Chemistry

23 (g) .

0.92 g/cm 3

Practice Problem A Given that 25 .0 rnL of mercury has a mass of 340 g, calculate (a) the density of

mercury and (b) the mass of 120 rnL of mercury. Practice Problem B Calculate (a) the density of a solid substance if a cube measuring 2.33 cm on

one side has a mass of 117 g and (b) the mass of a cube of the same substance measuring 7.41 cm on one side.

The box on page 14 illustrates the importance of using units carefully in scientific work.

Checkpoint 1.3 1.3.1

1.3.2

Scientific Measurement

The coldest temperature ever recorded on Earth was -128.6°F (recorded at Vostok Station, Antarctica, on July 21, 1983). Express this temperature in degrees Celsius and kelvin.

1.3.3

A sample of water is heated from room temperature to just below the boiling point. The overall change in temperature is 72°e. Express this temperature change in kelvins.

a) - 89 .2°C, -89.2 K

a) 345 K

b) - 289.1 °C, -15.9K

b) 7? K

c) - 89.2°C, 183.9 K

c) 0 K

d) - 173.9°C, 99.3 K

d) ?Ol K

e) -7.0°C, 266.2 K

e) ?73 K

What is the density of an object that has a volume of 34.2 cm3 and a mass of 19.6 g? a) 0.573 g/cm 3

1.3.4

,

Given that the density of gold is 19.3 glcm3, calculate the volume (in cm3) of a gold nugget with a mass of 5.98 g. a) 3.23 cm 3 ,

b) 1.74 g/cm 3

b) 5.98 cm' ,

c) 670 g/cm 3

c) 115 cm ' ,

d) 53.8 g/cm 3

d) 0.310 cm'

e) 14.6g/cm 3

e) 13.3 cm 3

13

How Important Are Units? ton, we would start with 1 Ib second law of motion,

On December 11, 1998, NASA launched the 125-million-dollar Mars Climate Orbiter, which was intended to be the Red Planet's first weather satellite. After a 416-million-mile (mi) journey, the spacecraft was supposed to go into Mars's orbit on September 23, 1999. Instead, it entered Mars's atmosphere about 100 krn (62 mi) lower than planned and was destroyed by heat. Mission controllers later determined that the spacecraft was lost because English measurement units were not converted to metric units in the navigation software. Engineers at Lockheed Martin Corporation, who built the spacecraft, specified its thrust in pounds, which is an English unit of force. Scientists at NASA's Jet Propulsion Laboratory, on the other hand, who were responsible for deployment, had assumed that the thrust data they were given were expressed in newtons, a metric unit. To carry out the conversion between pound and new-

=

0.4536 kg and, from Newton's

force = (mass)(acceleration) = (0.4536 kg)( 9.81 m/s

2

)

2

= 4.45 kg . m/s = 4.45 N 2

because 1 newton (N) = 1 kg . m/s . Therefore, instead of converting lIb ofjorce to 4.45 N, the scientists treated it as a force of 1 N. The considerably smaller engine thrust employed because of the engineers' failure to convert from English to metric units resulted in a lower orbit and the ultimate destruction of the spacecraft. Commenting on the failure of the Mars mission, one scientist said, "This is going to be the cautionary tale that will be embedded into introduction to the metric system in elementary school, high school, and college science courses till the end of time."

• •

.

~

I.

• -... •

,

•

.

=

)

.. '"

,

-

f

'-i ~

-;'

.

- . ,*,~ "'ft

---'"II i'

~I

••

••

-. •

/ Mars Climate Orbiter during preflight tests.

The Properties of Matter Substances are identified by their properties as well as by their composition. Properties of a substance may be quantitative (measured and expressed with a number) or qualitative (not requiring explicit measurement).

Physical Properties Color, melting point, boiling point, and physical state are all physical properties. A physical property is one that can be observed and measured without changing the identity of a substance. For example, we can determine the melting point of ice by heating a block of ice and measuring the temperature at which the ice is converted to water. Liquid water differs from ice in appearance 14

SECTION 1.5

Uncertainty in Measurement

15

but not in composition; both liquid water and ice are H 2 0. Melting is a physical change; one in which the state of matter changes, but the identity of the matter does not change. We can recover the original ice by cooling the water until it freezes. Therefore, the melting point of a substance is a physical property. Similarly, when we say that nitrogen dioxide gas is brown, we are referring to the physical property of color.

Chemical Properties The statement "Hydrogen gas bums in oxygen gas to form water" describes a chemical property of hydrogen, because to observe this property we must carry out a chemical change burning in oxygen (combustion), in this case. After a chemical change, the original substance (hydrogen gas in this case) will no longer exist. What remains is a different substance (water, in this case). We cannot recover the hydrogen gas from the water by means of a physical process, such as boiling or freezing. Every time we bake cookies, we bring about a chemical change. When heated, the sodium bicarbonate (baking soda) in cookie dough undergoes a chemical change that produces carbon dioxide gas. The gas forms numerous little bubbles in the dough during the baking process, causing the cookies to "rise." Once the cookies are baked, we cannot recover the sodium bicarbonate by cooling the cookies, or by any physical process. When we eat the cookies , we cause further chemical changes that occur during digestion and metabolism.

Extensive and Intensive Properties All properties of matter are either extensive or intensive. The measured value of an extensive property depends on the amount of matter. Mass is an extensive property. More matter means more mass. Values of the same extensive property can be added together. For example, two gold nuggets will have a combined mass that is the sum of the masses of each nugget, and the length of two city buses is the sum of their individual lengths. The value of an extensive property depends on the amount of matter. The value of an intensive property does not depend on the amount of matter. Density and temperature are intensive properties. Suppose that we have two beakers of water at the same temperature and we combine them to make a single quantity of water in a larger beaker. The density and the temperature of the water in the larger combined quantity will be the same as they were in the two separate beakers. Unlike mass and length, which are additive, temperature, density, and other intensive properties are not additive.

Uncertainty in Measurement Chemistry makes use of two types of numbers: exact and inexact. Exact numbers include numbers with defined values, such as 2.54 in the definition 1 inch (in) = 2.54 cm, 1000 in the definition 1 kg = 1000 g, and 12 in the definition 1 dozen = 12 objects. (The number 1 in each of these definitions is also an exact number.) Exact numbers also include those that are obtained by counting. Numbers measured by any method other than counting are inexact. Measured numbers are inexact because of the measuring devices that are used, the individuals who use them, or both. For example, a ruler that is poorly calibrated will result in measurements that are in error-no matter how carefully it is used. Another ruler may be calibrated properly but have insufficient resolution for the necessary measurement. Finally, whether or not an instrument is properly calibrated or has sufficient resolution, there are unavoidable differences in how different people see and interpret measurements.

em

1

2

3

4

em

1

2

3

4

Significant Figures An inexact number must be reported in such a way as to indicate the uncertainty in its value. This is done using significant figures. Significant figures are the meaningful digits in a reported number. Consider the measurement of the memory can! in Figure 1.10 using the ruler above it. The card's width is between 2 and 3 cm. We may record the width as 2.5 cm, but because there are no gradations between 2 and 3 cm on this ruler, we are estimating the second digit. Although we are certain about the 2 in 2.5 , we are not certain about the 5. The last digit in a measured number is referred to as the uncertain digit; and the uncertainty associated with a measured number is generally considered to be + 1 in the place of the last digit. Thus , when we report the width of the memory card to be 2.5 cm, we are implying that its width is 2.5 + 0.1 cm. Each of the digits in a

Figure 1.10 The width we report for the memory card depends on which ruler we use to measure it.

16

CHAPTER 1

Chemistry: The Central Science

It is important not to imply greater certainty in a measured number than is realistic. For example, it would be inappropriate to report the width of the memory card in Figure 1.10 as 2.4500 cm, because this would imply an uncertainty of :t o.0001.

measured number, including the uncertain digit, is a significant figure. The reported width of the circle, 2.5 cm, contains two significant figures. A ruler with millimeter gradations would enable us to be certain about the second digit in this measurement and to estimate a third digit. Now consider the measurement of the memory card using the ruler below it. We may record the width as 2.45 cm. Again, we estimate one digit beyond we can read. The reported width of 2.45 cm contains three significant figures. Reporting the . . . . .. .those . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. width as 2.45 cm implies that the width is 2.45 + 0.01 cm. The number of significant figures in any number can be determined using the following guidelines: 1. Any digit that is not zero is significant (112.1 has four significant figures). 2. Zeros located between nonzero digits are significant (305 has three significant figures, and 50.08 has four significant figures). 3. Zeros to the left of the first nonzero digit are not significant (0.0023 has two significant figures, and 0.000001 has one significant figure). 4. Zeros to the right of the last nonzero digit are significant if the number contains a decimal point (1.200 has four significant figures).

5. Zeros to the right of the last nonzero digit in a number that does not contain a decimal point

Appendix 1 reviews scientific notation.

mayor may not be significant (100 may have one, two, or three significant figures-it is to tell without additional information). To avoid ambiguity in such cases, it is ... . . . .. . . . . .impossible . . . . .. ...... ... ...... . .. ... . .. .. best to express such numbers using scientific notation. If the intended number of significant figures is one, the number is written as 1 X 102 ; if the intended number of significant figures 2 is two, the number is written as 1.0 X 10 ; and if the intended number of significant figures is three, the number is written as 1.00 X 102 . Sample Problem 1.4 lets you practice determining the number of significant figures in a number.

Sample Problem 1.4 Determine the number of significant figures in the following measurements: (a) 443 em, (b) 15.03 g, (c) 0.0356 kg, (d) 3.000 X 10-7 L, (e) 50 mL, (f) 0.9550 m. Strategy All nonzero digits are significant, so the goal will be to determine which of the zeros is

significant. Think About It Be sure that you

have identified zeros correctly as either significant or not significant. They are significant in (b), (d), and (f); they are not significant in (c); and it is not possible to tell in (e).

Setup Zeros are significant if they appear between nonzero digits or if they appear after a nonzero

digit in a number that contains a decimal point. Zeros mayor may not be significant if they appear to the right of the last nonzero digit in a number that does not contain a decimal point. . Solution (a) 3; (b) 4; (c) 3; (d) 4; (e) 1 or 2, an ambiguous case; (f) 4.

Practice Problem Determine the number of significant figures in the following measurements: (a) 1129 m, (b) 0.0003 kg, (c) 1.094 em, (d) 3.5 X 10 12 atoms, (e) 150 mL, (f) 9.550 km.

Calculations with Measured Numbers Because we often use one or more measured numbers to calculate a desired result, a second set of guidelines specifies how to handle significant figures in calculations.

1. In addition and subtraction, the answer cannot have more digits to the right of the decimal point than any of the original numbers. For example: 102.50

~

two digits after the decimal point

+ 0.231

~

three digits after the decimal point

102.731

~

round to 102.73

143.29

~

two digits after the decimal point

~

one digit after the decimal point

~

round to 123.2

-20.1 123.19

SECTION 1.5

Uncertainty in Measurement

17

The rounding procedure works as follows. Suppose we want to round 102.13 and 54.86 each to one digit to the right of the decimal point. To begin, we look at the digit(s) that will be dropped. If the leftmost digit to be dropped is less than 5, as in 102.13, we round down (to 102.1), meaning that we simply drop the digit(s). If the leftmost digit to be dropped is equal to or greater than 5, as in 54.86, we round up (to 54.9), meaning that we add 1 to the preceding digit. 2. In multiplication and division, the number of significant figures in the final product or quotient is determined by the Oliginal number that has the smallest number of significant figures. The following examples illustrate this rule: 1.4 X 8.011 = 11.2154 11.57/305.88 = 0.037825290964

~

round to 11 (limited by 1.4 to two significant figures) ~

round to 0.03783 (limited by 11.57 to four significant figures)

3. Exact numbers can be considered to have an infinite number of significant figures and do not limit the number of significant figures in a calculated result. For example, a penny minted after 1982 has a mass of 2.5 g. If we have three such pennies, the total mass is . .. .. . . . . .. .. . . ..... . . , .. .. . . . . . . . . . . . . 3 X 2.5 g = 7.5 g

The answer should not be rounded to one significant figure because 3 is an exact number. 4. In calculations with multiple steps, rounding the result of each step can result in "rounding error." Consider the following two-step calculation: First step:

A X B = C

Second step:

C X D

=E

Suppose that A = 3.66, B = 8.45, and D = 2.11. The value of E depends on whether we round off C prior to using it in the second step of the calculation.

Method 1 C = 3.66 X 8.45 = 30.9 E = 30.9 X 2.11 = 65.2

Method 2 C = 3.66 X 8.45 = 30.93 E = 30.93 X 2.11 = 65.3

In general, it is best to retain at least one extra digit until the end of a multistep calculation, as shown by method 2, to minimize rounding error. Sample Problems 1.5 and 1.6 show how significant figures are handled in arithmetic operations.

I

Perform the following arithmetic operations and report the result to the proper number of significant figures: ' (a) 317.5 mL + 0.675 mL, (b) 47.80 L - 2.075 L, (c) 13.5 g -7 45.18 L, (d) 6.25 cm X 1.175 cm, (e) 5.46 X 102 g + 4.991 X 10 3 g.

Strategy Apply the rules for significant figures in calculations, and round each answer to the appropriate number of digits.

Setup (a) The answer will contain one digit to the right of the decimal point to match 317.5, which has the fewest digits to the right of the decimal point. (b) The answer will contain two digits to the right of the decimal point to match 47.80. (c) The answer will contain three significant figures to match 13.5, which has the fewest number of significant figures in the calculation. (d) The answer will contain three significant figures to match 6.25. (e) To add numbers expressed in scientific notation, first write both numbers to the same power of 10. That is, 4.991 X 103 = 49.91 X 102 , so the answer 2 will contain two digits to the right of the decimal point (when multiplied by 10 ) to match both 5.46 and 49.91.

I

(Continued)

Note that it is the number of pennies (3), not the mass, that is an exact number.

18

CHAPTER 1

Chemistry: The Central Science

Solution

(a) Think About It It may look as

though the rule of addition has been violated in part (e) because the final answer (5.537 X 103 g) has three places past the decimal point, not two. However, the rule was applied ? to get the answer 55.37 X 10- g, which has four significant figures. Changing the answer to correct scientific notation doesn't change the number of significant figures, but in this case it changes the number of places past the decimal point.

3l7.5mL + 0.675 mL 318.l75 mL

(b) 47.80L -2.075 L 45.725 L (c)

~

~

round to 318.2 mL

round to 45.73 L

13.5 a 45.18 ~ = 0.298804781 giL

(d) 6.25 cm X 1.175 cm (e)

+

=

round to 0.299 gIL

~

7 .34375 cm2

~

?

round to 7.34 cm-

5.46 X 102 g 2 49.91 X 10 g

-

55.37 X 102 g = 5.537 X 103 g

Practice Problem A Perform the following arithmetic operations, and report the result to the proper

number of significant figures: (a) 105.5 L + 10.65 L, (b) 81.058 m - 0.35 m, (c) 3.801 X 10 + 1.228 X 10 19 atoms, (d) 1.255 dm X 25 dm, (e) 139 g -7- 275.55 mL.

21

atoms

Practice Problem B Perform the following arithmetic operations, and report the result to the proper

number of significant figures: (a) 1.0267 cm X 2.508 cm X 12.599 cm, (b) 15.0 kg -7- 0.036 m (c) 1.113 X 10 10 kg - 1.050 X 109 kg, (d) 25.75 mL + 15.00 mL, (e) 46 cm 3 + 180.5 cm3

3

,

Sample Problem 1.6 An empty container with a volume of 9.850 X 102 cm 3 is weighed and found to have a mass of 124.6 g. The container is filled with a gas and reweighed. The mass of the container and the gas is 126.5 g. Determine the density of the gas to the appropriate number of significant figures. Strategy This problem requires two steps: subtraction to determine the mass of the gas, and division

to determine its density. Apply the corresponding rule regarding significant figures to each step. Setup In the subtraction of the container mass from the combined mass of the container and the gas, the result can have only one place past the decimal point: 126.5 g - 124.6 g = 1.9 g. Thus,

in the division of the mass of the gas by the volume of the container, the result can have only two significant figures. Solution

mass of gas

126.5 g -124.6 g 1.9 g

Think About It In this case,

although each of the three numbers we started with has four significant figures, the solution has only two significant figures.

density =

~

one place past the decimal point (two significant figures)

1.9 g 2

3 =

0.00193 g/cm

3

~

round to 0.0019 g/cm

3

9 .850 X 10 cm The density of the gas is 1.9 X 10- 3 g/cm 3

Practice Problem A An empty container with a volume of 150.0 cm 3 is weighed and found to have a

mass of 72.5 g. The container is filled with a liquid and reweighed. The mass of the container and the liquid is 194.3 g. Determine the density of the liquid to the appropriate number of significant figures. Practice Problem B Another empty container with an unknown volume is weighed and found to have a mass of 81.2 g. The container is then filled with a liquid with a density of 1.015 g/cm 3

and reweighed. The mass of the container and the liquid is 177.9 g. Determine the volume of the container to the appropriate number of significant figures.

Accuracy and Precision Accuracy and precision are two ways to gauge the quality of a set of measured numbers. Although the difference between the two terms may be subtle, it is important. Accuracy tells us how close

SECTION 1.5

Uncertainty in Measurement

19

a measurement is to the true value. Precision tells us how closely multiple measurements of the same thing are to one another (Figure 1.11). Suppose that three students are asked to determine the mass of an aspirin tablet. Each student weighs the aspirin tablet three times. The results (in grams) are

Average value

Student A 0.335 0.331 0.333 0.333

Student C 0.369 0.373 0.371 0.371

Student B 0.357 0.375 0.338 0.357

(a)

The true mass of the tablet is 0.370 g. Student A's results are more precise than those of student B, but neither set of results is very accurate. Student C's results are both precise (very small deviation of individual masses from the average mass) and accurate (average value very close to the true value). Figure 1.12 shows all three students' results in relation to the true mass of the tablet. Highly accurate measurements are usually precise, as well, although highly precise measurements do not necessarily guarantee accurate results. For example, an improperly calibrated meterstick or a faulty balance may give precise readings that are significantly different from the correct value.

Checkpoint-1.5 1.5.1

(b)

1.5.3

What is the result of the following calculation to the correct number of significant figures?

63 .102 X 10.18 =

153.1

a) 642.3784

a) 29

b) 642.378

b) 28.9

c) 642.38

c) 28.89 d) 30

e) 642

e) 3 X 10 1

Which of the following is the sum of the following numbers to the correct number of significant figures?

1.5.4

-7-

5.3 =

(c)

d) 642.4

3.115

Figure 1.11

(6 .266 - 6.261)

a) 760.8431

a) 9.5785 X 10-6

b) 760.843

b) 9.579 X 10-6

c) 760.84

c) 9.58 X 10-6

d) 760.8

d) 9.6 X 10-6

e) 761

e) 1 X 10-5

0.380 Student A

Student B

The distribution of papers shows the difference between accuracy and precision. (a) Good accuracy and good precision.

What is the result of the following calculation to the correct number of significant figures ?

+ 0.2281 + 712.5 + 45 =

-

-

•

Uncertainty in Measurement

What is the result of the following calculation to the COlTect number of significant figures?

1.5.2

,

«

0 .370

Student C

•

-7-

(b) Poor accuracy but good precision. (c) Poor accuracy and poor precision.

522.0 =

+

•

0.360 Measurement 1

0.335 g

0.357 g

0.369 g

~

OIl

0.350

~

Measurement 2 Measurement 3

0.33 1 g 0.33 3 g

0.375 g 0.338 g

0.373 g 0.371 g

en en

:s'"

0.340 0.330 0 .320

0

0.310 0.300

,

,

A

B

•

Measured mass True mass

, C

Student

Figure 1.12

Graphing the students' data illustrates the difference between precision and accuracy. Student A's results are precise (values are close to one another) but not accurate because the average value is far from the true value. Student B's results are neither precise nor accurate. Student C's results are both precise and accurate.

20

CHAPTER 1

Chemistry: The Central Science

Using Units and Solving Problems Solving problems correctly in chemistry requires careful manipulation of both numbers and units. Paying attention to the units will benefit you greatly as you proceed through this, or any other, chemistry course.

Conversion Factors A conversion factor is a fraction in which the same quantity is expressed one way in the numerator and another way in the denominator. By definition, for example, 1 in = 2.54 cm. We can derive a conversion factor from this equality by writing it as the following fraction:

1 in 2.54 cm Because the numerator and denominator express the same length, this fraction is equal to 1; as a result, we can equally well write the conversion factor as 2.54 cm 1 in Because both forms of this conversion factor are equal to 1, we can multiply a quantity by either form without changing the value of that quantity. This is useful for changing the units in which a given quantity is expressed-something you will do often throughout this text. For instance, if we need to convert a length from inches to centimeters, we mUltiply the length in inches by the appropriate conversion factor. 12.00 %x 2.54 cm

1%

=

30.48 cm

We chose the form of the conversion factor that cancels the unit inches and produces the desired unit, centimeters. The result contains four significant figures because exact numbers, such as those obtained from definitions, do not limit the number of significant figures in the result of a calculation. Thus, the number of significant figures in the answer to this calculation is based on the number 12.00, not the number 2.54.

Dimensional Analysis

Tracking Units

The use of conversion factors in problem solving is called dimensional analysis or the factor-label method. Many problems require the use of more than one conversion factor. The conversion of 12.00 in to meters, for example, takes two steps: one to convert inches to centimeters, which we have already demonstrated; and one to convert centimeters to meters. The additional conversion factor required is derived from the equality 1 m = 100 cm and is expressed as either 100cm 1m

or

1m 100cm

We choose the conversion factor that will introduce the unit meter and cancel the unit centimeter (i.e. , the one on the right). We can set up a problem of this type as the following series of unit conversions so that it is unnecessary to calculate an intermediate answer at each step: 12.00 %X 2.54 ~ X 1 m 100 ~ 1 i«'

0.3048 m

Careful . . . . . tracking of units and their cancellation can be a valuable tool in checking your work. If we had accidentally used the reciprocal of one of the conversion factors, the resulting units would have been something other than meters, Unexpected or nonsensical units can reveal an error in your problem-solving strategy.

..... If we had accidentally used t he reciproca l of the conversion from centimeters to meters, the result would have been 3048 cm'/ m, which wou ld make no sense- both because the units are nonsensical and because the numerica l result is not reasonable. You know that 12 inches is a foot and that a foot is not equa l to thousands of meters!

=

How Can You Enhance Your Chances of Success in Chemistry Class? calculation. Always carry at least one extra significant fig ure in intermediate calculations. Make sure that the final answer has the correct number of significant figures.

Success in a chemistry class depends largely on problem-solving ability. The sample problems throughout the text are designed to help you develop problem-solving skills. Each is divided into four steps: Strategy, Setup, Solution, and Think About It.

Think About It: Consider your calculated result and ask yourself whether or not it makes sense. Compare the units and the magnitude of your result with your ballpark estimate from the Strategy step. If your result does not have the appropriate units, or if its magnitude or sign is not reasonable, check your solution for possible errors. A very important part of problem solving is being able to judge whether the answer is reasonable. It is relatively easy to spot a wrong sign or incorrect units, but you should also develop a sense of magnitude and be able to tell when an answer is either way too big or way too small. For example, if a problem asks how many molecules are in a sample and you calculate a number that is less than I , you should know that it cannot be correct.

Strategy: Read the problem carefully and determine what is being asked and what information is provided. The Strategy step is where you should think about what skills are required and layout a plan for solving the problem. Give some thought to what you expect the result to be. If you are asked to determine the number of atoms in a sample of matter, for example, you should expect the answer to be a whole number. Determine what, if any, units should be associated with the result. When possible, make a ballpark estimate of the magnitude of the correct result, and make a note of your estimate. Setup: Next, gather the information necessary to solve the problem. Some of the information will have been given in the problem itself. Other information, such as equations, constants, and tabulated data (including atomic masses) should also be brought together in this step. Write down and label clearly all of the information you will use to solve the problem. Be sure to write appropriate units with each piece of information.

Finally, each sample problem is followed by at least one practice problem. This typically is a very similar problem that can be solved using the same strategy. Most sample problems also have a second practice problem that tests the same skills, but requires an approach slightly different from the one used to solve the preceding sample and practice problems. Regular use of the sample problems and practice problems in this text can help you develop an effective set of problemsolving skills. They can also help you assess whether you are ready to move on to the next new concepts. If you struggle with the practice problems, then you probably need to review the corresponding sample problem and the concepts that led up to it.

Solution: Using the necessary equations, constants, and other information, calculate the answer to the problem. Pay particular attention to the units associated with each number, tracking and canceling units carefully throughout the calculation. In the event that multiple calculations are required, label any intermediate results, but don't round to the necessary number of significant figures until the final

Sample Problem 1.7 shows how to derive conversion factors and use them to do unit conversIOns. •

The Food and Drug Administration (FDA) recommends that dietary sodium intake be no more than 2400 mg per day. What is this mass in pounds (lb), if 1 lb = 453.6 g ?

1g 1000 mg

or

1000 mg 1g

and

1 lb

=

·• • • •

Strategy This problem requires a two-step dimensional analysis, because we must convert milligrams to grams and then grams to pounds. Assume the number 2400 has four significant figures. Setup The necessary conversion factors are derived from the equalities 1 g 453.6 g.

••

•

1000 mg and 1 lb

= ·• •

453.6 g

or 453.6 g l I b

•

From each pair of conversion factors, we select the one that will result in the proper unit cancellation. •

• •

2400 ~ X

Because pounds are much larger than milligrams, a given mass will be a much smaller number of pounds than of milligrams.

•

•

Solution

• •

• • •

1.% X 1 lb = 0.00529llb 1000;ag 453.6%

Think About It Make sure that the magnitude of the result is reasonable and that the units have canceled properly. If we had mistakenly multiplied by 1000 and 453.6 instead of dividing by them, the result (2400 mg X 1000 mglg X 453.6 glIb 2 9 =. . ..1.089 X 10 mg /lb) would be ......... . unreasonably large-and the units would not have canceled properly.

(Continued) 21

•

22

CHAPTER 1

Chem istry : The Centra l Science

Practice Problem A The American Heart Association recommends that healthy adults limit dietary cholesterol to no more than 300 mg per day. Convert this mass of cholesterol to ounces (1 oz =

28.3459 g). Assume 300 mg has just one significant figure. Practice Problem B A gold nugget has a mass of 24.98 g. What is its mass in ounces?

Sample Problem 1.8 shows how to handle problems in which conversion factors are squared or cubed in dimensional analysis.

Sample Problem 1.8 An average adult has 5.2 L of blood. What is the volume of blood in cubic meters ? Strategy There are several ways to solve a problem such as this. One way is to conveli liters to

cubic centimeters and then cubic centimeters to cubic meters. Setup 1 L = 1000 cm 3 and 1 cm = 1 X 10- m. When a unit is raised to a power, the corresponding 2

Think About It Based on the

preceding conversion factors, I L = I X 10- 3 m 3 Therefore, 5 L of blood would be equal to 5 X 10- 3 m 3, which is close to the calculated answer.

conversion factor must also be raised to that power in order for the units to cancel appropriately. Solution

3

2

5.2 L X 1000 cm X 1 X 10- m 3 = 5.2 X 10-3 m3 IL Icm

Practice Problem A The density of silver is 10.5 g/cm3 What is its density in kg/m' ?

3

Practice Problem B The density of mercury is 13.6 g/cm . What is its density in mg/mm3?

Checkpoint 1.6 1.6.1

Using Units and Solving Problems

The density of lithium metal is 535 kg/m3 What is this density in g/cm 3?

1.6.3

a) 0.000535 g/ cm 3

What is the volume of a 5.75-g object that has a density of 3.97 g/cm 3? a) 1.45 cm 3

b) 0.535 g/cm3

b) 0.690 cm3 c) 22.8 cm3

c) 0.0535 g/ cm 3

d) 0.0438 cm 3

d) 0.54 g/ cm 3

e) 5.75 cm 3

e) 53.5 g/ cm 3

1.6.4 1.6.2

Convert 43.1 cm3 to liters.

How many cubic centimeters are there in a cubic meter?

a) 43.1 L

a) 10

b) 43,100 L

b) 100

c) 0.0431 L

c) 1000

d) 4310L

d) I X 104

e) 0.043 L

e) 1 X 106

APPLYING WHAT YOU'VE LEARNED

23

Applying What You've learned Although naturally occurring smallpox was eradicated by a superbly coordinated effort including the World Health Organization and health-care providers worldwide, the classification of smallpox as a Category A bioterrorism agent has renewed interest in its treatment and prevention. Moreover, although vaccination against the disease is considered relatively safe for most individuals, it is not entirely without risk. The CDC estimates that 14 to 52 out of every million people who are vaccinated for smallpox will suffer serious, potentially life-threatening reactions to the vaccine. In these cases, immediate medical attention is required. The first course of treatment is with vaccinia immune globulin (VIG). If a p'atleni does not respond to treatment wi'th \liG, a second option is cidofovir, a drug that currently is approved by the Food and Drug of the eye in individuals with Administration (FDA) to treat specific viral infections . . . . . .. . . compromised immune systems. Cidofovir, marketed under the name Vistide, is distributed in vials containing 375 mg of the drug dissolved in 5 mL of water. The manufacturer specifies that the drug should be kept at room temperature (68 °F-7r F). The vial contents are first diluted with saline and then administered intravenously with a recommended dosage of 5 mg cidofovir per kilogram of body weight.

Smallpox vaccine is made from the vaccinia virus.

Both drugs are available for use in the treatment of a serious reaction to smal lpox vaccine only through the FDA's Investigational New Drug (IND) protocol.

NH2

Problems:

a)

b)

c)

Convert cidofovir's recommended storage-temperature range to the Celsius scale. [ ~~ Sample Problem 1.2] If the fluid in a single vial of cidofovir has a volume of 5.00 mL and a mass of 5.89 g, what is the density of the fluid? [ ~~ Sample Problem l.3] (Report the density to the appropriate number of significant figures. [ ~~ Sample Problem l.5 ]) What mass of cidofovir should be administered to a 177-lb man (lIb = 0.4536 kg) ? [ ~~ Sample Problem l.7]

N :Y'

o

N

OH Cidofovir

CHAPTER 1

24

Chemist ry: The Cent ral Sci ence

CHAPTER SUMMARY Section 1.1 •

Chemistry is the study of matter and the changes matter undergoes.

•

Chemists go about research using a set of guidelines and practices known as the scientific method, in which observation s give rise to laws, data give rise to hypotheses, hypotheses are tested with experiments, and successful hypotheses give rise to theories, which are further tested by experiment.

Section 1.2 •

•

Physical properties are those that can be determined without the matter in question undergoing a chemical change. A physical change is one in which the identity of the matter involved does not change.

•

Chemical properties are determined only as the result of a chemical change, in which the original substance is converted to a different substance. Physical and chemical properties may be intensive (independent of the amount of matter) or extensive (dependent on the amount of matter).

Section 1.5

All matter exists either as a substance or as a mixture of substances. Substances may be elements (containing only one kind of atom) or compounds (containing two or more kinds of atoms). A mixture may be homogeneous (a solution) or heterogeneous. Mixtures may be separated using physical processes. Compounds can be separated into their constituent elements using chemical processes. Elements cannot be separated into simpler substances.

•

Measured numbers are inexact. Numbers obtained by counting or that are part of a definition are exact numbers.

•

Significant figures are used to specify the uncertainty in a measured number or in a number calculated using measured numbers. Significant figures must be carried through calculations such that the implied uncertainty in the final answer is reasonable.

•

Accuracy refers to how close measured numbers are to a true value. Precision refers to how close measured numbers are to one another.

Section 1.3 •

Scientists use a system of units referred to as the International System of Units or SI units.

•

There are seven base SI units including the kilogram (for mass) and the kelvin (for temperature). SI units for such quantities as volume and density are derived from the base units.

Section 1.6

Section 1.4 •

•

A conversion factor is a fraction in which the numerator and denominator are the same quantity expressed in different units. Multiplying by a conversion factor is unit conversion.

•

Dimensional analysis is a series of unit conversions used in the solution of a multistep problem.

Substances are identified by their quantitative (involving numbers) and qualitative (not involving numbers) properties.

KEyWORDS Accuracy, 18

Element, 6

Law,S

Quantitative property, 14

Chemical change, 15

Extensive property, 15

Mass, 9

Scientific method, 5

Chemical property, 15

Heterogeneous mixture, 7

Matter, 4

SI unit, 9

Chemistry, 4

Homogeneous mixture, 7

Mixture, 7

Significant figures, 15

Compound, 7

Hypothesis, 5

Physical change, 15

Substance, 6

Conversion factor, 20

Intensive property, 15

Physical property, 14

Theory,S

Density, 12

International System of Units, 9

Precision, 19

Dimensional analysis, 20

Kelvin, 10

Qualitative property, 14

KEY EQUATIONS + 273.15

1.1

K = °C

1.2

temperature in degrees Celsius = (temperature in degrees Fahrenheit - 32°F) X

1.3

temperature in degrees Fahrenheit =

1.4

m d=V

~:~

X (temperature in degrees Celsius)

~:;

+ 32°F

25

QUESTIONS AND PROBLEMS

QUESTIONS AND PROBLEMS ======================~--~-=~

Section 1.1: The Study of Chemistry

1.14

Classify each of the following substances as an element or a compound: (a) hydrogen, (b) water, (c) gold, (d) sugar.

1.15

Classify each of the following as an element, a compound, a homogeneous mixture, or a heterogeneous mixture: (a) seawater, (b) helium gas, (c) sodium chloride (salt), (d) a bottle of soft drink, (e) a milkshake, (f) air in a bottle, (g) concrete.

1.16

Identify each of the diagrams shown here as a solid, liquid, gas, or mixture of two substances.

Review Questions 1.1

Define the terms chemistry and matter.

1.2

Explain what is meant by the scientific method.

1.3

What is the difference between a hypothesis and a theory?

Problems 1.4

1.5

1.6

Classify each of the following statements as a hypothesis, law, or theory. (a) Beethoven's contribution to music would have been much greater if he had malTied. (b) An autumn leaf gravitates toward the ground because there is an attractive force between the leaf and Earth. (c) All matter is composed of very small particles called atoms. Classify each of the followi ng statements as a hypothesis, law, or theory. (a) The force acting on an object is equal to its mass times its acceleration. (b) The universe as we know it started with a big bang. (c) There are many civilizations more advanced than ours on other planets.

• (a)

1.17

(b)

(c)

(d)

Identify each of the diagrams shown here as an element or a compound.

Identify the elements present in the following molecules (see Table 1.1 ).

(a)

(b)

(c)

(d)

Section 1.3: Scientific Measurement (a)

1.7

(b)

(c)

(d)

Identify the elements present in the following molecules (see Table 1.1 ).

(a)

(b)

(c)

Review Questions 1.18

Name the SI base units that are important in chemistry, and give the SI units for expressing the following: (a) length, (b) volume, (c) mass, (d) time, (e) temperature.

1.19

Write the numbers represented by the following prefixes : (a) mega-, (b) kilo-, (c) deci-, (d) centi-, (e) milli-, (f) micro-, (g) nano- , (h) pico-.

1.20

What units do chemists normally use for the density of liquids and solids? For the density of gas? Explain the differences.

1.21

What is the difference between mass and weight? If a person weighs 168 lb on Earth, about how much would the person weigh on the moon ? Describe the three temperature scales used in the laboratory and in everyday life: the Fahrenheit, Celsius, and Kelvin scales.

(d)

Section 1.2: Classification of Matter

Review Questions 1.8

Give an example for each of the following terms: (a) matter, (b) substance, (c) mixture.

1.9

Give an example of a homogeneous mixture and an example of a heterogeneous mixture.

1.22

1.10

Give an example of an element and a compound. How do elements and compounds differ?

Problems

1.11

1.13

Bromine is a reddish-brown liquid. Calculate its density (in g/mL) if 586 g of the substance occupies 188 mL.

1.24

The density of ethanol, a colorless liquid that is commonly known as grain alcohol, is 0.798 g/mL. Calculate the mass of 17.4 mL of the liquid .

1.25

Convert the following temperatures to degrees Celsius or Fahrenheit: (a) 95 °F, the temperature on a hot summer day; (b) 12°F, the temperature on a cold winter day; (c) a 102°F fever; (d) a furnace operating at 1852°F; (e) -273. 15°C (theoretically the lowest attainable temperature).

What is the number of known elements?

Problems 1.12

1.23

Give the names of the elements represented by the chemical symbols Li, F, P, Cu, As, Zn, CI, Pt, Mg, U, AI, Si, Ne (see the table inside the front cover). Give the chemical symbols for the following elements: (a) potassium, (b) tin, (c) chromium, (d) boron, (e) barium, (f) plutonium, (g) sulfur, (h) argon, (i) mercury (see the table inside the front cover).

CHAPTER 1

26 1.26

1.27

1.28

1.29

1.30

Chemistry: The Central Science

(a) Normally the human body can endure a temperature of 105°F for only short periods of time without permanent damage to the brain and other vital organs. What is this temperature in degrees Celsius? (b) Ethylene glycol is a liquid organic compound that is used as an antifreeze in car radiators. It freezes at -1l.5 °e. Calculate its freezing temperature in degrees Fahrenheit. (c) The temperature on the surface of the sun is about 6300°C. What is this temperature in degrees Fahrenheit?

Section 1.5: Uncertainty in Measurement Review Questions 1.40

The density of water at 40°C is 0.992 g/mL. What is the volume of 2.50 g of water at this temperature?

Comment on whether each of the following statements represents an exact number: (a) 50,247 tickets were sold at a sporting event, (b) 509.2 rnL of water was used to make a birthday cake, (c) 3 dozen eggs were used to make a breakfast, (d) 0.41 g of oxygen was inhaled in each breath, (e) Earth orbits the sun every 365.2564 days.

1.41

The density of platinum (Pt) is 21.5 g!cm 3 at 25 °C. What is the volume of 87.6 g of Pt at this temperature?

What is the advantage of using scientific notation over decimal notation?

1.42

Convert the following temperatures to kelvin: (a) 113°C, the melting point of sulfur; (b) 37°C, the normal body temperature; (c) 357°C, the boiling point of mercury.

Define significant figure. Discuss the importance of using the proper number of significant figures in measurements and calculations.

1.43

Convert the following temperatures to degrees Celsius: (a) 77 K, the boiling point of liquid nitrogen, (b) 4.2 K, the boiling point of liquid helium, (c) 601 K, the melting point oflead.

Distinguish between the terms accuracy and precision. In general, explain why a precise measurement does not always guarantee an accurate result.

Problems

Section 1.4: The Properties of Matter

1.44

Express the following numbers in scientific notation: (a) 0.000000027, (b) 356, (c) 47,764, (d) 0.096.

1.45

Express the following numbers as decimals: (a) 1.52 X 10- 2 , (b) 7.78 X 10-8.

1.46

Express the answers to the following calculations in scientific notation:

Review Questions 1.31

What is the difference between qualitative data and quantitative data?

1.32

Using examples, explain the difference between a physical property and a chemical property.

1.33

How does an intensive property differ from an extensive property?

1.34

Determine which of the following properties are intensive and which are extensive: (a) length, (b) volume, (c) temperature, (d) mass.

(a) 145.75 + (2.3 (b) 79,500 -7- (2.5 (c) (7 .0 X 10-3) (d) ( 1.0 X 104) X 1.47

Problems 1.35

1.36

1.37

Classify the following as qualitative or quantitative statements, giving your reasons. (a) The sun is approximately 93 million mi from Earth. (b) Leonardo da Vinci was a better painter than Michelangelo. (c) Ice is less dense than water. (d) Butter tastes better than margarine. (e) A stitch in time saves nine. Determine whether the following statements describe chemical or physical properties: (a) Oxygen gas supports combustion. (b) Fertilizers help to increase agricultural production. (c) Water boils below 100°C on top of a mountain. (d) Lead is denser than aluminum. (e) Uranium is a radioactive element. Determine whether each of the following describes a physical change or a chemical change: (a) The helium gas inside a balloon tends to leak out after a few hours. (b) A flashlight beam slowly gets dimmer and finally goes out. (c) Frozen orange juice is reconstituted by adding water to it. (d) The growth of plants depends on the sun's energy in a process called photosynthesis. (e) A spoonful of salt dissolves in a bowl of soup.

1.39

A student pours 44.3 g of water at 10°C into a beaker containing 115.2 g of water at 100 e. What are the final mass, temperature, and density of the combined water? The density of water at 10°C is 1.00 g/rnL. A 37.2-g sample oflead (Pb) pellets at 20°C is mixed with a 62.7-g sample of lead pellets at the same temperature. What are the final mass, temperature, and density of the combined sample? The density of Pb at 20°C is 11.35 g/cm3 .

Express the answers to the following calculations in scientific notation: (a) 0.0095 + (8.5 X 10- 3) (b) 653 -7- (5.75 X 10- 8) (c) 850,000 - (9.0 X 105) (d) (3.6 X 10-4) X (3.6 X 106)

1.48

Determine the number of significant figures in each of the following measurements: (a) 4867 mi, (b) 56 rnL, (c) 60,104 tons, (d) 2900 g, (e) 40.2 g/cm 3, (f) 0.0000003 cm, (g) 0.7 min, (h) 4.6 X 10 19 atoms.

1.49

Determine the number of significant figures in each of the following measurements: (a) 0.006 L, (b) 0.0605 dm, (c) 60.5 mg, (d) 605.5 cm 2 , (e) 9.60 X 103 g, (f) 6 kg, (g) 60 m.

1.50

Carry out the following operations as if they were calculations of experimental results, and express each answer in the correct units with the con-ect number of significant figures: (a) 5.6792 m + 0.6 m + 4.33 m (b) 3.70 g - 2.9133 g (c) 4.51 cm X 3.6666 cm

1.51 1.38

10-1 ) X 102) (8.0 X 10-4) (9.9 X 106) X

Cany out the following operations as if they were calculations of experimental results, and express each answer in the con-ect units with the con-ect number of significant figures: (a) 7.310 km -7- 5.70 km (b) (3.26 X 10- 3 mg) - (7.88 X 10-5 mg) (c) (4.02 X 106 dm) + (7.74 X 107 dm)

QUESTIONS AND PROBLEMS

1.52

1.53

Three students (A, B, and C) are asked to detennine the volume of a sample of ethanol. Each student measures the volume three times with a graduated cylinder. The results in milliliters are: A (87.1, 88.2, 87.6); B (86.9, 87.1, 87.2); C (87.6,87.8,87 .9). The true volume is 87.0 mL. Comment on the precision and the accuracy of each student's results.

1.67

The density of ammonia gas under certain conditions is 0.625 gIL. Calculate its density in g/cm3

1.68

(a) Carbon monoxide (CO) is a poisonous gas because it binds very strongly to the oxygen carrier hemoglobin in blood. A concentration of 8.00 X 102 ppm by volume of carbon monoxide is considered lethal to humans. Calculate the volume in liters occupied by carbon monoxide in a room that measures 17.6 m long, 8.80 m wide, and 2.64 m high at this concentration. (b) Prolonged exposure to mercury (Hg) vapor can cause neurological disorder and respiratory problems. For safe air quality control, the concentration of mercury vapor must be under 0.050 mg/m3. Convert this number to gIL. (c) The general test for type II diabetes is that the blood sugar (glucose) level should be below 120 mg per deciliter (mg/dL). Convert this number to micrograms per milliliter (fLg/mL).

1.69

The average time it takes for a molecule to diffuse a distance of x cm is given by

Three apprentice tailors (X, Y, and Z) are assigned the task of measuring the seam of a pair of trousers. Each one makes three measurements. The results in inches are X (31.5, 31.6, 31.4); Y (32.8,32.3,32,7); Z (3 1.9, 32.2, 32.1). The true length is 32.0 in. Comment on the precision and the accuracy of each tailor's measurements.

Section 1.6: Using Units and Solving Problems Problems 1.54

Carry out the following conversions: (a) 22.6 m to decimeters, (b) 25.4 mg to kilograms, (c) 556 mL to liters, (d) 10.6 kg/m3 to g/cm3.

1.55

Carry out the following conversions: (a) 242lb to milligrams, (b) 68.3 cm3 to cubic meters, (c) 7.2 m 3 to liters, (d) 28.3 fLg to pounds.

1.56

The average speed of helium at 25°C is 1255 mfs. Convert this speed to miles per hour (mph).

1.57

How many seconds are there in a solar year (365.24 days)?

1.58

How many minutes does it take light from the sun to reach Earth? (The distance from the sun to Earth is 93 million mi; the speed of light is 3.00 X 108 mfs.)

1.59

A slow jogger runs a mile in 13 min. Calculate the speed in (a) in/s, (b) mfmin, (c) kmJh (1 mi = 1609 m; 1 in = 2.54 cm).

1.60

A 6.0-ft person weighs 168 lb. Express this person's height in meters and weight in kilograms (lIb = 453.6 g; 1 m = 3.28 ft).

1.61

The current speed limit in some states in the United States is 55 mph. What is the speed limit in kilometers per hour (1 mi = 1609 m)?

1.62

For a fighter jet to take off from the deck of an aircraft carrier, it must reach a speed of 62 mfs. Calculate the speed in miles per hour.

1.63

The "normal" lead content in human blood is about 0.40 part per million (that is, 0.40 g of lead per million grams of blood). A value of 0.80 part per million (ppm) is considered to be dangerous. How many grams of lead are contained in 6.0 X 10 3 g of blood (the amount in an average adult) if the lead content is 0.62 ppm?

1.64

1.65

1.66

Carry out the following conversions: (a) 1.42 light-years to miles (a light-year is an astronomical measure of distance-the distance traveled by light in a year, or 365 days; the speed of light is 3.00 X 108 mfs), (b) 32.4 yd to centimeters, (c) 3.0 X 1010 cmfs to ftls. Carry out the following conversions: (a) 185 nm to meters, (b) 4.5 billion years (roughly the age of Earth) to seconds (assume 365 days in a year), (c) 71.2 cm 3 to cubic meters, (d) 88.6 m3 to liters. Aluminum is a lightweight metal (density = 2.70 g/cm3) used in aircraft construction, high-voltage transmission lines, beverage cans, and foils. What is its density in kg/m3?

27

2

t= x 2D where t is the time in seconds and D is the diffusion coefficient. Given that the diffusion coefficient of glucose is 5.7 X 10-7 cm2/s, calculate the time it would take for a glucose molecule to diffuse 10 fLm, which is roughly the size of a cell. 1.70

A human brain weighs about 1 kg and contains about 1011 cells. Assuming that each cell is completely filled with water (density = 1 g/mL), calculate the length of one side of such a cell if it were a cube. If the cells are spread out into a thin layer that is a single cell thick, what is the surface area in square meters?

Additional Problems 1.71

Which of the following statements describe physical properties and which describe chemical properties? (a) Iron has a tendency to rust. (b) Rainwater in industrialized regions tends to be acidic. (c) Hemoglobin molecules have a red color. (d) When a glass of water is left out in the sun, the water gradually disappears. (e) Carbon dioxide in air is converted to more complex molecules by plants during photosynthesis.

1.72

Give one qualitative and one quantitative statement about each of the following: (a) water, (b) carbon, (c) iron, (d) hydrogen gas, (e) sucrose (cane sugar), (f) salt (sodium chloride), (g) mercury, (h) gold, (i) air.

1.73

In 2004, about 95.0 billion lb of sulfuric acid were produced in the United States. Convert this quantity to tons.

1.74

In detennining the density of a rectangular metal bar, a student made the following measurements: length, 8.53 cm; width, 2.4 cm; height, 1.0 cm; mass, 52.7064 g. Calculate the density of the metal to the correct number of significant figure s.

1.75

Calculate the mass of each of the following: (a) a sphere of gold with a radius of 10.0 em (volume of a sphere with a radius r is V = 4/3TIr3; density of gold = 19.3 g/cm3), (b) a cube of platinum of edge length 0.040 mm (density = 21.4 g/cm 3), (c) 50.0 mL of ethanol (density = 0.798 g/mL).

1.76

A cylindrical glass tube 12.7 cm in length is filled with mercury (density = 13.6 g/mL). The mass of mercury needed to fill the tube is 105.5 g. Calculate the inner diameter of the tube (volume of a cylinder of radius r and length h is V = TIr2h).

28 1.77

CHAPTER 1

Chemistry: The Central Science

The following procedure was used to determine the volume of a flask. The flask was weighed dry and then filled with water. If the masses of the empty flask and filled flask were 56.12 g and 87.39 g, respectively, and the density of water is 0.9976 g/cm3, calculate the volume of the flask in cubic centimeters.

1.78

The speed of sound in air at room temperature is about 343 mls. Calculate this speed in miles per hour (1 mi = 1609 m).

1.79

A piece of silver (Ag) metal weighing 194.3 g is placed in a graduated cylinder containing 242.0 mL of water. The volume of water now reads 260.5 mL. From these data calculate the density of silver.

1.80

4

A lead sphere has a mass of 1.20 X 10 g, and its volume is 1.05 X 10 3 cm 3 Calculate the density of lead.

1.82

Lithium is the least den se metal known (density = 0.53 g/cm\ What is the volume occupied by 1.20 X 103 g of lithium?

1.83

The medicinal thermometer commonly used in homes can be read to +O.I °F, whereas those in the doctor's office may be accurate to +0.1 °C. Percent error is often expressed as the absolute val ue of the difference between the tme value and the experimental value, divided by the true value: true value - experimental value

percent error = - - - -- - - - -- - true value

1.86

1.87

1.88

1.89

Magnesium (Mg) is a valuable metal used in alloys, in batteries, and in the manufacture of chemicals. It is obtained mostly from seawater, which contains about 1.3 g of Mg for every kilogram of seawater. Referring to Problem l.89, calculate the volume of 4 seawater (in liters) needed to extract 8.0 X 10 tons of Mg, which is roughly the annual production in the United States.

1.91

A student is given a cmcible and asked to prove whether it is made of pure platinum. She first weighs the cmcible in air and then weighs it suspended in water (density = 0.9986 glmL). The readings are 860.2 g and 820.2 g, respectively. Based on these measurements and given that the density of platinum is 21.45 glcm 3, what should her conclusion be? (Hint: An object suspended in a fluid is buoyed up by the mass of the fluid displaced by the object. Neglect the buoyancy of air.)

1.92

The surface area and average depth of the Pacific Ocean are 1.8 X 108 krn 2 and 3.9 X 103 m, respectively. Calculate the volume of water in the ocean in liters.

1.93

The unit "troy ounce" is often used for precious metals such as gold (Au ) and platinum (Pt) (1 troy ounce = 31.103 g). (a) A gold coin weighs 2.41 troy ounces . Calculate its mass in grams. (b) Is a troy ounce heavier or lighter than an ounce (lIb = 160z; 1 lb = 453.6 g)?

1.94

Osmium (Os) is the densest element known (density = 22.57 g/cm\ Calculate the mass in pounds and in kilograms of an Os sphere 15 cm in diameter (about the size of a grapefruit) (volume of a sphere of radius r is 51T1)).

1.95

Calculate the percent error for the follow ing measurements: (a) The density of alcohol (ethanol) is found to be 0.802 g/mL (true value = 0.798 g/mL). (b) The mass of gold in an earring is analyzed to be 0.837 g (true value = 0.864 g).

1.96

The natural abundances of elements in the human body, expressed as percent by mass, are oxygen (0), 65 percent; carbon (C), 18 percent; hydrogen (H), 10 percent; nitrogen (N), 3 percent; calcium (Ca), 1.6 percent; phosphorus (P), 1.2 percent; all other elements, 1.2 percent. Calculate the mass in grams of each element in the body of a 62-kg person.

1.97

The men's world record for mnning a mile outdoors (as of 1997) is 3 min 44.39 s. At this rate, how long would it take to run a 1500-m race (1 mi = 1609 m)?

1.98

Venus, the second closest planet to the sun, has a surface temperature of 7.3 X 102 K. Convert this temperature to degrees Celsius and degrees Fahrenheit.

1.99

Chalcopyrite, the principal ore of copper (Cu), contains 34.63 percent Cu by mass. How many grams of Cu can be obtained from 5.11 X 103 kg of the ore?

X 100%

The vertical lines indicate absolute value. In degrees Celsius, express the percent error expected from each of these thermometers in measuring a person's body temperature of 38.9°C.

1.85

1.90

The experiment described in Problem 1.79 is a cmde but convenient way to determine the density of some solids. Describe a similar experiment that would enable you to measure the density of ice. Specifically, what would be the requirements for the liquid used in your experiment?

1.81

1.84

density is 1.03 g/mL. Calculate the total mass of sodium chloride in kilograms and in tons (1 ton = 2000 lb; lIb = 453.6 g) .

Vanillin (used to flavor vanilla ice cream and other food s) is the substance whose aroma the human nose detects in the smallest amount. The threshold limit is 2.0 X 1O- 11 g per liter of air. If the current price of 50 g of vanillin is $1 12, determine the cost to supply enough vanillin so that the aroma could be detected in a large aircraft hangar with a volume of 5.0 X 107 ft3 At what temperature does the numelical reading on a Celsius thermometer equal that on a Fahrenheit thermometer? Suppose that a new temperature scale has been devised on which the melting point of ethanol (-117 .3°C) and the boiling point of ethanol (78 .3 °C) are taken as OOS and 100 0 S, respectively, where S is the symbol for the new temperature scale. Derive an equation relating a reading on this scale to a reading on the Celsius scale. What would this thermometer read at 25 °C? A resting adult requires about 240 mL of pure oxygen per minute and breathes about 12 times every minute. If inhaled air contains 20 percent oxygen by volume and exhaled air 16 percent, what is the volume of air per breath? (Assume that the volume of inhaled air is equal to that of exhaled air.) (a) Referring to Problem 1.87, calculate the total volume (in liters) of air an adult breathes in a day. (b) In a city with heavy traffic, the air contains 2.1 X 10-6 L of carbon monoxide (a poisonous gas) per liter. Calculate the average daily intake of carbon monoxide in liters by a person. The total volume of seawater is 1.5 X 1021 L. Assume that seawater contains 3.1 percent sodium chloride by mass and that its

4

1.100

It has been estimated that 8.0 X 10 tons of gold (Au) have been mined. Assume gold costs $625 per ounce. What is the total worth of this quantity of gold?

1.101

A 1.0-mL volume of seawater contains about 4.0 X 10- 12 g of go ld. The total volume of ocean water is 1.5 X 1021 L. Calculate the total amount of gold (in grams) that is present in seawater and the worth of the gold in dollars (see Problem 1.100). With so much gold out there, why hasn't someone become rich by mining gold from the ocean?

QUESTIONS AND PROBLEMS

1.102

Measurements show that 1.0 g of iron (Fe) contains 1.1 X 1022 Fe atoms. How many Fe atoms are in 4.9 g of Fe, which is the total amount of iron in the body of an average adult?

1.103

The thin outer layer of Earth, called the crust, contains only 0.50 percent of Earth's total mass and yet is the source of almost all the elements (the atmosphere provides elements such as oxygen, nitrogen, and a few other gases). Silicon (Si) is the second most abundant element in Earth's crust (27.2 percent by mass). Calculate the mass of silicon in kilograms in Earth's crust (mass of Earth = 5.9 X 1021 tons; 1 ton = 2000 Ib; 1 lb = 453 .6 g).

1.104

1.105

One gallon of gasoline in an automobile's engine produces on the average 9.5 kg of carbon dioxide, which is a greenhouse gas; that is, it promotes the warming of Earth's atmosphere. Calculate the annual production of carbon dioxide in kilograms if there are 40 million cars in the United States and each car covers a distance of 5000 mi at a consumption rate of 20 miles per gallon.

A gas company in Massachu setts charges $ 1.30 for 15.0 fe of natural gas. (a) ,Convert thi s rate to dollars per liter of gas. (b) If it takes 0.304 ff of gas to boil a liter of water, starting at room temperature (25°C), how much would it cost to boil a 2.1-L kettle of water?

1.113

Pheromones are compounds secreted by females of many insect species to attract mates. Typically, 1.0 X 10-8 g of a pheromone is sufficient to reach all targeted males within a radius of 0.50 mi. Calculate the density of the pheromone (in grams per liter) in a cylindrical air space having a radius of 0.50 mi and a height of 40 ft. (Volume of a cylinder of radius r and height h is 'ITr2 h.)

1.114

A bank teller is asked to assemble $ 1 sets of coins for his clients. Each set is made up of three quarters, one nickel, and two dimes. The masses of the coins are quarter, 5.645 g; nickel, 4.967 g; and dime, 2.316 g . What is the maximum number of sets that can be assembled from 33.871 kg of quarters, 10.432 kg of ni ckels, and 7.990 kg of dimes? What is the total mass (in grams) of the assembled sets of coins?

1.115

A graduated cylinder is filled to the 40.00-mL m ark with a mineral oil. The masses of the cylinder before and after the addition of the mineral oil are 124.966 g and 159.446 g, respectively. In a separate experiment, a metal ball bearing of mass 18.7 13 g is placed in the cylinder and the cylinder is again filled to the 40.00-mL mark with the mineral oil. The combined mass of the ball bearing and mineral oil is 50.952 g. Calculate the density and radius of the ball bearing (volume of a sphere of radius r is 4/3'IT?).

1.116

Bronze is an alloy made of copper (Cu) and tin (Sn). Calculate the mass of a bronze cylinder of radius 6.44 cm and length 44.37 cm. The composition of the bronze is 79.42 percent Cu and 20.5 8 percent Sn and the densities of Cu and Sn are 8.94 g/cm 3 and 7.31 g/ cm3, respectively. What assumption should you make in this calculation?

1.117

A chemist in the nineteenth century prepared an unkn ow n substance. In general, do you think it would be more diffic ult to prove that it is an element or a compound ? Explain.

1.11 8

A chemist mixes two liquids A and B to form a homogeneous mixture. The densities of the liquids are 2.0514 g/mL for A and 2.6678 g/mL for B. When she drops a small object into the mixture, she find s that the obj ect becomes suspended in the liquid; that is, it neither sinks nor fl oats. If the mi xture is made of 41.37 percent A and 58.63 percent B by volume, what is the density of the object? Can thi s procedure be used in general to determine the densities of solids? What assumptions must be made in applying thi s method ?

1.119

You are given a liquid. Briefly describe the steps you would take to show whether it is a pure substance or a homogeneous mi xture.

1.120

TUMS is a popular remedy for acid indigestion . A typical TUMS tablet contains calci um carbonate plus some inert sub stances . When ingested, it reacts with the gastric juice (hydrochl oric acid) in the stomach to give off carbon dioxide gas. When a 1.328-g tablet reacted with 40.00 mL of hydrochloric acid (density = 1.1 40 g/mL), carbon dioxide gas was given off and the resulting solution weighed 46.699 g. Calculate the number of liters of carbon dioxide gas released if its density is 1.81 gIL.

1.121

A 250-mL glass bottle was filled with 242 mL of water at 20°C and tightly capped. It was then left outdoors overnight, where the average temperature was -5°C. Predict what would happen. The density of water at 20°C is 0 .998 g/ cm 3 and that of ice at _5 °C is 0 .916 g/ cm 3

2

1.106

A sheet of aluminum (AI) foil has a total area of 1.000 ft and a mass of 3.636 g. What is the thickness of the foil in millimeters (den sity of Al = 2.699 g/cm 3)?

1.107

Comment on whether each of the following is a homogeneous mixture or a heterogeneous mixture : (a) air in a closed bottle, (b) air over New York City.

1.108

Chlorine is used to disinfec t swimming pools. The accepted concentration for this purpose is 1 ppm chlorine, or 1 g of chlorine per million gram s of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 6.0 percent chlorine by 4 mass and there are 2.0 X 10 gallons (gal) of water in the pool (1 gal = 3.79 L; density of liquids = 1.0 g/mL).

1.109

The world's total petroleum reserve is estimated at 2.0 X 1022 joules [a joule (J) is the unit of energy where 1 J = 1 kg . m 2/s 2]. At the present rate of consumption, 1.8 X 1020 j oules per year (J/yr), how long would it take to exhaust the supply?

1,110

In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 2 0.10 mL of oil could spread over the surface of water about 40 m in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers (1 nm = 1 X 10- 9 m).

1.111

1.112

10

The radius of a copper (Cu) atom is roughly 1.3 X 10- m. How many times can you divide evenly a lO-cm-Iong piece of copper wire until it is reduced to two separate copper atoms? (Assume there are appropriate tools for this procedure and that copper atoms are lined up in a straight line, in contact with each other. Round off your answer to an integer. )

Fluoridation is the process of adding fluorine compounds to drinking water to help fight tooth decay. A concentration of 1 ppm of fluorine is sufficient for the purpose (1 ppm means one part per million, or 1 g of fluorine per 1 million g of water). The compound normally chosen for fluoridati on is sodium fluoride, which is also added to some toothpastes. Calculate the quantity of sodium flu oride in kilograms needed per year for a city of 50,000 people if the daily consumption of water per person is 150 gal. What percent of the sodium fluoride is "wasted" if each person uses only 6.0 L of water a day for drinking and cooking (sodium fluoride is 45.0 percent fluorine by mass; 1 gal = 3.79 L ; 1 year = 365 days; 1 ton = 2000 Ib; lIb = 453.6 g ; den sity of water = 1.0 g/mL)?

29

30

CHAPTER 1

Chemistry: The Central Science

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: VERBAL REASONING English writer and essayist Lady Mary Wortley Montagu (1689-1762) traveled extensively and was fascinated by the customs in other countries. While in Turkey, she observed the practice of "engrafting" wherein people were inoculated against smallpox by intentional exposure to a mild form of the disease. She was so convinced of the efficacy and the safety of engrafting, that she had both of her children inoculated. She herself had survived smallpox as a child. Lady Montagu campaigned for the practice when she returned to England, and despite opposition from doctors and religious leaders, inoculation came into common use. It remained the primary defense against the scourge of smallpox for decades-until Jenner developed the practice of vaccination.

1.

2.

a) a doctor. b) Turkish. c) severely scarred by smallpox. d) a member of a prominent British family. 3.

The author refers to Lady Montagu having survived smallpox in order to a) explain why Lady Montagu was fascinated by the practice of engrafting. b) compare Lady Montagu to the doctors and religious leaders in England. c) explain why Lady Montagu herself did not undergo the engrafting procedure. d) emphasize Lady Montagu's fascination with other cultures.

The main point of the passage is that a) Lady Montagu survived smallpox as a child. b) Lady Montagu brought the practice of engrafting from Turkey to England. c) Doctors in eighteenth-century England were opposed to the practice of engrafting. d) Jenner developed the practice of vaccination.

Based on the passage, Lady Montagu was most likely

4.

Based on the passage, the author most likely thinks that Lady Montagu was a) educated and infl uential. b) inconsequential in the prevention of smallpox in England. c) trained in science and medicine. d) married to the British ambassador to Turkey.

ANSWERS TO IN-CHAPTER MATERIALS ,

-".-.~~,

Practice Problems

Checkpoints

1.1A 273 K and 373 K, range = 100 K. LIB -270.5°C. 1.2233 °C. 1.3A (a) 14 g/mL, (b) 1.6 X 103 g. 1.3B (a) 9.25 g/cm 3, (b) 3.76 X 103 g. 1.4 (a) 4, (b) 1, (c) 4, (d) 2, (e) 2 or 3, (f) 4. 1.5A (a) 116.2 L, (b) 80.71 m, (c) 3.813 X 1021 atoms, (d) 31 dm2 , (e) 0.504 g/mL. 1.5B (a) 32.44 cm3, (b) 4.2 X 102 kg/m3, (c) 1.008 X 1010 kg, (d) 40.75 mL, (e) 227 cm 3 1.6A 0.8120 g/cm3 1.6B 95.3 cm 3 1.7 A 0.01 oz. 1.7B 0.88l3 oz. 1.8A 1.05 X 104 kg/m3 1.8B 13.6 mg/mm3.

1.3.1 c. 1.3.2 a. 1.3.3 b. 1.3.4 d. 1.5.1 d. 1.5.2 e. 1.5.3 a. 1.5.4 e. 1.6.1 b. 1.6.2 c. 1.6.3 a. 1.6.4 e.

Applying What You've Learned a) The recommended storage-temperature range for cidofovir is 20°C-25 °C. b) The density of the fluid in a vial is 1.18 g/mL. (The den sity should be reported to three significant figures.) c) The recommended dosage of cidofovir for a 177-lb man is 4 X 102 mg or 0.4 g.

•

toms, an ons

2.1 2.2

The Atomic Theory

100 (ALl1iJ!S SW1HG

The Structure of the Atom

• • • • •

o ecu es,

Discovery of the Electron Radioactivity The Proton and the Nucleus Nuclear Model of the Atom The Neutron

2.3

Atomic Number, Mass Number, and Isotopes

2.4 2.5

The Periodic Table

2.6

Molecules and Molecular Compounds

The Atomic Mass Scale and Average Atomic Mass

• • • •

2.7

Molecules Molecular Formulas Naming Molecular Compounds Empirical Formulas

Ions and Ionic Compounds

• • • • • •

Atomic Ions Polyatomic Ions Formulas of Ionic Compounds Naming Ionic Compounds Hydrates Familiar Inorganic Compounds

It'

Ie 1 4 3 1 .IDol Nl 14N"03

Think About It Use the law of conservation of mass to check your answers. Make sure that the combined mass of both products is equal to the mass of reactant you determined in part (a) . In this case (rounded to the appropriate number of significant figures), 10.0 g + 8.18 g = 18.2 g. Remember that small differences may arise as the result of rounding.

Thus, 18.2 g of ammonium nitrate must be heated in order to produce 10.0 g of nitrous oxide. (b) Starting with the number of moles of nitrous oxide determined in the first step of part (a), 0.227 ~ X 2 mol H 2 0 = 0.454 mol H 2 0 1 J!1Ol-N2V 0.454 ~

X

18.02 g H 20 = 8.18 g H 20

1~

Therefore, 8.18 g of water will also be produced in the reaction.

Practice Problem A Calculate the mass of water produced by the metabolism of 56.8 g of glucose. (See the box on page 80 for the necessary equation.) Practice Problem B What mass of glucose must be metabolized in order to produce 175 g of water?

..

-----------------------------------------------------Checkpoint 3.6 3.6.1

Calculations with Balanced Chemical Equations

How many moles of LiOH will be produced if 0.550 mol Li reacts according to the fo llowing equation? 2Li(s)

•

3.6.2

Determine the stoichiometric amount (in grams) of O 2 necessary to react with 5.71 g AI according to the following equation:

+ 2H 20(l) - _ . 2LiOH(aq) + H2(g)

a) 0.550 mol

a) 5.08 g

b) LlOmol

b) 9.03 g

c) 0.275 mol

c) 2.54 g

d) 2.20 mol

d) 4 .28 g

e) 2.00 mol

e) 7.61 g

92

CHAPTER 3

Stoichiometry: Ratios of Combination

Limiting Reactants

Limiting reactants and excess reactants are also referred to as limiting reagents and excess

reagents.

[W

....,

Multimed ia

Limiting reagent in reaction of NO and O2

When a chemist carries out a reaction, the reactants usually are not present in stoichiometric amounts. Because the goal of a reaction is usually to produce the maximum quantity of a useful compound from the starting materials, an excess of one reactant is commonly supplied to ensure that the more expensive or more important reactant is converted completely to the desired product. Consequently, some of the reactant supplied in excess will be left over at the end of the reaction. The reactant used up first in a reaction is called the limiting reactant, of this reactant limits the amount of product that can form. When all the . . . . . . .because . . . . . . . . . . . the . . . . . .amount . . . .. ....... ...... ...................... ....... ... ........................ .... . limiting reactant has been consumed, no more product can be formed. Excess reactants are those present in quantities greater than necessary to react with the quantity of the limiting reactant. The concept of a limiting reactant applies to everyday tasks, too, such as making ham sandwiches. Suppose you want to make the maximum number of ham sandwiches possible, each of which will consist of two slices of bread and one slice of ham. If you have eight slices of bread and six slices of ham, how many sandwiches can you make? The answer is four, because after making four sandwiches you will be out of bread. You will have two slices of ham left over, but without additional bread you will be unable to make any more sandwiches. In this case, bread is the limiting reactant and ham is the excess reactant.

Determining the Limiting Reactant In problems involving limiting reactants, the first step is to detennine which is the limiting reactant. After the limiting reactant has been identified, the rest of the problem can be solved using the approach outlined in Section 3.6. Consider the formation of methanol (CH 30H) from carbon monoxide and hydrogen: CO(g)

+ 2Hig) - _ . CH30H(l)

Suppose that initially we have 5 moles of CO and 8 moles of H 2, the ratio shown in Figure 3.6(a). We can use the stoichiometric conversion factors to determine how many moles of H2 are necessary in order for all the CO to react. From the balanced equation, we have 1 mol CO === 2 mol H2. Therefore, the amount of H2 necessary to react with 5 mol CO is moles ofH 2 = 5~X 2 mol H2

=

10 mol H2

1~

,

Because there are only 8 moles of H2 available, there is insufficient H2 to react with all the CO. Therefore, H? is the limiting reactant and CO is the excess reactant. H2 will be used up first, and when it is gone, the formation of methanol will cease and there will be some CO left over, as shown in Figure 3.6(b). To determine how much CO will be left over when the reaction is complete, we must first calculate the amount of CO that will react with all 8 moles of H 2 :

t

moles of CO = 8 ~ X 1 mol CO = 4 mol CO 2~

- --" Media Player/MPEG Animation: Figure 3.7, Limiting Reactant Problems, pp. 94- 95.

Thus, there will be 4 moles of CO consumed and 1 mole (5 mol - 4 mol) left over. Sample Problem 3.12 illustrates how to combine the concept of a limiting reactant with the conversion between mass and moles. Figure 3.7 (pp. 94-95) illustrates the steps for this type of calculation .

•

Ca)

Figure 3.6

(b)

The reaction of (a) H2 and CO to fOlm (b) CH30H. Each molecule represents 1 mole of substance.

In this case, H2 is the limiting reactant and there is 1 mole of CO remaining when the reaction is complete.

SECTION 3.7

Lim iting Reactants

93

Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHC0 3) and citric acid (H3C6HS0 7) react to form carbon dioxide gas, among other products.

The formation of COz causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 g of citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO 2 forms.

• • "

. .,• ',> .. . 't< •

.,..

t

".• •

?

•

. ""

. .'

•

Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the

necessary stoichiometric conversion factor and determine which reactant is limiting. Again, using the balanced equation, write the stoichiometric conversion factors to determine the number of moles of excess reactant remaining and the number of moles of CO 2 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO 2 to grams.

•

Setup The required molar masses are 84.01 g/mol for NaHC0 3, 192.12 g/mol for H 3C6H 50 7, and

44.01 g/mol for COz. From the balanced equation we have 3 mol NaHC0 3 "'" 1 mol H 3C6H s0 7, 3 mol NaHC0 3 "'" 3 mol COlo and 1 mol H3C6HS07 "'" 3 mol CO 2 , The necessary stoichiometric conversion factors are therefore:

3 mol NaHC0 3 1 mol H3C6HS07

1 mol H3C6HS07 3 mol NaHC0 3

3 mol CO 2 3 mol NaHC0 3

3 mol CO 2 The reaction of sodium bicarbonate and citric acid produces Alka-Seltzer's effervescence.

Solution

1.700 g NaIleu) X 1 mol NaHC0 3 = 0.02024 mol NaHC0 3 84.01 g NaIICUi 1.000 g H3C6H5U~ X

1 mol H3C6 HS0 7 = 0.005205 mol H3C6H S0 7 192.12g H3C6IIsO;

(a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with 0.02024 mol sodium bicarbonate. 0.02024..mgl NaMCU 3 X I mol H3C 6HSO! = 0.006745 mol H3C 6H S0 7 3jl101 NaIle03 The amount of H3C6HS07 required to react with 0.02024 mol of NaHC0 3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant. (b) To determine the mass of excess reactant (NaHC0 3) left over, first calculate the amount of NaHC0 3 that will react: 0.005205 mol H3C6H SO;

X

Think About It In a problem

such as this, it is a good idea to check your work by calculating the amounts of the other products in the reaction. According to the law of conservation of mass, the combined starting mass of the two reactants (1.700 g + 1.000 g = 2.700 g) should equal the sum of the masses of products and leftover excess reactant. In this case, the masses of HzO and Na3C6Hs07 produced are 0.2815 g and 1.343 g, respectively. The mass of CO 2 produced is 0.6874 g [from part (c)] and the amount of excess NaHC0 3 is 0.388 g [from part (b)]. The total, 0.2815 g + 1.343 g + 0.6874 g + 0.388 g, is 2.700 g, identical to the total mass of the reactants.

3 mol NaHC0 3 = 0.01562 mol NaHC0 3 I mol R ,C6IIS07

Thus, 0.01562 mole of NaHC0 3 will be consumed, leaving 0.00462 mole unreacted. Convert the unreacted amount to grams as follows: 0.000462.moi NaMeO; X 84.01 g NaHCO: = 0.388 g NaHC0 3 1 nJOi NaIleU3 (c) To determine the mass of COz produced, first calculate the number of moles of CO 2 produced from the number of moles of limiting reactant (H 3C6H s0 7) consumed:

Convert this amount to grams as follows: 0.01562 ~ X 44.01 g CO 2 = 0.6874 g CO 2 1~

To summarize the results: (a) citric acid is the limiting reactant, (b) 0.388 g sodium bicarbonate remains unreacted, and (c) 0.6874 g carbon dioxide is produced.

Practice Problem Ammonia is produced by the reaction of nitrogen and hydrogen according to the

• 2NH3(g). Calculate the mass of ammonia produced when 35.0 g of equation, Nz(g) + 3H2 (g) nitrogen react with 12.5 g of hydrogen. Which is the excess reactant and how much .of it will be left over when the reaction is complete?

..

--------------------------------------------------------

Figure 3.7

•

••

• Determine what mass of NH3 fOlIllS

Convert to moles.

when 84.06 g N2 and 22.18 g H2 are combined and react according to the equation:

N2 + 3H2

84.06 g N2 = 3.000 mol N2 28.02 g/mol 22.18 g H2 2.016 g/mol

•

Determine moles NH 3 .

= 11.00 mol H2

+

I

before reaction

84.06 g N2

+ 22.18 g H2

106.24 g

I

after reaction

102.2 g NH3

+ 4.03 g H2

106.2 g

Compare the total mass after the reaction with the total mass before the reaction. The small difference between the masses before and after is due to rounding.

94

Use coefficients as conversion factors.

3.000 UH'l}

11.00

r~i X 2 mol NH3 1 UJ(~ l Hi

··L

X

2 mol NH3

-3ii~~"-

3 ruQ} Hi

=

6.000 mol NH3

= 7.333 mol NH3 Either way, the smaller amount of product is correct.

or

3.000 N2

Rewrite the balanced equation using actual amounts.

3.667 N2

+ +

9.000 H2

--+.

6.000 NH3

11.000 H2

--+.

7.333 NH3

6.000 mol NH3

Convert to grams.

6.000 WQl NH3 X 17.03 g Nl!3 = 1022 NH 1 UJQl NH3 . g 3

t I

-= "

N2 was the limiting reactant. Calculate how much H2 is left over.

2.00 0191 Hi X

2.016 g H2

1 ruQ} Iii

11.00 mol initially - 9.00 mol consumed 2.00 mol H2 remaining

= 4.03 g H2 Convert to grams.

