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7: Other Ways to Classify Reactions

7: Other Ways to Classify Reactions

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Chapter Review

F 40. The “Chemistry in Focus” segment Garden-Variety



Acid–Base Indicators discusses acid–base indicators

found in nature. What colors are exhibited by red

cabbage juice under acid conditions? Under basic

conditions?

PROBLEMS

41. Calculate the pH corresponding to each of the hydrogen ion concentrations given below, and indicate

whether each solution is acidic or basic.

a.

b.

c.

d.



ϩ



Ϫ3



[H ] ϭ 4.02 ϫ 10 M

[Hϩ] ϭ 8.99 ϫ 10Ϫ7 M

[Hϩ] ϭ 2.39 ϫ 10Ϫ6 M

[Hϩ] ϭ 1.89 ϫ 10Ϫ10 M



a.

b.

c.

d.



[Hϩ] ϭ 9.35 ϫ 10Ϫ2 M

[Hϩ] ϭ 3.75 ϫ 10Ϫ4 M

[Hϩ] ϭ 8.36 ϫ 10Ϫ6 M

[Hϩ] ϭ 5.42 ϫ 10Ϫ8 M



[OHϪ] ϭ 4.73 ϫ 10Ϫ4 M

[OHϪ] ϭ 5.99 ϫ 10Ϫ1 M

[OHϪ] ϭ 2.87 ϫ 10Ϫ8 M

[OHϪ] ϭ 6.39 ϫ 10Ϫ3 M



44. Calculate the pH corresponding to each of the hydroxide ion concentrations given below, and indicate

whether each solution is acidic or basic.

a.

b.

c.

d.



[OHϪ] ϭ 8.63 ϫ 10Ϫ3 M

[OHϪ] ϭ 7.44 ϫ 10Ϫ6 M

[OHϪ] ϭ 9.35 ϫ 10Ϫ9 M

[OHϪ] ϭ 1.21 ϫ 10Ϫ11 M



45. Calculate the pH corresponding to each of the pOH

values listed, and indicate whether each solution is

acidic, basic, or neutral.

a. pOH ϭ 4.32

b. pOH ϭ 8.90



c. pOH ϭ 1.81

d. pOH ϭ 13.1



46. Calculate the pOH value corresponding to each of

the pH values listed, and tell whether each solution is

acidic or basic.

a. pH ϭ 9.78

b. pH ϭ 4.01



c. pH ϭ 2.79

d. pH ϭ 11.21



47. For each hydrogen ion concentration listed, calculate

the pH of the solution as well as the concentration of

hydroxide ion in the solution. Indicate whether each

solution is acidic or basic.

a.

b.

c.

d.



ϩ



Ϫ8



[H ] ϭ 4.76 ϫ 10 M

[Hϩ] ϭ 8.92 ϫ 10Ϫ3 M

[Hϩ] ϭ 7.00 ϫ 10Ϫ5 M

[Hϩ] ϭ 1.25 ϫ 10Ϫ12 M



a.

b.

c.

d.



[Hϩ] ϭ 1.91 ϫ 10Ϫ2 M

[Hϩ] ϭ 4.83 ϫ 10Ϫ7 M

[Hϩ] ϭ 8.92 ϫ 10Ϫ11 M

[Hϩ] ϭ 6.14 ϫ 10Ϫ5 M



49. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

values.

c. pH ϭ 1.02

d. pH ϭ 7.00



50. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

values.

a. pH ϭ 11.21

b. pH ϭ 4.39



43. Calculate the pH corresponding to each of the hydroxide ion concentrations given below, and indicate

whether each solution is acidic or basic.

a.

b.

c.

d.



48. For each hydrogen ion concentration listed, calculate

the pH of the solution as well as the concentration of

hydroxide ion in the solution. Indicate whether each

solution is acidic or basic.



a. pH ϭ 9.01

b. pH ϭ 6.89



42. Calculate the pH corresponding to each of the hydrogen ion concentrations given below, and indicate

whether each solution is acidic or basic.



