphere. The temperature at which
the equilibrium vapor pressure of a liquid equals 1 atmosphere is called its normal
boiling-point temperature or just its normal boiling point. If a liquid has an external

546 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 19 pressure of 1 atmosphere acting on it, the normal boiling point is the temperature at
which the liquid will boil.

For example, if you are told that the boiling point of hexane is 69 °C, you can
assume that this is the temperature at which the equilibrium vapor pressure of the
hexane reaches 1 atm. If the pressure acting on the liquid hexane is 1 atm, it will boil
at 69 °C. If the pressure acting on the hexane is less than 1 atm, it will boil at a lower
temperature. If the pressure acting on the hexane is greater than 1 atm, it will not boil
until it is heated to a temperature above 69 °C.

Relative Boiling-Point Temperatures and Strengths of Attractions

We can predict which of two substances has a higher boiling point by comparing
the relative strengths of the attractions between the particles. We know that increased
strength of attractions leads to decreased rate of evaporation, decreased rate of
condensation at equilibrium, decreased concentration of vapor, and decreased vapor
pressure at a given temperature. This leads to an increased temperature necessary to
reach a vapor pressure of 1 atmosphere. Thus stronger attractions holding the particles
in the liquid lead to higher normal boiling points. For example, because the attractions
between water molecules are stronger than the attractions between acetone molecules,
the normal boiling point of water (100 °C) is higher than the normal boiling point of
acetone (56.5 °C). Figure 14.15 summarizes this reasoning, and Figure 14.16 shows
graphs of the change in equilibrium vapor pressure with changes in temperature for
acetone and water.

Figure 14.15 Figure 14.16 Objective 19
Strengths of Attractions and Boiling-Point Temperatures
Boiling-Point Temperatures for Acetone and Water

*ODSFBTFE TUSFOHUI PG BUUSBDUJPOT

%FDSFBTFE SBUF PG FWBQPSBUJPO "DFUPOF
%FDSFBTFE SBUF PG DPOEFOTBUJPO

BU FRVJMJCSJVN 1WBQ L1B
L1B 8BUFS
-PXFS DPODFOUSBUJPO PG WBQPS OFDFTTBSZ BUN ɨF OPSNBM
UP SFBDI MPXFS SBUF PG DPOEFOTBUJPO
CPJMJOH QPJOU
-PXFS WBQPS QSFTTVSF BU PG BDFUPOF JT
BOZ HJWFO UFNQFSBUVSF ¡$
)JHIFS UFNQFSBUVSF OFDFTTBSZ UP CSJOH UIF ɨF OPSNBM
WBQPS QSFTTVSF UP UIF FYUFSOBM QSFTTVSF CPJMJOH QPJOU
PG XBUFS JT
*ODSFBTFE CPJMJOH QPJOU UFNQFSBUVSF ¡$


5FNQFSBUVSF ¡$

14.3 Particle-Particle Attractions 547

14.3 Particle-Particle Attractions

When you look at a glass of water or a teaspoon of salt, what do you see? A colorless
liquid? A mound of white crystals? A chemist sees much, much more. Chemists may
think of water, for example, as a vast crowd of jostling H2O molecules, constantly
breaking away from and attracting one another as they tumble throughout their
container. They may see a salt crystal as a system of alternating cations and anions
held together by ionic bonds. The purpose of this section is to expand your ability to
visualize the particle nature of substances by introducing you to the ways that particles
are bound together in the liquid form.

Dipole-Dipole Attractions

We know from Chapter 3 that because hydrogen chloride consists of two nonmetallic Objective 20
elements, it is a molecular compound: it is composed of molecules, which are groups of
atoms held together by covalent bonds. Because chlorine atoms attract electrons more
strongly than hydrogen atoms do, the covalent bond in each HCl molecule is a polar
covalent bond. The chlorine atom has a partial minus charge, and the hydrogen atom
has a partial positive charge. A molecule that contains a partial positive charge at one
end and a partial negative charge at the other is called a molecular dipole, or often just
a dipole.