Whatls the point? There is more than one correct approach to solving many types of problems. This limiting reactant problem shows two different routes to the correct answer, and shows how the result can be compared .to the infOImation given in the problem to determine whether or not it is reasonable and correct. 95

7 96

CHAPTER 3

Stoichiometry: Ratios of Combination

Reaction Yield When you use stoichiometry to calculate the amount of product formed in a reaction, you are calculating the theoretical yield of the reaction. The theoretical yield is the amount of product that forms when all the limiting reactant reacts to form the desired product. It is the maximum obtainable yield, predicted by the balanced equation. In practice, the actual yield the amount of product actually obtained from a reaction is almost always less than the theoretical yield. There are many reasons for the difference between the actual and theoretical yields. For instance, some of the reactants may not react to form the desired product. They may react to form different products, in something known as side reactions, or they may simply remain unreacted. In addition, it may be difficult to isolate and recover all the product at the end of the reaction. Chemists often determine the efficiency of a chemical reaction by calculating its percent yield, which tells what percentage the actual yield is of the theoretical yield. It is calculated as follows: Equation 3.2

actual yield

% yield =

.

theoretical YIeld

X 100%

Percent yields may range from a tiny fraction to 100 percent. (They cannot exceed 100 percent.) Chemists try to maximize percent yield in a variety of ways. Factors that can affect percent yield, including temperature and pressure, are discussed in Chapter 15. Sample Problem 3.13 shows how to calculate the percent yield of a pharmaceutical manufacturing process.

Aspirin, acetylsalicylic acid (C9 H g 0 4 ), is the most commonly used pain reliever in the world. It is produced by the reaction of salicylic acid (C 7H60 3) and acetic anhydride (C 4H60 3) according to the following equation:

+ C7H60 3 salicylic acid

+

+

• C4H60 3

•

C9H s04

acetic anhydride

acetyl salicylic acid

acetic acid

In a certain aspirin synthesi s, 104.8 g of salicylic acid and 110.9 g of acetic anhydride are combined. Calculate the percent yield of the reaction if 105.6 g of aspirin are produced. Strategy Convert reactant grams to moles, and determine which is the limiting reactant. Use the

balanced equation to determine the number of moles of aspirin that can be produced, and convert this number of moles to grams for the theoretical yield. Use the actual yield (given in the problem) and the calculated theoretical yield to calculate the percent yield. Setup The necessary molar masses are 138.12 glmol for salicylic acid, 102.09 glmol for acetic

anhydride, and 180.15 glmol for aspirin. Solution

104.8 ~

X

1 mol C7 H60 3

=

138.12~

110.9 ~ X

1 mol C4 H60 3

=

102.09~

0.7588 mol C 7H 60 3 1.086 mol C 4H60 3

Because the two reactants combine in a 1: I mole ratio, the reactant present in the smallest number of moles (in this case, salicylic acid) is the limiting reactant. According to the balanced equation, one mole of aspirin is produced for every mole of salicylic acid consumed. 1 mol salicylic acid (C 7H60 3)

"'"

1 mol aspirin (C 9Hs0 4 )

Therefore, the theoretical yield of aspirin is 0.7588 mol. We convert this to grams using the molar mass of aspirin: 0.7588 !.!JOI C9IIgO~ X 180.15 g C9Hg0 4 = 136.7 g C9H g0 4 Think About It Make sure you

have used the proper molar masses and remember that percent yield can never exceed 100 percent.

1 -mol

C~H804

Thus, the theoretical yield is 136.7 g. If the actual yield is 105.6 g, the percent yield is 105.6 g 6 X 100% = 77 .25% yield % yield = 13 .7 g

SECTION 3.7

Lim iti n g React ants

97

Practice Problem A Diethyl ether is produced from ethanol according to the following equation :

Calculate the percent yield if 68 .6 g of ethanol reacts to produce 16. 1 g of ether. Practice Problem B What mass of ether will be produced if 221 g of ethanol reacts with a 73.2

percent yield?

..

~.--------------------------------------------------------~ Limiting Reactants

Checkpoint 3.7 3.7.1

What mass of CaS04 is produced according to the given equation when 5.00 g of each reactant are combined? CaFis)

+ H 2 S04 (aq)

• CaS04(s)

How many moles of NH3 can be produced by the combination of 3.0 mol N2 and 1.5 mol H2?

+ 2HF(g)

a) 2.0 mol

a) 10.0 g

b) 1.5 mol

b) 11.6 g

c) 0.50 mol

c) 6.94 g

d) 6.0 mol

d) 8.72 g

e) 1.0 mol

e) 5.02 g

3.7.2

3.7.3

3.7.4

What mass of water is produced by the reaction of 50.0 g CH 30H with an excess of O 2 when the yield is 53 .2%?

What is the percent yield for a process in which 10.4 g CH3 0H reacts and 10.1 g CO 2 form s according to the following equation? . 2CH 30H(l)

+ 30 2 (g)

• 2C0 2 (g)

2CH 30H(g)

+ 30 2 (g) -

a) 28 .1 g

+ 4H2 O(l)

b) 56.2 g

a) 97 .1%

c) 29.9 g

b) 70.7%

d) 15.0 g

c) 52.1 %

e) 26.6 g

d) 103% e) 37.9%

•

•

--+. 2C0 2 (g)

+ 4H 20(I)

How Am I Supposed to Remember All These Reactions? As you continue to study chemistry, you will encounter a wide variety of chemical reactions. The sheer number of different reactions can seem daunting at times, but most of them fall into a relatively small number of categories. Becoming familiar with several reaction types and learning to recognize patterns of reactivity will help you make sense out of the reactions in this book. Three of the most commonly encountered reaction types are combination, decomposition, and combustion.

Combination. A reaction in which two or more reactants combine to form a single product is known as a combination reaction. Examples include the reaction of ammonia and hydrogen chloride to form ammonium chloride, NH3(g)

+ HCI(g) - - NH4CI(s)

and the reaction of nitrogen and hydrogen gases to form ammonia,

and the decomposition of hydrogen peroxide to produce water and oxygen gas,

+

2H z02(aq) - - 2HzO(l)

Combustion. As you learned in Section 3.3, a combustion reaction is one in which a substance burns in the presence of oxygen. Combustion of a compound that contains C and H (or C, H, and 0) produces carbon dioxide gas and water. By convention, we will consider the water produced in a combustion reaction to be liquid water. Examples of this type of combustion are the combustion of formaldehyde, CHzOel)

+

Nz(g)

--

+ 3H2(g) - - 2NH3(g)

Decomposition. A reaction in which two or more products form from a single reactant is known as a decomposition reaction. A decomposition reaction is essentially the opposite of a combination reaction. Examples of this type of reaction include the decomposition of calcium carbonate to produce calcium oxide and carbon dioxide gas, CaC03 (s) ~ CaO(s)

+ Oz(g)

+ Oig) - - COig) + HzO(l)

and the combustion of methane,

+

CHig)

+

+ 20z(g) - - COig) + 2HzO(l)

Although these combustion reactions are shown here as balanced equations, oxygen is generally supplied in excess in such processes to ensure complete combustion.

+ CO 2 (g)

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) 2CO(g) + 02(g) - - 2C0 2(g) , (b) 2HC0 2H(l) + 02(g) - - 2C0 2(g) + 2H20(l), (c) 2KCl0 3 (s) - - 2KCI(s) + 30 2(g). Strategy Look at the reactants and products in each balanced equation to see if two or more reactants combine into one product (a combination reaction), if one reactant splits into two or more products (a decomposition reaction), or if the main products formed are carbon dioxide gas and water (a combustion reaction).

Think About It Make sure that a reaction identified as a combination has only one product [as in part (a)], a reaction identified as a decomposition has only one reactant [as in part (b)], and a reaction identified as a combustion produces only COz and H20 [as in part (c)].

Setup The equation in part (a) depicts two reactants and one product. The equation in part (b) represents a combination with O2 of a compound containing C, H, and 0 to.produce CO 2 and H2 0. The equation in part (c) represents two products being formed from a single reactant. Solution These equations represent (a) a combination reaction, (b) a combustion reaction, and (c) a decomposition reaction.

98

Practice Problem Identify each of the following as a combination, decomposition, or combustion reaction: (a) C2H 40 Z(l) + 20 z(g) - - 2COzCg) + 2H zO(l) , (b) 2Na(s) + Clz(g) - - 2NaCI(s), (c) 2NaH(s) - - 2Na(s) + Hz(g).

APPLYING WHAT YOU'VE LEARNED

99

Applying What You've Learned Cisplatin [Pt(NH3h Ch] is sometimes called "the penicillin of cancer drugs" because it is effective in the treatment of a wide variety of cancers. It is prepared by the reaction of ammonium tetrachloroplatinum(II) with ammonia: +

2-

+

+

+

•

• This is a very expensive process because (NH4hPtCI4 contains platinum, a precious metal that historically has sold for roughly twice the price of gold. Using a large excess of ammonia helps manufacturers maximize conversion of the high-priced reactant to the desired product.

Problems: a)

Determine the molecular mass and percent composition by mass for ammonium tetrachloroplatinum(II) and for cisplatin. [ 11~ Sample Problems 3.1 and 3.2]

b)

Balance the equation for the production of cisplatin from ammonium tetrachIOl;oplatinum(II) and ammonia. [ ~~ Sample Problem 3.3]

c)

Determine the number of each type of atom in 50.0 g of cisplatin. Problem 3.7]

d)

In a particular process, 172.5 g (NH4)2PtCI4 is combined with an excess of NH 3 .

[I~~

Sample

Assuming all the limiting reactant is converted to product, how many grams of Pt(NH3)2Cl2 will be produced? [ ~~ Sample Problem 3.11] e)

Note that the equation is not balanced .

If the actual amount of Pt(NH 3h CI 2 produced in part (d) is 129.6 g, what is the percent yield? [ ~~ Sample Problem 3.13]

+

+

CHAPTER 3

Stoichiometry: Ratios of Combination

CHAPTER SUMMARY •

Section 3.1 •

Molecular mass is calculated by summing the masses of all atoms in a molecule. Molecular weight is another term for molecular mass.

•

For ionic compounds, we use the analogous terms formula mass and formula weight.

•

Molecular masses, molecular weights, formula masses, and formula weights are expressed in atomic mass units (amu).

Section 3.2 •

Molecular or formula mass can be used to determine percent composition by mass of a compound.

•

A chemical equation is a written representation of a chemical reaction or a physical process. Chemical species on the left side of the equation are called reactants, whlle those on the right side of the equation are called products.

•

The physical state of each reactant and product is specified in parentheses as (s), (I), (g), or (aq) for solid, liquid, gas, and aqueous (dissolved in water), respectively.

•

Section 3.5 •

Combustion analysis is used to determine the empirical formula of a compound. The empirical formula can be used to calculate percent composition.

•

The empirical formula and molar mass can be used to determine the molecular formula.

Section 3.6 •

Section 3.3

Chemical equations are balanced only by changing the stoichiometric coefficients of the reactants and/or products, and never by changing the formulas of the reactants and/or products (i.e., by changing their subscripted numbers).

•

A mole is the amount of a substance that contains 6.022 X 1023 [Avogadro's number (NA)] of elementary particles (atoms, molecules, ions, formula units, etc.).

Molar mass (At) is the mass of one mole of a substance, usually expressed in grams. The molar mass of a substance in grams is numerically equal to the atomic, molecular, or formula mass of the substance in amu.

A balanced chemical equation can be used to determine how much product will form from given amounts of reactants, how much of one reactant is necessary to react with a given amount of another, or how much reactant is required to produce a specified amount of product. Reactants that are combined in exactly the ratio specified by the balanced equation are said to be "combined in stoichiometric amounts."

Section 3.7 •

The limiting reactant is the reactant that is consumed completely in a chemical reaction. An excess reactant is the reactant that is not consumed completely. The maximum amount of product that can form depends on the amount of limiting reactant.

•

The theoretical yield of a reaction is the amount of product that will form if all the limiting reactant is consumed by the desired reaction.

•

The actual yield is the amount of product actually recovered.

•

Percent yield [(actual/theoretical) X 100%] is a measure of the efficiency of a chemical reaction.

•

Combustion (in which a substance burns in the presence of oxygen), combination (in which two or more reactants combine to form a single product), and decomposition (in which a reactant splits apart to form two or more products) are three types of reactions that are commonly encountered.

Section 3.4 •

Molar mass and Avogadro's number can be used to interconvert among mass, moles, and number of particles (atoms, molecules, ions, formula units, etc.).

KEyWORDS Actual yield, 96

Combustion reaction, 98

Molar mass, 84

Product, 77

Aqueous, 77

Decomposition reaction, 98

Mole, 82

Reactant, 77

Avogadro's number, 82

Excess reactant, 92

Molecular mass, 74

Stoichiometric amount, 90

Chemical equation, 76

Formula mass, 74

Molecular weight, 74

Stoichlometric coefficients, 77

Combination reaction, 98

Formula weight, 74

Percent composition by mass, 75

Theoretical yield, 96

Combustion analysis, 87

Limiting reactant, 92

Percent yield, 96

KEY EQUATIONS 3.1

n X atomic mass of element f I f d X 100% percent by mass of an element = molecular or ormu a mass 0 compoun

3.2

actual yield X 100% % yield = theoretical yield

QUESTIONS AND PROBLEMS

101

QUESTIONS AND PROBLEMS Section 3.1: Molecular and Formula Masses

Section 3.3: Chemical Equations

Review Questions

Review Questions

3.1

3.2

What is meant by the term molecular mass, and why is the molecular mass that we calculate generally an average molecular mass?

3.14

Use the formation of water from hydrogen and oxygen to explain the following terms: chemical reaction, reactant, and product.

3.15

What is the difference between a chemical reaction and a chemical equation?

3.16

Why must a chemical equation be balanced? What law is obeyed by a balanced chemical equation?

3.1 7

Write the symbols used to represent gas, liquid, solid, and the aqueous phase in chemical equations.

Explain the difference between the terms molecular mass and formula mass. To what type of compound does each term refer?

Problems 3.3

3.4

Calculate the molecular mass (in amu) of each of the following substances: (a) CH3Cl, (b) N 10 4 , (c) SOl, (d) C 6H 1l , (e) H10 l , (f) CI1H22011, (g) NH 3 · Calculate the molecular mass (in amu) of each of the following substances: (a) C6H60, (b) H 1 S0 4 , (c) C 6H 6, (d) C 6H I2 0 6, (e) BCI3, (f) N 10S, (g) H 3P0 4.

3.5

Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) CH 4, (b) N01 , (c) S03, (d) C 6H 6, (e) NaI, (f) K 1S0 4 , (g) Ca3(P04) l '

3.6

Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) Li l C0 3, (b) C1H 6 , (c) NFl , (d) AI10 3, (e) Fe(N0 3)3, (f) PCIs, (g) Mg 3N 2.

Problems 3.18

Write an unbalanced equation to represent each of the following reactions: (a) nitrogen and oxygen react to form nitrogen dioxide, (b) dinitrogen pentoxide reacts to form dinitrogen tetroxide and oxygen, (c) ozone reacts to form oxygen, (d) chlorine and sodium iodide react to form iodine and sodium chloride, and (e) magnesium and oxygen react to form magnesium oxide.

3.19

Write an unbalanced equation to represent each of the following reactions: (a) potassium hydroxide and phosphoric acid react to form potassium phosphate and water; (b) zinc and silver chloride react to fonn zinc chloride and silver; (c) sodium hydrogen carbonate reacts to fonn sodium carbonate, water, and carbon dioxide; (d) ammonium nitrite reacts to form nitrogen and water; and (e) carbon dioxide and potassium hydroxide react to form potassium carbonate and water.

3.20

For each of the following unbalanced chemical equations, write the corresponding chemical statement.

Section 3.2: Percent Composition of Compounds

Review Questions 3.7

.8

Use ammonia (NH3) to explain what is meant by the percent composition by mass of a compound.

(a) S8 + O 2 - _ . S02 (b) CH4 + Oz • COz + H zO (c) N2 + H l • NH3 (d) P40 IQ + H zO • H 3P04 (e) S + RN0 3 • H ZS0 4 + NO l

Describe how the knowledge of the percent composition by mass of an unknown compound can help us identify the compound.

Problems

.9

.10

3.11

Tin (Sn) exists in Earth's crust as SnOz. Calculate the percent composition by mass of Sn and 0 in SnOz.

3.21

For many years chloroform (CHCI 3) was used as an inhalation anesthetic in spite of the fact that it is also a toxic substance that may cause severe liver, kidney, and heart damage. Calculate the percent composition by mass of this compound. All the substances listed here are fertilizers that contribute nitrogen to the soil. Which of these is the richest source of nitrogen on a mass percentage basis? (a) Urea [(NH2)lCOl (b) Ammonium nitrate (NH4N0 3 ) (c) Guanidine [RNC(NH Z)2l (d) Ammonia (NH 3)

.:' .1 2

The anti caking agent added to Morton salt is calcium silicate (CaSi0 3). This compound can absorb up to 2.5 times its mass of water and still remain a free-flowing powder. Calculate the percent composition of CaSi0 3.

3.13

Tooth enamel is Cas(P04MOH). Calculate the percent composition of the elements present.

+ H 20

For each of the following unbalanced chemical equations, wri te the corresponding chemical statement. • KOH + H z (a) K + H 20 (b) Ba(OH)2 + HCI • BaCIl + H 20 (c) Cu + HN0 3 • CU(N0 3) 2 + NO + H 20 • AI2(S04)3 + H2 (d) Al + H 2S0 4 (e) HI • H l + 12

3.22

Balance the following equations using the method outlined in Section 3.3. (a) C + O 2 - _ . CO (b) CO + O 2 - _ . CO 2 (c) H2 + Brz - _ . RBr (d) K + H 20 - _ . KOH + H z (e) Mg + O 2 --+. MgO • O2 (f) 0 3 (g) H10 l • H zO + 0 1 (h) N2 + H2 • NH3 (i) Zn + AgCl • ZnCl2 + Ag • S02 (j) S8 + 0 1 (k) NaOH + H ZS0 4 • Na2S04 + H 20 (I) CI l + NaI • NaCI + 12 • K 3P0 4 + H 20 (m) KOH + H 3P04 (n) CH4 + Brl • CBr4 + HBr

CHAPTER 3

3.23

Stoichiometry: Ratios of Combination

(a) NzOs -----+. N Z0 4 + Oz (b) KN0 3 • KNO z + Oz (c) NH4 N0 3 • NzO + HzO (d) NH 4 NO z • N z + HzO (e) NaHC0 3 • NaZC0 3 + H 20 + COz (f) P 4 0 lO + HzO • H 3P0 4 (g) HCl + CaC0 3 • CaClz + HzO + COz • Alz(S04)3 + Hz (h) Al + H 2 S04 (i) COz + KOH • K ZC0 3 + H 20 (j) CH4 + O2 • COz + HzO (k) BezC + H 20 • Be(OH)z + CH4 (1) Cu + RN0 3 • Cu(N0 3 )z + NO + H 20 • H 2S04 + NO z + HzO (m) S + RN0 3 (n) NH3 + CuO • Cu + N z + HzO 3.24

Which of the following equations best represents the reaction shown in the diagram?

(a) 8A + 4B -----+. C + D (b) 4A + 8B • 4C + 4D (c) 2A + B • C +D (d) 4A + 2B • 4C + 4D (e) 2A + 4B • C +D

•

•

•

•

•

3.25

it take to count 6.0 X lO z3 particles? Assume that there are 365 days in a year.

Balance the following equations using the method outlined in Section 3.3.

•

•

3.31

The thickness of a piece of paper is 0.0036 in. Suppose a certain book has an Avogadro's number of pages; calculate the thickness of the book in light-years. (Hint: See Problem 1.64 for the definition of light-year.)

3.32

How many atoms are there in 5.10 moles of sulfur (S)?

3.33

How many moles of cobalt (Co) atoms are there in 6.00 X 109 (6 billion) Co atoms?

3.34

How many moles of calcium (Ca) atoms are in

3.35

How many grams of gold (Au) are there in 15.3 moles of Au?

3.36

What is the mass in grams of a single atom of each of the following elements: (a) Hg, (b) Ne?

3.37

What is the mass in grams of a single atom of each of the following elements: (a) As, (b) Ni?

3.38

What is the mass in grams of 1.00 X 10 12 lead (Pb) atoms?

3.39

How many atoms are present in 3.14 g of copper (Cu)?

3.40

Which of the following has more atoms: 1.10 g of hydrogen atoms or 14.7 g of chromium atoms?

3.41

Which of the following has a greater mass: two atoms of lead or 5.1 X 10- 23 mole of helium?

3.42

Calculate the molar mass of the following substances: (a) LizC0 3, (b) CS z, (c) CHCl 3 (chloroform), (d) C 6H g0 6 (ascorbic acid, or vitamin C), (e) KN0 3, (f) Mg 3Nz.

3.43

Calculate the molar mass of a compound if 0.372 mole of it has a mass of 152 g.

3.44

How many molecules of ethane (C zH 6 ) are present in 0.334 g of

• •

Which of the following equations best represents the reaction shown in the diagram?

+ B ----+. C + D (b) 6A + 4B • C +D (c) A + 2B • 2C + D (a) A

(d) 3A + 2B (e) 3A + 2B

• 2C + D • 4C + 2D

• •

C2H6?

• •• ) , • (

----+.

A

B

• C _ D

Section 3.4: The Mole and Molar Masses

3.45

Calculate the number of C, H, and 0 atoms in 1.50 g of glucose (C 6H 1Z0 6), a sugar.

3.46

Urea [(NHz)zCOj is used for fertilizer and many other things . Calculate the number of N, C, 0, and H atoms in 1.68 X 104 g of urea.

3.47

Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula C I9H 3g O. Normally, the amount of this pheromone secreted by a female insect is about 1.0 X 10- 12 g. How many molecules are there in this quantity?

3.48

The density of water is 1.00 g/mL at 4 DC. How many water molecules are present in 2.56 mL of water at this temperature?

3.49

Cinnarnic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H100. (a) Calculate the percent composition by mass of C, H, and 0 in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?

3.50

Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; 0: 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g?

Review Questions 3.26

3.27

Define the term mole. What is the unit for mole in calculations? What does the mole have in common with the pair, the dozen, and the gross? What does Avogadro's number represent? What is the molar mass of an atom? What are the commonly used units for molar mass?

3.28

What does the word empirical in empirical formula mean?

3.29

If we know the empirical formula of a compound, what additional information do we need to determine its molecular formula ?

Problems 3.30

77.4 g of Ca?

Earth's population is about 6.5 billion. Suppose that every person on Earth participates in a process of counting identical particles at the rate of two particles per second. How many years would

QUESTIONS AND PROBLEMS 3.51

Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of C, H, N, and O. Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8 percent C, 2.50 percent H, 11.6 percent N. What is its molecular formula given that its molar mass is about 120 g?

3.52

The formula for rust can be represented by Fe20 3' How many moles of Fe are present in 24.6 g of the compound?

3.53

How many grams of sulfur (S) are needed to react completely with 246 g of mercury (Hg) to form HgS?

3 .54

Calculate the mass in grams of iodine (12) that will react completely with 20.4 g of aluminum (Al) to form aluminum iodide (AII3)'

3.55

Tin(H) fluoride (SnF2 ) is often added to tootbpaste as an ingredient to prevent tooth decay. What is the mass of F in grams in 24.6 g of the compound?

3 .56

Determine the empirical formulas of the compounds with the following compositions: (a) 2.1 percent H, 65.3 percent 0, 32.6 percent S, (b) 20.2 percent AI, 79.8 percent Cl.

3.57

-. -8

The empirical formula of a compound is CH. If the molar mass of this compound is about 78 g, what is its molecular formula?

3.59

The molar mass of caffeine is 194.19 g. Is the molecular formula of caffeine C 4H sN zO or CSHION40z?

3.60

3.61

.).62

Problems 3.65

Menthol is a flavoring agent extracted from peppermint oil. It contains C, H, and O. In one combustion analysis, 10.00 mg of the substance yields 11.53 mg H 20 and 28.16 mg CO 2 , What is the empirical formula of menthol?

3.66

Determine the empirical formula of an organic compound containing only C and H given that combustion of a 1.50-g sample of the compound produces 4.71 g CO 2 and 1.93 g H 20 .

3.67

Lactic acid, which consists of C, H , and 0 , has long been tbought to be responsible for muscle soreness following strenuous exercise. Determine the empirical formula of lactic acid given that combustion of a 1O.0-g sample produces 14.7 g CO 2 and 6.00 g H 20.

3.68

Perchloroetbylene, also known as "perc," is the solvent used in most dry cleaning. It consists of C and Cl. Determine the empirical formula of perchloroethylene given that a 2.58-g sample yields 1.37 g CO 2 in a combustion analysis experiment.

3.69

Paradichlorobenzene, a common ingredient in solid air fresheners, contains only C, H, and Cl and has a molar mass of about 147 g. Given that combustion of 1.68 g of this compound produces 3.02 g CO 2 and 0.412 g H 20, determine its empirical and molecular formulas.

3.70

Ascorbic acid (vitamin C) contains C, H, and O. In one combustion analysis, 5.24 g of ascorbic acid yields 7.86 g CO 2 and 2.14 g H 20. Calculate the empirical formula and molecular formula of ascorbic acid given that its molar mass is about 176 g.

3.71

The amino acid cysteine plays an important role in the threedimensional structure of proteins by forming "disulfide bridges." The percent composition of cysteine is 29.74 percent C, 5.82 percent H, 26.41 percent 0 , 11.56 percent N, and 26.47 percent S. Wbat is tbe molecular formula if its molar mass is approximately 121 g?

Determine the empirical formula s of the compounds with the following compositions: (a) 40.1 percent C, 6.6 percent H, 53.3 percent 0, (b) 18.4 percent C, 21.5 percent N, 60.1 percent K.

.J.)

Monosodium glutamate (MSG), a food-flavor enhancer, has been blamed for "Chinese restaurant syndrome," the symptoms of which are headaches and chest pains. MSG has the following composition by mass: 35.51 percent C, 4.77 percent H, 37.85 percent 0, 8.29 percent N, and 13.60 percent Na. Wbat is its molecular formula if its molar mass is about 169 g? Toxicologists use the term LDso to describe the number of grams of a substance per kilogram of body weight that is a lethal dose for 50 percent of test animals. Calculate the number of arsenic(VI) oxide molecules corresponding to an LDso value of 0.015 for a 184-lb man, ass uming that the test animals and humans have the same LDso. Chemical analysis shows that the oxygen-carrying protein hemoglobin is 0.34 percent Fe by mass. Wbat is the minimum possible molar mass of hemoglobin? The actual molar mass of hemoglobin is about 65,000 g. How would you account for the discrepancy between your minimum value and the experimental value?

Section 3.6: Calculations with Balanced Chemical Equations Review Questions 3.72

On what law is stoichiometry based? Why is it essential to use balanced equations in solving stoichiometric problems ?

3.73

Describe the steps involved in balancing a chemical equation.

Problems 3.74

Explain why, in combustion analysis, we cannot determine the amount of oxygen in the sample directly from the amount of oxygen in the products H 2 0 and CO 2 ,

+ Oz(g) --+. 2COzCg)

Starting witb 3.60 moles of CO, calculate the number of moles of CO 2 produced if there is enough oxygen gas to react with all the CO .

•~eview Questions

~.64

Consider the combustion of carbon monoxide (CO) in oxygen gas: 2CO(g)

Section 3.5: Combustion Analysis

Explain how tbe combined mass of CO 2 and H 20 produced in combustion analysis can be greater than the mass of the sample that is combusted.

103

3.75

Silicon tetrachloride (SiCI4) can be prepared by heating Si in chlorine gas: SiCs)

+ 2Clz (g) - _ . SiC Ii

I)

In one reaction, 0.507 mole of SiCl4 is produced. How many moles of molecular chlorine were used in the reaction ?

-----_.....-.

_

104

3.76

_-,_ _ _ _ _,,"rr.,. . .. ; . . . . - -

CHAPTER 3

-

Stoichiometry: Ratios of Combination

In a particular reaction, 6.0 moles of NH3 were produced. How many moles of Hz and how many moles of N z were consumed to produce this amount of NH3?

3.77

3.86

The fertilizer ammonium sulfate [(NH4)2S04] is prepared by the reaction between ammonia (NH3) and sulfuric acid: How many kilograms of NH3 are needed to produce 1.00 X 105 kg of (NH4hS04?

Consider the combustion of butane (C 4H IO ) :

In a particular reaction, 5.0 moles of C4H lO react with an excess of Oz. Calculate the number of moles of CO 2 formed. 3.78

(NH 4 N0 3). The other product is HzO . (a) Write a balanced equation for this reaction. (b) How many grams of NzO are formed if 0.46 mole of NH 4N0 3 is used in the reaction?

Ammonia is a principal nitrogen fertilizer. It is prepared by the reaction between hydrogen and nitrogen:

3.87

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is

A common laboratory preparation of oxygen gas is the therm al decomposition of potassium chlorate (KCI0 3 ). Assuming complete decomposition, calculate the number of grams of O 2 gas that can be obtained from 46.0 g of KCl0 3. (The products are KCI and O 2,)

Section 3.7: Limiting Reactants How much sulfur (in tons), present in the original materials, would result in that quantity of SOz?

3.79

3.80

When baking soda (sodium bicarbonate or sodium hydrogen carbonate, NaHC0 3) is heated, it releases carbon dioxide gas, which is responsible for the rising of cookies, doughnuts, and bread. (a) Write a balanced equation for the decomposition of the compound (one of the products is NaZC0 3). (b) Calculate the mass of NaHC0 3 required to produce 20.5 g of CO 2 , When potassium cyanide (KCN) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation:

Review Questions 3.88

Define limiting reactant and excess reactant. What is the significance of the limiting reactant in predicting the amount of the product obtained in a reaction? Can there be a limiting reactant if only one reactant is present?

3.89

Give an everyday example that illustrates the limiting reactant concept.

3.90

Why is the theoretical yield of a reaction determined only by the amount of the limiting reactant?

3.9 1

Why is the actual yield of a reaction almost always smaller than the theoretical yield?

KCN(aq) + HCI(aq) ----+. KCl(aq) + HCN(g)

If a sample of 0.140 g of KCN is treated with an excess of HCI, calculate the amount of HCN formed, in grams.

3.81

Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide:

Problems 3.92

2A +B ----+. C (a) In the diagram here that represents the reaction, which reactant, A or B, is the limiting reactant? (b) Assuming a complete reaction, draw a molecular-model representation of the amounts of reactants and products left after the reaction. The atomic arrangement in C is ABA.

C6 H 12 0 6 ----+. 2C zH sOH + 2C0 2 glucose

ethanol

Starting with 500.4 g of glucose, what is the maximum amount of ethanol in grams and in liters that can be obtained by this process (density of ethanol = 0.789 g/mL)? 3.82

•• • •

Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate (CUS04 . 5HzO). When this compound is heated in air above 100°C, it loses the water molecules and also its blue color:

If 9.60 g of CUS04 are left after heating 15.01 g of the blue compound, calculate the number of moles of HzO originally present in the compound.

3.83

Consider the reaction

3.93

Consider the reaction

Assuming each model represents one mole of the substance, show the number of moles of the product and the excess reactant left after the complete reaction.

For many years the extraction of gold-that is, the separation of gold from other materials-involved the use of potassium cyanide: 4Au + 8KCN + Oz + 2HzO ----+. 4KAu(CNh + 4KOH What is the minimum amount of KCN in moles needed to extract 29.0 g (about an ounce) of gold?

3.84

3.85

Limestone (CaC0 3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 1.0 kg of limestone. Nitrous oxide (N20) is also called "laughing gas." It can be prepared by the thermal decomposition of ammonium nitrate

3.94

Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (N0 2 ), a dark-brown gas:

QUESTIONS AND PROBLEMS

In one experiment 0.886 mole of NO is mixed with 0.503 mole of O 2 , Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of N0 2 produced.

3.95

The depletion of ozone (0 3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from high-altitude jet planes. The reaction is 0 3 + NO

Disulfide dichloride (S2CI2) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine:

What is the theoretical yield of S2C12 in grams when 4.06 g of Ss are heated with 6.24 g of C12? If the actual yield of S2Cl2 is 6.55 g, what is the percent yield?

• O 2 + N02

If 0.740 g of 0 3 reacts with 0.670 g of NO, how many grams of N0 2 will be produced? Which compound is the limiting reactant? Calculate the number of moles of the excess reactant remaining at the end of the reaction.

3.97

3.103

105

3.104

Propane (C3HS) is a minor component of natural gas and is used in domestic cooking and heating. (a) Balance the following equation representing the combustion of propane in air:

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion • Na2C03 + CO 2 + H20, (b) NH3 reaction: (a) 2NaHC0 3 • 2C0 2 + + HCl • NH4Cl, (c) 2CH 30H + 302 4H2 0.

3.105

(b) How many grams of carbon dioxide can be produced by burning 3.65 moles of propane? Assume that oxygen is the excess reactant in this reaction.

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion • • 3C02 + 4H20, (b) 2NF2 reaction: (a) C3HS + 50 2 N 2F4, (c) CUS04 . 5H20 • CUS04 + 5H 20.

Additional Problems 3.106

The diagram represents the products (C0 2 and H20) formed after the combustion of a hydrocarbon (a compound containing only C and H atoms). Write an equation for the reaction. (Hint: The molar mass of the hydrocarbon is about 30 g.)

3.107

Consider the reaction of hydrogen gas with oxygen gas:

Consider the reaction Mn02

+ 4HCl - _ . MnCl2 + Cl 2 + 2H20

If 0.86 mole of Mn02 and 48.2 g of HCI react, which reactant will be used up first? How many grams of Cl 2 will be produced? .J.98

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction

In one process 6.00 kg of CaF2 is treated with an excess of H 2S0 4 and yields 2.86 kg of HF. Calculate the percent yield of HF.

3.99

2H2(g)+ 0 2(g) - _ . 2H20(g)

Nitroglycerin (C 3H 5 N 30 9 ) is a powerful explosive. Its decomposition may be represented by

This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of O 2 in grams that can be obtained from 2.00 X 102 g of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of O2 generated is found to be 6.55 g. : .100

Assuming a complete reaction, which of the diagrams (a-d) shown here represents the amounts of reactants and products left after the reaction?

Titanium(IV) oxide (Ti0 2) is a white substance produced by the action of sulfuric acid on the mineral ilmenite (FeTi0 3):

Its opaque and nontoxic properties make it suitable as a pigment in plastics and paints. In one process 8.00 X 103 kg of FeTi03 yielded 3.67 X 103 kg of Ti0 2. What is the percent yield of the reaction? (a)

3.101

Ethylene (C 2H 4), an important industrial organic chemical, can be prepared by heating hexane (C 6H I4) at 800°C: . C6Hl4

• C 2H 4 + other products

If the yield of ethylene production is 42.5 percent, what mass of hexane must be used to produce 481 g of ethylene? .:: .102

When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s)

+ Nig) -_I 2Li3N(s)

What is the theoretical yield of Li3N in grams when 12.3 g of Li are heated with 33.6 g of N2? If the actual yield of Li3N is 5.89 g, what is the percent yield of the reaction?

3.108

(b)

(c)

(d)

Cysteine, shown here, is one of the 20 amino acids found in proteins in humans. Write the molecular formula of cysteine, and calculate its molar mass.

106

3.109

CHAPTER 3

Stoichiometry: Ratios of Combination

Isofiurane, shown here, is a cornmon inhalation anesthetic. Write its molecular formula, and calculate its molar mass.

3.120

A certain metal oxide has the formula MO where M denotes the metal. A 39A6-g sample of the compound is strongly heated in an atmosphere of hydrogen to remove oxygen as water molecules. At the end, 31.70 g of the metal is left over. If 0 has an atomic mass of 16.00 amu, calculate the atomic mass of M and identify the element.

3.121

An impure sample of zinc (Zn) is treated with an excess of sulfuric acid (H 2S0 4) to form zinc sulfate (ZnS04) and molecular hydrogen (H2). (a) Write a balanced equation for the reaction. (b) If 0.0764 g of H2 is obtained from 3.86 g of the sample, calculate the percent purity of the sample. (c) What assu mptions must you make in part (b)?

3.122

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is

CI 3.110

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: 4NH 3(g) + 502(g) 2NO(g) + 0 2(g ) 2N0 2(g) + H 20(l)

--+. 4NO(g) + 6H20(I) • 2N0 2(g) • HN0 3(aq)

+ HN0 2(aq)

Suppose that 1.64 X 103 kg of Fe is obtained from a 2.62 X 103-kg sample ofF~03 . Assuming that the reaction goes to completion, what is the percent purity of Fe203 in the original sample?

What mass of NH3 (in grams) must be used to produce 1.00 ton of HN0 3 by the Ostwald process, assuming an 80 percent yield in each step (1 ton = 2000 Ib; 1 Ib = 453.6 g)?

3.111

A sample of a compound of CI and 0 reacts with an excess of H2 to give 0.233 g of HCI and 00403 g of H 20. Determine the empirical formula of the compound.

3.112

The atomic mass of element X is 33042 amu. A 27.22-g sample of X combines with 84.10 g of another element Y to form a compound XY. Calculate the atomic mass ofY.

3.113

How many moles of 0 are needed to combine with 0.212 mole of C to form (a) CO and (b) CO 2?

3.114

The aluminum sulfate hydrate [Ah(S04)3 . xH 20] contains 8.10 percent Al by mass. Calculate x, that is, the number of water molecules associated with each Ah(S04)3 unit.

3.115

The explosive nitroglycerin (C3HsN309) has also been used as a drug to treat heart patients to relieve pain (angina pectoris). We now know that nitroglycerin produces nitric oxide (NO), which causes muscles to relax and allows the arteries to dilate. If each nitroglycerin molecule releases one NO per atom of N, calculate the mass percent of NO in nitroglycerin.

3.116

3.117

3.118

3.119

The compound 2,3-dimercaptopropanol (HSCH2CHSHCH20H), commonly known as British Anti-Lewisite (BAL), was developed during World War I as an antidote to arsenic-containing poison gas. (a) If each BAL molecule binds one arsenic (As) atom, how many As atoms can be removed by 1.0 g of BAL ? (b) BAL can also be used to remove poisonous heavy metals like mercury (Hg) and lead (Pb). If each BAL binds one Hg atom, calculate the mass percent of Hg in a BAL-Hg complex. (An H atom is removed when a BAL molecule binds an Hg atom.) Mustard gas (C4H sCI2S) is a poisonous gas that was used in World War I and banned afterward. It causes general destruction of body tissues, resulting in the formation of large water blisters. There is no effective antidote. Calculate the percent composition by mass of the elements in mustard gas. The carat is the unit of mass used by jewelers. One carat is exactly 200 mg. How many carbon atoms are present in a 2-carat diamond? An iron bar weighed 664 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (F~03 ). Calculate the final mass of the iron bar and rust.

3.123

Carbon dioxide (C0 2) is the gas that is mainly responsible for global warming (the greenhouse effect). The burning of fossil fuels is a major cause of the increased concentration of CO 2 in the atmosphere. Carbon dioxide is also the end product of metabolism (see Sample Problem 304). Using glucose as an example of food, calculate the annual human production of CO 2 in grams, assuming that each person consumes 5.0 X 102 g of glucose per day, that the world 's population is 6.5 billion, and that there are 365 days in a year.

3.124

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2:1. A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is 178 g.

3.125

Which of the following has the greater mass : 0.72 g of O 2 or 0.0011 mole of chlorophyll (CssH n MgN40s)?

3.126

Analysis of a metal chloride XCI 3 shows that it contains 67.2 percent Cl by mass. Calculate the molar mass of X, and identify the element.

3.127

Hemoglobin (C29s2H4664N8I20832SgFe4) is the oxygen carrier in blood. (a) Calculate its molar mass. (b) An average adult has about 5.0 L of blood. Every milliliter of blood has approximately 5.0 X 109 erythrocytes, or red blood cells, and every red blood cell has about 2.8 X lOs hemoglobin molecules. Calculate the mass of hemoglobin molecules in grams in an average adult.

3.128

Myoglobin stores oxygen for metabolic processes in muscle. Chemical analysis shows that it contains 0.34 percent Fe by mass. What is the molar mass of myoglobin? (There is one Fe atom per molecule.)

3.129

Calculate the number of cations and anions in each of the following compounds: (a) 8.38 g of KBr, (b) SAO g of Na2S04, (c) 7 AS g of Ca3(P04) 2.

3.130

A mixture of NaBr and Na2S04 contains 29.96 percent Na by mass. Calculate the percent by mass of each compound in the mixture.

QUESTIONS AND PROBLEMS

3.131

3.144

The formula of a hydrate of barium chloride is BaCI2 • xH 20. If 1.936 g of the compound gives 1.864 g of anhydrous BaS04 upon treatment with sulfuric acid, calculate the value of x.

3.145

It is estimated that the day Mt. St. Helens erupted (May 18, 1980), about 4.0 X 105 ton s of S02 were released into the atmosphere. If all the S02 were eventually converted to sulfuric acid, how many tons of H 2S0 4 were produced?

3.146

A mixture of CUS04 . 5H 20 and MgS04 . 7H20 is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of CUS04 . 5H 20 in the mixture?

Calculate the percent composition by mass of all the elements in calcium phosphate [Ca3(P04)2], a major component of bone.

3.147

Lysine, an essential amino acid in the human body, contains C, H, 0, and N. In one experiment, the complete combustion of 2.175 g oflysine gave 3.94 g CO 2 and 1.89 g H 20 . In a separate experiment, 1.873 g oflysine gave 0.436 g NH3. (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is 150 g. What is the molecular formula of the compound?

When 0.273 g of Mg is heated strongly in a nitrogen (N2) atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Calculate the empirical formula of the compound containing Mg and N. Name the compound.

3.148

A mixture of methane (CH 4) and ethane (C 2 H 6) of mass 13.43 g is completely burned in oxygen. If the total mass of CO 2 and H 20 produced is 64.84 g, calculate the fraction of CH4 in the mixture.

3.149

Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, "leaded gasoline"). When 51.36 g of this compound are burned in an apparatus such as that shown in Figure 3.5, 55.90 g of CO 2 and 28.61 g of H20 are produced. Determine the empirical formula of the gasoline additive.

3.150

Because of its detrimental effect on the environment, the lead compound described in Problem 3.149 has been replaced in recent years by methyl tert-butyl ether (a compound of C, H, and 0 ) to enhance the performance of gasoline. (As of 1999, this compound is also being phased out because of its contamination of drinking water.) When 12.1 g of the compound is burned in an apparatus like the one shown in Figure 3.5, 30.2 g of CO 2 and 14.8 g of H 20 are formed. What is the empirical formula of the compound?

3.151

Suppose you are given a cube made of magnesium (Mg) metal of edge length 1.0 cm. (a) Calculate the number of Mg atoms in the cube. (b) Atoms are spherical in shape. Therefore, the Mg atoms in the cube cannot fill all the available space. If only 74 percent of the space inside the cube is taken up by Mg atoms, calculate the radius in picometers of an Mg atom. (The density of Mg is 1.74 g/cm3, and the volume of a sphere of radius ris ~ 1Tr3.)

3.152

A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide (CaO) to form calcium sulfite (CaS03)' Calculate the daily mass (in kilograms) of CaO needed by a power plant that uses 6.60 X 106 kg of coal per day.

3.153

Air is a mixture of many gases. However, in calculating its molar mass we need consider only the three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78.08 percent nitrogen, 20.95 percent oxygen, and 0.97 percent argon, what is the molar mass of air?

3.154

A die has an edge length of 1.5 cm. (a) What is the volume of one mole of such dice? (b) Assuming that the mole of dice could be packed in such a way that they were in contact with one another, forming stacking layers covering the entire surface of Earth, . calculate the height in meters the layers would extend outward. [The radius (I') of Earth is 6371 km, and the area of a sphere is 41Tr2.]

Aspirin or acetyl salicylic acid is synthesized by combining salicylic acid with acetic anhydride: C7H 60 3 + C 4H 60 3 salicylic acid acetic anhydride

• C 9Hg0 4 + HC2H 30 2 aspIrIn

acetic acid

(a) How much salicylic acid is required to produce 0.400 g of aspirin (about the content in a tablet), assuming acetic anhydride is present in excess? (b) Calculate the amount of salicylic acid needed if only 74.9 percent of salicylic is converted to aspirin. (c) In one experiment, 9.26 g of salicylic acid reacts with 8.54 g of acetic anhydride. Calculate the theoretical yield of aspirin and the percent yield if only 10.9 g of aspirin is produced. 3.132

3.133

3.134

Does 1 g of hydrogen molecules contain as many H atoms as I g of hydrogen atoms?

3.135

Avogadro's number has sometimes been described as a conversion factor between amu and grams. Use the fluorine atom (19.00 amu) as an example to show the relationship between the atomic mass unit and the gram.

3. 136

The natural abundances of the two stable isotopes of hydrogen (hydrogen and deuterium) are 99.99 percent: Hand 0.01 percent Assume that water exists as either H20 or D 20. Calculate the number of D 20 molecules in exactly 400 mL of water (density 1.00 g/mL).

TH. 3.137

In the formation of carbon monoxide, CO, it is found that 2.445 g of carbon combine with 3.257 g of oxygen. What is the atomic mass of oxygen if the atomic mass of carbon is 12.01 amu?

3. 138

What mole ratio of molecular chlorine (CI 2) to molecular oxygen (0 2) would result from the breakup of the compound ClzO? into its constituent elements?

3.139

Which of the following substances contains the greatest mass of chlorine: (a) 5.0 g C12, (b) 60.0 g NaCI0 3, (c) 0.10 mol KCI, (d) 30.0 g MgCI 2, (e) 0.50 mol C1 2?

3.140

A compound made up of C, H, and CI contains 55.0 percent CI by mass. If 9.00 g of the compound contain 4.19 X 1023 H atoms, what is the empirical formula of the compound?

3.141

Platinum forms two different compounds with chlorine. One contains 26.7 percent CI by mass, and the other contains 42.1 percent CI by mass. Determine the empirical formulas of the two compounds.

3. 142

3.143

107

Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is 1.68 g. From the data given, show that the simplest formula of the oxide is X20 3 and write a balanced equation for the reaction. A compound X contains 63.3 percent manganese (Mn) and 36.7 percent 0 by mass. When X is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent Mn and 28.0 percent o is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of X to Y.

CHAPTER 3

108

Stoichiometry: Ratios of Combination

3.155

The following is a crude but effective method for estimating the o rder of magnitude of Avogadro's number using stearic acid (ClsH360 2)' When stearic acid is added to water, its molecules collect at the surface and form a monolayer; that is, the layer is only one molecule thick. The cross-sectional area of each stearic acid molecule has been measured to be 0.21 nm2 In one experiment it is found that 1.4 X 10- 4 g of stearic acid is needed to form a monolayer over water in a dish of diameter 20 cm. Based on these measurements, what is Avogadro 's number? (The area of a circle of radius r is 1T?)

3.159

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major so urces of potash are potassium chloride (KCI) and potassium sulfate (K2S04 ) . Potash production is often reported as the potassium oxide (K20) equivalent or the amount of K 20 that could be made from a given mineral. (a) If KCI costs $0.55 per kg, for what price (do llar per kg) must K2 S04 be sold in order to supply the same amount of potassium on a per dollar basis? (b) What mass (in kg) of K20 contains the same number of moles of K atoms as 1.00 kg of KCI?

3.156

Octane (CsH 1s) is a component of gasoline. Complete combustion of octane yields H 20 and CO 2 , Incomplete combustion produces H 20 and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run , 1.000 gallon (gal) of octane is burned in an engine. The total mass of CO, CO 2, and H 20 produced is 11.53 kg. Calculate the efficiency of the process; that is, calculate the fraction of octane converted to CO 2 , The density of octane is 2.650 kg/gal.

3.160

A 21.496-g sample of magnesium is burned in air to form magnesium oxide and magnesium nitride. When the products are treated with water, 2.813 g of gaseous ammonia is generated. Calculate the amounts of magnesium nitride and magnesium oxide formed.

3.161

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

3.162

A sample of iron weighing 15.0 g was heated with potassium chlorate (KCI0 3) in an evacuated container. The oxygen generated from the decomposition of KCI0 3 converted some of the Fe to Fez03' If the combined mass of Fe and Fe203 was 17.9 g, calculate the mass of Fe20 3 formed and the mass of KCI0 3 decomposed.

3.163

A sample containing NaCl, Na2S04, and NaN0 3 gives the following elemental analysis: 32.08 percent Na, 36.01 percent 0, 19.51 percent Cl. Calculate the mass percent of each compound in the sample.

3.157

Industrially, hydrogen gas can be prepared by combining propane gas (C3Hs) with steam at about 400°C. The products are carbon monoxide (CO) and hydrogen gas (H2) ' (a) Write a balanced equation for the reaction. (b) How many kilograms of H2 can be obtained from 2.84 X 10 3 kg of propane?

3.158

A reaction having a 90 percent yield may be considered a successful experiment. However, in the synthesis of complex molecules such as chlorophyll and many anticancer drugs , a chemist often has to carry out multiple-step syntheses. What is the overall percent yield for such a synthesis, ass uming it is a 30-step reaction with a 90 percent yield at each step ?

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES The first step in producing phosphorus fertilizer is the treatment of fluorapatite, a phosphate rock, with sulfuric acid to yield calcium dihydrogen phosphate, calcium sulfate, and hydrogen fluoride gas. In one experiment, a chemist combines 1.00 kg of each reactant. I.

+ 2H2 SOi

b) 2CaP04F(s) +

c) CaS(P04)3F(S)

aq)

4.

- _ . CaH 2 P0 4(aq) H 2 S04(aq)

+ CaS04(aq) + HF(g)

- _ . CaH 2P04(aq) 3H2 S04 (aq)

+ CaS04(aq) + HF(g)

+

- _ . 3Ca(H 2P0 4)z(aq)

d) 2CaS(P04)3F(S)

+

2CaS04(aq)

+

HF(g)

+ 7H 2S0 4(aq) - _ . 3Ca(H 2P04)z(aq)

+ 7CaS04(aq) + 2HF(g)

Assuming that all the limiting reactant is converted to products, what mass of calcium dihydrogen phosphate is produced? a) 464 g b) 696 g c) 92.8 g d) 199 g

What mass of the excess reactant remains unreacted? a) 319 g b) 68 1 g c) 406 g d) 490 g

Select the correct balanced equation to represent the reaction. a) 2CaP04F(s)

2.

3.

What is the percent yield of the reaction if 197 g of calcium dihydrogen phosphate is produced? a) 99 .0 percent b) 42.5 percent c) 28.3 percent d) 212 percent

ANSWERS TO IN-CHAPTER MATERIALS

109

ANSWERS TO IN-CHAPTER MATERIALS Answers to Practice Problems

Answers to Applying What You've Learned

3.1A (a) 95 .21 amu, (b) 98.09 amu, (c) 206.27 amu. 3.1B (a) 100.09 amu, (b) 47.02 amu, (c) 194.20 amu. 3.2A 63 .56 % C, 6 .001 % H , 9.268% ,2 1.17% 0. 3.2B 57.13% C, 6.165 % H , 9.521 % N , 27.18 % 0. 3.3A C3Hs(g) + 50 2(g) • 3COig) + 4H20(l). 3.3B H 2S0 4(aq) + 2NaOH(aq) • Na2S0iaq) + 2H 20(l). 3.4 ClsH320 2(aq) + 2502(g) --+. 18C0 2(g) + 16H20(l). 3.5A (a) 4.40 X 1024 atoms K, (b) 1.48 X 102 mol K. 3.5B (a) 6.32 X 10 17 atoms He, (b) 3.87 X 10- 3 mol He. 3.6A 495 g. 3.6B 0.704 mol. 3.7 A 6.68 X 1023 O 2 molecules. 3.7B 985 g S03' 3.8A C 2H 6 0. 3.8B CH 2· 3.9A C 3H 4 0 3 and C 6H s0 6 . 3.9B CH0 2 and C2H 20 4 . 3.10 0.264 mol H 2 , 0.176 mol NH 3. 3.llA 34.1 g. 3.llB 292 g. 3.12 42.6 g ammonia; nitrogen is limiting reagent and 4.945 g hydrogen left over. 3.13A 29.2%. 3.13B 130 g. 3.14 (a) combustion, (b) combination, (c) decomposition.

a) The molecular masses of ammonium tetrachloroplatinum(II) and cisplatin are 373.0 and 300.1 amu, respectively. The percent composition by mass of ammonium tetrachloroplatinum(II) is 7 .51 2% N , 2.162% H, 52.31 % Pt, and 38.02% Cl; and the percent composition by mass of cisplatin is 9.337% N, 2.015 % H, 65.01 % Pt, and 23 .63 % Cl. b) The balanced equation is (NH4)2PtCI4(aq) + 2NH 3(aq) • Pt(NH3)2Clis) + 2NH 4 Cl(aq) . c) 50.00 g of cisplatin contains 1.003 X 1023 Pt atoms, 2.007 X 1023 N atoms, 6.020 X 1023 H atoms, and 2.007 X 1023 Cl atoms. d) If all the limiting reactant is converted to product, 138.8 g of cisplatin will be produced. e) An actual yield of 129.6 g constitutes a 93.37% yield.

Answers to Checkpoints 3.1.1 a. 3.1.2 d. 3.2.1 b. 3.2.2 e. 3.3.1 e. 3.3.2 d. 3.3.3 e. 3.3.4 d. 3.4.1 b. 3.4.2 d. 3.4.3 a. 3.4.4 e. 3.5.1 e. 3.5.2 d. 3.6.1 a. 3.6.2 a. 3.7.1 c. 3.7.2 b. 3.7.3 e. 3.7.4 c.

•

•

eactlons In • ueous o utlons

4.1

General Properties of Aqueous Solutions

• •

4.2

Electrolytes and Nonelectrolytes Strong Electrolytes and Weak Electrolytes

.precipitation Reactions

• • •

•

4.3

Solubility Guidelines for Ionic Compounds in Water Molecular Equations Ionic Equations Net Ionic Equations

Acid-Base Reactions

• • •

4.4

Strong Acids and Bases Bnmsted Acids and Bases Acid-Base Neutralization

Oxidation-Reduction Reactions

• • • •

4.5

Oxj,dation Numbers Oxidation of Metals in Aqueous Solutions Balancing Simple Redox Equations Other Types of Redox Reactions

Concentration of Solutions

• • •

4.6

Molarity Dilution Solution Stoichiometry

Aqueous Reactions and Chemical Analysis

• •

Gravimetric Analysis Acid-Base Titrations

Prevention of Drunk Driving Every year in the United States tens of thousands of people are killed and half a million more are injured as a result of drunk driving. In recent years, most states have lowered . . .. . the legal limit of blood alcohol concentration (BAC) from 0.10 to 0.08 percent. Despite stiffer penalties for drunk-driving offenses and high-profile campaigns to educate the public about the dangers of driving while intoxicated, law enforcement agencies still must devote a great deal of work to removing drunk drivers from America's roads. The police often use a device called a Breathalyzer to test drivers suspected of being drunk. In one type of device the breath of a driver suspected of driving under the influence of alcohol is bubbled through an orange solution containing potassium dichromate (K2Cr20 7) and sulfuric acid (H 2S04), The alcohol in the driver's breath reacts with the dichromate ion to produce acetic acid (HC 2H 30 ?), which is colorless, and green chromium(III) sulfate [Cr2(S04)2]. The degree of color change from orange to green indicates the alcohol concentration in the breath sample, which is used to estimate the BAC. The basis for the Breathalyzer test is a relatively simple chemical reaction called an oxidation-reduction reaction. This is one of several important types of reactions that can occur in aqueous solution.

A blood alcohol concentration of 0.08 percent means that 100 mL of blood contai ns 0.08 g of ethanol.

•

A Breathalyzer has two ampoules containing identical solutions. The driver's breath is bubbled through the solution in one ampoule, and the solution in the other ampoule remains unchanged. The device contains a calibrated meter that compares the colors in the two ampoules.

In This Chapter, You Will Learn

about some of the properties of aqueous solutions and about several different types of reactions that can occur between dissolved substances. You will also learn how to express the concentration of a solution and how concentration can be useful in solving quantitative problems.

Before you begin, you should review [ ~~

Media Player/ MPEG Content

Sections 2.6 and 2.7]

•

Identifying compounds as either molecular or ionic

•

Names, formulas, and charges of the common polyatomic ions

[ ~~

Table 2.9]

Chapter in Review

A traditional sobriety test for a driver suspected of being intoxicated may have included instructing the driver to walk a straight line or touch his or her own nose. Today it is common for the more quantitative method of the Breathalyzer test to be used. 111

=

112

CHAPTER 4

Reactions in Aqueous Solutions

General Properties of Aqueous Solutions

--.. -

_

Multimedia

Solutions-strong, weak, and nonelectrolytes.

A solution is a homogeneous mixture [ ~~ Section 1.2] of two or more substances. Solutions may be gaseous (such as air), solid (such as brass), or liquid (such as saltwater). Usually, the substance present in the largest amount is referred to as the solvent and any substance present in a smaller amount is called the solute. For example, if we dissolve a teaspoon of sugar in a glass of water, water is the solvent and sugar is the solute. In this chapter, we will focus on the properties of aqueous solutions those in which water is the solvent. Throughout the remainder of this chapter, unless otherwise noted, solution will refer specifically to an aqueous solution.

Electrolytes and Nonelectrolytes You have probably heard of electrolytes in the context of sports drinks such as Gatorade. Electrolytes in body fluids are necessary for the transmission of electrical impulses, which are critical to physiological processes such as nerve impulses and muscle contractions. In general, an electrolyte is a substance that dissolves in water to yield a solution that conducts electricity. By contrast, a is ... a substance that dissolves in water to yield a solution that does not conduct elec.. . . . . .nonelectrolyte . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ., ................. ... . tricity. Every water-soluble substance fits into one of these two categories. The difference between an aqueous solution that conducts electricity and one that does not is the presence or absence of ions. As an illustration, consider solutions of sugar and salt. The physical processes of sugar (sucrose, CI2H22011) dissolving in water and salt (sodium chloride, NaCl) dissolving in water can be represented with the following chemical equations: '"

A substance that dissolves in a particular solvent is said to be "soluble" in that solvent. In this chapter, we will use the word soluble to mean "water-soluble."

and Note that while the sucrose molecules remain intact upon dissolving, becoming aqueous sucrose molecules, the sodium chloride dissociates, producing aqueous sodium ions and aqueous chloride ions. Dissociation is the process by which an ionic compound, upon dissolution, breaks apart into its constituent ions. It is the presence of ions that allows the solution of sodium chloride to conduct electricity. Thus, sodium chloride is an electrolyte and sucrose is a nonelectrolyte. Like sucrose, which is a molecular compound [ ~~ Section 2.6], many water-soluble molecular compounds are nonelectrolytes. Some molecular compounds are electrolytes, however, because they ionize on dissolution. Ionization is the process by which a molecular compound forms ions when it dissolves. Recall from Chapter 2 that acids are compounds that dissolve in water to produce hydrogen ions (H+) [ ~~ Section 2.6] . HCI, for example, ionizes to produce H+ ions and CI- ions.

6

one of two important classes of molecular compounds that are electrolytes. . . . . . .Acids . . . . ... constitute . . Bases may be molecular, like ammonia (NH3 ), or ionic, like sodium hydroxide (NaOH).

Molecular bases constitute the other one. A base is a compound that dissolves in water to produce hydroxide ions (OH - ). Ammonia (NH3), for example, ionizes in water to produce ammonium (NHt) and hydroxide (OH- ) ions.

Strong Electrolytes and Weak Electrolytes In a solution of sodium chloride, all the dissolved compound exists in the form of ions. Thus, NaCl, which is an iOrllc compound [ ~~ Section 2.7] , is said to have dissociated completely. An electrolyte that dissociates completely is known as a strong electrolyte. All water-soluble ionic compounds dissociate completely upon dissolving, so all water-soluble ionic compounds are strong electrolytes. The list of molecular compounds that are strong electrolytes is fairly ShOlto It complises the seven strong acids, which are listed in Table 4.1. A strong acid iOrllzes completely, resulting in a solution that contains hydrogen ions and the cOlTesponding arllons but essentially no acid molecules. Most of the molecular compounds that are electrolytes are weak electrolytes. A weak electrolyte is a compound that produces ions upon dissolving but exists in solution predominantly as molecules that are not ionized. Most acids (except those listed in Table 4.1) are weak electrolyte . Acetic acid (HC 2H 30?) is not one of the strong acids listed in Table 4.1, so it is a weak acid. Its ionization in water is represented by the following chemical equation:

SECTION 4.1

General Properties of Aqueous Solutions

113

Ionization Equation

Acid

• H +(aq)

+ CI- (aq)

Hydrochloric acid

HCI(aq)

Hydrobromic acid

HBr(aq) ---+. H \aq) + Br-(aq)

+ I-(aq) + N0 3(aq) + CI0 3(aq) + CI0 4(aq)

HI(aq)

• H +(aq)

Nitric acid

HN0 3(aq)

• H +(aq)

Chloric acid

HCI0 3 (aq)

• H +(aq)

Perchloric acid

HCI0 4 (aq)

• H +(aq)

Sulfuric acid *

H 2 S04 (aq)

• H +(aq) + HS0 4(aq)

HS0 4 (aq).

• H +(aq)

Hydroiodic acid

+ SO~ - (aq)

*Note that although each sulfuric acid molecule has two ionizable hydrogen atoms, it only undergoes the first ionization completely, effectively producing one H+ ion and one HSO. ion per H 2S0 4 molecule. The second ionization happens only to a very small extent.

Not~he

use of the double arrow,. • , in this equation and in two earlier equations, including one in Table 4.1. This denotes a reaction that occurs in both directions and does not result in all the reactant(s) (e.g., acetic acid) being converted permanently to product(s) (e.g., hydrogen ions and • acetate ions). Instead, forward and reverse reactions both occur, and a state of dynamic chemical equilibrium is established. Although acetic acid molecules ionize, the resulting ions have a strong tendency to recombine to form acetic acid molecules again. Eventually, the ions produced by the ionization will be recombining at the same rate at which they are produced, and there will be no further change in the numbers of acetic acid molecules, hydrogen ions, or acetate ions. Because there is a stronger tendency for the ions to recombine than for the molecules to ionize, at any given point in time, most of the dissolved acetic acid exists as molecules that are not ionized (reactant). Only a very small percentage exists in the form of hydrogen ions and acetate ions (products). The ionization of a weak base, while similar in many ways to the ionization of a weak acid, requires some additional explanation. Ammonia (NH3) is a common weak base. The ionization of ammonia in water is represented by the equation

ote that the ammonia molecule does not ionize by breaking apart into ions. Rather, it does so by ionizing a water molecule. The H+ ion from a water molecule attaches to an ammonia molecule, producing an ammonium ion (NHt) and leaving what remains of the water molecule, the OHion, in solution. +

+

•

•

+

•

•

+

NH t (aq)

+

OH-(aq)

As with the ionization of a weak acid, the reverse process predominates and at any given point in time, there will be far more NH3 molecules present than there will be NH t and OH- ions. We can distinguish between electrolytes and nonelectrolytes experimentally using an apparatus like the one pictured in Figure 4.1. A lightbulb is connected to a battery using a circuit that includes the contents of the beaker. For the bulb to light, electric current must flow from one electrode to the other. Pure water is a very poor conductor of electricity because H 2 0 ionizes to only a minute extent. There are virtually no ions in pure water to conduct the current, so H 20 is considered a nonelectrolyte. If we add a small amount of salt (sodium chloride), however, the lightbulb will begin to glow as soon as the salt dissolves in the water. Sodium chloride dissociates completely in water to give Na + and CI- ions. Because the NaCI solution conducts electricity, we say that NaCI is an electrolyte. If the solution contains a nonelectrolyte, as it does in Figure 4.1(a), the bulb will not light. If the solution contains an electrolyte, as it does in Figure 4.1 (b) and (c), the bulb will light. The

•

•

• • •

In a state of dynamic chemk:al equilibrium, or simply equilibrium, both forward and reverse reactions continue to occur. However, because they are occurring at the same rate, no net change is observed over time in the amounts of reactants or products. Chemical equilibrium is the subject of Chapters 15 to 17.

114

CHAPTER 4

Reactions in Aqueous Solutions

-

•

......

_-

•

•

•

•

•

•

(a)

•

•

To (+)

To (-)

To (+)

To ( -)

electrode

electrode

electrode

electrode

(b)

(c)

Figure 4.1

An apparatus for distinguishing between electrolytes and nonelectrolytes, and between weak electrolytes and strong electrolytes. A solution's ability to conduct electricity depends on the number of ions it contains . (a) Pure water contains almost no ions and does not conduct electricity, therefore the lightbulb is not lit. (b) A weak electrolyte solution such as HF(aq) contains a small number of ions, and the lightbulb is dimly lit. (c) A strong electrolyte solution such as NaCI(aq) contains a large number of ions, and the lightbulb is brightly lit. The molar amounts of dissolved substances in the beakers in (b) and (c) are equal.

cations in solution are attracted to the negative electrode, and the anions are attracted to the positive electrode. This movement sets up an electric current that is equivalent to the flow of electrons along a metal wire. How brightly the bulb burns depends upon the number of ions in solution. In Figure 4.1 (b), the solution contains a weak electrolyte and therefore a rela- "-.J tively small number of ions, so the bulb lights only weakly. The solution in Figure 4.1(c) contain s a strong electrolyte, which produces a relatively large number of ions, so the bulb lights brightly.

Bringing Chemistry to life The Invention of Gatorade In 1965 , University of Florida (UF) assistant coach Dwayne Douglas was concerned about the health of Gators football players. He noted that during practices and games in hot weather the players (1) lost a great deal of weight, (2) seldom needed to urinate, and (3) had limited stamina, especially during the second half of a practice or game. He consulted Dr. Robert Cade, researcher and kidney-disease specialist at UP's medical college, who embarked on a project to identify the cause of the athletes' lack of endurance. It was found that after a period of intense activity accompanied by profuse sweating, the players had low blood sugar, low blood volume, and an imbalance of electrolytes-all of which contributed to heat exhaustion. Cade and his research fellows theorized that the depletion of sugar, water, and electrolytes might be remedied by having the athletes drink a solution containing just the right amounts of

SECTION 4.1

General Properties of Aqueous Solutions

each. Using this theory, they developed a beverage containing water, sugar, and sodium and potassium salts similar to those present in sweat. By all accounts, the beverage tasted so bad that no one would drink it. Mary Cade, Robert Cade's wife, suggested adding lemon juice to make the concoction more palatable and the drink that would become Gatorade was born. In their 1966 season the Gators earned a reputation as the "second-half' team, often coming from behind in the third or fourth quarter. Gators coach Ray Graves attributed his team's newfound late-in-the-game strength to the newly developed sideline beverage that replenished blood sugar, blood volume, and electrolyte balance. Sports drinks are now a multibillion dollar industry, and there are several popular brands, although Gatorade still maintains a large share of the market.

----------------------------------------.. Sports drinks typically contain sucrose (C 12H 22 0 11 ), fructose (C 6H 120 6), sodium citrate (Na3C6Hs07), potassium citrate (K3C 6H s0 7), and ascorbic acid (H2 C6H 60 6), among other ingredients. Classify each of these ingredients as a nonelectrolyte, a weak electrolyte, or a strong electrolyte. Strategy Identify each compound as ionic or molecular; identify each molecular compound as acid, base, or neither; and identify each acid as strong or weak. Setup Sucrose and fructose contain no cations and are therefore molecular compounds-neither is an acid or a base. Sodium citrate and potassium citrate contain metal cations and are therefore ionic compounds. Ascorbic acid is an acid that does not appear on the list of strong acids in Table 4.1, so ascorbic acid is a weak acid. Solution Sucrose and fructose are nonelectrolytes. Sodium citrate and potassium citrate are strong electrolytes. Ascorbic acid is a weak electrolyte.

I

Practice Problem A so-called enhanced water contains citric acid (H3C 6H s0 7), magnesium lactate [Mg(C 3H s0 3)2], calcium lactate [Ca(C 3H s0 3)2], and potassium phosphate (K3P0 4). Classify each of these compounds as a nonelectrolyte, a weak electrolyte, or a strong electrolyte.

------------------------------------------------------------..

'-._. - -

General Properties of Aqueous Solutions

Checkpoint 4.1 4.1.1

4.1.2

4.1.3

Soluble ionic compounds are

Which of the following compounds is a weak electrolyte?

a) always nonelectrolytes

a) LiCl

b) always weak electrolytes

b) (C2H s)2 NH

c) always strong electrolytes

c) KN0 3

d) never strong electrolytes

d) NaI

e) sometimes nonelectrolytes

e) HN0 3

Soluble molecular compounds are •

4.1.4

Which of the following compounds is a strong electrolyte?

a) always nonelectrolytes

a) HF

b) always weak electrolytes

b) H 2C0 3

c) always strong electrolytes

c) NaF

d) never strong electrolytes

d) NH3

e) sometimes strong electrolytes

e) H 2 O •

Think About It Remember that any soluble ionic compound is a strong electrolyte, whereas most molecular compounds are nonelectrolytes or weak electrolytes. The only molecular compounds that are strong electrolytes are the strong acids listed in Table 4.1.

115

How Can I Tell if a Compound Is an Electrolyte? While the experimental method described in Figure 4.1 can be useful, often you will have to characterize a compound as a nonelectrolyte, a weak electrolyte, or a strong electrolyte just by looking at its formula. A good first step is to determine whether the compound is ionic or molecular. An ionic compound contains a cation (w hich is either a metal ion or the ammonium ion) and an anion (which may be atomic or polyatomic). A binary compound that contains a metal and a nonmetal is almost always ionic. This is a good time to review the polyatomic anions in Table 2.8 [ ~. Section 2.7] . You will need to be able to recognize them in the formulas of compounds. Any ionic compound that dissolves in water is a strong electrolyte. If a compound does not contain a metal cation or the ammonium cation, it is molecular. In this case, you will need to determine whether or not the compound is an acid. Acids generally can be recognized by the way their formulas are written, with the ionizable hydrogens written first. HC 2H 30 2 , H 2 C0 3 , and H 3P0 4 are acetic acid, carbonic acid, and phosphoric acid,

respectively. Formulas of carboxylic acids, such as acetic acid, often are written with their ionizable hydrogen atoms last in order to keep the functional group together in the formula. Thus, either HC zH 3 0 ? or CH 3COOH is con'ect for acetic acid. To make it easier to identify compounds as acids, in this chapter we will write all acid formulas with the ionizable H atom(s) first. If a compound is an acid, it is an electrolyte. If it is one -of the acids listed in Table 4.1 , it is a strong acid and therefore a strong electrolyte. Any acid not listed in Table 4.1 is a weak acid and therefore a weak electrolyte. If a molecular compound is not an acid, you must then consider whether or not it is a weak base. Many weak bases are related to ammonia in that they consist of a nitrogen atom bonded to hydrogen and/or carbon atoms. Examples include methylamine (CH 3NH?), pyridine (CsHsN), and hydroxylamine (NH 20H). Weak bases are weak electrolytes. If a molecular compound is neither an acid nor a weak base, it is a nonelectrolyte.

Classify each of the following compounds as a nonelectrolyte, a weak electrolyte, or a strong electrolyte: (a) methanol (CH 30H), (b) sodium hydroxide (NaOH), (c) ethylamine (C2HsNHz), and (d) hydrofluoric acid (HF). Strategy Classify each compound as ionic or molecular. Soluble ionic compounds are

Acetic acid

strong electrolytes. Classify each molecular compound as an acid, base, or neither. Molecular compounds that are neither acids nor bases are nonelectrolytes. Molecular compounds that are bases are weak electrolytes . Finally, classify acids as either strong or weak. Strong acids are strong electrolytes, and weak acids are weak electrolytes. Setup (a) Methanol contains neither a metal cation nor the ammonium ion. It is therefore

molecular. Its formula does not begin with H, so it is probably not an acid, and it does not contain a nitrogen atom, so it is not a weak base. Molecular compounds that are neither acids nor bases are nonelectrolytes. Think About It Make sure that

you have correctly identified compounds that are ionic and compounds that are molecular. Remember that strong acids are strong electrolytes, weak acids and weak bases are weak electrolytes, and strong bases are strong electrolytes (by virtue of their being soluble ionic compounds). Molecular compounds, with the exceptions of acids and weak bases, are nonelectrolytes.

(b) Sodium hydroxide contains a metal cation (Na +) and is therefore ionic. It is also one of the strong bases. (c) Ethylarnine contains no cations and is therefore molecular. It is also a nitrogen-containing base, similar to ammonia. (d) Hydrofluoric acid is, as its name suggests, an acid. However, it is not on the list of strong acids in Table 4.1 and is, therefore, a weak acid. Solution (a) Nonelectrolyte

(b) Strong electrolyte (c) Weak electrolyte (d) Weak electrolyte

Practice Problem A Identify the following compounds as nonelectrolytes, weak electrolytes,

or strong electrolytes: ethanol (C 2HsOH), nitrous acid (HN0 2), and sodium hydrogen carbonate (NaHC0 3, also known as bicarbonate). Practice Problem B Identify the following compounds as nonelectrolytes, weak electrolytes,

or strong electrolytes: phosphorous acid (H 3P0 3), hydrogen peroxide (HZ0 2), and ammonium sulfate [(NH4)2S04l .

116

SECTION 4.2

Precipitation Reactions

117

Precipitation Reactions When an aqueous solution of lead(II) nitrate [Pb(N0 3)?] is added to an aqueous solution of sodium iodide (NaI), a yellow insoluble solid lead(U) iodide (PbI 2 ) forms. Sodium nitrate (NaN0 3), the other reaction product, remains in solution. Figure 4.2 shows this reaction in progress. An insoluble solid product that separates from a solution is called a precipitate, and a chemical reaction in which a precipitate forms is called aprecipitation reaction. Precipitation reactions usually involve ionic compounds, but a precipitate does not form every time two solutions of electrolytes are combined. Instead, whether or not a precipitate forms when two solutions are mixed depends on the solubility of the products.

Solubility Guidelines for Ionic Compounds in Water

•

. . When an ionic substance such as sodium chloride dissolves in water, the water molecules remove Multimed ia individual ions from the three-dimensional solid structure and sUlTound them. This process, called Precipitation of BaS04. hydration, is shown in Figure 4.3. Water is an excellent solvent for ionic compounds because H 20 is a polar molecule; that is, its electrons are distributed such that there is a partial negative charge on the oxygen atom, denoted by the 8- ... symbol, and partial positive charges, denoted by the 8+ . ... ... ... ...... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... . . . symbol, on each of the hydrogen atoms. The oxygen atoms in the sUlTounding water molecules The partial charges on the oxygen atom and the hydrogen atoms sum to zero. Water molecules, are attracted to the cations, while the hydrogen atoms are attracted to the anions. These attractions although polar, have no net charge. You will explain the orientation of water molecules around each of the ions in solution. The surrounding learn more about partial charges and molecular polarity in Chapters 8 and 9. water molecules prevent the cations and anions from recombining. Solubility is defined as the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Not all ionic compounds dissolve in water. Whether or not If the water molecules' attraction for the ions exceeds the ions' attraction to one another, an ionic compound is water soluble depends on the relative. .magnitudes of the water molecules' . . ... . . . . . . . . . ... .. ............. .. . . . . . . . then the ionic compound will dissolve. If the attraction to the ions, and the ions' attraction for each other. We willieam more about the magniions' attraction to each other exceeds the water tudes of attractive forces in ionic compounds in Chapter 8, but for now it is useful to learn some molecules' attraction to the ions, then the ,

compound won't dissolve.

o o

The addition of a colorless NaI(aq) solution ...

-/---N03'

to a colorless Pb(N0 3 h(aq) solution ...

Figure 4.2

produces PbI2 (s), a yellow precipitate ...

which settles out of solution. The remaining solution contains Na+ and NO) ions.

A colorless aqueou s solution of NaI is added to a colorless aqueous solLition of Pb(N0 3)2' A yellow precipitate, PbI2, fonns. Na + and

0 :3 ions remain in solution.

118

CHAPTER 4

Reactions in Aqueous Solutions

Figure 4.3

Hydration of anions and cations of a soluble ionic compound. Water molecules surround each anion with their partial positive charges (H atoms) oriented toward the negatively charged anion; and they surround each cation with their partial negative charges (0 atoms) oriented toward the positively charged cation.