539



c. pH ϭ 7.44

d. pH ϭ 1.38



51. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pOH

values.

a.

b.

c.

d.



pOH ϭ 4.95

pOH ϭ 7.00

pOH ϭ 12.94

pOH ϭ 1.02



52. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

or pOH values.

a.

b.

c.

d.



pOH ϭ 4.99

pH ϭ 7.74

pOH ϭ 10.74

pH ϭ 2.25



53. Calculate the pH of each of the following solutions

from the information given.

a.

b.

c.

d.



[Hϩ] ϭ 4.78 ϫ 10Ϫ2 M

pOH ϭ 4.56

[OHϪ] ϭ 9.74 ϫ 10Ϫ3 M

[Hϩ] ϭ 1.24 ϫ 10Ϫ8 M



54. Calculate the pH of each of the following solutions

from the information given.

a.

b.

c.

d.



[Hϩ] ϭ 4.39 ϫ 10Ϫ6 M

pOH ϭ 10.36

[OHϪ] ϭ 9.37 ϫ 10Ϫ9 M

[Hϩ] ϭ 3.31 ϫ 10Ϫ1 M



16.5 Calculating the pH of Strong Acid Solutions

QUESTIONS

55. When 1 mole of gaseous hydrogen chloride is dissolved in enough water to make 1 L of solution, approximately how many HCl molecules remain in the

solution? Explain.



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



540 Chapter 16 Acids and Bases

56. A bottle of acid solution is labeled “3 M HNO3.” What

are the substances that are actually present in the solution? Are any HNO3 molecules present? Why or

why not?

PROBLEMS

57. Calculate the hydrogen ion concentration and the

pH of each of the following solutions of strong acids.

a. 1.04 ϫ 10Ϫ4 M HCl

b. 0.00301 M HNO3



c. 5.41 ϫ 10Ϫ4 M HClO4

d. 6.42 ϫ 10Ϫ2 M HNO3



58. Calculate the pH of each of the following solutions of

strong acids.

a.

b.

c.

d.



1.21 ϫ 10Ϫ3 M HNO3

0.000199 M HClO4

5.01 ϫ 10Ϫ5 M HCl

0.00104 M HBr



a. Write a chemical equation indicating how HCl behaves as an acid in liquid ammonia.

b. Write a chemical equation indicating how OHϪ

behaves as a base in liquid ammonia.

66. Strong bases are bases that completely ionize in water

to produce hydroxide ion, OHϪ. The strong bases include the hydroxides of the Group 1 elements. For example, if 1.0 mole of NaOH is dissolved per liter, the

concentration of OHϪ ion is 1.0 M. Calculate the

[OHϪ], pOH, and pH for each of the following strong

base solutions.

a.

b.

c.

d.



16.6 Buffered Solutions



0.10 M NaOH

2.0 ϫ 10Ϫ4 M KOH

6.2 ϫ 10Ϫ3 M CsOH

0.0001 M NaOH



67. Which of the following conditions indicate an acidic

solution?



QUESTIONS

59. What characteristic properties do buffered solutions

possess?

60. What two components make up a buffered solution?

Give an example of a combination that would serve

as a buffered solution.

61. Which component of a buffered solution is capable

of combining with an added strong acid? Using your

example from Exercise 60, show how this component

would react with added HCl.

62. Which component of a buffered solution consumes

added strong base? Using your example from Exercise 60, show how this component would react with

added NaOH.

PROBLEMS

63. Which of the following combinations would act as

buffered solutions?

a. HCl and NaCl

b. CH3COOH and KCH3COO



base theory can be extended easily to other solvents.

One such solvent that has been investigated in depth

is liquid ammonia, NH3.



c. H2S and NaHS

d. H2S and Na2S



64. A buffered solution is prepared containing acetic

acid, HC2H3O2, and sodium acetate, NaϩC2H3O2Ϫ,

both at 0.5 M. Write a chemical equation showing

how this buffered solution would resist a decrease in

its pH if a few drops of aqueous strong acid HCl solution were added to it. Write a chemical equation

showing how this buffered solution would resist an

increase in its pH if a few drops of aqueous strong

base NaOH solution were added to it.