Under normal pressure, HCl gas condenses to a liquid at -84.9 °C. This is the
temperature at which the attractions between the HCl molecules are strong enough
to keep the molecules close together. The attraction between one HCl molecule and
another is a dipole-dipole attraction, which is an attraction between the partial
negative end of one molecule and the partial positive end of another molecule (Figure
14.17).

δ+ δ− δ+ δ− Figure 14.17
( #L ( #L Dipole-Dipole Attractions
Between HCl Molecules

Objective 20

If you were riding on a molecule in a sample of liquid hydrogen chloride, you would
see several other HCl molecules close by, and your particle would be attracted to each
of them. You would move constantly from place to place, breaking old dipole-dipole
attractions and making new ones (Figure 14.18).

D

548 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

When a polar molecular substance, such as hydrogen chloride, is heated to convert
it from liquid to gas, dipole-dipole attractions are broken. The molecules themselves
remain unchanged. For example, when liquid HCl is boiled, the dipole-dipole
attractions between HCl molecules are broken, but the covalent bonds between the
hydrogen atoms and the chlorine atoms within the HCl molecules are unaffected.

Predicting Bond Type

To predict whether molecules are polar or nonpolar, we need first to predict whether
the bonds within them are polar or nonpolar. Remember that atoms form nonpolar
covalent bonds, polar covalent bonds, or ionic bonds depending on the relative electron
attracting ability of the atoms in the bond (Section 3.2). If the two atoms forming a
bond have very different electron-attracting abilities, one or more electrons will be
transferred between the atoms, forming ions and an ionic bond. At the other extreme,
if there is no significant difference in the electron-attracting ability of the two atoms
forming a chemical bond, the bond they form is a nonpolar covalent bond, with no
significant transfer of electrons from one atom to another and therefore no significant
charges on either of the atoms. Polar covalent bonds lie between these two extremes.
The atoms differ just enough in their electron attracting ability to result in a transfer
that is significant but not complete. This leads to partial charges on the atoms and a
polar covalent bond.

Electronegativity, EN, is a term chemists use to describe the electron-attracting
ability of an atom in a chemical bond. Figure 14.19 shows the electronegativities for
many of the elements. The higher an element’s electronegativity, the greater its ability
to attract electrons from other elements. For example, because chlorine has a higher
electronegativity (3.16) than hydrogen (2.20), we expect chlorine atoms to attract
electrons more strongly than hydrogen atoms do. Comparing electronegativities allows
us to predict whether a covalent bond is nonpolar covalent, polar covalent, or ionic.

Figure 14.19
A Table of Average Electronegativities "


" " ) " " " " "


-J #F # $ /0 '

# # # # # # # # # # "M 4J 1 4 $M
/B .H


, $B 4D 5J 7 $S .O 'F $P /J $V ;O (B (F "T 4F #S ,S


3C 4S : ;S /C .P 5D 3V 3I 1E "H $E *O 4O 4C 5F * 9F


$T #B -V )G 5B 8 3F 0T *S 1U "V )H 5M 1C #J 1P "U


'S 3B

Objective 22a If the difference in electronegativity (DEN) between the atoms in the bond is less than
0.4, we expect them to share electrons more or less equally, forming a nonpolar covalent
bond. For example, hydrogen has an electronegativity of 2.20, and carbon has an
electronegativity of 2.55. The difference in electronegativity (DEN) is 0.35, which is

14.3 Particle-Particle Attractions 549

less than 0.4. We expect the bond between carbon and hydrogen to be a nonpolar Objective 22a
covalent bond. Objective 22a

If the difference in electronegativity (DEN) between two atoms is between 0.4 and 1.7,

we expect the bond between them to be a polar covalent bond. For example, hydrogen
has an electronegativity of 2.20, and chlorine has an electronegativity of 3.16. The
difference in electronegativity (DEN) is 0.96, so the bond is a polar covalent bond,
which is sometimes called a bond dipole.