+

•

•••

Some books list fewer exceptions to these solubil ity rules. In fact, ionic compounds listed here as "insoluble" are actually very slightly sol uble. It is how soluble a compound must be to be called" soluble" that may vary from book to book.

Water-Soluble Compounds

Insoluble Exceptions

Compounds containing an alkali metal cation (Li +, Na+, K +, Rb +, Cs +) or the ammonium ion (NHt) · . . . . ...

Ionic compounds often are classified according to the anions they contain. Compounds that contain the chloride ion are called chlorides, compounds containing the nitrate ion are called nitrates, and so on.

·Compounds containing the nitrate ion (NO }), acetate ion (C 2H 30 2 ), or chlorate ion (CIO 3 ) Compounds containing the chloride ion (CI-), bromide ion (Br-), or iodide ion (1-)

ComRounds containing Ag +, Hg~+ , or Pb2+

Compounds containing the sulfate ion (SO~ - )

Compounds containing Ag +, Hg~ + , Pb2+ , Ca2 +, Sr2+ , or Ba2+

Water-Insoluble Compounds

Soluble Exceptions

Compounds containing the carbonate ion (CO ~- ), phosphate ion (PO~ - ), chromate ion (CrOJ- ), or sulfide ion (S2- )

Compounds containing Li +, N a +, K +, Rb +, Cs +, or NH t

Compounds containing the hydroxide ion (OH-)

Compounds containing Li +, N a +, K +, Rb +, Cs +, or Ba2+

· . . .... Note the reoccurrence of the same three groups of ions in the exceptions columns in Tables 4.2 and 4.3: Group 1A or the ammonium cation; Ag+, Hg ~+, or Pb'+; and the heavier Group 2A cations.

...

.. .

~

-

-

Multimedia

Chemical reactions-predicting precipitation reactions (interactive).

guidelines that enable us to predict the solubility of ionic compounds. Table 4.2 lists groups of compounds that are soluble and shows the insoluble exceptions, Table 4.3 lists groups of compounds that are insoluble and shows the soluble exceptions, Sample Problem 4.3 gives you some practice applying the solubility guidelines.

SECTION 4.2

Precipitation Reactions

119

Sample Problem 4.3 Classify each of the following compounds as soluble or insoluble in water: (a) AgN0 3, (b) CaS04, (c) K 2C0 3 . Strategy Use the guidelines in Tables 4.2 and 4.3 to determine whether or not each compound is

expected to be water soluble. Setup (a) AgN0 3 contains the nitrate ion (NO }). According to Table 4.2, all compounds containing

the nitrate ion are soluble. (b) CaS04 contains the sulfate ion (SO ~- ). According to Table 4.2, compounds containing the sulfate ion are soluble unless the cation is Ag +, Hgi +, Pb z+, Ca z+, Sr2+, or Ba2 +. Thus, the Caz+ ion is one of the insoluble exceptions. (c) KZC03 contains an alkali metal cation (K+) for which, according to Table 4.2, there are no insoluble exceptions. Alternatively, Table 4.3 shows that most compounds containing the carbonate ion (CO~- ) are insoluble- but compounds containing a Group lA cation such as K+ are soluble exceptions. Solution (a) Soluble, (b) Insoluble, (c) Soluble.

Think About It Check the ions

in each compound against the information in Tables 4.2 and 4.3 to confirm that you have drawn the right conclusions.

Practice Problem A Classify each of the following compounds as soluble or insoluble in water:

(a) PbCl z, (b) (NH4)3P04, (c) Fe(OH)3' Practice Problem B Classify each of the following compounds as soluble or insoluble in water:

(a) MgBrz, (b) Ca3(P04)z, (c) KCl0 3.

Molecular Equations The reaction shown in Figure 4.2 can be represented with the chemical equation Pb(N0 3Maq)

+ 2NaI(aq) - _ . 2NaN0 3 (aq) + PbI2 (s)

Based on this chemical equation, the metal cations seem to exchange anions. That is , the Pb2+ ion, originally paired with NO ) ions, ends up paired with 1- ions; similarly, each Na + ion, originally paired with an 1- ion, ends up paired with an NO ) ion: 'equatIoii; 'writteii; 'Is . a molecular equation, which is a chemical equation written with all compounds represented by their chemical formulas, making it look as though they exist in solution as molecules or formula units. You now know enough chemistry to predict the products of this type of chemical reaction! Simply write the formulas for the reactants, and then write formulas for the compounds that would form if the cations in the reactants were to trade anions. For example, if you want to write the equation for the reaction that occurs when solutions of sodium sulfate and barium hydroxide are combined, you would first write the formulas of the reactants [ ~~ Section 2.7] :

'Thls'

'as'

called ..... .

Then you would write the formula for one product by combining the cation from the first reactant (Na +), with. the anion from the second reactant (OH - ); you would then write the formula for the other product by combining the cation from the second reactant (Ba2+) with the anion from the first (SO~ - ). Thus, the equation is Na2S0iaq)

+ Ba(OHMaq) - _ . 2NaOH + BaS04

Although we have balanced the equation [ ~~ Section 3 .3], we have not yet put phases in parentheses for the products. The final step in predicting the outcome of such a reaction is to determine which of the products, if any, will precipitate from solution. We do this using the solubility guidelines for ionic compounds (Tables 4.2 and 4.3). The first product (NaOH) contains a Group lA cation (Na +) and will therefore be soluble. We indicate its phase as (aq). The second product (BaS04) contains the 2 sulfate ion (SO~ - ). Sulfate compounds are soluble unless the cation is Ag+, Hg ~ + , Pb +, Ca2+ , Sr2 +, or Ba2 +. BaS04 is therefore insoluble and will precipitate. We indicate its phase as (s) : Na2S04(aq)

+ Ba(OHMaq) - _ . 2NaOH(aq) + BaSOi

s)

Reactions in which compounds exchange ions are sometimes called metathesis or double replacement reactions.

,

,

;

i)

;

II

I'

120

CHAPTER 4

Reactions in Aqueous Sol utions

Ionic Equations Although molecular equations are useful, especially from the standpoint of knowing which solutions to combine in the laboratory, they are in a sense unrealistic. Soluble ionic compounds are strong electrolytes [ ~~ Section 4 . 1] . As such, they exist in solution as hydrated ions, rather than as formula units. Thus, it would be more realistic to represent the aqueous species in the reaction of Na2S04(aq) with Ba(OH)z(aq) as follows: Na2S04(aq) --+. 2Na+(aq) Ba(OH)z(aq)

• Ba2+(aq)

NaOH(aq)

• Na+(aq)

•

+ SO~-(aq)

+ 20H-(aq) + OH-(aq)

If we were to rewrite the equation, representing the dissolved compounds as hydrated ions, would be

~t

This version of the equation is called an ionic equation, a chemical equation in which any compound that exists completely or predominantly as ions in solution is represented as those ions. Species that are insoluble or that exist in solution completely or predominantly as molecules are represented with their chemical formulas, as they were in the molecular equation.

Net Ionic Equations Na + (aq) and OH-(aq) both appear as reactants and products in the ionic equation for the reaction of Na2S04(aq) with Ba(OH)z(aq). Ions that appear on both sides of the equation arrow are called spectator ions because they do not participate in the reaction. Spectator ions cancel one another, just as identical terms on both sides of an algebraic equation cancel one another, so we need not show spectator ions in chemical equations. >'-"" +

aq)

+ SO~-(aq) + Ba2+(aq) + 2

- aq) --+. ~ + aq)

+2

- aq)

+ BaSOis)

Eliminating the spectator ions yields the following equation: . . . . . . . . . .. .. Although the reactants may be written in either order in the net ionic equation, it is common for the cation to be shown first and the anion second.

••• • ••••.• • • • • • • • •••• • .•• . .••. • • • • ' ') :.f , • . • • .. • •• •.

Ba- (aq)

.

'2 ······· ········· ··· · · · · ···

+ S04-(aq)

• BaS04(s)

This version of the equation is called a net ionic equation, which is a chemical equation that includes only the species that are actually involved in the reaction, The net ionic equation, in effect, tells us what actually happens when we combine solutions of sodium sulfate and barium hydroxide. The steps necessary to determine the molecular, ionic, and net ionic equations for a precipitation reaction are as follows: 1. Write and balance the molecular equation, predicting the products by assuming that the cations trade anions. 2. Write the ionic equation by separating strong electrolytes into their constituent ions. 3. Write the net ionic equation by identifying and canceling spectator ions on both sides of the equation, If both products of a reaction are strong electrolytes, all the ions in solution are spectator ions. In this case, there is no net ionic equation and no reaction takes place, Sample Problem 4.4 illustrates the stepwise determination of molecular, ionic, and net ionic equations.

Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of lead acetate [Pb(C 2H30 2h]' and calcium chloride (CaCl 2), are combined.

Strategy Predict the products by exchanging ions and balance the equation. Determine which product will precipitate based on the solubility guidelines in Tables 4.2 and 4.3. Rewrite the equation showing strong electrolytes as ions. Identify and cancel spectator ions.

SECTION 4.3

Acid-Base Reactions

121

Setup The products of the reaction are PbCl 2 and Ca(C 2H 30 2)2' PbCl 2 is insoluble, because Pb2+

I

is one of the insoluble exceptions for chlorides, which are generally soluble. Ca(C 2H 30 2)2 is soluble because all acetates are soluble. Think About It Remember that

Solution Molecular equation:

the charges on ions in a compound must sum to zero. Make sure that you have written correct formulas for the products and that each of the equations you have written is balanced. If you find that you are having trouble balancing an equation, check to make sure you have correct formulas for the products.

!

I

Ionic equation: Pb 2+(aq)

+ 2C 2H 30 2" (aq) + Ca2+(aq) + 2Cqaq) - _ . PbCI 2(s) + Ca2+ (aq) + 2C 2H 30 2 (aq)

Net ionic equation: .

Pb2+ (aq)

+ 2Cqaq) - _ . PbClis)

I Practice Problem A Write the molecular, ionic, and net ionic equations for the combination of Sr(N0 3)iaq) and Li 2S0 4 (aq).

I

Practice Problem B Write the molecular, ionic, and net ionic equation s for the combination of KN0 3(aq) and BaCI2 (aq).

~'-----------------------------------------------------------"

Checkpoint 4.2 4 .2 .1

Precipitation Reactions

Which of the following are water soluble? (Choose all that apply.)

4.2.3

a) Na2S

a) CO ~ - and OW

b) Ba(C 2H 30 2h

b) Li + and OH-

c) CaC0 3

c) Li + and Ba2+

d) CuBr2

d) Ba2+ and OHe) Ba2+ and CO ~-

e) Hg 2CI 2

4.2.2

What are the spectator ions in the ionic equation for the combination of Li 2C0 3(aq) and Ba(OH)iaq)?

Which of the following are water insoluble? (Choose all that apply.)

4.2.4

Select the correct net ionic equation for the combination of Fe(N0 3Maq) and Na2C03(aq).

a) Ag 2Cr04

a) Na+(aq )

+ CO ~ -(aq)

• NaC0 3(s)

b) Li 2C0 3

b) Fe 2 +(aq) + CO ~ - (aq)

• FeC0 3(s)

c) Ca3(P04)2

c) 2Na+(aq) + CO ~-(aq)

• Na2C03(S)

d) BaS04

d) Fe2+ (aq)

e) ZnCl 2

e) Na+(aq ) + N0 3" (aq )

+ 2N0 3(aq )

Acid-Base Reactions Another type of reaction occurs when two solutions, one containing an acid and one containing a base, are combined. We frequently encounter acids and bases in everyday life (Figure 4.4). Ascorbic acid, for instance, is also known as vitamin C, acetic acid is the component responsible for the sour taste and characteristic smell of vinegar, and hydrochloric acid is the acid in muriatic acid and is also the principal ingredient in gastric juice (stomach acid). Ammonia, found in many cleaning products, and sodium hydroxide, found in drain cleaner, are common bases. Acid-base chemistry is extremely important to biological processes. Let's look again at the properties of acids and bases, and then look at acid-base reactions.

• Fe(N0 3M s) • NaN0 3(s)

! 122

Figure 4.4

CHAPTER 4

Reacti ons in Aqueous Solutions

Some common acids

and bases. Clockwise from the top left: Acetic acid (HC2H 3 0 2), ascorbic add (C 6H s0 6 or, with its ionizable hydrogens written first, H 2 C 6H 60 6), hydrochloric acid (HCI), ammonia (NH 3), and sodium hydroxide (NaOH). HCI and NaOH are both strong electrolytes and exist in solution entirely as ions. Water molecules are not shown.

o

•,

~

•

••

I

e;;

< .

+2 .+6,II - 8'

d) The molar concentration of a 0.025 percent weight by volume solution of K2Cr20 7 is 8.5 X 10- 4 M. 4 e) 15 mL of a 0.014 M stock solution of K2Cr20 7 would be required to prepare 250 mL of 8.5 X 10- M KZCr20 7' [K+) = 1.7 X 10- 3 M; [Cr20 ~ - ) = 8.5 X 10- 4 M.

o

+

!

. +2 : - 2 i !

i!

!

•

•

ermoc emlstr

5.1

Energy and Energy Changes

• •

•

5.2

Introduction to Thermodynamics

• • •

5.3

States and State Functions The First Law of Thermodynamics Work and Heat

Enthalpy

• • •

5.4

Reactions Carried Out at Constant Volume or at Constant Pressure Enthalpy and Enthalpy Changes Thermochemical Equations

Calorimetry

• •

•

5.5 5.6

Forms of Energy Energy Changes in Chemical Reactions Units of Energy

Specific Heat and Heat Capacity Constant-Pressure Calorimetry Constant-Volume Calorimetry

Hess's Law Standard Enthalpies of Formation

Calories in Food

Nutrition Facts Serving Size Abou t 24 biSCUits

The Food and Drug Administration (FDA) and the U.S. Department of Agriculture (USDA) require "Nutrition Facts" labels on nearly all prepared foods. These labels indicate such things as serving size, grams of fat per serving, and Calories per serving. Health-conscious consumers typically are avid readers of food labels and use them to make informed dietary choices. Furthermore, because dieting to lose weight is something of a national obsession, many Americans are willing to pay premium prices for packaged foods that both taste good and are low in fat and Calories. For this reason, it is important that the labels on packaged foods contain accurate information. In 1997, spurred by numerous complaints from consumers and health-care professionals, the FDA launched an investigation into the practices of a weight-loss product company with packaging facilities in Kentucky and Illinois. The company sold "low-fat, ................. . . . . . . . .. ....... ... .. . ... . . . . . . . . . . . . . . . . . carob-coated" doughnuts that, according to package labeling, each contained just 3 g of fat and 135 Calories. According to an analysis by the FDA, the doughnuts were• actually chocolate-coated and each contained 18 g of fat and 530 Calories! As the FDA investigation revealed, the company had simply purchased ordinary doughnuts from a Chicago bakery and repackaged them as "diet" doughnuts, marking up the price by as much as 200 percent. The investigation led to a climinal indictment. Experimental determination of the caloric content of food is done using the techniques ... .. . . .. . . of thennochemistry. A weighed sample of the food is combusted in a bomb calorimeter, and the heat generated by the combustion, which is equal to the energy content of the food, is determined by measuring the resulting temperature increase in the calorimeter.

(59g/2 1 oz.)

About 12

'1'1 Cup Vitamins A&D Amount Per Serving

Calories Catorles Irom Fat

Cereal

Fat Free Milk

200 10

240 10

% Daily Valu e'""

Total Fat 19 ' Saturated Fat 09 Monounsaturalod Fat 09 PolyullS(:lIurated Fal 0.59 Trans Fat Oq Cholesterol Om- L~ser pulses

Cornea -----:flattened

Flap replaced Central cornea flattened

In This Chapter, You Will Learn

about some of the properties of electromagnetic radiation or light and how these properties have been used to study and elucidate the electronic structure of atoms. You will also learn how to determine the arrangement of electrons in a particular atom.

Before you begin, you should review [ ~~

•

Tracking units

•

The nuclear model of the atom

•

Section 1.6] [ ~~

...-... , r ..

Media Player/ MPEG Content Chapter in Review

Section 2.2]

Lasers are used in a variety of surgical procedures, including cosmetic procedures. Laser resurfacing, shown here, is done to rejuvenate the appearance of the face. 193

194

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

The Nature of Light When we say "light," we generally mean visible light, which is the light we can detect with our eyes. Visible light, however, is only a small part of the continuum of radiation that comprises the electromagnetic spectrum. In addition to visible light, the electromagnetic spectrum includes radio waves, microwave radiation, infrared and ultraviolet radiation, X rays, and gamma rays, as shown in Figure 6.1. Some of these terms may be familiar to you. For instance, the danger of exposure to ultraviolet radiation is why you need to use sunscreen. You may have used microwave radiation from a microwave oven to reheat food or to pop popcorn; you may have had X rays during a routine dental checkup or after breaking a bone; and you may recall from Chapter 2 that gamma rays are emitted from some radioactive materials. Although these phenomena may seem very different from each other and from visible light, they all are the transmission of energy in the form of waves.

Properties of Waves

The speed of light is an exact number and usually does not limit the number of significant figures in a calculated result. In most calculations, however, the speed of light is rounded to three significant figures: c = 3.00 X 108 mls.

The fundamental properties of waves are illustrated in Figure 6.2. Waves are characterized by their wavelength, frequency, and amplitude. Wavelength A (lambda) is the distance between identical points on successive waves (e.g., successive peaks or successive troughs). The frequency JJ (nu) is the number of waves that pass through a particular point in 1 second. Amplitude is the vertical distance from the midline of a wave to the top of the peak or the bottom of the trough. The speed of a wave depends on the type of wave and the nature of the medium through which the wave is traveling (e.g., air, water, or a vacuum). The speed of light through a vacuum, ..... c,- lii 2.99792458 X 108 mls. The speed, wavelength, and frequency of a wave are related by the equation Equation 6.1

C

=

AV

•

Frequency is expressed as cycles per second, or simply reciprocal seconds (s -1), which is also known as hertz (Hz).

.. ... ~h~r~' ~ '~~i ~ .~~ ~x'p~~s~~d i~ '~~t~rs' (~) a~d' ~eciprocal seconds (S- I), respectively. While wavelength in meters is convenient for this equation, the units customarily used to express the wavelength of electromagnetic radiation depend on the type of radiation and the magnitude of the corresponding wavelength. The wavelength of visible light, for instance, is on the order of nanometers (nm, or 10- 9 m), and that of microwave radiation is on the order of centimeters (cm, or

10- 2 m).

Wavelength (nm)

10- 3

10- 1

1..1_ _ _ _ _-'-I_ _ _ __

10

103

105

107

109

1011

1013

..l.I_ _ _ _ _....lI_ _ _ _ _...JI_ _ _ _ _ _IL-_ _ _ _--L 1 _ _ _ _ _LI_ _ _ _ _1

Frequency (Hz)

Type of radiation

Xray

400nm

Figure 6.1

450

Sun lamps

500

Heat lamps

550

Microwave ovens, police radar, satellite stations

600

UHF TV, cellular telephones

650

FM radio, VHF TV

AM radio

700

Electromagnetic spectrum. Each type of radiation is spread over a specific range of wavelengths (and frequencies). Visible light ranges from 400 run (violet) to 700 run (red).

SECTION 6.1

The Nature of Light

195

Figure 6.2

Characteristics of waves: wavelength, amplitude, and frequency.

Wavelength = distance between peaks AA= 2AB = 4Ac Wavelength AA

Amplitude

A

B

---'--~--- 7~.--~----~-;~~-~-~----;-- '

c Frequency = cycles (waves) per second -I" VA -- l"2v B - 4YC

The Electromagnetic Spectrum In 1873 James Clerk Maxwell proposed that visible light consisted of electromagnetic waves. According to Maxwell's theory, an electromagnetic wave has an electric field component and a magnetic field component. These two components have the same wavelength and frequency, and hence the same speed, but they travel in mutually perpendicular planes (Figure 6.3). The significance of Maxwell's theory is that it provides a mathematical description of the general behavior of light. In particular, his model accurately describes how energy in the form of radiation can be propagated through space as oscillating electric and magnetic fields.

The Double-Slit Experiment A simple yet convincing demonstration of the wave nature of light is the phenomenon of inteiference. When a light source passes through a narrow opening, called a slit, a bright line is generated in the path of the light through the slit. When the same light source passes through two closely paced slits, however, as shown in Figure 6.4, the result is not two bright lines, one in the path of each slit, but rather a series of light and dark lines known as an inteiference pattern. When the light sources recombine after passing through the slits, they do so constructively where the two waves are in phase (giving rise to the light lines) and destructively where the waves are out of phase (giving rise to the dark lines). Constructive interference and destructive interference are properties of waves. The various types of electromagnetic radiation in Figure 6.1 differ from one another in wavelength and frequency. Radio waves, which have long wavelengths and low frequencies, are

z

Figure 6.3

Electric field and magnetic field components of an electromagnetic wave. These two components have the same wavelength, frequency, and amplitude, but they vibrate in two mutually perpendicular planes.

Electric field component

v Magnetic field component

\

x

196

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

•

So

H -+-+-I

First screen Second screen

Max imum -- 0- - Minimum o

(b)

(a)

Figure 6.4

Double-slit experiment. (a) Red lines conespond to the maximum intensity resulting from constructive interference. Dashed blue lines correspond to the minimum intensity resulting from destructive interference. (b) Interference pattern with alternating bright and dark lines.

emitted b y large antennas, such as those used by broadcasting stations. The shorter, visible light wave s are produced b y the motions of electrons within atoms and molecules. The shortest waves, which also have the highest frequency, are 'Y (gamma) rays , which result from nuclear processes [ ~~ Section 2.2] . A s we will see shortly, the higher the frequency, the more energetic the radiation. Thus, ultraviolet radiation, X ray s, and 'Y ray s are high-energy radiation, whereas infrared radiation, microwave radiation, and radio waves are low-energy radiation. Sample Problem 6.1 illustrates the conversion between wavelength and frequency.

~

' Samp' l~ :P;oblein6: n-::i;c~ ~

'r'';__ " ,

_:~

~. ____ ~

"~.

1 is called an excited state. Each excited state is higher in energy than the ground state. In the hydrogen atom, an electron for which n is greater than 1 is said to be in an excited state. 2 The radius of each circular orbit in Bohr's model depends on n . Thus, as n increases from 1 to 2 to 3, the orbit radius increases very rapidly. The higher the excited state, the farther away the electron is from the nucleus (and the less tightly held it is by the nucleus) . Bohr's theory enables us to explain the line spectrum of the hydrogen atom. Radiant energy absorbed by the atom causes the electron to move from the ground state (n = 1) to an excited state (n > 1). Conversely, radiant energy (in the form of a photon) is emitted when the electron moves . . . . . . . . . . . .. . . . . . . . . from a higher-energy excited state to a lower-energy excited state or the ground state. The quantized movement of the electron from one energy state to another is analogous to the movement of a tennis ball either up or down a set of stairs (Figure 6.10, p. 206). The ball can be on any of several steps but never between steps. The journey from a lower step to a higher one is an energy-requiring process, whereas movement from a higher step to a lower step is an energyreleasing process. The quantity of energy involved in either type of change is determined by the distance between the beginning and ending steps. Similarly, the amount of energy needed to move an electron in the Bohr atom depends on the difference in energy levels between the initial and final states. To apply Equation 6.5 to the emission process in a hydrogen atom , let us suppose that the electron is initially in an excited state characterized by n j. During emission, the electron drops to a lower energy state characterized by nf (the subscripts i and f denote the initial and final states, respectively). This lower energy state may be the ground state, but it can be any state lower then the initial excited state. The difference between the energies of the initial and final states is I:1E

=

Ef

-

It is important to recognize that the electron in a hydrogen atom can move from a higher-energy state to any lower-energy state. It does not necessarily move from a higher-energy state to the ground state.

Ej

From Equation 6.5 , X 1O- 18 J

1.-

E j = - 2.18 X 10- 18 J

~

Ef

= -2.18

n~

and

n~ I

Therefore, I:1E

=

-2.18 X 10- 18 J

- 2.18

n~

10- 18 J

X 2

n I·

•

= - 2.18

X

10-

18

J _1

1

nf

nj

?

?

••

•

• ••••• •••••••

•

••••••••

•• • • ••

•••

••

When n, > nf, tlE is negative, indicating energy is emitted. When nf > n;, tl E is positive, indicating energy is absorbed.

n=2 n=3

n=2 n=6

,

n

n

=

=1

1 e

e-

n= 1 n= 1

n=2 n=5

n=2 n=4

What's the point? Each line in the visible emission spectrum of hydrogen is the result of an electronic transition from a higher excited state (n = 3,4, 5, or 6) to a lower excited state (n = 2). The energy gap between the initial and final states determines the wavelength of the light emitted.

206

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

Because this transition results in the emission of a photon of frequency v and energy hv , we can write Equation 6.6

b.E = hv = -2.18 X 10- 18 J

\ - ~ nf nj

A photon is emitted when nj > nf. Consequently, the term in parentheses is positive, making b.E negative (energy is lost to the surroundings). A photon is absorbed when nr > ni, making the term in parentheses negative, so b.E is positive. Each spectral line in the emission spectrum of hydrogen corresponds to a particular transition in a hydrogen atom. When we study a large number of hydrogen atoms, we observe all possible transitions and hence the corresponding spectral lines. The brightness of a spectral line depends on how many photons of the same wavelength are emitted. To calculate the wavelength of an emission line, we substitute ciA for v and then divide both sides of Equation 6.6 by hc. In addition, because wavelength can only have positive values, \ye :" 'take 'the 'absoilite vaiue' orthe 'fIght 'sIde 'of the'equation. (In this case, we do so simply by eliminatFigure 6.10 Mechanical analogy for the emission processes. The ball can : ing the negative sign.) •

rest on any step but not between steps. ... Because 2. 18 x 10- 18 Jlhc = 1.096 x 10' mwhich to three significant figures is equal to Roo, this equation is essentially the same as the Rydberg equation (Equation 6.4).

Figure 6.11

• • • • • • •

1 2.18 X 10- 18 J 1 -A hc

Equation 6.7

'

1 ,

Energy levels in the hydrogen atom and the various emission series. Each series terminates at a different value of n.

1

The emission spectrum of hydrogen includes a wide range of wavelengths from the infrared to the ultraviolet. Table 6.1 lists the series of transitions in the hydrogen spectrum, each with a different value of nr. The series are named after their discoverers (Lyman, Balmer, Paschen, and Brackett). The Balmer series was the first to be studied because some of its lines occur in the visible region. Figure 6.11 shows transitions associated with spectral lines in each of the emission series. Each horizontal line represents one of the allowed energy levels for the electron in a hydrogen atom. The energy levels are labeled with their n values. Sample Problem 6.4 illustrates the use of Equation 6.7.

Series

nf

Lyman

1

2,3,4, ...

Ultraviolet

Balmer

2

3,4,5, ...

Visible and ultraviolet

Paschen

3

4,5,6, ...

Infrared

Brackett

4

5,6,7, ...

Infrared

n·I

Spectrum Region

00

7

6

5

4

-

~c

1- -

-

-

- -

--C:..T;";

-- -

-

-

-

3

Paschen • senes

Balmer . senes

n = 1

Lyman • senes

Brackett senes

•

SECTION 6.3

Bohr's Theory of the Hydrogen Atom

207

"Sample Problem 6.4 Calculate the wavelength (in nm) of the photon emitted when an electron transitions from the n = 4 state to the n = 2 state in a hydrogen atom. Strategy Use Equation 6.7 to calculate A. -

Setup According to the problem, the transition is from n = 4 to n

= 2, so nj = 4 and nt

=

2. The

required constants ate h = 6.63 X 10- 34 J . sand e = 3.00 X 108 rn/s. • • • •

Solution

• •

·

1

2.18 X 10he

18

J

1

• • • • • •

1 n-2I

·•

18

- (6.63

(1 2.18 X 10- J 10- 34 J . s)(3 .00 X 108 rn/s) 22

X

1 42

·

•

:

. . . . . . . . . . . . - .. . . . . . . . . . .. . . .. . . . . . . . . . . . . . . . . . . .. . . .

Therefore, A = 4.87 X 10- 7 m

Remember to keep at least one extra digit in intermediate answers to avoid rounding error in the final result [ ~~ Section 1.5] .

=

487 nm

Think About It Look again at the

.................... : line spectrum of hydrogen in Figure 6.7 and make sure that your result matches one of them. Note that for an emission, nj is always greater than nf, and Equation 6.7 gives a positive result.

Practice Problem A What is the wavelength (in nm) of a photon emitted during a transition from the

n

= 3 state to the n =

1 state in the H atom?

Practice Problem B What is the value of nj for an electron that emits a photon of wavelength 93.14

nm when it returns to the ground s tate in the H atom?

Bringing Chemistry to life Lasers

I

The word laser is an acronym for light amplification by stimulated emission of radiation. It is a special type of emission that may involve electronic transitions in atoms or molecules. The first laser, developed in 1960, was a ruby laser. Ruby is a deep-red mineral containing corundum (AI 20 3 ) , in which some of the A1 3+ ions have been replaced by Cr3+ ions. A cylindrical ruby crystal is positioned between two perfectly parallel mirrors, one of which is only partially reflective. A light source called a flashlamp is used to excite the chromium atoms to a higher energy level. The excited atoms are unstable, so at a given instant some of them return to the ground state by emitting a photon in the red region of the spectrum (A = 694.3 nm). Emission of photons occurs in all directions, but photons emitted directly at either mirror will be reflected back. As the reflected photons pass back through the ruby crystal, they stimulate the emission of more photons in the same direction. These photons, in tum, are reflected back through the crystal, stimulating still more emissions, and so on. Because the light waves are in phase that is , their maxima coincide and their minima coincide the photons enhance one another, increasing their power with each passage between the mirrors. When the light reaches a certain intensity, it emerges fro m the partially reflective mirror as a laser beam. Laser light is characterized by three properties: It is intense, . . . . . . .. . . . . . . . . . . . . . . . . . .. ... it has a precisely known wavelength and therefore energy, and it is coherent.

Coherent means that the light waves are all in

phase. Totally reflectiog mirror

FJashlamp

•

Laser beam

A = 694.3 om

Ruby rod

Partially reflecting mirror

208

CHAPTER 6

Quantum Theory and the Electron ic Structure of Atoms

The applications of lasers are quite numerous. In addition to their use in surgery, their high intensity and ease of focus make them suitable for drilling holes in metals, welding, and carrying out nuclear fusion. The fact that they are highly directional and have precisely known wavelengths makes them very useful for telecommunications. Lasers are also used in isotope separation, in chemical analysis, in holography (three-dimensional photography), in compact disc players, and in supermarket scanners.

Bohr's Theory of the Hydgrogen Atom

Checkpoint 6.3 6.3.1

Calculate the energy of an electron in the n = 3 state in a hydrogen atom.

6.3.3

a) 2.42 x 10- 19 J

b) -2.42

X

c) 7.27 X 10-

d) - 7.27 e) -6.54

X X

a) 4.87

10- 19 J 19

J 18

X

10- 7 m

b) 6.84 X 10- 7 m

c) 1.28

10- 19 J 10-

What is the wavelength of light emitted when an electron in a hydrogen atom goes from n = 5 to n = 3?

X

10- 6 m

d) 3.65 X 10- 7 m

J

e) 1.02 X 10- 7 m

6.3.2

Calculate t1E of an electron that goes from n = 1 to n = 5. a) 8.72

X

b) - 8.72 c) 5.45 d) 2.09

X X

e) -2.09

6.3 .4

10- 20 J X

10- 20 J

10- 17 J 10X

18

J

10- 18 J

Which wavelength corresponds to the transition of an electron in a hydrogen atom from n = 2 to n = I? a) 182 nm b) 91.2nm

c) 724nm d) 812nm e) 122nm

Wave Properties of Matter Bohr's theory was both fascinating and puzzling. It fit the experimental data, but physicists did not understand the underlying principle. Why, for example, was an electron restricted to orbiting the nucleus at certain fixed distances? For a decade no one, not even Bohr himself, could offer a logical explanation. In 1924 Louis de Broglie6 provided a solution to this puzzle. De Broglie reasoned that if energy (light) can, under certain circumstances, behave like a stream of particles (photons), then perhaps particles such as electrons can, under certain circumstances, exhibit wavelike properties.

The de Broglie Hypothesis In developing his revolutionary theory, de Broglie incorporated his observations of macroscopic phenomena that exhibited quantized behavior. For example, a guitar string has certain discrete frequencies of vibration, like those shown in Figure 6.12(a). The waves generated by plucking a guitar string are standing or stationary waves because they do not travel along the string. Some points on the string, called nodes, do not move at all; that is, the amplitude of the wave at these points is zero. There is a node at each end, and there may be one or more nodes between the ends . The greater the frequency of vibration, the shorter the wavelength of the standing wave and the greater the number of nodes. According to de Broglie, an electron in an atom behaves like a standing wave. However, as Figure 6.12 shows, only certain wavelengths are possible or allowed. De Broglie argued that if an electron does behave like a standing wave in the hydrogen atom, the wavelength must fit the circumference of the orbit exactly; that is, the circumference of the orbit must be an integral multiple of the wavelength, as shown in Figure 6.12(b). Otherwise, the wave would partially cancel itself by destructive interference on each successive orbit, quickly reducing its amplitude to zero. .

6. Louis Victor Pierre Raymond Duc de Broglie (1892-1977). French physicist. A member of an old and noble family in France, he held the title of a prince. In his doctoral dissertation, he proposed that matter and radiation have the properties of both wave and particle. For this work, de Broglie was awarded the Nobel Prize in Physics in 1929.

SECTION 6.4

Wave Properties of Matter

Figure 6.12 Standing waves of a vibrating guitar string. The length of the string must be equal to a whole number times one-half the wavelength (AI2). (b) In a circular orbit, only whole number multiples of wavelengths are allowed. Any fractional number of wavelengths . would result in cancellation of the wave due to destructive interference.

I~~----~--~-L----------~'I

1

I 1 1

n=3 1 1 1

eK;,,?~

1 1

n = 1 1---. 1

I I

n=2

/ /1

I half-wavelength

209

n = 5

1

~I

1

1

1

1

1

1L

=

2( ~)

1 1

n=3

2 half-wavelengths

1

1

1

1

1

1

1

Forbidden

I =3(;) L

1

I

n = 33

3 half-wavelengths

L=n(;) (a)

(b)

The relationship between the circumference of an allowed orbit (27Tr) and the wavelength (A) of the electron is given by

27Tr = nA

Equation 6.8

where r is the radius of the orbit, A is the wavelength of the electron wave, and n is a positive integer (1,2, 3, ... ). Because n is an integer, r can have only certain values (integral multiples of A) as n increases from 1 to 2 to 3 and so on. And, because the energy of the electron depends on the size of the orbit (or the value of r), the energy can have only certain values, too. Thus, the energy of the electron in a hydrogen atom, if it behaves like a standing wave, must be quantized. De Broglie's reasoning led to the conclusion that waves can behave like particles and particles can exhibit wavelike properties. De Broglie deduced that the particle and wave properties are related by the following expression:

-

A= h mu

Equation 6.9 ~

. . . . . . . . . . . . . . .. . . . . . . . ..

where A, m, and u are the wavelength associated with a moving particle, its mass, and its velocity, respectively. Equation 6.9 implies that a particle in motion can be treated as a wave, and a wave can exhibit the properties of a particle. To help you remember this important point, notice that the left side of Equation 6.9 involves the wavelike property of wavelength, whereas the right ide involves mass, a property of particles. A wavelength calculated using Equation 6.9 is usually referred to specifically as a de Broglie wavelength. Likewise, we will refer to a mass calculated using Equation 6.9 as a de Broglie mass. Sample Problem 6.5 illustrates how de Broglie's theory and Equation 6.9 can be applied.

Calculate the de Broglie wavelength of the "particle" in the fo llowing two cases: (a) a 25-g bullet traveling at 612 m/s, and (b) an electron (m = 9.109 X 10- 31 kg) moving at 63 .0 m/s . . (Continued)

Mass, m, must be expressed in kilograms for units to cancel properly in Equation 6.9.

210

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

Strategy Use Equation 6.9 to calculate the de Broglie wavelengths. Remember that the mass in Think About It While you are new

at solving these problems, always write out the units of Planck's constant (J . s) as kg . m2 /s . This will enable you to check your unit cancellations and detect cornmon errors such as expressing mass in grams rather than kilograms. Note that the calculated wavelength of a macroscopic object, even one as small as a bullet, is extremely small. An object must be at least as small as a subatomic particle in order for its wavelength to be large enough for us to observe.

Equation 6.9 must be expressed in kilograms in order for the units to cancel properly. Setup Planck's constant, h, is 6.63 X 10: .... '~b~i~'~~ : '6.63 X 10- 34 kg . m2/s.

34

J . s or, for the purpose of making the unit cancellation

• •

: ·•

Solution

•

1 ko

• • • •

(a) 25 g X 100;g = 0.025 kg

• •

h iI. = mu

• •

• • • •

·• • • •

X 10- 34 kg . m 2/s - - - - - - -=

6.63

(b) iI.

• •

=

h mu

(0.025 kg)(612 mls)

6.63 X 10- 34 kg . m2/s

=

(9.109

X

10-

31

kg)(63 .0 mls)

= 1.16

4.3

_

X 10- 3)

m

5

X

10- m

·• •

·• • • • • I

Practice Problem A Calculate the de Broglie wavelength (in nm) of a hydrogen atom (m = l.674 X

10- 27 kg) moving at 1500 emls. Practice Problem B Use Equation 6.9 to calculate the de Broglie mass associated with a photon of

radiation of wavelength 810 nm. The velocity of a photon is the speed of light, c.

Diffraction of Electrons Shortly after de Broglie introduced his equation and predicted that electrons should exhibit wave properties, successful electron diffraction experiments were carried out by Clinton Davisson? and 8 9 Lester Germer in the United States and G. P. Thomson in England. These experiments demonstrated that electrons do indeed possess wavelike properties. By directing a beam of electrons (which are most definitely particles) through a thin piece of gold foil, Thomson obtained a set of concentric rings on a screen, similar to the diffraction pattern observed when X rays (which are most definitely waves) were used. Figure 6.13 shows X-ray and electron diffraction patterns for aluminum.

Checkpoint 6.4 6.4.1

(a)

Wave Properties of Matter

Calculate the de Broglie wavelength associated with a helium-4 atom (4.00 amu) moving at 3.0 X 106 mls. a) 2.0

X

10- 20 m

b) 3.3 X 10-

11

c) 3.3

10-

14

10-

19

10-

27

d) l.8 e) 6.6

X X X

6.4.2

What is the de Broglie mass of a photon of light with iI. = 122 nm? (The velocity of a photon is the speed of light, c.)

m

a) l.81 X 10- 35 kg

m

b) 2.21 X 10- 42 kg

m

c) 2.00

m

d) 2.88 X 10- 19 kg

X

10- 25 kg

e) 9.04 X 10- 36 kg

Quantum Mechanics

(b)

The discovery that waves could have matterlike properties and that matter could have wavelike properties was revolutionary. Although scientists had long believed that energy and matter were distinct entities, the distinction between them, at least at the atomic level, was no longer clear. Bohr's theory was tremendously successful in explaining the line spectrum of hydrogen, but it failed to explain the spectra of atoms with more than one electron. The electron appeared to behave as a particle in some circumstances and as a wave in others. Neither description could completely

Figure 6.13

(a) X-ray diffraction pattern of aluminum foil. (b) Electron diffraction of aluminum foil. The similarity of these two patterns shows that electrons can behave like X rays and display wave properties.

7. Clinton Joseph Davisson (1881-1958). American physicist. He and G. P. Thom son shared the Nobel Prize in Physics in 1937 for demonstrating the wave properties of electrons. 8. Lester Halbert Germer (1896-1972). American physicist. Discoverer (with Davisson) of the wave properties of electrons. 9. George Paget Thomson (1892-1975). English physicist. Son of J. J. Thomson, he received the Nobel Prize in Physics in 1937, along with Clinton Davisson, for demon strating the wave properties of electrons.

SECTION 6.S

Quantum Mechanics

211

,

explain the behavior of electrons in atoms. This left scientists frustrated in their quest to understand exactly where the electrons in an atom are.

The Uncertainty Principle To describe the problem of trying to locate a subatomic particle that behaves like a wave, Werner lO Heisenberg formulated what is now known as the Heisenberg uncertainty principle: It is impossible to know simultaneously both the momentum p (defined as mass times velocity, m X u) and the position x of a particle with certainty. Stated mathematically,

Ll.x • t::..p For a particle of mass m,

>

h

Equation 6.10

41T

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Ll.x • mt::..u

> _h_

Equation 6.11

41T

Like the de Broglie wavelength equation, Equation 6.11 requires that mass be expressed in kilograms. Unit cancellation will be more obvious if you express Planck's constant in kg • m2/s rather than J • s.

where Ll.x and t::..u are the uncertainties in measuring the position and velocity of the particle, respectively. The > signs have the following meaning. If the measured uncertainties of position and velocity are large (say, in a crude experiment), their product can be substantially greater than h141T (hence the> sign). The significance of Equation 6.11 is that even in the most favorable conditions for measuring position and velocity, the product of the uncertainties can never be less than h141T (hence the = sign). Thus, making measurement of the velocity of a particle more precise (i.e., making t::..u a sl1Ulll quantity) means that the position must become correspondingly less precise (i.e., Ll.x will become larger). Similarly, if the position of the particle is known more precisely, its velocity measurement must become less precise. If the Heisenberg uncertainty principle is applied to the hydrogen atom, we find that the electron cannot orbit the nucleus in a well-defined path, as Bohr thought. If it did, we could determine precisely both the position of the electron (from the radius of the orbit) and its speed (from its kinetic energy) at the same time. This would violate the uncertainty principle. Sample Problem 6.6 shows how to use the Heisenberg uncertainty principle.

An electron in a hydrogen atom is known to have a velocity of 5 X 106 m/s + 1 percent. Using the uncertainty principle, calculate the minimum uncertainty in the position of the electron and, given that the diameter of the hydrogen atom is less than I angstrom (A), comment on the magnitude of this uncertainty compared to the size of the atom. Strategy The uncertainty in the velocity, I percent of 5 X 106 mJs, is t:. u. Using Equation 6.11 , calculate t:.x and compare it with the diameter of the hydrogen atom.

Think About It A common error is expressing the mass of the particle in grams instead of kilograms, but you should discover this inconsistency if you check your unit cancellation carefully. Remember that if one uncertainty is small, the other must be large. The uncertainty principle applies in a practical way only to submicroscopic particles. In the case of a macroscopic object, where the mass is much larger than that of an electron, small uncertainties, relative to the size of the object, are possible for both position and velocity.

Setup The mass of an electron (from Table 2.1 , rounded to three significant figures and converted to kilograms) is 9.11 X 10- 3 1 kg. Planck's constant, h, is 6.63 X 10- 34 kg· m2/s. Solution

h t:.x=--,-------'-'-----,--41T • mt:.u

Therefore,

t:.x =

6.63 X 10- 34 kg . m 2/s

41T(9.11 X 10- 31 kg)(5 X 104 m/s)

The minimum uncertainty in the position x is I X 10- 9 m = 10 position is 10 times larger than the atom!

>

I X 10- 9 m

A. The uncertainty in the electron's

Practice Problem Calculate the minimum uncertainty in the position of the 25-g bullet from Sample Problem 6.5 given that the uncertainty in its velocity is + I percent.

..

----------------------------------------------------------------------~

10. Werner Karl Heisenberg (1 901-1976). German physicist. One of the founders of modem quantum theory. Heisenberg received the Nobel Prize in Physics in 1932.

•

212

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms ,

The Schrodinger Equation

•

Figure 6.14

Representation of the electron den sity distributi on surrounding the nucleus in the hydrogen atom. It shows a higher probability of finding the electron closer to the nucleus.

Bohr made a significant contribution to our understanding of atoms, and his suggestion that the energy of an electron in an atom is quantized remains unchallenged, but his theory did not provide a complete description of the behavior of electrons in atoms. In 1926 the Austrian physicist Erwin Schrodinger, II using a complicated mathematical technique, formulated an equation that describes the behavior and energies of submicroscopic particles in general, an equation analogous to Newton 's laws of motion for macroscopic objects. The Schrodinger equation requires advanced calculus to solve, and we will not discuss it here. The equation, however, incorporates both particle behavior, in terms of mass m, and wave behavior, in terms of a wave function Ij; (psi), which depends on the location in space of the system (such as an electron in an atom). The wave function itself has no direct physical meaning. However, the probability of finding the electron in a certain region in space is proportional to the square of the wave function, 1j;2. The idea of relating 1j;2 to probability stemmed from a wave theory analogy. According to wave theory, the intensity of light is proportional to the square of the amplitude of the wave, or 1j;2. The most likely place to find a photon is where the intensity is greatest-that is, where the value of 1j;2 is greatest. A similar argument associates 1j;2 with the likelihood of finding an electron in regions sun'ounding the nucleus. Schrodinger's equation launched an entirely new field, called quantum mechanics (or wave mechanics), and began a new era in physics and chemistry. We now refer to the developments in quantum theory from 1913-when Bohr presented his model of the hydrogen atom-to 1926 as "old quantum theory."

The Quantum Mechanical Description of the Hydrogen Atom The Schrodinger equation specifies the possible energy states the electron can occupy in a hydrogen atom and identifies the corresponding wave functions (Ij;). These energy states and wave functions are characterized by a set of quantum numbers (to be discussed shortly), with which we can construct a comprehensive model of the hydrogen atom. Although quantum mechanics does not allow us to specify the exact location of an electron in an atom, it does define the region where the electron is most likely to be at a given time. The concept of electron density gives the probability that an electron will be found in a particular region of an atom. The square of the wave function, Ij;?, defines the distribution of electron density in three-dimensional space around the nucleus. Regions of high electron density represent a high probability of locating the electron (Figure 6.14). To distinguish the quantum mechanical description of an atom from Bohr's model, we speak of an atomic orbital, rather than an orbit. An atomic orbital can be thought of as the wave function of an electron in an atom. When we say that an electron is in a celiain orbital, we mean that the distribution of the electron density or the probability of locating the electron in space is described by the square of the wave function associated with that orbital. An atomic orbital, therefore, has a characteristic energy, as well as a characteristic distribution of electron density.

Checkpoint 6.S 6.5.1

Quantum Mechanics

What is the minimum uncertainty in the position of an electron moving at a speed of 4 X 106 m/s + 1 percent? (The mass of an electron is 9.11 X 10- 3 1 kg.)

6.5.2

What is the minimum uncertainty in the position of a proton moving at a speed of 4 X 106 m/s + 1 percent? (The mass of a proton is l.67 X 10- 27 kg. )

a) 2 X 10- 8 m

a) 1 X 1O -

13

m

10- 9 m

b) 8 X lO-

lO

m

c) 6 X 10- 9 m

c) 4 X lO-

ll

m

d) 7 X 10- 8 m

d) 3 X lO-

12

m

e) 8 X 1O -

13

m

b) 1

e) I

X

X

10-

12

m

II . Erwin Schriidinger ( 1887-1961). Austrian phys ici st. Schriidinger formul ated wave mechanics, whi ch laid the founda· ti on for modern quantum theory. He received the Nobel Prize in Physics in 1933.

SECTION 6.6

Quantum Numbers

213

Quantum Numbers In Bohr's model of the hydrogen atom, only one number, n, was. .necessary the location .. . . . . to describe . ... . . of the electron. In quantum mechanics, three quantum numbers are required to describe the distribution of electron density in an atom. These numbers are derived from the mathematical solution of SchrOdinger's equation for the hydrogen atom. They are called the principal quantum number, the angular momentum quantum number, and the magnetic quantum number. Each atomic orbital in an atom is characterized by a unique set of these three quantum numbers. -

Principal Quantum Number (n) The principal quantum number (n) designates the size of the orbital. The larger n is, the greater the average distance of an electron in the orbital fro m the nucleus and therefore the larger the orbital. The principal quantum number can have integral values of 1,2, 3, and so forth, and it corresponds to the quantum number in Bohr' s model of the hydrogen atom. Recall from Equation 6.5 that in a hydrogen atom, the value of n determines the energy of an orbital. (As we will see shortly, this is not the case for an atom that contains more than one electron.)

Angular Momentum Quantum Number (e) The angular momentum quantum number (.e) describes the shape of the atomic orbital (see Section 6.7). The values of.e are integers that depend on the value of the principal quantum number, n. For a given value of n, the possible values of.e range from 0 to n -. 1. If n = 1, there is only one possible value of .e; that is, 0 (n - 1 where n = 1). If n = 2, there are two values of .e: 0 and 1. If n = 3, there are three values of .e: 0, 1, and 2. The value of .e is designated by the letters s, p, d, andfas follows: 12

Orbital designation

o

1

2

3

s

p

d

f

Thus, if.e = 0, we have an s orbital; if.e = 1, we have a p orbital; and so on. A collection of orbitals with the same value of n is frequently called a shell. One or more orbitals with the same nand .e values are referred to as a subs hell. For example, the shell designated by n = 2 is composed of two subshells: .e = 0 and .e = 1 (the allowed values of.e for n = 2). These subshells are called the 2s and 2p subshells where 2 denotes the value of n, and sand p denote the values of .e.

Magnetic Quantum Number (me) The magnetic quantum number (m e) describes the orientation of the orbital in space (see Section 6.7). Within a subshelI , the value of me depends on the value of .e. For a certain value of.e, there are (2.e + 1) integral values of me as follows :

- .e, ... 0, ... +.e If .e = 0, there is only one possible value of me: O. If .e = 1, then there are three values of me: -1, 0, and + 1. If .e = 2, there are five values of me, namely, - 2, -1, 0, + 1, and +2, and so on. The number of me values indicates the number of orbitals in a subshell with a particular .e value; that is, each me value refers to a different orbital. Table 6.2 summarizes the allowed values of the three quantum numbers, n, .e, and m e, and Figure 6.15 illustrates schematically how the allowed values of quantum numbers give rise to the number of subshells and orbitals in each shell of an atom. The number of subshells in a shell is 2 equal to n, and the number of orbitals in a shell is equal to n . Sample Problem 6.7 gives you some practice with the allowed values of quantum numbers.

P . The unusual sequence of letters (s, p, d, and!) has a historical origin. Physicists who studied atomic emission spectra

tried to correlate their observations of spectral lines with the energy states involved in the transitions. They described the emission lines as sharp, principal, diffuse, andfimdamental.

e.

The three quantum numbers n, and me specify the size, shape, and orientation of an orbital, respectively.

214

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

When n

e can

•

IS

be

When

e is

me can be

1

only 0

0

only 0

2

oor 1

0 1

only 0 -1 ,0,or+1

3

0,1,or2

0 1 2

only 0 -1,0,or+1 -2, -1,0,+1,or+2

0, 1, 2, or 3

0 1 2 3

only 0 -1,0,or+1 -2, -1,0, +1, or +2 -3, -2, -1,0, +1, +2, or +3

•

•

•

•

•

•

•

•

•

•

•

•

,

4

3----------------------------BBB[D1 +1 ~ +2~ +3 ~e = 31fsubshenl e 2-------------1-2 ~ -llQ]EJEJ---BBCDI +1 ~ +2 ~---e 21 dsubShell e = 1----BI 0~ +1 ~ ---BQ]EJ---------B[DB------e = 11 subshell e 00----[D---------Q]---------------[D---------e 01 subshenl e= =

=

p

=

=

n

=1

n=2

Figure 6.15

n=3

S

n=4

Illustration of how quantum numbers designate shells, subshells, and orbitals.

. · S~mplePr~blem6.7 ...... •

_

.

'

-

-0

•

What are the possible values for the magnetic quantum number (me) when the principal quantum number (n) is 3 and the angular momentum quantum number (C) is 17 Think About It Consult Table 6.2 to make sure your answer is correct. Table 6.2 confirms that it is the value of C, not the value of n, that determines the possible values of me.

Strategy Use the rules governing the allowed values of me. Recall that the possible values of me depend on the value of C, not on the value of n. Setup The possible values of me are - C, ... ,0, ... , +C. Solution The possible values of me are -1, 0, and + 1.

Practice Problem A What are the possible values for me when the principal quantum number (n) is 2 and the angular momentum quantum number (C) is 07 Practice Problem B What are the possible values for me when the principal quantum number (n) is 3 and the angular momentum quantum number (C) is 27

Electron Spin Quantum Number (ms) Whereas three quantum numbers are sufficient to describe an atomic orbital, an additional quantum number becomes necessary to describe an electron that occupies the orbital. Experiments on the emission spectra of hydrogen and sodium atoms indicated that each line in the emission spectra could be split into two lines by the application of an external magnetic field. The only way physicists could explain these results was to assume that electrons act like tiny magnets. If electrons are thought of as spinning on their own axes, as Earth does, their magnetic properties can be accounted for. According to electromagnetic theory, a spinning charge

Quantum Numbers

SECTION 6.6

generates a magnetic field, and it is this motion that causes an electron to behave like a magnet. Figure 6.16 shows the two possible spinning motions of an electron, one clockwise and the other counterclockwise. To specify the electron's spin, we use the electron spin quantum number (m s). Because there are two possible directions of spin, opposite each other: possible vaiiies:" +~ and -~. Conclusive proof of electron spin was established by Otto Stern 13 and Walther Gerlach 14 in 1924. Figure 6.17 shows the basic experimental arrangement. A beam of gaseous atoms generated in a hot furnace passes through a nonuniform magnetic field. The interaction between an electron and the magnetic field causes the atom to be deflected from its straight-line path. Because the direction of spin is random, the electrons in half of the atoms will be spinning in one direction. Those atoms will be deflected in one way. The electrons in the other half of the atoms will be spin- . ning in the opposite direction. Those atoms will be deflected in the other direction. Thus, two spots of equal intensity are observed on the detecting screen. To summarize, we can designate an orbital in an atom with a set of three quantum numbers. These three quantum numbers indicate the size (n), shape (€), and orientation (me) of the orbital. A fourth quantum number (ms) is necessary to designate the spin of an electron in the orbital.

m;has 'two

Checkpoint 6.6 6.6.1

Two electrons in the same orbital with opposite spins are referred to as "paired."

Quantum Numbers

Which of the following is a legitimate set of three quantum numbers: n, and me7 (Select all that apply.)

e,

a) 1, 0,

215

6.6.3

How many subshells are there in the shell designated by n = 37 a) 1

°

,

b) 2

b) 2,0,0

c) 3

c) 1, 0, +1

d) 6

d) 2, 1, +1

e) 9

e) 2,2, - 1

6.6.4 6.6.2

How many orbitals are there in a subshell designated by the quantum numbers n = 3, = 27

What is the total number of orbitals in the shell designated by n = 37

(a)

a) 1

e

Figure 6.16

(a) Clockwise and (b) counterclockwise spins of an electron. The magnetic fields generated by these two spinning motions are analogous to those from the two magnets. The upward and downward arrows are used to denote the direction of spin.

b) 2

a) 2

c) 3

b) 3

d) 6

c) 5

e) 9

d) 7 e) 10

Figure 6.17

Atom beam

Detecting screen

(b)

Magnet Slit screen

Oven

Experimental arrangement for demonstrating the spinning motion of electrons. A beam of atoms is directed through a magnetic field. When a hydrogen atom, with a single electron, passes through the field, it is deflected in one direction or the other, dependin g on the direction of the electron's spin. In a stream consisting of many atoms, there will be equal distributions of the two kinds of spins, so two spots of equal intensity are detected on the screen.

13. Otto Stern (1888-1969). German physicist. He made important contributions to the study of the magnetic properties of atoms and the kinetic theory of gases. Stern was awarded the Nobel Prize in Physics in 1943 . l ·t Walther Gerlach (1889-1979). German physicist. Gerlach's main area of research was in quantum theory.

•

21 6

CHAPTER 6

Qu antum Theory and th e Electronic Structure of Atoms

Atomic Orbitals Strictly speaking, an atomic orbital does not have a well-defined shape because the wave function characterizing the orbital extends from the nucleus to infinity. In that sense, it is difficult to say what an orbital looks like. On the other hand, it is certainly useful to think of orbitals as having specific shapes. Being able to visualize atomic orbitals is essential to understanding the formation of chemical bonds and molecular geometry, which are discussed in Chapters 8 and 9. In this section, we will look at each type of orbital separately.

s Orbitals

The radial probability distribution can be thought of as a map of "where an electron spends most of its time."

For any value of the principal quantum number (n), the value 0 is possible for the angular momentum quantum number (f), corresponding to an s subshell. Furthermore, when f = 0, the magnetic quantum number (me) has only one possible value, 0, corresponding to an s orbital. Therefore, there is an s subshell in every shell, and each s subshell contains just one orbital, an s orbital. Figure 6.18 illustrates three ways to represent the distribution of electrons: the probability ...... denSIty,' 'die 'spheriC' 2, and each p subshell contains three p orbitals. These three p orbitals are labeled 2px, 2py, and 2pz (Figure 6.19), with the subscripted letters indicating the axis along which each orbital is oriented. These three p orbitals are identical in size, shape, and energy; they differ from one another only in orientation. Note, however, that there is no simple relation between the values of me and the x, y, and z directions. For our purpose, you need only remember that because there are three possible values of me, there are three p orbitals with different orientations. The boundary surface diagrams of p orbitals in Figure 6.19 show that each p orbital can be thought of as two lobes on opposite sides of the nucleus. Like s orbitals, p orbitals increase in size from 2p to 3p to 4p orbital and so on.

d Orbitals and Other Higher-Energy Orbitals When the principal quantum number (n) is 3 or greater, the value 2 is possible for the angular momentum quantum number (C), corresponding to a d subshell. When C = 2, the magnetic quantum number (me) has five possible values, - 2, - 1, 0, + I , and +2, each corresponding to a different d orbital. Again there is no direct correspondence between a given orientation and a particular me value. All the 3d orbitals in an atom are identical in energy and are labeled with subscripts denoting their orientation with respect to the x, y, and .G axes and to the planes defined by them. The d orbitals that have higher principal quantum numbers (4d, 5d, etc.) have shapes similar to those shown for the 3d orbitals in Figure 6.20.

z

z

Figure 6.20

z

z

z

the d orbitals.

x

Y

x

Y

J

x

Y

y

x

'-

y

x

Boundary surfaces for

218

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

The! orbitals are important when accounting for the behavior of elements with atomic numbers greater than 57, but their shapes are difficult to represent. In general chemistry we will not concern ourselves with the shapes of orbitals having values greater than 2. Sample Problem 6.8 shows how to label orbitals with quantum numbers.

e

Sample Problem 6.8 List the values of n,

e, and me for each of the orbitals in a 4d subshell.

Strategy Consider the significance of the number and the letter in the 4d designation and determine

the values of nand the value of

e.

e. There are multiple possible values for me, which will have to be deduced from

Setup The integer at the beginning of an orbital designation is the principal quantum number (n).

The letter in an orbital designation gives the value of the angular momentum quantum number (e). The magnetic quantum number (ms) can have integral values of -e, ... , 0, . .. , +e. Solution The values of nand Think About It Consult Figure

e are 4 and 2, respectively, so the possible values of me are -2, -1,0,

+1, and +2.

6.15 to verify your answers. Practice Problem A Give the values of n,

e, and me for the orbitals in a 3d subshell.

Practice Problem B Using quantum numbers, explain why there is no 2d subshell.

Energies of Orbitals The energies of orbitals in the hydrogen atom depend only on the value of the principal quantum number (n), and energy increases as n increases. For this reason, orbitals in the same shell have the same energy regardless of their subshell (Figure 6.21). Is < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f Thus, all four orbitals (one 2s and three 2p) in the second shell have the same energy; all nine orbitals (one 3s, three 3p, and five 3d) in the third shell have the same energy; and all sixteen orbitals (one 4s, three 4p, five 4d, and seven 4j) in the fourth shell have the same energy. The energy picture is more complex for many-electron atoms than' it is for hydrogen, as is discussed in Section 6.8.

Figure 6.21

Orbital energy levels in the hydrogen atom. Each box represents one orbital. Orbitals with the same principal quantum number (n) all have the same energy.

-~-14P ~ 4p ~@ -14d ~ 4d ~ 4d ~ 4d ~B -1 4f ~

4f

~

4f

~

4f

~0~~-

~

3p

~~-13d ~

-~-12P ~

2p

~~-----------------------------------------

- G - 1 3P

3d

~ 3d ~ 3d ~~------------------------

-~----------------------------------------------------

SECTION 6.8

Checkpoint 6.7 6 .7.1

Electron Configuration

219

Atomic Orbitals

How many orbitals are there in the Sf subshell?

6 .7.3

a) 5

In a hydrogen atom, which orbitals are higher in energy than a 3s orbital? (Select all that apply.)

b) 7

a) 3p

c) 14

b) 4s

d) 16

c) 2p

e) 28

d) 3d e) 4p

6 .7 .2

The energy of an orbital in the hydrogen atom depends on

6 .7.4

a) n,

Which of the following sets of quantum numbers, n, f, and me, corresponds to a 3p orbital?

b)

a) 3,0,0

•

e, and me nand e

b) 3,1 , 0

c) n only d)

e only

c) 3, 2, - 1

d) 1, 1, -2

e) me only

e) 1, 3, 1

Electron Configuration The hydrogen atom is a particularly simple system because it contains only one electron. The electron may reside in the Is orbital (the ground state), or it may be found in some higher-energy orbital (an excited state). With many-electron systems, we need to know the ground-state electron configuration that is, how the electrons are distributed in the various atomic orbitals. To do this, we need to know the relative energies of atomic orbitals in a many-electron system, which differ from those in a one-electron system such as hydrogen.

Energies of Atomic Orbitals in Many-Electron Systems Consider the two emission spectra shown in Figure 6.22. The spectrum of helium contains more lines than that of hydrogen. This indicates that there are more possible transitions, corresponding to emission in the visible range, in a helium atom than in a hydrogen atom. This is due to the splitting' ecl'by' heilum;s' Figure 6.23 shows the general order of orbital energies in a many-electron atom. In contrast to the hydrogen atom, in which the energy of an orbital depends only on the value of n (Figure 6.21), the energy of an orbital in a many-electron system depends on both the value of n and the value of e. For example, 3p orbitals all have the same energy, but they are higher in energy than the 3s orbital and lower in energy than the 3d orbitals. In a many-electron atom, for a given value of n, the energy of an orbital increases with increasing value of e. One important consequence of me splitting of energy levels is the relative energies of d orbitals in one shell and the s orbital in the next higher shell. As Figure 6.23 shows, the 4s orbital is lower in energy than the 3d orbitals. Likewise, the 5s orbital is lower in energy than the 4d orbital, and so on. This fact becomes important when we determine how the electrons in an atom populate the atomic orbitals.

o(eiiergy 'ieveis'c'aus

eiec'tros'tatlc'interactIons 'betwe'en

two ·eiectron·s.'· ...... .

Figure 6.22

H ydrogen

I

400nm

500

Comparison of the emission spectra of H and He. I

I

600

700

I

I

600

700

Helium

I

400nm

500

" Splitting" of energy levels refers to the splitting of a shell into subshells of different energies, as shown in Figure 6.23.

220

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

Figure 6.23

Orbital energy levels in many-electron atoms. For a given value of n, orbital energy increases with the value of e.

----------------~ 4d ~ 4d r~~B - ~- ---------------------------------~~~- ------- - --------n

______________

~

3d

~

3d

r~~~ -

-~-----------------------------

------~~~------------------ ~- ---------------------------------~~~------------------~-----------------------------

-~-----------------------------

The Pauli Exclusion Principle According to the Pauli 15 exclusion principle, no two electrons in an atom can have the same four and me values (meaning that quantum numbers. If two electrons in an atom have the same n, they occupy the same orbital), then they must have different values of ms; that is, one must have ms = +~ and the other must have ms = - ~. Because there are only two possible values for ms, and no two electrons in the same orbital may have the same value for ms, a maximum of two electrons may occupy an atomic orbital, and these two electrons must have opposite spins. We can indicate the arrangement of electrons in atomic orbitals with labels that identify each orbital (or subshell) and the number of electrons in it. Thus, we could describe a hydrogen atom in ...... ... . . . . . .. . ... . . . .. .. I the ground state using Is .

e,

.

15' is read as " one 5 one."

~ denotes

lsI denotes the principal / quantum number n

the number of electrons in the orbital or subshell

~ denotes the angular momentum quantum number e

We can also represent the arrangement of electrons in an atom using orbital diagrams, in which each orbital is represented by a labeled box. The orbital diagram for a hydrogen atom in the ground state is

The ground state for a many-electron atom is the one in which all the electrons occupy orbitals of the lowest possible energy.

The upward arrow denotes one of the two possible spins (one of the two possible ms values) of the electron in the hydrogen atom (the other possible spin is indicated with a downward arrow). Under certain circumstances, as we will see shortly, it is useful to indicate the explicit locations of electrons. . . . . .. ... .. .. .. .. .... .. . . . . . . . . . . . .. .. ..... ..... The orbital diagram for a helium atom in the ground state is

15. Wolfgang Pauli (1900-195 8). Austrian physicist. One of the founders of quantum mechanics, Pauli was awarded the Nobel Prize in Physics in 1945.

SECTION 6.8

The label Ii ·i~di~~t~~· th~;~ .~~ .t;;~. ~l~~tr~~~ ·i~· the i; ·~~bit~l·. · N~te· ~so that·th~ ~~~~·i ·i~ · the ·bo~ · . . . point in opposite directions, representing opposite electron spins. Generally when an orbital diagram includes an orbital with a single electron, we represent it with an upward arrow although we could represent it equally well with a downward arrow. The choice is arbitrary and has no effect on the energy of the electron.

The Aufbau Principle We can continue the process of writing electron configurations for elements based on the order of orbital energies and the Pauli exclusion principle. This process is based on the Aujbau principle, which makes it possible to "build" the periodic table of the elements and determine their electron configurations by steps. Each step involves adding one proton to the nucleus and one electron to the appropriate atomic orbital. Through this process we gain a detailed knowledge of the electron configurations of the elements. As we will see in later chapters, knowledge of electron configurations helps us understand and predict the properties of the elements. It also explains why the elements fit into the periodic table the way they do. After helium, the next element in the periodic table is lithium, which has three electrons. Because of the restrictions imposed by the Pauli exclusion principle, an orbital can accommodate no more than two electrons. Thus, the third electron cannot reside in the Is orbital. Instead, it must reside in the next available orbital with the lowest possible energy. According to Figure 6.23, this is me 2s orbital. Therefore, the electron configuration of lithium is l i2s1, and the orbital diagram is

irnilarly, we can write the electron configuration of beryllium as li2i and represent it with the orbital diagram

ith both the Is and the 2s orbitals filled to capacity, the next electron, which is needed for the electron configuration of boron, must reside in the 2p subshell. Because all three 2p orbitals are of ~ ual energy, or degenerate, the electron can occupy anyone of them. By convention, we usually ,how the first electron to occupy the p subshell in the first empty box in the orbital diagram.

11 1

Hund's Rule • Till the sixth electron, which is needed to represent the electron configuration of carbon, reside in iIe 2p orbital that is already half occupied, or will it reside in one of the other, empty 2p orbitals? .-\ccording to Hund'sl 6 rule, the most stable arrangement of electrons in orbitals of equal energy l.S me one in which the number of electrons with the same spin is maximized. As we have seen, no 0':0 electrons in any orbital may have the same spin, so maximizing the number of electrons with iIe same spin requires putting the electrons in separate orbitals. Accordingly, in any subshell, an ::lectron will occupy an empty orbital rather than one that already contains an electron. The electron configuration of carbon is, therefore, l i2i2p2, and its orbital diagram is

' . Frederick Hund (1896-1997). German physici st. Hund' s work was mainly in quantum mechanics. He also helped to .:e--elop the mo lecular orbital theory of chemical bonding.

Electron Configuration 1S2 is read as "one s two, " not as "one s squared. "

221

222

CHAPTER 6

Quantum Theory and the Electronic Structure of Atoms

Similarly, the electron configuration of nitrogen is l i2i2p 3, and its orbital diagram is Is

111111 1 2

2p

2s

p3

3s

3p

3d

4s

4p

4d

5p

5s

5d

Once all the 2p orbitals are singly occupied, additional electrons will have to pair with those already in the orbitals. Thus, the electron configurations and orbital diagrams for 0, F, and Ne are 4f

0

6p

7s

7p

0]

0]

ls2

2s2

0]

0]

ls 2

2s2

IHI12p411 I

5f

F 6s

ls22s22p4

ls 22s 22 p 5

6d

IH11~11 2p 5

I

IHIHIHI 2

Ne

p6

Figure 6.24

A simple way to remember the order in which orbitals fill with electrons.

General Rules for Writing Electron Configurations Based on the preceding examples we can formulate the following general rules for determining the electron configuration of an element in the ground state:

Remember that in this context, degenerate means "of equal energy. " Orbitals in the same subshell are degenerate.

••

•

.".,.,.~==~.

1. Electrons will reside in the available orbitals of the lowest possible energy. 2. Each orbital can accommodate a maximum of two electrons. 'wiii degenerate orbitals if an empty orbital is available. 4. Orbitals will fill in the order indicated in Figure 6.23. Figure 6.24 provides a simple way for you to remember the proper order.

"...... "':'( .Eiectrons

'not 'paIr 'i"n'

Sample Problem 6.9 illustrates the procedure for determining the ground· state electron con· figuration of an atom . Multimedia

Atomic Structure

electron configurations.

Sample Problem 6.9 Write the electron configuration and give the orbital diagram of a calcium (Ca) atom (Z = 20) . Strategy Use the general rules given and the Aufbau principle to "build" the electron configuration

of a calcium atom and represent it with an orbital diagram. Setup Because Z = 20, we know that a Ca atom has 20 electrons. They will fill orbitals in the order

designated in Figure 6.23, obeying the Pauli exclusion principle and Hund's rule. Orbitals will fill in the following order: Is, 2s, 2p, 3s, 3p, 4s. Each s subshell can contain a maximum of two electrons, whereas each p subshell can contain a maximum of six electrons. Think About It Look at Figure

6.23 again to make sure you have filled the orbitals in the right order and that the sum of electrons is 20. Remember that the 4s orbital fills before the 3d orbitals.

Solution

0] 0] ls2

2s2

IHIHIHI 2p 6

0]2 3s

IHIHIHI 3p 6

0]2 4s

Practice Problem A Write the electron configuration and give the orbital diagram of a rubidium

(Rb) atom (Z = 37). Practice Problem B Write the electron configuration and give the orbital diagram of a bromine (Br) atom (Z = 35).

SECTION 6.9

Checkpoint 6.8 6.8.1

223

Electron Configuration

Which of the following electron configurations correctly represents the Ti atom?

6.8.3

Which orbital diagram is correct for the ground state S atom?

a) Is2 2S2 2p 6 3i 3l3ct 2 2 b) I s2 2i 2l3s 3l4i 3d c) ls2 2i 2p 6 3i 3l4s2 3dIO

a)

d) Ii 2i 2p 6 3s2 3p6 3dIO e) I i 2s2 2p 6 3i 3p6 4s

6.8.2

Electron Configurations and the Periodic Table

b)

4

What element is represented by the following electron configuration? I i 2S2 2p 6 3s2 3p 6 4s2 3d 10 4p4

c)

a) Br b) As

d)

c) S d) Se e) Te

e)

IHIHIHI 2

[IT]

p6

3s 2

IHIHIHI 2

[IT]

[IT]

[IT]

ls2

2s2

[IT]

[IT]

ls2

2s2

[IT]

[IT]

l s2

2s2

[IT]

[IT]

ls2

2s2

IH12p4 111 I

[IT]

[IT]

1 1~IHIH I

ls2

2s2

2p 6

p6

3s 2

IHIHI 3p4 IH113p411 I

IHIHI 2p4

Electron Configurations and the Periodic Table The electron configurations of all elements except hydrogen and helium can be represented using a noble gas core, which shows in brackets the electron configuration of the noble gas element that most recently precedes the element in question, followed by the electron configuration in the outermost occupied subshells. Figure 6.25 gives the ground-state electron configurations of elements from H (Z = 1) through Rg (Z = 111). Notice the similar pattern of electron configurations in the elements lithium (Z = 3) through neon (Z = 10) and those of sodium (Z = 11) through argon 1 (Z = 18). Both Li and Na, for example, have the configuration ns in their outermost occupied subshells. For Li, n = 2; for Na, n = 3. Both F and CI have electron configuration nin/, where n = 2 for F and n = 3 for CI, and so on. As mentioned in Section 6.8, the 4s subshell is filled before the 3d subshell in a manyelectron atom (see Figure 6.23). Thus, the electron configuration of potassium (Z = 19) is l i2i2p 63s 23p 64s1. Because 1i2i2p 63s 23p 6 is the electron configuration of argon, we can simplify the electron configuration of potassium by writing [Ar]4s1, where [Ar] den otes the "argon core."

..

~

[Ar]

•

[Ar] 4s

1

The placement of the outermost electron in the 4s orbital (rather than in the 3d orbital) of potassium is strongly supported by experimental evidence. The physical and chemical properties of potassium are very similar to those of lithium and sodium, the first two alkali metals. In both lithium and sodium, the outermost electron is in an s orbital (there is no doubt that their outermost electrons occupy s orbitals because there is no 1d or 2d subshell). Based on its similarities to the other alkali metals, we expect potassium to have an analogous electron onfiguration; that is, we expect the last electron in potassium to occupy the 4s rather than the 3d orbital. The elements from Group 3B through Group 1B are transition metals [ ~~ Section 2.4] . Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells. In the first transition metal series, from scandium (Z = 21) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ......... ...... ............................. ... .. . through copper (Z = 29), additional electrons are placed in the 3d orbitals according to Hund's rule. However, there are two anomalies. The electron configuration of chromium (Z = 24) is [Ar]4S13d5 and not [Ar]4i3d 4 , as we might expect. A similar break in the pattern is observed for

Although zinc and the other elements in Group 2B sometimes are included under the heading "transition metals," they neither have nor readily acquire partially filled d subshells. Strictly speaking, they are not transition metals.

224

CHAPTER 6

Quantum Theory and the Electron ic Structure of Atoms

IA

8A 18 2

I

1

1 Core

2 [He]

3 [Ne]

4 [Ar]

5 [Kr]

6 [Xe]

7 [Rn]

H lsi

He

3

2A 2 4

2s1

2s2

11

12

o Metals o Metalloids o Nonmetals

Li Be Na1 Mg2 19

20

3B 3 21

4s

4s

3d 4s

3s

3s

4B 4 22

5B 5 23

3d 4s

3d 4s

3A 13 5

4A 14 6

SA 15 7

2s22pl

2s22p2

2s 2p3

13

14

15

6A 16 8

Is2

7A

I

17

10

9

Ne F B C N 0 2 2 2 Zs22 6 2s22p4

6B 6 24

7B 7 25

1

IB

8 26

8B 9 27

10 28

3d 4s

3d 4s2

3d 4s

1

II

29

2B 12 30

Si P Al 2 2 2

3s 3pl

31

2s Zp5

16

p

18

17

Ar S2 CI 2 2

3s 3p 2 3s 3p3 3s 3p4

3s 3p5 3s 3p6

32

33

34

35

36

3d 104s 4p 2

3d 104s 4p 3

3d 4s 4p4

3d 4s 4p 5

3d 4s 4 p6

50

51

52

53

54

4d Ss Spl

Sp2

4d Ss Sp3

4d Ss Sp4

81

82

83

84

3

Cu Ga Ge As Se Br Kr V Cr Fe Co Ni Zn K1 Ca2 Sc Ti Mn 4 2 2 2 lO 2 lO 2 lO 2 l 2 2 2 5 2 6 2 7 8 2 3 2 3d54s I 3d 4s

37

38

39

40

41

42

43

44

45

Ssl

Ss2

4d Ss

4d Ss

4d Ss

4d Ss

4d Ss

4d 7Ss

4d ss 1

46

3d 104s I 3d 104s 2 3d 104s

47

48

4dlOSs

4dlOSs

4p l 49

Pd Cd Sb Te Xe Rb Sr Yl 2 Zr Nb Mo Tc Ru Ag In Sn I Rh 5 i lO i 2 lO 2 4d IOSs2 lO 2 lO 2 l l 2 5 i 3 2 2 2 8 5 2 55

56

6s

6s

87

88

71

72

.