Additional Problems

65. The concepts of acid–base equilibria were developed

in this chapter for aqueous solutions (in aqueous solutions, water is the solvent and is intimately involved

in the equilibria). However, the Brønsted–Lowry acid–



a.

b.

c.

d.



pH ϭ 3.04

[Hϩ] Ͼ 1.0 ϫ 10Ϫ7 M

pOH ϭ 4.51

[OHϪ] ϭ 3.21 ϫ 10Ϫ12 M



68. Which of the following conditions indicate a basic

solution?

a.

b.

c.

d.



pOH ϭ 11.21

pH ϭ 9.42

[OHϪ] Ͼ [Hϩ]

[OHϪ] Ͼ 1.0 ϫ 10Ϫ7 M



69. Buffered solutions are mixtures of a weak acid and its

conjugate base. Explain why a mixture of a strong acid

and its conjugate base (such as HCl and ClϪ) is not

buffered.

70. Which of the following acids are classified as strong

acids?

a.

b.

c.

d.

e.



HNO3

CH3COOH (HC2H3O2)

HCl

HF

HClO4



71. Is it possible for a solution to have [Hϩ] ϭ 0.002 M

and [OHϪ] ϭ 5.2 ϫ 10Ϫ6 M at 25 °C? Explain.

72. Despite HCl’s being a strong acid, the pH of

1.00 ϫ 10Ϫ7 M HCl is not exactly 7.00. Can you suggest a reason why?

73. According to Arrhenius, bases are species that produce

ion in aqueous solution.

74. According to the Brønsted–Lowry model, a base is a

species that

protons.

75. A conjugate acid–base pair consists of two substances related by the donating and accepting of a(n)

.



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



Chapter Review

76. Acetate ion, C2H3O2Ϫ, has a stronger affinity for protons than does water. Therefore, when dissolved in

water, acetate ion behaves as a(n)

.

77. An acid such as HCl that strongly conducts an electric current when dissolved in water is said to be a(n)

acid.

78. Draw the structure of the carboxyl group, OCOOH.

Show how a molecule containing the carboxyl group

behaves as an acid when dissolved in water.

79. Because of

, even pure water contains measurable quantities of Hϩ and OHϪ.

80. The ion-product constant for water, Kw, has the value

at 25 °C.

81. The number of

in the logarithm of a number is equal to the number of significant figures in the

number.

82. A solution with pH ϭ 4 has a (higher/lower) hydrogen ion concentration than a solution with pOH ϭ 4.

83. A 0.20 M HCl solution contains

M hydrogen

ion and

M chloride ion concentrations.

84. A buffered solution is one that resists a change in

when either a strong acid or a strong base is

added to it.

85. A(n)

solution contains a conjugate acid–base

pair and through this is able to resist changes in its pH.

86. When sodium hydroxide, NaOH, is added dropwise

to a buffered solution, the

component of

the buffer consumes the added hydroxide ion.

87. When hydrochloric acid, HCl, is added dropwise to a

buffered solution, the

component of the

buffer consumes the added hydrogen ion.

88. Which of the following represent conjugate acid–

base pairs? For those pairs that are not conjugates,

write the correct conjugate acid or base for each

species in the pair.

a.

b.

c.

d.



89. In each of the following chemical equations, identify

the conjugate acid–base pairs.

a. CH3NH2 ϩ H2O 3

4 CH3NH3ϩ ϩ OHϪ

b. CH3COOH ϩ NH3 3

4 CH3COOϪ ϩ NH4ϩ

Ϫ

c. HF ϩ NH3 3

4 F ϩ NH4ϩ

90. Write the conjugate acid for each of the following.

c. H2O

d. OHϪ



a. NH3

b. NH2Ϫ



91. Write the conjugate base for each of the following.

a.

b.

c.

d.



H3PO4

HCO3Ϫ

HF

H2SO4



92. Write chemical equations showing the ionization

(dissociation) in water for each of the following acids.

a.

b.

c.

d.



CH3CH2COOH (Only the last H is acidic.)

NH4ϩ

H2SO4

H3PO4



93. Which of the following bases have relatively strong

conjugate acids?

a.

b.

c.

d.





ClϪ

HSO4Ϫ

NO3Ϫ



94. Calculate [Hϩ] in each of the following solutions, and

indicate whether the solution is acidic, basic, or

neutral.

a.

b.

c.

d.