If the difference in electronegativity (DEN) between two atoms is greater than 1.7, we

expect the bond between them to be ionic. For example, sodium has an electronegativity of
0.93, and chlorine has an electronegativity of 3.16. The difference in electronegativity
(DEN) is 2.23, so we predict their bond to be ionic.

There are exceptions to these guidelines. For example, the electronegativity difference
is 1.78 in the H−F bond and 1.79 in the P−F bond. Our electronegativity guidelines
lead us to predict that these would be ionic bonds, but experiments show them to be
covalent bonds. Exceptions such as these will not confuse you if you remember that the
bond between two nonmetallic atoms is always a covalent bond. (Figure 14.20).

Objective 22A Figure 14.20
Identification of Bond Type

$PWBMFOU CPOE *POJD CPOE
OPONFUBM OPONFUBM BMXBZT DPWBMFOU
NFUBM OPONFUBM VTVBMMZ JPOJD


/POQPMBS DPWBMFOU CPOE D D

$&/
1PMBS DPWBMFOU CPOE $&/
$&/

Electronegativities also give us a simple way of predicting which atom in a polar Objective 22b
covalent bond has the partial negative charge and which has the partial positive charge. Objective 22c
In a polar covalent bond, the atom with the higher electronegativity has the partial negative

charge, and the atom with the lower electronegativity has the partial positive charge. For
example, the electronegativity of chlorine (3.16) is higher than for hydrogen (2.20). In
an H−Cl bond, we predict that the chlorine atom attracts electrons more strongly than
the hydrogen atom and gets a partial negative charge. Due to its lower electronegativity,
the hydrogen atom has a partial positive charge.

Similar logic allows us to predict which atom in an ionic bond forms the positive
cation and which forms the negative anion. In an ionic bond between monatomic ions,
the nonmetallic element has the higher electronegativity and forms the anion, and the

metallic element has the lower electronegativity and forms the cation.

550 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 22d We can use another simple guideline to predict which of two bonds is more polar.
The bond with the greater difference in electronegativity (DEN) is likely to be the more polar

bond.

For example, DEN for the H−Cl bond is 0.96. The DEN for the H−F bond is
1.78. Because there is a greater difference between the electron-attracting abilities of
hydrogen atoms and fluorine atoms than between hydrogen atoms and chlorine atoms,
we expect larger partial charges in the H−F bond. That is, we expect the H−F bond to
be more polar than the H−Cl bond.

The following sample study sheet summarizes the guidelines for using electronegativity
to make predictions about bonds.

Sample Study Tip-off – You wish to (1) classify a chemical bond as nonpolar covalent, polar
Sheet 14.1
covalent, or ionic, (2) identify which element in a polar covalent bond is partially
Electronegativity, negative and which is partially positive, (3) identify which element in an ionic bond
Types of forms the anion and which forms the cation, or (4) identify which of two bonds is
Chemical Bonds, more polar.
and Bond
Polarity General Steps

Objective 22 Use the following guidelines to identify a chemical bond as ionic, nonpolar
covalent, or polar covalent.

DEN < 0.4 → nonpolar covalent
DEN 0.4-1.7 → polar covalent
DEN > 1.7 → ionic
Use the following guidelines to identify which element in a polar covalent bond is
partially negative and which is partially positive.
Higher electronegativity → partial negative charge
Lower electronegativity → partial positive charge
Use the following guidelines to identify which element in an ionic bond between
monatomic ions forms the anion and which forms the cation.
Nonmetal, which has a higher electronegativity → anion
Metal, which has the lower electronegativity → cation
Use the following guideline to decide which of two bonds is likely to be more
polar.
The greater the DEN, the more polar the bond.
Example See Examples 14.1 and 14.2.