73

74

75

76

4l sd 6s 2

4fl4Sd 6s 2

4fl4Sd 6s 2

4l4Sd 6s 2

105

106

107

108

109

Sl46d 7

77

4d

78

79

80

4d OSs 2 4d OSs Sp5

Sp6

85

86

W Re Pt Au Pb Bi Po At Rn Cs1 Ba2 Lui Hf2 Ta Os Ir Hg TI 6 4 4 3 4 9 iO iO i0 i0 5 6 4l4Sd 4fl4Sd 6s 2 6s 2

103

104

4l4Sd 7 4l sd 4l4SdIO 4l4Sd IO 4fl4Sd 4l4Sd 4l4SdIO 4fl4Sd IO 4fl4Sd 4fl4Sd 6s 26pl 6s 26p2 6s 26p3 6s 26p4 6s 26p5 6s 26p6 6s 1 6s 2 6s 2 6s 1

110

III

Fr1 Ra2 Lr l Rf2 Db3 Sg 4 Bh5 Hs 6 Mt Ds 8 Rg9 7s

7.1

Sfl46d 5fl46d 7s 2 7s 2

Sl46d 7s 2

Sfl46d 7s 2

Sfl46d 7s 2

Sfl46d 7s 2

57

58

59

60

Sd 16,,2

4lsd 6s 2

4f 6s

89

90

91

Lan thani des 6 [Xe]

Actinides 7 [Rn]

Figure 6.25

112

11 3

114

115

116

(117)

Sfl46d Sl46d 10 Sl46d l0 Sl46d l0 Sl46d 10 Sl46d 10 7s 27p l 7s 27p2 7s 27p3 7s 27p4 7s 2 7s 2

7s 2

Sl46d 7s 2

61

62

63

64

65

66

4f 46s

4f 56s

4f 6s

4f 76s

4lsd 6s 2

4f 96s

4l06s

92

93

94

95

96

97

98

Sl7s

Sf 76d

67

68

69

U8 Sl46d 10 7s 27p6

7

70

La Cel Pr Nd Eu Gd Pm Sm Tb Dy Ho Er Tm Yb 6 3 2 2 2 2 l 2 l1 2 2 3 6 2 4f 6s 2 4l 26s 2 4l 6s 2 4fl46s 100

99

101

102

Ac Pa Th Pu Cm U Np Am Bk Cf Es Fm Md No l 2 2 l 4 l 2 3 l 2 l 2 l1 2 13 2 6 2

6d 7s

2

6d 7s

st6d 7s 2

Sf 6d 7s 2

Sf 6d 7s 2

Sf 7s

7s 2

Sl7s

Sf l07s

Sf 7s

2

Sfl27s

Sf 7s

Sl47s2

7

Ground-state electron configurations for the known elements.

Electron configurations such as these may also be written with the d subshell first. For example, [Arl4s'3d' o can also be written as [ArI3d104s'. Either way is acceptable.

'cc)Ji'flguratlon'

... copper: whose 'eiectron '{ i '[Ar]4S13dlO rather than [Ar]4s23Jl. The reason for these anomalies is that a slightly greater stability is associated with the half-filled (3d5) and completely filled (3dlO) subshells.

Cr

[Ar] [ ]

4s

11 11 11 11 11 1 3d 3d 3d 3d 3d

Cu

[Ar] [ ] 4s

IHIHIHIHIHI 3d 3d 3d 3d 3d

For elements Zn (Z = 30) through Kr (Z = 36), the 3d, 4s, and 4p subshells fill in a straightforward manner. With rubidium (Z = 37), electrons begin to enter the n = 5 energy level. Some of the electron configurations in the second transition metal series [yttrium (Z = 39) through silver (Z = 47)] are also irregular, but the details of many of these irregularities are beyond the scope of this text and we will not be concerned with them. The sixth period of the periodic table begins with cesium (Z = 55) and barium (Z = 56), whose electron configurations are [Xe]6s 1 and [Xe ]6i, respectively. Following barium, there is a gap in the periodic table where the lanthanide (rare earth) series belongs. The lanthanides are a

SECTION 6.9

Electron Configurations and the Periodic Table

225

series of 14 elements that have incompletely filled 4f subshells or that readily give rise to cations that have incompletely filled 4fsubshells. The lanthanides (and the actinides, to be discussed next) are shown at the bottom of the periodic table to keep the table from being too wide. lA 1

2A 2

1

3A 4A 5A 6A 7A 13 14 15 16 17

2

8A 18

2

r--;-

3B 4B 5B 6B 7B 8B ~ IB 2B 3 4 5 6 7 8 9 10 11 12

3

1

3

4

4 5

5 ,

6

6

7

7

Lanthanides 6 La Ce Pr Nd Pm Sm Eu Od Th Dy Ho Er Tm Yb 6 Actinides 7 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No 7

lA 1

1

2A 2

3A 4A SA 6A 7A 13 14 15 16 17

8A 18 1 2

?

3B 4B 5B 6B 7BiB8B~IB 2B 3 4 5 6 7 8 9 10 11 12

~

,)

3

4

4

5

5

6

La Ce Pr Nd Pm Sm Eu Od Th Dy Ho Er Tm Yb

6

7

Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No

7 . . . . .. . . .

... ..

When n = 4, ecan equal 3, corresponding to In theory, the lanthanides arise from the filling of the seven degenerate 4f orbitals. In reality, an f subshell. There are seven possible values for 30wever, the energies of the 5d and 4f orbitals are very close and the electron configurations of me when e = 3: - 3, -2, -1 ,0, +1, +2, and +3, :hese elements sometimes involve 5d electrons. For example, in lanthanum itself (Z = 57) the 4f Therefore, there are seven f orbitals, orbital is slightly higher in energy than the 5d orbital. Thus, lanthanum' s electron configuration is ;:XeJ6s25d 1 rather than [Xe J6s 24l. After the 4f subs hell is completely filled, the next electron enters the 5d subshell of lutetium Z = 71). This series of elements, including lutetium and hafnium (Z = 72) and extending through ;nercury (Z = 80), is characterized by the filling of the 5d subshell. The 6p subshells are filled jext, which takes us to radon (Z = 86). 1 The last row of elements begins with francium (Z = 87; electron configuration [RnJ7s ) :::nd radium (Z = 88; electron configuration [RnJ7 i), and then continues with the actinide series, . , , . .. .. ... ...... . . .. ... . ... . Most of these elements are not found in nature ... ·hich starts at actinium (Z = 89) and ends with nobelium (Z = 102). The actinide series has but have been synthesized in nuclear reactions, ;>artially filled 5fand/or 6d subshells. The elements lawrencium (Z = 103) through darmstadtium which are the subject of Chapter 20, Z = 110) have a filled Sf subshell and are characterized by the filling of the 6d subshell. With few exceptions, you should be able to write the electron configuration of any ele=ent, using Figure 6.23 (or Figure 6.24) as a guide. Elements that require particular care are the =

energies of Na and Mg are 496 and 738 kJ/mol, respectively. The second ionization energies of Na and Mg are 4562 and 1451 kJ/ mol , respectively.

lEI (Na) because Mg is to the right of Na in the periodic table (i.e., Mg has the greater effective nuclear charge, so it is more difficult to remove its electron). lEz(Na) > IE2 (Mg)

because the second ionization of Mg removes a valence electron, whereas the second ionization of Na removes a core electron.

Practice Problem A Which element, Mg or AI, will have the higher first ionization energy and

which will have the higher third ionization energy? Practice Problem B Explain why Rb has a lower fEI than Sr, but Sr has a lower fEz than Rb.

..

--------------------------------------------------------------~

250

CHAPTER 7

Electron Configurat ion and the Period ic Table

Electron Affinity Electron affinity (EA) is the energy released (the negative of the enthalpy change fJ.H) when an atom in the gas phase accepts an electron. Consider the process in which a gaseous chlorine atom accepts an electron:

Cl(g)

Although IE and EA both increase from left to right across a period, an increase in IE means that it is less likely that an electron will be removed from an atom. An increase in fA, on the other hand, means that it is more likely that an electron will be accepted by an atom.

+ e- - _ . Cl - (g)

fJ.H

= -349.0 kJ/mol

A negative value of fJ.H indicates an exothermic process [ ~~ Section S.3], so 349.0 kJ/mol of energy are released (the definition of electron affinity) when a mole of gaseous chlorine atoms accepts a mole of electrons. A positive electron affinity indicates a process that is energetically favorable. In general, the larger and more positive the EA value, the more favorable the process and the more apt it is to occur. Figure 7.10 shows electron affinities for the main group elements. . . . . . . . .. ....... ........................ .. ......................................... .. Like ionization energy, electron affinity increases from left to right across a period. This trend in EA is due to the increase in effective nuclear charge from left to right (i.e., it becomes progressively easier to add a negatively charged electron as the positive charge of the element's nucleus increases). There are also periodic interruptions of the upward trend of EA from left to right, similar to those observed for lEb although they do not occur for the same elements. For example, the EA of a Group 2A element is lower than that for the corresponding Group lA element, and the EA of a Group SA element is lower than that for the corresponding Group 4A element. These exceptions to the trend are due to the electron configurations of the elements involved. lA 1

Figure 7.10 (a) Electron affinities (in kllmol) of the main group elements. (b) Electron affinity as a function of atomic number.

H + 72.8

8A 18 2A 2

3A 13

4A 14

5A 15

6A 16

N

0 +141

7A 17

He (0.0)

Li +59.6

Be EA] (P) because although P is to the right of Si in the third period of the periodic table (giving P the larger Zeff), adding an electron to a P atom requires placing it in a 3p orbital that is partially occupied. The energy cost of pairing electrons outweighs the energy advantage of adding an electron to an atom with a larger effective nuclear charge.

Practice Problem A Would you expect Mg or Al to have the higher EAl? Practice Problem B Why is the EA] for Ge greater than the EA] for As ?

~'--------------------------------------------------------------------~"

Think About It The first electron

affinities of AI, Si, and Pare 42.5, 134, and 72.0 kJ/mol, respectively.

252

CHAPTER 7

Electron Configuration and t he Periodic Tab le

Metallic Character .. Malleability is the property that allows metals to be pounded into thin sheets. Ductility, the capacity to be drawn out into wi res, is another characteristic of metals.

Metals tend to . . .. .. . . ... . . . ..• Be shiny, lustrous, and malleable • Be good conductors of both heat and electricity • Have low ionization energies (so they commonly form cations) • Form ionic compounds with chlorine (metal chlorides) • Form basic, ionic compounds with oxygen (metal oxides) Nonmetals, on the other hand, tend to • • • • •

/

Vary in color and lack the shiny appearance associated with metals Be brittle, rather than malleable Be poor conductors of both heat and electricity Form acidic, molecular compounds with oxygen Have high electron affinities (so they commonly form anions)

Metallic character increases from top to bottom in a group and decreases from left to right within a period. Metalloids are elements with properties intermediate between those of metals and nonmetals. Because the definition of metallic character depends on a combination of properties, there may be some variation in the elements identified as metalloids in different sources. Astatine (At), for example, is listed as a metalloid in some sources and a nonmetal in others.

Checkpoint 7.4 7.4.1

Periodic Trends in Properties of Atoms

An'ange the elements Ca, Sr, and Ba in order of increasing IE ].

7.4.3

a) Ca < Sr < Ba

For each of the following pairs of elements, indicate which will have the greater EA ]: Rb or Sr, C or N , 0 or F:

< Sr < Ca

a) Rb, C, 0

c) Ba < Ca < Sr

b) Sr, N , F

d) Sr < Ba < Ca

c) Sr, C, F

e) Sr < Ca < Ba

d) Sr, N , 0

b) Ba

e) Rb, C, F

7.4.2

Arrange the elements Li, Be, and B in order of increasing JE 2.

7.4.4

a) Li < Be < B b) Li

< B < Be

Which element, K or Ca, will have the greater IEb which will have the greater lE2 , and which will have the greater EA ]?

c) Be < B < Li

a) Ca, K, K

d) Be < Li < B

b) K,K, Ca

e) B < Be < Li

c). K, Ca, K d) Ca, Ca, K e) Ca, Ca, Ca

Electron Configuration of Ions Because many ionic compounds are made up of monatomic anions and cations, it is helpful to know how to write the electron configurations of these ionic species. Just as for atoms, we use the Pauli exclusion principle and Hund 's rule to write the ground-state electron configurations of cations and anions. Recall from Chapter 2 that we can use the periodic table to predict the charges on many of the ions formed by main group elements. Elements in Groups lA and 2A, for example, form ions with charges of + 1 and + 2, respectively. Elements in Groups 6A and 7 A form ions with charges of - 2 and -1 , respectively. Knowing something about electron configurations enables us to explain these charges.

SECTION 7.5

Electron Configuration of Ions

253

Ions of Main Group Elements In Section 7.4, we learned about the tendencies of atoms to lose or gain electrons. In every period of the periodic table, the element with the highest lE I is the Group 8A element, the noble gas. [See Figure 7.8(b).] Also, Group 8A is the only group in which none of the members has any tendency to accept an electron; that is, they all have negative EA values. [See Figure 7.1O(b).] High ionization energies and low electron affinities make the noble gases almost completely unreactive. Ultimately, the cause of this lack of reactivity is electron configuration. The I s2 configuration of He 2 and the ns np 6 (n :> 2) valence electron configurations of the other noble gases are extraordinarily stable. Other main group elements tend to either lose or gain the number of electrons needed to achieve the same number of electrons .as the nearest noble gas. Species with identical electron . . . . . . . .. ..... .................... . ..... ... . . , . . . . . . . .. . .. . . . . . . configurations are called isoelectronic. To write the electron configuration of an ion formed by a main group element, we first write the configuration for the atom and either add or remove the appropriate number of electrons. Electron configurations for the sodium and chloride ions are Na: 1s22s22p 6 3s I 2

6

CI: ls22s 2p 3i3 p 5

•

•

Na+: l s22s2 2p 6

It is a common error to mistake species with the same valence electron configuration for isoelectronic species. For example. F- and Ne are isoelectronic. F- and (1- are not.

(10 electrons total, isoelectronic with Ne)

CI - : l i2i2p 63i3 p 6

(18 electrons total, isoeJectronic with Ar)

We can also write electron configurations for ions using the noble gas core. Na: [Ne]3s 1 --+. Na+: [Ne]

Sample Problem 7.6 gives you some practice writing electron configurations for the ions of main group elements.

Write electron configurations for the following ions of main group elements: (a) N3 - , (b) Ba2+ , and (c) Be2+ Strategy First write electron configurations for the atoms. Then add electron s (for anions) or remove

electrons (for cations) to account for the charge. Setup (a) N3 - forms when N (ls22s22p 3 or [He]2s2 2p\

a main group nonmetal, gains three

electrons. (b) Ba2+ forms when Ba (ls22i2p63s23p64i3d 104p6SS24d 1OSp66s2 or [Xe]6s 2) loses two electrons.

(c) Be2+ forms when Be (1s22s2 or [He]2i) loses two electrons. Solution (a) [He]2i2p 6 or [Ne]

(b) [Kr]Si4d IO Sp6 or [Xe] (c)

Think About It Be sure to add

Ii or [He]

electrons to form an anion, and remove electrons to form a cation.

Practice Problem A Write electron configurations for (a) 0 2 - , (b) Ca2+ , and (c) Se

2

- .

Practice Problem B List all the species (atoms and/or ions) that are likely to have the following

electron configuration: l i 2s22p6

..

------------------------------------------------------~

Ions of d-Block Elements i{ecall from Section 6.8 that the 4s orbital fills before the 3d orbitals for the elements in the first ::-ow of the d-block (Sc to Zn) [ ~~ Section 6 .8] . Following the pattern for writing electron con-gurations for main group ions, then , we might expect the two electrons lost in the formation of :he Fe 2 + ion to come from the 3d subshell. It turns out, though, that an atom always loses electrons ••••••• - t from the shell with the highest value of n. In the case of Fe, that would be the 4s subshell. Fe: [Ar]4i3d 6

•

Fe2+ : [Ar]3d 6

::-on can also form the Fe 3 + ion, in which case the third electron is removed from the 3d subshell.

• •

•

This explains. in part. why many of the transition metals can form ions with a + 2 charge.

254

CHAPTER 7

Electron Configuration and the Periodic Table

In general, when a d-block element becomes an ion, it loses electrons first from the ns subshell and then from the (n - l)d subshell. Sample Problem 7.7 gives you some practice writing electron configurations for the ions of d-block elements.

.

Sample Problem 7.7 ...•.. Write electron configurations for the following ions of d-block elements: (a) Zn2+, (b) Mn2+, and (c) Cr3+ ••••

Remember that the eledron configuration of Cr is anomalous in that it has only one 4s eledron, making its d subs hell half filled [ ~~ Section 6 .9] .

·

• •

Strategy First write electron configurations for the atoms. Then add electrons (for anions) or remove

•

electrons (for cations) to account for the charge. The electrons removed from a d-block element must come first from the outermost s subshell, not the partially filled d subshell.

• • •

·• • • • • • • •

Think About It Be sure to add

electrons to form an anion and remove electrons to form a cation. Also, double-check to make sure that electrons removed from a d-block element come first from the ns subshell and then, if necessary, from the (n - l)d subshell.

• • ~

Setup (a) Zn 2+ forms when Zn ( l s22s22l3i3p64i3dIO or [ArJ4s23io) loses two electrons.

(b) Mn2+ forms when Mn (li2s 22p63i3p64i3d s or [ArJ4s23ds) loses two electrons. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. (c) Cr3+ forms when Cr (li2s22p 63s23l4s13d s or [ArJ4SI3d 5) loses three electrons- one from the 4s subshell and two from the 3d subs hell. Solution (a) [ArJ3dlO

(b) [ArJ3d s (c) [ArJ3d 3

Practice Problem A Write electron configurations for (a) C0 3 +, (b) Cu 2 +, and (c) Ag +. Practice Problem B What common d-block ion (see Figure 2.14) is isoelectronic with Zn2+ ?

Checkpoint 7.5 7.5.1

Electron Configuration of Ions

Which of the following ions are isoelectronic with a noble gas? (Select all that apply.)

7.5.3

Select the correct ground-state electron configuration for Ti2+ . a) [ArJ4S23d2

a) Mn2+ b) Ca2+

b) [ArJ4s23ct c) [ArJ4s2

c) Br-

d) [ArJ3d2

d) 0 2+

e) [ArJ4s13d l

e) F -

7.5.4 7.5.2

Which of the following pairs are isoelectronic with each other? (Select all that apply.) a) Ca2+ and Sr2+ b) 0 2- and Mg2+ c)

r- and Kr

d) S2- and Cl-

Select the correct ground-state electron configuration for S2- . a) [NeJ3l b) [NeJ3i3 p 6 c) [NeJ3i3p2 d) [NeJ3p 6 e) [NeJ

e) He and H+

Ionic Radius When an atom gains or loses one or more electrons to become an ion, its radius changes. The ionic radius, the radius of a cation or an anion, affects the physical and chemical properties of an ionic compound. The three-dimensional structure of an ionic compound, for example, depends on the relative sizes of its cations and anions .

SECTION 7.6 •

Comparing Ionic Radius with Atomic Radius When an atom loses an electron and becomes a cation, its radius decreases due in part to a reduction in electron-electron repulsions (and consequently a reduction in shielding) in the valence shell. A significant decrease in radius occurs when all of an atom's valence electrons are removed. This is the case with ions of most main group elements, which are isoelectronic with the noble gases preceding them. Consider Na, which loses its 3s electron to become Na +:

••

Ionic Radius

255

•

0,""",,""' _"'" M uIt imed ia Periodic Table

atomic and ionic radii.

Valence orbitals in Na and in Na+

The valence electron of Na has a principal quantum number of n = 3. When it has been removed, the resulting Na + ion no longer has any electrons in the n = 3 shell. The outermost electrons of the Na + ion have a principal quantum number of n = 2. Because the value of n determines the distance from the nucleus, this corresponds to a smaller radius. When an atom gains one or more electrons and becomes an anion, its radius increases due to increased electron-electron repulsions. Adding an electron causes the rest of the electrons in the valence shell to spread out and take up more space in order to maximize the distance between them. Figure 7.12 shows the ionic radii for those ion s of main group elements that are isoelectronic with noble gases and compares them to the radii of the parent atoms. Note that the ionic radius, like the atomic radiu s, increases from top to bottom in a group.

Isoelectronic Series An isoelectronic series is a series of two or more species that have identical electron configurations, but different nuclear charges. For example, 0 2 -, F - , and Ne constitute an isoelectronic

Figure 7.12

Group

lA

2A

3A

4A

Li . 2

5A

6A

7A

N

0

F

1+

. 2-

152176

75/146

Na 1+

3

2+

186/102

160172

K

Ca

"0

1+

.S: 4 1-0 ~

p

Al

Mg

73/140 72/133 S

CI

3+ 143/54

11 0/212 10311 84 99/ 181

Br

+

227/ 13 8

1971100

1141196

Rb

Sr

I

+

1+

5

248/152

215/1 18

Cs

Ba

6

1+ 265/167

2+

2221135

133/220

A comparison of atomic and ionic radii (in picometers) for main group elements and their common ions (those that are isoelectronic with noble gases).

256

CHAPTER 7

Electron Configuration and the Periodic Table

series. Although these three species have identical electron configurations, they have different radii. In an isoelectronic series, the species with the smallest nuclear charge (i.e., the smallest atomic number, Z) will have the largest radius. The species with the largest nuclear charge (i.e., the largest Z) will have the smallest radius.

+8

- 10

- 10

F-

Ne

Sample Problem 7.8 shows how to identify members of an isoelectronic series and how to arrange them according to radius.

Sample Problem.7.8 Identify the isoelectronic series in the following group of species, and arrange them in order of increasing radius: K+, Ne, Ar, Kr, p 3- , S2-, and Cl- . Think About It Consult Figure

Strategy Isoelectronic series are species with identical electron configurations but different

7.12 to check your result. With identical electron configurations, the attractive force between the valence electrons and the nucleus will be strongest for the largest nuclear charge. Thus, the larger the nuclear charge, the closer in the valence electrons will be pulled and the smaller the radius will be.

nuclear charges. Determine the number of electrons in each species. The radii of isoelectronic series members decreases with increasing nuclear charge. Setup The number of electrons in each species is as follows: 18 (K+), 10 (Ne), 18 (Ar), 36 (Kr),

18 (p 3-), 18 (S2-), and 18 (Cl-). The nuclear charges of the species with 18 electrons are + 19 (K+), + 18 (Ar) , + 15 (p 3-), + 16 (S2-), and + 17 (cq. Solution The isoelectronic series includes K+, Ar, p 3-, SZ- , and CI - . In order of increasing radius:

K+ < Ar < Cl- < Sz- < p 3- .

Practice Problem A Arrange the followin g isoelectron ic series in order of increasing radius: Se 2 -,

Br- , Kr, and Rb +. Practice Problem B List all the common ions that are isoelectronic with Ne.

Checkpoint 7.6 7.6.1

Ionic Radius

Which of the following species are isoelectronic with Kr? (Select all that apply. )

7.6.2

Which of the following are arranged correctly in order of increasing radius ? (Select all that apply. )

r-

a) He

a) F- < C]- < Br-


b:' ':0' ••

or

••

:o=c=o: ,

'

,

C has 8 e-

H

Each H has 2 e - ~ ~

~ ':C:H

or

\

C=C / \

H

Each C has 8 e-

/

H H

A triple bond arises when two atoms share three pairs of electrons, as in molecules such as nitrogen (N2) and acetylene (C 2H 2) : Each N has 8 e'N"'N'

or

e--@c(pH

or

e

Each H has 2

•••

•

H-C

C-H

Each C has 8 eIn ethylene and acetylene, all the valence electrons are used in bonding; there are no lone pairs on the carbon atoms. In fact, with the important exception of carbon monoxide (CO), most stable molecules containing carbon do not have lone pairs on the carbon atoms, Multiple bonds are shorter than single bonds. Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule (Figure 8.4), Table 8.3 li sts some experimentally determined bond lengths. For a given pair of atoms , such as carbon and nitrogen, triple bonds are shorter than double bonds, and double bonds are shorter than single bonds, The shorter multiple bonds are also stronger than single bonds, as we will see in Section 8.9.

Bond Type

Bond Length (pm)

C-H

107

O-H

96

C-O

143

C=O

121

C=O

113

C-C

154

C=C

133

C-C

120

C-N

143

C=N

138

C-N

116

N-N

147

N=N

124

N-N

110

N-O

136

N=O

122

0-0

148

0=0

121

A

286

CHAPTER 8

Chemical Bonding I: Basic Concepts

(CCI4 Property

Na(1

((1 4

Appearance

White solid

Colorless liquid

Melting point (DC)

801

- 23

Molar heat of fusion "' (kJ/mol)

30.2

2.5

Boiling point (DC)

1413

76.5

Molar heat of vaporization* (kl l mol)

600

30

Density (g/cm 3 )

2.17

1.59

Solubility in water

High

Very low

Solid

Poor

Poor

Liquid

Good

Poor

Aqueous

Good

Poor

Electrical conductivity

*The molar heat of fusion and molar heat of vaporization are the amounts of heat needed to melt 1 mole of the solid and to vaporize 1 mole of the liquid, respectively.

Comparison of Ionic and Covalent Compounds Ionic and covalent compounds differ markedly in their general physical properties because of differences in the nature of their bonds. There are two types of attractive forces in covalent compounds, the intramolecular bonding force that holds the atoms together in a molecule, and the intermolecular forces between molecules. Bond enthalpy, discussed in Section 8.9, can be used to quantify the intramolecular bonding force. Intermolecular forc~s are usually quite weak compared to the forces holding atoms together within a molecule, so molecules of a covalent compound are not held together tightly. As a result, covalent compounds are usually gases, liquids, or lowmelting solids. On the other hand, the electrostatic forces holding ions together in an ionic compound are usually very strong, so ionic compounds are solids at room temperature and have high melting points. Many ionic compounds are soluble in water, and the resulting aqueous solutions conduct electricity because the compounds are strong electrolytes. Most covalent compounds are insoluble in water, or if they do dissolve, their aqueous solutions generally do not conduct electricity, because the compounds are nonelectrolytes. Molten ionic compounds conduct electricity because they contain mobile cations and anions; liquid or molten covalent compounds do not conduct electricity because no ions are present. Table 8.4 compares some of the properties of a typical ionic compound, sodium chloride (NaCl), with those of a covalent compound, carbon tetrachloride (CCI4) .

Electronegativity and Polarity So far, we have described chemical bonds as either ionic, when they occur between a metal and a nonmetal, or covalent, when they occur between nonmetals. In fact, ionic and covalent bonds are simply the extremes in a spectrum of bonding. Bonds that fall between these two extremes are polar, meaning that electrons are shared but are not shared equally. Such bonds are referred to as polar covalent bonds. The following shows a comparison of the different types of bonds, where M and X represent two different elements:

[

I

M:X "

Pure covalent bond Neutral atoms held together by equally shared electrons

Polar covalent bond Partially charged atoms held together by unequally shared electrons

Ionic bond Oppositely charged ions held together by electrostatic attraction

SECTION 8.4

El ectronegativity and Po larity

287

Figure 8.5

HF

Electron density maps show the distribution of charge in a covalent species (H2)' a polar covalent species (HF), and an ionic species (NaP) . The most electron-rich regions are red; the most electron-poor regions are blue.

NaP

To illustrate the spectrum of bonding, let's consider three substances: H 2, HF, and NaP. In the H2 molecule, where the two bonding atoms are identical, the electrons are shared equally. That is, the electrons in the covalent bond spend roughly the same amount of time in the vicinity of each' H atom. In the HF molecule, on the other hand, where the two bonding atoms are different, the electrons are not shared equally. They spend more time in the vicinity of the F atom than in the In NaP, the vicinity of the H atom. (The 0 symbol is used to denote partial charges on the atoms.) . . . . .. . ............ ..... . electrons are not shared at all but rather are transferred from sodium to fluorine. One way to visualize the distribution of electrons in species such as H 2 , HF, and NaF is to use electrostatic potential models (Figure 8.5). These models show regions where electrons spend a lot of time in red, and regions where electrons spend very little time in blue. (Regions where electrons spend a moderate amount of time appear green.)

In fact, the transfer of electrons in NaF is nearly complete. Even in an ionic bond, the electrons in question spend a small amount of time near the cation.

Electronegativity Electronegativity is the ability of an atom in a compound to draw electrons to itself. It determines where electrons in a compound spend most of their time. Elements with high electronegativity have a greater tendency to attract electrons than do elements with low electronegativity. Electronegativity is related to electron affinity and ionization energy. An atom such as fluorine, which has a high electron affinity (tends to accept electrons) and a high ionization energy (does not lose electrons easily), has a high electronegativity. Sodium, on the other hand, has a low electron affinity, a low ionization energy, and therefore a low electronegativity. Electronegativity is a relative concept, meaning that an element's electronegativity can be 2 measured only in relation to the electronegativity of other elements. Linus Pauling devised a method for calculating the relative electronegativities of most elements. These values are shown in Figure 8.6. In general, electronegativity increases from left to right across a period in the periodic table, as the metallic character of the elements decreases. Within each group, electronegativity decreases with increasing atomic number and increasing metallic character. The transition metals do not follow these trends. The most electronegative elements (the halogens, oxygen, nitrogen, and sulfur) are found in the upper right-hand corner of the periodic table, and the least electronegative

8A 18

1 H 2.1

2A 2

3A 13

4A 14

5A 15

6A 16

7A 17

Li 1.0

Be 1.5

B 2.0

C 2.5

N 3.0

0 3.5

F 4.0

Na 0.9

Mg 1.2

3B 3

4B 4

5B 5

6B 6

AI 1.5

Si 1.8

P 2.1

S 2 .5

CI 3.0

K 0.8

Ca 1.0

Sc 1.3

Ti 1.5

V

Rb 0 .8

Sr 1.0

Y

Cs 0.7

Ba 0.9

Fr 0.7

Ra 0.9

1.2

Lu 1.3

8

9

10

11

2B 12

1.6

Cr Mn 1.6 1.5

Fe 1.8

Co 1.9

Ni 1.9

Cu 1.9

Zn 1.6

Ga 1.6

Ge 1.8

As 2.0

Se 2.4

Br 2.8

Kr 3.0

Zr 1.4

Nb 1.6

Mo 1.8

Tc 1.9

Ru 2.2

Rb 2.2

Pd 2.2

Ag 1.9

Cd 1.7

In 1.7

Sn 1.8

Sb 1.9

Te 2.1

I 2.5

Xe 2.6

ill 1.3

Ta 1.5

W 1.7

Re 1.9

Os 2.2

Ir

Pt

2 .2

2.2

Au 2.4

Hg 1.9

TI 1.8

Pb 1.9

Bi 1.9

Po 2.0

At ?2

7

==::.

~."'."",

Multimedia

Periodic Table

eledronegativity.

Electronegativities of common elements.

lA

7B~8B~ IB

,

Figure 8.6

Increasing electronegativity

.->

...".....'"

? Linus Carl Pauling (1901- 1994). American chemist. Regarded by many as the most in fl uential chemist of the twentieth

century. Pauling received the Nobel Prize in Chemi stry in 1954 for his work on protein structure, and'the Nobel Peace Prize in 1962 for his tireless campaign against the testing and proliferation of nuclear arms . He is the only person ever to have received two unshared Nobel Prizes.

288

CHAPTER 8

Chemical Bonding I: Basic Concepts

Figure 8.7

Variation of electronegativity with atomic number. F

4

.-o

Cl

Br

3

I

>

Ru

Mn

Zn

1

Li

Na

Rb

K

O +---------.-----~--.---~~--~~~~--r=--~~·=···=···r_~~

o

10

20

30 Atomic Number

40

50

elements (the alkali and alkaline earth metals) are clustered near the lower left-hand corner. These trends are readily apparent in the graph in Figure 8.7. Electronegativity and electron affinity are related but distinct concepts. Both indicate the tendency of an atom to attract electrons. Electron affinity, however, refers to an isolated atom's ability to attract an additional electron in the gas phase, whereas electronegativity refers to the ability of an atom in a chemical bond (with another atom) to attract the shared electrons. Electron affinity, moreover, is an experimentally measurable quantity, whereas electronegativity is an estimated number that cannot be measured directly. Atoms of elements with widely different electronegativities tend to form ionic compounds with each other because the atom of the less electronegative element gives up its electrons to the atom of the more electronegative element. Atoms of elements with comparable electronegativities tend to form polar covalent bonds, or simply polar bonds, with each other because the shift in electron density from one atom to the other is usually small. Only atoms of the same element, which have the same electronegativity, can be joined by a pure covalent bond. There is no sharp distinction between nonpolar covalent and polar covalent or between polar covalent and ionic, but the following guidelines can help distinguish among them:

r=-

, _ ....... Multimedia

Chemical Bonding- ionic covalent and polar covalent bonds.

• A bond between atoms whose electronegativities differ by less than 0.5 is generally considered purely covalent or nonpolar. • A bond between atoms whose electronegativities differ by the range of 0.5 to 2.0 is generally considered polar covalent. • A bond between atoms whose electronegativities differ by 2.0 or more is generally considered •

•

lOme.

Sometimes chemists describe bonds using the term percent ionic character. A purely ionic bond would have 100 percent ionic character (although no such bond is known). A purely covalent, nonpolar bond has a percent ionic character. Sample Problem 8.4 shows how to use electronegativities to determine whether a chemical bond is nonpolar, polar, or ionic.

Samp'ie Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond in CsBr, and (c) the carbon-carbon double bond in C 2H 4 .

Strategy Using the information in Figure 8.6, determine which bonds have identical, similar, and widely different electronegativities.

Setup Electronegativity values from Figure 8.6 are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5).

SECTION 8.4

Electronegativity and Polarity

289

Solution (a) The difference between the electronegativities of F and Cl is 4.0 - 3.0 = 1.0, making

the bond in CIF polar. Think About It By convention,

(b) In CsBr, the difference is 2.8 - 0.7 = 2.1 , making the bond ionic.

the difference in electronegativity is always calculated by subtracting the small er number from the larger one, so the result is always positive.

(c) In C 2H 4 , th e two atoms are identical. (Not onl y are they the same element, but each C atom is bonded to two H atoms.) The carbon-carbon double bond in C 2 H 4 is nonpolar.

Practice Problem A Classify the fo llowing bonds as nonpolar, polar, or ionic: (a) the bonds in H 2 S,

(b) the H-O bonds in H 2 0 2 , and (c) the 0-0 bond in H 20 2 . Practice Problem B Using data from Figure 8.6, list all the main group elements that can form

purely ionic compounds with N.

Dipole Moment and Partial Charges The shift of electron density in a polar bond is symbolized by placing a crossed arrow (a dipole arrow) above the Lewis structure to indicate the direction of the shift. Por example,

•..

H-P: ••

The consequent charge separation can be represented as 8+

.. 8-

H-P: ..

A quantitative measure of the polarity of a bond is its dipole moment (/1,), whi ch is .... calculated .. as the product of the charge (Q) and the distance (I') between the charges: jJ- =QX r

Equation 8.1

. ...

The distance, r, between partial charges in a polar diatomic molecule is the bond length expressed in meters. Bond lengths are usually given in angstroms (A), or picometers (pm), so it is generally necessary to convert to meters.

In order for a diatomic molecule containing a polar bond to be electrically neutral, the partial positive and partial negative charges must have the same magnitude. Therefore, the Q term in Equation 8.1 refers to the magnitude of the partial charges and the calculated value of jJ- is always positive. 3 Dipole moments are usually expressed in debye units (D), named for Peter Debye. In terms of more familiar SI units, 1 D = 3.336 X 10- 30 C . m where C is coulombs '--N \ H-C-H H

+ F- (g)

8.103

6,

N

• F+(g)

Which is the preferred di ssociation for F2 , energetically speaking?

Based on changes in enthalpy, which of the following reactions will occur more readily? _. CH 3CI(g) + H (g) • CH 3 (g) + HCl(g)

H

F 2 (g)

Using the following information and the fact that the average C- H bond enthalpy is 414 kllmol, estimate the standard enthalpy of formation of methane (CH 4 ).

Which of the following molecules has the shortest nitrogen-tonitrogen bond: N 2H 4 , NzO, N z, N 2 0 4 ? Explain.

.97

H

8.98

8.93

.96

I

Draw four reasonable resonance structures for the P0 3F z- ion. The central P atom is bonded to the three 0 atoms and to the F atom. Show formal charges.

(a) CI(g) + CHig) (b) CI(g) + CH4 (g)

8.95

H

Give an example of an ion or molecule containing Al that (a) obeys the octet rule, (b) has an expanded octet, and (c) has an incomplete octet.

C(S) 2H 2 (g)

8.92

309

H

The following is a simplified (skeletal) structure of the amino acid tryptophan. Draw a complete Lewis structure of the molecule.

using (a) Equation 8.3 and (b) Equation 5.19, given that i1H'ffor 12 (g) is 61.0 kllmo1.

310

8.110

CHAPTER 8

Chemical Bonding I: Basic Concepts

Draw Lewis structures for the following organic molecules: (a) methanol (CH 30H); (b) ethanol (CH 3CHzOH); (c) tetraethyl lead [Pb(CHzCH 3)4], which is used in "leaded gasoline"; (d) methylamine (CH 3NH 2), which is used in tanning; (e) mustard gas (CICH2CHzSCHzCH zCI), a poisonous gas used in World War I; (f) urea [(NHz)zCO], a fertilizer; and (g) glycine (NHzCHzCOOH), an amino acid.

8.111

Write Lewis structures for the following four isoelectronic species: (a) CO, (b) NO +, (c) CN- , (d) N2. Show formal charges .

8.11 2

Oxygen forms three types of ionic compounds in which the anions are oxide (0 2- ), peroxide (O~ - ), and superoxide (O z). Draw Lewis structures of these ions.

8.113

Comment on the correctness of the statement, "All compounds containing a noble gas atom violate the octet rule."

8.114

Write three resonance structures for (a) the cyanate ion (NCO - ) and (b) the isocyanate ion (CNO - ). In each case, rank the resonance structures in order of increasing importance.

8.115

(a) From the following data calculate the bond enthalpy of the F z ion. F 2(g) F- (g) F z (g)

• 2F(g) • F(g) + e• F 2 (g) + e-

t1H ~xn =

156.9 kJ/mol

(d) The radical is generated when sunlight hits water vapor. Calculate the maximum wavelength (in nm) required to break an O-H bond in HzO. 8.122

Vinyl chloride (C zH3CI) differs from ethylene (C ZH 4) in that one of the H atoms is replaced with a CI atom. Vinyl chloride is used to prepare poly(vinyl chloride), which is an important polymer used in pipes. (a) Draw the Lewis structure of vinyl chloride. (b) The repeating unit in poly(vinyl chloride) is -CH2-CHCI-. Draw a portion of the molecule showing three such repeating units. (c) Calculate the enthalpy change when 1.0 X 103 kg of vinyl chloride forms poly(vinyl chloride).

8.123

Experiments show that it takes 1656 kJ/mol to break all the bonds in methane (CH4 ) and 4006 kJ/mol to break all the bonds in propane (C3HS)' Based on these data, calculate the average bond enthalpy of the C-C bond .

8.124

In 1998 scientists using a special type of electron microscope were able to measure the force needed to break a single chemical bond. If 2.0 X 10- 9 N was needed to break a C-Si bond, estimate the bond enthalpy in kJ/mol. Assume that the bond has to be stretched by a distance of 2 A (2 X 10- 10 m) before it is broken.

8.125

The American chemist Robert S. Mulliken suggested a different definition for the electro negativity (EN) of an element, given by

t1H ~xn =

333kJ/moi t1H ~xn = 290 kJ/mol

EN=

(b) Explain the difference between the bond enthalpies ofFz and F 2 .

8.11 6

IE -.:I_+~EA_ _

2

where lEI is the first ionization energy and EA is the electron affinity of the element. Calculate the electronegativities of 0, F, and CI using the preceding equation. Compare the electronegativities of these elements on the Mulliken and Pauling scales . (To convert to the Pauling scale, divide each EN value by

The resonance concept is sometimes described by analogy to a mule, which is a cross between a horse and a donkey. Compare this analogy with the one used in this chapter, that is, the description of a rhinoceros as a cross between a griffin and a unicorn. Which description is more appropriate? Why?

230 kJ/mo!.)

8.117

The N - 0 bond distance in nitric oxide is 115 pm, which is intermediate between a triple bond (106 pm) and a double bond (120 pm). (a) Draw two reso nance structures for NO, and comment on their relative importance. (b) Is it possible to draw a resonance structure having a triple bond between the atoms?

8.118

Although nitrogen dioxide (NO z) is a stable compound, there is a tendency for two such molecules to combine to form dinitrogen tetroxide (NZ0 4 ). Why? Draw four resonance structures of N 20 4 , showing formal charges.

8.119

In the gas phase, aluminum chloride exists as a dimer (a unit of two) with the formula AlzC1 6 . Its skeletal structure is given by Cl CI "'-/"'-/ AI Al /"'-/"'Cl Cl Cl

8.126

Halothane (CF 3CHClBr) Isoflurane (CF3CHClOCHFz) Enflurane (CHFCICFzOCHF2) Methoxyflurane (CHClzCFzOCH3 ) Draw Lewis structures of these molecules. .

8.127

A student in your class claims that magnesium oxide actually consists of Mg + and 0 - ions, not Mg2+ and 0 2 - ions. Suggest some experiments one could do to show that your classmate is wrong.

8.128

From the lattice energy of KCI in Table 8.6, and the ionization energy of K and electron affinity on Cl in Figures 7.8 and 7.10, respectively, calculate the t1Ho for the reaction

Cl

K (g)

8.129

Complete the Lewis structure and indicate the coordinate covalent bonds in the molecule. 8.120

8.121

The hydroxyl radical (OH) plays an important role in atmospheric chemistry. It is highly reactive and has a tendency to combine with an H atom from other compounds, causing them to break up. Thus OH is so metimes called a "detergent" radical because it helps to clean up the atmosphere. (a) Draw the Lewis structure for the radical. (b) Refer to Table 8.6 and explain why the radical has a high affinity for H atoms . (c) Estimate the enthalpy change for the following reaction:

+ CHig) ---+-. CH 3(g) + H 20 (g)

.

+ CI(g) ---+. KCl(s)

The species H ~ is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information 2H

Draw a Lewis structure for nitrogen pentoxide (N2 0 S) in which each N is bonded to three 0 atoms.

OH(g)

Among the common inhaled anesthetics are:

+ H+ Hz

• H~ • 2H

t1W = - 849 kJ/mol t1Ho = 436.4 kJ/mol

calculate t1Ho for the reaction H + + H2

• H~

8.130

The bond enthalpy of the C - N bond in the amide group of proteins (see Problem 8.84) can be treated as an average of C-N and C=N bonds. Calculate the maximum wavelength of light needed to break the bond.

8.131

In 1999 an unusual cation containing only nitrogen (Nr) was prepared. Draw three resonance structures of the ion, showing formal charges. (Hint: The N atoms are joined in a linear fashion.)

ANSWERS TO IN-CHAPTER MATERIALS

311

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES Nitrous oxide (N20) is an anesthetic commonly used for dental procedures. Because of the euphoria caused by inhaling it, N 20 is commonly known as "laughing gas." It is licensed for use as a food additive and as an aerosol propellant. It is used to displace air from potato chip bags in order to extend shelf life and as the propellant in whipped cream canisters. In recent years N 20 has become popular as a recreational drug, due in part to its ready availability to consumers. Although N 2 0 is legal, it is regulated by the FDA; its sale and distribution for the purpose of human consumption are not permitted.

Bond

Use formal charges to choose the best of the resonance structures shown.

a) I b) II c) III d) IV

3.

Using the best resonance structure and the average bond enthalpies given, determine the /::"H'ffor N 20.

Bond Enthalpy (l,J/mol) 176

N-O N=O N-N N=N

a) 73 kJ/mol

6m

b) -73 kJ/mol c) 166 kJ/mol d) -166 kJ/mol

193 418 94l.4 498.7

N=N 0=0 l.

2.

4.

Why does the calculated /::,.Hf'value differ from the tabulated value of 8l.56 kJ/mol?

Which of the following Lewis structures are possible for N 2 0? ••

:N-N-O: ..

•

•

•

'N=N=O' • ••

•

I a) I only b) I and II c) I, II, and III d) I, II, III, and IV

•

••

••

••

••

:N=O=N: • •

:N-O -N:

III

IV

II

a) The tabulated value is wrong. b) None of the resonance structures depicts the bonds realistically. c) To a reasonable number of significant figures, the calculated and tabulated values are the same. d) Using bond enthalpies gives only an estimate of /::"H ~xn '

ANSWERS TO IN-CHAPTER MATERIALS Practice Problems

••

D:J - .

8.1A (a) Ca 2 +, (b) ~B~ 3- , (c) 8.lE (a) 2 -, (b) +, (c) 3 - . 8.2A MgCI 2 . 8.2B NaF < MgO < AIN. 8.3A 629 kJ/mo!. 8.3B -841 kJ/mo!. 8.4A (a) nonpolar, (b) polar, (c) nonpolar. 8.4B Li, Na, K, Rb, Cs,

:Cl:

.. .1 ..

..

..

I

..

Fr, Ca, Sr, Ba, and Ra. 8.SA 0.12. 8.SB 0.11 D. 8.6A :F-N-F: •• •• ••

.. '0' .. I .. •

•

8.6B :O-CI-O: 8.7A C atom = 0, double-bonded 0 atom = 0, •• •• •• single-bonded 0 atoms = -l. 8.7B S atom = + 1, 0 atoms = -1 , •

II

..

•

8.9A :O=N-O: • •• 8.9B

o 0 -1 [N=C-~:]-

8.10A

'0'

..

II

0

••

..

I

•

:O-N = O:• •• -?

0

:C]:" .

••

:p: ..

.. I ;I:: 8.11B :Cl-B-Cl: 8.12A :CI-P, . 8.12B :F-Sb, . •• •• .• I ... · •. j"

..

I

..

..

I /C,r

.Cl. :C]: . .

..

+1

· . [B-c=s:] -

8.11A :P-Be-P: •• •• ••

••

• •

• • '0'

0

[N=C=~:] -

'O- H 8.10B :S=N-S: • •• :Cl:

..

:O-N-O: •• •• - 1

• •

I' .. :Cl: ..

8.1.1 b. 8.1.2 e. 8.2.1 c. 8.2.2 a. 8.4.1 b. 8.4.2 c. 8.5.1 d. 8.5.2 c. 8.6.1 e . 8.6.2 c. 8.7.1 a, b, d. 8.7.2 b. 8.8.1 e. 8.8.2 a, b, e. 8.8.3 e. 8.8.4 d. 8.9.1 a. 8.9.2 e.

••

. F.. :F: .

..

'N'

.g.

b) The difference in electronegativity is (3.5 - 3.0) = 0.5, making this a polar covalent bond. c) The partial charges on Nand o 0 o in NO are +0.029 and - 0.029, respectively. d) :N=Q: and l ' '. '. e) Formal charges are indicated on each - '.0,,-- -;:::-0: Lewis structure in part (d). f) /::"Ho for the N 1 1+ explosive decomposition of nitroglycerin H :0: H is -l.94 X 103 kJ/mo!. a)

..

• •

• •

••

Answers to Checkpoints

••

overall charge = -2. 8.8A - C - Q -H 8.8B CI- N-Cl •

•

Answers to Applying What You've Learned

•

'0'

.. '0' . I ..

•

8.13B :F---=Kr=---F: 8.14 -557 kJ. •• •• ••

:CI~I-;-CI:

8.13A

..

• • 'F'

••

I I I H-C--C--C-H I I I :0:

:0/ • ••

-1

I +1 N

~O'. ••

H

:0:

.:0: •

-1

I 1 N+

~O" ••

I

11

11

lea

on In

Molecular Geometry and Bonding Theories 9.1

Molecular Geometry

• •

• •

The VSEPR Model Electron-Domain Geometry and Molecular Geometry Deviation from Ideal Bond Angles Geometry of Molecules with More Than One Central Atom

9.2

Molecular Geometry and Polarity

9.3 9.4

Valence Bond Theory Hybridization of Atomic Orbitals

• •

Hybridization of s and p Orbitals Hybridization of s, p, and d Orbitals

9.5

Hybridization in Molecules Containing Multiple Bonds

9.6

Molecular Orbital Theory

• • • • •

9.7

Bonding and Antibonding Molecular Orbitals (T Molecular Orbitals Bond Order TT Molecular Orbitals Molecular Orbital Diagrams

Bonding Theories and Descriptions of Molecules with Delocalized Bonding

Mole,ular Shape and Our Sense of Smell The 2004 Nobel Prize in Physiology or Medicine was awarded to Richard Axel and Linda Buck for their discoveries of odorant receptors and the organization of the olfactory system. Their work shed new light on how our sense of smell works. Although it is somewhat more complicated than can be explained in detail here, the odor of a molecule depends in large part on which of the olfactory receptors it stimulates. Olfactory receptors are specialized proteins, located on hairlike cilia in the back of the nose. Stimulation of a receptor triggers an electrical signal that travels along nerve fibers to the brain, where the scent is identified. Which olfactory receptors a molecule stimulates depends on its shape. We are thought to have receptors for at least seven different, fundamental scents: floral, pepperminty, musky, pungent, camphorlike, ethereal, and putrid. Different parts of a larger molecule may have distinctly different shapes, enabling it to stimulate more than one type of olfactory receptor. This results in the perception of a combination of odors. For example, parts of the benzaldehyde molecule are shaped such that they stimulate the camphorlike, floral, and pepperminty receptors. This is a combination that we perceive as the smell of almonds. Such combinations enable us to recognize thousands of different smells.

o~ H

Benzaldehyde

This discussion illustrates the importance of molecular geometry to the most enigmatic of our senses smell. Many biochemical processes are specific in that they depend on the shapes of the molecules involved. Chemical bonding theories help us to predict and/ or explain these shapes.

In This Chapter, You Will Learn

how to determine the three-dimensional shape of a molecule and how the interactions of atomic orbitals give rise to chemical bonds.

Before you begin, you should review

• Shapes of atomic orbitals

[ ~~

Section 6.7]

• Electron configurations of atoms • Drawing Lewis structures

[ ~~

[ ~~

Section 6.8]

Section 8.5]

---

Media Player/

MPEG Content Chapter in Review

Volatile aromatic compounds give coffee its characteristic rich aroma. Molecules responsible for the odor of a substance have · distinctive shapes, which enables them to stimulate our sense of smell. 313

314

CHAPTER 9

Chemica l Bonding II: Molecul ar Geomet ry and Bond ing Theories

Molecular Geometry

Any atom that is bonded to two or more other atoms can be considered a "central" atom.

••

H-N-H I H Lewis structure of NH3

r-'" , ,... _

Many familiar chemical and biochemical processes depend heavily on the three-dimensional shapes of the molecules and/or ions involved. Our sense of smell is one example; the effectiveness of a particular drug is another. Although the shape of a molecule or poly atomic ion must be determined experimentally, we can predict their shapes reasonably well using Lewis structures [ ~~ Section 8.5] and the valence-shell electron-pair repulsion (VSEPR) model. In this section, we will focus primarily on determining the shapes of molecules of the general type ABx, where A is it central atom surrounded by x B atoms and x can have integer values of 2 to 6. For example, NH3 is an AB3 molecule in which A is nitrogen, B is hydrogen, and x = 3. Its Lewis structure is shown in the margin. Table 9.1' lists examples of each type of ABx molecule and polyatomic ion that we will consider. Throughout this chapter, we will discuss concepts that apply both to molecules and to polyatomic ions, but we will usually refer to them collectively as "molecules." Having the molecular formula alone is insufficient to predict the shape of a molecule. For instance, AB? molecules may be linear or bent:

Multimedia

Chemical bonding- valence-shell electron-pair repulsion theory.

Linear

Bent

Moreover, AB3 molecules may be planar, pyramidal, or T-shaped:

/""'1, Trigonal planar

Pyramidal

T-shaped

In order to determine shape, we must start with a correct Lewis structure and apply the VSEPR model.

The VSEPR Model . . . .. . , ... Recall that the electrons in the valence shell are the ones involved in chemical bonding [ ~~ Section 8. 1] .

Electron Domains Lone pair Single bond Double bond Triple bond

... ..

..

.. . . . . . . . . .. .. .. ..... The basis of the VSEPR model is that electron pairs in the valence shell of an atom repel one another. As we learned in Chapter 8, there are two types of electron pairs: bonding pairs and nonbonding pairs (also known as lone pairs). Furthermore, bonding pairs may be found in single bonds For .clarity, . .. . or. . in. multiple . . ... bonds. . . .. . . we will refer to electron domains instead of electron pairs when we use the VSEPR model. An electron domain in this context is a lone pair or a bond, regardless of whether the bond is single, double, or triple. Consider the following examples: .....

co?

NH3 •

••

••

:0=0-0:

:O=C=O: •

••

•

•

H-N-H

••

I

H

PC Is ••

••

:CI: •• .. /Cl: :Cl-p/ .. .. I "--:CI: ..

ci

:F:

.. .. I .. :F- Xe;-F: .. I' .. :F: ••

••

2 double bonds Total number 2 electron of electron domains domains on central atom

+

1 single bond 1 double bond 1 lone pair

3 electron domains

bonds 1 lone pair

4 single bonds + 2 lone 5 single bonds paIrs

4 electron domains

5 electron domains

3 single

+

•

6 electron domains

SECTION 9.1

Molecular Geometry

315

BeCI20 S02, H 2 0, NO z BF3 , NH 3 , ClF3 , SO ~ CCI4 , NH t , SF4 , XeF4 , CI0 4 PCI5 , IFs, SbF5 , BrFs SF6 , UF6, TiCI ~ -

Note the number of electron domains on the central atom in each molecule. The VSEPR model predicts that because these electron domains repel one another, they will arrange themselves to be as far apart as possible, thus minimizing the repulsive interactions between them. We can visualize the arrangement of electron domains using balloons, as shown in Figure 9.1. Like the B atoms in our ABx molecules, the balloons are all connected to a central, fixed point, which represents the central atom (A). When they are as far apart as possible, they adopt the five geometries shown in the figure. When there are only two balloons, they orient themselves to point in opposite directions [Figure 9.I(a)]. With three balloons the arrangement is a trigonal plane [Figure 9.1(b)]. With four balloons the arrangement adopted is a tetrahedron [Figure 9.1 (c)]. With five balloons, three of them adopt positions in a trigonal plane whereas the other two point opposite to each other, forming an axis that is perpendicular to the trigonal plane [Figure 9. 1(d)]. This geometry is called a trigonal bipyramid. Finally, with six balloon s the arrangement is an octahedron, which is essentially a square bipyramid [Figure 9.1 (e)]. Each of the ABx molecules we consider will have one of these five electron-domain geometries: linear, trigonal planar', tetrahedral, trigonal bipyramidal, or octahedral.

It is essential that you are able to draw the correct Lewis structure and that you count the electron domains on the central atom carefully. This will be your " blueprint" for determining the shape of a molecule or polyatomic ion.

Figure 9.1

The arrangements adopted by (a) two, (b) three, (c) four, (d) five, and (e) six balloons.

(a)

(b)

(c)

(d)

(e)

CHAPTER 9

316

Chemical Bonding II: Molecular Geometry and Bonding Theories

EleGUQn=ggffiiin

Media Player/ MPEG Content Chemical bonding- VSEPR and molecular geometry (interactive).

Geom@trY and

iVh:JU~ti.dir

ti@Om@iry

It is important to distinguish between the electron-domain geometry, which is the arrangement of electron domains (bonds and lone pairs) around the central atom, and the molecular geometry, which is the an'angement of bonded atoms. Figure 9.2 illustrates the molecular geometries of ABx molecules in which all the electron domains are bonds- that is, there are no lone pairs on any of the central atoms. In these cases, the molecular geometry is the same as the electron-domain geometry. In an ABx molecule, a bond angle is the angle between two adjacent A-B bonds. In an AB2 molecule there are only two bonds and therefore only one bond angle, and, provided that there are no lone pairs on the central atom, the bond angle is 180°. AB3 and AB4 molecules have three and four bonds, respectively. However, in each case there is only one bond angle possible between any two A-B bonds. In an AB3 molecule, the bond angle is 120°, and in an AB4 molecule, the bond angle is 109.So-again, provided that there are no lone pairs on the central atoms. Similarly, in an AB 6 molecule the bond angles between adjacent bonds are all 90°. (The angle between any two A - B bonds that point in opposite directions is 180°.) AB s molecules contain two different bond angles between adjacent bonds. The reason for this is that, unlike those in the other ABx molecules , the positions occupied by bonds in a trigonal bipyramid are not all equivalent. The three bonds that are arranged in a trigonal plane are referred to as equatorial. The bond angle between any two of the three equatorial bonds is 120°. The two bonds that form an axis perpendicular to the trigonal plane are referred to as axial. The bond angle between either of the axial bonds and anyone of the equatorial bonds is 90°. (As in the case of the AB6 molecule, the angle between any two A-B bonds that point in opposite directions is 180°.) Figure 9.2 illustrates all these bond angles. The angles shown in the figure are the bond angles that are observed when all the electron domains on the central atom are identicaL As we will see later in this section, the bond angles in many molecules will differ slightly from these ideal values. When the central atom in an ABx molecule bears one or more lone pairs, the electron-domain geometry and the molecular geometry are no longer the same. However, we still use the electrondomain.geometry as a first step in determining . . the molecular geometry. The first step in determining the molecular geometry of 0 3 (or any species), for example, is to draw its Lewis structure. Two different resonance structures can be drawn for 0 3: '

It would be impossible to overstate the importance of being able to draw Lewis structures correctly, especially for students who will go on to study organic chemistry.

:.0=0-0: ..< --+. :0-0=0: ••

••

•

Either one can be used to determine its geometry. The next step is to count the electron domains on the central atom. In this case, there are three: one single bond, one double bond, and one lone pair. Using the VSEPR model, we first determine the electron-domain geometry. According to the information in Figure 9.2, three electron domains on the central atom will be arranged in a trigonal plane. Molecular geometry, however, is dictated by the arrangement of atoms. If we consider only the positions of the three atoms in this molecule, the molecular geometry (the molecule's shape) is bent. ••

Electron-domain geometry: trigonal planar

Molecular geometry: bent

In addition to the five basic geometries depicted in Figure 9.2, you must be familiar with how molecular geometry can differ from electron-domain geometry. Table 9.2 shows the common ~igl:!f@

@:j

Five geometries of ABx molecules in which all the electron domains are bonds.

I

•

~ Trigonal planar

Linear

90 0

~

~

1200 •

•

1800 •

•

Tetrahedral

•

109S Trigonal bipyramidal

Octahedral

SECTION 9.1

Total Number of Electron Domains

Type of Molecule

Electron-Domain Geometry

Number of Lone Pairs

Placement of Lone Pairs

I

Mo lecular Geometry

Molecular Geometry

317

Example

••

1

3 Trigonal planar

Bent

••

1

4

Tetrahedral

4

- I ..;;;;.

Trigonal pyramidal

2

••

••

Tetrahedral

Bent

•

1

5

I

I .,. .

~I •

Seesaw-shaped

Trigonal bipyrarnidal

-

5

••

2

Trigonal bipyramidal

T-shaped

••

3

5

Trigonal bipyramidal

6

ABs

, -...;:r ----1"

f

IF2

Linear

1

I ----- ?

r ..: : : . ,.

~Ir

r-.;::

Square pyramidal Octahedral

_____

2

6

••

P'

~::::::: ••

Square planar Octahedral

BrFs

318

CHAPTER 9

Chemical Bonding II: Molecular Geometry and Bonding Theories

molecular geometries where there are one or more lone pairs on the central atom. Note the positions occupied by the lone pairs in the trigonal bipyramidal electron-domain geometry. When there are lone pairs on the central atom in a trigonal bipyramid, the lone pairs preferentially occupy equatorial positions because repulsion is greater when the angle between electron domains is 90° or less. Placing a lone pair in an axial position would put it at 90° to three other electron domains. Placing it in an equatorial position puts it at 90° to only two other domains, thus minimizing the number of strong repulsive interactions. All positions are equivalent in the octahedral geometry, so one lone pair on the central atom can occupy any of the positions. If there is a second lone pair in this geometry, though, it must occupy the position opposite the first. This arrangement minimizes the repulsive forces between the two lone pairs (they are 180° apart instead of 90° apart). In summary, the steps to determine the electron-domain and molecular geometries are as follows: 1. Draw the Lewis structure of the molecule or polyatomic ion.

2. Count the number of electron domains on the central atom. 3. Determine the electron-domain geometry by applying the VSEPR model. 4. Determine the molecular geometry by considering the positions of the atoms only. Sample Problem 9.1 shows how to determine the shape of a molecule or poly atomic ion.

Sample Problem 9.1 . Determine the shapes of (a) 50 3 and (b) ICI 4 .

Strategy Use Lewis structures and the VSEPR model to determine first the electron-domain geometry and then the molecular geometry (shape).

Setup (a) The Lewis structure of 50 3 is: ••

•

• • '0' I •.

:O=S-O: • •• There are three electron domains on the central atom: one double bond and two single bonds. (b) The Lewis structure ofICl 4 is: ••

:CI: ·. .1 .. :CI-·I-. CI: · .

J.

. .

:CI: •• There are six electron domains on the central atom in ICl 4 : four single bonds and two lone pairs. Solution

(a) According to the VSEPR model, three electron domains will be aITanged in a trigonal plane. Since there are no lone pairs on the central atom in 503 , the molecular geometry is the same as the electron-domain geometry. Therefore, the shape of S03 is trigonal planar.

Electron-domain geometry: trigonal planar - _ . Molecular geometry: trigonal planar (b) Six electron domains will be aITanged in an octahedron. Two lone pairs on an octahedron will be located on opposite sides of the centr'al atom, making the shape of ICl 4 square planar.

Think About It Compare these

results with the information in Figure 9.2 and Table 9.2. Make sure that you can draw Lewis structures cOITectly. Without a correct Lewis structure, you will be unable to determine the shape of a molecule.

Electron-domain geometry: octahedral

•

Molecular geometry: square planar

SECTION 9.1

Molecular Geometry

319

Practice Problem A Determine the shapes of (a) CO2 and (b) SCI 2 . Practice Problem B Determine the shapes of (a) SbF5 and (b) BrF5'

Deviation from Ideal Bond Angles Some electron domains are better than others at repelling neighboring domains. As a result, the bond angles may be slightly different from those shown in Figure 9.2. For example, the electrondomain geometry of ammonia (NH3) is tetrahedral, so we might predict the H-N - H bond angles to be 109.5°. In fact, the bond angles are about 107°, slightly smaller than predicted. The lone pair on the nitrogen atom repels the N - H bonds more strongly than the bonds repel one another. It therefore "squeezes" them closer together than the ideal tetrahedral angle of 109.5°. In effect, a lone pair takes up more space than the bonding pairs. This can be understood by considering the attractive forces involved in determining the location of the electron pairs. A lone pair on a central atom is attracted only to the nucleus of that atom. A bonding pair of electrons, on the other hand, is simultaneously attracted by the nuclei of both of the bonding atoms. As a result, the lone pair has more freedom to spread out and greater capacity to repel other electron domains. Also, because they contain more electron density, multiple bonds repel more strongly than single bonds. Consider the bond angles in each of the following examples:

10 1. 5 ° -+--1 84.8 °

BrFS

Geometry of Molecules with More Than One Central Atom Thus far we have considered the geometries of molecules having only one central atom. We can determine the overall geometry of more complex molecules by treating them as though they have multiple central atoms. Methanol (CH 30 H) , for example, has a central C atom and a central 0 atom, as shown in the following Lewis structure:

H

I .. H-C-O-H I .. H Both the C and the 0 atoms are sUlTounded by four electron domain s. In the case of C, they are three C-H bonds and one C-O bond. In the case of 0 , they are one O - C bond, one O-H bond, and two lone pairs. In each case, the electron-domain geometry is tetrahedral. However, the molecular geometry of the C part of the molecule is tetrahedral, whereas the molecular geometry of the 0 part of the molecule is bent. Note that although the Lewis structure makes it appear as though there is a 180° angle between the O-C and O-H bonds, the angle is actually approximately 109.5°, the angle in a tetrahedral arrangement of electron domains. Sample Problem 9.2 shows how to determine when bond angles differ from ideal values.

Acetic acid, the substance that gives vinegar its characteristic smell and sour taste, is sometimes used in combination with corticosteroids to treat certain types of ear infections . Its Lewis structure is •

•

H"O' I II ..

H-C-C-O-H

I

..

H Determine the molecular geometry about each of the central atoms, and determine the approximate value of each of the bond angles in the molecule. Which if any of the bond angles would you expect to be smaller than the ideal values? (Continued)

When we specify the geometry of a particular portion of a molecule, we refer to it as the geometry" about" a particular atom. In methanol, for example, we say that the geometry is tetrahedral about the C atom and bent about the 0 atom.

320

CHAPTER 9

Chem ical Bo nd ing II : Mo lecular Geomet ry and Bond ing Theor ies

Strategy Identify the central atoms and count the number of electron domains around each of them.

Use the VSEPR model to determine each electron-domain geometry, and the information in Table 9.2 to determine the molecular geometry about each central atom. Setup The leftmost C atom is surrounded by four electron domains: one C-C bond and three C- H

bonds. The middle C atom is surrounded by three electron domains : one C - C bond, one C-O bond, and one C=O (double) bond. The 0 atom is surrounded by four electron domains: one O - C bond, one O-H bond, and two lone pairs. Solution The electron-domain geometry of the leftmost C is tetrahedral. Because all four electron

domains are bonds, the molecular geometry of this part of the molecule is also tetrahedral. The electrondomain geometry of the middle C is trigonal planar. Again, because all the domains are bonds, the molecular geometry is also trigonal planar. The electron-domain geometry of the 0 atom is tetrahedral. Because two of the domains are lone pairs, the molecular geometry about the 0 atom is bent. Bond angles are determined using electron-domain geometry. Therefore, the approximate bond angles about the leftmost Care 109.5°, those about the middle Care 120°, and those about the 0 are 109.5°. The angle between the two single bonds on the middle carbon will be less than 120° because the double bond repels the single bonds more strongly than they repel each other. Likewise, the bond angle between the two bonds on the 0 will be less than 109.5° because the lone pairs on 0 repel the single bonds more strongly than they repel each other and push the two bonding pairs closer together. The angles are labeled as follows:

- 109.5°

Think About It Compare these

.. > 120° '0'

H

H

answers with the information in Figure 9.2 and Table 9.2.

< 120°

< 109.5°

Practice Problem A Ethanolamine (HOCH 2CH 2NH 2) has a smell similar to ammonia and is

commonl y found in biological tissues. Its Lewis structure is

H H .. I I .. H- O-C-C- N- H ..

I

I

I

H H H Determine the molecular geometry about each central atom and label all the bond angles . Cite any expected deviations from ideal bond angles. ,

Practice Problem B Determine the molecular geometry about each central atom in H 2S0 4 , Label

any expected deviations from ideal bond angles. Its Lewis structure is

..

• • '0' •• I •. H-O-S .. I - O..

H

:0:

..

Molecular Geometry 9.1.1

What are the electron-domain geometry and molecular geometry of CO ~-?

9.1.2

What are the electron-domain geometry and molecular geometry ofCIO;-?

a) tetrahedral, trigonal planar

a) tetrahedral, trigonal planar

b) tetrahedral, trigonal pyramidal

b) tetrahedral, trigonal pyramidal

c) trigonal pyramidal, trigonal pyramidal

c) trigonal pyramidal, trigonal pyramidal

d) trigonal planar, trigonal planar

d) trigonal planar, trigonal planar

e) tetrahedral, tetrahedral

e) tetrahedral, tetrahedral

SECTION 9.2

9.1.3

9.1.4

What is the approximate value of the bond angle indicated?

H

I I H-C=C~H < b) < c) > d) > e)
c) < d) > e)
0 3, CO, CO 2, N0 2, N 20, CH4 , CFCI 3?

9.88

The bond angle of S02 is very close to 120 0 , even though there is a lone pair on S. Explain.

9.89

The compound 3 ' -azido-3 ' -deoxythymidine, commonly known as AZT, is one of the drugs used to treat AIDS. What are the hybridization states of the C and N atoms in this molecule?

9.90

The following molecules (AX4 Y2) all have an octahedral geometry. Group the molecules that are equivalent to each other.

N

Cyclopropane (C 3H 6) has the shape of a triangle in which a C atom is bonded to two H atoms and two other C atoms at each corner. Cubane (CsHs) has the shape of a cube in which a C atom is bonded to one H atom and three other C atoms at each corner. (a) Draw Lewis structures of these molecules. (b) Compare the CCC angles in these molecules with those predicted for an Sp3 _ hybridized C atoID. (c) Would you expect these molecules to be easy to make? The compound 1,2-dichloroethane (C 2H4 Ci 2) is nonpolar, while cis-dichloroethylene (C 2H2Ci2) has a dipole moment: The reason for the difference is that groups connected by a single bond can rotate with respect to each other, but no rotation occurs when a double bond connects the groups. On the basis of bonding considerations, explain why rotation occurs in 1,2-dichloroethane but not in cis-dichloroethylene.

H

CI

Cl

I C I

I C I

H

CI H

H

1,2-dichloroethane

Cl

9.87

(a) What is the hybridization of N in the molecule? (b) Which structure is polar?

9.85

/

H

F

F

9.84

Does the following molecule have a dipole moment? Explain.

Briefly compare the VSEPR and hybridization approaches to the study of molecular geometry. Draw Lewis structures and give the other information requested for the following molecules: (a) BF3. Shape: planar or nonplanar? (b) Cl0 3. Shape: planar or nonplanar? (c) HCN. Polar or nonpolar? (d) OF2. Polar or nonpolar? (e) N0 2. Estimate the ONO bond angle.

359

\

y

y

y

X

(a)

(b)

x

x

x

X (d)

(c)

Cl

/

9.91

The compounds carbon tetrachloride (CC14) and silicon tetrachloride (SiCI4 ) are si milar in geometry and hybridization. However, CCl 4 does not react with water but SiCl 4 does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the Si atom in SiCi 4 .)

9.92

Write the ground-state electron configuration for B 2. Is the molecule diamagnetic or paramagnetic?

/ \ H H cis-dichloroethylene

360

9.93

CHAPTER 9

Chemical Bonding II: Molecular Geometry and Bonding Theories

What is the hybridization of C and of N in this molecule?

NH?

I -

-?,C"" ,/H

N

C

I

II

-?,C"" /C"" o N H

I

H

9.94

The stable allotropic form of phosphorus is P4 , in which each P atom is bonded to three other P atoms. Draw a Lewis structure of this molecule and describe its geometry. At high temperatures, P4 dissociates to form P 2 molecules containing a P=P bond. Explain why P 4 is more stable than P2.

9.95

Use molecular orbital theory to explain the difference between the bond enthalpies of F2 and F 2 .

9.96

Use molecular orbital theory to explain the bonding in the azide ion (N :;-). (The alTangement of atoms is NNN.)

9.97

The ionic character of the bond in a diatomic molecule can be estimated by the formul a

!!:...ed

X

9.99

9.102

The molecule benzyne (C 6 H4 ) is a very reactive species. It resembles benzene in that it has a six-membered ring of carbon atoms. Draw a Lewis structure of the molecule and account for the molecule's high reactivity.

9.103

Assume that the third-period element phosphorus forms a diatomic molecule, P2, in an ana logous way as nitrogen does to form N 2 . (a) Write the electronic configuration for P 2· Use [Ne2J to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

9.104

Consider an N2 molecule in its first excited electronic state, that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular orbital. (a) Identify the molecular orbitals involved, and sketch a diagram to show the transition. (b) Compare the bond order and bond length ofN ; with N 2, where the asterisk denotes the excited molecule. (c) Is N ; diamagnetic or paramagnetic? (d) When N ; loses its excess energy and converts to the ground state Nb it emits a photon of wavelength 470 nm, which makes up part of the auroras' lights. Calculate the energy difference between these levels.

9.105

The Lewis structure for O 2 is

100%

where J.L is the experimentally measured dipole moment (in C . m), e is the electronic charge, and d is the bond length (in meters) . (The quantity ed is the hypothetical dipole moment for the case in which the transfer of an electron from the less electronegative to the more electronegative atom is complete.) Given that the dipole moment and bond length of HF are 1.92 D and 91.7 pm, respectively, calculate the percent ionic character of the molecule. ( 1D = 3.336 X 10- 30 Cm) 9.98

o

•

:0=0: • •

Draw three Lewis structures for compounds with the formula C 2 H 2F 2. Indicate which of the compounds are polar. Greenhouse gases absorb (and trap) outgoing infrared radiation (heat) from Earth and contribute to global warming. A molecule of a greenhouse gas either possesses a permanent dipole moment or has a changing dipole moment during its vibrational motions. Consider three of the vibrational modes of carbon dioxide O=C=O

O=C=O

t

t

O=C=O

~

Use molecular orbital theory to show that the structure actually corresponds to an excited state of the oxygen molecule. 9.106

Draw the Lewis structure of ketene (C 2H 20) and describe the hybridization states of the C atoms. The molecule does not contain O-H bonds. On separate diagrams, sketch the formation of the sigma and pi bonds.

9.107

The compound TCDD, or 2,3,7,8-tetrachlorodibenzo-p-dioxin, is highly toxic:

where the alTOWS indicate the movement of the atoms. (During a complete cycle of vibration, the atoms move toward one extreme position and then reverse their direction to the other extreme position .) Which of the preceding vibrations are responsible for CO 2 behaving as a greenhouse gas? 9.100

9.101

Progesterone is a hormone responsible for female sex characteristics. In the usual shorthand structure, each point where lines meet represents a C atom, and most H atoms are not shown. Draw the complete structure of the molecule, showing all C and H atoms. Indicate which C atoms are s/- and sp3-hybridized.

Cl

o

Cl

o

It gained considerable notoriety in 2004 when it was implicated in the attempted murder of a Ukrainian politician. (a) Describe its geometry, and state whether the molecule has a dipole moment. (b) How many pi bonds and sigma bonds are there in the molecule?

Aluminum trichloride (AICI 3) is an electron-deficient molecule. It has a tendency to form a dimer (a molecule made up of two AlCl 3 units):

(a) Draw a Lewis structure for the dimer. (b) Describe the hybridization state of Al in AICl 3 and A12C1 6 . (c) Sketch the geometry of the dimer. (d) Do these molecules possess a dipole moment?

•

9.108

Write the electron configuration of the cyanide ion (CN- ). Name a stable molecule that is isoelectronic with the ion.

9.109

Carbon monoxide (CO) is a poisonous compound due to its ability to bind strongly to Fe2+ in the hemoglobin molecule. The molecular orbitals of CO have the same energy order as those of the N2 molecule. (a) Draw a Lewis structure of CO and assign formal charges. Explain why CO has a rather small dipole moment of 0.12 D. (b) Compare the bond order of CO with that from molecular orbital theory. (c) Which of the atoms (C or 0) is more likely to form bonds with the Fe2+ ion in hemoglobin?

ANSWERS TO IN-CHAPTER MATERIALS

36 1

PRE-PROFESSIONAL PRACTICE EXAM QUESTIONS: PHYSICAL SCIENCES These questions are not based on a descriptive passage.

1.

Which of the following has a bond order of 2?

What is the shape ofthe ICl 3 molecule? a) Trigonal planar b) Trigonal pyramidal c) T-shaped d) Tetrahedral

2.