[OHϪ] ϭ 4.22 ϫ 10Ϫ3 M

[OHϪ] ϭ 1.01 ϫ 10Ϫ13 M

[OHϪ] ϭ 3.05 ϫ 10Ϫ7 M

[OHϪ] ϭ 6.02 ϫ 10Ϫ6 M



95. Calculate [OHϪ] in each of the following solutions,

and indicate whether the solution is acidic, basic, or

neutral.

a.

b.

c.

d.



[Hϩ] ϭ 4.21 ϫ 10Ϫ7 M

[Hϩ] ϭ 0.00035 M

[Hϩ] ϭ 0.00000010 M

[Hϩ] ϭ 9.9 ϫ 10Ϫ6 M



96. For each pair of concentrations, tell which represents

the more basic solution.

a. [Hϩ] ϭ 0.000013 M or [OHϪ] ϭ 0.0000032 M

b. [Hϩ] ϭ 1.03 ϫ 10Ϫ6 M or [OHϪ] ϭ 1.54 ϫ 10Ϫ8 M

c. [OHϪ] ϭ 4.02 ϫ 10Ϫ7 M or [OHϪ] ϭ 0.0000001 M

97. Calculate the pH of each of the solutions indicated

below. Tell whether the solution is acidic, basic, or

neutral.

a.

b.

c.

d.



H2O, OHϪ

H2SO4, SO42Ϫ

H3PO4, H2PO4Ϫ

HC2H3O2, C2H3O2Ϫ



541



[Hϩ] ϭ 1.49 ϫ 10Ϫ3 M

[OHϪ] ϭ 6.54 ϫ 10Ϫ4 M

[Hϩ] ϭ 9.81 ϫ 10Ϫ9 M

[OHϪ] ϭ 7.45 ϫ 10Ϫ10 M



98. Calculate the pH corresponding to each of the hydroxide ion concentrations given below. Tell whether

each solution is acidic, basic, or neutral.

a.

b.

c.

d.



[OHϪ] ϭ 1.4 ϫ 10Ϫ6 M

[OHϪ] ϭ 9.35 ϫ 10Ϫ9 M

[OHϪ] ϭ 2.21 ϫ 10Ϫ1 M

[OHϪ] ϭ 7.98 ϫ 10Ϫ12 M



99. Calculate the pOH corresponding to each of the pH

values listed, and indicate whether each solution is

acidic, basic, or neutral.

a.

b.

c.

d.



pH ϭ 1.02

pH ϭ 13.4

pH ϭ 9.03

pH ϭ 7.20



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



542 Chapter 16 Acids and Bases

100. For each hydrogen or hydroxide ion concentration

listed, calculate the concentration of the complementary ion and the pH and pOH of the solution.

a. [Hϩ] ϭ 5.72 ϫ 10Ϫ4 M

b. [OHϪ] ϭ 8.91 ϫ 10Ϫ5 M

c. [Hϩ] ϭ 2.87 ϫ 10Ϫ12 M

d. [OHϪ] ϭ 7.22 ϫ 10Ϫ8 M



a.

b.

c.

d.



101. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

values.

a. pH ϭ 8.34

b. pH ϭ 5.90



c. pH ϭ 2.65

d. pH ϭ 12.6



102. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

or pOH values.

a.

b.

c.

d.



pH ϭ 5.41

pOH ϭ 12.04

pH ϭ 11.91

pOH ϭ 3.89



103. Calculate the hydrogen ion concentration, in moles

per liter, for solutions with each of the following pH

or pOH values.

pOH ϭ 0.90

pH ϭ 0.90

pOH ϭ 10.3

pH ϭ 5.33



104. Calculate the hydrogen ion concentration and the

pH of each of the following solutions of strong acids.

a.

b.

c.

d.



1.4 ϫ 10Ϫ3 M HClO4

3.0 ϫ 10Ϫ5 M HCl

5.0 ϫ 10Ϫ2 M HNO3

0.0010 M HCl



105. Write the formulas for three combinations of weak

acid and salt that would act as buffered solutions. For

each of your combinations, write chemical equations

showing how the components of the buffered solution would consume added acid and base.



All even-numbered Questions and Problems have answers in the back of this book and solutions in the Solutions Guide.