14.3 Particle-Particle Attractions 551

Example 14.1 - Using Electronegativities Objective 22

Classify the following bonds as nonpolar covalent, polar covalent, or ionic. If a bond
is polar covalent, identify which atom has the partial negative charge and which has
the partial positive charge. If the bond is ionic, identify which element is negative and
which is positive.

a. O bonded to H
b. C bonded to S
c. Li bonded to Cl
d. H bonded to F

Solution

a. Both O and H represent nonmetallic elements, so the bond is covalent. Because
oxygen’s electronegativity is 3.44 and hydrogen’s is 2.20, the difference in
electronegativity (DEN) is 1.24, so this is a polar covalent bond. The higher
electronegativity of the oxygen tells us that oxygen atoms attract electrons more
strongly than hydrogen atoms. Therefore, the oxygen atom in the O−H bond is
d-, and the hydrogen atom in the O−H bond is d+.

b. Both C and S represent nonmetallic elements, so the bond is covalent. The
electronegativity of carbon is 2.55, and the electronegativity of sulfur is 2.58. The
DEN is only 0.03, so the bond is a nonpolar covalent bond.

c. Lithium is a metal, and chlorine is a nonmetal. Bonds between metallic atoms and
nonmetallic atoms are expected to be ionic. Because lithium’s electronegativity is
0.98 and chlorine’s is 3.16, the DEN is 2.18, so again we would predict the bond
to be best described as an ionic bond. The nonmetallic chlorine has the negative
charge, and the metallic lithium has the positive charge.

d. On the basis of the hydrogen’s electronegativity being 2.20 and fluorine’s being
3.98, for a DEN of 1.78, we might expect these elements to form an ionic bond.
However, because both of these elements are nonmetallic, the bond is instead a
very polar covalent bond. The higher electronegativity of the fluorine tells us that
fluorine atoms attract electrons more strongly than hydrogen atoms. The fluorine
atom in the H−F bond is d-, and the hydrogen atom in the H−F bond is d+.

Example 14.2 - Using Electronegativities Objective 22

Which bond would you expect to be more polar, H−S or H−Cl?

Solution
The DEN for the H−S bond is 0.38, and the DEN for the H−Cl bond is 0.96. We
expect the bond with the greater difference, the H−Cl bond, to be more polar. The
Cl in the H−Cl bond would have a larger d- charge than the S in the H−S bond.

552 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 22 Exercise 14.1 - Using Electronegativities
Objective 22
Classify the following bonds as nonpolar covalent, polar covalent, or ionic. If a bond
is polar covalent, identify which atom has the partial negative charge and which has
the partial positive charge. If a bond is ionic, identify which atom is negative and
which is positive.

a. N bonded to H
b. N bonded to Cl
c. Ca bonded to O
d. P bonded to F

Exercise 14.2 - Comparing Bond Polarities

Which bond would you expect to be more polar, P−H or P−F? Why?

Objective 23 Predicting Molecular Polarity

Now that we can predict whether covalent bonds are polar or nonpolar, we can continue
our discussion of polar molecules. Three questions will help you predict whether
substances are composed of polar or nonpolar molecules. First,

Is the substance molecular?

If a substance is composed of all nonmetallic elements and does not contain the
ammonium ion, you can assume that the answer to the first question is yes.

For reasons that we explore below, molecules are polar if they have an asymmetrical
arrangement of polar covalent bonds, so another question is

If the substance is molecular, do the molecules contain polar covalent bonds?

When there are no polar bonds in a molecule, there is no permanent charge
difference between one part of the molecule and another, and the molecule is nonpolar.
For example, the Cl2 molecule has no polar bonds because the two atoms in each
Cl−Cl bond attract electrons equally. It is therefore a nonpolar molecule. None of the
bonds in hydrocarbon molecules, such as hexane, C6H14, are significantly polar, so
hydrocarbons are nonpolar molecular substances. Lastly,

If the molecules contain polar covalent bonds, are these bonds asymmetrically
arranged?

A molecule can possess polar bonds and still be nonpolar. If the polar bonds are
evenly (or symmetrically) distributed, the bond dipoles cancel and do not create a
molecular dipole. For example, the two bonds in a molecule of CO2 are significantly
polar, but they are symmetrically arranged around the central carbon atom. No one
side of the molecule has more negative or positive charge than another side, and so the
molecule is nonpolar:

D
0 $D 0 D

14.3 Particle-Particle Attractions 553

A water molecule is polar because (1) its O−H bonds are significantly polar, and (2) Objective 23
its bent geometry makes the distribution of those polar bonds asymmetrical. The side
of the water molecule containing the more electronegative oxygen atom is partially Objective 24
negative, and the side of the molecule containing the less electronegative hydrogen Objective 25
atoms is partially positive.