3.

N2 II a) I only b) II only

4.

c) III only d) I and III

Which of the following is paramagnetic?

Which of the following molecules is nonpol ar?

O~ -

a) NCI 3 b) BCI 3

O2 II

I a) I and II b) II and III

c) PCI 3 d) BrCI 3

O~ III

O ~+

IV

c) III and IV d) I and III

ANSWERS TO IN-CHAPTER MATERIALS Answers to Sample Problems

9.9A

9.1A (a) linear, (b) bent. 9.1B (a) trigonal bipyramidal, (b) square pyramidal. 9.2A Bent about 0, tetrahedral about each C, trigonal pyramidal about N. All bond angles are - 109.5°.

H ~ I H-O-C.. I H

H I ~ C-N-H I I H H

9.2B Te trahedral about S , bent about each central O. All bond angles are - 109.50. Angles labeled in blue are < 109.5°. Angles labeled in blue are --CH=CH2 + HCI

•

>--CHCH 3

I

CI

QUESTIONS AND PROBLEMS

10.57

(a) Acetylide ion undergoes nucleophilic addition to aldehydes and ketones to give the species shown. Subsequent addition of water yields an acetylenic alcohol. Add curved arrows to the equations to show how the reaction occurs. H •'0' • '0;• • I H- O: ••

..

10.62

A compound has the empirical formula CsH I2 O. Upon controlled oxidation, it is converted into a compound of empirical formula C5H100, which behaves as a ketone. Draw possible structures for the original compound and the final compound.

10.63

Isopropanol is prepared by reacting propylene (CH 3CHCH2) with sulfuric acid, followed by treatment with water. (a) Show the sequence of steps leading to the product. What is the role of sulfuric acid? (b) Draw the structure of an alcohol that is an isomer of isopropanol. (c) Is isopropanol a chiral molecule?

•

_ ..

+ :O-H •• (b) A nucleophilic site on a molecule can substitute for a halogen elsewhere in the same molecule to form a ring. Use curved arrows to show how the following cyclization is related to a conventional nucleophilic substitution.

+

• 10.58

10.59

.. -

:Br: ••

Complete the following reaction by including essential unshared electron pairs and formal charges. Use curved arrows to show electron flow.

The product of the reaction of chlorobenzene with NaNH2 is a very reactive species called benzyne. Add necessary electron pairs and formal charges to the net ionic equation, and use curved arrows to show how benzyne is formed. To what general type of reaction does this belong?

Section 10.6: Organic Polymers Revie w Questions 10.64

Define the following terms : monomer, polymer, copolymer.

10.65

Name 10 objects that contain synthetic organic polymers.

10.66

Calculate the molar mass of a particular polyethylene sample, -fCH2-CH27t" where n = 4600.

10.67

Describe the two major mechanisms of organic polymer synthesis.

10.68

What are the steps involved in polymer formation by chain reaction?

10.69

Polys accharides, proteins, and nucleic acids comprise the three main classes of biopolymers. Compare and contrast them with respect to structure and function. What are the building block units for each? What are the key functional groups involved in linking the units together? In which biopolymer is there the greatest variety of building block structure? In which is there the least?

H

H

H

H 2N- +

•

409

Problems

H

CI H

10.70

Teflon is formed by a radical addition reaction involving the monomer tetrafluoroethylene. Show the mechanism for this reaction.

10.71

Vinyl chloride (H 2C==CHCI), undergoes copolymerization with l,l-dichloroethylene, (H 2C==CCI 2), to form a polymer commercially known as Saran. Draw the structure of the polymer, showing the repeating monomer units.

10.72

Deduce plausible monomers for polymers with the following repeating units:

H H H2N - H

+ CI-

+ H H

10.60

Consider the following reactions of butanal:

o II

CH 3CH 2CH 2CH

'/

(b)

CO---
--CONH---
--NH-+

n 10.73 In which reaction is butanal oxidized? In which reaction is it reduced? 10.61

Esters can be prepared by the acid-catalyzed condensation of a carboxylic acid and an alcohol:

Is this an oxidation-reduction reaction? If so, identify the species being oxidized and the species being reduced.

Deduce plausible monomers for polymers with the fo llowing repeating units: (a) -fCH 2-CH==CH -CH 27t, (b) -fCO-fCH276 NH 7t,

10.74

Draw the structures of the dipeptides that can be formed from the reaction between the amino acids glycine and alanine.

10.75

Draw the structures of the dipeptides that can be formed from the reaction between the amino acids glycine and lysine.

410

10.76

CHAPTER 10

Organic Chemistry

From the C 4H 9 alkyl groups in Problem 10.77, find the ones that cOlTespond to "R" among the amino acids listed in Figure 10.14. Match the name of the amino acid with the name of the group.

From among the given nucleotides, identify those that occur naturally in RNA and those that occur in DNA. Do any not occur in either RNA or DNA? 10.80

N--...../


-,

2500 -

~

•

2000-

2000

'"

1500

-~

1000

~

"8 ;::l

1000 500 0 -150

..0

.~

1500 -

1

1

-100

-50

.

1

0

.

1

I

I

50

100

150

2500

500 0

. I

200

-273.15 -200

-100

0

100

200

Temperature CC)

Temperature (0C)

(a)

(b) ~

Figure 11 .10

(a) Plot of the volume of a sample of gas as a function of temperature. (b) Plot of the volume of a sample of gas as a function of temperature at three different pressures.

Remember that a kelvin and a degree Celsius have the same magnitude. Thus, while we add 273.15 to the temperature in °C to get the temperature in K, a change in temperature in Celsius is equal to the change in temperature in K. A temperature of 20°C is the same as 293.15 K. A change in temperature of 20°C, however, is the same as a change in temperature of 20 K.

volume of a gas sample increases when heated and decreases when cooled. Figure 11.1O(a) shows a plot of data typical of Charles's and Gay-Lussac's experiments. Note that with pressure held constant, the volume of a sample of gas plotted as a function of temperature yields a straight line. These experiments were carried out at several different pressures [Figure 11.1 O(b)], each yielding a different straight line. Interestingly, if the lines are extrapolated to zero volume, they all meet at the x axis at the temperature -273 .15 dc. The implication is that a gas sample occupies zero volume at -273.15 °C. This is not observed in practice, however, because all gases condense to form liquids or solids before -273.15 °C is reached. In 1848 Lord KelvinS realized the significance of the extrapolated lines all meeting at -273. 15 °C. He identified -273.15 °C as absolute zero, theoretically the lowest attainable temperature. Then he set up an absolute temperature scale, now called the Kelvin temperature scale, with absolute zero as the lowest point [ ~. Section 1.3] . On the Kelvin scale, 1 kelvin (K) is equal in magnitude to 1 degree Celsius. The difference is simply an offset of 273.15. We obtain the abso273.15 to the temperature expressed in Celsius, although we often use . . . . .lute . . . . . temperature .. .. ,.... ... '"by .adding . simply 273 instead of 273.15. Several important points on the two scales match up as follows:

Kelvin Scale (K)

Celsius Scale (0C)

OK

-273.1SoC

Freezing point of water

273.15 K

O°C

Boiling point of water

373.15 K

100°C

Absolute zero

The dependence of the volume of a sample of gas on temperature is given by Equation 11.3(a)

.. . .. ....... ... .. ....... Don't forget that volume is proportional to absolute temperature. The volume of a sample of gas at constant pressure doubles if the temperature increases from 100 K to 200 K-but not if the temperature increases from 100°C to 200°C I

..

v = k2T ....... .. . . . .. . . . . . .

..

(at constant pressure) .. ..

or Equation 11.3(b)

V T

=

k2 (at constant pressure)

where k2 is the proportionality constant. Equations 11.3(a) and (b) are expressions of Charles's and Gay-Lussac's law, often refelTed to simply as Charles's law, which states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas.

5. William Thomson, Lord Kelvin (1824-1907). Scottish mathematician and physicist. Kelvin did important work in many branches of physics.

SECTION 11.2

The Gas Laws

Just as we did with the pressure-volume relationship at constant temperature, we can compare two sets of volume-temperature conditions for a given sample of gas at constant pressure. From Equation 11.3 we can write

or (at constant pressure)

Equation 11.4

where VI is the volume of the gas at T] and V2 is the volume of the gas at T2 • Sample Problem 11.3 shows how to use Charles's law.

~Sample~ Problemli.3 I

I

. ••..

A sample of argon gas that originally occupied 14.6 L at 25.0°C was heated to 50.0°C at constant pressure. What is its new volume? Strategy Use Equation 11.4 to solve for V2. Remember that temperatures must be expressed in kelvin. Setup TI = 298.15 K, VI = 14.6L, and T2 = 323.15 K. Solution

Think About It When temperature increases at constant pressure, the volume of a gas sample increases.

14.6 L X 323.15 K = 15.8 L 298.15 K

I

Practice Problem A A sample of gas originally occupies 29 .1 L at O.O°C. What is its new volume when it is heated to 15.0°C? (Assume constant pressure.)

I I

Practice Problem B At what temperature (in 0c) will a sample of gas occupy 82.3 L if it occupies 50.0 L at 75.0°C? (Assume constant pressure.)

I~----------------------------------------------------------"

Avogadro's law: The Amount-Volume Relationship In 1811 , the Italian scientist Amedeo Avogadro proposed that equal volumes of different gases contain the same number of particles (molecules or atoms) at the same temperature and pressure. This hypothesis gave rise to Avogadro's law, which states that the volume of a sample of gas is directly proportional to the number of moles in the sample at constant temperature and pressure:

Vo:n or

V = k3n

(at constant temperature and pressure)

Equation 11.5

Avogadro's law makes it possible to predict the volumes of gaseous reactants and products. Consider the reaction of H2 and N2 to form NH3:

The balanced equation reveals the ratio of combination of reactants in terms of moles [ ~~ Section 3.4]. However, because the volume of a gas (at a given temperature and pressure) is directly proportional to the number of moles, the balanced equation also reveals the ratio of combination in terms of volume. Thus, if we were to combine three volumes (liters, milliliters, etc.) of hydrogen gas with one volume of nitrogen gas, assuming they react completely according to the balanced equation, we would expect two volumes of ammonia gas to be produced (Figure 11.11). The ratio of combination of H2 and N2 (and production of NH3)' whether expressed in moles or units of volume, is 3: 1:2.

425

426

CHAPTER 11

Gases

+ + 3 molecules

,

3 moles 3 volumes

+ + +

1 molecule

2 molecules

1 mole

2 moles

1 volume

2 volumes

Figure 11 .11

Illustration of Avogadro's law. The volume of a sample of gas is directly proportional to the number of moles.

Sample Problem 11.4 shows how to apply Avogadro's law.

Samp'le Think About It Remember

that the coefficients in balanced chem ical equations indicate ratios in molecules or moles . Under conditions of constant temperature and pressure, the volume of a gas' is proportional to the number of moles. Therefore, the coefficients in balanced equations containing only gases also indicate ratios in liters, provided the reactions occur at constant temperature and pressure.

It is important to recognize that coefficients indicate ratios in liters only in balanced equations in which ali of the reactants and products are gases. We cannot apply the same approach to reactions in which there are solid, liquid, or aqueous species.

Checkpoint 11.2 11.2.1

If we combine 3.0 L of NO and 1.5 L of O2 , and they react according to the balanced equation 2NO(g) + 0 2(g) • 2N0 2(g), what volume of N02 will be produced? (Assume that the reactants and product are all at the same temperature and pressure.) Strategy Apply Avogadro's law to determine the volume of a gaseous prod uct. Setup Because volume is proportional to the number of moles, the balanced equation determines

in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometric amounts [ ~~ Section 3.6] . Solution According to the balanced equation, the volume of N02 formed will be equal to the

volume of NO that reacts. Therefore, 3.0 L of N02 will foml.

Practice Problem A What volume (in liters) of water vapor will be produced when 34 L of H2 and 17 L of O 2 react according to the equation 2H2(g) + 0 2(g) • 2H 20(g)? Practice Problem B What volumes (in liters) of carbon monoxide and oxygen gas must react according to the equation 2CO(g) + OzCg) , 2C0 2 (g) to form 3.16 L of carbon dioxide?

The Gas Laws

Given PI = 1.50 atm, VI = 37.3 mL, and P2 = 1.18 atm, calculate V2 . Assume that nand T are constant. a) 0.0211 mL

At what temperature will a gas sample occupy 100.0 L if it originally occupies 76 .1 L at 89.5°C? Assume constant P.

b) 0.0341 mL

a) 276°C

c) 29.3 mL

b) 118°C

d) 12.7 mL

c) 203 °C

e) 47.4 mL

d) 68.1 °C

11 .2 .3

e) 99.6°C 11 .2 .2

Given TI = 21.5°C, VI = 50.0 mL, and T2 = 316°C, calculate V2 . Assume that nand P are constant. a) 100 mL

11 .2 .4

What volume ofNH3 will be produced when 180 mL ofH2 reacts with 60.0 mL of N2 according to the following equation:

b) 73.5 mL c) 25.0 mL

Assume constant T and P for reactants and products.

d) 3.40 mL

a) 120 mL

e) 26.5 mL

b) 60 mL

c) 180 mL d) 240 mL e) 220 mL

The Ideal Gas Equation

SECTION 11.3

427

The Ideal Gas Equation Recall that the state of a sample of gas is described completely using the four variables T, P, V, and n. Each of the gas laws introduced in Section 11.2 relates one variable of a sample of gas to another while the other two variables are held constant. In experiments with gases, however, there are usually changes in more than just two of the variables. Therefore, it is useful for us to combine the equations representing the gas laws into a single equation that will enable us to account for changes in any or all of the four variables.

Deriving the Ideal Gas Equation from the Empirical Gas Laws

.

Summarizing the gas law equations from Section 11.2:

Multimedia Gas Laws- ideal gas law (interactive) .

1 V ex -

Boyle's law :

~

P

V ex T

Charles's law: Avogadro 's law:

Vex n

We can combine these equations into the following general equation that describes the physical behavior of all gases:

Vex nT P

or

nT V= R P where R is the proportionality constant. This equation can be rearranged to give:

PV = nRT

Equation 11.6

Equation 11.6 is the most commonly used form of the ideal gas equation, which describes the relationship among the four variables P, V, n, and T. An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation. Although the behavior of real gases generally differs slightly from that predicted by Equation 11.6, in most of the cases we will encounter, the differences are usually small enough for us to use . . the ideal gas equation to make reasonably good predictions about the behavior of gases. The proportionality constant, R, in Equation 11.6 is called the gas constant. Its value and units depend on the units in which P and V are expressed. (The variables nand T are always expressed in mol and K, respectively.) Recall from Section 11. 1 that pressure is commonly expressed in atmospheres, rnmHg (torr), pascals, or bar. Volume is typically expressed in liters or 3 . Table 11.4 lists several different milliliters, but can also be expressed. . .in other units, such as m .. . ... . expressions of the gas constant, R.

Numerical Value

Unit

0.08206

L . atmIK . mol

62.36

L . torr/K . mol

0.08314

L . barlK . mol

8.314

m 3 • PaiK . mol

8.314

11K· mol

1.987

callK · mol

Note that the produ ct of volu me and pressure gives units of energy (i.e ., joules and calories).

.

We will discuss the conditions that result in deviation from ideal behavior in Section 11.7.

.

.

When you use R in a calculation, use the version that facilitates proper cancellation of units.

•

428

CHAPTER 11

Gases

Keep in mind that all these expressions of R are equal to one another, just as I yard is equal to 3 ft. They are simply expressed in different units. One of the simplest uses of the ideal gas equation is the calculation of one of the variables when the other three are already known. For example, we can calculate the volume of 1 mole of an ideal gas at ooe and 1 atm, conditions known as standard temperature and pressure (STP). In this case, n, T, and P are given. R is a constant, leaving Vas the only unknown. We can reanange Equation 11.6 to solve for V,

•

In thermochemistry we often used 25°( as the" standard " temperature although temperature is not actually part of the definition of the standard state [ ~~ Section 5 .6] . The standard temperature for gases is defined specifically as O°C.

V= nRT P

enter the information that is given, and calculate V. Remember that in calculations using the ideal gas equation, temperature must always be expressed in kelvins. In this problem, because they are specified rather than measured, O°C, 1 mole, and 1 atm are exact numbers and do not affect the number of significant figures in the result [ ~~ Section 1.5].

V

=

(1 mol)(O.08206L . atrnJK . mol)(273.15K)

1 atm

= 22.41L

Thus, the volume occupied by 1 mole of an ideal gas at STP is 22.41 L, a volume slightly less than 6 gal. Sample Problem 11.5 shows how to calculate the molar volume of a gas at a temperature other than O°e.

Calculate the volume of a mole of ideal gas at room temperature (25°C) and 1 atm. Strategy Convert the temperature in °C to temperature in kelvins, and use the ideal gas equation to solve for the unknown volume.

Think About It With the pressure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (O°C), so the molar volume at room temperature (25 °C) should be higher than the molar volume at O°C-and it is.

Setup The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = 0.08206 L . atmlK . mol in order to solve for volume in liters. Solution (1 mol)(0.08206 L . atmlK . mol)(298.1S K) V= = 24.5 L 1 atm

Practice Problem A What is the volume of 5.12 moles of an ideal gas at 32°C and 1.00 atm? Practice Problem B At what temperature (in 0c) would 1 mole of ideal gas occupy 50.0 L (P = 1.00 atm)?

Applications of the Ideal Gas Equation Using some simple algebraic manipulation, we can solve for variables other than those that appear explicitly in the ideal gas equation. For example, if we know the molar mass of a gas (g/mol), we can determine its density at a given temperature and pressure. Recall from Section 11.1 that the density of a gas is generally expressed in units of gIL. We can reanange the ideal gas equation to solve for mol!L;

P V RT

-n -

If we then multiply both sides by the molar mass, .AIt, we get .. . n

.....

. . . . .. . .

X At = m, where m is mass in grams.

Another way to arrive at Equation 11.7 is to substitute mlJiIt for n in the ideal gas equation and rearrange to solve for mlV (denSity):

m m PV = -RT and At V

=

d

=

'PAt "RT "==

' "

. . ..

.AIt X !!:. = P x.AIt V RT

• • •

where.Alt X nlV gives gIL or density, d. Therefore, •

Equation 11.7

•

d=P.AIt RT

SECTION 11.3

The Ideal Gas Equation

429

Figure 11 .12

(a) Evacuated flask. (b) Flask filled with gas. The mass of the gas is the difference between the two masses. The density of the gas is determined by dividing mass by volume .

•

----= 92.0 13

----=

JiI.

92.029 "

(a)

(b)

Conversely, if we know the density of a gas, we can determine its molar mass:

.M = dRT P

Equation 11.S

In a typical experiment, in which the molar mass of a gas is determined, a flask of known volume is evacuated and weighed [Figure 11.12(a)]. It is then filled (to a known pressure) with the gas of unknown molar mass and reweighed [Figure 11.12(b)]. The difference in mass is the mass of the gas sample. Dividing by the known volume of the flask gives the density of the gas, and the molar mass can then be determined using Equation ll.S. Similarly, the molar mass of a volatile liquid can be determined by placing a small volume of it in the bottom of a flask, the mass and volume of which are known. The flask is then immersed in a hot-water bath, causing the volatile liquid to completely evaporate and its vapor to fill the escapes. When no more vapor escapes, flask. Because the flask is open, some of the excess vapor .. . . .. . .. . .. the flask is capped and removed from the water bath. The flask is then weighed to determine the mass of the vapor. (At this point, some or all of the vapor has condensed but the mass remains the same.) The density of the vapor is determined by dividing the mass of the vapor by the volume of the flask. Equation II.S is then used to calculate the molar mass of the volatile liquid. Sample Problems 11.6 and 11.7 illustrate the use of Equations 11.7 and 11 .S.

Because the flask is open to the atmosphere while the volatile liquid vaporizes, we can use atmospheric pressure as P. Also, because the flask is capped at the water bath temperature, we can use the water bath temperature as T.

Sample Problem 11.6 Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so CO 2 can smother the flames by depriving them of oxygen. (Air has a density of approximately 1.2 gIL at room temperature and 1 atm.) Calculate the density of CO 2 at room temperature (2S °C) and 1.0 atm. Strategy Use Equation 11.7 to solve for density. Because the pressure is expressed in atm, we Think About It The calculated

should use R = 0.OS206 L . atmIK . mol. Remember to express temperature in kelvins.

density of CO 2 is greater than that of air under the same conditions (as expected). Although it may seem tedious, it is a good idea to write units for each and every entry in a problem such as this. Unit cancellation is very useful for detecting errors in your reasoning or your solution setup.

Setup The molar mass of CO 2 is 44.01 g/mol. Solution

44.01g (1 atm)-d=P.M= mol =1. SglL RT (0.OS206 L . atm)(29S.1SK) K'mol

Practice Problem A Calculate the density of heli um in a helium balloon at 2S.0°C. (Assume that the

pressure inside the balloon is 1.10 atm.) Practice Problem B Calculate the density of air at O°C and 1 atm. (Assume that air is SO percent N2

and 20 percent O2 ,)

•

..

~------------------------------------------------------~

430

CHAPTER 11

Gases

Samp'le A company has just patented a new synthetic alcohol for alcoholic beverages. The new product is said to have all the pleasant properties associated with ethanol but none of the undesirable effects such as hangover, impairment of motor skills, and risk of addiction. The chemical formula is proprietary. You analyze a sample of the new product by placing a small volume of it in a roundbottomed flask with a volume of 511.0 rnL and an evacuated mass of 131.918 g. You submerge the flask in a water bath at 100.0°C and allow the volatile liquid to vaporize. You then cap the flask and remove it from the water bath. You weigh it and determine the mass of the vapor in the flask to be 0.768 g. What is the molar mass of the volatile liquid, and what does it mean with regard to the new product? (Assume the pressure in the laboratory is 1 atm.) Strategy Use the measured mass of the vapor and the given volume of the flask to determine the

density of the vapor at 1 atm and 100.0°C, and then use Equation 11.8 to determine molar mass. Setup P

= 1 atm, V = 0.5110 L, R = 0.8206 L . atm/mol . K, and T = 373 .15 K.

Solution

0.768 g d = = 1.5029 aIL 0.5110 L b

Think About It Because more

than one compound can have a particular molar mass, this method is not definitive for identification. However, in this circumstance, further testing of the proprietary formu la certainly would be warranted.

oM =

1.5029 g )( 0.08206 ]0'· atnf)(373.15){) ( Y v/ . mol ~

1 Jlli1f

= 46.02 glmol

The result is a molar mass suspiciously close to that of ethanol!

Practice Problem A Determine the molar mass of a gas with a density of 1.905 gIL at 80.0°C and

1.00 atm. Practice Problem B Determine the molar mass of a gas with a density of 5.14 gIL at 73.0°C and

1.00 atm .

Checkpoint 11.3 11.3.1

The Ideal Gas Equation

Calculate the volume occupied by 8.75 moles of an ideal gas at STP.

11.3.3

a) 196 L

Determine the density of a gas with oM = 146.07 glmol at 1.00 atm and 100.0°C.

b) 268 L

a) 6.85 X 10- 3 giL

c) 0.718 L

b) 4.77 gIL

d) 18.0 L

c) 146 gIL

e) 2.56 L

d) 30.6 gIL e) 17 .8 gi L

11.3.2

Calculate the press ure exerted by 10.2 moles of an ideal gas in a 7.5-L vessel at 150°C.

11.3.4

a) 17 atm

Determine the molar mass of a gas with d = 1.963 gIL at 1.00 atm and 100.0°C.

b) 31 atm

a) 0.0166 g/mol

c) 0.72 atm

b) 60 .1 g/mol

d) 1.3 atm

c) 16.1 glmol

e) 47 atm

d) 6.09 X 103 glmol e) 1.63 X 103 gl mol

Reactions with Gaseous Reactants and Products In Chapter 3 we used balanced chemical equations to calculate amounts of reactants and/or products in chemical reactions expressing those amounts in mass (usually grams). However, in the

SECTION 11.4

Reactions with Gaseous Reactants and Products

case of reactants and products that are gases, it is more practical to measure and express amounts in volume (liters or milliliters). This makes the ideal gas equation useful in the stoichiometric analysis of chemical reactions that involve gases.

Calculating the Required Volume of a Gaseous Reactant According to Avogadro's law, the volume of a gas at a given temperature and pressure is proportional to the number of moles. Moreover, balanced chemical equations give the ratio of combination of gaseous reactants in both moles and volume (Figure 11.11). Therefore, if we know the volume of one reactant in a gaseous reaction, we can determine the required amount of another reactant (at the same temperature and pressure). For example, consider the reaction of carbon monoxide and oxygen to yield carbon dioxide: 2CO(g)

+ 02(g) --+. 2C02(g)

The ratio of combination of CO and O 2 is 2: 1, whether we are talking about moles or units of volume. Thus, if we want to determine the stoichiometric amount [ ~~ Section 3.6] of O 2 required to combine with a particular volume of CO, we simply use the conversion factor provided by the balanced equation, which can be expressed as any of the following: 1 mol O 2 2 mol CO

or

1 L0 2 2LCO

or

1 mL0 2 2mLCO

Let's say we want to determine what volume of O 2 is required to react completely with 65.8 mL of CO at STP. We could use the ideal gas equation to convert the volume of CO to moles, use the stoichiometric conversion factor to convert to moles O 2 , and then use the ideal gas equation again to convert moles O 2 to volume. But this method involves several unnecessary steps. We get the same result simply by using the conversion factor expressed in milliliters:

65.8.rnL-eO X = 1 mL O 2 = 32.9 mL 0 2~

2

In cases where only one of the reactants is a gas, we do need to use the ideal gas equation in our analysis. Recall, for example, the reaction of sodium metal and chlorine gas used to illustrate the Born-Haber cycle [I~~ Section 8.2]: 2Na(s) + ClzCg) --+. 2NaCl(s) Given moles (or more commonly the mass) of Na, and information regarding temperature and pressure, we can determine the volume of C12 required to react completely: X

grams Na

1 mol Na 22.99 g Na

X

=

..

X RT P

1 mol Cl2 = 2 mol Na

moles Na

..

moles C12

..

liters C12

Sample Problem 11.8 shows how to use the ideal gas equation in a stoichiometric analysis. . .

.

Sample Problem 11.8 Sodium peroxide (Na202) is used to remove carbon dioxide from (and add oxygen to) the air supply in spacecrafts. It works by reacting with CO 2 in the air to produce sodium carbonate (Na2C03) and O 2,

What volume (in liters) of CO 2 (at STP) will react with a kilogram of Na202?

Strategy Convert the given mass of Na202 to moles, use the balanced equation to determine the stoichiometric amount of CO 2, and then use the ideal gas equation to convert moles of CO 2 to liters.

,

Setup The molar mass of Na202 is 77.98 glmol (1 kg as an exact number.)

=

1000 g). (Treat the specified mass of Na202

(Continued)

431

432

CHAPTER 11

Gases Solution

~ "T~

1 mol Na20 2 1000 P-J-'ffi2V2 X 8 ~ = 12.82 mol Na20 2 77 .9 2

Think About It The answer seems

like an enormous volume of CO 2, If you check the cancellation of units carefully in ideal gas equation problems, however, then with practice you will develop a sense of whether such a calculated volume is reasonable.

12.82

Veo

= ,

2 2X

2 mol CO 2 = 12.82 mol CO 2 2m

(1 2.82 mol CO 2)(0.08206 L· atmlK· mol)(273.15 K) 1 atm

8

= 2 7.4 L

CO 2

Practice Problem A What volume (in liters) of CO 2 can be consumed at STP by 525 g NaZ02? Practice Problem B What mass (in grams) of Na20 Zis necessary to consume 1.00 L COz at STP?

Determining the Amount of Reactant Consumed Using Change in Pressure Although none of the empirical gas laws focuses on the relationship between nand P explicitly, we can rearrange the ideal gas equation to show that n is directly proportional to P at constant V and T: Equation 11.9(a)

n

=

P (:T) (at constant Vand T) X

Therefore, we can use the change in pressure in a reaction vessel to determine how many moles of a gaseous reactant are consumed in a chemical reaction:

!:::..n = !:::..P

Equation 11.9(b) . . .. This refers to a reaction in which there is only one gaseous reactant and in which none of the products is a gas, such as the reaction described in Sample Problem 11.9. In reactions involving multiple gaseous species, I:l.n refers to the net change in number of moles of gas- and the analysis gets somewhat more complicated.

. ..

X

(:T) (at constant V and T)

... . where !:::..n is the number of moles of gas consumed and !:::..P is the change in pressure in the reaction vessel. Sample Problem 11.9 shows how to use Equation 11.9(b). .

.

..

Another air-purification method for enclosed spaces involves the use of "scrubbers" containing aqueous lithium hydroxide, which reacts with carbon dioxide to produce lithium carbonate and water: 2LiOH(aq)

+ COzeg) - _ . LizC0 3(s) + H 20(l)

Consider the air supply in a submarine with a total volume of 2.5 X 105 L. The pressure is 0.9970 atm, and the temperature is 25°C. If the pressure in the submarine drops to 0.9891 atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many moles of CO 2 are consumed? Strategy Use Equation 11.9(b) to determine

Think About It Careful

cancellation of units is essential. Note that this amount of CO 2 corresponds to 162 moles or 3.9 kg of LiOH. (It's a good idea to verify this yourself.)

!:::..n, the number of moles of CO 2 consumed.

Setup t:.P = 0.9970 atm - 0.9891 atm = 7.9 X 10- 3 atm. According to the problem statement, V = 2.5 X 105 L and T = 298.15 K. For problems in which P is expressed in atmospheres and V in liters, use R = 0.08206 L . atmlK . mol. Solution

t:.neo,

= 7.9

3

X 10- atm X (0.08206 L .

a~~ .1~:~ X (298.15 K) = 81 moles CO2 consumed

Practice Problem A Using all the same conditions as those described in Sample Problem 11.9,

calculate the number of moles of CO 2 consumed if the pressure drops by 0.010 atm. Practice Problem B By how much would the pressure in the submarine drop if 2.55 kg of LiOH

were completely consumed by reaction with CO Z? (Assume the same starting P, V, and T as in Sample Problem 11.9.)

SECTION 11.4

Reactions w ith Gaseous Reactants and Products

Predicting the Volume of a Gaseous Product Using a combination of stoichiometry and the ideal gas equation, we can calculate the volume of gas that we expect to be produced in a chemical reaction. We first use stoichiometry to determine the number of moles produced, and then apply the ideal gas equation to determine what volume will be occupied by that number of moles under the specified conditions. Sample Problem 11.10 shows how to predict the volume of a gaseous product. _..

.

.

• ',Sample Problem 11.10 The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3 ):

A typical driver-side air bag contains about SO g of NaN3 . Determine the volume of N2 gas that would be generated by the decomposition of 50.0 g of sodium azide at 85°C and 1 atm. Strategy Convert the given mass of NaN 3 to moles, use the ratio of the coefficients from the balanced chemical equation to determine the corresponding number of moles of N2 produced, and then use the ideal gas equation to determine the volume of that number of moles at the specified temperature and pressure. Setup The molar mass of NaN3 is 65.02 g/mol. Solution mol NaN3

=

50.0 g NaN 3 65 .02 g/mol

Think About It The calculated volume represents the space between the driver and the steering wheel and dashboard that must be filled by the air bag in order to prevent injury. Air bags also contain an oxidant that consumes the sodium metal produced in the reaction.

'

= 0.769 mol NaN3

3 mol No 0.769 mol NaN 3 X 2 - = 1.15 mol No mol NaN3 VN

2

(1.15 mol N2)(0.08206 L . atm/mol . K)(3S8.IS K) = = 33.9 L N2 1 atm

Practice Problem A The chemical equation for the metabolic breakdown of glucose (C2H 12 0 6) is the same as that for the combustion of glucose [ ~~ Section 3 .3- Bringing Chemistry to Life box] :

Calculate the volume of CO 2 produced at normal human body temperature (37°C) and 1.00 atm when 10.0 g of glucose is consumed in the reaction. Practice Problem B The passenger-side air bag in a typical car must fill a space approximately four times as large as the driver-side air bag in order to be effective. Calculate the mass of sodium azide required to fill a l2S-L air bag.

.

Checkpoint 11.4 11 .4.1

Reactions with Gaseous Reactants and Products

Determine the volume of Cl2 gas at STP that will react with 1.00 mole of Na solid to produce NaCl according to the equation, 2Na(s)

+ CI2 (g)

• 2NaCI(s)

11 .4.2

Determine the mass of NaN 3 required for an air bag to produce 100.0 L of N2 gas at 85.0°C and 1.00 atm according to the equation, 2NaN 3(s)

a) 22.4 L

a) 332 g

b) 44.8 L

b) 148 g

c) 15.3 L

c) 221 g

d) 11.2 L

d) 664 g

e) 30.6 L

e) 442g

• 2Na(s)

+ 3N2 (g)

433

•

434

CHAPTER 11

Gases

Gas Mixtures So far our discussion of the physical propelties of gases has focused on the behavior of pure gaseous substances, even though the gas laws were all developed based on observations of samples of air, which is a mixture of gases. In this section, we will consider gas mixtures and their physical behavior. We will restrict our discussion in this section to gases that do not react with one another; that is, ideal gases.

Dalton's Law of Partial Pressures When two or more gaseous substances are placed in a container, each gas behaves as though it occupies the container alone. For example, if we place 1.00 mole of N2 gas in a 5.00-L container at ooe, it exerts a pressure of P

=

(1.00 mol)(0.08206 L . atmlK . mol)(273.15 K) 5.00L

= 4.48 atm

If we then add a mole of another gas, such as O 2, the pressure exerted by N2 does not change. It remains at 4.48 atm. The O2 gas exerts its own pressure, also 4.48 atm. Neither gas is affected by the presence of the other. In a mixture of gases, the pressure exerted by each gas is known as the partial pressure (PJ of the gas. We use subscripts to denote partial pressures: P N2

P02

=

(1.00 mol)(0.08206 L . atmlK . mol)(273 .15 K) 5.00 L

= 4.48 atm

=

(1.00 mol)(0.08206 L . atmlK . mol)(273 .15 K) 5.00 L

= 4.48 atm

and we can solve the ideal gas equation for each component of any gas mixture: nRT p. = _Ie:-I

V

Dalton's law of partial pressures states that the total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture:

Thus, the total pressure exerted by a mixture of 1.00 mol N2 and 1.00 mol O2 in a 5.00-L vessel at ooe is

Ptotal = PN 2

+ Po2 = 4.48 atm + 4.48 atm =

Figure 11.13 illustrates Dalton's law of partial pressures.

Figure 11.13

Schematic illustration of Dalton's law of partial pressures. Total pressure is equal to the sum of partial pressures.

•

I

,

8.96 atm

SECTION 11.5

Gas Mixtures

435

Sample Problem 11.11 shows how to apply Dalton's law of partial pressures.

~

.~

~

Sample

Problem 11.11

A 1.00-L vessel contains 0.215 mole of N2 gas and 0~0118 mole of H2 gas at 25.5 °C. Determine the partial pressure of each component and the total pressure in the vessel. Strategy Use the ideal gas equation to find the partial pressure of each component of the mixture,

and sum the two partial pressures to find the total pressure. Setup T = 298.65 K. Solution

(0.215 mOI)( 0.08206 L . atm)(298.65 K) K'mol 1.00 L = 5.27 atm (0.01 18 mOI)(0.08206 L . atm)(298.65 K) K . mol 1.00L P tota! = P N

2

+ PH2 =

5.27 atm

+ 0.289 atm =

=

0.289 atm

5.56 atm

Think About It The total pressure in the vessel can also be determined by summing the number

of moles of mixture components (0.215 + 0.0118 = 0.227 mol) and solving the ideal gas equation for P IOIa!:

P tota !

(0.227 mOI)(0.08206 L . atm )(298.65 K) K'mol = 00 = 5.56 atm 1. L

Practice Problem A Determine the partial pressures and the total pressure in a 2.50-L vessel

containing the following mixture of gases at 15.8°C: 0.0194 mol He, 0.0411 mol H 2, and 0.169 mol Ne. Practice Problem B Determine the number of moles of each gas present in a mixture of CH4 and

C2H 6 in a 2.00-L vessel at 25.0°C and 1.50 atm, given that the partial pressure of CH4 is 0.39 atm.

L -_________________________________________________

~

Mole Fractions The relative amounts of the components of a gas mixture can be specified u sing mole fractions. The mole fraction (Xi) of a component of a mixture is the number of moles of the component divided by the total number of moles in the mixture:

X·I

=

n·I n total

Equation 11.10

There are three things to remember about mole fractions: 1. The mole fraction of a mixture component is always less than 1. 2. The sum of mole fractions for all components of a mixture is always 1. 3. Mole fraction is dimensionless. In addition, nand P are proportional [Equation 11.9(a)] at a specified T and V, so we can determine mole fraction by dividing the partial pressure of a component by the total pressure:

PI X· = I P lOta!

Equation 11.11

Rearranging Equations 11.10 and 11.11 gives Equation 11.12

Mole fractions do not refer only to gas mixtures. They can be used to specify the concentrations of components of mixtures in any phase. Mole fractions are used extensively in Chapter 12~

436

CHAPTER 11

Gases

and Equation 11.13 Sample Problem 11.12 lets you practice calculations involving mole fractions, partial pressures, and total pressure. .

.

Sample Problem 11.12 In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung disease, which occurs commonly in premature infants. The nitric oxide used in this therapy is supplied to hospitals in the form of a N 2INO mixture. Calculate the mole fraction of NO in a 1O.00-L gas cylinder at room temperature (25 °C) that contains 6.022 mol N2 and in which the total pressure is 14.75 atm. Strategy Use the ideal gas equation to calculate the total number of moles in the cylinder. Subtract

moles of N2 from the total to determine moles of NO . Divide moles NO by total moles to get mole fraction (Equation 11.11). Setup The temperature is 298.15 K. Solution Think About It To check your

total moles

work, determine X N2 by subtracting X NO from 1. Using each mole fraction and the total pressure, calculate the partial pressure of each component using Equation 11.13 and verify that they sum to the total pressure.

= PV = RT

mol NO

(14.75 atm)( lO.OO L) ( 0.08206 L . atm )(298 .15 K) K· mol

= total moles _

X NO -

n NO _

n total

-

= 6.029 mol

- mol N2 = 6.029 - 6.022

= 0.007 mol NO

0.007 mol NO - 0001 . 6.029 mol

Practice Problem A Determine the mole fractions and partial pressures of CO 2, CH4 , and He in a

sample of gas that contains 0.250 mole of CO 2, 1.29 moles of CH4 , and 3.51 moles of He, and in which the total pressure is 5.78 atm. Practice Problem B Determine the partial pressure and number of moles of each gas in a 15.75-L

vessel at 30.0°C containing a mixture of xenon and neon gases only. The total pressure in the vessel is 6.50 atm, and the mole fraction of xenon is 0.761.

Using Partial Pressures to Solve Problems The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown in Figure 11.14. Dalton's law of partial pressures is usefulin the analysis of these kinds

Cylinder filled with water ready to be placed in the plastic basin

Cylinder being filled with oxy,

KCI0 3 and Mn02 ,~

-

-'"'

P total

-

.-

Fig ure 11.14

'""

--

.~

(a)

--

(b)

(a) Apparatus for measuring the amount of gas produced in a chemical reaction. (b) When the water levels inside and outside the collection vessel are the same, the pressure inside the vessel is equal to atmospheric pressure.

SECTION 11.5

T(°e)

P (torr)

T(°e)

0

4.6

35

5

6.5

10

P (torr)

T(°e)

P (torr)

42.2

70

233.7

40

55.3

75

289.1

9.2

45

71.9

80

355.1

15

12.8

50

92.5

85

433.6

20

17.5

55

118.0

90

525.8

25

23.8

60

149.4

95

633.9

30

31.8

65

187.5

100

760.0

.

",

.

•

,==.

of experimental results. For example, the decomposition of potassium chlorate (KCl0 3), the reaction used to generate emergency oxygen supplies on airplanes, produces potassium chloride and oxygen: 2KCl0 3(s) --+. 2KCl(s)

'

Gas Mixtures

Multimedia

Gas Laws-colleding a gas over water.

+ 30ig)

The oxygen gas is collected over water, as shown in Figure 11.14(a) . The volume of water displaced by the gas is equal to the volume of gas produced. (Prior to reading the volume of the gas, the level of the graduated cylinder must be adjusted such that the water levels inside and outside the cylinder are the same. This ensures that the pressure inside the graduated cylinder is the same as atmospheric pressure [Figure 11.14(b)].) However, because the measured volume contains both the oxygen produced by the reaction and water vapor, the pressure exerted inside the graduated cylinder is the sum of the two partial pressures: Ptotal

= Po2 + PH20

By subtracting the partial pressure of water from the total pressure, which is equal to atmospheric pressure, we can determine the partial pressure of oxygen and thereby determine how many moles are produced by the reaction. We get the partial pressure of water, which depends on temperature, from a table of values. Table 11.5 lists the partial pressure (also known as the vapor pressure) of water at various temperatures. Sample Problem 11.13 shows how to use Dalton's law of partial pressures to determine the amount of gas produced in a chemical reaction and collected over water.

" "Sample Problem 11.13 Calcium metal reacts with water to produce hydrogen gas

[~

Section 7.7] :

Determine the mass of H2 produced at 25°C and 0.967 atm when 525 mL of the gas is collected over water as shown in Figure 11.14. •

Strategy Use Dalton's law of partial pressures to detemrine the partial pressure of H2 , use the ideal gas equation to determine moles of H 2 , and then use the molar mass of H2 to convert to mass . (Pay careful attention to units. Atmospheric pressure is given in atmospheres, whereas the vapor pressure of water is tabulated in torr.) Setup V = 0.525 Land T = 298.15 K. The partial pressure of water at 25 °C is 23.8 torr (Table 11.5) or 23.8 torr (1 atml760 torr) = 0.0313 atm. The molar mass ofH2 is 2.016 g/mo!. Solution P~

= P lOla )

moles ofH2 =

-

P~o

Think About It Check unit

= 0.967 atm - 0.0313 atm = 0.936 atm

(0.9357 atm)(0.525 L)

201

=.

X

10- 2

mo

cancellation carefully, and remember that the densities of gases are relatively low. The mass of approximately half a liter of hydrogen at or near room temperature and 1 atm should be a very small number.

1

0.08206 L . atm )(298.15 K) ( K'mol

mass of H2 = (2.008 X 10-2 mol)(2.016 g/mol)

= 0.0405 g H2 (Continued)

437

438

CHAPTER 11

Gases

Practice Problem A Calculate the mass of O 2 produced by the decomposition of KCl0 3 when 821

rnL of O2 is collected over water at 30.0°C and 1.015 atm. Practice Problem B Determine the volume of gas collected over water when 0.501 g O2 is produced

by the decomposition of KCl0 3 at 35.0°C and 1.08 atm.

Bringing Chemistry to Life Hyperbaric Oxygen Therapy

One unfortunate group of patients was undergoing treatment when the power to the chamber was sh ut off accidentally. All the patients died. At the time, their deaths were attributed to influenza, but they almost certainly died as the result of the unintended rapid decompression .

In 1918, during the Spanish flu epidemic that claimed tens of millions of lives worldwide, physician Orville Cunningham noted that people living at lower elevations appeared to have a greater chance of surviving the flu than those living at higher elevations. Believing this to be the result of increased air pressure, he developed a hyperbaric chamber to treat flu victims. One of Cunningham's earliest and most notable successes was the recovery of a flu-stricken colleague who had built a hyperbaric chamber large enough to accombeen near death. Cunningham subsequently . . modate dozens of patients and treated numerous flu victims, most with success. In the decades following the Spanish flu epidemic, hyperbaric therapy fell out of favor with the medical community and was largely discontinued. Interest in it was revived when the U.S. military ramped up its underwater activities in the 1940s and hyperbaric chambers were constructed to treat military divers suffering from decompression sickness (DCS), also known as "the bends." Significant advancement in hyperbaric methods began in the 1970s when the Undersea Medical Society (renamed the Undersea and Hyperbaric Medical Society in 1976) became involved in the clinical use of hyperbaric chambers. Today, hyperbaric oxygen therapy (HBOT) is used to treat a wide variety of conditions, including carbon monoxide poisoning, anemia caused by critical blood loss, severe bums, and life-threatening bacterial infections. Once considered an "alternative" therapy and viewed with skepticism, HEaT is now covered by most insurance plans.

/

o

.~

•

The Catalina Hyperbaric Chamber at the University of Southern California's Wrigley Marine Science Center treats mostly victims of diving accidents. In one treatment protocol, the chamber is pressurized to 6.0 atm with compressed air and the patient breathes a mixture of gases that is 47 percent O2 by volume. In another protocol, the chamber is pressurized with compressed air to 2.8 atm and the patient breathes pure O2 , Determine the partial pressure of O2 in each treatment protocol and compare the results. Strategy Because the gas the patient breathes is inside the hyperbaric chamber, its total pressure is

Catalina Hyperbaric Chamber

the same as the chamber pressure. To obtain the partial pressure of O2 , multiply the mole fraction of O2 in the breathing gas in each protocol by the total pressure (Equation 11.13). Setup In the first protocol, the breathing gas is 47 percent O2 , so the mole fraction of O2 is 0.47.

In the second protocol, where pure O2 is used, the mole fraction of O2 is 1. Use Equation 11.13 to calculate the partial pressure of O2 in each protocol.

Solution In the first protocol, the pressure of O2 is

0.47 X 6.0 atm = 2.8 atm Think About It Monoplace

hyperbaric chambers, which are large enough to accommodate only one person, commonly are pressurized to 2.8 atm with pure O2 ,

In the second protocol, the pressure of O2 is

1

X

2.8 atm = 2.8 atm

Both protocols produce the same pressure of O2 ,

Practice Problem What mole fraction of O2 is necessary for the partial pressure of O 2 to be 2.8 atm

when the total pressure is 4.6 atm?

---

SECTION 11.6

Checkpoint 11.5 11.5.1

The Kinetic Molecular Theory of Gases

439

Gas Mixtures

What is the partial pressure of He in a 5.00-L vessel at 25°C that contains 0.0410 mole of He, 0.121 mole of Ne, and 0.0922 mole of Ar?

11 .5.3

a) 1.24 atm

What is the partial pressure of oxygen in a gas mixture that contains 4.10 moles of oxygen, 2.38 moles of nitrogen, and 0.917 mole of carbon dioxide and that has a total pressure of 2.89 atm?

b) 0.248 atm

a) 1.60 atm

c) 0.117 atm

b) 3.59 atm

d) 2.87 atm

c) 0.391 atm

e) 0.201 atm

d) 0.705 atm e) 0.624 atm

11.5.2

What is the mole fraction of CO 2 in a mixture of 0.756 mole of N 2 , 0.189 mole of 0 20 and 0.0132 mole of CO 2 ?

11.5.4

a) 0.789

What mass of acetylene (C 2H2 ) is produced by the reaction of calcium carbide (CaC 2) and water,

b) 0.0138 if 425 mL of the gas is collected over water at 30°C and a pressure of 0.996 atm?

c) 0.0140 d) 1.003

a) 0.016 g

e) 0.798

b) 16.3 g c) 0.019 g d) 0.424 g e) 0.443 g

The Kinetic Molecular Theory of Gases The gas laws were delived empirically, and they enable us to predict the macroscopic behavior of gases. They do not explain, however, why gases behave as they do. The kinetic molecular theory, which was put forth in the nineteenth century by a number of physicists, notably Ludwig Boltzmann6 and James Maxwell,7 explains how the molecular nature of gases gives rise to their macroscopic properties. The basic assumptions of the kinetic molecular theory are as follows: 1. A gas is composed of particles that are separated by relatively large distances. The volume occupied by individual molecules is negligible. 2. Gas molecules are constantly in random motion, moving in straight paths, colliding with the walls of their container and with one another in perfectly elastic collisions. (Energy is transferred but not lost in the collisions.) 3. Gas molecules do not exert attractive or repulsive forces on one another. 4. The average kinetic energy, Eb of gas molecules in a sample is proportional to the absolute temperature:

Ek rx T

Recall that kinetic energy is the energy associated with motion

[ ~~

Section 5.1] :

1 2 E k -- zmu

Thus, the kinetic energy of an individual gas molecule is proportional to its mass and its velocity squared. When we talk about a group of gas molecules, we determine the average kinetic energy using the mean square speed, u2 , which is the average of the speed squared for all the molecules in the sample: 2 2 2 ? u2 = Uj + U2 + U3 + ... + U N N where N is the number of molecules in the sample. 6. Ludwig Eduard Boltzmann (1844--1906). Austrian physicist. Although Boltzmann was one of the greatest theoretical physicists of all time, his work was not recognized by other scientists in his own lifetime. He suffered from poor health and severe depression and committed suicide in 1906. 7. James Clerk Maxwell (1831-1879). Scottish physicist. Maxwell was one of the great theoretical physicists of the nineteenth century. His work covered many areas in physics, including the kinetic theory of gases, thermodynamics , and electricity and magnetism.

440

CHAPTER 11

Gases

Figure 11 .15 Gases can be compressed by decreasing their volume. (a) Before volume decrease. (b) After volume decrease, the increased frequency of collisions between molecules and the walls of their container constitutes higher pressure.

\\

•

•

(b)

(a)

Application to the Gas laws Kinetic molecular theory enables us to understand some of the properties and behavior of gases in the following ways.

Compressibility Gases are compressible because molecules in the gas phase are separated by large distances (assumption 1) and can be moved closer together by decreasing the volume occupied by a sample of gas (Figure 11.15).

Boyle's Law (V

IX

1/P)

The pressure exerted by a gas is the result of the collisions of gas molecules with the walls of their container (assumption 2). The magnitude of the pressure depends on both the frequency of collision and the speed of molecules when they collide with the walls. Decreasing the volume occupied by a sample of gas increases the frequency of these collisions, thus increasing the pressure (Figure 11.15).

Charles's Law (V

IX

T)

Heating a sample of gas increases its average kinetic energy (assumption 4). Because the masses of the molecules do not change, an increase in average kinetic energy must be accompanied by an increase in the mean square speed of the molecules. In other words, heating a sample of gas makes the gas molecules move faster. Faster-moving molecules collide more frequently and with greater speed at impact, thus increasing the pressure. If the container can expand (as is the case with a balloon or a cylinder with a movable piston), the volume of the gas sample will increase, thereby decreasing the frequency of collisions until the pressure inside the container and the pressure outside are again equal (Figure 11.16).

Avogadro's Law (V

IX

n)

Because the magnitude of the pressure exerted by a sample of gas depends on the frequency of the collisions with the container wall, the presence of more molecules would cause an increase in pressure. Again, the container will expand if it can. Expansion of the container will decrease the frequency of collisions until the pressures inside and outside the container are once again equal (Figure 11.17).

Dalton's Law of Partial Pressures (Ptotal =

IPi)

Gas molecules do not attract or repel one another (assumption 3), so the pressure exerted by one gas is unaffected by the presence of another gas. Consequently, the total pressure exerted by a

The Kinetic Molecular Theory of Gases

SECTION 11.6

441

Heating or cooling a gas at constant pressure

\

p

~ Lower temperature

r

p

Higher temperature

• (Volume increases)

(Volume

• • •

(a)

I-Charles's Law

Heating or cooling a gas at a constant volume

\

p

Lower temperature

Higher temperature

..

~

(Pressure decreases)

(Pressure increases)

•

•

p

I

rI

• p = (

n: ) n;: is constant T

(b)

Figure 11.16

Charles's law. (a) The volume of a sample of gas at constant pressure is proportional to its absolute temperature. (b) The pressure of a sample of gas at constant volume is proportional to its absolute temperature.

Dependence of volume on amount of gas at constant temperature and pressure !

I

..

~ p

~ P

"

,

I

Gas cylinder

,

Add gas molecules

Remove gas

• (Volume decreases)

"

"

"

..

..

(Volume increases)

Valve Avogadro's Law

V

Figure 11.17

= (

RJ) RJ is constant n

The volume of a gas at constant temperature and pressure is proportional to the number of moles.

442

CHAPTER 11

Gases

Volwne and temperature are constant

I

,

Combining the gases

+

Figure 11 .18

Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the individual components' partial pressures.

mixture of gases is simply the sum of the partial pressures of the individual components in the mixture (Figure 11.18).

Molecular Speed One of the important outcomes of the kinetic molecular theory is that the total kinetic energy of a mole of gas (any gas) is equal to ~RT. With assumption 4 we saw that the average kinetic energy 2 of one molecule is "imu . For 1 mole of the gas we write

1 -2 3 NA (-mu = -RT 2 2 where NA is Avogadro's number. Because m X NA = .AIt, we can rearrange the preceding equation as follows: 3RT

.AIt Taking the square root of both sides gives

~ u2 = 3RT .AIt where ~ u is the root-mean-square (rms) speed (u rms )' The result, 2

Equation 11.14

Remember that the average kinetic energy of a gas depends on its absolute temperature. Therefore. any two gas samples at the same temperature have the same average kinetic energy.

.AIt

gives us the root-mean-square speed, which is the speed of a molecule with the average kinetic . energy in a gas sample. Equation 11.14 indicates two important things: (1) The root-mean-square speed is directly proportional to the square root of the absolute temperature, and (2) the rootmean-square speed is inversely proportional to the square root of .Alt. Thus, for any two samples ... . orgas at the same temperature, the gas with the larger molar mass will have the lower root-meansquare speed, U rms' Keep in mind that most molecules will have speeds either higher or lower than urms-and that Urms is temperature dependent. James Maxwell studied extensively the behavior of gas molecules at various temperatures. Figure 11.19(a) shows typical Maxwell speed distribution curves for nitrogen gas at three different temperatures. At a given temperature, the distribution curve tells us the number of molecules moving at a certain speed. The maximum of each curve represents the •

In Equation 11.14, R must be expressed as 8.314 J/mol • K and oM must be expressed in kg/moL

3RT

The Kinetic Molecular Theory of Gases

SECTION 11.6

443

T= 300K

N2 (28 .02 g/mol) --------------~

el2 (70.90 g/mol)

/lOOK

-"

-'"oa is related to the experimentally measured pressure, P rea ), by the equation an 2 Pideal

=

P real

+

V

2

where a is a constant and n and V are the number of moles and volume of the gas, respectively. The correction term for pressure (an2/V2) can be understood as follows. The intermolecular interaction that gives rise to nonideal behavior depends on how frequently any two molecules encounter each other. The number of such encounters increases with the square of the number of molecules per unit volume (n/V), and a is a proportionality constant. The quantity Pi deal is the pressure we would measure if there were no intermolecular attractions. The other correction concerns the volume occupied by the gas molecules. In the ideal gas equation , V represents the volume of the container. However, each molecule actually occupies a very small but nonzero volume. We can correct for the volume occupied by the gas molecules by subtracting a term, nb, from the volume of the container: Vreal =

Videa I -

nb

where nand b are the number of moles and the proportionality constant, respectively. Incorporating both corrections into the ideal gas equation gives us the van der Waals equation, with which we can analyze gases under conditions where ideal behavior is not expected. 2

Equation 11.16

P

+ an2 V

--_ . corrected

~-

pressure term

(V - nb)

=

nRT

- .---

-.

corrected

volume tenn

The van der Waals constants a and b for a number of gases are listed in Table 11.6. The magnitude of a indicates how strongly molecules of a particular type of gas attract one another. The magnitude of b is related to molecular (or atomic) size, although the relationship is not a simple one. Sample Problem 11.1 6 shows how to use the van der Waals equation.

-

Deviation from Ideal Behavior

SECTION 11.7

A sample of 3.50 moles of NH3 gas occupies 5.20 L at 47°C. Calculate the pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der Waals equation . Strategy (a) Use the ideal gas equation, PV = nRT.

(b) Use Equation 11 .16 and a and b values for NH3 from Table 11.6. Setup T = 320.15 K, a = 4.17 atm' Llmof, and b = 0.0371 LImo!.

(3.50 mOl)( 0.08206 L . atm )(320.15 K) . nRT K· mol SolutIon (a) P = V = 5.20 L = 17.7 atm (b) Evaluating the correcti on terms in the van der Waals equation, we get ?

4.17 at~ . L- (3 .50 mOI)2 mol-

-'-----'=-=-=--- - ' - , - - - - - =

1. 89 atm

(5.20 L)2 nb = (3 .50 mOl)(0.0371 L) = 0.130 L mol

Finall y, substituting these results into Eq uation 11.16, we have (P

+ 1.89 atm)(5.20 L - 0.130 L)

= (3 .50 mOI)(0.08206 L · atm )(320. 15 K)

K'mol

Think About It As is often the

case, the pressure exerted by the real gas sample is lower than predicted by the ideal gas equation .

P = 16.2 atm

.

Practice Problem A Using data from Table 11.6, calculate the pressure exerted by 11.9 moles of

neon gas in a volume of 5.75 L at 25 °C using (a) the ideal gas equation and (b) the van der Waals equation (Eq uation 11.16). Compare your results. Practice Problem B Calculate the pressure exerted by 0.35 mole of oxygen gas in a volume of 6.50

L at 32°C using (a) the ideal gas equation and (b) the van der Waals equation.

I

..

--------------------------------------------------------_

_.

Checkpoint 11. 7 11 .7 .1

Deviation from Ideal Behavior

Whi ch of the following conditions

cause deviation from ideal behavior? (Select all that appl y.) a) High pressure b) Low pressure c) High temperature d) Low temperature e) High volu me

11.7.2

Using the van der Waals equation, calculate the pressure exerted by 1.5 moles of carbon dioxide in a 3.75-L vessel at JO°C. a) 9.3 atm b) 8.9 atm c) 10 atm

d) 8.6 atm e) -2 .4 atm

447

448

CHAPTER 11

Gases

Applying What You've Learned

The Gamow Bag earned the praises even of Sir Edmund Hillary, leader of the first expedition to the summit of Mount Everest.

Scuba divers are not the only athletes who can suffer the detrimental effects of sudden changes in pressure. Mountain climbers, too, are susceptible to the dangers of rapid ascent. At high elevation, air pressure is significantly lower than at sea level. A lower total pressure means a lower partial pressure of oxygen, and insufficient oxygen or hypoxia can cause altitude sickness. Early symptoms of altitude sickness include headache, dizziness, and nausea. In severe cases, climbers may suffer hallucinations, seizure, coma, and even death. In 1990, Igor Gamow, a professor of microbiology at the University of Colorado, , ., patented it portabie device 'for 'iiigh-aitltude ' treatment 'of ' rateuneatalyzed Figure 14.14 shows the potential energy profiles for both reactions. The total energies of the reactants (A and B) and those of the products (C and D) are unaffected by the catalyst; the only difference between the two is a lowering of the activation energy from Ea to E~. Because the activation energy for the reverse reaction is also lowered, a catalyst enhances the rates of the forward and reverse reactions equally. There are three general types of catalysis, depending on the nature of the rate-increasing substance: heterogeneous catalysis, homogeneous catalysis, and enzyme catalysis. •

Heterogeneous Catalysis In heterogeneous catalysis, the reactants and the catalyst are in different phases. The catalyst is usually a solid, and the reactants are either gases or liquids. Heterogeneous catalysis is by far the most important type of catalysis in industrial chemistry, especially in the synthesis of many important chemicals. Heterogeneous catalysis is also used in the catalytic converters in automobiles. At high temperatures inside a car's engine, nitrogen and oxygen gases react to form nitric oxide:

When released into the atmosphere, NO rapidly combines with O 2 to form N0 2. Nitrogen dioxide and other gases emitted by automobiles, such as carbon monoxide (CO) and various unburned hydrocarbons, make automobile exhaust a major source of air pollution. Most new cars are equipped with catalytic converters [Figure 14.1S(a)]. An efficient catalytic converter serves two purposes: It oxidizes CO and unburned hydrocarbons to CO 2 and H 20, and it converts NO and N0 2 to N2 and O2, Hot exhaust gases into which air has been injected are passed through the first chamber of one converter to accelerate the complete burning of hydrocarbons and to decrease CO emissions. Because high temperatures increase NO production, however, a second chamber containing a different catalyst (a transition metal or a transition metal oxide such as CuO or Cr20 3) and operating at a lower temperature is required to dissociate NO into N2 and O 2 before the exhaust is discharged through the tailpipe [Figure 14.1S(b)].

SECTION 14.6 Internal combustion engine

Figure 14.15

Exhaust

Tailpipe Air compressor

Catalytic converters (a)

(b)

Homogeneous Catalysis In homogeneous catalysis the reactants and the catalyst are dispersed in a single phase, usually liquid. Acid and base catalyses are the most important types of homogeneous catalysis in liquid solution. For example, the reaction of ethyl acetate with water to form acetic acid and ethanol normally occurs too slowly to be measured.

In the absence of the catalyst, the rate law is given by

The reaction, however, can be catalyzed by an acid. Often a catalyst is shown above the arrow in a chemical equation:

In the presence of acid, the rate is faster and the rate law is given by rate

=

kc [CH 3 COOC 2H 5 ][H +]

Because kc > k in magnitude, the rate is determined solely by the catalyzed portion of the reaction. Homogeneous catalysis has several advantages over heterogeneous catalysis. For one thing, the reactions can often be can-ied out under atmospheric conditions, thus reducing production costs and minimizing the decomposition of products at high temperatures. In addition, homogeneous catalysts can be designed to function selectively for particular types of reactions, and homogeneous catalysts cost less than the precious metals (e.g., platinum and gold) used in heterogeneous catalysis.

Enzymes: Biological Catalysts Of all the intricate processes that have evolved in living systems, none is more striking or more essential than enzyme catalysis. Enzymes are biological catalysts. The amazing fact about enzymes is that not only can they increase the rate of biochemical reactions by factors ranging from 106 to 10 18, but they are also highly specific. An enzyme acts only on certain reactant molecules, called substrates, while leaving the rest of the system unaffected. It has been estimated that an average living cell may contain some 3000 different enzymes, each of them catalyzing a specific reaction in which a substrate is converted into the appropriate product(s). Enzyme catalysis is usually homogeneous because the substrate and enzyme are present in aqueous solution. An enzyme is typically a large protein molecule that contains one or more active sites where interactions with substrates take place. These sites are structurally compatible with specific substrate molecules, in much the same way that a key fits a particular lock. In fact, the notion of a rigid enzyme structure that binds only to molecules whose shape exactly matches that of the active

Catalysis

575

(a) A two-stage catalytic converter for an automobile. (b) In the second stage, NO molecules bind to the surface of the catalyst. The N atoms bond to each other and the 0 atoms bond to each other, producing N2 and O2 , respectively.

576

CHAPTER 14

Chemical Kinetics

Figure 14.16

The lock-and-key model of an enzyme's specifi city for substrate molecules.

Substrate

Products

+

+

Enzyme

Enzyme-substrate complex

Enzyme

Figure 14.17

Left to right: The binding of glucose molecule (red) to hexokinase (an enzyme in the metabolic pathway). Note how the region at the active site closes around glucose after binding. Often, the geometries of both the substrate and the active site are altered to fit each other.

site was the basis of an early theory of enzyme catalysis, the so-called lock-and-key theory devell oped by Emil Fischer in 1894 (Figure 14.16). Fischer's hypothesis accounts for the specificity of enzymes, but it contradicts research evidence that a single enzyme binds to substrates of different sizes and shapes. Chemists now know that an enzyme molecule (or at least its active site) has a fair amount of structural flexibility and can modify its shape to accommodate more than one type of substrate. Figure 14.17 shows a molecular model of an enzyme in action. The mathematical treatment of enzyme kinetics is quite complex, even when we know the basic steps involved in the reaction. A simplified scheme is given by the following elementary steps: E+S

kl

• L, •

ES

ES where E, S, and P represent enzyme, substrate, and product, respectively, and ES is the enzymesubstrate intermediate. It is often assumed that the formation of ES and its decomposition back to enzyme and substrate molecules occur rapidly and that the rate-determining step is the formation of product. Figure 14.18 shows the potential energy profile for the reaction. In general, the rate of such a reaction is given by the equation

MP] rate = - !1t

= k[ES] The concentration of the ES intermediate is itself proportional to the amount of the substrate present, and a plot of the rate versus the concentration of substrate typically yields a curve like that shown in Figure 14.19. Initially the rate rises rapidly with increasing substrate concentration.

1. Emil Fischer (1852-19 19). German chemist. Regarded by many as the greatest organic chemist of the nineteenth century.

Fischer made many significant contributions in the synthesis of sugars and other important molecules. He was awarded the Nobel Prize in Chemistry in 1902.

SECTION 14.6

Catalysis

577

Figure 14.18

>.

>.

bJl

2.'l

'""'

""

-

OJ

~

~

c

"0

~

f\

"c " .-'"

--" .

Comparison of (a) an uncatalyzed reaction and (b) the same reaction catalyzed by an enzyme. The plot in (b) assumes that the catalyzed reaction has a two-step mechanism, in which the second step (ES • E + P) is rate determining .

"0"

S

~

0..

E+S

/'

0..

V

ES

p

E+P

Reaction progress

Reaction progress

(a)

(b)

,

Figure 14.19

c

o

All active sites are occupied at and beyond this substrate concentration.

Plot of the rate of product formation versus substrate concentration in an enzyme-catalyzed reaction.

[S]

Above a certain conoentration, however, all the active sites are occupied, and the reaction becomes zeroth order in the substrate. In other words, the rate remains the same even though the substrate concentration increases. At and beyond this point, the rate of formation of product depends only on how fast the ES intermediate breaks down, not on the number of substrate molecules present.

Bringing Chemistry to Life Catalysis and Hangovers We learned at the beginning of the chapter that alcohol dehydrogenase (AD H) and aldehyde dehydrogenase (ALDH) catalyze the metabolism of methanol. In analogous reactions, ADH and ALDH catalyze the metabolism of ethanol, the alcohol in alcoholic beverages. The first reaction converts ethanol to acetaldehyde, which is far more toxic than ethanol:

It is acetaldehyde (CH 3CHO) that causes the misery associated with a hangover including headache, nausea, and vomiting. In the second reaction, acetaldehyde is converted to acetic acid, which is harmless:

An effective but painful part of the treatment for alcohol abuse is to administer disulfiram, marketed under the name Antabuse. Disulfiram blocks the action of ALDH, preventing the conversion of acetaldehyde to acetic acid. The resulting buildup of acetaldehyde causes the patient to feel very sick almost immediately, making the next cocktail far less appealing. The action of disulfiram was discovered accidentally when Danish pharmaceutical researchers who were taking the drug as an experimental treatment for parasitic diseases became very ill every time they consumed alcohol.

•

578

CHAPTER 14

Chemical Kinetics

Applying What You've Learned

1

It takes as little as 5 mL (l tsp) of methanol to cause permanent blindness or death; and unlike ethanol, methanol can be absorbed in toxic amounts by ingestion, inhalation of vapor, or absorption through the skin. Nevertheless, methanol is present in a number of common household products including antifreeze, windshield-washing fluid, and paint remover. One method that has been used to synthesize methanol is the combination of carbon monoxide and hydrogen gases at 100 o e: CO(g)

1

+ 2H2(g) ---+. CH3 0H(g)

This reaction is catalyzed by a nickel compound.

Problems: [I~~

a)

Write the expression for the rate of this reaction in terms of [CO]. Problem 14.1]

b)

Write the expression for the rate of this reaction in terms of [H2 ] and in terms of [CH 3 0H]. [ ~~ Sample Problem 14.2]

c)

Sample

I

Given the following table of experimental data at 100°C, determine the rate law and the rate constant for the reaction. Then, determine the initial rate of the reaction when the starting concentration of CO is 16.5 M. [ ~. Sample Problem 14.3]

Experiment 1 2 3

[CO] (M) 5.60 5.60 11.2

[H2 ] (M)

11.2 22.4 11.2

Initial Rate (Mis) 0.952 0.952 1.90

d)

Calculate the time required for the concentration of CO to be reduced from 16.5 M to 1.91 M. [ ~~ Sample Problem 14.4]

e)

Calculate

f)

Given that k is 3.0 S-1 at 200°C, calculate Ea of the reaction. [ ~. Sample Problem 14.9]

g)

Use the calculated value of Ea to determine the value of k at 180°C. [I•• Sample Problem 14.10]

t ll2

of the reaction.

[ ~.

Sample Problem 14.6]

)IW I~

.t ; "

KEY WORDS

579

CHAPTER SUMMARY Section 14.1 •

•

The rate of a chemical reaction is the change in concentration of reactants or products over time. Rates may be expressed as an average rate over a given time interval or as an instantaneous rate. The rate constant (k) is a proportionality constant that relates the rate of reaction with the concentration(s) of reactant(s). The rate constant k for a given reaction changes only with temperature.

between temperature and the rate constant is expressed by the Arrhenius equation.

• Reactions occur when molecules of sufficient energy (and appropriate orientation) collide. Effective collisions are those that result in the formation of an activated complex, also called a transition state. Only effective collisions can result in product formation.

• The activation energy (Ea) is the minimum energy that colliding molecules must possess in order for the collision to be effective.

Section 14.2 • The rate law is an equation that expresses the relationship between rate and reactant concentration(s). In general, the rate law for the reaction of A and B is rate = k[Ay[BY

Section 14.5 • A reaction mechanism may consist of a series of steps, called elementary reactions. Unlike rate laws in general, the rate law for an elementary reaction can be written from the balanced equation, using the stoichiometric coefficient for each reactant species as its exponent in the rate law.

• The reaction order is the power to which the concentration of a given reactant is raised in the rate law equation. The overall reaction order is the sum of the powers to which reactant concentrations are raised in the rate law.

• A species that is produced in one step of a reaction mechanism and subsequently consumed in another step is called an intermediate. A species that is first consumed and later regenerated is called a catalyst. Neither intermediates nor catalysts appear in the overall balanced equation.

• The initial rate is the instantaneous rate of reaction when the reactant concentrations are starting concentrations. •

The rate law and reaction order must be determined by comparing changes in the initial rate with changes in starting reactant concentrations. In general, the rate law cannot be determined solely from the balanced equation.

•

The rate law of each step in a reaction mechanism indicates the molecularity or overall order of the step. A unimolecular step is first order, involving just one molecule; a bimolecular step is second order, involving the collision of two molecules; and a termolecular step is third order, involving the collision of three molecules. Termolecular processes are relatively rare.

•

If one step in a reaction is much slower than all the other steps, it is the rate-determining step. The rate-determining step has a rate law identical to the experimental rate law.

Section 14.3 •

•

•

The integrated rate law can be used to determine reactant concentrations after a specified period of time. It can also be used to determine how long it will take to reach a specified reactant concentration. The rate of afirst-order reaction is proportional to the concentration of a single reactant. The rate of a second-order reaction is proportional to the product of two reactant concentrations ([A][B]) , or on the concentration of a single reactant squared ([A] 2 or [B]z). The rate of a zeroth-order reaction does not depend on reactant concentration. The half-life (t1l2) of a reaction is the time it takes for half of a reactant to be consumed. The half-life is constant for first-order reactions, and it can be used to determine the rate constant of the reaction.

Section 14.4 •

Section 14.6 •

A catalyst speeds up a reaction, usually by lowering the value of the activation energy. Catalysis refers to the process by which a catalyst increases the reaction rate.

•

Catalysis may be heterogeneous, in which the catalyst and reactants exist in different phases, or homogeneous, in which the catalyst and reactants exist in the same phase.

•

Enzymes are biological catalysts with high specificity for the reactions that they catalyze.

Collision theory explains why the rate constant, and therefore the reaction rate, increases with increasing temperature. The relationship

KEyWORDS Activated complex, 564

Enzymes, 575

Intermediate, 568

Second-order reaction, 560

Activation energy (Ea)' 563

First-order reaction, 555

Molecularity, 568

Termolecular, 569

Arrhenius equation, 564

Half-life (t IlZ), 558

Rate constant (k), 547

Transition state, 564

Bimolecular, 569

Heterogeneous catalysis, 574

Rate-determining step, 569

Unimolecular, 569

Catalyst, 573

Homogeneous catalysis, 575

Rate law, 551

Zeroth-order reaction, 561

Collision theory, 562

Initial rate, 551

Rate of reaction, 548

Effective collision, 562

Instantaneous rate, 546

Reaction mechanism, 568

Elementary reaction, 568

Integrated rate law, 556

Reaction order, 551

580

CHAPTER 14

Chemical Kinetics

KEY EQUATIONS ----

-- ..

-. -- ~~

14.1

1 Ll[Al rate = - a !:::..t

14.2

rate

14.3

I [All = -kt n [Alo

14.4

In [All = -kt

=

!:::.. [B 1 !:::.. t

1 Ll[ C] c!:::..t

1 Ll[Dl d !:::..t

k[Ay[B Y

+ In [Alo

0.693

14.5

t1/2 =

14.6

1 = kt [All

14 .7

1 b

k

+

1 [Alo

1 t1/2

=

k[Alo

14.8

k = Ae -E.jRT

14.9

Ea In k = InA ----=RT

14.10 14.11

QUESTIONS AND PROBLEM=S= = = == = = = = = = Section 14.1: Reaction Rates

14.7

Consider the reaction 2NO(g)

Review Questions 14.1

What is meant by the rate of a chemical reaction? What are the units of the rate of a reaction?

14.2

Distinguish between average rate and instantaneous rate. Which of the two rates gives us an unambiguous measurement of reaction rate ? Why?

14.3

What are the advantages of measuring the initial rate of a reaction?

14.4

Identify two reactions that are very slow (take days or longer to complete) and two reactions that are very fast (reactions that are over in minutes or seconds).

+

0 2(g) - - + . 2N0 2(g)

Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.066 MIs. (a) At what rate is N0 2 being formed? (b) At what rate is molecular oxygen reacting? 14.8

Consider the reaction

Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0.082 MIs. (a) At what rate is ammonia being formed? (b) At what rate is molecular nitrogen reacting?,

Problems

Section 14.2: Dependence of Reaction Rate on Reactant Concentration

14.5

Review Questions

Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products: (a) H 2 (g) + I2(g) • 2HI(g) (b) 5Br-(aq) + Br0 3(aq) + 6H+(aq) - _ . 3Brzeaq)

14.6

+

3H2 0(l)

Write the reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products: (a) 2H2(g) + 0 2(g) - _ . 2H 20(g) (b) 4NH 3(g) + 50 2(g) • 4NO(g)

+

6H zO(g)

14.9

Explain what is meant by the rate law of a reaction.

14.10

Explain what is meant by the order of a reaction.

14.11

What are the units for the rate constants of first-order and secondorder reactions?

14.12

• product. (a) Write Consider the zeroth-order reaction: A the rate law for the reaction. (b) What are the units for the rate constant? (c) Plot the rate of the reaction versus [A l.

14.13

The rate constant of a first-order reaction is 66 rate constant in units of minutes?

S- l

What is the

QUESTIONS AND PROBLEMS

14.14

On which of the following properties does the rate constant of a reaction depend: (a) reactant concentrations, (b) nature of reactants, (c) temperature?

14.22

581

The following gas-phase reaction was studied at 290°C by observing the change in pressure as a function of time in a constant-volume vessel:

Problems

14.15

Determine the order of the reaction and the rate constant based on the following data.

The rate law for the reaction NH ~ (aq)

+ NO;-(aq) ----+. N 2(g) + 2H 20(l)

Time (s)

o

is given by rate = k[NH ~ ][NO;-]. At 25°C, the rate constant is 3.0 X 1O-4 IM' s. Calculate the rate of the reaction at this temperature if [NH ~ l = 0.36 M and [N0 2 l = 0.075 M. 14.16

Use the data in Table 14.2 to calculate the rate of the reaction at the time when [F2l = 0.020 M and [CI0 2l = 0.035 M.

14.17

Consider the reaction A

From the following data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant.

14.18

[Bl (M)

1.50 1.50 3.00

1.50 2.50 1.50

Rate (Mis) 3.20 X 10- 1 3.20 X 10- 1 6.40 X 10- 1

Consider the reaction

X +Y

From the following data, obtained at 360 K , (a) determine the order of the reaction, and (b) determine the initial rate of disappearance of X when the concentration of X is 0.30 M and that ofY is 0.40 M.