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17

17.1 How Chemical Reactions

Occur

17.2 Conditions That Affect

Reaction Rates

17.3 The Equilibrium Condition

17.4 Chemical Equilibrium:

A Dynamic Condition

17.5 The Equilibrium Constant:

An Introduction

17.6 Heterogeneous Equilibria

17.7 Le Châtelier’s Principle

17.8 Applications Involving the

Equilibrium Constant

17.9 Solubility Equilibria



Equilibrium

Equilibrium can be analogous to traffic flowing

both ways on a bridge, such as San Francisco’s

Golden Gate Bridge. (James Martin/Stone/Getty

Images)



17.1 How Chemical Reactions Occur



545



C



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hemistry is mostly about reactions—processes in which groups of

atoms are reorganized. So far we have learned to describe chemical reactions by using balanced equations and to calculate amounts of reactants

and products. However, there are many important characteristics of reactions that we have not

yet considered.

For example, why

do refrigerators prevent food from spoiling? That is, why do the

chemical reactions that

cause food to decompose occur more slowly Refrigeration prevents food spoilage.

at lower temperatures? On the other hand, how can a chemical manufacturer speed up a chemical reaction that runs too slowly to be economical?

Another question that arises is why chemical reactions carried out in

a closed vessel appear to stop at a certain point. For example, when the reaction of reddish-brown nitrogen dioxide to form colorless dinitrogen

tetroxide,

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2NO2(g)

Reddish-brown



S



N2O4(g)

Colorless



is carried out in a closed container, the reddish-brown color at first fades

but stops changing after a time and then stays the same color indefinitely

if left undisturbed (see Figure 17.1). We will account for all of these important observations about reactions in this chapter.



17.1 How Chemical Reactions Occur

To understand the collision model of how chemical reactions occur.



OBJECTIVE:



Ken O’Donoghue



In writing the equation for a chemical reaction, we put the reactants on the

left and the products on the right with an arrow between them. But how do

the atoms in the reactants reorganize to form the products?



a



Figure 17.1



A sample containing a

large quantity of reddishbrown NO2 gas.



b



As the reaction to form

colorless N2O4 occurs,

the sample becomes

lighter brown.



c



After equilibrium is

reached [2NO2(g) I

N2O4(g)], the color

remains the same.



546 Chapter 17 Equilibrium



O

N



O

N



O

N



O

N



Br



Br



Br



Br



a



O

N



Br



The collision occurs.



Figure 17.2

Visualizing the reaction

2BrNO(g) S 2NO(g) ϩ Br2(g).



O

N



O

N



Br



Br



Br



c



b



Two BrNO molecules

approach each other at

high speeds.



O

N



d



The energy of the

collision causes

the Br—N bonds

to break and

allows the Br—Br

bond to form.



The products:

one Br2 and

two NO

molecules.



Chemists believe that molecules react by colliding with each other.

Some collisions are violent enough to break bonds, allowing the reactants to

rearrange to form the products. For example, consider the reaction

2BrNO(g) 3

4 2NO(g) ϩ Br2(g)

which we picture to occur as shown in Figure 17.2. Notice that the BrON

bonds in the two BrNO molecules must be broken and a new BrOBr bond

must be formed during a collision for the reactants to become products.

The idea that reactions occur during molecular collisions, which is

called the collision model, explains many characteristics of chemical reactions. For example, it explains why a reaction proceeds faster if the concentrations of the reacting molecules are increased (higher concentrations

lead to more collisions and therefore to more reaction events). The collision

model also explains why reactions go faster at higher temperatures, as we

will see in the next section.



17.2

OBJECTIVES:



Recall from Section 13.8 that

the average kinetic energy of

a collection of molecules is

directly proportional to the

temperature (K).



Conditions That Affect Reaction Rates

To understand activation energy. • To understand how a catalyst speeds

up a reaction.

It is easy to see why reactions speed up when the concentrations of reacting

molecules are increased: higher concentrations (more molecules per unit

volume) lead to more collisions and so to more reaction events. But reactions

also speed up when the temperature is increased. Why? The answer lies in the

fact that not all collisions possess enough energy to break bonds. A minimum energy called the activation energy (Ea) is needed for a reaction to

occur (see Figure 17.3). If a given collision possesses an energy greater than

Ea, that collision can result in a reaction. If a collision has an energy less than

Ea, the molecules will bounce apart unchanged.