) D

D
0 )D

You can see a more complete description of the procedure used to predict whether
molecules are polar or nonpolar at the web address below. For now, it will be useful to
remember that the following substances have polar molecules: water, H2O; ammonia,
NH3; oxyacids (such as nitric acid, HNO3); hydrogen halides (HCl, HF, HBr, and HI);
methanol, CH3OH; and ethanol, C2H5OH. You will also be expected to recognize the
following as composed of nonpolar molecules: the elements composed of molecules
(H2, N2, O2, F2, Cl2, Br2, I2, P4, S8, and Se8); carbon dioxide, CO2; and hydrocarbons
(such as pentane, C5H12, and hexane, C6H14).

A description of a procedure for determining whether molecules are polar
can be found at the textbook’s Web site.

Hydrogen Bonds Objective 26

Now, let’s return to our discussion of attractions between molecules (intermolecular
attractions). Hydrogen fluoride, HF, is used to make uranium hexafluoride, UF6,
a chemical that plays an important role in the production of nuclear reactor fuel.
Chemists have learned that gaseous hydrogen fluoride condenses to a liquid at
19.54 °C, a much higher temperature than the one at which hydrogen chloride
condenses to a liquid (-84.9 °C). This difference suggests that the attractions between
HF molecules are much stronger than the attractions between HCl molecules. (It is
common to consider room temperature to be 20 °C, or 68 °F, so HF is a liquid at just
slightly below room temperature.)

One reason the attractions are so much stronger between HF molecules than
between HCl molecules is that the HF molecules have much higher partial positive and
negative charges on the atoms. The difference in electronegativity between hydrogen
and chlorine is 0.96, while the difference in electronegativity between hydrogen and
fluorine is 1.78. As a result, the H−F bond is far more polar, with a very large partial
positive charge on each hydrogen atom and a very large partial negative charge on
each fluorine atom. These relatively large charges on HF molecules lead to such strong
intermolecular attractions that chemists place them in a special category and call them
hydrogen bonds. Hydrogen bonds are attractions that occur between a nitrogen,
oxygen, or fluorine atom of one molecule and a hydrogen atom bonded to a nitrogen,

554 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

oxygen, or fluorine atom in another molecule. Hydrogen fluoride is the only fluorine
containing compound that is capable of hydrogen bonding (Figure 14.21).

Figure 14.21 Hydrogen bond
Two Ways to Illustrate Hydrogen Bonding in HF
In HF, the hydrogen bond is the attraction between δ+H F δ− δ+H F δ−
the partial negative charge of a fluorine atom in
one HF molecule and the partial positive charge δ+ δ− δ+ δ−
of a hydrogen atom attached to a fluorine atom in
another HF molecule. Fluorine atom

Objective 26

Hydrogen atom

Hydrogen bonds are very important in the chemistry of life. The strands of DNA
that carry our genetic information are held together by hydrogen bonds, and hydrogen
bonds help to give the enzymes that regulate our body chemistry their characteristic
shapes. Hydrogen bonding also explains the physical characteristics of water.

Life on Earth, if it existed at all, would be far different if water were a gas at normal
temperatures, like other similar substances. For example, hydrogen sulfide, H2S (the
substance that gives rotten eggs their distinctive smell) has a structure that is very similar
to water, but it is a gas above -60.7 °C. The fact that water is a liquid suggests that the
attractions between H2O molecules are stronger than the attractions in H2S. Hydrogen
sulfide molecules have relatively weak dipole-dipole attractions between them, but
H2O molecules have strong hydrogen bonds between them owing to the attraction
between the partial negative charge on an oxygen atom of one water molecule and the
partial positive charge on a hydrogen atom of another molecule (Figure 14.22).