14.19

14.20

where P is the total pressure.

Review Questions 14.23

Write an equation relating the concentration of a reactant A at t = 0 to that at t = t for a first-order reaction. Define all the terms, and give their units. Do the same for a second-order reaction.

14.24

Define half-life. Write the equation relating the half-life of a first-order reaction to the rate constant.

14.25

Write the equations relating the half-life of a second-order reaction to the rate constant. How does it differ from the equation for a first-order reaction?

14.26

For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in telms of the half-life (tll2) and in terms of the rate constant k.

•Z

Initial Rate of Disappearance of X (Mis) 0.053 0.127 1.02 0.254 0.509

[Xl (M) 0.10 0.20 0.40 0.20 0.40

[Yl (M) 0.50 0.30 0.60 0.60 0.30

Determine the overall orders of the reactions to which the following rate laws apply: (a) rate = k[N0 2l 2 , (b) rate = k, (c) rate = k[H2lz[Brzlll2, (d) rate = k[NOlz[Ozl.

Problems

14.27

What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in 60 min? Assume first-order kinetics.

14.28

The thermal decomposition of phosphine (PH 3 ) into phosphorus and molecular hydrogen is a first-order reaction:

The half-life of the reaction is 35.0 s at 680°e. Calculate (a) the first-order rate constant for the reaction and (b) the time required for 95 percent of the phosphine to decompose.

Consider the reaction •

A--+'B The rate of the reaction is 1.6 X lO- z Mis when the concentration of A is 0.15 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.

14.21

14.29

Determine the order of the reaction and the rate constant based on the following pressures, which were recorded when the reaction was carried out at 430°C in a constant-volume vessel. P C 4H S (mmHg)

0 2,000 4,000 6,000 8,000 10,000

400 316 248 196 155 122

The rate constant for the second-order reaction 2NOBr(g) --+. 2NO(g)

+ BrzCg)

is 0.801M . s at 10 0 e. (a) Starting with a concentration of 0.086 M, calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when [NOBrlo = 0.072 M and [NOBrlo = 0.054 M.

Cyclobutane decomposes to ethylene according to the equation

Time (s)

15.76 18.88 22.79 27.08

Section 14.3: Dependence of Reactant Concentration on Time

+ B --+. products

[Al (M)

181 513 1164

P(mmHg)

14.30

The rate constant for the second-order reaction

is 0.541M . sat 300°e. How long (in seconds) would it take for the concentration of NO z to decrease from 0.65 M to 0.18 M?

582

14.31

CHAPTER 14

Chemical Kinetics

The second-order rate constant for the dimerization of a protein (P)

p +p

Problems 14.41

• P2

is 6.2 X 1O - 31M . s at 25°C. If the concentration of the protein is 2.7 X 10- 4 M , calculate the initial rate (M is) of formation of P2 . How long (in seconds) will it take to decrease the concentration of P to 2.7 X 10- 5 M ? 14.32

is given in the following table. Determine graphically the activation energy for the reaction.

T(K)

Consider the first-order reaction X • Y shown here. (a) What is the half-life of the reaction? (b) Draw pictures showing the number of X (red) and Y (blue) molecules at 20 s and at 30 s.

•• ••• • ••• • • •• •• •• ••• • •••• ••• • •

14.42

(ii)

14.43

The rate constant of a first-order reaction is 4.60 X 10- 4 S-1 at 350°C. If the activation energy is 104 kJ/mol, calculate the temperature at which its rate constant is 8.80 X 10- 4 s - 1.

14.45

The rate constants of some reactions double with every 10° rise in temperature. Assume that a reaction takes place at 295 K and 305 K. What must the activation energy be for the rate constant to double as described?

14.46

The rate at which tree crickets chirp is 2.0 X 102 per minute at 27°C but only 39.6 per minute at 5°C. From these data, calculate the "activation energy" for the chirping process. (Hint: The ratio of rates is equal to the ratio of rate constants.)

14.47

The rate of bacterial hydrolysis of fish muscle is twice as great at 2.2°C as at -1.1 0C. Estimate an Ea value for this reaction. Is there any relation to the problem of storing fish for food?

14.48

The activation energy for the denaturation of a protein is 39.6 kJ/mol. At what temperature will the rate of denaturation be 20 percent greater than its rate at 25 °C?

14.49

Diagram A describes the initial state of reaction

(iii)

14.34

Define activation energy. What role does activation energy play in chemical kinetics?

14.35

Write the Arrhenius equation, and define all terms.

14.36

Use the Arrhenius equation to show why the rate constant of a reaction (a) decreases with increasing activation energy and (b) increases with increasing temperature.

14.37

The burning of methane in oxygen is a highly exothermic reaction. Yet a mixture of methane and oxygen gas can be kept indefinitely without any apparent change. Explain.

14.38

Sketch a potential energy versus reaction progress plot for the following reactions:

14.40

For the reaction

14.44

Review Questions

14.39

Given the same reactant concentrations, the reaction

the frequency factor A is 8.7 X 10 12 S - l and the activation energy is 63 kJ/mol. What is the rate constant for the reaction at 75°C?

Section 14.4: Dependence of Reaction Rate on Temperature

(a) S(s) + 0 2(g) • S02(g) (b) CI2 (g) • CI(g) + CI(g)

k (8 - 1) 1.74 X 10- 5 6.61 X 10- 5 2.51 X 10- 4 7.59 X 10- 4 2.40 X 10- 3

at 250°C is 1.50 X 103 times as fast as the same reaction at 150°C. Calculate the activation energy for this reaction. Assume that the frequency factor is constant.

The reaction A - -.... B shown here follows first-order kinetics. Initially different amounts of A molecules are placed in three containers of equal volume at the same temperature. (a) What are the relative rates of the reaction in these three containers? (b) How would the relative rates be affected if the volume of each container were doubled? (c) What are the relative half-lives of the reactions in (i) to (iii)?

(i)

298 308 318 328 338

t = 10 s

t = 0s

14.33

Variation of the rate constant with temperature for the first-order reaction

I1Ho = -296 kJ/mol I1W = 243 kJ/mol

The reaction H + H2 • H2 + H has been studied for many years. Sketch a potential energy versus reaction progress diagram for this reaction. Over the range of about +3 °C from normal body temperature, the metabolic rate, M T, is given by M T = M 37 ( 1.1 )';T, where M 37 is the normal rate (at 37°C) and I1T is the change in T. Discuss this equation in terms of a possible molecular interpretation.

Diagram A Suppose the reaction is carried out at two different temperatures as shown in diagram B. Which picture represents the result at the higher temperature? (The reaction proceeds for the same amount of time at both temperatures.)

QUESTIONS AND PROBLEMS

14.59

583

The rate law for the reaction 2NO(g)

+ CI2 (g) -----. 2NOCI(g)

is given by rate = k[NO][CI 2]. (a) What is the order of the reaction? (b) A mechanism involving the following steps has been proposed for the reaction: (b)

(a)

NO(g) + Clig) -----. NOCI 2 (g) • 2NOCI(g) NOCI 2(g) + NO(g)

Diagram B

If this mechanism is correct, what does it imply about the relative rates of these two steps ?

Section 14.5: Reaction Mechanisms Review Questions 14.50

What do we mean by the mechanism of a reaction?

14.51

What is an elementary step? What is the molecularity of a reaction?

14.52

Classify the following elementary reactions as unimolecular, bimolecular, or termolecular: (a) 2NO + Br2 - _ . 2NOBr (b) CH3NC • CH 3CN (c) SO + O 2 • SOz + 0

14.53

14.54

14.60

For the reaction X z + Y + Z • XY + XZ, it is found that doubling the concentration of X 2 doubles the reaction rate, tripling the concentration ofY triples the rate, and doubling the concentration of Z has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of Z has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

14.61

The rate law for the reaction

is rate = k[Hz][NOf Which of the following mechanisms can be ruled out on the basis of the observed rate expression?

Reactions can be classified as unimolecular, bimolecular, and so on. Why are there no zero-molecular reactions? Explain why termolecular reactions are rare.

Mechanisml H2 + NO - _ . H20 + N N + NO • Nz + 0 0+ Hz • HzO

Determine the molecularity, and write the rate law for each of the following elementary steps:

Mechanism II Hz + 2NO - _ . NzO + HzO • N2 + H2 0 NzO + Hz

(a) X - _ . products (b) X + Y • products (c) X + Y + Z • products • products (d) X + X (e) X + 2Y • products 14.55

14.56

(slow) (fast) (fast)

Mechanism III 2NO :;:.==' N 20 2 N20 2 + H 2 • N 2 0 + HzO N 20 + Hz • N z + H2 0

What is the rate-determining step of a reaction? Give an everyday analogy to illustrate the meaning of rate determining. The equation for the combustion of ethane (C zH 6 ) is

14.62

(slow) (fast)

(fast equilibrium) (slow) (fast)

The rate law for the decomposition of ozone to molecular oxygen

•

IS

Explain why it is unlikely that this equation also represents the elementary step for the reaction.

rate

[0 3]2

=

k [0

] 2

14.57

Specify which of the following species cannot be isolated in a reaction: activated complex, product, intermediate.

The mechanism proposed for this process is k,

Problems

0 3 ' k-I '0+0 2

14.58

0 + 03

Classify each of the following elementary steps as unimolecular, bimolecular, or termolecular.

+

+

•

20 z

Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing O 2 concentration.

+

•

k2

(a)

Section 14.6: Catalysis

+

•

Review Questions

(b)

+

• (c)

+

14.63

How does a catalyst increase the rate of a reaction?

14.64

What are the characteristics of a catalyst?

14.65

A certain reaction is known to proceed slowly at room temperature. Is it possible to make the reaction proceed at a faster rate without changing the temperature?

584

CHAPTER 14

Chemical Kinetics

14.66

Distinguish between homogeneous catalysis and heterogeneous catalysis.

14.67

Are enzyme-catalyzed reactions examples of homogeneous or heterogeneous catalysis? Explain.

14.68

The concentrations of enzymes in cells are usually quite small. What is the biological significance of this fact ?

14.69

When fruits such as apples and pears are cut, the exposed areas begin to turn brown. This is the result of an enzyme-catal yzed reaction. Often the browning can be prevented or slowed by adding a few drops of lemon juice. What is the chemical basis of this treatment?

14.70

14.77

The following pictures represent the progress of the reaction A • B where the red spheres represent A molecules and the green spheres represent B molecules. Calculate the rate constant of the reaction.

•• • •• • • • • ••• t

14.78

The first-order rate constant for the dehydration of carbonic acid

=

• •

• ••

t = 20 s

0s

t

=

40 s

The following pictures show the progress of the reaction • A2. Determine whether the reaction is first order or 2A second order, and calculate the rate constant.

is about 1 X 102 S - I. In view of this rather high rate constant, explain why it is necessary to have the enzyme carbonic anhydrase to enhance the rate of dehydration in the lungs.

•

Problems 14.71

14.72

Most reactions, including enzyme-catalyzed reactions, proceed faster at higher temperatures. However, for a given enzyme, the rate drops off abruptly at a certain temperature. Account for this behavior.

ES

Use the data in Sample Problem 14.5 to determine graphically the half-life of the reaction.

14.80

The following data were collected for the reaction between hydrogen and nitric oxide at 700°C: 2H2(g)

(fast equilibrium) k2

•

E + P

(slow)

Derive an expression for the rate law of the reaction in terms of the concentrations of E and S. (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.)

Additional Problems List four factors that influence the rate of a reaction.

14.74

Suggest experimental means by which the rates of the following reactions could be followed: (a) CaC0 3 (s) • CaO(s) + CO 2(g) (b) CI 2 (g) + 2Br-(aq) • Br2(aq) + 2Cqaq) (c) C 2H 6 (g) • C 2H 4 (g) + H 2 (g) (d) C2HsI(g) + H 20(l) • C 2H sOH(aq) + H +(aq) + r-(aq)

• 2H 20(g)

Experiment

[H2 ] (M)

[NO] (M)

1 2 3

0.010 0.0050 0.010

0.025 0.025 0.0125

+ N 2(g) Initial Rate (MIs) 24 X 10- 6 1.2 X 10- 6 0.60 X 10- 6

When methyl phosphate is heated in acid solution, it reacts with water:

If the reaction is carried out in water enriched with 18 0, the oxygen-18 isotope is found in the phosphoric acid product but not in the methanol. What does this tell us about the mechanism of the reaction? 14.82

The rate of the reaction

'The rate constant for the reaction N0 2 (g)

CH3COOH(aq)

+ CO(g) ---+. NO(g) + CO 2 (g)

is 1.64 X 10- 1M . s." What is incomplete about this statement? In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.0 cm 3 Calculate the surface area of the catalyst. If the sphere is broken down into eight smaller spheres, each having a volume of 1.25 cm 3, what is the total surface area of the spheres? Which of the two geometric configurations of the catalyst is more effective? (The surface area of a sphere is 41Tr2, where r is the radius of the sphere.) Based on your analysis here, explain why it is sometimes dangerou s to work in grain elevators.

+ C 2HsOH(aq)

shows first-order characteristics-that is, rate = k[CH 3COOC 2H s]- even though this is a second-order reaction (first order in CH 3COOC 2H s and first order in H 2 0). Explain.

6

14.76

+ 2NO(g)

(a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume that the oxygen atom is the intermediate. ) 14.81

14.73

t = 30 min

14.79

Consider the following mechanism for the enzyme-catalyzed reactIOn:

E+S

14.75

t = 15 min

t=Omin

14.83

Explain why most metals used in catalysis are transition metals.

14.84

The reaction 2A + 3B • C is first order with respect to A and B. When the initial concentrations are [A] = 1.6 X 10- 2 M and [BJ = 2.4 X 10- 3 M, the rate is 4.1 X 10- 4 MIs. Calculate the rate constant of the reaction.

QUESTIONS AND PROBLEMS

14.85

The bromination of acetone is acid-catalyzed:

CH 3COCH 3

+ Brz

14.93

The reaction of G2 with ~ to form 2EG is exothermic, and the reaction of Gz with X z to form 2XG is endothermic. The activation energy of the exothermic reaction is greater than that of the endothermic reaction. Sketch the potential energy profile diagrams for these two reactions on the same graph.

14.94

In the nuclear industry, workers use a rule of thumb that the radioactivity from any sample will be relatively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after this time period. (Hint: Radioactive decays obey first-order kinetics.)

14.95

Briefly comment on the effect of a catalyst on each of the following: (a) activation energy, (b) reaction mechanism, (c) enthalpy of reaction, (d) rate of forward reaction, (e) rate of reverse reaction.

14.96

When 6 g of granulated Zn is added to a solution of 2 M HCI in a beaker at room temperature, hydrogen gas is generated. For each of the following changes (at constant volume of the acid) state whether the rate of hydrogen gas evolution will be increased , decreased, or unchanged: (a) 6 g of powdered Zn is used, (b) 4 g of granulated Zn is used, (c) 2 M acetic acid is used instead of 2 M HCl, (d) temperature is rai sed to 40°C.

14.97

Strictly speaking, the rate law derived for the reaction in Problem 14.80 applies only to certain concentrations of Hz. The general rate law for the reaction takes the form

catalyst H+

--=.:_. CH COCH2Br + H + + Br3

The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine, and H + ions at a certain temperature: [H+] Rate of Disappearance [CH3 COCH 3] [Br 2] (M) (M) (M) ofBr2 (Mis) 5.7 X lO- s (1) 0.30 0.050 0.050 5.7 X lO- s (2) 0.30 0.10 0.050 1.2 X 10- 4 (3) 0.30 0.050 0.10 3.1 X 10- 4 (4) 0.40 0.050 0.20 7.6 X lO- s (5) 0.40 0.050 0.050 (a) What is the rate law for the reaction? (b) Determine the rate constant. (c) The following mechanism has been proposed for the reaction:

585

(fast equilibrium)

(slow)

fa te =

Show that the rate law deduced from the mechanism is consistent with that shown in part (a). 14.86

14.87

The decomposition of N 20 to N2 and O2 is a first-order reaction. At 730°C the half-life of the reaction is 3.58 X 103 min. If the initial pressure of N 2 0 is 2.10 atm at 730°C, calculate the total gas pressure after one half-life. Assume that the volume remains constant. The reaction S20~- + 21• 2S0~ - + 12 proceeds slowly in aqueous solution, but it can be catalyzed by the Fe 3 + ion. Given that Fe 3 + can oxidize 1- and Fe 2+ can reduce SzO ~-, write a plausible two-step mechanism for this reaction. Explain why the uncatalyzed reaction is slow.

14.88

What are the units of the rate constant for a third-order reaction?

14.89

The integrated rate law for the zeroth-order reaction A •B is [Al, = [Ala - kt. (a) Sketch the following plots: (i) rate versus [Al, and (ii) [Al, versus t. (b) Derive an expression for the halflife of the reaction. (c) Calculate the time in half-lives when the integrated rate law is no longer valid, that is, when [Al, = o.

14.90

14.91

14.92

A flask contains a mixture of compounds A and B. Both compounds decompose by first -order kinetics. The half-lives are 50.0 min for A and 18.0 min for B. If the concentrations of A and B are equal initially, how long will it take for the concentration of A to be four times that of B ? Referring to Sample Problem 14.5, explain how you would measure the partial pressure of azomethane experimentally as a function of time. The rate law for the reaction 2N0 2 (g) • N 20 4 (g) is rate = k[N0 2f Which of the following changes will change the value of k? (a) The pressure of NO z is doubled. (b) The reaction is run in an organic solvent. (c) The volume of the container is doubled. (d) The temperature is decreased. (e) A catalyst is added to the container.

k)[NO]z[Hzl --'-'--=-=-==-~ 1 + kz[Hzl

where k) and k2 are constants. Derive rate law expressions under the conditions of very high and very low hydrogen concentrations. Does the result from Problem 14.80 agree with one of the rate expressions here? 14.98

A certain first-order reaction is 35.5 percent complete in 4.90 min at 25 °C. What is its rate constant?

14.99

The decomposition of dinitrogen pentoxide has been studied in carbon tetrachloride solvent (CCI 4 ) at a certain temperature: 2N2O S

[NzOs] (M) 0.92 l.23 1.79 2.00 2.21

• 4N0 2 + O 2 Initial Rate (Mis) 0.95 X lO- s l.20 X lO- s l.93 X lO - s 2.10 X lO- s 2.26 X 10- 0

Determine graphically the rate law for the reaction, and calculate the rate constant. 14.100 The thermal decomposition of NzOs obeys first-order kinetics. At 45°C, a plot of In [N20 sl versus t gives a slope of - 6.18 X 10- 4 min -1 What is the half-life of the reaction? 14.101 When a mixture of methane and bromine is exposed to light, the following reaction occurs slowly:

Suggest a reasonable mechanism for this reaction . (Hint: Bromine vapor is deep red; methane is colorless.) 14.102 The rate of the reaction between H2 and 12 to form HI (discussed on page 571 ) increases with the intensity of visible light. (a) Explain why this fact suppo'rts the two-step mechanism given. (1z vapor is purple.) (b) Explain why the visible light has no effect on the formation of H atoms.

586

CHAPTER 14

Chemical Kinetics

14.103 To prevent brain damage, a standard procedure is to lower the body temperature of someone who has been resuscitated after suffering cardiac arrest. What is the physiochemical basis for this procedure? 14.104 In Lewis Carroll's Through the Looking Glass, Alice wonders whether looking-glass milk on the other side of the mirror would be fit to drink. What do you think?

14.105 Consider the following elementary step:

x+

2Y--· XY 2

(a) Write a rate law for this reaction. (b) If the initial rate of formation of XY2 is 3.8 X 10- 3 Mis and the initial concentrations of X and Yare 0.26 M and 0.88 M , respectively, what is the rate constant of the reaction? 14.106 In recent years, ozone in the stratosphere has been depleted at an alarmingly fast rate by chlorofluorocarbons (CFCs). A CFC molecule such as CFCl3 is first decomposed by UV radiation: CFCI 3 - _ . CFCI2 + Cl

14.111 (a) What can you deduce about the activation energy of a reaction if its rate constant changes significantly with a small change in temperature? (b) If a bimolecular reaction occurs every time an A and a B molecule collide, what can you say about the orientation factor and activation energy of the reaction? 14.112 The rate law for the following reaction

CO(g)

+ N0 2(g) --+. COig) + NO(g)

is rate = k[N0 2f Suggest a plausible mechanism for the reaction, given that the unstable species N0 3 is an intermediate.

14.113 Radioactive plutonium-239 (t' /2 = 2.44 X 105 yr) is used in nuclear reactors and atomic bombs. If there are 5.0 X 102 g of the isotope in a small atomic bomb, how long will it take for the substance to decay to 1.0 X 102 g, too small an amount for an effective bomb? 14.114 Many reactions involving heterogeneous catalysts are zeroth order; that is, rate = k. An example is the decomposition of phosphine (PH)) over tungsten (W):

The chlorine radical then reacts with ozone as follows: Cl + 0 3 - _ . CIO+ O2 CIO + 0 • Cl + O2 (a) Write the overall reaction for the last two steps. (b) What are the roles of CI and CIO? (c) Why is the fluorine radical not important in this mechanism? (d) One suggestion to reduce the concentration of chlorine radicals is to add hydrocarbons such as ethane (C 2H 6) to the stratosphere. How will this work? (e) Draw potential energy versus reaction progress diagrams for the uncatalyzed and catalyzed (by Cl) destruction of ozone: 0 3 + o • 20 2, Use the thermodynamic data in Appendix 2 to determine whether the reaction is exothermic or endothermic.

14.107 Chlorine oxide (ClO), which plays an important role in the depletion of ozone (see Problem 14.106), decays rapidly at room temperature according to the equation

From the following data, determine the reaction order and calculate the rate constant of the reaction.

Time (s) 0.12 X 10- 3 0.96 X 10- 3 2.24 X 10- 3 3.20 X 10- 3 4.00 X 10- 3

It is found that the reaction is independent of [PH 3] as long as phosphine's pressure is sufficiently high (> I atm). Explain.

14.115 Thallium(I) is oxidized by cerium(IV) as follows:

The elementary steps, in the presence of Mn(II), are as follows: Ce4+ + Mn2+ --+. Ce3+ + Mn3+ Ce4+ + Mn3+ TI + + Mn4+

(a) Identify the catalyst, intermediates, and the rate-determining step if the rate law is rate = k[Ce 4 +][Mn2+ ]. (b) Explain why the reaction is slow without the catalyst. (c) Classify the type of catalysis (homogeneous or heterogeneous). 14.116 Sucrose (C' 2H2201l )' commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose (C 6H I2 0 6) and glucose (C6H'206): fructose

[CIO] (M)

8.49 7.10 5.79 5.20 4.77

X 10- 6

10- 6 X 10- 6 X 10- 6 X 10- 6 X

14.108 A compound X undergoes two simultaneous first-order reactions as follows: X • Y with rate constant k, and X • Z with rate constant k 2. The ratio of k,lk2 at 40°C is 8.0. What is the ratio at 300°C? Assume that the frequency factors of the two reactions are the same.

14.109 Consider a car fitted with a catalytic converter. The first 5 min or so after it is started are the most polluting. Why? 14.110 The following scheme in which A is converted to B, which is then converted to C, is known as a consecutive reaction. A--· B--. C

Assuming that both steps are first order, sketch on the same graph the variations of [A] , [B], and [C] with time.

• Ce3+ + Mn4+ • TI3+ + Mn2 +

glucose

This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include [H20 ] even though water is a reactant.

Time (min)

o 60.0 96.4 157.5

0.500 0.400 0.350 0.280

14.117 The first-order rate constant for the decomposition of dimethyl ether is 3.2 X 10- 4 s-' at 450°C. The reaction is carried out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is 0.350 atm. What is the pressure of the system after 8.0 min? Assume ideal behavior.

QUESTIONS AND PROBLEMS

14.118 At 25°C, the rate constant for the ozone-depleting reaction is 7.9 X 10- 15 cm 3/molecule . s. Express the rate constant in units of I IM . s.

14.119 Consider the following elementary steps for a consecutive reaction: A k,. B k, . C (a) Write an expression forthe rate of change of B. (b) Derive an expression for the concentration of B under "steady-state" conditions; that is, when B is decomposing to C at the same rate as it is formed from A. 14.120 Ethanol is a toxic substance that, when consumed in excess, can impair respiratory and cardiac functions by interference with the neurotransmitters of the nervous system. In the human body, ethanol is metabolized by 'the enzyme alcohol dehydrogenase to acetaldehyde, which causes hangovers. Based on your knowledge of enzyme kinetics, explain why binge drinking (that is, consuming too much alcohol too fast) can prove fatal.

14.121 Strontium-90, a radioactive isotope, is a major product of an atomic bomb explosion. It has a half-life of 28.1 yr. (a) Calculate the first-order rate constant for the nuclear decay. (b) Calculate the fraction of 90Sr that remains after 10 half-lives. (c) Calculate the number of years required for 99 .0 percent of 90Sr to disappear. 14.122 Consider the potential energy profiles for the following three reactions (from left to right). (1) Rank the rates (slowest to fastest) of the reactions. (2) Calculate t:.H for each reaction, and determine which reaction(s) are exothermic and which reaction(s) are endothermic. Assume the reactions have roughly the same frequency factors. 50 kJ/mol

40 kJ/mol

30 kJ/mol

.--'" ~

p.

14.125 The activation energy for the decomposition of hydrogen peroxide

is 42 kJ/mol, whereas when the reaction is catalyzed by the enzyme catalase, it is 7.0 kJ/mol. Calculate the temperature that would cause the uncatalyzed decomposition to proceed as rapidly as the enzyme-catalyzed decomposition at 20°e. Assume the frequency factor A to be the same in both cases. 14.126 The activity of a radioactive sample is the number of nuclear disintegrations per second, which is equal to the first-order rate constant times the number of radioactive nuclei present. The fundamental unit of radioactivity is the curie (Ci), where 1 Ci cotTesponds to exactly 3.70 X 10 10 disintegrations per second. This decay rate is equivalent to that of 1 g of radium-226. Calculate the rate constant and half-life for the radium decay. Starting with 1.0 g of the radium sample, what is the activity after 500 yr? The molar mass of Ra-226 is 226.03 g/mol.

14.127 To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to fotTll oxyhemoglobin (Hb0 2) according to the simplified equation Hb(aq)

+ 02(aq)

k . Hb0 2(aq)

where the second-order rate constant is 2.1 X 1061M' sat 37°C. For an average adult, the concentrations of Hb and O2 in the blood at the lungs are 8.0 X 10- 6 M and 1.5 X 10- 6 M, respectively. (a) Calculate the rate of formation of Hb0 2 . (b) Calculate the rate of consumption of Oz. (c) The rate of formation of HbO z increases to 1.4 X 10- 4 Mis during exercise to meet the demand of the increased metaboli sm rate. Assuming the Hb concentration to remain the same, what must the oxygen concentration be to sustain this rate of Hb0 2 formation? 14.128 At a certain elevated temperature, ammonia decomposes on the surface of tungsten metal as follow s:

t

t

7

c

1l o

587

20 kJ/mol

- 20 kJ/mol ">.

~ c K

The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds in the forward direction (from left to right). The initial concentrations are equilibrium concentrations. The system is already at equilibrium, and there will be no net movement in either direction. The ratio of initial concentrations of products to reactants is too large. To reach equilibrium, products must be converted to reactants, The system proceeds in the reverse direction (from right to left).

Sample Problem 15.7 shows how the value of Q is used to determine the direction of a reaction that is not at equilibrium.

Sample Problem 15.7 At 375°C, the equilibrium constant for the reaction

is 1.2. At the start of a reaction, the concentrations of N2 , Hz, and NH3 are 0.071 M, 9.2 X 10- 3 M, and 1.83 X 10- 4 M, respectively. Determine whether this system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.

Strategy Use the initial concentrations to calculate Qc, and then compare Qc with Kc.

\

\

SECTION 15.4

Using Equilibrium Expressions to Solve Problems

Setup Think About It In proceeding

(1.83 X 1O-4 )z

-

-

------,---,- = 0.61

(0.071)(9.2

X

10- 3) 3

Solution The calculated value of Qc is less than Kc. Therefore, the reaction is not at equilibrium and

must proceed to the right to establish eqUilibrium.

Practice Problem A The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric

oxide and chlorine,

+ Clz(g) +=,~. 2NOCl(g)

2NO(g)

is 6.5 X 104 at 35°C. In which direction will the reaction proceed to reach equilibrium if the starting concentrations of NO, Clz, and NOCI are 1.1 X 10- 3 M, 3.5 X 10- 4 M, and 1.9 M, respectively? •

Practice Problem B Calculate Kp for the formation of nitrosyl chloride from nitric oxide and

chlorine at 35°C, and determine whether the reaction will proceed to the right or the left to achieve equilibrium when the starting pressures are P NO = 1.01 atm, P CI = 0.42 atm, and P NOCI = 1.76 atm. 2

Calculating Equilibrium Concentrations If we know the equilibrium constant for a reaction, we can calculate the concentrations in the equi1ibrium mixture from the initial reactant concentrations. Consider the following system involving two organic compounds, cis- and trans-stilbene:

,

/

,

H

l =C H

cis-Stilbene

trans-Stilbene

The equilibrium constant (Ke) for this system is 24.0 at 200°C. If we know that the starting concentration of cis-stilbene is 0.850 M, we can use the equilibrium expression to determine the equilibrium concentrations of both species. The stoichiometry of the reaction tells us that for every mole of cis-stilbene converted, 1 mole of trans-stilbene is produced. We will let x be the equilibrium concentration of trans-stilbene in mollL; therefore, the equilibrium concentration of cisstilbene must be (0.850 - x) mollL. It is useful to summarize these changes in concentrations in an equilibrium table: cis-stilbene +,==' trans-stilbene

0.850

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

o

-x

+x

0.850 - x

x

We then use the equilibrium concentrations, defined in terms of x, in the equilibrium expression:

Ke =

24.0 = x

[trans-stilbene]

.

[cis-stIlbene]

x 0.850 - x

= 0.816 M

to the right, a reaction consumes reactants and produces more products. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium will be established.

607

608

CHAPTER 15

Chemical Equilibrium

Having solved for x, we calculate the equilibrium concentrations of cis- and trans-stilbene as follows: [cis-stilbene] = (0.850 - x) M = 0.034 M [trans-stilbene]

= x M = 0.816 M

A good way to check the answer to a problem such as this is to use the calculated equilibrium concentrations in the equilibrium expression and make sure that we get the correct Kc value.

K = 0.816 = 24 c 0.034 Sample Problems 15.8 and 15.9 provide additional examples of this kind of problem.

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,

is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and 12') Strategy Construct an equilibrium table to determine the equilibrium concentration of each species in terms of an unknown (x ); solve for x, and use it to calculate the equilibrium molar concentrations. Setup Insert the starting concentrations that we know into the equilibrium table:

H 2(g)

Initial concentration (M):

+ I 2 (g) :;::.=::!:' 2HI(g)

0.240

0.240

0

Change in concentration (M): Equilibrium concentration (M): Solution We define the change in concentration of one of the reactants as x. Because there is no

product at the start of the reaction , the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations:

+ Initial concentration (M):

•

,

2HI(g)

o

0.240

--------~----------~--------

Change in concentration (M):

-x

-x

0.240 - x

0.240 - x

+ 2x

----------+---------------~-------

Equilibrium concentration (M):

2x

•

Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x . ?

[HIl~q

K

= c

Think About It Always check

your answer by inserting the calculated concentrations into the equilibrium expression:

[HIl~q

(0.378)2

[H2leq[I2leq

(0.05 1)2 = 54.9

= Kc

The small difference between the calculated Kc and the one given in the problem statement is due to rounding.

54.3

=-::---c=-':--

[H2leq[I2leq (2X) 2

= --c::--~--....,..-c::-::-cc:--­

(0.240 - x)(0.240 - x)

" 54.3 =

(0.240 - X)2

2x 0.240 - x

x = 0.189

Using the calculated value of x, we can determine the equilibrium concentration of each species as follows: [H2leq = (0.240 - x) M = 0.051 M [I2leq = (0.240 - x) M = 0.051 M [HIleq

=

2x

=

0.378 M

SECTION 15.4

Using Equilibrium Expressions to Solve Problems

Practice Problem A Calculate the equilibrium concentrations of H 2, 12, and HI at 430°C if the initial concentrations are [H2l; = [I2li = 0 M, and [HIli = 0.525 M. Practice Problem B Calculate the equilibrium concentrations of H2o 120 and HI at 430°C if the initial concentrations are [H2]; = [I2l; = 0.100 M, and [HIl; = 0.200 M.

For the same reaction and temperature as in Sample Problem 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2l; = 0.00623 M, [12]; = 0.00414 M, and [HIl; = 0.0424 M. Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the value of Kc (given in the problem statement of Sample Problem 15.8) to determine which direction the reaction will proceed in order to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations. · Setup

(0.0424i (0.00623)(0.00414) = 69.7

[HIl! [H 2];[I2l;

Therefore, Qc > Kc, so the system will have to proceed to the left (reverse) to reach equilibrium. The equilibrium table is H2(g)

Initial concentration (M):

+ I2(g) :;:,==' 2HI(g)

0.00623 0.00414

0.0424

Change in concentration (M): Equilibrium concentration (M): Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and 12 will increase. Therefore, the table should be filled in as follows:

+

Hig)

Initial concentration (M):

I 2(g)

,

2HI(g)

0.00623

0.00414

0.0424

+x

+x

-2x

Change in concentration (M): Equilibrium concentration (M):

,

0.00623

+x

0.00414

+x

0.0424 - 2x

Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.

54 .3 =

(0.0424 - 2xi --c::-::-::-::-'--::---,--,-::---::-::--'-:--c---c-

(0.00623 + x)(0.00414 + x)

It isn't possible to solve this equation the way we did in Practice Problem 15.8 (by taking the square root of both sides) because the concentrations of H2 and 12 are unequal. Instead, we have to carry out the multiplications, 54.3(2.58 X 10- 5

+

1.04 X 1O- 2x

+ x2)

3

= 1.80 X 10- -

1.70 X 1O- 1x

+ 4x2

Collecting terms we get

50.3x2

+ 0.735x

This is a quadratic equation of the form ax2 (see Appendix 1) is

- 4.00 X 10- 4 = 0

+ bx + c

=

O. The solution for the quadratic equation

-b ± ~b2 - 4ac x= 2a (Continued)

609

610

CHAPTER 15

Chemical Equilibrium Here we have a = 50.3, b = 0.735, and c = -4.00 X 1O~4, so

x=

-0.735 ± ~(0.735)2 -. 4(50.3)( - 4.00 X 1O ~4) 2(50.3)

x = 5.25

Think About It Checking this result gives

X 1O~4

or

x = -0.0151

Only the first of these values, 5.25 X 1O~4, makes sense because concentration cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows: [H2]eq = (0.00623+ x) M = 0.00676 M

[HI]2 Kc = [H 2][12]

[12]eq = (0.00414 + x) M = 0.00467 M

(0.0414i --:-.,.-:-::--:c---:---:-::--:-::-- = 54. 3 (0.00676)(0.00467)

[HI]eq = (0.0424 - 2x) = 0.0414 M

Practice Problem At 1280°C the equilibrium constant Kc for the reaction

Br2(g) :;::.=::t, 2Br(g) is 1.1 X 1O ~3 . If the initial concentrations are [Br2] = 6.3 X 1O~2 M and [Br] calculate the concentrations of these two species at equilibrium.

=

1.2 X 1O~2 M,

Here is a summary of the use of initial reactant concentrations to determine equilibrium concentrations:

If Q < K the reaction will occur as written. If Q > K the reverse reaction will occur.

1. Construct an equilibrium table, and fill in the initial concentrations (including any that are zero). 2. Use initial concentrations to calculate the reaction quotient, Q, and compare Q to K to deter.............. mine 'the 'cfirectloi11ll'whiCh 'th'e'reacdoil will proceed. 3. Define x as the amount of a particular species consumed, and use the stoichiometry of the reaction to define (in terIllS of x) the amount of other species consumed or produced. 4. For each species in the equilibrium, add the change in concentration to the initial concentration to get the equilibrium concentration. 5. Use the equilibrium concentrations and the equilibrium expression to solve for x. 6. Using the calculated value of x, determine the concentrations of all species at equilibrium. 7. Check your work by plugging the calculated equilibrium concentrations into the equilibrium expression. The result should be very close to the Kc stated in the problem. The same procedure applies to Kp. Sample Problem 15.10 shows how to solve an equilibrium problem using partial pressures .

.

Sample Problem 15.10 For this reaction, 1M to Kp.

=

0; therefore, Kc is equal

A mixture of 5.75 atm of H2 and 5.75 atm OfI2 is contained in a 1.0-L vessel at 430°C. The .......... eqiiiili:iii'uiii con·s·t'ant' (Kp) for the reaction

at this temperature is 54.3. Determine the equilibrium partial pressures of Hb Ib and HI. Strategy Construct an equilibrium table to determine the equilibrium partial pressures. Setup The equilibrium table is

H 2(g) Ini tial partial pressure (atm): Change in partial pressure (atm): Equilibrium partial pressure (atm):

+

12(g)

•

,

2HI(g)

5.75

5.75

0

- x

- x

+2x

5.75 - x

5.75 - x

2x

SECTION 15.5

Factors That Affect Chemical Equilibrium

611

Solution Setting the equilibrium expression equal to K p , 54.3 =

(2X)2

-~~--,

xi

(5 .75 Taking the square root of both sides of the equation gives --./54.3 = 7.369 =

2x 5.75 - x

Think About It Plugging the calculated partial pressures into the equilibrium expression gives

2x 5.75 - x

7.369(5.75 - x) = 2x

(9.04)2

-----:c =

42.37 - 7.369x = 2x

(1.23)2

42.37 = 9.369x

x

=

4.52

The equilibrium partial pressures are PH = PI = 5.75 - 4.52 = 1.23 atm, and Pm = 9.04 atm. 2

•

2

54.0

The small difference between this result and the equilibrium constant given in the problem statement is due to rounding.

Practice Problem A Determine the equilibrium partial pressures of H 2 , 120 and HI if we begin the experiment with 1.75 atm each of H2 and 12 at 430°C. ·················· · ············ ········ ····························· When a reaction starts with reactants and Practice Problem B. Determine the equilibrium partial pressures of H 2 , 12 , and HI if we begin the experiment with 2.75 atm HI at 430°C.

Checkpoint 15.4

Using Equilibrium Expressions to Solve Problems

Use the following information to answer questions 15.4.1 and 15.4.2: Kc for the reaction

A + B.

• 2C

is 1.7 X 10- 2 at 250°C. 15.4.1

What will the equilibrium concentration of A, B, and C be at this temperature if [Ali = [Bli = 0.750 M

. ([Cl

=

OJ?

15.4.2

What will the equilibrium concentrations of A , B, and C be at this temperature if [C]i = 0. 875 M ([Al i =

[Bl

=

OJ?

a) 6.1 X 10- 3 M, 6.1 X 10- 3 M, 0.092M

b) 0.41 M, 0.41 M , 0.054 M

b) 0.046 M, 0.046 M, 0.092 M

c) 0.43 M, 0.43 M , 0.0074 M

c) 0.70 M, 0.70 M, 0.046 M

d) 0.43 M , 0.43 M, 0.44 M

d) 0.70 M, 0.70 M, 0.092 M

e) 0.43 M, 0.43 M, 0.43 M

a) 0.41 M, 0.41 M, 0.82 M

e) 0.087 M, 0.087 M, 0.66 M

Factors That Affect Chemical Equilibrium One of the interesting and useful features of chemical equilibria is that they can be manipulated in specific ways to maximize production of a desired substance. Consider, for example, the industrial production of ammonia from its constituent elements by the Haber process:

More than 100 million tons of ammonia is produced annually by this reaction, with most of the resulting ammonia being used for fertilizers to enhance crop production. Clearly it would be in the best interest of industry to maximize the yield of NH3. In this section, we willleam about the vari0us ways in which an equilibrium can be manipulated in order to accomplish this goal. Le Chiitelier's principle states that when a stress is applied to a system at equilibrium, the system will respond by shifting in the direction that minimizes the effect of the stress. In this context, "stress" refers to a disturbance of the system at equilibrium by any of the following means:

products, be sure to calculate Q and compare it to K to determine which direction the reaction will proceed to reach equilibrium.

612

CHAPTER 15

Chemical Equilibrium

• The addition of a reactant or product • The removal of a reactant or product • A change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products • A change in temperature "Shifting" refers to the occunence of either the forward or reverse reaction such that the effect of the stress is partially offset as the system reestablishes equiliblium. An equilibrium that shifts to the right is one in which more products are produced by the forward reaction. An equi'librium that shifts to the left is one in which more reactants are produced by the reverse reaction. Using Le Chatelier's principle, we can predict the direction in which an equilibrium will shift, given the specific stress that is applied.

Addition or Removal of a Substance Again using the Haber process as an example,

consider a system at 700 K, in which the equiliblium concentrations are as follows:

Remember that at equilibrium, the reaction quotient, Q" is eq ual to the equilibrium constant, K,.

. . . . . . . . , . . . . . . . . . . . . . .. ......... .. .......... .. ... .. .... ......,............................................................... Using these concentrations in the reaction quotient expression, we can calculate the value of Kc for the reaction at this temperature as follows: [NH3f [N2] [H2] 3

( 1.52)2

--------=- = 0.297 = Kc (2.05)(1.56)3

If we were to apply stress to this system by adding more N 2, increasing its concentration from 2.05 M to 3.51 M, the system would no longer be at equilibrium. To see that this is true, use the new concentration of nitrogen in the reaction quotient expression. The new calculated value of Qc (0.173) is no longer equal to the value of Kc (0.297).

[NH 3]2 [N 2] [H 2] 3

While reestablishing equilibrium causes a decrease in the N2 concentration, the final concentration will still be higher than that in the original equilibrium mixture. The system responds to the stress of the added reactant by consuming part of it.

( 1.52)2

- - ------.,. =

0.173

=I=-

Kc

(3.51)(1.56)3

For this system to reestablish equilibrium, the net reaction will have to shift in such a way that Qc is again equal to Ke, which is constant at a given temperature. Recall from Section 15.4 that when Q is less than K, the reaction proceeds to the light in order to achieve equilibrium. Likewise, an equilibrium that is stressed in such a way that Q becomes less than K will shift to the right in order This means that the forward reaction, the consumption .of N2 and H2 to . . . . . . .to . . . reestablish . . . . . . . . . . . . . . . .equiliblium. . . . . . . . . .. .............................................................................. produce NH 3, will occur. The result will be a net decrease in the concentrations of N2 and H2 (thus making the denominator of the reaction quotient smaller), and a net increase in the concentration of NH3 (thus making the numerator larger). When the concentrations of all species are such that Qc is again equal to Kc, the system will have established a new equilibrium position, meaning that it will have shifted in one direction or the other, resulting in a new equilibrium concentration for each species. Figure 15.5 shows how the concentrations of N 2 , H 2, and NH3 change when N2 is added to the original equiliblium mixture. Conversely, if we were to remove N2 from the original equiliblium mixture, the lower concentration in the denominator of the reaction quotient would result in Qc being greater than K e . In this case the reaction will shift to the left. That is, the reverse reaction will take place, thereby increasing the concentrations of N2 and H2 and decreasing the concentration of NH3 until Qc is once again equal to K c. The addition or removal of NH3 will cause a shift in the equilibrium, too. The addition of NH3 will cause a shift to the left; the removal of NH3 will cause a shift to the right. Figure 15.6(a) shows the additions and removals that cause this equiliblium to shift to the right. Figure 15.6(b) shows those that cause it to shift to the left. In essence, a system at equiliblium will respond to addition of a species by consuming some of that species , and it will respond to the removal of a species by producing more of that species. It is important to remember that addition or removal of a species from an equilibrium mixture does not change the value of the equiliblium constant, K. Rather, it changes temporarily the value of the reaction quotient, Q. Furthermore, in order to cause a shift in the equilibrium, the species added or removed must be one that appears in the reaction quotient expression. In the case of a heteroge-

SECTION 1S.5

Factors That Affect Chemica l Equilibrium

Figure 15.5

Adding more of a reactant to a system at eq uilibrium causes the equilibrium position to shift toward product. The system responds to the addition of Nz by consuming some of the added N z (and so me of the other reactant, H z) to produce more NH3.

4.0 3.5 ~

S c

Nz

3.0 2.5 -

0

.~

~

'"c u "

l:J

0 " U

2.0 NH3

1.5 -

Hz

1.0 .

0 .5 0.0

613

I

I

I

Time [NH3f 3

[Nzl[HzI • . Original equili brium mi xture

[NHJz

( 1.52)z - - - - - : ; = 0.173

(3.51)(1.56)3

( 164)z -

[NzJ[HzI3

Immedi atel y after addition of Nz

-

-

---0

(3.45)(1 .38)3

= 0.297

After equilibrium has been reestablished

Addition Addition

~

-

Addi tion

~

Nz(g) + 3H z(g)

,

'

+ 3Hig)

2NH3(g)

Nz(g)

~

/ / Removal Removal

Removal

(a)

/ +: , :::::;,:-

2NH 3 (g)

(b)

neous equilibrium, altering the amount of a solid or liquid species does not ch ange the p ositio n of the equilibrium because doing so does not change the value of Q. Sample Problem 15.11 shows the effects of stress on a sys tem at equilibrium .

.

. Sample Problem 15.11 Hydrogen sulfide (H zS) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur. 2H zS(g)

+ O zCg) :;:,==' 2S(s) + 2H zO(g)

For each of the fo llowing scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O z(g), (b) removal of H 2 S(g), (c) removal of H zO(g), and (d) addition of S(s). Strategy Use Le Chiitelier's principle to predict the direction of shift for each case. Remember that the position of the equilibrium is onl y changed by the addition or removal of a species that appears in the reaction quotient expression. Setup Begin by writing the reaction quotient expression: [HzO] z

Qc =

?

[J:lzSnOz]

Because sulfur is a solid, it does not appear in the expression. Changes in the concentration of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression for Qc wi ll shift the equilibrium to the right: Addition

\

2H2S(g)

Addition

+

\

0 2(g)

-

,

2S(s)

+

2H20(g)

\ Removal (Continued)

Figure 15.6

(a) Addition of a reactant or removal of a product will cause an equilibrium to shift to the right. (b) Addition of a product or removal of a reactant will cause an equilibrium to shift to the left.

614

CHAPTER 15

Chemical Equilibrium

Think About It In each case,

analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its concentration increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall denominator-giving the overall fraction a smaller value. Thus, Q will temporarily be smaller than K and the reaction will have to shift to the right, consuming some of the added O 2 (along with some of the H 2 S in the mixture) in order to reestablish equilibrium.

Removal of a reactant or addition of a product that appears in the expression for Qc will shift the equilibrium to the left:

Addition N2(g)

I Removal

+

jH2(g)

•

I Removal

-

I 2NH3(g)

Solution

(a) Shift to the right (b) Shift to the left (c) Shift to the right (d) No change

Practice Problem For each change indicated, determine whether the equilibrium

PCI 3 (g)

+ CI 2 (g) :;:.==' PCls(g)

will shift to the right, shift to the left, or neither: (a) addition of PCI 3(g), (b) removal of PCI 3 (g), (c) removal of PCIs(g), and (d) removal of CI2 (g) .

Changes in Volume and Pressure If we were to start with a gaseous system at equilibrium in a cylinder with a movable piston, we could change the volume of the system, thereby changing the concentrations of the reactants and products. Consider again the equilibrium between N 20 4 and N02: At 25 °C the equilibrium constant for this reaction is 4.63 X 10- 3. Suppose we have an equilibrium mixture of 0.643 M N 20 4 and 0.0547 M N02 in a cylinder fitted with a movable piston. If we push down on the piston, the equilibrium will be disturbed and will shift in the direction that minimizes the effect of this disturbance. Consider what happens to the concentrations of both species if we decrease the volume of the cylinder by half. Both concentrations are initially doubled: [N20 4J = 1.286 M and [NOzJ 0.1094 M. If we plug the new concentrations into the reaction quotient expression, we get (0.1094)2 ---=-1.-=-:28=-=6-

3

= 9.3 1

X

10-

which is not equal to Ke, so the system is no longer at equilibrium. Because Qe is greater than Ke, the equilibrium will have to shift to the left in order for equilibrium to be reestablished (Figure 15.7). In general, a decrease in volume of a reaction vessel will cause a shift in the equilibrium in the direction that minimizes the total number of moles of gas. Conversely, an increase in volume will cause a shift in the direction that maximizes the total number of moles of gas.

Figure 15.7

The effect of a volume decrease (pressure increase) on the N 20 4 (g) • • 2N0 2 (g) equilibrium. When volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas.

SECTION 1S.5

Factors That Affect Chemical Equil ibrium

615

Sample Problem 15.12 shows how to predict the equilibrium shift that will be caused by a volume change.

~

Sample Problem 15.12 For each reaction , predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased. (a) PCIs(g) • (b) 2PbS(s) (c) H 2(g)

' PCI 3 (g)

+ 30z(g).

+ Iz(g).

+ Clz(g) ' 2PbO(s) 2S0 z(g)

' 2HI(g)

Strategy Determine which direction minimized the number of moles of gas in the reaction. Count . ... .. " . . . . . . . . . .. . . ... ... . .. " . . . .. .... . . .. only moles of gas.

.. . ...

Setup We have (a) 1 mole of gas on the reactant side and 2 moles of gas on the product side, (b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, and (c) 2 moles of gas on each side. Solution

(a) Shift to the left (b) Shift to the right (c) No shift

Practice Problem A For each reaction, predict the direction of shift caused by increasing the volume of the reaction vessel.

(a) 2NOCI(g)

===' 2NO(g)+Clz(g)

+=.

(b) CaC0 3 (s) .

' CaO(s)+C0 2 (g)

Practice Problem B For the following equilibrium, give an example of a stress that will cause a shift to the right, a shift to the left, and one that will cause no shift:

It is possible to change the total pressure of a system without changing its volume by adding an inert gas such as helium to the reaction vessel. Because the total volume remains the same, the concentrations of reactant and product gases do not change. Therefore, the equilibrium is not disturbed and no shift will occur.

Changes in Temperature A change in concentration or volume may alter the position of an equilibrium (i.e. , the relative amounts of reactants and products), but it does not change the value of the equilibrium constant. Only a change in temperature can alter the value of the equilibrium constant. To understand why, consider the following reaction: The forward reaction is endothermic (absorbs heat, (t:..H > 0):

t:..HO= 58.0 kJ/mol If we treat heat as though it were a reactant, we can use Le Chil.teber's principle to predict what will happen if we add or remove heat. Increasing the temperature (adding heat) will shift the reaction in the forward direction because heat appears on the reactant side. Lowering the temperature (removing heat) will shift the reaction in the reverse direction. Consequently, the equilibrium constant, given by ?

[N0 2]eq

K c

= -:----:-[N20 4] eq

increases when the system is heated and decreases when the system is cooled (Figure 15.8). A similar argument can be made in the case of an exothermic reaction, where heat can be considered

It is a common error to count all the species in a readion to determine which side has fewer moles. To determine what diredion shift a volume change w ill cause, it is only the number of moles of gas that matters.

Think About It When there is no difference in the number of moles of gas, changing the volume of the reaction vessel will change the concentrations of reactant(s) and product(s)-but the system will remain at equilibrium. (Q will remain equal to K.)

616

CHAPTER 15

Chemical Equilibrium

Figure 15.8 (a) NZ0 4-NO z equilibrium. (b) Because the reaction is endothermic, at higher temperature, the N Z0 4(g). • 2NO z(g) equilibrium shifts toward product, making the reaction mixture darker.

mL 5,.

......_ _-.......... ...... .-------

(b)

(a)

.

to be a product. Increasing the temperature of an exothermic reaction causes the equilibrium constant to decrease, shifting the equilibrium toward reactants. Another example of this phenomenon is the equilibrium between the following ions: CoCl ~ ~ Blue

+ 6H20.

• CO(H20)~ +

+ 4CI~ + Heat

Pink

The reaction as written, the formation of CO(H20)~ + , is exothermic. Thus, the reverse reaction, the formation of CoCl ~~, is endothermic. On heating, the equilibrium shifts to the left and the solution turns blue. Cooling favors the exothermic reaction [the formation of CO(H20)~+] and the solution turns pink (Figure 15.9). In summary, a temperature increase favors an endothermic reaction, and a temperature decrease favors an exothermic process. Temperature affects the position of an equilibrium by changing the value of the equilibrium constant. Figures 15.10 and 15.11 illustrate the effects of various stresses on systems at equilibrium.

Catalysis Media Player/MPEG Animation : Figures 15.10 and 15.11, Le Chiltelier's Principle, pp. 618~62 1.

A catalyst speeds up a reaction by lowering the reaction's activation energy [ ~~ Section 14.6] . However, a catalyst lowers the activation energy of the forward and reverse reactions to the same extent (see Figure 14.14). The presence of a catalyst, therefore, does not alter the equilibrium constant, nor does it shift the position of an equilibrium system. Adding a catalyst to a reaction mix-

\

(a)

(b)

(c)

Figure 15.9 (a) An equilibrium mixture of CoCll~ ions and Co(H20) ~ + ions appears violet. (b) Heating favors the formation of CoCll~, making the solution look more blue. (c) Cooling favors the formation of Co(HzO)~ + , making the solution look more pink.

SECTION 15.5

Factors That Affect Chem ica l Equilibrium

617

ture that is not at equilibrium will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but it might take a much longer time.

Bringing Chemistry to Life Hemoglobin Production at High Altitude . As we learned at the end of Chapter 11 , rapid ascent to a high altitude can cause altitude sickness. The symptoms of altitude sickness, including dizziness, headache, and nausea, are caused by hypoxia, an insufficient oxygen supply to body tissues. In severe cases, without prompt treatment, a victim may slip into a coma and die. And yet a person staying at a high altitude for weeks or months can recover gradually from altitude sickness, adjust to the low oxygen content in the atmosphere, and live and function nOllnally. The combination of oxygen with the hemoglobin (Hb) molecule, which carries oxygen through the blood, is a complex reaction, but for our purposes it can be represented by the following simplified equation: * Hb(aq) + 02(aq) += , ~ . Hb0 2(aq ) where Hb0 2 is oxyhemoglobin, the hemoglobin-oxygen complex that actually transports oxygen to tissues. The equilibrium expression for this process is [Hb0 2] Kc = -[Hb - ] [-O?]At an altitude of 3 km, the partial pressure of oxygen is only about 0.14 atm, compared with 0.20 atrn at sea level. Accorcling to Le Chfttelier's principle, a decrease in oxygen concentration will shift the hemoglobin-oxyhemoglobin equiliblium from right to left. This change depletes the supply of oxyhemoglobin, causing hypoxia. Over time, the body copes with this problem by producing more hemoglobin molecules. As the concentration of Hb increases, the equilibrium gradually shifts back toward the right (toward the formation of oxyhemoglobin). It can take several weeks for the increase in hemoglobin production to meet the body's oxygen needs adequately. A return to full capacity may require several years to occur. Studies show that long-time residents of high-altitude areas have high hemoglobin levels in their blood-sometimes as much as 50 percent more than individuals living at sea level. The production of more hemoglobin and the resulting increased capacity of the blood to deliver oxygen to the body have made high-altitude training and hypoxic tents popular among some athletes. *Biological processes such as this are not true equilibria, but rather they are steady-state situations. In a steady state, the co nstant concentrations of reactants and products are not the result of forward and reverse reactions occurring at the same rate. Instead, reactant concentration is replenished by a previous reaction and product concentration is maintained by a subsequent reaction. Nevertheless, many of the principles of equilibrium, including Le Chiitelier's principle, still apply.

Checkpoint 15.5 15.5.1

Factors That Affect Chemical Equilibrium

Which of the following equilibria will shift to the right when H2 is added? (Select all that apply.) a) 2H2 + O 2 • b) 2HI •

• 2H 20

• H2 + 12

c) H2 + CO 2 • d) 2NaHC0 3 • e) 2CO

+

15.5.3

O2 •

• H 20

+

CO

• Na2C0 3 + H 20

+

Which of the following equilibria will shift to the left when the temperature is increased? [L'.H (kJ/mol) values are given in parentheses.] (Select all that apply.) a) S

+ H2 •

b) C

+ H 20

•

CO 2

d) MgO(s)

• 2C0 2

+

L'.H (-20)

+ H2 H 20 + CO

• CO •

c) H2 + CO 2 • e) 2CO

15.5.2

• H2S

+ CO 2

•

O2 •

• 2C0 2

• MgC0 3

L'.H (131)

Lill(41) L'.H (-ll7) L'.H (-566)

Which of the following will cause the equilibrium C(s)

+ CO 2(g)

+=.~ .

2CO(g )

to shift to the right? (Select all that apply. ) a) Decreasing the volume b) Increasing the volume c) Adding more C(s) d) Adding more CO 2(g) e) Removing CO(g) as it forms

15.5.4

For which of the following reactions will a change in volume not affect the position of the equilibrium? (Select all that apply.) a) MgO(s) + CO 2 (g) , b) H 2 (g)

+ CI 2(g)

c) BaC0 3(s) • d) Br(l)

e) C(s)

•

• MgC0 3(s)

• 2HCl(g)

• BaO(s)

+ Hig) , + CO 2(g) •

+ CO 2 (g)

• 2HBr(g) • 2CO (g)

Figure 15.10 Le Chiitelier's principle.

•

"/ [Hz] = 0.112 M [Iz] = 0.112M [HI] = 0 .825 M

Q = c

[HI]2 = (0.825)2 = 54.3 [H 21[Izl (0. 112)(0.112)

LH2 ] = 0.112M [I2] = 0 .112 M [HI] = 0.825 M

[Hz]=0.112M LIz] = 0.112M [HI] = 0.825 M

Q = c

[HI]z = (0.825)z = 54.3 LH 2][Iz] (0. 112)(0.112)

Q = c

z [HIl = (0.825)2 = 54.3 (0 .112)(0.112) [H 2] [12]

Qc = Kc

Add a reactant- Iz(g)

Add a product-HI(g)

--

,•

[H2 ] = 0.112 M LIz] = 0.499 M [HI] = 0.825 M

_

(0.825)2 _ Qc - (0.11 2)(0 .499) - 12 Qc eft Kc

Equilibrium shifts toward product

[Hz] = 0.0404 M LI2] = 0.4?7 M [HI] = 0.968 M

Q = c

(0.968)2 = 54.3 (0.0404)(0.427)

Qc = Kc

618

[Hz] = O.I!? M [Iz] = 0.112 M [HI]=3.17M

(3.17)2 Qc = (0. 112)(0. 112) = 801

Qc eft Kc

Qc = Kc

Add a species not involved in the equilibrium - He(g)


boat conversion is 41 kJ/mol. If the energy for the chair frequency factor is 1.0 X 10 12 S- I, what is kl at 298 K? The equi librium constant Ke for the reaction is 9.83 X 103 at 298 K.

where C is a constant. The following table gives the equilibrium constant (Kp) for the reaction at various temperatures. 2NO(g)

T(K)

+ 0 2(g)

====' 2N0 2(g)

+=.

15.123 Consider the following reaction at a certain temperature

138

5.12

0.436

0.0626

0.0130

600

700

800

900

1000

A2 + B2 :;:.==' 2AB The mixing of 1 mole of A2 with 3 moles of B2 gives rise to x mole of AB at equilibrium. The addition of 2 more moles of A2 produces another x mole of AB. What is the equilibrium constant for the reaction?

Determine graphically the 6Ho for the reaction.

15.119 Consider the reaction between N0 2 and N 20 4 in a closed container: Initially, 1 mole of N 20 4 is present. At equilibrium, x mole of N 20 4 has dissociated to form N0 2. (a) Derive an expression for Kp in terms of x and P, the total press ure. (b) How does the expression in part (a) help you predict the shift in equilibrium due to an increase in P? Does your prediction agree with Le Chiitelier's principle?

15 .124 Iodine is sparingly soluble in water but much more so in carbon tetrachloride (CCI4 ) . The equilibrium constant, also called the partition coefficient, for the distribution of 12 between these two phases

is 83 at 20°C. (a) A student adds 0.030 L of CC1 4 to 0.200 L of an aqueous solution containing 0.032 g of 12, The mixture at 20°C is shaken, and the two phases are then allowed to separate. Calculate the fraction of 12 remaining in the aqueous phase. (b) The student now repeats the extraction OfI2 with another 0.030 L of CCI 4 . Calculate the fraction of the 12 from the original solution that remains in the aqueous phase. (c) Compare the result in part (b) with a single extraction using 0.060 L of CCI 4 . Comment on the difference.

15.120 (a) Use the van't Hoff equation in Problem 15.] 18 to derive the following expression, wh ich relates the equilibrium constants at two different temperatures:

(J... _ J...)

In KI = t:..Ho K2 R T2

TI

How does this equation support the prediction based on Le Chiitelier's principle about the shift in equilibrium with temperature? (b) The vapor pressures of water are 31.82 mmHg at 30°C and 92.51 mmHg at 50°C. Calculate the molar heat of vaporization of water.

15.121 The Kp for the reaction

==' S02(g) + CI 2(g)

S02CI2(g) :;:.

is 2.05 at 648 K. A sample of S02Cl2 is placed in a container and heated to 648 K, while the total pressure is kept constant at 9.00 atm. Calculate the partial pressures of the gases at equilibrium.

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES Lime (CaO) is used to prevent S02 from escaping from the smokestacks of coal-burning power plants via the formation of solid CaS04 . 2H20 (gypsum). One of the important reactions in the overall process is the decomposition of CaC0 3: CaC0 3 (s) :;:.==> CaO(s)

nsoc.

d) 3 .7 X 10- 8

If a 12.0-g sample of solid CaC0 3 is placed in an evacuated vessel what will the press ure of CO 2 be when the system reaches at equilibrium?

nsoc,

c) 1.7 X 10- 3 atm d) 1.0 atrn

Which of the following actions will cause an increase in the pressure of CO 2 in the vessel described in Question 2? a) Addition of He gas b) Addition of S02 gas c) Addition of more CaC0 3 solid d) Increasing the volume of the vessel

Calculate the value of Kp for the decomposition of CaC0 3 at 725°C. a) 3.0 X 10- 6 c) 2.0 X 10- 2 b) 2.5 X 10- 4

2.

3.

+ CO 2(g)

which has an equilibrium constant (Ke) of 3.0 X 10- 6 at 1.

it) 3.0 X 10- 6 atm b) 3.3 X 105 atm

4.

nsoc

If a 12.0-g sample of solid CaC0 3 is placed in a vessel at in which the pressure of CO 2 is 2.5 atm, what mass of CaO will form? a) 6.n g b) 12.0 g

c) 6.00 g d) None

ANSWERS TO IN-CHAPTER MATERIALS

633

ANSWERS TO IN-CHAPTER MATERIALS (b) CH4 + 2HzO. ' COZ + 4H z, (c) Iz + Hz • ' 2HI. lS.6A 8.1 X 10- 6 lS.6B 0.104. 15.7A Left. lS.7B Right. lS.8A [Hz] = [I z] = 0.056 M; [HI] = 0.413 M. lS.8B [Hz] = [I z] = 0.043 M; [HI] = 0.314 M. 15.9 [Br] = 8.8 X 10- 3 , [Brz] = 6.5 X 10- 2 lS.10A [PH] = 2 [PI] = 0.37 atm, PHI = 2.75 atm. lS.10B PH = PI = 0.294 atm, PHI = 2 2 2 2.16 atm.1S.11A (a) Right, (b) left, (c) right, (d) left.1S.12A (a) Right, (b) right, (c) right. lS.12B Right shift: remove HF; left shift: remove Hz; no shift: reduce volume of container.

lS.2B (a) Hz + Clz • ' 2HCl, (b) H + + F - :;::.==:' HF, (c) CrH + 40H- . ' Cr(OH); , (d) HC10 • ' H+ + ClO - , (e) H ZS0 3 • ' H + + HS0 3, (f) 2NO + Brz . ' 2NOBr. lS.3A (a) Kc =

[HClt

, (b) Kc =

I

[ Aa(NH3)~ ][Cl-] 2 (b) Kc = [Ag+][Cl- ], (c) Kc = [Ba +][F- f , (d) Kc = b -2 • [NH3] lS.4A (a) 2.3 X 1024 , (b) 6.6 X 10- 13 , (c) 2.8 X 10- 15 lS.4B (a) 1.5 X 29

38

10 , (b) 1.3 X 10 , (c) 8.5 X 10 . lS.SA (a) Kp = (P N H)

(b) Kp = P eo , (c) Kp = 2

3'

(P N )(P H 2

) 2

(Peal

,

(Peoi (P o ) 2

2

15.2.1 e. 15.2.2 a. 15.3.1 d. 15.3.2 a. 15.3.3 c. 15.3.4 c. 15.4.1 d. 15.4.2 b. 15.5.1 a, c. 15.5.2 b, d, e. 15.5.3 a, d, e. 15.5.4 b.

Answers to Applying What You've learned

,

[SiCl4] [H2f [Hg2+] [C1- f [Z 2+] [NH+][OW] [N'(CO) ] (c)Kc= 1 4,(d)Kc = n .lS.3B (a) Kc = 4 , 2 [Cot [Fe +] [ NH 3]

34

Answers to Checkpoints

lS.SB (a) 2NO z + O 2 •

' 2N0 3 ,

[OD17X] . b) 3.05 X 10- 5. c) [ODl7X-A77] 30.5. [ODlA] [OF2A] [ODlA][OF2A][A77IAj' d) Qc = 3.6 X 10- 5 ; therefore, the mixture is not at equilibrium. It will have to proceed to the left (reverse reaction) in order to achieve equilibrium. e) At equilibrium, [ODIA] = 0.05 M, [OF2A] = 0.05 M, [ODl 7X] = 0.95 M. a)

s an

16.1 16.2 16.3 16.4 • •

16.5 • • •

16.6 • • •

16.7 • •

Bf0nsted Acids and Bases The Acid-Base Properties of Water The pH Scale Strong Acids and Bases Strong Acids Strong Bases

Weak Acids and Acid Ionization Constants The Ionization Constant, Ka Calculating pH from Ka Using pH to Determine Ka

Weak Bases and Base Ionization Constants · The Ionization Constant, Kb Calculating pH from Kb Using pH to Determine Kb

Conjugate Acid-Base Pairs The Strength of a Conjugate Acid or Base The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair

16.8

Diprotic and Polyprotic Acids

16.9

Molecular Structure and Acid Strength

• •

•

16.10 • • • •

16.11 • •

16.12

Hydrohalic Acids Oxoacids Carboxylic Acids

Acid-Base Properties of Salt Solutions Basic Salt Solutions Acidic Salt Solutions Neutral Salt Solutions Salts in Which Both the Cation and the Anion Hydrolyze

Acid-Base Properties of Oxides and Hydroxides Oxides of Metals and of Nonmetals Basic and Amphoteric Hydroxides

Lewis Acids and Bases

ases

An Acid Essential to Good Health Scurvy is a devastating disease caused by a deficiency of vitamin C. It has a host of terrible symptoms including extreme fatigue and weakness, hemorrhaging, and loss of teeth due to inflamed gums. Untreated, it results in death, usually from pneumonia or other acute infection-or from heart failure. Most mammals make their own vitamin C, but humans and other primates, as well as fruit bats and guinea pigs, must obtain it from their diet. Since ancient times, people were at risk for contracting scurvy when their diets did not include an adequate supply of fresh fruits and vegetables. One segment of the population historically deprived of an appropriate diet was sailors. During the fourteenth and fifteenth centuries, as the development of ships and sails made voyages on the open ocean possible, and .returns to shore became less frequent, scurvy became .. ...... . . . .. . . . . . . commonplace among sailors. -

Although reports of cures for scurvy date back to the sixteenth century, it was not until 1747 that the effectiveness of citrus was proven with the experiments of Scottish naval surgeon James Lind. It was another 40 years, though, before the British navy began supplying citrus juice to its crews. It was the compulsory issue of the juice of lemons or limes that gave rise to the term limey, referring to a British sailor.

One hundred of Portuguese explorer Vasco de Gama's 160-member crew died from scurvy by the time they had sailed around the southern tip of Africa in 1497.

Vitamin C, or ascorbic acid (H 2C6H 60 6), acts as an antioxidant, reacting with oxidizing species such as the ·OH radical, which can damage an organism's DNA. It is one of many acids that are important to human health.

•

Ascorbic acid

•

In This Chapter, You Will Learn

more about the properties of acids and bases and how those properties are

related to molecular structure.

Before you begin, you should review •

The list of strong acids and the list of strong bases

•

How to solve equilibrium problems

[ ~.

[~.

Section 4.3, Table 4.4]

Section 15.4]

Media Player/ MPEG Content Chapter in Review

Acids such as citric acid and ascorbic acid (vitamin C) are responsible for the sour taste of citrus fruits. 635

636

CHAPTER 16

Acids and Bases

Br¢nsted Acids and Bases In Chapter 4 we learned that a Br¢nsted acid is a substance that can donate a proton and a Br¢nsted base is a substance that can accept a proton [ ~~ Section 4.3]. In this chapter we extend our di scussion of Br¢nsted acid-base theory to include conjugate acids and conjugate bases. When a Br¢nsted acid donates a proton, what remains of the acid is known as a conjugate base. For example, in the ionization of HCI in water, HCI (aq)

+ H 20(l) +=.~. H30 +(aq) + Cl - (aq)

acid

conjugate base

HCI donates a proton to water, producing the hydronium ion CH30 +) and the chloride ion CCI-), which is the conjugate base of HCl. The two species, HCI and Cl- , are known as a conjugate acid-base pair or simply a conjugate pair. Table 16.1 lists the conjugate bases of several familiar species. Conversely, when a Br¢nsted base accepts a proton, the newly formed protonated species is known as a conjugate acid. When ammonia (NH3) ionizes in water,

Species CH3COOH

NH3(aq)

Conjugate Base CH3COO -

H 2O

OH-

NH3

NH2"

H 2SO 4

HS0 4

+ H 20(l)

~. NH t Caq)

+=.

base

+ OH- (aq)

conjugate acid

NH3 accepts a proton from water to become the ammonium ion (NH)t . The ammonium ion is the conjugate acid of ammonia. Table 16.2 lists the conjugate acids of several common species. Any reaction that we describe using Br¢nsted acid-base theory involves an acid and a base. The acid donates the proton, and the base accepts it. Furthermore, the products of such a reaction are always a conjugate base and a conjugate acid. It is useful to identify and label each species in a Br¢nsted acid-base reaction. For the ionization of HCl in water, the species are labeled as follows: loses a proton 1----

-----,

gains a proton HCI(aq)

+

H 30 +(aq)

NHt

H 2O

H30 +

OH-

H 2O

+

CI-(aq)

conjugate base

r. .. ;.:., ~d..... ~

•

Conjugate Acid

NH3

H 2NCONH2 (urea)

• •

base

acid

Species

H2 0 (l)

1

ell!

And for the ionization of NH3 in water, gains a proton loses a proton

H 2NCONH!

NH3(aq)

+

base

H 2°(l)

acid

• •

NHt(aq) • ~

. r Ck

.

+

OH-(aq) •

,

conjugate base

Sample Problems 16.1 and 16.2 let you practice identifying conjugate pairs and the species in a Br¢nsted acid-base reaction. "

Sample Problem 16.1 ".""" Think About It A species does

not need to be what we think of as an acid in order for it to have a conjugate base. For example, we would not refer to the hydroxide ion (OH- ) as an acid- but it does have a conjugate base, the oxide ion (0 2- ) . Furthermore, a species that can either lose or gain a proton, such as HC0 .1 , has both a conjugate base (CO~-) and a conjugate acid (H 2C0 3).

What is (a) the conjugate base of HN0 3, (b) the conjugate acid of 0 2 - , (c) the conjugate base of HS0 4 , and (d) the conjugate acid of HC0 .1 ? Strategy To find the conjugate base of a species, remove a proton from the formula. To find the conjugate acid of a species, add a proton to the formula. Setup The word proton, in this context, refers to H+. Thus, the formula and the charge will both be affected by the addition or removal of H+. Solution (a) NO.1 (b)

OW

(c) SO~ ­ (d) H 2C0 3

SECTION 16.2

The Acid-Base Properties of Water

Practice Problem A What is (a) the conjugate acid of Cl0 4 , (b) the conjugate acid of S2- , (c) the conjugate base of H 2S, and (d) the conjugate base of H 2C 20 4 ? Practice Problem B HS0 3 is the conjugate acid of what species? HS0 3 is the conjugate base of what species?

Label each of the species in the following equations as an acid, base, conjugate base, or conjugate acid:

+ NH3(aq). • F - (aq) + NHr (aq) (b) CH3COO - (aq) + H20(l). • CH 3COOH(aq) + OW(aq) (a) HF(aq)

Strategy In each equation, the reactant that loses a proton is the acid and the reactant that gains a proton is the base. Each product is the conjugate of one of the reactants. Two species that differ only by a proton constitute a conjugate pair. . Setup (a) HF loses a proton and becomes F- ; NH 3 gains a proton and becomes NHr .

(b) CH 3COO- gains a proton to become CH3COOH; H 20 loses a proton to become OH-. Solution (a) HF(aq)

+ NH 3(aq)

acid

::0:.=~.

base

(b) CH3COO - (aq)

+ H 20(l)

base

F-(aq) conjugate

conjugate

base

acid

::o:.=~'

acid

+ Nm(aq)

CH 3COOH(aq) conjugate acid

+ OH-(aq) conjugate base

Practice Problem A Identify and label the species in each reaction. (a) NHr(aq)

(b) CN- (aq)

+ H 20(l). + H 20(l).

+ H30 +(aq) HCN(aq) + OW(aq)

• NH3(aq) •

Practice Problem B (a) Write an equation in which HS0 4 reacts (with water) to form its conjugate base. (b) Write an equation in which HS0 4 reacts (with water) to form its conjugate acid.

Checkpoint 16.1 16.1.1

Bnmsted Acids and Bases

Which of the following pairs of species are conjugate pairs? (Select all that apply.)

16.1.2

Which of the following species does not have a conjugate base? (Select all that apply.)

a) H 2S and S2-

a) HC 20 4

b) NH2 and NH3

b) OW

c) O 2 and H 20 2

c) 0 2 -

d) HBr and Br-

d) CO ~-

e) HCI and OH-

e) HCIO

The Acid-Base Properties of Water Water is often referred to as the "universal solvent," because it is so common and so important to life on Earth. In addition, most of the acid-base chemistry that you will encounter takes place in aqueous solution. In this section, we take a closer look at water's ability to act as either a Br¢nsted acid (as in the ionization of NH 3) or a Brlilnsted base (as in the ionization of Hel). A species that can behave either as a Br¢nsted acid or a Br¢nsted base is called amphoteric. Water is a very weak electrolyte, but it does undergo ionization to a small extent:

Th ink About It In a Br¢nsted acid-base reaction, there is always an acid and a base, and whether a substance behaves as an acid or a base depends on what it is combined with. Water, for example, behaves as a base when combined with HCI but behaves as an acid when combined with NH 3 .

637

638

CHAPTER 16

Acids and Bases

This reaction is known as the autoionization o/water. Because we can represent the aqueous proton as either H + or H30 + [ ~~ Section 4.3] , we can also write the auto ionization of water as

+

H30 +(aq)

." t.

+

~

...~.

OH-(aq)

+ Mu lti media

Chemical Equilibrium (interactive).

Recall that in a heterogeneous equilibrium such as this, liquids and solids do not appear in the equilibrium expression [ ~~ Section 15.3] .