The reason that a reaction occurs faster as the temperature is increased

is that the speeds of the molecules increase with temperature. So at higher

temperatures, the average collision is more energetic. This makes it more

likely that a given collision will possess enough energy to break bonds and

to produce the molecular rearrangements needed for a reaction to occur.



Higher

temperatures



Higher

speeds



More high-energy

collisions



More collisions

that break bonds



Faster

reaction



17.2 Conditions That Affect Reaction Rates



O

N



O

N



Energy



Br



Br



547



Is it possible to speed up a reaction without changing the temperature

or the reactant concentrations? Yes, by using something called a catalyst, a

substance that speeds up a reaction without being consumed. This may sound too

good to be true, but it is a very common occurrence. In fact, you would not

be alive now if your body did not contain thousands of catalysts called enzymes. Enzymes allow our bodies to speed up complicated reactions that

would be too slow to sustain life at normal body temperatures. For example,

the enzyme carbonic anhydrase speeds up the reaction between carbon dioxide and water

CO2(g) ϩ H2O(l) 4

3 Hϩ(aq) ϩ HCO3Ϫ(aq)



Ea

2BrNO

Reactant



2NO + Br2

Products



Reaction progress



Figure 17.3

When molecules collide, a certain

minimum energy called the

activation energy (Ea) is needed

for a reaction to occur. If the

energy contained in a collision of

two BrNO molecules is greater

than Ea, the reaction can go

“over the hump” to form

products. If the collision energy is

less than Ea, the colliding

molecules bounce apart

unchanged.



Although O atoms are too

reactive to exist near the earth’s

surface, they do exist in the

upper atmosphere.



to help prevent an excess accumulation of carbon dioxide in our blood.

Although we cannot consider the details here, a catalyst works because

it provides a new pathway for the reaction—a pathway that has a lower activation energy than the original pathway, as illustrated in Figure 17.4. Because of the lower activation energy, more collisions will have enough energy to allow a reaction. This in turn leads to a faster reaction.

A very important example of a reaction involving a catalyst occurs in

our atmosphere; it is the breakdown of ozone, O3, catalyzed by chlorine

atoms. Ozone is one constituent of the earth’s upper atmosphere that is especially crucial, because it absorbs harmful high-energy radiation from the

sun. There are natural processes that result in both the formation and the destruction of ozone in the upper atmosphere. The natural balance of all these

opposing processes has resulted in an amount of ozone that has been relatively constant over the years. However, the ozone level now seems to be decreasing, especially over Antarctica (Figure 17.5), apparently because chlorine atoms act as catalysts for the decomposition of ozone to oxygen by the

following pair of reactions:

Cl ϩ O3 S ClO ϩ O2

O ϩ ClO S Cl ϩ O2

Sum: Cl ϩ O3 ϩ O ϩ ClO S ClO ϩ O2 ϩ Cl ϩ O2

When species that appear on both sides of the equation are canceled,

the end result is the reaction

O ϩ O3 S 2O2



Catalyzed

reaction

pathway



Cutaways of catalytic converters

used in automobiles.



Energy



Delphi Automotive Services



Uncatalyzed

reaction

pathway



Ea



Figure 17.4



EЈa

Products

Reactants



Reaction progress



Comparison of the activation

energies for an uncatalyzed

reaction (Ea ) and for the same

reaction with a catalyst present

(EЈa ). Note that a catalyst works

by lowering the activation

energy for a reaction.



C H E M I S T R Y I N F OCUS

Protecting the Ozone



trucks sold in the United States have air conditioners that employ HFC-134a. Converting the

140 million autos currently on the road in the

hlorofluorocarbons (CFCs) are ideal compounds United States that use CF2Cl2 will pose a major

for refrigerators and air conditioners because they headache, but experience suggests that replaceare nontoxic and noncorrosive. However, the ment of Freon-12 with HFC-134a is less expensive

chemical inertness of these substances, once than was originally feared. For example, Volvo

thought to be their major virtue, turns out to be Cars of North America estimates that a Volvo can

their fatal flaw. When these compounds leak into be converted from Freon-12 to HFC-134a for

the atmosphere, as they inevitably do, they are so around $300.

unreactive they persist there for decades. EventuA related environmental issue involves really these CFCs reach altitudes where ultraviolet placing the halons for firefighting applications.