Figure 14.22 Hydrogen bond
Hydrogen Bonding Between Water Molecules Hδ+ Hδ+
In water, the hydrogen bond is the attraction O
between the partial negative charge of an δ+ H δ− δH+ O
oxygen atom in one molecule and the partial δ+ δ−
positive charge of a hydrogen atom attached
to an oxygen atom in another molecule. δ+

Objective 26 δ+ Hydrogen atom
Oxygen atom
δ− δ+
δ−

As you discovered in Chapter 3, the three most common alcohols are methanol,
CH3OH, ethanol, C2H5OH, and 2-propanol, C3H7OH.

)

)$ )
) )) )

)$ 0) )$ $ 0) )$ $ 0)

) )) ))
Methanol, CH3OH Ethanol, C2H5OH 2-propanol, C3H7OH

14.3 Particle-Particle Attractions 555

Like water, alcohol molecules contain the O−H bond and therefore experience
hydrogen bonding of their molecules in the liquid and solid form (Figure 14.23).

Hydrogen bond Figure 14.23
Hydrogen Bonding in Methanol
H H D HHC H In methanol, the hydrogen bond is the attraction
C O between the partial negative charge of an
D

556 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

London Forces

Are you old enough to remember tincture of iodine? Before the introduction of

antibacterial ointments and sprays for the treatment of minor cuts and burns, tincture

of iodine, a mixture of iodine and alcohol, was a common disinfectant for the skin.

The pure iodine used to make this solution consists of diatomic molecules, I2, whose
mutual attractions are strong enough to form a solid at room temperature. Because the

only bond within the I2 molecule is the nonpolar I−I bond, we do not expect normal
dipole-dipole attractions between iodine molecules. Why, then, do they attract each

other strongly?

Although we expect the I−I bond to be nonpolar, with a symmetrical distribution

of the electrons, this arrangement is far from static. Even though the most probable

distribution of charge in an isolated I2 molecule is balanced, a sample of iodine that
contains many billions of molecules will always include, by chance, some molecules

whose electron clouds are shifted more toward one iodine atom than toward the other.

Objective 27 The resulting dipoles are often called instantaneous dipoles because they may be short-

lived (Figure 14.25). The constant collisions between molecules contribute to the

formation of instantaneous dipoles. When I2 molecules collide, the repulsion between
their electron clouds will distort the clouds and shift them from their nonpolar state.

An instantaneous dipole can create a dipole in the molecule next to it. For example,

the negative end of one instantaneous dipole will repel the negative electron cloud of a

nonpolar molecule next to it, pushing the cloud toward the far side of the neighboring

molecule. The new dipole is called an induced dipole. The induced dipole can then

induce a dipole in another neighboring molecule. This continues until there are many

polar molecules in the system. The resulting partial charges on instantaneous and

induced dipoles lead to attractions between the opposite charges on the molecules

(Figure 14.25). These attractions are called London dispersion forces, London forces,

or dispersion forces.

Objective 28 The larger the molecules of a substance, the stronger the London forces between

them. A larger molecule has more electrons and a greater chance of having its electron

cloud distorted from its nonpolar shape. Thus instantaneous dipoles
'(BT are more likely to form in larger molecules. The electron clouds in

larger molecules are also larger, so the average distance between the

*ODSFBTJOH TJ[F (BT $ M nuclei and the electrons is greater; as a result, the electrons are held less
PG NPMFDVMFT tightly and shift more easily to create a dipole.
MFBET UP
The increased strength of London forces with increased size of
JODSFBTJOH
TUSFOHUIT PG -JRVJE # S molecules explains a trend in the properties of halogens. Fluorine, F2,
-POEPO GPSDFT and chlorine, Cl2, which consist of relatively small molecules, are gases
at room temperature and pressure. Bromine, Br2, with larger molecules
4PMJE * and therefore stronger London forces between them, is a liquid. Iodine,
I2, with still larger molecules, is a solid.