+ conjugate acid

base

acid

equilibrium

•

•

conjugate base

.

where one water molecule acts as an acid and the other acts as a base . .............. As ·lildkatecn:'y"ih·e ·doubie ·arrov,i"iu· the·e·ci.uatlOii: the·i-e·a ctlon ISail equiiibii"um: The equilibrium expression for the autoionization of water is or

The constant Kw is sometimes referred to as the

ion-product constant.

Recall that we disregard the units when we substitute concentrations into an equilibrium expression [ ~~ Section 15.5].

Because the autoionization of water is an important equilibrium that you will encounter frequently the study of acids and bases, we use the subscript w to indicate that the equilibrium constant is · . . . .. .in . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . , that specifically for the autoionization of water. It is important to realize, though, that Kw is simply a Kc for a specific reaction. We will frequently replace the c in Kc expressions with a letter or a series of letters to indicate the specific type of reaction to which the Kc refers. For example, Kc for the ionization of a weak acid is called Ka, and Kc for the ionization of a weak base is called K b. In Chapter 17 we will encounter Ksp, where "sp" stands for "solubility product." Each specially subscripted K is simply a Kc for a specific type of reaction. In pure water, autoionization is the only source ofH30 + and OH-, and the stoichiometry of the reaction tells us that their concentrations are equal. At 25°C, the concentrations of hydronium 7 and hydroxide ions in pure water are [H30 +] = [OH- ] = 1.0 X 10- M. Using the equilibrium ·. . . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . expression, we can calculate the value of Kw at 25 °C as follows:

Kw = [H30 +][OH-]

=

(1.0

X

10- 7)(1.0

X

10- 7 )

=

1.0

X

10- 14

FUltherrnore, in any aqueous solution at 25°C, the product of H30 + and OH - concentrations is equal to 1.0 X 10- 14 . Equation 16.1 Although their product is a constant, the individual concentrations of hydronium and+ hydroxide can be influenced by the addition of an acid or a base. The relative amounts of H30 and OH determine whether a solution is neutral, acidic, or basic.

· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Because the product of H30+ and OW concentrations is a constant, we cannot alter the concentrations independently. Any change in one also affects the other.

• When [H30 +] • When [H30 +] • When [H30 +]

= [OH- ], the solution is neutral. > [OH- ], the solution is acidic. < [OH- ], the solution is basic.

Sample Problem 16.3 shows how to use Equation 16.1.

Sample Problem 16.3 •.. The concentration of hydronium ions in stomach acid is 0.10 M. Calculate the concentration of hydroxide ions in stomach acid at 25 °C. Strategy Use the value of Kw to determine [OH - ] when [H30 +] = 0.10 M. Setup Kw = [H 30 +][OH- ]

= 1.0

X 10- 14 at 25°C. Rearranging Equation 16.1 to solve for [OH- ] , [OH- ] = 1.0 X 10[H30 +]

Think About It Remember

that equilibri urn constants are temperature dependent. The value of Kw is 1.0 X 10- 14 only at 25°C.

14

Solution [OH- ] = 1.0 X 10-

14

=

1.0 X 10- 13 M

0.10

Practice Problem A The concentration of hydroxide ions in the antacid milk of magnesia is 5.0 X 10- 4 M. Calculate the concentration of hydronium ions at 25°C. Practice Problem B The value of Kw at normal body temperature (37°C) is 2.8 X 10- 14 . Calculate the concentration of hydroxide ions in stomach acid at body temperature. ([H30 +] = 0.10 M)

SECTION 16.3

Checkpoint 16.2 16.2.1

16.2.2

Calculate [H30 +] in a solution in which [OH-] = 0.25 M at 25°C. a) 4.0 X 10- 14 M

b) 8.3 X 10- 17 M 14

639

The Acid-Base Properties of Water

Calculate [OH-] in a solution in which [H30 +] = 0.0012 M at 25 °C. a) l.2 X 10- 3 M

b) 1.0 X 10-

14

M

M

c) 2.5 X 10 13 M

d) 8.3 X 10- 12 M

d) 1.0 X 10- 7 M

e) 1.2XlO II M

e) 4.0 X 10- 7 M

c) l.0 X 10-

The pH Scale

The pH Scale The acidity of an aqueous solution depends on the concentration of hydronium ions , [H30 +]. This concentration can range over many orders of magnitude, which can make reporting the numbers cumbersome. To describe the acidity of a solution, rather than report the molar concentration of hydronium ions, we typically use the more convenient pH scale. The pH of a solution is defined as the negative base-l 0 logarithm 'of the' hydi-oni'u m '{on 'coricenii-iti'on '(In' motA.)". ·········· ..................... . or

Equation 16.2

The pH of a solution is a dimensionless quantity, so the units of concentration must be removed from [H30 +] before taking the logarithm. Because [H30 +] = [OH- ] = 1.0 X 10- 7 M in pure water at 25 °C, the pH of pure water at 25 °C is -log (1.0 X 10- 7) = 7.00 " ............... ................................ .. Remember, too, that a solution in which [H30 +] = [OH- ] is neutral. At 25 °C, therefore, a neutral solution has pH 7.00. An acidic solution, one in which [H30 +] > [OH-] , has pH < 7.00, whereas a basic solution, in which [H30 +] < [OH-], has pH > 7.00. Table 16.3 shows the calculation of pH for solutions ranging from 0.10 M to 1.0 X 10- 14 M. In the laboratory, pH is measured with a pH meter (Figure 16.1 ). Table 16.4 lists the pH values of a number of common fluids. Note that the pH of body fluids varies greatly, depending on the location and function of the fluid. The low pH (high acidity) of gastric juices is vital for digestion of food, whereas the higher pH of blood is required to facilitate the transport of oxygen.

[H30+](M)

.......

0.10

-log [H30+]

pH

- log (1.0 X 10- 1)

1.00

2

0.010

-log (1.0 X 10-

)

2.00

1.0 X 10 3

-log (1.0 X 10-3)

3.00

1.0 X 10 4 -log (1.0 X 10- 4) 1.0 X 10- 5;;;''''I====~l= o= g =(l=.0= X=I=0~5)

4.00

1.0 X 10- 6

-log (1.0 X 10- 6)

6.00

Acidic

1.0 X 10- 7

- log (1.0 X 10- 7 )

7.00

Neutral

1.0 X 10- 8

-log (1.0 X 10- 8)

8.00

Basic

1.0 X 10- 9

-log (1.0 X 10- 9 )

9.00

1.0 X 10- 10

-log (1.0 X 10- 1°)

10.00

1.0 X 10-

11

-log (1.0 X 10-

11

A word about significant figures: When we take the log of a number with two significant figures, we report the result to two places past the decimal point. Thus, pH 7.00 has two significant figures, not three.

5.00

)

11.00 12.00

1.0 X 10- 13

-log (1.0 X 10- 12) - log (1.0 X 10- 13 )

1.0 X 10- 14

-log (1.0 X 10- 14 )

14.00

1.0 X 10- 12

Equation 16.2 converts numbers that can span an enormous range (-10 ' to 10- 14) to numbers generally ranging from -1 to 14.

13.00

Figure 16.1

ApHmeteris commonly used in the laboratory to determine the pH of a solution. Although many pH meters have a range of 1 to 14, pH values can actually be less than 1 and greater than 14.

640

CHAPTER 16

Acids and Bases

Fluid

pH

Fluid

pH

Stomach acid

l.0

Saliva

6.4-6.9

Lemon juice

2.0

Milk

6.5

Vinegar

3.0

Pure water

7.0

Grapefruit juice

3.2

Blood

7.35 -7.45

Orange juice

3.5

Tears

7.4

Urine

4.8 - 7.5

Milk of magnesia

10.6

Rainwater (in clean air)

5.5

Household ammonia

11.5

A measured pH can be used to determine+ experimentally the concentration of hydronium ion in solution. Solving Equation 16.2 for [H30 ] gives

. ... . .. .. .. . . ... . .. ... .. . . 10' is the inverse function of log. (it is usually the second function on the same key.) You must be comfortable performing these operations on your calculator.

Equation 16.3 Sample Problems 16.4 and 16.5 illustrate calculations involving pH.

Sample Problem 16.4 Determine the pH of a solution at 25 °C in which the hydronium ion concentration is (a) 3.5 10- 4 M, (b) 1.7 X 10- 7 M, and (c) 8.8 X 10- 11 M.

X

Strategy Given [HP +], use Equation 16.2 to solve for pH. Setup

Solution

(a) pH = -log (3 .5 X 10- 4)

(a) pH = 3.46

(b) pH = - log (1.7 X 10- 7)

(b) pH = 6.77

(c) pH = -log (8 .8 X 10- 11 )

(c) pH = 10.06

Think About It When a hydronium ion concentration falls between two "benchmark" concentrations in

Table 16.3, the pH falls between the two corresponding pH values. In part (c), for example, the hydronium ion concentration (8.8 X 10- 11 M) is greater than 1.0 X 10- 11 Mbut less than 1.0 X 10- 10 M. Therefore, we expect the pH to be between 11.00 and 10.00.

pH 1.0 X 10- 10 8.8 X 10- 11* 1.0 X 10- 11

-log (1.0 X 10- 10) - log (8 .8 X 10- 11) -log (1.0 X 10- 11 )

10.00 1O.06t 11.00

*[H30 +] between two benchmark values I pH between two benchmark values

Recogni zing the benchmark concentrations and corresponding pH values is a good way to detennine whether or not your calculated result is reasonable. Practice Problem A Determine the pH of a solu tion at 25 °C in which the hydronium ion concentration is (a) 3.2 X 10- 9 M, (b) 4.0 X 10- 8 M, and (c) 5.6 X 10- 2 M. Practice Problem B Detennine the pH of a solution at 25°C in which the hydronium ion concentration is (a) 1.2 M, (b) 3.0 X 10- 11 M, and (c) 8.6 X 10- 12 M. \

Sample Problem 16.5 Calculate the hydronium ion concentration in a solution at 25°C in which the pH is (a) 4.76, (b) 11.95, and (c) 8.01. Strategy Given pH, use Equation 16.3 to calculate [H30 +]. Setup

(a) [H30 +] = 10- 476

SECTION 16.3

(b) [H30 +] = 10- 1195 (c) [H30 +] = 10-

Think About It If you use the calculated hydronium ion concentrations to recalculate pH, you will get numbers slightly different from those given in the problem. In part (a), for example, -log (1.7 X 10- 5) = 4.77. The small difference between this and 4.76 (the pH given in the problem) is due to a rounding error. Remember that a concentration derived from a pH with two digits to the right of the decimal point can have only two significant figures. Note also that the benchmarks can be used equally well in this circumstance. A pH between 4 and 5 corresponds to a hydronium ion concentration between 1 X 10- 4 lvf and 1 X 10- 5 lvf.

80 1

Solution

(a) [H30 +] = 1.7 (b) [H30 +] = 1.1 (c) [H30 +] = 9.8

X

10- 5 lvf

X

10- 12 lvf

X

10- 9 lvf

Practice Problem A Calculate the hydronium ion concentration in a solution at 25°C in which the pH is (a) 9.90, (b) 1.45, and (c) 7.01. Practice Problem B Calculate the hydronium ion concentration in a solution at 25°C in which the pH is (a) 2.11, (b) 11.59, and (c) 6.87.

ApOH scale analogous to the pH scale can be defined using the negative base-lO logarithm of the hydroxide ion concentration of a solution, [OH-].

pOH

=-

log [OH-]

Equation 16.4

Rearranging Equation 16.4 to solve for hydroxide ion concentration gives Equation 16.5 Now consider again the Kwequilibrium expression for water at 25 °C: [H30 +][OH- ]

l.0 X 10- 14

=

Taking the negative logarithm of both sides, we obtain -log ([H30 +][OH-])

=

- log (l.0 X 10- 14 )

-(log [H30 +] + log [OH - ]) = 14.00 - log [H30 +] - log [OH- ] = 14.00 (-log [H30 +])

+ (-log [OH - ])

= 14.00

And from the definitions of pH and pOH, we see that at 25 °C pH

+ pOH

= 14.00

Equation 16.6

Equation 16.6 provides another way to express the relationship between the hydronium ion concentration and the hydroxide ion concentration. On the pOH scale, 7.00 is neutral, numbers greater than 7.00 indicate that a solution is acidic, and numbers less than 7.00 indicate that a solution is basic. Table 16.5 lists pOH values for a range of hydroxide ion concentrations at 25 °C.

[OH-] (M)

pOH

0.10

l.00

1 X 10- 3

3.00

X 10- 5

5.00

Basic

1 X 10- 7

7.00

Neutral

1 X 10

9.00

Acidic

1

9

The pH Scale

1 X 1O- 1l

1l.00

1 X 10- 13

13.00

64 1

642

CHAPTER 16

Acids and Bases

Sample Problems 16.6 and 16.7 illustrate calculations involving pOH. Think About It Remember that

the pOH scale is, in essence, the reverse of the pH scale. On the pOH scale, numbers below 7 indicate a basic solution, whereas numbers above 7 indicate an acidic solution. The pOH benchmarks (abbreviated in Table 16.5) work the same way the pH benchmarks do. In part (a), for example, a hydroxide ion concentration between 1 X 10- 4 M and 1 X 10- 5 M conesponds to a pOH between 4 and 5: [OH- ] (M)

pOH

4

4.00 4.43 t 5.00

1.0 X 103.7 X 10- 5 * 1.0 X 10- 5

Determine the pOH of a solution at 25 °C in which the hydroxide ion concentration is (a) 3.7 X 10 - 5 M, (b) 4.1 X 10- 7 M, and (c) 8.3 X 10- 2 M. Strategy Given [OH- ], use Equation 16.4 to calculate pOH. Setup

(a) pOH

=

-log (3.7 X 10- 5)

(b) pOH

=

-log (4. 1 X 10- 7 )

(c) pOH

=

-log (8 .3 X 10- 2)

Solution

(a) pOH = 4.43 (b) pOH = 6.39 (c) pOH = 1.08

*[OH - ] between two benchmark values tpOH between two benchmark values

Practice Problem A Determine the pOH of a solution at 25°C in which the hydroxide ion

concentration is (a) 5.7 X 10- 12 M, (b) 7.3 X 10- 3 M, and (c) 8.5 X 10- 6 M. Practice Problem B Determine the pOH of a solution at 25°C in which the hydroxide ion 9

11

C . ._co_n_c_e_n t_ra_t_io_n_i_S_ca_)_2_.8__X__ 10__8_M __ ,C_b_)_9_.9__ X_l_0___M_,_a_n_d_ _C)_ 1_._0 _X__l _0 _ _ _M_.____________

------~I

Sample Problem 16.7 Calculate the hydroxide ion concentration in a solution at 25°C in which the pOH is (a) 4.91, Cb) 9.03, and (c) 10.55. Strategy Given pOH, use Equation 16.5 to calculate [OH- ]. Setup

Think About It Use the

benchmark pOH values to determine whether these solutions are reasonable. In part (a), for example, the pOH between 4 and 5 corresponds to [OH- ] between 1 X 10- 4 M and 1 X 10- 5 M.

(a) [OH-]

=

10- 491

(b) [OH- ]

=

10- 903

(c) [OH-] = 10-

1055

Solution

(a) [OW] = 1.2 X 10- 5 M

(b) [OH-]

= 9.3 X

10- 10 M

(c) [OH- ] = 2.8 X 10- 11 M

Practice Problem A Calculate the hydroxide ion concentration in a solution at 25 °C in which the

pOH is Ca) 13.02, (b) 5.14, and (c) 6.98. Practice Problem B Calculate the hydroxide ion concentration in a solution at 2YC in which the

pOH is Ca) 11.26, (b) 3.69, and (c) 1.60.

Bringing Chemistry to Life Antacids and the pH Balance in Your Stomach An average adult produces between 2 and 3 L of gastric juice daily. Gastric juice is an acidic digestive fluid secreted by glands in the mucous membrane that lines the stomach. It contains hydrochloric acid (Hel), among other substances. The pH of gastric juice is about 1.5, which

SECTION 16.3

corresponds to a hydrochloric acid concentration of 0.03 M a concentration strong enough to dissolve zinc metal! The inside lining of the stomach is made up of parietal cells, which are fused together to form tight junctions. The interiors of the cells are protected from the surroundings by cell membranes. These membranes allow water and neutral molecules to pass in and out of the stomach, but they usually block the movement of ions such as H +, Na +, K +, and CI- . The H + ions come from carbonic acid (H2C03) formed as a result of the hydration of CO 2, an end product of metabolism: CO?(g)

+ H 20(l) :;::.====' H 2C0 3(aq)

H 2C0 3(aq).

• H +(aq) + HC0 3 (aq)

These reactions take place in the blood plasma bathing the cells in the mucosa. By a process known as active transport, H + ions move across the membrane into the stomach interior. (Active transport processes are aided by enzymes. ) To maintain electrical balance, an equal number of CIions also move from the blood plasma into the stomach. Once in the stomach, most of these ions are prevented by cell membranes from diffusing back into the blood plasma. The purpose of the highly acidic medium within the stomach is to digest food and to activate certain digestive enzymes. Eating stimulates H + ion secretion. A small fraction of these ions normally are reabsorbed by the mucosa, causing a number of tiny hemorrhages. About half a million cells are shed by the lining every minute, and a healthy stomach is completely relined every few days. However, if the acid content is excessively high, the constant influx of H + ions through the membrane back to the blood plasma can cause muscle contraction, pain, swelling, inflammation, and bleeding. One way to temporarily reduce the H + ions concentration in the stomach is to take an antacid. The major function of antacids is to neutralize excess HCI in gastric juice. The table below lists the active ingredients of some popular antacids. The reactions by which these antacids neutralize stomach acid are as follows: NaHC0 3(aq)

+ HCI(aq) -

-+. NaCI(aq) + H 20(l) + CO 2(g)

CaC0 3(aq)

+ 2HCI(aq)

• CaCI 2(aq)

+ H 20(l) + CO 2(g)

MgC0 3 (aq)

• MgCI 2(aq)

Mg(OH)z(s)

+ 2HCI(aq) + 2HCl(aq)

• MgCI2 (aq)

+ H 2 0(l) + CO 2 (g) + 2H20 (l)

Al(OH)2NaC03(s)

+ 4HCl(aq)

• AlCl3 (aq)

+ NaCl(aq) + 3H 20(l) + CO 2 (g)

Active Ingredients in Some Common Antacids Commercial Name Active Ingredients Alka-Seltzer Milk of magnesia Rolaids TUMS Maalox

Aspirin, sodium bicarbonate, citric acid Magnesium hydroxide Dihydroxyaluminum sodium carbonate Calcium carbonate Sodium bicarbonate, magnesium carbonate

The CO 2 released by most of these reactions increases gas pressure in the stomach, causing the person to belch. The fizzing that takes place when Alka-Seltzer dissolves in water is caused by carbon dioxide, which is released by the reaction between citric acid and sodium bicarbonate:

This effervescence helps to disperse the ingredients and enhances the palatability of the solution.

Checkpoint 16.3 16.3.1

The pH Scale

Determine the pH of a solution at 25°C in which [H+] = 6.35 X 10- 8 M.

16.3.2

Determine [H+] in a solution at 25 °C if pH = 5.75.

a) 7.65

a) 1.8 X 10- 6 M

b) 6.80

b) 5.6 X 10- 9 M

d) 6.35

6 c) 5.8 X 10- M d) 2.4 X 10- 9 M

e) 8.00

6M 1.0 X 10e)

c) 7.20

The p H Scale

643

644

CHAPTER 16

Acids and Bases

16.3.3

16.3.4

Determine the pOH of a solution at 25°C in which [OH- ] = 4.65 X 10- 3 M.

Determine [OH- ] in a solution at 25°C if pH = 10.50. a) 3.2 X 10- 11 M

a) 11.67

b) 3.2 X 10- 4 M

b) 13.68

c) 1.1 X 10- 2 M

c) 0.32

d) 7.1 X 10- 8 M

d) 4.65

e) 8.5 X 10- 7 M

e) 2.33

Strong Acids and Bases

We indicate that ionization of a strong acid is complete by using a single arrow ( •) instead of the double, equilibrium arrow • ) in the equation. (.

Most of this chapter and Chapter 17 deal with equilibrium and the application of the principles of equilibrium to a variety of reaction types. In the context of our discussion of acids and bases, however, it is necessary to review the ionization of strong acids and the dissociation of strong bases. · . . . . . . . . . . . . . . .. ..................................................................................................... . These reactions generally are not treated as equilibria but rather as processes that go to completion. This makes the determination of pH for a solution of strong acid or strong base relatively simple.

Strong Acids

[ ,

There are many different acids, but as we learned in Chapter 4, relatively few qualify as strong. Multi m ed ia

Acids and Bases- the dissociation of strong and weak acids (interactive).

•

Remember that although sulfuric acid has two ionizable protons, only the first ionization is complete.

As we will see in Section 16.5, a solution of equal concentration but containing a weak acid has a higher pH.

Ionization Reaction

Strong Acid

••

• • • • • • • • • •

HCI(aq) HBr(aq) HI(aq) HN0 3(aq) HCI0 3(aq) HClOiaq) H 2 S0 4 (aq)

Hydrochloric acid Hydrobrorruc acid Hydroiodic acid Nitric acid Chloric acid Perchloric acid Sulfuric acid

+ H 20(l) + H z0(l) + H z0(l) + H 2 0(!) + H 20(l) + H 20(l)

• • • • • • •

+ H 20(!)

H30 +(aq) H 30 +(aq) H 30 +(aq) H 30 +(aq) H 30 +(aq) H 30 + (aq) H 30+(aq)

+ HS0 4 (aq)

It is a good idea to comrrtit this short list of strong acids to memory. Because the ionization of a strong acid is complete, the concentration of hydronium ion at equilibrium is equal to the starting concentration of the strong acid. For instance, if we prepare a 0.10 M solution of HCl , the concentration of hydronium ion in the solution is 0.10 M. All the HCI ionizes, and no HCl molecules remain. Thus, at equilibrium (when the ionization is complete), [HCI] = 0 M and [H30 +] = [Cl-] = 0.10 M. Therefore, the pH of the solution (at 25 °C) is pH = - log (0.10) = 1.00 · . . . . . . . . , . . . .. . . . . . . . . . . . .. .................................. ..... ................. ....................... ............... . This is a very low pH, which is consistent with a relatively concentrated solution of a strong acid. Sample Problems 16.8 and 16.9 let you practice relating the concentration of a strong acid to the pH of an aqueous solution.

Think About It Again, note that when a hydronium ion concentration falls between two of the benchmark concentrations in Table 16.3, the pH falls between the two corresponding pH values. In part (b), for example, the hydronium ion concentration of 1.2 X 10- 4 M is greater than 1.0 X 10- 4 M and less than 1.0 X 10- 3 M. Therefore, we expect the pH to be between 4.00 and 3.00.

Calculate the pH of an aqueous solution at 25°C that is (a) 0.035 M in Hl, (b) 1.2 X 10- 4 Min HN0 3, and (c) 6.7 X 10- 5 Min HCl04 .

Strategy Hl, HN0 3, and HCl0 4 are all strong acids, so the concentration of hydronium ion in each solution is the same as the stated concentration of the acid. Use Equation 16.2 to calculate pH.

pH 1.0 X 10- 3 1.2 X 10- 4 * 1.0 X 10- 4

+ CI (aq) + Br-(aq) + I - (aq) + N0 3(aq) + CI0 3(aq) + CI0 4 (aq)

- log (1.0 X 10- 3) -log (1.2 X 10- 3) - log (1.0 X 10- 4 )

3.00 3.92t 4.00

Setup (a) [H30 +] = 0.035 M (b) [H30 +]

=

1.2 X 10- 4 M

(c) [H30 +]

=

6.7 X 10- 5 M

*[H30 +] between two benchmark values I pH between two benchmark values

Solution (a) pH

Being comfortable with the benchmark hydronium ion concentrations and the corresponding pH values will help you avoid some of the common errors in pH calculations.

(b) pH

=

-log (1.2 X 10- 4 )

=

3.92

(c) pH

=

- log (6.7 X 10- 5)

=

4.17

=

-log (0.035)

=

1.46

--

Strong Acids and Bases

SECTION 16.4

------

645

Practice Problem A Calculate the pH of an aqueous solution at 25°C that is (a) 0.081 M in HI, (b) 8.2 X 10- 6 M in HN0 3, and (c) 5.4 X 10- 3 Min HCI0 4 . Practice Problem B Calculate the pH of an aqueous solution at 25 °C that is (a) 0.011 Min HN0 3, (b) 3.5 X 10- 3 MinHBr, and (c) 9.3 X 10- 10 Min HCI.

Calculate the concentration of HCI in a solution at 25 °C that has pH (a) 4.95 , (b) 3.45, and (c) 2.78. Strategy Use Equation 16.3 to convert from pH to the molar concentration of hydronium ion. In a

strong acid solution, the molar concentration of hydronium ion is equal to the acid concentration. Setup (a) [HCl]

(b) [HCI]

=

=

[H3 0 +]

[HP+] =

=

10- 495

10- 345

(c) [HCI] = [H3 0 +] = 10-

•

·.....

When we take the inverse log of a number with two digits to the right of the decimal point, the result has two significant figures.

,

278

• • • • •

Solution (a) 1.1 X 10- 5 M

• •

(b) 3.5 X 10-4 M • • (c)1.7Xlo- 3 M ·········································· ................................................................

!• I

Practice Problem A Calculate the concentration of HN0 3 in a solution at 25°C that has pH (a) 2.06,

I

(b) 1.82, and (c) 3.04.

,

Th ink About It As pH decreases, acid concentration increases.

(b) 1.77, and (c) 6.01.

Practice Problem B Calculate the concentration of HBr in a solution at 25 °C that has pH (a) 4.81 ,

I~------------------------------------------------------~.. •

• •

F

Strong Bases

•

The list of strong bases is also fairly short. It consists of the hydroxides of alkali metals (Group 1A) and the hydroxides of the heaviest alkaline earth metals (Group 2A). The dissociation of a strong base is, for practical purposes, complete. Equations representing dissociations of the strong bases are as follows:

.

-

.

--~

oS

~

Multimedia

Acids and Bases- ionization of a strong base and weak base (interadive).

Group lA hydroxides LiOH(aq) NaOH(aq) KOH(aq) RbOH(aq) CsOH(aq)

• • • • •

Li+(aq) + OH (aq) Na+(aq) + OH-(aq) K +(aq) + OH- (aq) Rb+(aq) + OH-(aq) Cs +(aq) + OH-(aq) . . .. . , . . . . . . . . . .. ............................................... ... .

Group 2A hydroxides •

• Ca2+ (aq) + 20H (aq) • Sr2+ (aq) + 20H - (aq) • Ba2+ (aq) + 20H- (aq)

Ca(OHh(aq) Sr(OHh(aq)

Ba(OHhCaq)

Again, because the reaction goes to completion, the pH of such a solution is relatively easy to calculate. In the case of a Group 1A hydroxide, the hydroxide ion concentration is simply the starting concentration of the strong base. In a solution that is 0.018 M in NaOH, for example, [OH-] = 0.018 M. Its pH can be calculated in two ways. We can either use Equation 16.1 to determine hydronium ion concentration, [H 3 0 +][OH-] = l.0

X

10- 14

[H 0 +] = l.0 X 103 [OH ]

14

= l.0

100.018 X

14

= 5.56

and then Equation 16.2 to determine pH: pH = -log (5.56 X 10- 13 M) = 12.25

X

10- 13 M

Recall that Ca(OHh and Sr(OHh are not very soluble, but what does dissolve dissociates completely [ ~ Section 4 .3, Table 4.4]

646

CHAPTER 16

Acids and Bases

or we can calculate the pOH with Equation 16.3 , pOH = - log (0.018) = 1.75 and use Equation 16.6 to convert to pH: pH + pOH = 14.00 pH = 14.00 - 1.75 = 12.25 Both methods give the same result. In the case of a Group 2A metal hydroxide, we must be careful to account for the reaction 4 stoichiometry. For instance, if we prepare a solution that is 1.9 X 10- M in barium hydroxide, 4 the concentration of hydroxide ion at equilibrium (after complete dissociation) is 2(1.9 X 10- M) 4 or 3.8 X 10- M twice the original concentration of Ba(OH)2' Once we have determined the hydroxide ion concentration, we can determine pH as before: [H 0 +] = 1.0 X 103 [OH ]

14

= 1.0

14

103.8 X 10- 4 X

2.63

=

X

10- 11 M

and pH = - log (2.63 X 10-

11

M) = 10.58

or pOH = -log (3 .8 X 10pH

+ pOH =

4

)

=

3.42

14.00

pH = 14.00 - 3.42 = 10.58 Sample Problems 16.10 and 16.11 illustrate calculations involving hydroxide ion concentration, pOH, and pH. .

--

_.

-

Sample Problem 16.10 Calculate the pOH of the following aqueous solutions at 25°C: (a) 0.013 M LiOH, (b) 0.013 M Ba(OH)2, (c) 9.2 X 10- 5 M KOH. Strategy LiOH, Ba(OHh, and KOH are all strong bases. Use reaction stoichiometry to determine

hydroxide ion concentration and Equation 16.4 to determine pOH. Setup (a) The hydroxide ion concentration is simply equal to the concentration of the base.

Therefore, [OH- ]

=

[LiOH]

= 0.013 M.

(b) The hydroxide ion concentration is twice that of the base: Ba(OHhCaq) - _ . Ba2+(aq)

Think About It These are basic

pOH values, which is what we should expect for the solutions described in the problem. Note that while the sol utions in parts (a) and (b) have the same base concentration, they do not have the same hydroxide concentration and therefore do not have the same pOH.

+ 20H - (aq)

Therefore, [OW] = 2 X [Ba(OH)2] = 2(0.013 M) = 0.026 M. (c) The hydroxide ion concentration is equal to the concentration of the base. Therefore, [OW] = [KOH] = 9.2 X 10- 5 M. Solution (a) pOH

=

- log (0.013)

=

1.89

(b) pOH = - log (0.026) = 1.59 (c) pOH = -log (9.2 X 10- 5) = 4.04

Practice Problem A Calculate the pOH of the following aqueous solutions at 25°C:

(a) 0.15 M NaOH, (b) 8.4 X 10- 3 M RbOH, (c) 1.7 X 10- 5 M CsOH. Practice Problem B Calculate the pOH of the following aqueous solutions at 25°C:

(a) 9.5 X 10- 8 MNaOH, (b) 6.1 X 10- 2 MLiOH, (c) 6.1 X 10- 2 MBa(OHh.

Sample Problem 16.11 An aqueous solution of a strong base has pH 8.15 at 25°C. Calculate the original concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH)2'

SECTION 16.5

Weak Aci ds and Acid Ionization Co nstants

Strategy Use Equation 16.6 to convert from pH to pOH and Equation 16.5 to determine the hydroxide ion concentration. Consider the stoichiometry of dissociation in each case to determine the concentration of the base itself.

• • • • • •

• • • •

Setup

647

•

Remember to keep an additional significant figure or two until the end of the problem- to avoid roundingerror [ H~ Section 1.5] .

• • •

pOH = 14.00 - 8.15 = 5.85

• •

(a) The dissociation of I mole of NaOH produces 1 mole of OH- . Therefore, the concentration of the base is equal to the concentration of hydroxide ion. (b) The dissociation of 1 mole of Ba(OHh produces 2 moles of OH- . Therefore, the concentration of the base is only one-half the concentration of hydroxide ion.

[OH- ] (a) [NaOH]

=

[OW ]

=

= 10- 585 =

1.41 X 10- 6 M

. . . . .... ..... ... . . . . . . . .. . . . . . . . . .. ...... .

1.4 X 10- 6 M

(b) [Ba(OHh] = HOW] = 7.1 X 10- 7 M

Practice Problem A An aqueous solution of a strong base has pH 8.98 at 25°C. Calculate the concentration of base in the solution (a) if the base is LiOH and (b) if the base is Ba(OH)2' Practice Problem B An aqueous solution of a strong base has pH 12.24 at 25°C. Calculate the concentration of base in the solution (a) if the base is NaOH and (b) if the base is Ba(OH)2'

Checkpoint 16.4

16.4.3

• •

Strong Acids and Bases

Calculate the pH of a 0.075 M solution of perchloric acid (HCl0 4 ) at 25 °C.

16.4.4

What is the concentration of KOH in a solution at 25 °C that has pOH 3.31?

a) 12.88

a) 2.0 X lO- II M

b) 7.75

b) 3.3 X 10- 1 M

c) 6.25

c) 3.3 X 10- 7 M

d) 1.12

d) 4.5 X 10- 4 M

e) 7.00

e) 4.9 X 10- 4 M

What is the concentration of HBr in a solution with pH 5.89 at 25°C?

16.4.5

What is the pH of a solution at 25°C that is 0.0095 Min LiOH?

a) 7.8 X 10- 9 M

a) 11.68

b) 1.3 X 10- 6 M

b) 2.02

c) 5.9 X 10- 14 M

c) 11.98

d) 8.1 X 10- 7 M

d) 1.72

e) 1.0 X 10- 7 M

e) 12.28

What is the pOH of a solution at 25 °C that is 1.3 X 10- 3 Min Ba(OHh ?

16.4.6

What is the concentration of Ca(OH)2 in a solution at 25 °C if the pH is 9.01?

a) 2.89

a) 1.0 X 10- 5 M

b) 2.59

b) 5.1 X 10- 6 M

c) 3.19

c) 2.0 X 10- 5 M

d) 11.11

d) 9.8 X 10- 9 M

e) 11.41

e) 4.9 X 10- 9 M

Weak Acids and Acid Ionization Constants Most acids are weak acids, w hich ionize only to a limited extent in water. At equilibrium, an aqueous solution of a weak acid contains a mixture of aqueous acid molecules, hydronium ions, and the corresponding conj ugate base. The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization . •

Think About It Alternatively, we could determine the hydroxide ion concentration using Equation 16.3,

• •

[H30 +]

=

10- 81 5 = 7.1 X 10- 9 M

and Equation 16.1,

•

14

[OH-] =

1.0 X 107.1 X 10- 9 M

=

1.4 X 10- 6 M

Once [OH- ] is known, the solution is the same as shown previously. ,

16.4.2

•

• • •

Solution

16.4.1

•

648

CHAPTER 16

Acids and Bases

The Ionization Constant, Ka Consider a weak monoprotic acid HA. Its ionization in water is represented by

or by

+ A -(aq)

RA(aq) :;:::.~. H +(aq)

The equilibrium expression for this reaction is .. Remember that H30+ and H+ are used interchangeably.

Ka =

[H~b + ][A- ]

[RA]

.

Remember that a Ka is a K,. The subscript a simply stands for " acid. "

For comparison, the pH of a 0.1 M solution of a strong acid such as Hel or HN0 3 is 1.0.

or ..

..

.

..

.

where Ka is the equilibrium constant for the reaction. More specifically, Ka is called the acid ionization constant. Although all weak acids ionize less than 100 percent, they vary in strength. The magnitude of Ka indicates how strong a weak acid is. A large Ka value indicates a stronger acid, whereas a small Ka value indicates a weaker acid. For example, acetic acid (CH 3COOH) and hydrofluoric acid (HF) are both weak acids, but HF is the stronger acid of the two, as evidenced by its larger Ka value. Solutions of equal concentration of the two acids do not have the same pH. The pH of the HF solution is lower.

Solution (at 25°C) 0.10MHF 0.10 M CH3COOH

pH 2.09 2.87

7.1 X 10- 4 1. 8 X 10- 5

Table 16.6 lists a number of weak acids and their Ka values at 25 °C in order of decreasing acid strength.

Calculating pH from Ka Using the method outlined in Chapter 15, we can use acid concentration and Ka to determine the equilibrium concentration of H30 +. From [H30 +], we can determine pH. Suppose we want to determine the pH of a 0.50 M HF solution at 25 °C. The ionization of HF is represented by

The equilibrium expression for this reaction is K

Name of Acid

a

=

[H O +][F-] 3 [HF]

=

7 1 X 10- 4 .

Formula

Structure

Ka

Hydrofluoric acid

HF

H-F

7 .1 X 10- 4

Nitrous acid

HN0 2

O=N - O-H

4.5 X 10- 4

0 Formic acid

HCOOH

II

l.7 X 10- 4

H-C-O-H 0 /

Benzoic acid

C 6H 5 COOH

\ ,'-..

I

"

C

0

H

6.5 X 10- 5

/,

o

, Acetic acid Hydrocyanic acid Phenol

II

CH3-C-O

H

l.8 X 10- 5 4.9 X 10- 10

H-C = N >---0

H

1.3 X 10- 10

SECTION 16.5

Weak Acids and Acid Ionization Constants ;

We construct an equilibrium table and enter the starting concentrations of all species in the equilibrium expression. HF(aq)

Initial concentration (M):

+ H 20(l) :;:::.===:!:' H 30 +(aq) + F - (aq)

o

0.50

o

Change in concentration (M): Equilibrium concentration (M):

,-

Using the reaction stoichiometry, we determine the changes in all species: HF(aq)

Initial concentration (M):

+ H 20(l) :;:::.===:!:' H 30 +(aq) + F - (aq)

0.50

o

o

-x

+x

+x

Change in concentration (M): Equilibrium concentration (M):

Finally, we express the equilibrium c0ncentration of each species in terms of x. HF(aq )

Initial concentration (M):

0.50

0

0

-x

+x

+x

0.50 - x

x

x

Change in concentration (M): Equilibrium concentration (M):

+ H 20 (l) :;:::.===:!:' H 30 +(aq) + F - (aq)

These equilibrium concentrations are then entered into the equilibrium expression to give Ka =

7.1

X

10- 4

1O- 4x - 3.55

X

10- 4 = 0

(x ) (x )

=

0.50 - x

Rearranging this expression, we get

+ 7.1

x2

X

This is a quadratic equation, which we can solve using the quadratic formula given in Appendix 1. In the case of a weak acid, however, often we can use a shortcut to simplify the calculation. Because HF is a weak acid, and weak acids ionize only to a slight extent, x must be small compared to 0.50. Therefore, we can make the following approximation: 0.50 - x = 0.50

Now the equilibrium expression becomes 1

x -,----,---=

0.50 - x

x

2

0.50

=

10- 4

7.1

X

=

3.55

Rearranging, we get x2

=

(0.50)(7 .1

X

10- 4 )

X

10- 4

x = ~3.55 X 10- 4 = 1.9 X 10- 2 M Thus, we have solved for x without having to use the quadratic equation. At equilibrium we have [HF]

=

(0.50 - 0.019) M

=

0.48 M

[H30 +] = 0.019 M

[F- ] = 0.019 M

and the pH of the solution is pH = -log (0.019) = 1.72 This shortcut gives a good approximation as long as the magnitude of x is significantly smaller than the initial acid concentration. As a rule, it is acceptable to use this shortcut if the calculated value of x is less than 5 percent of the initial acid concentration. In this case, the approximation is acceptable because 0.019 M X 100% = 3.8% 0.50M

-

... ... Remember that solids and pure liquids do not appear in the equili brium expression [ ~~

Section 15.3] .

649

650

CHAPTER 16

Acids and Bases •

This is the formula for the percent ionization of the acid [ ~~ Section 13.6] . Recall that the percent ionization of a weak electrolyte, such as a weak acid, depends on concentration. Consider a more dilute solution of HF, one that is 0.050 M. Using the preceding procedure to solve for x, we would get 6.0 X 10- 3 M. The following test shows, however, that this answer is not a valid approximation because it is greater than 5 percent of 0.050 M: 3

6.0 X 10- M X 100% = 12% 0.050 M

... Tri'this 'case',' we'must'soivd'o r 'X' using' the' quadratic equation [ ~~

Sample Problem 15 ,9] . Sample Problem 16.12 shows how to use Ka to determine the pH of a weak acid solution.

In many cases, use of the quadratic equation can be avoided with a method called successive approximation, which is presented in Appendix 1.

The Ka of hypochlorous acid (HClO) is 3.5 X 10- 8. Calculate the pH of a solution at 25°C that is 0.0075 Min HClO. Strategy Construct an equilibrium table, and express the equilibrium concentration of each

species in tenns of x. Solve for x using the approximation shortcut, and evaluate whether or not the approximation is valid. Use Equation 16.2 to detennine pH. Setup

HClO (aq)

Initial concentration (M):

+

==' H 30 +(aq) + ClO- (aq) o o

H 20(l) +:.

0.0075

-x

+x

+x

0.0075 - x

x

x

Change in concentration (M): Equilibrium concentration (M):

Solution These equilibrium concentrations are then substituted into the equilibrium expression to

Ka =

= 3.5 X

(x) (x)

10- 8

0.0075 - x Assuming that 0.0075 - x = 0.0075, ?

x-

= 3.5 X

10- 8

0.0075 x 2 = (3 .5 X 10- 8)(0.0075) ..

Solving for x, we .get . . . ... . ... .. . . ..

Applying the 5 percent test indicates that the approximation shortcut is valid in this case: (1.62

x

10 - 510.0075) x 100 %

< 5%.

x = -12.625 X 10 10 = 1.62 X 10- 5 M According to the equilibrium table, x = [H30 +]. Therefore, pH = - log (1.62 X 10- 5 ) = 4.79

Think About It We learned in Section 16.2 that the concentration of hydronium ion in pure water at 25°C is 1.0 X

10- 7 M, yet we use 0 M as the

starting concentration to solve for the pH of a solution of weak acid:

==' H +(aq) + A - (aq)

HA(aq) +:.

Initial concentration (M) :

o

o

Change in concentration (M): Equilibrium concentration (M): The reason for this is that the actual concentration of hydronium ion in pure water is insignificant compared to the amount produced by the ionization of the weak acid. We could use the actual concentration of hydronium as the initial concentration, but doing so would not change the result because 7 (x + 1.0 X 10- ) M = x M. In solving problems of this type, we neglect the small concentration of H + due to the autoionization of water. Practice Problem A Calculate the pH at 25°C of a 0.18 M solution of a weak acid that has Ka =

9.2 X 10- 6 Practice Problem B Calculate the pH at 25 °C of a 0.065 M solution of a weak acid that has Ka =

1.2 X 10- 5

SECTION 16.5

Weak Acids and Acid Ionization Constants

651

Using pH to Determine Ka In Chapter 15 we learned that we can determine the value of an equilibrium constant using equilibrium concentrations [ ~. Section 15.2]. Using a similar approach, we can use the pH of a weak acid solution to determine the value of Ka. Suppose we want to determine the Ka of a weak acid (HA) and we know that a 0.25 M solution of the acid has a pH of 3.47 at 25 °C. The first step is to use pH to determine the equilibrium hydronium ion concentration. Using Equation 16.3, we get . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . .. [H30 +] = 10- 3.47 = 3.39 X 10- 4 M We use the starting concentration of the weak acid and the equilibrium concentration of the hydronium ion to construct an equilibrium table and determine the equilibrium concentrations of all three species.

+ Initial concentration (M): Change in concentration (M):

A -(aq)

0.25

0

0

-3.39 X 10- 4

+ 3.39 X 10- 4

+ 3.39 X 10- 4

0.2497

3.39 X 10- 4

3.39 X 10- 4

Equilibrium concentration (M):

Remember that we keep extra signifi cant figures until the end of a mUlti-step problem to minimize rounding error. [ H~ Section 1.5]

These equilibrium concentrations are substituted into the equilibrium expression to give

=

K a

(3.39 X 100.2497

4 )2

= 4.6

X

10- 7 ,

Therefore, the Ka of this weak acid is 4.6 X 10- 7 . Sample Problem 16.13 shows how to determine Ka using pH.

Sample Problem 16.13 Aspirin (acetylsalicylic acid, HC 9H70 4 ) is a weak: acid. It ionizes in water according to the equation

O~ OH A 0.10 M aqueous solution of the aspirin has a pH of 2.27 at 25°C. Determine the Ka of aspirin.

o

Strategy Determine the hydronium ion concentration from the pH. Use the hydronium ion concentration to determine the equilibrium concentrations of the other species, and plug the equilibrium concentrations into the equilibrium expression to evaluate Ka.

o Aspirin

Setup Using Equation 16.3, we have [H30 +]

= 1O- z.z7

=

5.37 X 10- 3 M

To calculate Ka, though, we also need the equilibrium concentrations of C 9H70 4 and HC 9H70 4 . The stoichiometry of the reaction tells us that [C 9H70 4 ] = [H30 +]. FUlthermore, the amount of aspirin that has ionized is equal to the amount of hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is (0.10 - 5.37 X 10- 3) M = 0.095 M. HC 9H 70 iaq)

Initial concentration (M):

+ HzO (l) +.==' H30 +(aq) + C 9H 7 0 4 (aq)

0.10

Change in concentration (M): Equilibrium concentration (M):

- 0.005 0.095

0

0

+ 5.37 X 10- 3 + 5.37 X 10- 3 5.37

X

10- 3

5.37

X

10

3

Solution Substitute the equilibrium concentrations into the equilibrium expression as follows: Think About It Check your work

The Ka of aspirin is 3.0

X

10- 4 .

Practice Problem A Calculate the Ka of a weak acid if a 0.065 M solution of the acid has a pH of 2.96 at 25°C. Practice Problem B Calculate the Ka of a weak acid if a 0.015 M solution of the acid has a pH of 5.03 at 25°C.

by using the calculated value of Ka to solve for the pH of a 0.10 M solution of aspirin.

652

CHAPTER 16

Acids and Bases

Checkpoint 16.5 16.5.1

Weak Acids and Acid Ionization Constants

The Ka of a weak acid is 5.5 X 10- 4 What is the pH of a 0.63 M solution of this acid at 25 °C?

16.5.2

A 0.042 M solution of a weak acid has pH 4.01 at 25°C. What is the Ka of this acid?

a) 3.26

a) 9.5 X 10- 9

2 b) 1.83 X 10-

b) 9.8 X 10- 5

c) 1.73

c) 0.91

d) 1.63

d) 4.2 X 10- 2

e) 0.201

e) 2.3 X 10- 7

Weak Bases and Base Ionization Constants Just as most acids are weak, most bases are also weak. The ionization of a weak base is incomplete and is treated in the same way as the ionization of a weak acid. In this section, we will see how the ionization constant for a weak base, Kb , is related to the pH of an aqueous solution.

The Ionization Constant, Kb ,

The ionization of a weak base can be represented by the equation

where B is the weak base and HB + is its conjugate acid. The equilibrium expression for the ionization is

where Kb is the equilibrium constant specifically known as the base ionization constant. Like weak acids, weak bases vary in strength. Table 16.7 lists a number of common weak bases and their ionization constants. The ability of anyone of these substances to act as a base is the result of the lone pair of electrons on the nitrogen atom. The presence of this lone pair is what enables a compound to accept a proton, which is what makes a compound a Br0nsted base.

Name of Base

Ethylamine

Formula C2H sNH2

Structure ••

CH-CH -N-H 3

I

2

Kb 5 .6 X 10- 4

H

Methylamine

CH3NH?

••

4.4 X 10- 4

CH ' -N-H

I

3

H

Ammonia

NH3

••

1.8 X 10- 5

H-N-H

I

H

1.7 X 10- 9

Pyridine

'/

••

>--N-H

Aniline

3.8

X

10- 10

I

./

H

Urea

..

o II

..

H-N- C-N-H

I H

1.5 X 10- 14

I H •

SECTION 16.6

Weak Bases and Base Ionization Constants

653

Cakulating pH from Kb Solving problems involving weak bases requires the same approach we used for weak acids. It is important to remember, though, that solving for x in a typical weak base problem gives us the hydroxide ion concentration rather than the hydronium ion concentration. Sample Problem 16.14 shows how to use Kb to calculate the pH of a weak base solution.

What is the pH of a 0.040 M ammonia solution at 25°C ? Strategy Construct an equilibrium table, and express equilibrium concentrations in terms of the unknown x. Plug these equilibrium concentrations into the equilibrium expression, and solve for x. From the value of x, determine the pH. Setup N H 3(aq)

Initial concentration (M):

0.040

o

o

-x

+x

+x

0.040 - x

x

x

Change in concentration (M): Equilibrium concentration (M):

+ H 20 (l) :;::.=~. N H ;(aq) + OW (aq)

.

Solution The equilibrium concentrations are substituted into the equilibrium expression to give = 1.8

(x) (x)

X 10- 5

0.040 - x Assuming that 0.040 - x = 0.040 and solving for x gives (x)(x) 0.040 - x x2

=

Applying the 5 percent test indicates that the approximation shortcut is valid in this case: (8.49 x 10- 4/0.040) x 100% = 2%.

(x)(x) = 1. 8 X 10-5

0.040

= ( 1.8

X

10- 5 )(0.040)

= 7.2

10- 7

X .

x = ')7.2 X 10

7

= 8.5

X

.

..

. ...

... ..

.

...... .

10- 4 M

According to the equilibrium table, x = [OH- j. Therefore, pOH = -log (x) : -log (8.5 X 10- 4 ) = 3.07 and pH = 14.00 - pOH = 14.00 - 3.07 = 10.93. The pH of a 0.040 M solution of NH3 at 25°C is 10.93.

Think About It It is a common error in Kb problems to forget that x is the hydroxide ion concentration rather than the hydronium ion concentration. Always make sure that the pH you calculate for a solution of base is a basic pH, that is, a pH greater than 7.

Practice Problem A Calculate the pH at 25 °C of a 0.0028 M solution of a weak base with a Kb of 6.8 X 10- 8 . Practice Problem B Calculate the pH at 25°C of a 0.1 6 M solution of a weak base with a Kb of 2.9 X 10- 11 .

Using pH to Determine Kb Just as we can use pH to determine the Ka of a weak acid, we can also use it to determine the Kb of a weak base. Sample Problem 16.15 demonstrates this procedure.

~o

Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the equation

N

N

I

A 0.15 M solution of caffeine at 25°C has a pH of 8.45. Determine the Kb of caffeine.

o ( Continued)

Caffeine

'-...CH

3

654

CHAPTER 16

Acids and Bases

Strategy Use pH to detennine pOH, and pOH to detennine the hydroxide ion concentration. From the hydroxide ion concentration, use reaction stoichiometry to detennine the other equilibrium concentrations and plug those concentrations into the equilibrium expression to evaluate Kb • Setup

pOH [OH- l

=

=

14.00 - 8.45

10- 5 55

= 2.82

= 5.55 X

10- 6 M

Based on the reaction stoichiometry, [HC 8 H lON 4 0 r l = [OH- ], and the amount of hydroxide ion in solution at equilibrium is equal to the amount of caffeine that has ionized. At equilibrium, therefore, [C 8H lO N 4 0 2 J = (0.15 - 2.82 C 8H ION 40 2(aq )

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M): Think About It Check your answer by using the calculated Kb to detennine the pH of a 0.15 M solution.

x

10- 6 ) M

= 0.15 M

+ H 20 (I) :;:,=::!:' HCgHlON40; (aq) +

OH- (aq)

0.15

0

0

-2.82 X 10- 6

+2.82 X 10- 6

+2.82 X 10- 6

0.15

2.82 X 10- 6

2.82 X 10- 6

Solution Plugging the equilibrium concentrations into the equilibrium expression gives

Practice Problem A Detennine the Kb of a weak base if a 0.50 M solution of the base has a pH of 9.59 at 25 °C. Practice Problem B Detennine the Kb of a weak base if a 0.35 M solution of the base has a pH of 1l.84 at 25°C.

Checkpoint 16.6 16.6.1

Weak Bases and Base Ionization Constants

What is the pH of a 0.63 M solution of weak base at 25°C if Kb = 9 .5 X 1O- 7 ?

16.6.2

A 0.12 M solution of a weak base has a pH of 10.76 at 25°C. Determine Kb .

a) 3.11

a) 2.5 X 10- 21

b) 10.89

b) 8.3 X 108

c) 7.00

c) 4.0 X 10- 8

d) 1.12

d) 2.8 X 10- 6

e) 12.88

e) l.0

X

10- 14

Conjugate Acid-Base Pairs At the beginning of this chapter, we introduced the concept of conjugate acids and conjugate bases. In this section, we examine the properties of conjugate acids and bases, independent of their parent compounds.

The Strength of a Conjugate Acid or Base When a strong acid such as HCI dissolves in water, it ionizes completely because its conjugate base (CI-) has essentially no affinity for the H+ ion in solution. HCI(aq) --+, H +(aq)

The products HCI and OW do not actually form when CI- and H2 0 are combined.

+ CI-(aq)

Because the chloride ion has no affinity for the H+ ion, it does not act as a Br¢nsted base in water. If we dissolve a chloride salt such as NaCI in water, for example, the Cl- ions in solution would not accept protons from the water: . ... .................. ...... ....... .............................. ... . CI-(aq) + H 20(l) X ' HCI(aq) + OH- (aq) The chloride ion, which is the conjugate base of a strong acid, is an example of a weak conjugate base.

SECTION 16.7

Conjugate Acid-Base Pairs

655

Now consider the case of a weak acid. When HF dissolves in water, the ionization happens only to a limited degree because the conjugate base, F- , has a strong affinity for the H + ion: HF(aq) :;:.====' H +(aq)

+ F -(aq)

This equilibrium lies far to the left (Ka = 7.1 X 10- 4) . Because the fluoride ion has a strong affinity for the H + ion, it acts as a BrQnsted base in water. If we were to dissolve a fluoride salt, such as . . . . . . .. . . . . . . . . . . . N aF in water, the F - ions in solution would, to some extent, accept protons from water:

Note that one of the products of the reaction of a conjugate base with water is always the corresponding weak acid .

The fluoride ion, which is the conjugate base of a weak acid, is an example of a strong conjugate base. Conversely, a strong base has a weak conjugate acid and a weak base has a strong conjugate acid. For example, H 2 0 is the weak conjugate acid of the strong base OH - , whereas the ammonium ion (NHt) is the strong conjugate acid of the weak base ammonia (NH3)' When an ammonium salt is dissolved in water, the ammonium ions donate protons to the water molecules:

In general, there is a reciprocal relationship between the strength of an acid or base 'ind i1it~· siren·gin ······ of its conjugate.

Conjugate Example base Formula

Acid strong

weak

II

weak conjugate

HN0 3

, HCN

N0 3

Base strong

Conjugate Example acid Formula

strong conjugate

weak conjugate

OH-

•

I

I CN-

weak

The Group 1A and heavy Group 2A metal hydroxides are classified as strong bases. It is the hydroxide ion itself, however, that accepts a proton and is therefore the Br0nsted base. Soluble metal hydroxides are simply sources of the hydroxide ion.

I strong conjugate

NH3

NH+ 4

•

I

I

...... ........ ..... ............... . It is important to recognize that the words strong and weak do not mean the same thing in the context of conjugate acids and conjugate bases as they do in the context of acids and bases in general. A strong conjugate reacts with water-either accepting a proton from it or donating a proton to it to a small but measurable extent. A strong conjugate acid acts as a weak BrQnsted acid in water; and a strong conjugate base acts as a weak BrQnsted base in water. A weak conjugate does not react with water to any measurable extent. . . . . . . . . . . . . ..

A strong conjugate acid is a weak Br0nsted acid. A strong conjugate base is a weak Br0nsted base.

The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair Because it accepts a proton from water to a small extent, what we refer to as a "strong conjugate base" is actually a weak BrQnsted base. Therefore, every strong conjugate base has an ionization constant, Kb • Likewise, every strong conjugate acid, because it acts as a weak BrQnsted base, has an ionization constant, Ka. A simple relationship between the ionization constant of a weak acid (Ka) and the ionization constant of its conjugate base (Kb) can be derived as follows , using acetic acid as an example: CH3COOH(aq) :;:.===:!:' H +(aq)

+ CH 3COO-(aq)

acid

. ...

conjugate base

. . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... .., .........,. .... ...... ..... ....... ...... . ...... ..... .

The conjugate base, CH3COO- , reacts with water according to the equation

and the base ionization equilibrium expression is written as [CH3COOH] [OH-] Kb = [CH3COO]

. . . , . . ..

A conjugate base such as the acetate ion is introduced into a solution by dissolving a solu ble salt containing acetate. Sodium acetate, for example, can be used to supply the acetate ion. The sodium ion does not take part in the reaction-it is a spectator ion [ ~~ Section

4.2].

656

CHAPTER 16

Acids and Bases

As for any chemical equations, we can add these two equilibria and cancel identical terms:

+ _CH3COO (aq) S::H3COOII(aq) + OH- Caq)

_CH 3COOII(aq) :;:::,====:' H +(aq) DA"'1(-;;'aq;;))

+ H )O(l)

,

'

The sum is the autoionization of water. In fact, this is the case for any weak acid and its conjugate base:

. . , ..

.. . .. .

. .. . .. .

. .. . . .

. . .. . . .

.

1M :;: :,====:' H + + Pi" + K + H 20 , ' 1M + OH ... H 2 0 :;::,=::z:' H +

+ OH-

ar for any weak base and its conjugate acid .

.B + H2 0

• •

..

Remember that the hydronium ion can be expressed as either H+ or H30+ These are two equivalent ways to represent the autoionization of water.

... . ..

..

.............. .. ..... ..

:;::,====:'

+ ~ + H20. . . . .. . ..

'

.wr + 0 H J1 + H30 +

2H2 0 :;::.====:' H 30 +

+ OH -

Recall that when we add two equilibria, the equilibrium constant for the net reaction is the product of the equilibrium constants for the individual equations [ ~~ Section 15 .3]. Thus, for any conjugate acid-base pair, Equation 16.7 Equation 16.7 gives the quantitative basis for the reciprocal relationship between the strength of an acid and that of its conjugate base Cor between the strength of a base and that of its conjugate acid). Because Kwis a constant, Kb must decrease if Ka increases, and vice versa. Sample Problem 16.16 shows how to determine ionization constants for conjugates.

Sample Problem 16.16 Determine (a) Kb of the acetate ion (CH 3COO-), (b) Ka of the methylammonium ion (CH3NH~ ) , (c) Kb of the fluoride ion (F -), and (d) Ka of the anunonium ion (NHt) . Strategy Each species listed is either a conjugate base or a conjugate acid . Determine the identity

of the acid correspondi ng to each conjugate base and the identity of the base corresponding to each conjugate acid ; then, consult Tables 16.6 and 16.7 for their ioni zation constants. Use the tabulated ionization constants and Equation 16.7 to calculate each indicated K value. Setup (a) A Kb value is requested, indicatin g that the acetate ion is a conjugate base. To identify the

corresponding BrSil nsted acid, add a proton to th e formula to get CH 3COOH (acetic acid). The Ka of acetic acid (from Table 16.6) is 1.8 X 10-'. (b) A Ka value is requested, indicating that the methylammonium ion is a conjugate acid. Determine the identity of the corresponding BrSilnsted base by removing a proton from the formula to get CH3NH2 (methylamine). The Kb of methylamine (from Table 16.7) is 4.4 X 10- 4 (c) F- is the conjugate base of HF; K a = 7.1 X 10- 4 (d) NHt is the conjugate acid ofNH 3; Kb = 1.8 X 10- 5 Solving Equation 16.7 separately for Ka and Kb gives, respectively, and 14

Solution (a) Conjugate base CH 3COO- : Kb = 1.0 X 10- = 5.6 X 10- 10 5

1.8 X 10-

Think About It Because the

conjugates of weak acids and bases have ionization constants, salts containing these ions have an effect on the pH of a solution. In Section 16.10 we will use the ioni zation constants of conjugate acids and conjugate bases to calculate pH for solutions containing dissolved salts.

---'-'---'--' -~-~------

-

14

(b) Conjugate acid CH 3NHi : Ka = 1.0 X 104.4 X 10- 4 14

(c) Conjugate base F - : Kb = 1.0 X 107 .1 X 10- 4

=

= 2.3

X

10- 11

1.4 X 10- 11

(d) Conjugate acid NHt : Ka = 1.0 X 1O- 1~ = 5.6 X 10- 10 1.8 X 10- )

SECTION 16.8

,I I

Diprotic and Polyprotic Acids

657

Practice Problem A Determine (a) Kb of the benzoate ion (C 6H sCOO - ), (b) Kb of the ascorbate ion (HC 6H 60 6 ), and (c) Ka of the ethyl ammonium ion (C2HsNH~ ). Practice Problem B Determine (a) Kb of the weak base B whose conjugate acid HB + has Ka. = 8.9 X 10- 4 and (b) Ka of the weak acid HA whose conjugate base has Kb = 2.1 X 10- 8

I ____________________________________________________________________ ~i

Checkpoint 16.7 16.7.1

~EI

Conjugate Acid-Base Pairs 16.7.2

Calculate the Kb of the cyanide ion (CW).

Which of the anions listed is the strongest base? (See Table 16.6.)

a) 4.9 X 10- 10 b) 2.0 X lO- s

b) Benzoate ion (C 6H sCOO - )

c) 4.9 X 10- 24

c) Nitrite ion (N0 2 )

d) 1.0 X 10- 7 e) 2.2 X lO- s

d) Phenolate ion (C 6H sO- )

a) Ascorbate ion (HC6H60 (j )

e) Formate ion (HCOO- )

Diprotic and Polyprotic Acids ,

Diprotic and polyprotic acids undergo successive ionizations, ..losing one proton at a time .. [ ~~ Sec. ..,. . . . . . . . . . . . .. . . . . . .. tion 4 .3], and each ionization has a Ka associated with it Ionization constants for a diprotic acid are designated Ka J and Ka2 . We write a separate equilibrium expression for each ionization, and we may need two or more equilibrium expressions to calculate the concentrations of species in solution at equilibrium. For carbonic acid (H 2C03 ), for example, we write ,

[H+][HC0 3]

K

= - -- - -

[H 2 C0 3]

aJ

HC0 3 (aq)

+.==' H+(aq) +

CO~ - (aq)

K a,

=

[H + ][CO ~ -]

[HC0 3 ]

Note that the conjugate base in the first ionization is the acid in the second ionization, Table 16,8 shows the ionization constants of several diprotic acids and one polyprotic acid, For a given acid, the first ionization constant is much larger than the second ionization constant, and so on, This trend makes sense because it is easier to remove a proton from a neutral species than from one that is negatively charged, and it is easier to remove a proton from a species with a single negative charge than from one with a double negative charge, Sample Problem 16.17 shows how to calculate equilibrium concentrations of all species in solution for an aqueous solution of a diprotic acid.

Oxalic acid (H 2C 2 0 4) is a poisonous substance used mainly as a bleaching agent. Calculate the concentrations of all species present at equilibrium in a 0.10 M solution at 25°C.

Strategy Follow the same procedure for each ionization as for the determination of equilibrium concentrations for a monoprotic acid. The conjugate base resulting from the first ionization is the acid for the second ionization, and its starting concentration is the equilibrium concentration from the first ionization.

Setup The ionizations of oxalic acid and the corresponding ionization constants are H 2C 20 4 (aq)

+.==' H +(aq) + HC 20 4 (aq )

HC 2 0 4 (aq)'

' H +(aq)

+ C20~-(aq)

KaJ = 6.5

Ka2

=

6,1

X 10- 2 X 10- 5

Construct an equilibrium table for each ionization, using x as the unknown in the first ionization and y as the unknown in the second ionization.

(Continued)

A triprotic acid has KaI' Ka1 , and Ka3 .

658

CHAPTER 16

Acids and Bases

Structure

Formula

Name of Acid

Ka 3

Ka 2

Ka 1

0 Sulfuric acid

II

H-O - S-O-H

H2SO 4

II

x 10- 2

Very large

1.3

6.5 X 10- 2

6.1 X 10- 5

1.3 X 10- 2

6.3 X 10- 8

8.0 X 10- 5

1.6 X 10- 12

0

o 0 II II

H-O-C-C-O-H

Oxalic acid

o II

H-O-S-O-H

Sulfurous acid

H-O

O-H

"c

C/

H " / \ iC""-O'/C=O

Ascorbic acid (vitamin C)

CHOH

I

CH?OH

-

•

o

II

Carbonic acid

H-O-C-O-"H

4.2 X 10- 7

4.8 X 10- 11

Hydrosulfuric acid*

H-S-H

9.5

X 10- 8

1 X 10- 19

o

II

H-O-P-O-H

Phosphoric acid

7.5 X 10- 3

6.2 X 10- 8

4.2 X 10- 13

I

o I

H * Th e second ionization constant of H2 5 is very low and difficult to measure. The va lue in this table is an est im ate.

H 2 C 2 0iaq) :;:.=::!:' H +(aq)

0.10

o

o

-x

+x

+x

0.10 - x

x

x

Initial concentration (M): Change in concentration (M) : Equilibrium concentration (M):

+ HC 20 4 (aq)

The equilibrium concentration of the hydrogen oxalate ion (HC 2 0 4 ) after the first ionization becomes the starting concentration for the second ionization. Additionally, the equilibrium concentration of H+ is the starting concentration for the second ionization.

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

x

x

o

- y

+y

+y

x - y

x+y

y

Solution K

=

[H+][HC2 0 4"l [H2C20 4 l

a,

6.5 X 10- 2 =

.

2

x 0.10 - x

Applying the approximation and neglecting x in the denominator of the expression gives 2

6.5 X 10- 2 = Ox 10 6.5

X

10- 3

x = 8.1

X

10- 2 M

X

2

=

SECTION 16.8

Diprotic and Po lyprotic Acids

Testing the approximation, 2

8.1 X 10- M X 100% = 81% O.lOM Clearly the approximation is not valid, so we must solve the following quadratic equation: X

Z

+ 6.5 X lO- z x - 6.5 X 10- 3

= 0

The result is x = 0.054 M. Thus, after the first ioni zation, the concentrations of species in solution are [H+l = 0.054 M [HC2 0 4 l = 0.054M [HZC2 0 4l = (0.10 - 0.054) M = 0.046 M Rewriting the equilibrium table for the second ionization, using the calculated value of x, gives the following: HC Z0 4 (aq) +.=::!:' H +(aq) + C20 ~ - (aq) Initial concentration (M):

0.054

0.054

o

-y

+y

+y

Change in concentration (M) :

0.054 - y

Equilibrium concentration (M) :

+Y

0.054

y

K = [H+ ][C20~ - l a, [HC2 0 4 l 6.1 X 10- 5 = _(0_.0_S4_+-=--y)---,(y_) 0.054 - y Assuming that y is very small and applying the approximations 0.054 0.054 gives

(0.OS4)(y) ---=--=-=,--'-'0.054

=

y

=

6. 1

X

+ Y = 0.054 and 0.054

- y=

10 - ) M

We must test the approximation as follows to see if it is valid :

Think About It Note that

5

6.1 X 1O- M X 100'7l = 011 '7l 0.054 M O. 0 This time, because the ionization constant is much smaller, the approximation is valid. At equilibrium, the concentrations of all species are [H 2C20 4 l = 0.046 M [HC20 4 l = (0.054 - 6.1 X 10- 5) M = 0.054 M [H+l = (0.054

+ 6.1

X 10- 5) M = 0.054 M

[CzO~- l = 6. 1 X 10- 5 M

Practice Problem A Calculate the concentrations of H ZC20 4 , HC 2 0

4,

C20~- , and H + ions in a

0.20 M oxalic acid solution at 2S°C. Practice Problem B Calculate the concentrations of H2S0 4 , HS0 4 , SO ~- , and H + ions in a 0.14 M

sulfuric acid solution at 25°C.

Checkpoint 16.8 16.8 .1

Diprotic and Polyprotic Acids

Calculate the equilibrium concentration of CO ~ - in a 0.050 M solution of carbonic acid at 25°C.

16.8.2

What is the pH of a 0.40 M solution of phosphoric acid at 25°C? a) 5.48

a) 4.2 X 10- 7 M

b) 1.26

b) 4. 8 X 10- 11 M

c) 3.98

c) 1.5 X 10- 4 M

d) 12.74

d) O.OSOM

e) 0.80

e) 0.049 M

the second ionization did not contribute significantly to the H + concentration. Therefore, we could determine the pH of this solution by considering only the first ionization. This is true in general for polyprotic acids where K a , is at least 1000 X K a, . [It is necessary to consider the second ionization to determine the concentration of oxalate ion (CzO ~-). l

659

660

CHAPTER 16

Ac ids and Bases

Molecular Structure and Acid Strength The strength of an acid is measured by its tendency to ionize: • H+ + X-

HX

Two factors influence the extent to which the acid undergoes ionization. One is the strength of the H- X bond. The stronger the bond, the more difficult it is for the HX molecule to break up and hence the weaker the acid. The other factor is the polarity of the H - X bond. The difference in the electronegativities between H and X results in apolar bond like

0+ 0H-X If the bond is highly polarized (i.e. , if there is a large accumulation of positive and negative charges on the H and X atoms, respectively), HX will tend to break up into H + and X - ions. A high degree of polarity, therefore, gives rise to a stronger acid. In this section, we consider the roles of bond strength and bond polarity in determining the strength of an acid.

Hydrohalic Acids

The polarity of the H- X bond actually decreases from H- F to H-I, largely because F is the most electronegative element. This would suggest that HF would be the strongest of the hydrohalic acids. Based on the data in Table 16.9, however, bond enthalpy is the more important factor in determining the strengths of these acids.

The halogens form a series of binary acids called the hydrohalic acids (HF, HCI, HBr, and HI). Table 16.9 shows that of this series only HF is a weak acid (Ka = 7.1 X 10- 4) . The data in the table indicate that the predominant factor in determining the strength of the hydrohalic acids is . ... bond strength. HF has the largest bond enthalpy, making its bond the most difficult to break. In this series of binary acids, acid strength increases as bond strength decreases. The strength of the acids increases as follows: HF < < HCI < HBr < HI

Oxoacids An oxoacid, as we learned in Chapter 2, contains hydrogen, oxygen, and a central, nonmetal atom [ ~~ Section 2 .7] . As the Lewis structures in Figure 16.2 show, oxoacids contain one or more 0- H bonds. If the central atom is an electronegative element, or is in a high oxidation state, it will attract electrons, causing the O - H bond to be more polar. This makes it easier for the hydrogen to be lost as H +, making the acid stronger.

Acid Strengtlis

Bond

Bond Enthalpy (kJ/mol)

Acid Strength

H-F

562.8

Weak

H-CI

431.9

Strong

H-Br

366.1

Strong

H-I

298.3

Strong

Figure 16.2

Lewis structures of some common oxoacids.

• •

• •

'0'

'0 ' oo

II

oo

••

•

oo

•

II

oo

oo

H-OC-O-H •• ••

H-O-N=O:

H-O-N-O: •• ••

Carboni c acid

N itrous acid

N itric acid • •

• •

'0 '

'0'

• •

'0' II H- O-P-O•. I . • oo

II H-O . . -P-O-H I .•

H-O-S-OH II

'0' • •

.0 . • •

oo

oo

H

H Phosphorous acid

oo

I H Phosphoric acid

oo

II

oo

oo

oo

Sulfuric acid

SECTION 16.9 ••

••

••

••

Molecular Structure and Acid Strength

••

H-O-CI: •• ••

H-O-C1-0: •• •• ••

Hypochlorous acid (+ 1)

Chlorous acid (+3 )

661

Figure 16.3

Lewis structures of the oxoacids of chlorine. The oxidation number of the Cl atom is shown •

in parentheses. Note that although hypochlorous acid is written as HClO,

••

:0:

••

.. I .. H-O-Cl-O: .. I ..

:0: •.

I

..

H-O-CI-O: •• •• ••

the H atom is bonded to the

:0: •• Perchloric acid (+7)

Chloric acid (+5)

°

atom.

To compare their strengths, it is convenient to divide the oxoacids into two groups: 1. Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number. Two examples are ••

••

• • '0'

I

••

H-O ••

•

Cl ••

• • '0'

••

0: ••

••

H-O ••

I

Br ••

••

0: ••

Within this group, acid strength increases with increasing electronegativity of the central atom. Cl and Br have the same oxidation number in these acids, + S. However, because Cl is more electronegative than Br, it attracts the electron pair it shares with oxygen (in the CI-O-H group) to a greater extent than Br does (in the corresponding Br-O-H group). Consequently, the O-H bond is more polar in chloric acid than in bromic acid and ionizes more readily. The relative acid strengths are HCI0 3 > HBr03

2. Oxoacids having the same central atom but different numbers of oxygen atoms. Within this •

group, acid strength increases with increasing oxidation number of the central atom. Consider the oxoacids of chlorine shown in Figure 16.3. In this series the ability of chlorine to draw electrons away from the OH group (thus making the O-H bond more polar) increases with the number of electronegative 0 atoms attached to Cl. Thus, HCI0 4 is the strongest acid because it has the largest number of oxygen atoms attached to Cl. The acid strength decreases as follows: HCIQ4

As the number of attached oxygen atoms

increases. the oxidation number of the central atom also increases [ ~~ Section 4.4] .

> HCI0 3 > HCI0 2 > HCIO

Sample Problem 16.18 , compares acid strengths based on molecular structure.

Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HErO, and HIO; (b) HNO} and RN02. Strategy In each group, compare the electronegativities or oxidation numbers of the central atoms to determine which O-H bonds are the most polar. The more polar the O-H bond, the more readily it is broken and the stronger the acid., Setup (a) In a group with different central atoms, we must compare electronegativities, The electronegativities of the central atoms in this group decrease as follows: Cl > Br > L

.'

, , ,

(b) These two acids have the same central atom but differ in the number of attached oxygen atoms. In a group such as this, the greater the number of attached oxygen atoms, the higher the oxidation number of the central atom and the stronger the acid,

, •

Another way to compare the strengths of these two is to remember that HN03 is one of the seven strong acids, HN0 2 is not.

,

Solution (a) Acid strength decreases as follows: HCIO > HErO> RIO, . . . . ... . . ..... . .. (b) RN0 3 is a stronger acid than HN02 .

·. ...

.

..... . . . . . . , . . . .. ........

Practice Problem Indicate which is the stronger acid: (a) HBr03 or HBr04; (b) H2Se04 or H 2 S04, •

.

.. .

Think About It Four of the strong acids are oxoacids: RN0 3, HCI0 4, HCI0 3, and H 2 S04,

662

CHAPTER 16

Acids and Bases

Carboxylic Acids So far our discussion has focused on inorganic acids. A particularly important group of organic acids is the carboxylic acids, whose Lewis structures can be represented by • •

"0' II ..

R-C-O-H .. You learned in Chapter 4 [ ~~ Section 4 . 1] that carboxylic acid formulas are often written with the ionizable H atom last, in order to keep the functional group together. You should recognize the formulas for organic acids written either way. For example, acetic acid may be written as HC 2 H3 0 2 or as CH 3COOH.

...... where'ids'part'Of the 'acid'iiioiecuie 'aiid'the 'shaded portioii represents'fue'carijoxyi" group:-COOH. The conjugate base of a carboxylic acid, called a carboxylate anion; RCOO- , can be represented by more than one resonance structure:

.. -

• •

• • '0'

U

R

C

.. -

0: ..

,

•

R

•

C

0:•

In terms of molecular orbital theory [ ~~ S ecti on 9 .6] , we attribute the stability of the anion to its ability to spread out or delocalize the electron density over several atoms. The greater the extent of electron delocalization, the more stable the anion and the greater the tendency for the acid to undergo ionization that is, the stronger the acid. The strength of carboxylic acids depends on the nature of the R group. Consider, for example, acetic acid and chloroacetic acid: • •

H

I C I

H

• •

'0'

II

CI "0'

.. 0..

C

I H-C I

H

H

II

C

..

0..

H

H Chloroacetic acid (Ka = l.4 x 10-3)

Acetic acid (Ka = l.8 x 10- 5)

The presence of the electronegative CI atom in chloroacetic acid shifts the electron density toward the R group, thereby making the 0- H bond more polar. Consequently, there is a greater tendency for chloroacetic acid to ionize:

..

..

• •

:CI: U

:CI: U H-

I C I

II

C

• •

..

O-H ..

•

,

H

I C I

II

C

.. 0: ..

+

H+

H

H

Chloroacetic acid is the stronger of the two acids. •

Acid-Base Properties of Salt Solutions In Section 16.7,. . .we the conjugate base of a weak acid acts as a weak Br\?lnsted base in . . . . .saw . . . . . .that . . - . . .. . . . . water. Consider a solution of the salt sodium fluoride (NaF). Because NaF is a strong electrolyte, it dissociates completely in water to give a solution of sodium cations (Na +) and fluoride anions (F- ). The fluoride ion, which is the conjugate base of hydrofluoric acid, reacts with water to produce hydrofluoric acid and hydroxide ion:

. . . . . . . . .. . . . . . . . . . . . . . . . .