light causes them to decompose, producing chlo- In particular, scientists are seeking an effective

rine atoms that promote the destruction of the replacement for CF3Br (halon-1301), the nonozone in the stratosphere (see discussion in Sec- toxic “magic gas” used to flood enclosed spaces

tion 17.2). Because of this probsuch as offices, aircraft, race

lem, the world’s industrialized nacars, and military tanks in case

tions have signed an agreement

of fire. The compound CF3I,

(called the Montreal Protocol)

which appears to have a lifethat banned CFCs in 1996 (with a

time in the atmosphere of only

10-year grace period for developa few days, looks like a promising nations).

ing candidate but much more

Worldwide production of

research on the toxicology and

CFCs has already decreased to

ozone-depleting properties of

half of the 1986 level of 1.13 milCF3I will be required before it

lion metric tons. One strategy for

receives government approval

replacing the CFCs has been to

as a halon substitute.

switch to similar compounds that

The chemical industry has

contain carbon and hydrogen

responded amazingly fast to the

atoms substituted for chlorine

ozone depletion emergency. It is

atoms. For example, the U.S. apencouraging that we can act

pliance industry has switched

rapidly when an environmental

from Freon-12 (CF2Cl2) to the An Amana refrigerator, one of

crisis occurs. Now we need to

many appliances that now use

compound CH2FCH3 (called HFC- HFC-134a. This compound is

get better at keeping the envi134a) for home refrigerators, replacing CFCs, which lead to the

ronment at a higher priority as

and most of the new cars and destruction of atmospheric ozone. we plan for the future.

Courtesy, Amana



C



Notice that a chlorine atom is used up in the first reaction but a chlorine

atom is formed again by the second reaction. Therefore, the amount of chlorine does not change as the overall process occurs. This means that the chlorine atom is a true catalyst: it participates in the process but is not consumed.

Estimates show that one chlorine atom can catalyze the destruction of about one

million ozone molecules per second.

The chlorine atoms that promote this damage to the ozone layer are

present because of pollution. Specifically, they come from the decomposition of compounds called Freons, such as CF2Cl2, which have been widely

used in refrigerators and air conditioners. The Freons have leaked into the at-



548



17.3 The Equilibrium Condition



1979



549



2008



Figure 17.5

Total Ozone (Dobson Units)



NASA



A photo showing the ozone

“hole” over Antarctica.



110



220



330



440



550



mosphere, where they are decomposed by light to produce chlorine atoms

and other substances. As a result, the manufacture of Freons was banned by

agreement among the nations of the world as of the end of 1996. Substitute

compounds are now being used in newly manufactured refrigerators and air

conditioners.



17.3 The Equilibrium Condition

OBJECTIVE:



To learn how equilibrium is established.

Equilibrium is a word that implies balance or steadiness. When we say that

someone is maintaining his or her equilibrium, we are describing a state of

balance among various opposing forces. The term is used in a similar but

more specific way in chemistry. Chemists define equilibrium as the exact

balancing of two processes, one of which is the opposite of the other.

We first encountered the concept of equilibrium in Section 14.4, when

we described the way vapor pressure develops over a liquid in a closed container (see Figure 14.9). This equilibrium process is summarized in Figure 17.6. The equilibrium state occurs when the rate of evaporation exactly

equals the rate of condensation.

So far in this textbook we have usually assumed that reactions proceed

to completion—that is, until one of the reactants “runs out.” Indeed, many

reactions do proceed essentially to completion. For such reactions we can assume that the reactants are converted to products until the limiting reactant

is completely consumed. On the other hand, there are many chemical

reactions that “stop” far short of completion when they are allowed to

take place in a closed container. An example is the reaction of nitrogen

dioxide to form dinitrogen tetroxide.

NO2(g) ϩ NO2(g) S N2O4(g)

Reddish-brown



Evaporation



Condensation



Colorless



The reactant NO2 is a reddish-brown gas, and the product N2O4 is

a colorless gas. Imagine an experiment where pure NO2 is placed

in an empty, sealed glass vessel at 25 °C. The initial dark brown

color will decrease in intensity as the NO2 is converted to colorless

N2O4 (see Figure 17.1). However, even over a long period of time,



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