The effect of molecular size on the strengths of London forces

can also be used to explain the different states observed among the byproducts of

Objective 29 crude oil. Crude oil is a mixture of hydrocarbon molecules that can be separated into

natural gas, gasoline, kerosene, lubricating oils, and asphalt by heating it and isolating

14.3 Particle-Particle Attractions 557

$IBODF PS DPMMJTJPOT DBVTF OPOQPMBS NPMFDVMFT UP GPSN JOTUBOUBOFPVT EJQPMFT

HPFT UP δ+ δ−

/POQPMBS NPMFDVMF *OTUBOUBOFPVT EJQPMF

*OTUBOUBOFPVT EJQPMFT JOEVDF EJQPMFT JO PUIFS OPOQPMBS NPMFDVMFT

δ+ δ− HPFT UP δ+ δ− δ+ δ−

3FQVMTJPO CFUXFFO UIF QBSUJBM OFHBUJWF DIBSHF PG UIF JOTUBOUBOFPVT *OEVDFE EJQPMF
EJQPMF BOE UIF OFHBUJWF DIBSHF PG UIF FMFDUSPOT JO UIF OPOQPMBS
NPMFDVMF QVTIFT UIF FMFDUSPOT JO UIF OPOQPMBS NPMFDVMF UP UIF SJHIU

GPSNJOH BO JOEVDFE EJQPMF
*OEVDFE EJQPMFT DBO JOEVDF EJQPMFT JO PUIFS OPOQPMBS NPMFDVMFT
SFTVMUJOH JO NBOZ
NPMFDVMFT XJUI QBSUJBM DIBSHFT -POEPO GPSDFT BSF UIF BUUSBDUJPOT CFUXFFO UIF QBSUJBM QPTJUJWF
BOE QBSUJBM OFHBUJWF DIBSHFT JO UIFTF JOTUBOUBOFPVT BOE JOEVDFE EJQPMFT

δ+ δ− δ+ δ− Objective 27

-POEPO GPSDF Figure 14.25
London Dispersion Forces

different substances as they evaporate from the mixture. The smallest hydrocarbons,

with up to four carbon atoms, have the weakest London forces between molecules,

so they evaporate from the crude oil mixture first and can be

piped to our homes as natural gas. Hydrocarbons with 5 to 25

carbons have strong enough London forces to remain liquids

at room temperature and pressure. After removal of the natural

gas molecules, further heating of the remaining mixture drives

off molecules with 6 to 12 carbon atoms, which can be used

as gasoline. Kerosene, a mixture of hydrocarbon molecules with

11 to 16 carbon atoms, evaporates next. Lubricating oils, which

contain hydrocarbon molecules with 15 to 25 carbon atoms, are

then driven off at an even higher temperature. Hydrocarbons

with more than 25 carbons remain solid throughout this process. Hydrocarbons have many uses. Natural gas
When the liquid components of crude oil have been separated, is used to heat our homes, cars require liquid
gasoline and lubricating oils, and solid asphalt is
the solid asphalt that is left behind can be used to build roads. used to make roads and driveways.

558 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

We know that polar molecules are attracted to each other by dipole-dipole
attractions or hydrogen bonds. Experiments have shown, though, that the
actual strengths of the attractions between polar molecules are greater than
we would predict from the polarity of the isolated molecules. A description
of why this is true can be found at the textbook’s Web site.

Objective 30 Particle Interaction in Pure Elements
Objective 30
Objective 30 As we have noted, the I2 molecules in solid iodine are held together by London forces,
but what about the solid forms of other elements? Let’s review what we discovered in
Section 2.5 about the particles that form the structures of the pure elements and find
out more about the attractions that hold them together.

Most of the elements are metallic and their pure solids are held together by metallic
bonds. According to the sea of electrons model described in Section 2.5, the fundamental
structure of a solid metal consists of cations that have lost one or more electrons. The
lost electrons are free to move throughout the solid, so our image of the particles in
a solid metal is one of island-like cations in a sea of electrons. Metallic bonds can be
described as the attractions between the positive metal cations and negative electrons
that surround them.

The nonmetallic elements exhibit more variety in the types of particles their solids
are made of and the types of attractions between them. The particles making up the
solid structure of diamond are carbon atoms, and the attractions that hold them in the
solid form are covalent bonds (Section 2.5). The noble gases are composed of separate
atoms. When a noble gas is cooled enough to form a liquid, London forces hold the
particles together.

Pure samples of all the other nonmetallic elements consist of molecules. Hydrogen,
nitrogen, oxygen, and all of the halogens (group 17) are composed of diatomic
molecules: H2, N2, O2, F2, Cl2, Br2, and I2. Sulfur and selenium can take different
forms, but the most common form for each is a molecule with eight atoms, S8 and
Se8. Phosphorus also has different forms, but the most important forms consist of P4
molecules. All of the molecules of these elements are nonpolar, and the attractions
holding their solids together are London forces.

Summary of the Types of Particles and the Attractions Between Them

Table 14.1 summarizes the kinds of particles that make up different substances and
the attractions between them. Figure 14.26 shows a scheme for predicting the types of
attractions between particles.

14.3 Particle-Particle Attractions 559

Table 14.1 Objective 30
The Particles in Different Types of Substances and the Attractions that Hold Them in
Attraction
the Solid and Liquid Form
Metallic bond
Type of substance Particles Examples
London forces
Elements Covalent bonds
London forces
Metal Cations in a sea of Au
electrons Ionic bond

Noble gases Atoms Xe London forces
Dipole-dipole forces
Carbon (diamond) Carbon atoms C(dia)
Hydrogen bonds
Other nonmetal elements Molecules H2, N2, O2, F2, Cl2,
Ionic compounds Cations and anions Br2, I2, S8, Se8, P4
NaCl

Molecular compounds

Nonpolar molecular Molecules CO2 and hydrocarbons
Polar molecules without Molecules HF, HCl, HBr, and HI
H−F, O−H, or N−H bond

Molecules with H−F, O−H Molecules HF, H2O, alcohols,
or N−H bond NH3

Objective 30 Figure 14.26
Scheme for Predicting
8IBU LJOE PG TVCTUBODF
Types of Attractions

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560 Chapter 14 Liquids: Condensation, Evaporation, and Dynamic Equilibrium

Objective 30 Example 14.3 - Types of Particles and Types of Attractions

For (a) C8H18, (b) aluminum, (c) Br2, (d) LiBr, (e) ethanol, and (f ) CH3NH2,
specify (1) the type of particle that forms the substance’s fundamental structure and
(2) the name of the type of attraction holding these particles in the solid and liquid
form.

Solution

a. The compound C8H18 is a hydrocarbon, so it is composed of C8H18 molecules
held together by London forces.

b. Aluminum is a metallic element, so it is composed of Al cations in a sea of
electrons. The particles are held together by metallic bonds.

c. The formula Br2 describes bromine, a nonmetallic element. Except for carbon
and the noble gases, the nonmetallic elements are composed of molecules held
together by London forces.

d. Because LiBr consists of a metal and a nonmetal, we assume it is an ionic
compound with cations and anions held together by ionic bonds.

e. The formula for ethanol is C2H5OH. Ethanol, like all alcohols, has a structure
formed by molecules held together by hydrogen bonds.

f. Because all of the symbols in the formula CH3NH2 represent nonmetallic
elements, we assume that CH3NH2 has all covalent bonds and therefore is
a molecular substance composed of CH3NH2 molecules. The -NH2 in the
formula suggests that the molecules have an N−H bond, which leads to hydrogen
bonding between molecules.

Objective 30 Exercise 14.3 - Types of Particles and Types of Attractions

For (a) iron, (b) iodine, (c) CH3OH, (d) NH3, (e) hydrogen chloride, (f ) KF, and
(g) carbon in the diamond form, specify (1) the type of particle that forms the
substance’s fundamental structure and (2) the name of the type of attraction that holds
these particles in the solid and liquid form.

If you can predict the types of attractions between particles in two different
substances, you can also predict the relative strengths of those attractions,
and then the relative boiling-point temperatures for the substances.
The textbook’s Web site shows a general procedure for making these
predictions.