Recall that a salt is an ionic compound formed by the reaction between an acid and a base [ ~~ Section 4.3] . Salts are strong electrolytes that dissociate completely into ions.

F - (aq)

+ H 20(l) :;:.:=:z;' HF(aq) + OH- (aq)

This is a specific example of salt hydrolysis, in which ions produced by the dissociation of a salt react with water to produce either hydroxide ions or hydronium ions thus impacting pH. Using our knowledge of how ions from a dissolved salt interact with water, we can determine (based on the identity of the dissolved salt) whether a solution will be neutral, basic, or acidic . Note in the preceding example that sodium ions (Na +) do not hydrolyze and thus have no impact on the pH of the solution.

Basic Salt Solutions Sodium fluoride is a salt that dissolves to give a basic solution. In general, an anion that is the conjugate base of a weak acid reacts with water to produce hydroxide ion. Other examples include the acetate ion (CH 3COO - ), the nitrite ion (N0 2 ), the sulfite ion (SO~ - ), and the hydrogen carbonate

SECTION 16.10

Acid-Base Properties of Salt Solutions

... .... .. . . .. . ... . . ion (HCO)). Each of these anions undergoes hydrolysis to produce the corresponding weak acid and hydroxide ion: . .. ...

663

....... ..... .......

A - (aq)

+ H 20(l) :;::.=::=:" HA(aq) + OH- (aq)

We can therefore make the qualitative prediction that a solution of a salt in which the anion is the conjugate base of a weak acid will be basic. We calculate the pH of a basic salt solution the same way we calculate the pH of any weak base solution, using the Kb value for the anion. The necessary Kb value is calculated usi ng the tabulated Ka value of the corresponding weak acid (see Table 16.6). Sample Problem 16.19 shows how to calculate the pH of a basic salt solution.

HC03" has an ionizable proton and can also act as a Br0nsted acid. However, its tendency to accept a proton is stronger than its tendency to donate a proton:

=z·

HC0 3 + H20 :;:.

H2C0 3 + OW

Kb = 10- 8

Remember that for any conjugate acid-base pair (Equation 16.7):

Calculate the pH of a 0.10 M solution of sodium fluoride (NaF) at 25°C. Strategy A solution of NaF contains Na + ions and F - ions. The F - ion is the conjugate base of the weak acid, HF. Use the Ka value for HF (7.1 X 10- 4 , from Table 16.6) and Equation 16.7 to determine Kb for F- : 14

1.0 X 10- = 1.4 X 107.1 X 10- 4

11

,

Then, solve this pH problem like any equilibrium problem, using an equilibrium table. Setup It's always a good idea to write the equation corresponding to the reaction that takes place along with the equilibrium expression: F-(aq)

+ H zO(I) ::;:.==" HF(aq) + OW(aq)

Construct an equilibrium table, and determine, in terms of the unknown x, the equilibrium concentrations of the species in the equilibrium expression: F - (aq)

0.10

o

o

-x

+x

+x

0.10 - x

x

x

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

+ HzO(l) ::;:.==" HF(aq) + OH - (aq)

Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve for x, we get 1.4 X 10- 11 =

x =

z X

0.10 - x

=

0.10

~( 1.4 X 10- 11 )(0.10)

= 1.2 X 10- 6 M

According to our equilibrium table, x = [OH- j. In this case, the autoionization of water makes a significant contribution to the hydroxide ion concentration so the total concentration will be the sum of 1.2 X 10- 6 M (from the ionization of F-) and 1.0 X 10- 7 M (from the autoionization of water). Therefore, we calculate the pOH first as pOH

=

-log (1.2 X 10- 6 + 1.0 X 10- 7)

= 5.95

and then the pH, pH

= 14.00 -

pOH

= 14.00 -

5.95

= 8.05

The pH of a 0.10 M solution of NaF at 25°C is 8.05.

Practice Problem A Determine the pH of a 0.15 M solution of sodium acetate (CH 3COONa) at 25 °C. Practice Problem B Determine the concentration of a solution of sodium fluoride (NaF) that has pH 8.51 at 25°C.

Think About It It's easy to mix up pH and pOH in this type of problem. Always make a qualitative prediction regarding the pH of a salt solution first, and then check to make sure that your calculated pH agrees with your prediction. In this case, we would predict a basic pH because the anion in the salt (F- ) is the conjugate base of a weak acid (HF). The calculated pH, 8.05, is indeed basic.

664

CHAPTER 16 Acids and Bases

Acidic Salt Solutions When the cation of a salt is the conjugate acid of a weak base, a solution of the salt will be acidic. For example, when ammonium chloride dissolves in water, it dissociates to give a solution of ammonium ions and chloride ions:

The ammonium ion is the conjugate acid of the weak base ammonia (NH3)' It acts as a weak Br0nsted acid, reacting with water to produce hydronium ion:

We would therefore predict that a solution containing the ammonium ion is acidic. To calculate the pH, we must deterllline the Ka for NHt using the tabulated Kb value for NH3 and Equation 16.7. Because Cl- is the weak conjugate base ofthe strong acid HCI, Cl- does not hydrolyze and therefore has no impact on the pH of the solution. Sample Problem 16.20 shows how to calculate the pH of an acidic salt solution.

Sample Problem 16.20 ·. Calculate the pH of a 0.10 M solution of ammonium chloride (NH4 Cl) at 25°e. Strategy A solution ofNH4 Cl contains NHt cations and Cl- anions. The NH t ion is the conjugate acid of the weak base NH3. Use the K b value for NH3 (1.8 X lO - s from Table 16.7) and Equation 16.7 to determine Ka for NH t . 14

1.0 X 101.8 X lO- s

=

5.6 X 10- 10

Setup Again, we write the balanced chemical equation and the equilibrium expression:

K

=

[NH 3] [H3 0 +]

~--==----,:----=.

[NH t ]

a

Next, construct a table to determine the equilibrium concentrations of the species in the equilibrium • expresslOn: NH t(aq)

Initial concentration (M) : Change in concentration (M): Equilibrium concentration (M):

+ H zO(l) ::;:,=~. NH3(aq) + H30+(aq)

0.10

o

o

-x

+x

• -rx

x

x

0.10 - x

,_._--

Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve for x, we get 5.6 X 10- 10 =

2

X

0.10 -x

=

0.10

x = ~(5 .6 X 10 10)(0.10) = 7.5 X 10- 6 M

Think About It In this case, we would predict an acidic pH because the cation in the salt (NHt) is the conjugate acid of a weak base (NH3). The calculated pH is acidic.

According to the equilibrium table, x pH

=

[H30 +]. The pH can be calculated as follows:

=

- log (7 .5 X 10- 6)

=

5.12

The pH of a 0.10 M solution of ammonium chloride (at 25°C) is 5.12.

. •

Practice Problem A Determine the pH of a 0.25 M solution of pyridinium nitrate (C sH 6NN0 3 ) at 2Ye. [Pyridinium nitrate dissociates in water to give pyridinium ions (CsH6N+), the conjugate acid of pyridinium (see Table 16.7), and nitrate ions (N0 3).] Practice Problem B Determine the concentration of a solution of ammonium chloride (NH4 Cl) that has pH 5.37 at 25°e. The metal ion in a dissolved salt can also react with water to produce an acidic solution. The extent of hydrolysis is greatest for the small and highly charged metal yations such as AI3+, Cr3 +, Fe3+, Bi3 +, and Be2+. For example, when aluminum chloride dissolves in water, each AI3+ ion becomes associated with six water molecules (Figure 16.4).

SECTION 16.10

Acid-Base Properties of Salt Solutions

..........

Al(OH)(H20)~ +

+

+

665

Figure 16.4 The six H20 molecules surround the AIH ion in an octahedral arrangement. The attraction of the small AIH ion for the lone pairs on the oxygen atoms is so great that the 0-H bonds in an H 20 molecule attached to the metal cation are weakened, allowing the loss of a proton (H +) to an incoming H 20 molecule. This hydrolysis of the metal cation makes the solution acidic.

Consider one of the bonds that forms between the metal ion and an oxygen atom from one of the six water molecules in Al(H20)~+: H •

AI '

I

cfl

~H

3

The positively charged A1 + ion draws electron density toward itself, increasing the polarity of the 0- H bonds. Consequently, the H atoms have a greater tendency to ionize than those in water molecules not associated with the Al3+ ion. The resulting ionization process can be written as

or as

The equilibrium constant for the metal cation hydrolysis is given by

Ka =

[Al(OH)(H20) ~+ ][H+ ] [Al(H20)~ + ]

= 1.3

X

10

-5

Al(OH)(H20)~+ can undergo further ionization:

and so on. It is generally sufficient, however, to take into account only the first stage of hydrolysis when determining the pH of a solution that contains metal ions.

Neutral Salt Solutions The extent of hydrolysis is greatest for the smallest and most highly charged metal ions because a compact, highly charged ion is more effective in polarizing the O-H bond and facilitating ionization. This is why relatively large ions of low charge, including the metal cations of Groups lA and . . ..................... . .. . ............ . ............... ··2:.f · ···························· 2A (the cations of the strong bases), do not undergo significant hydrolysis (Be is an exception). Thus, most metal cations of Groups lA and 2A do not impact the pH of a solution. Similarly, anions that are conjugate bases of strong acids do not hydrolyze to any significant degree. Consequently, a salt composed of the cation of a strong base and the anion of a strong acid, such as NaCl, produces a neutral solution. To summarize, the pH of a salt solution can be predicted qualitatively by identifying the ions in solution and determining which of them, if any, undergoes significant hydrolysis.

Examples A cation that will make a solution acidic is • The conjugate acid of a weak base • A small, highly charged metal ion (other than from Group lA or 2A) An anion that will make a solution basic is • The conjugate base of a weak acid

CN- , N0 2 , CH 3COO-

A cation that will not affect the pH of a solution is 2 • A Group lA or heavy Group 2A cation (except Be +) An anion that will not affect the pH of a solution is • The conjugate base of a strong acid

•

Cl- , N0 3 , CI0 4

The metal cations of the strong bases are those ofthe alkali metals: (Li+, Na+, K+, Rb+, and (s+) and those of the heavy alkaline earth metals (51"'+ and Ba2+).

666

CHAPTER 16

Acids and Bases

Sample Problem 16.21 lets you practice predicting the pH of salt solutions.

Predict whether a 0.10 M solution of each of the following salts will be basic, acidic, or neutral: (a) LiI, (b) NH4 N0 3 , (c) Sr(N0 3h , (d) KN0 2 , (e) NaCN. Strategy Identify the ions present in each solution, and determine which, if any, will impact the pH of the solution. Setup (a) Ions in solution: Li+ and r. Li+ is a Group lA cation; 1- is the conjugate base of the strong acid HI. Therefore, neither ion hydrolyzes to any significant degree. (b) Ions in solution: NH1 and NO) . NH1 is the conjugate acid of the weak base NH3 ; N0 3" is the conjugate base of the strong acid HN0 3 . In this case, the cation will hydrolyze, making the pH acidic:

(c) Ions in solution: Sr2 + and N03" . Sr2+ is a heavy Group 2A cation; NO ) is the conjugate base of the strong acid HN0 3 . Neither ion hydrolyzes to any significant degree. (d) Ions in solution: K+ and N0 2 . K+ is a Group I A cation; N0 2 is the conjugate base of the weak acid HN0 2 . In this case, the anion hydrolyzes, thus making the pH basic: N0 2 (aq)

+ H 20(l) :;::.==' HN0 2(aq) + OW(aq)

(e) Ions in solution: Na + and CN- . Na + is a Group 1A cation; CN- is the conjugate base of the weak acid HCN. In this case, too, the anion hydrolyzes, thus making the pH basic: CW(aq)

Think About It It's very important that you be able to identify the ions in solution correctly. If necessary, review the formulas and charges of the common polyatomic ions [ ~ Secti on 2 .7, Table 2 .8] .

+ H 20(l)

:;:::.==:!:' H CN(aq)

+ OH - (aq)

Solution (a) Neutral (b) Acidic (c) Neutral (d) Basic (e) Basic

Practice Problem A Predict whether a 0.10 M solution of each of the following salts will be basic, acidic, or neutral: (a) CH3COOLi, (b) CsHsNHCI, (c) KF, (d) KN0 3 , (e) KCl04 . Practice Problem B In addition to those given in Sample Problem 16.21 and Practice Problem A, identify two salts that will dissolve to give (a) an acidic solution, (b) a basic solution, and (c) a neutral solution.

•

Salts in Which Both the Cation and the Anion Hydrolyze So far we have considered salts in which only one ion undergoes hydrolysis. In some salts, both the cation and the anion hydrolyze. Whether a solution of such a salt is basic, acidic, or neutral depends on the relative strengths of the weak acid and the weak base. Although the process of calculating the pH in these cases is more complex than in cases where only one ion hydrolyzes, we can make qualitative predictions regarding pH using the values of Kb (of the salt's anion) and Ka (of the salt's cation): • When Kb > K a, the solution is basic. • When Kb < Ka, the solution is acidic. • When Kb = K a, the solution is neutral or nearly neutral. The salt NH4 NO z, for example, dissociates in solution to give NHt (Ka = 5.6 X 10- 1°) and 11 NO z (Kb = 2.2 X 10- ). Because Ka for the ammonium ion is larger than Kb for the nitrite ion, we would expect the pH of an ammonium nitrite solution to be slightly acidic.

SECTION 16.11

Checkpoint 16.10

Acid-Base Properti es of Oxides and Hydroxides

Acid-Base Properties of Salt Solutions

16.10.1

Calculate the pH of a 0.075 M solution of NH4N0 3 at 25°C.

a)

5.19

b)

16.10.3

Which of the following salts will produce a basic solution when dissolved in water? (Select all that apply.)

8.81

a)

Sodium hypochlorite (NaCIO)

c)

7.00

b)

Potassium fluoride (KF)

d)

2.93

c)

Lithium carbonate (LizC0 3)

e)

11.07

d)

Barium chloride (BaCl z)

e)

Ammonium iodide (NH4I)

16.10.4

Which of the following salts will produce a neutral solution when dissolved in water? (Select all that apply.)

a)

Calcium chlorite [Ca(CIOzh]

b)

Potassium iodide (KI)

c)

Lithium nitrate (LiN0 3)

d)

Barium cyanide [Ba(CN)z]

e)

Ammonium iodide (NH4I)

16.10.2

Calculate the pH of a 0.082 M solution of NaCN at 25 °C.

a)

5.20

b)

8.80

c)

d) e)

7.00

2.89 lLlI

Acid-Base Properties of Oxides and Hydroxides As we saw in Chapter 8, oxides can be classified as acidic, basic, or amphoteric. Thus, our discussion of acid-base reactions would be incomplete if we did not examine the properties of these compounds.

Oxides of Metals and Nonmetals Figure 16.5 shows the formulas of a number of oxides of the main group elements in their highest oxidation states. All alkali metal oxides and all alkaline earth metal oxides except BeO are basic. Beryllium oxide and several metallic oxides in Groups 3A and 4A are amphoteric. Nonmetallic oxides in which the oxidation number of the main group element is high are acidic (e.g., N1 0 s , S03, and CI 20 7), but those in which the oxidation number of the main group element is low (e.g., CO and NO) show no measurable acidic properties. No nonmetallic oxides are known to have basic properties.

Basic oxide

lA 1

8A 18 Acidic oxide

2A 2

Li20

BeO

Na20 MgO

3A 13 B 20 3

Amphoteric oxide

3B 3

4B 4

5B 5

6B 6

7B 7

I

8

8B 9

10

I

1B 11

2B 12

4A 14

SA 15

CO 2 N 20

6A 16

7A 17 OF2

S

Al 20 3 SiOz P 40 10

S03

Clz07

K 20

CaO

Ga203 Ge02 As20

S

Se03 Br207

Rb 20

SrO

InZ0 3 Sn02 Sb 2 0

S

Te03

Cs 20

BaO

TI 20 3 Pb02 BizOs Po0 3 Atz°7

Figure 16.5

Oxides of the main group elements in their highest oxidation states.

12°7

667

668

CHAPTER 16

Acids and Bases

The basic metallic oxides react with water to form metal hydroxides: Na20(S) + H 20(l) - _ . 2NaOH(aq) BaO(s) + H?O(I)

• Ba(OHMaq)

The reactions between acidic oxides and water are as follows: CO 2 (g)

+ H 20(I)

S03(g)

+ H 2 0(l). + H 2 0(I).

N 20 S(g) P4 0

IO

(g)

==' H 2C0 3(aq)

~.

• H 2 SOiaq) • 2HN0 3(aq)

+ 6H2 0(l).

• 4H 3P0 4 (aq)

+ H 20(l).

• 2HCIOiaq)

CI 20 7(g)

The reaction between CO 2 and H 2 0 explains why pure water gradually becomes acidic when it is exposed to air, which contains CO 2, The pH of rainwater exposed only to unpolluted air is about 5.5. The reaction between S03 and H?O is largely responsible for acid rain. Reactions between acidic oxides and bases and those between basic oxides and acids resemble normal acid-base reactions in that the products are a salt and water: CO 2(g)

BaO(s)

+ 2NaOH(aq) - _ . Na2C03(aq) + H 20(l) + 2HN03(aq)

• Ba(N0 3Maq)

+ H 20(l)

Aluminum oxide (Al2 0 3) is amphoteric. Depending on the reaction conditions, it can behave either as an acidic oxide or as a basic oxide. For example, Ah03 acts as a base with hydrochloric acid to produce a salt (AICI 3) and water:

and acts as an acid with sodium hydroxide:

Only a salt, sodium aluminum hydroxide [NaAI(OH)4, which contains the Na + and AI(OH)4 ions] is formed in the reaction with sodium hydroxide no water is produced. Nevertheless, the reaction is still classified as an acid-base reaction because Al 2 0 3 neutralizes NaOH. Some transition metal oxides in which the metal has a high oxidation number act as acidic oxides. Two examples are manganese(VII) oxide (Mn20 7) and chromium(VI) oxide (Cr03), both of which react with water to produce acids: permanganic acid

Cr03(S)

+ H 20(l) - _ . H 2Cr0 4(aq) chromic acid

Basic and Amphoteric Hydroxides All the alkali and alkaline earth metal hydroxides, except Be(OHh, are basic. Be(OH)2, as well as AI(0H)3, Sn(OH)2> Pb(OH)z, Cr(OH)3, CU(OH)2, Zn(OHh, and Cd(OH)2> is amphoteric. All amphoteric hydroxides are insoluble, but beryllium hydroxide reacts with both acids and bases as follows: Be(OHMs) Be(OH)is)

+ 2H+(aq) - _ . Be2+(aq) + 2H20(l)

+ 20H- (aq)

• Be(OH)~- (aq)

Aluminum hydroxide reacts with both acids and bases in a similar fashion: AI(OHMs)

+ 3H+(aq) - _ . AI3+ (aq) + 3H20(l)

AI(OHMs) + OH-(aq)

• AI(OH)4 (aq)

Lewis Acids and Bases So far we have discussed acid-base properties in terms of the Br0nsted theory. For example, a Br0nsted base is a substance that must be able to accept protons. By this definition, both the hydroxide ion and ammonia are bases:

SECTION 16.12

+

••

- :O-H ••

••

•

H-O-H ••

H

+

+

H

:N-H

I

H-N-H

•

I

H

H

In each case, the atom to which the proton becomes attached possesses at least one unshared pair of electrons. This characteristic property of OH-, NH 3, and other Br0nsted bases suggests a more general definition of acids and bases. In 1932 G. N. Lewis defined what we now call a Lewis base as a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons. In the protonation of ammonia, for example, NH3 acts as a Lewis base because it donates a pair of electrons to the proton H +, which acts as a Lewis acid by accepting the pair of electrons. A Lewis acid-base reaction, therefore, is one that involves the donation of a pair of electrons from one species to another. The significance of the Lewis concept is that it is more general than other definitions. Lewis acid-base reactions include many reactions that do not involve Br0nsted acids. Consider, for example, the reaction between boron trifluoride (BF3) and ammonia to form an adduct compound: F

F

I B I

H

I :N I

+

F acid

•

H

F

H

F

H

I B N H I I

F

H base

The B atom in BF3 is sp2-hybridized [ ~~ Section 9.4] . The vacant, un hybridized 2pz orbital accepts the pair of electrons from NH3. Thus, BF3 functions as an acid according to the Lewis definition, even though it does not contain an ionizable proton. A coordinate covalent bond [ ~~ Section 8.8] is formed between the Band N atoms. In fact, every Lewis acid-base reaction results in the formation of a coordinate covalent bond. Boric acid is another Lewis acid containing boron. Boric acid (a weak acid used in eyewash) is an oxoacia with the following structure: H

I ·0· I

•

•

••

H

0• •

••

0••

B

H

Boric acid does not ionize in water to produce H +. Instead, it produces H + in solution by taking a hydroxide ion away from water.

In this Lewis acid-base reaction, boric acid accepts a pair of electrons from the hydroxide ion that is derived from the water molecule, leaving behind the hydrogen ion. The hydration of carbon dioxide to produce carbonic acid,

•

can be explained in tenns of Lewis acid-base theory as well. The first step involves the donation of a lone pair on the 0 atom in H 20 to the C atom in CO 2 . An orbital is vacated on the C atom to accommodate the lone pair by relocation of the electron pair in one of the C-O pi bonds, changing the hybridization of the oxygen atom from Sp2 to Sp 3:

.. -

·0· • • .. + I H-O-C

•

I

II

H

.0. • •

As a result, H 20 is a Lewis base and CO 2 is a Lewis acid. Finally, a proton is transferred onto the o atom bearing the negative charge to form H 2C03 :

.. -

.---- :0: .. + I

H-O-C

I

H

II

.0. • •

••

H-O:

•

.. I :O-C I II H

.0. • •

Lewis Acids and Bases

669

670

CHAPTER 16

Acids and Bases

Other examples of Lewis acid-base reactions are Ag+(aq)

+ 2NH3(aq) :;::,~. Ag(NH3)~(aq)

Cd 2 +(aq) + 4I-(aq) ,

Ni(s) + 4CO(g) '

• CdI~-(aq) • Ni(COMg)

The hydration of metal ions is in itself a Lewis acid-base reaction. When copper(Il) sulfate (CUS04) dissolves in water, each Cu2+ ion becomes associated with six water molecules as Cu(H20)~+. In this case, Cu 2+ acts as the acid, accepting electrons, whereas H20 acts as the base, donating electrons. Sample Problem 16.22 shows how to classify Lewis acids and bases.

Identify the Lewis acid and Lewis base in each of the following reactions: (a) C 2H 5 0C 2H 5 (b) Hg2+(aq)

+

+ AlCl 3

•

4CW(aq) '

•

(C2 H 5)zOAlCI3 • Hg(CN)~-(aq)

Strategy Determine which species in each reaction accepts a pair of electrons (Lewis acid) and which species donates a pair of electrons (Lewis base). Setup (a) It can be helpful to draw Lewis structures of the species involved:

• •

An electron-deficient molecule is one with less than a complete octet around the central atom.

Think About It In Lewis acid-base reactions, the acid is usually a cation or an electron-deficient molecule, whereas the base is an anion or a molecule containing an atom with lone pairs.

••

• •••

(b) Metal ions act as Lewis acids, accepting electron pairs from anions or molecules with lone pairs. . . . . .. ·························2 .... . . .................. .......................... .... .. . Solution (a) The AI is sp -hybridized in AICl3 with an empty 2pz orbital. It is electron deficient, sharing only six electrons. Therefore, the AI atom has the capacity to gain two electrons to complete its octet. This property makes AlCl 3 a Lewis acid. On the other hand, the lone pairs on the oxygen atom in C 2H 5 0C 2H s make the compound a Lewis base:

(b) Hg2+ accepts four pairs of electrons from the CN- ions. Therefore, Hg2 + is the Lewis acid and CN- is the Lewis base.

Practice Problem Identify the Lewis acid and Lewis base in the following reaction:

Checkpoint 16.12 16.12.1

Lewis Acids and Bases

Which of the following cannot act as a Lewis base? (Select all that apply.) a) NH3

16.12.2

Which of the following is a Lewis acid but not a Bn2jnsted acid? (Select all that apply.)

b) OW

a) H2O

c) CH4

b) BCl 3

d) Fe2+

c) OW

e) AI3+

d) AI3+ e) NH3

APPLYING WHAT YOU'VE LEARNED

Applying What You've Learned Although it is a diprotic acid (Ka = 8.0 X 10- 5 and Ka, = 2.S X 10- 12), ascorbic acid only loses one proton under biological conditions to give-H+ and the hydrogen ascorbate • IOn:

A person who takes vitamin C supplements but who has a sensitive stomach may be advised to take calcium ascorbate [Ca(HC6H60 6)2], which is less acidic than ascorbic acid and contains the hydrogen ascorbate ion.

Problems: a)

Give the formulas for the conjugate acid and the conjugate base of the hydrogen ascorbate ion. [ ~ Sample Problem 16.1] •

b)

,

Determine [H+] and pH for a solution of ascorbic acid prepared by dissolving a 1000-mg tablet in SO.O mL of water at 2S°C. [ ~ Sample Problem 16.12] [ ~~

c)

Calculate [OH- ] in the solution in part (b).

d)

Calculate the pOH of the solution in part (b).

e)

Sample Problem 16.3]

[~

Sample Problem 16.6]

What concentration of HCI would have the same pH as the solution in part (b)? [ ~~ SampleProblem 16.9]

f)

Detelmine Kb for the ascorbate ion (C6H60~-). [ ~~ Sample Problem 16.16]

g)

Calculate the concentrations of all species in the solution in part (b). [ ~~ Sample Problem 16.l7]

• )

Calcium Ascorbate 10()% Pure Buffered Vitamin C powder Net Wt. 8 oz. (227 A Dlata., Sb:JIpl.mpl

67 1

672

CHAPTER 16

Acids and Bases

CHAPTER SUMMARY Section 16.1

Section 16.7

•

A Br12lilsted acid donates a proton; a Br\
•

•

When a Br\
The conjugate base of a strong acid is a weak conjugate base, meaning that it does not react with water.

•

The conjugate base of a weak acid is a strong conjugate base, meaning that it acts as a weak Br\
conjugate acid.

•

The combination of a Br\
The conjugate base of a strong base is a weak conjugate acid, meaning that it does not react with water.

•

The conjugate acid of a weak base is a strong conjugate acid, meaning that it acts as a weak Br¢nsted acid in water.

•

For any conjugate acid-base pair, Ka X Kb = Kw.

conjugate base. • •

When a Br\
conjugate pair.

Section 16.2 •

Water is amphoteric, meaning it can act both as a Br\
•

Pure water undergoes auto ionization (to a very small extent), resulting in concentrations of H + and OH- of l.0 X 10- 7 Mat 25 °e.

•

Kw is the eqUilibrium constant for the autoionization of water, also called the ion product constant. Kw = [H+] [OH- ] = l.0 X 10- 14 at 2Ye.

Section 16.8 •

Diprotic and polyprotic acids have more than one proton to donate. They undergo stepwise ionizations. Each ionization has a Ka value associated with it.

•

The Ka values for stepwise ionizations become progressively smaller.

•

In most cases, it is only necessary to consider the first ionization of an acid to determine pH. To determine the concentrations of other species at eqUilibrium, it may be necessary to consider subsequent ionizations.

Section 16.3 • • •

The pH scale measures acidity: pH = - log [H +]. pH = 7.00 is neutral, pH

< 7.00 is acidic, and pH >

7.00 is basic.

The pOH scale is analogous to the pH scale, but it measures basicity: pOH = -log [OW].

< 7.00 is basic, and pOH > 7.00 is acidic.

•

pOH = 7.00 is neutral, pOH

•

pH + pOH = 14.00 (at 2YC).

Section 16.4 •

There are seven strong acids: HCI, RBr, HI, RN0 3 , HCI0 3 , HCI0 4 , and H 2S0 4 .

•

Strong acids ionize completely in aqueous solution. (Only the first ionization of the diprotic acid H 2S04 is complete.)

•

The strong bases are the Group IA and the heaviest Group 2A hydroxides: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OHh, Sr(OHh, and Ba(OH)z.

Section 16.5 • • •

A weak acid ionizes only partially. The acid ionization constant, Ka , is the equilibrium constant that indicates to what extent a weak acid ionizes. We solve for the pH of a solution of weak acid using the concentration of the acid, the Ka value, and an equilibrium table. We can also determine the Ka of a weak acid if we know the initial acid concentration and the pH at equilibrium.

Section 16.9 •

The strength of an acid is affected by molecular structure.

•

Polar and weak bonds to the ionizable hydrogen lead to a stronger acid.

•

Resonance stabilization of the conjugate base favors the ionization process, resulting in a stronger acid.

Section 16.10 •

Salts dissolve in water to give neutral, acidic, or basic solutions depending on their constituent ions. Salt hydrolysis is the reaction of an ion with water to produce hydronium or hydroxide ions.

•

Cations that are strong conjugate acids such as NH1 make a solution more acidic.

•

Anions that are strong conjugate bases such as F - make a solution more basic. Anions that are conjugate bases of strong acids have no effect on pH.

•

Small, highly charged metal ions hydrolyze to give acidic solutions.

Section 16.11 •

Oxides of metals generally are basic; oxides of nonmetals generally are acidic.

•

Metal hydroxides may be basic or amphoteric.

Section 16.12 Section 16.6

•

Lewis theory provides more general definitions of acids and bases.

•

A weak base ionizes only partially. The base ionization constant, K", is the equilibrium constant that indicates to what extent a weak base ionizes.

•

A Lewis acid accepts a pair of electrons; a Lewis base donates a pair of electrons.

•

We solve for the pH of a solution of weak base using the concentration of the base, the Kb value, and an equilibrium table.

•

A Lewis acid is generally electron-poor and need not have a hydrogen atom.

•

We can also determine the Kb of a weak base if we know the concentration and the pH at equilibrium.

•

A Lewis base is an anion or a molecule with one or more lone pairs of electrons.

QUESTIONS AND PROBLEMS

673

KEyWORDS Acid ionization constant (Ka), 648

Conjugate base, 636

pH, 639

Weak acid, 647

Amphoteric, 637

Conjugate pair, 636

pOH, 641

Weak base, 652

Autoionization of water, 638

Ion-product constant, 638

Salt hydrolysis, 662

Weak conjugate acid, 655

Base ionization constant (Kb)' 652

Lewis acid, 669

Strong conjugate acid, 655

Weak conjugate base, 654

Conjugate acid, 636

Lewis base, 669

Strong conjugate base, 655

KEY EQUATIONS 16.2 •

16.3

pH = - log [H30 +] [H30 +] = lO - pH

16.5

pOH = - log [OW] [OH- ] = lO- poH

16.6

pH

16.7

Ka X Kb = Kw

16.4

+ pOH

or

= 14.00

QUESTIONS AND PROBLEMS Section 16.1: Br0nsted Acids and Bases

16.8

Oxalic acid (HZC Z0 4) has the following structure: O=C-OH

Review Questions 16.1

Define Br(llnsted acids and bases . Give an example of a conjugate pair in an acid-base reaction.

16.2

In order for a species to act as a Br(llnsted base, an atom in the species must possess a lone pair of electrons. Explain why this is so.

I

O=C-OH An oxalic acid solution contains the following species in varying concentrations: HZC Z0 4, HCZ0 4 , C20 ~-, and H +. (a) Draw Lewis structures of HC Z0 4 and CzO~ - . (b) Which of the four species can act only as acids, which can act only as bases, and which can act as both acids and bases?

Problems

Section 16.2: The Acid-Base Properties of Water

16.3

Review Questions

16.4

Classify each of the following species as a Br(llnsted acid or base, or both: (a) H20, (b) OH- , (c) H30 +, (d) NH3, (e) NH1 , (f) NH z , (g) NO;-, (h) co j-, (i) HEr, (j) HCN. Identify the acid-base conjugate pairs in each of the following reactions: • CH 3COOH + CW (a) CH 3COO- + HCN • (b) HCO;- + HCO;- ( • H ZC0 3 + coj(c) H ZP0 4 + NH3 ( • HPO~ - + NH1 (d) HCIO + CH3NHz ( • CH3NH~ + CIO(e) co j- + H 20 • • HCO;- + OW

16.5

Write the formulas of the conjugate bases of the following acids: (a) HNO z, (b) H 2 S04 , (c) HzS, (d) HCN, (e) HCOOH (formic acid).

16.6

Write the formula for the conjugate acid of each of the following bases: (a) HS - , (b) HCO;- , (c) CO~- , (d) H 2P0 4 , (e) HPO~ - , (f) PO~ - , (g) HS0 4 , (h) SO~ - , (i) soj- .

16.9

Write the equilibrium expression for the autoionization of water.

16.10

Write an equation relating [H+] and [OH-] in solution at 25°C.

16.11

The equilibrium constant for the autoionization of water H zO(I) •

Write the formula for the conjugate base of each of the following acids: (a) CH2CICOOH, (b) HI04 , (c) H3 P0 4 , (d) H2 P0 4 , (e) HPO~-, (f) H2 S0 4 , (g) HS0 4 , (h) RI0 3 , (i) HSO;- , (j) NH1 , (k) HzS, (I) HS-, (m) HCIO.

+ OH-(aq)

is 1.0 X 10- 14 at 25°C and 3.8 X 10- 14 at 40°C. Is the forward process endothermic or exothermic? 16.12

Define the term amphoteric.

16.13

Compare the magnitudes of [H+] and [OH-] in aqueous solutions that are acidic, basic, and neutral.

Section 16.3: The pH Scale

Review Questions 16.14

16.7

• H +(aq)

Define pH. Why do chemists normally choose to discuss the acidity of a solution in terms of pH rather than hydrogen ion concentration [H+]?

674

16.15

16.16

CHAPTER 16

Acids and Bases

16.31

Why are ionizations of strong acids and strong bases generally not treated as equilibria?

The pH of a solution is 6.7. From this statement alone, can you conclude that the solution is acidic? If not, what additional information would you need? Can the pH of a solution be zero or negative? If so, give examples to illustrate these values.

Problems

Define pOH. Write the equation relating pH and pOH.

16.32

Calculate the pH of an aqueous solution at 25°C that is 4 (a) 0.12 Min HC1, (b) 2.4 Min HN0 3, and (c) 3.2 X 10- M in HCI04 .

16.33

Calculate the pH of an aqueous solution at 25 °C that is (a) 1.02 M in HI, (b) 0.035 M in HCI04 , and (c) 1.5 X 10- 6 M in HCl.

Problems 16.17

Calculate the concentration of OH- ions in a 1.4 X 10- 3 M HCl solution.

16.18

Calculate the concentration of H+ ions in a 0.62 M NaOH solution.

16.34

Calculate the concentration of HBr in a solution at 25°C that has a pH of (a) 0.12, (b) 2.46, and (c) 6.27.

Calculate the pH of each of the following solutions: (a) 0.0010 M HC1, (b) 0.76 M KOH.

16.35

Calculate the concentration of HN0 3 in a solution at 25°C that has a pH of (a) 4.21, (b) 3.55, and (c) 0.98.

Calculate the pH of each of the following solutions: (a) 2.8 X 10- 4 M Ba(OH)z, (b) 5.2 X 10- 4 M HN0 3.

16.36

Calculate the pOH and pH of the following aqueous solutions at 25 °C: (a) 0.066 M KOH, (b) 5.43 M NaOH, (c) 0.74 M Ba(OH)z.

16.37

Calculate the pOH and pH of the following aqueous solutions at 25°C: (a) 1.24 M LiOH, (b) 0.22 M Ba(OH)z, (c) 0.085 M NaOH.

16.38

An aqueous solution of a strong base has a pH of 9.78 at 25 °C. Calculate the concentration of the base if the base is (a) LiOH and (b) Ba(OHh-

16.39

An aqueous solution of a strong base has a pH of 11.04 at 25°C. Calculate the concentration of the base if the base is (a) KOH and (b) Ba(OH)z.

16.19

16.20

16.21

16.22

16.23

Calculate the hydrogen ion concentration in mol/L for solutions with the following pH values: (a) 2.42, (b) 11.21, (c) 6.96, (d) 15.00. Calculate the hydrogen ion concentration in mol/L for each of the following solutions: (a) a solution whose pH is 5.20, (b) a solution whose pH is 16.00, (c) a solution whose hydroxide concentration is 3.7 X 10- 9 M. Complete the following table for a solution: pH

Solution is

[H+]

7; solution is _ _ _ _ _ _ . (b) pOH = 7; solution is _ _ _ _ _ _ . (c) pOH < 7; solution is _ _ _ _ _ _ _ .

Review Questions 16.40

Explain what is meant by the strength of an acid.

16.41

What does the ionization constant tell us about the strength of an acid?

16.42

List the factors on which the Ka of a weak acid depends.

16.43

Why do we normally not quote Ka values for strong acids such as HCl and HN0 3? Why is it necessary to specify temperature when giving Ka values?

16.25

The pOH of a solution is 9.40 at 25 °C. Calculate the hydrogen ion concentration of the solution.

16.26

Calculate the number of moles of KOH in 5.50 mL of a 0.360 M KOH solution. What is the pOH of the solution at 25 °C?

16.44

How much NaOH (in grams) is needed to prepare 546 mL of solution with a pH of 10.00 at 25 °C?

Which of the following solutions has the highest pH: (a) 0.40 M HCOOH, (b) 0.40 M HC104 , (c) 0.40 M CH3COOH?

16.45

Without referring to the text, write the formulas of four weak acids.

16.27

16.28

A solution is made by dissolving 18.4 g ofHCl in enough water to make 662 mL of solution. Calculate the pH of the solution at 25°C.

Problems 16.46

Classify each of the following species as a weak or strong acid: (a) HN0 3 , (b) HF, (c) H 2S0 4 , (d) HS0 4 , (e) H 2C0 3 , (f) HC0 3, (g) HC1, (h) HCN, (i) HN0 2 .

16.47

Classify each of the following species as a weak or strong base: (a) LiOH, (b) CW, (c) H 2 0, (d) CI0 4 , (e) NH 2 .

16.48

Which of the following statements are true for a 0.10 M solution of a weak acid HA? (Choose all that apply.)

Section 16.4: Strong Acids and Bases Review Questions 16.29

16.30

Without referring to the text, write the formulas offour strong acids and four strong bases. Which of the following statements are true regarding a 1.0 M solution of a strong acid HA at 25°C? (Choose all that apply.) (a) (b) (c) (d)

[A - ] > [H+] The pH is 0.00. [H+] = 1.0 M [HA] = 1.0 M

(a) The pH is 1.00. (b) [H+] » [A - ]. (c) [H+] = [A- l (d) The pH is less than 1.

675

QUESTIONS AND PROBLEMS

16.49

The Ka for benzoic acid is 6.5 X 10- 5. Calculate the pH of a 0.10 M aqueous solution of benzoic acid at 25°C.

16.50

The Ka for hydrofluoric acid is 7.1 X 10- 4. Calculate the pH of a 0.15 M aqueous solution of hydrofluoric acid at 25°C.

16.51

Calculate the pH of an aqueous solution at 25 °C that is 0.095 M in hydrocyanic acid (HCN). (Ka for hydrocyanic acid = 4.9 X 10- 10 .)

16.52

16.53

Calculate the pH of an aqueous solution at 25°C that is 0.34 M in phenol (C6H50H). (Ka for phenol = 1.3 X 10- 10 .) Calculate the Ka of a weak acid if a 0.19 M aqueous solution of the acid has a pH of 4.52 at 25°C.

16.54

The pH of an aqueous acid solution is 6.20 at 25°C. Calculate the Ka for the acid. The initial acid concentration is 0.010 M.

16.55

What is the original molarity of a solution of formic acid (HCOOH) whose pH is 3.26 at 25°C? (Ka for formic acid = 1.7 X 10- 4 .)

16.56

What is the original molarity of a solution of a weak acid whose Ka is 3.5 X 10- 5 and whose pH is 5.26 at 25°C?

16.57

In biological and medical applications, it is often necessary to study the autoionization of water at 37°C instead of 25°C. Given that Kw for water is 2.5 X 10- 14 at 37°C, calculate the pH of pure water at this temperature.

= A- , B-, orC -

HA, HE, or HC

=

HA

HE

HC

Section 16.7: Conjugate Acid-Base Pairs

Review Questions 16.66

Write the equation relating Ka for a weak acid and Kb for its conjugate base. U se NH3 and its conjugate acid NHt to derive the relationship between Ka and K b .

16.67

From the relationship KaKb =Kw, what can you deduce about the relative strengths of a weak acid and its conjugate base?

Problems 16.68

The following diagrams represent solutions of three salts NaX (X = A , B , or C). (a) Which X - has the weakest conjugate acid? (b) Arrange the three X - anions in order of decreasing base strength. The Na + ion and water molecules have been omitted for clarity. =

A- , B - , or C-

= OH-

__ =

HA, HE, or HC

Section 16.6: Weak Bases and Base Ionization Constants

Review Questions 16.58

16.59

Compare the pH values for 0.10 M solutions of NaOH and of NH3 to illustrate the difference between a strong base and a weak base. Which of the following has a higher pH: (a) 1.0 M NH3 , (b) 0.20 M NaOH (Kb for NH3 = 1.8 X 1O- 5)?

Problems 16.60

Calculate the pH for each of the following solutions at 25 °C: (a) 0.10 M NH3 , (b) 0.050 M C5H5N (pyridine). (Kb for pyridine = 1.7 X 10- 9 .)

16.61

The pH of a 0.30 M solution of a weak base is 10.66 at 25°C. What is the Kb of the base?

16.62

What is the original molarity of an aqueous solution of ammonia (NH 3) whose pH is 11.22 at 25°C? (Use the Kb for ammonia provided in Problem 16.59.)

16.63

Calculate the pH at 25°C of a 0.61 M aqueous solution of a weak 4 base B with a Kb of 1.5 X 10- .

16.64

Determine the Kb of a weak base if a 0.19 M aqueous solution of the base at 25°C has a pH of 10.88.

16.65

The following diagrams represent aqueous solutions of three different monoprotic acids: HA, HE, and HC. (a) Which conjugate base (A -, B - , or C - ) has the smallest Kb value? . (b) Which anion is the strongest base? The water molecules have been omitted for clarity.

NaA

16.69

NaB

NaC

Calculate Kb for each of the following ions: CN- , F - , CH 3COO- , HCO )" .

Section 16.8: Diprotic and Polyprotic Acids

Review Questions 16.70

Write all the species (except water) that are present in a phosphoric acid solution. Indicate which species can act as a Br!15nsted acid, which as a Br!15nsted base, and which as both a Br!15nsted acid and a Br!15nsted base.

Problems 16.71

(1) Which of the following diagrams represents a solution of a weak diprotic acid? (2) Which diagrams represent chemically implausible situations? (The hydrated proton is shown as a hydronium ion. Water molecules are omitted for clarity.)

(a)

16.72

(b)

(c)

(d)

The first and second ionization constants of a diprotic acid H2A are K a, and Ka2 at a certain temperature. Under what conditions will [A2 -] = Ka? ,

676

16.73

CHAPTER 16

Acids and Bases

Compare the pH of a 0.040 M HCI solution with that of a 0.040 M H 2S0 4 solution. (Hint: H 2S0 4 is a strong acid; Ka for HS0 4 = 1.3 X 10- 2.)

16.74

What are the concentrations of HS0 4 , SO ~-, and H + in a 0.20 M KHS04 solution? (Hint: H2S04 is a strong acid; Ka for HS0 4 = l.3 X 10- 2 )

16.75

Calculate the concentrations of H +, HC0 3 , and CO ~- in a 0.025 M H2C0 3 solution.

16.76

Calculate the pH at 25°C of a 0.25 M aqueous solution of phosphoric acid (H 3P0 4). (Ka , Ka." and Ka for phosphoric acid are 7.5 X 10- 3,6.25 X 10- 8, 'and -4.8 X 10- 13 , respectively.)

16.77

Calculate the pH at 25°C of a 0.25 M aqueous solution of oxalic acid (H2C20 4)._ (Ka, and Ka2 for oxalic acid are 6.5 X 10- 2 and 6.1 X 10- ', respectively.)

16.89

Predict whether the following solutions are acidic, basic, or nearly neutral: (a) NaBr, (b) K 2S03, (c) ~N02' (d) Cr(N0 3h

16.90

A certain salt, MX (containing the M + and X- ions), is dissolved in water, and the pH of the resulting solution is 7.0. What can you say about the strengths of the acid and the base from which the salt is derived?

16.91

In a certain experiment, a student finds that the pHs of 0.10 M solutions of three potassium salts KX, KY, and KZ are 7.0, 9.0, and 11.0, respectively. Arrange the acids HX, HY, and HZ in order of increasing acid strength.

16.92

Predict whether a solution containing the salt K 2HP04 will be acidic, neutral, or basic.

16.93

Predict the pH (>7, (e) NH3, (f) OW, (g) H +, (h) BCI 3 . 16.107 Describe the following reaction in terms of the Lewis theory of acids and bases: AICI 3(s) + Cqaq) - _ . AICl 4 (aq) 16.108 Which would be considered a stronger Lewis acid: (a) BF3 or BCI3, (b) Fe2+ or Fe3+? Explain. 16.109 All Br\iinsted acids are Lewis acids, but the reverse is not true. Give two examples of Lewis acids that are not Br\iinsted acids. 16.110 Identify the Lewis acid and the Lewis base in the following reactions:

16.121 Like water, liquid ammonia undergoes autoionization:

(a) Identify the Br\iinsted acids and Br\iinsted bases in this reaction. (b) What species correspond to H + and OH- , and what is the condition for a neutral solution? 16.122 HA and HB are both weak acids although HB is the stronger of the two. Will it take a larger volume of a 0.10 M NaOH solution to neutralize 50.0 mL of 0.10 M HB than would be needed to neutralize 50.0 mL of 0.10 M HA? 16.123 A solution contains a weak monoprotic acid HA and its sodium salt NaA both at 0.1 M concentration. Show that [OH- ] = KwiKa' 16.124 The three common chromium oxides are CrO, Cr203' and Cr03' If Cr203 is amphoteric, what can you say about the acid-base properties of CrO and Cr03?

(a) Fe(s) + 5CO(g) - _ . Fe(CO)5(1) (b) BCI3(g) + NH 3 (g) • CI3BNH3(s) (c) Hg2+(aq) + 4r-(aq) • HgI~-(aq) 16.111 Identify the Lewis acid and the Lewis base in theJollowing reactions:

16.125 Use the data in Table 16.6 to calculate the equilibrium constant for the following reaction: HCOOH(aq)

(a) AlBr3(s) + Br- (aq) • AlBr4(aq) (b) Cr(s) + 6CO(g) • Cr(COMs) (c) Cu2+(aq) + 4CN- (aq) • Cu(CN)~-(aq)

+ OW(aq) :;:::.===' HCOO- (aq) + H 20(l)

16.126 Use the data in Table 16.6 to calculate the equilibrium constant for the following reaction:

Additional Problems 16.112 H 2S04 is a strong acid, but HS0 4 is a weak acid. Account for the difference in strength of these two related species. 16.113 Which of the following diagrams best represents a strong acid, such as HCI, dissolved in water? Which represents a weak acid? Which represents a very weak acid? (The hydrated proton is shown as a hydronium ion. Water molecules are omitted for clarity.)

(a)

(b)

(d)

(c)

16.114 Predict the direction that predominates in this reaction: F- (aq)

+

H 20(l)

+.==' HF(aq) + OW(aq)

16.115 Predict the products and tell whether the· following reaction will occur to any measurable extent: CH3COOH(aq)

+ Cqaq)

,

16.116 In a 0.080 M NH3 solution, what percent of the NH3 is present as NH1? 16.117 Calculate the pH and percent ionization of a 0.88 M HN0 2 solution at 25°C. 16.118 A typical reaction between an antacid and the hydrochloric acid in gastric juice is NaHC0 3(s)

+

HCI(aq)

+.==' NaCI(aq) + H 2 0(l) + CO 2(g)

Calculate the volume (in liters) of CO 2 generated from 0.350 g of NaHC0 3 and excess gastric juice at 1.00 atm and 37.0°C. 16.119 To which of the following would the addition of an equal volume of 0.60 M NaOH lead to a solution having a lower pH: (a) water, (b) 0.30 M HCI, (c) 0.70 M KOH, Cd) 0.40 M NaN0 3?

16.127 Most of the hydrides of Group 1A and Group 2A metals are ionic (the exceptions are BeH2 and MgH2> which are covalent compounds). (a) Describe the reaction between the hydride ion (H- ) and water in terms of a Br\iinsted acid-base reaction. (b) The same reaction can also be classified as a redox reaction. Identify the oxidizing and reducing agents. 16.128 Calculate the pH of a 0.20 M ammonium acetate (CH 3COONH4) solution. 16.129 Novocaine, used as a local anesthetic by denti sts, is a weak base (Kb = 8.91 X 10- 6). What is the ratio of the concentration of the base to that of its acid in the blood plasma (pH = 7.40) of a patient? (As an approximation, use the Ka values at 25 °C.) 16.130 Which of the following is the stronger base: NF3 or NH3? (Hint: F is more electronegative than H.) 16.131 Which of the following is a stronger base: NH3 or PH 3? (Hint: The N-H bond is stronger than the P-H bond.) 16.132 The ion product of D 20 is 1.35 X 10- 15 at 25°C. (a) Calculate pD where pD = -log [D +]. (b) For what values of pD will a solution be acidic in D 20? (c) Derive a relation between pD and pOD. 16.133 Give an example of (a) a weak acid that contains oxygen atoms, (b) a weak acid that does not contain oxygen atoms, (c) a neutral molecule that acts as a Lewis acid, (d) a neutral molecule that acts as a Lewis base, (e) a weak acid that contains two ionizable H atoms, (f) a conjugate acid-base pair, both of which react with HCI to give carbon dioxide gas .. 16.134 What is the pH of 250.0 mL of an aqueous solution containing 0.616 g of a strong acid? 16.135 (a) Use VSEPR to predict the geometry of the hydronium ion (H30+). (b) The 0 atom in H 20 has two lone pairs and in principle can accept two H + ions. Explain why the species H4 0 2 does not exist. What would be its geometry if it did exist?

678

CHAPTER 16

Acids and Bases

16.l36 HF is a weak acid, but its strength increases with concentration. Explain. (Hint: F- reacts with HF to form HF 2. The equilibrium constant for this reaction is 5.2 at 25°C.)

16.137 When chlorine reacts with water, the resulting solution is weakly acidic and reacts with AgN0 3 to give a white precipitate. Write balanced equations to represent these reactions. Explain why manufacturers of household bleaches add bases such as NaOH to their products to increase their effectiveness. 16.138 When the concentration of a strong acid is not substantially higher than l.0 X 10- 7 M, the ionization of water must be taken into account in the calculation of the solution's pH. (a) Derive an expression for the pH of a strong acid solution, including the contribution to [H+] from H20. (b) Calculate the pH of a 1.0 X 10- 7 M HCl solution.

16.139 Calculate the pH of a 2.00 M NH4CN solution. 16.140 Calculate the concentrations of all species in a 0.100 M H 3P0 4 solution.

16.141 In the vapor phase, acetic acid molecules associate to a certain extent to form dimers: At 51 °C, the pressure of a certain acetic acid vapor system is 0.0342 atm in a 360-mL flask. The vapor is condensed and neutralized with l3.8 mL of 0.0568 M NaOH. (a) Calculate the degree of dissociation (a) of the dimer under these conditions: (CH 3COOH)2 •

• 2CH 3COOH (b) Calculate the equilibrium constant Kp for the reaction in part (a). 16.142 Calculate the concentrations of all the species in a 0.100 M Na2C03 solution.

16.143 Henry's law constant for CO 2 at 38°C is 2.28 X 10- 3 mol/L . atm. Calculate the pH of a solution of CO 2 at 38°C in equilibrium with the gas at a partial pressure of 3.20 atm. 16.144 Hydrocyanic acid (HCN) is a weak acid and a deadly poisonous compound-in the gaseous form (hydrogen cyanide), it is used in gas chambers. Why is it dangerous to treat sodium cyanide with acids (such as HCI) without proper ventilation?

16.145 How many grams of NaCN would you need to dissolve in enough water to make exactly 250 mL of solution with a pH of 1O.00? 16.146 A solution offormic acid (HCOOH) has a pH of2.53. How many grams of formic acid are there in 100.0 mL of the solution?

16.147 Calculate the pH of a 1-L solution containing 0.150 mole of CH3COOH and 0.100 mole ofHCl. 16.148 A 1.87-g sample of Mg reacts with 80.0 mL of a HCI solution whose pH is -0.544. What is the pH of the solution after all the Mg has reacted? Assume constant volume.

16.149 You are given two beakers, one containing an aqueous solution of strong acid (HA) and the other an aqueous solution of weak acid (HB) of the same concentration. Describe how you would compare the strengths of these two· acids by (a) measuring the pH, (b) measuring electrical conductance, and (c) studying the rate of hydrogen gas evolution when these solutions are combined with an active metal such as Mg or Zn. 16.150 Use Le Chatelier's principle to predict the effect of the following changes on the extent of hydrolysis of sodium nitrite (NaN0 2) solution: (a) HCl is added, (b) NaOH is added, (c) NaCI is added, (d) the solution is diluted.

16.151 A 0.400 M formic acid (HCOOH) solution freezes at -0.758°C. Calculate the Ka of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry out your calculations to three significant figures and round off to two for Ka.) 16.152 The disagreeable odor of fish is mainly due to organic compounds (RNH2) containing an amino group, - NH 2, where R is the rest of the molecule. Arnines are bases just like ammonia. Explain why putting some lemon juice on fish can greatly reduce the odor.

16.153 A solution of methylamine (CH 3NH2) has a pH of 10.64. How many grams of methylamine are there in 100.0 mL of the solution? 16.154 Describe the hydration of S02 as a Lewis acid-base reaction.

16.155 Both the amide ion (NH2) and the nitride ion (N 3- ) are stronger bases than the hydroxide ion and hence do not exist in aqueous solutions. (a) Write equations showing the reactions of these ions with water, and identify the Br\1Snsted acid and base in each case. (b) Which of the two is the stronger base? 16.156 The atmospheric sulfur dioxide (S02) concentration over a certain region is 0.12 ppm by volume. Calculate the pH of the rainwater due to this pollutant. Assume that the dissolution of S02 does not affect its pressure. (Ka for H 2S03 = 1.3 X 10- 2.)

16.157 Explain the action of smelling salt, which is ammonium carbonate [(NH4 hC0 3]. (Hint: The thin film of aqueous solution that lines the nasal passage is slightly basic.) 16.158 About half of the hydrochloric acid produced annually in the United States (3.0 billion pounds) is used in metal pickling. This process involves the removal of metal oxide layers from metal surfaces to prepare them for coating. (a) Write the overall and net ionic equations for the reaction between iron(lli) oxide, which represents the rust layer over iron, and HCI. Identify the Br\1Snsted acid and base. (b) Hydrochloric acid is also used to remove scale (which is mostly CaC0 3) from water pipes. Hydrochloric acid reacts with calcium carbonate in two stages; the first stage forms the bicarbonate ion, which then reacts further to form carbon dioxide. Write equations for these two stages and for the overall reaction. (c) Hydrochloric acid is used to recover oil from the ground. It dissolves rocks (often CaC0 3) so that the oil can flow more easily. In one process, a 15 percent (by mass) HCI solution is injected into an oil well to dissolve the rocks. If the density of the acid solution is 1.073 g/mL, what is the pH of the solution?

16.159 Which of the folIowing does not represent a Lewis acid-base reaction? (a) H 20 + H + • H30 + (b) NH3 + BF3 • H3NBF3 • PFs (c) PF3 + F2 (d) AI(OH)3 + OW • AI(OH)4 16.160 Determine whether each of the following statements is true or false. If false, explain why the statement is wrong. (a) All Lewis acids are Br\1Snsted acids. (b) The conjugate base of an acid always carries a negative charge. (c) The percent ionization of a base increases with its concentration in solution. (d) A solution of barium fluoride is acidic.

16.161 How many milliliters of a strong monoprotic acid solution at pH = 4.12 must be added to 528 mL of the same acid solution at pH = 5.76 to change its pH to 5.34? Assume that the volumes are additive.

ANSWERS TO IN-CHAPTER MATERIALS

16.162 Hemoglobin (Hb) is a blood protein that is responsible for transporting oxygen. It can exist in the protonated form as HbH +. The binding of oxygen can be represented by the simplified equation

(a) What form of hemoglobin is favored in the lungs where oxygen concentration is highest? (b) In body tissues, where the cells release carbon dioxide produced by metabolism, the blood is more acidic due to the formation of carbonic acid. What form of hemoglobin is favored under this condition? (c) When a person hyperventilates, the concentration of CO 2 in his or her blood decreases. How does this action affect the given equilibrium? Frequently a person who is hyperventilating is advised to breathe into a paper bag. Why does this action help the individual? 16.163 A 20.27-g sample of a metal carbonate (MC0 3) is combined with 500 mL of a 1.00 M HCl solution.• The excess HCl acid is then neutralized by 32.80 mL of 0.588 M NaOH. Identify M.

679

16.164 Calculate the pH of a solution that is 1.00 M HCN and 1.00 M HF. Compare the concentration (in molarity) of the CN- ion in this solution with that in a 1.00 M HCN solution. Comment on the difference. 16.165 Tooth enamel is largely hydroxyapatite [Ca3(P04hOH]. When it dissolves in water (a process called demineralization), it dissociates as follows:

The reverse process, called remineralization, is the body's natural defense against tooth decay. Acids produced from food remove the OH- ions and thereby weaken the enamel layer. Most toothpastes contain a fluoride compound such as NaF or SnF2 . What is the function of these compounds in preventing tooth decay? 16.166 Use the van't Hoff equation (see Problem 15.118) and the data in Appendix 2 to calculate the pH of water at its normal boiling point.

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES The following questions are not based on a descriptive passage. 1.

Which of the following best describes the titration of a weak acid with a strong base? a) Low starting pH, and pH "'" 7 at equivalence point b) High starting pH, and pH "'" 7 at equivalence point c) High starting pH, and pH < 7 at equivalence point d) Low starting pH, and pH > 7 at equivalence point

2.

3.

What is the pH of a solution made by combining 150 mL of 0.125 M HCl and 150 mL of 0.120 M Ba(OH)z at 25°C? a) 1.240 'b) 12.760 c) 2.602 d) 11.398

What volume of 0.085 M NaOH is necessary to titrate to the equivalence point 25.0 mL of 0.15 M HCN at 25 °C? a) 44 mL b) 88 mL c) 22 mL d) 25 mL

4.

Determine the pH at the equivalence point for the titration in Question 3. a) 2.98 b) 11.02 c) 8.71 d) 5.29

ANSWERS TO IN-CHAPTER MATERIALS Answers to Practice Problems 16.1A (a) HCl0 4 , (b) HS - , (c) HS - , (d) HC2 0 4 . 16.1B sot , H 2S0 3. 16.2A (a) NHt acid, H 20 base, NH3 conj ugate base, H30 + conjugate acid; (b) CN- base, HzO acid, HCN conjugate acid, OH- conjugate base. 16.2B (a) HS0 4 + H 20 • H30 + + SO~- , (b)HS04 + H 20 • H2S0 4 + OH- . 16.3A 2.0 X 10- 11 M. 16.3B 2.8 X 10- 13 M. 16.4A (a) 8.49, (b) 7.40, (c) 1.25. 16.4B (a) -0.08, (b) 10.52, (c) 11.07. 16.5A (a) 1.3 X 10- 10 M, (b) 3.5 X 10- 2 M, (c) 9.8 X 10- 8 M. 16.5B (a) 7.8 X 10- 3 M, (b) 2.6 X 10- 12 M, (c) 1.3 X 10- 7 M. 16.6A (a) 11.24, (b) 2.14, (c) 5.07. 16.6B (a) 7.55, (b) 8.00, (c) 11.00. 16.7A (a) 9.5 X 10- 14 M, (b) 7.2 X 10- 6 M, (c) 1.0 X 10- 7 M. 16.7B (a) 5.5 X 1O- 1Z M, (b) 2.0 X 10- 4 M, (c) 2.5 X lO- z M. 16.8A (a) 1.09, (b) 5.09, (c) 2.27. 16.8B (a) 1.96, (b) 2.46, (c) 7.00. 16.9A (a) 8.7 X 10- 3 M, (b) 1.7 X 10- 2 M, (c) 9.8 X 10- 7 M. 16.9B (a) 1.5 X 10- 6 M, (b) 1.5 X 10- 5 M, (c) 1.5 X 10- 2 M. 16.10A (a) 0.82, (b) 2.08, (c) 4.77. 16.10B (a) 6.71, (b) 1.21 , (c) 0.91. 16.11A (a) 9.5 X 10- 6 M, (b) 4.8 X 10- 6 M. 16.11B (a) 1.7 X lO- z M, (b) 8.7 X 10- 3 M. 16.12A 2.89. 16.12B 3.05. 16.13A 1.9 X 10- 5 . 16.13B 5.8 X 10- 9 . 16.14A 9.14. 16.14B 8.32. 16.15A 3.0 X 10- 9 . 16.15B 1.4 X 10- 4. 16.16A (a) 1.5 X 10- 10 , (b) 1.2 X 10- 10 , (c) 1.8 X

10- 11 . 16.16B (a) 1.1 X 10- 11 , (b) 4.8 X 10- 7 16.17A [H ZC Z0 4] = 0.11 M, [HC20 4 ] = 0.086 M, [C20~-] = 6.1 X 10- 5 M, [H+] = 0.086 M. 16.17B [H2S04] = 0 M, [HS0 4] = 0.13 M, [SO~-] = 0.011 M, [H+] = 0.15 M. 16.18 (a) HBr04, (b) H ZS0 4. 16.19A 8.96. 16.19B 0.75 M. 16.20A 2.92. 16.20B 0.032 M. 16.21A (a) Basic, (b) acidic, (c) basic, (d) neutral, (e) neutral. 16.21B (a) NH4Br and CH3NH3I, (b) NaHP0 4 and KOBr, (c) Nal and KEr. 16.22 Lewis acid: Co3+, Lewis base: NH3 .

Answers to Checkpoints 16.1.1 b, d. 16.1.2 c, d. 16.2.1 d. 16.2.2 a. 16.3.1 c. 16.3.2 a. 16.3.3 e. 16.3.4 b. 16.4.1 d.16.4.2 b. 16.4.3 b. 16.4.4 e.16.4.5 c. 16.4.6 b. 16.5.1 c. 16.5.2 e. 16.6.1 b. 16.6.2 d. 16.7.1 a. 16.7.2 d. 16.8.1 b. 16.8.2 c. 16.10.1 a. 16.10.2 e. 16.10.3 a, c. 16.10.4 b, c. 16.12.1 c, d, e. 16.12.2 b, d.

Answers to Applying What You've Learned a) Conjugate acid: H ZC6H 60 6, conjugate base: C6H60~- . b) [H+] = 0.003 M, pH = 2.52. c) [OH-] = 3.3 X lO- IZ M. d) pOH = 11.48. e) 3.0 X 10- 3 M. f) 4.0 X 10- 3. g) [HZC 6H60 6] "'" 0.110 M, [HC 6H 60 6 ] = [H+] = 3.0 X 10- 3 M, [C6H60~-] = 2.5 X lO- IZ M.

- ase

17.1 17.2 • •

17.3 • •

• •

17.4 •

• •

17.5 • • •

17.6 • •

•

•

•

UII rIa • • • • • UII rIa I It

The Common Ion Effect Buffer Solutions Calculating the pH ofa Buffer Preparing a Buffer Solution with a Specific pH

Acid-Base Titrations Strong Acid-Strong Base Titrations Weak Acid- Strong Base Titrations Strong Ac-id-Weak Base Titrations Aeid-Base Indicators

Solubility Equilibria Solubility Product Expression and Ksp Calculations Involving K,p and Solubility Predicting Precipitation Reactions

Factors Affecting Solubility The Common Ion Effect pH Complex Ion Formation

Separation of Ions Using Differences in Solubility Fractional Precipitation Qualitative Analysis of Metal Ions in Solution

•

Equilibrium and Tooth Decay Teeth are protected by a hard enamel layer about 2 mm thick that is composed of a mineral called hydroxyapatite [CaS(P04)30H]. Demineralization is the process by which hydroxyapatite dissolves in the saliva. Because phosphates of alkaline earth metals such as calcium are insoluble [I~~ Section 4.2, Table 4.3] , this process happens only to a very small extent:

... . . .. .. . . . . . . . The reverse process, called remineralization, is the body's natural defense against tooth decay: ...

.

..

In children, the growth of the enamel layer (mineralization) occurs faster than demineralization; in adults, demineralization and remineralization occur at roughly equal rates.

Bacteria in the mouth break down some of the food we eat to produce organic acids such as acetic acid and lactic acid. Acid production is greatest from foods that are high in sugar. Thus, after a sugary snack, the H + concentration in the mouth increases, causing OH- ions in the saliva to be consumed.

Removal of OH- from the saliva draws the CaS(P04)30H dissolution equilibrium to the right, promoting demineralization. Once the protective enamel layer is weakened, tooth decay begins. The best way to prevent tooth decay is to eat a diet low in sugar and brush immediately after every meal.

In This Chapter, You Will Learn

more about the behavior of acids and bases and about the factors that affect the

aqueous solubility of ionic compounds.

.--' -

Before you begin. you should review [~~

"-

Section 16.3]

•

Determination of pH

•

Manipulating equilibrium expressions

[~~

Section 15.3]

Media Player/ MPEG Content Chapter in Review

•

Saliva becomes more acidic when high-sugar foods are eaten. Brushing regularly with fluoride toothpaste helps prevent cavities by making tooth enamel less soluble in acid. f

681

682

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

The Common Ion Effect Up until now, we have discussed the properties of solutions containing a single solute. In this section, we will examine how the properties of a solution change when a second solute is introduced. Recall that a system at equilibrium will shift in response to being stressed and that stress can be applied in a variety of ways, including the addition of a reactant or a product [I•• Section 15.5] . Consider a liter of solution containing 0.10 mole of acetic acid. Using the Kafor acetic acid (1.8 X 10- 5) and an equilibrium table [ ~. Section 16.5 ], the pH of this solution at 25 °C can be deter mined: CH 3COOH(aq) +:.===z:. H +(aq)

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

+

CH 3COO- (aq)

0.10

0

0

-x

+x

+x

0.10 - x

x

x 3

Assuming that (0.10 - x) M = 0.10 M and solving for x, we get 1.34 X 10- . Therefore, 3 [CH 3COOH] = 0.09866 M, [H+] = [CH 3COO - ] = 1.34 X 10- M and pH = 2.87. Now consider what happens when we add 0.050 mole of sodium acetate (CH3COONa) to the solution. Sodium acetate dissociates completely in aqueous solution to give sodium ions and acetate ions:

. . . . . . . . . . . . . . . .. ....... ............... . ..............,........ ., .....

The percent ionization of acetic acid is

13~~~~3 M

X 100% = 1.3%

CH3COONa(aq) H 20 , Na \aq) + CH3COO - (aq) . . . . . . . . . . . . . . . . . . .. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .,.........By adding sodium acetate, we also add sodium ions to the solution, However, sodium ions do not interact with water or w ith any of the other species present [ ~ Section [6 .10].

Thus, by adding sodium acetate, we have increased the concentration of acetate ion. Because acetate ion is a product in the ionization of acetic acid, the addition of acetate ion causes the equilibrium to shift to the left. The net result is a reduction in the percent ionization of acetic acid. Addition H + (aq)

+

/

CH3COO - (aq)

Equilibrium is driven toward reactant. Shifting the equilibrium to the left consumes not only some of the added acetate ion, but also some ofthe hydrogen ion. This causes the pH to change (in this case the pH increases). Sample Problem 17.1 shows how an equilibrium table can be used to calculate the pH of a solution of acetic acid after the addition of sodium acetate.

Sample Problem 17.1 ", " Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) Strategy Construct a new eqUilibrium table to solve for the hydrogen ion concentration . Remember that prior to the ionization of a weak acid, the concentration of hydrogen ion in water at 25°( is 1.0 x 10- 7 M. However, because this concentration is insignificant compared to the concentration resulting from the ionization, we can neglect it in our equilibrium table.

.......... Setup' We' ~se' the' s'tat~d concei;t~~iion ci ~~etic' aci'd: o.io M:~nd [H+] = 0 M as the initial concentrations in the table:

CH 3COOH(aq) :.;::.=::t, H+(aq) 0.10

o

0.050

-x

+x

+x

0.10 - x

x

0.050 + x

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

+ CH3COO - (aq)

Solution Substituting the equilibrium concentrations, in terms of the unknown x, into the equilibrium expression gives 1.8 X 10-5 = (x)~o.~o + x) .1 - x Because we expect x to be very small (even smaller than 1.34 X 10- 3 M- see above), because the • ionization of CH3COOH is suppressed by the presence of CH 3COO-, we assume ' (0.10 - x) M

= 0.10 M

and

(0.050 - x) M = 0.050 M

SECTION 17.2

683

In this case, the percent ionization of acetic acid is

Therefore, the equilibrium expression simplifies to ~5

1.8 X 10

Buffer Solutions

(x)(0.050) =

3.6 OX1~0~s M x 100% = 0.036 %

0.10 •

and x = 3.6 X 1O~5 M. According to the equilibrium table, [H+] = x, so pH = - log (3.6 X 1O ~5) = 4.44 . .... .

•••

Think About It The equilibrium concentrations of CH 3COOH, CH3COO ~, and H + are the same

This is considerably smaller than the percent ionization prior to the addition of sodiu m acetate.

regardless of whether we add sodium acetate to a solution of acetic acid, add acetic acid to a solution of sodium acetate, or dissolve both species at the same time. We could have constructed an equilibrium table starting with the equilibrium concentrations in the 0.10 M acetic acid solution:

0.09866

Initial concentration (M): Change in concentration (M) : Equilibrium concentration (M):

1.34 X 1O~3

5.134 X 1O ~2

• • • • • • • • • • • •• • • • • • • • •

•

• • • • •

This is the sum of the equilibrium concentratioof acetate ion in a 0.10 M solution of acetic aoc

- - - -- - t - - - - - - - I - - - - - - - -

(1.34 x 10~3 M) and the added acetate ion (0.050 M).

- - -+ - .::.y--+----- y"----,-----t------..:o.y___ 0.09866

+Y

1.34 X 1O ~3 - Y 5.134 X 1O~2 - Y

In this case, the reaction proceeds to ,the left. (The acetic acid concentration increases, and the concentrations of hydrogen and acetate ions decrease.) Solving for y gives 1.304 X 1O~3 M. [H+] = 1.34 X 1O~ 3 - Y = 3.6 X 1O ~ 5 M and pH = 4.44. We get the same pH either way . .......... .

• • • • • ••

•••

Practice Problem A Determine the pH at 25 °C of a solution prepared by dissolving 0.075 mole of

sodium acetate in 1.0 L of 0.25 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.)

Treating the problem as though both CH 3COmand CH3 COO- are added at the same time and • the reaction proceeds to the right simplifies the solution.

Practice Problem B Determine the pH at 25°C of a solution prepared by dissolving 0.35 mole of

sodium acetate in 1.0 L of 0.25 M acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution .)

An aqueous solution of a weak: electrolyte contains both the weak: electrolyte and its ionization products , which are ions. If a soluble salt that contains one of those ions is added, the equilibrium shifts to the left, thereby suppressing the ionization of the weak: electrolyte. In general, when a compound containing an ion in common with a dissolved substance is added to a solution . . . . . at equilibrium, the equilibrium shifts to the left. This phenomenon is known as the common ion

effect.

Checkpoint 17.1 17.1.1

The Common Ion Effect

Which of the following would cause a decrease in the percent ionization of nitrous acid (HN0 2) when added to a solution of nitrous acid at equilibrium? (Select all that apply.) a) NaN0 2

17.1.2

The common ion can also be H+ or OH ~ . For example, addition of a strong acid to a solution of a weak acid suppresses ionization of the ~ acid. Similarly, addition of a strong base to a solution of weak base suppresses ionization of the weak base.

What is the pH of a solution prepared by adding 0.05 mole of NaF to 1.0 L of 0.1 M HF at 2YC? (Assume that the addition of NaF does not change the volume of the solution.) (Ka for HF = 7.1 X 1O ~4 . ) a) 2.1

b) H 20

b) 2.8

c) Ca(N0 2h

c) 1.4

d) HN0 3

d) 4.6

e) NaN0 3

e) 7.3

Buffer Solutions A solution that contains a weak acid and its conjugate base (or a weak: base and its conjugate B'uffer'soiutioiis; by 'vli-tiie' o(ihe1r 'comPOSItIon; acid) is a buffer solution' or 'slmpiy a changes in pH upon addition of small amounts of either an acid or a base. The ability to resist pH change is very important to chemical and biological systems, including the human body. The pH of blood is about 7.4, whereas that of gastric juices is about 1.5. Each of these pH values is crucial for proper enzyme function and the balance of osmotic pressure, and each is maintained within a very narrow pH range by a buffer.

'buffer:

'r(dst' .. .

Any solution of a weak acid contains some conjugate base. In a buffer solution, though, ;;-:: amounts of weak acid and conjugate base m= be comparable, meaning that the conjugate hax must be supplied by a dissolved salt.

684

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

Calculating the pH of a Buffer Remember that sodium acetate is a strong electrolyte [ ~~ Section 4 ,1]. so it dissociates completely in water to give sodium ions and acetate ions:

., .. C'o nsider' a;;olution th'at i's 'i :6 M Ii; aceti'c 'aClC!" and 'i :6 Msodium acetate. If a small amount of acid is added to this solution, it is consumed completely by the acetate ion,

thus converting a strong acid (H+) to a weak acid (CH 3COOH). Addition of a strong acid lowers the pH of a solution. However, a buffer's ability to convert a strong acid to a weak acid minimizes the effect of the addition on the pH. Similarly, if a small amount of a base is added, it is consumed completely by the acetic acid,

thus converting a strong base (OH-) to a weak base (CH 3COO-). Addition of a strong base increases the pH of a solution. Again, however, a buffer's ability to convert a strong base to a weak base minimizes the effect of the addition on pH. To illustrate the function of a buffer, suppose that we have 1 L of the acetic acid-sodium acetate solution described previously. We can calculate the pH of the buffer using the procedure in Section 17.1: . ,, '

.

CH3COOH(aq) Multimed ia

Acids and Bases effect of addition of a strong acid and a strong base on a buffer,

+:.=::z:'

H+(aq) + CH3COO-(aq)

Initial concentration (M):

1.0

0

1,0

Change in concentration (M):

-x

+x

+x

1.0 - x

x

Equilibrium concentration (M):

1.0

+x

The equilibrium expression is

+ x) ,---::-- -

(x)(l.O

K = a

. . . . . . . . . . . . . . . . . . . . . . . . ..

The forward reaction is suppressed by the presence of the common ion, CH 3COO- , and the reverse process is suppressed by the presence of CH 3COOH.

. ......... .. ........ .......

1.0 - x

.............. .

.,

Because it is reasonable to assume that x will be very small, (1.0 - x) M = 1.0 M

and

(1.0

+ x) M

= 1.0 M

Thus, the equilibrium expression simplifies to 1.8 X 10- 5 = (x)(l .O) = x 1.0 At equilibrium, therefore, [H+] = 1.8 X 10- 5 M and pH = 4.74. Now consider what happens when we add 0.10 mole of HCI to the buffer. (We assume that the addition of HCI causes no change in the volume the solution.) The reaction that takes place ........... ....................... , ......................... ...... .. of .............. . when we add a strong acid is the conversion of H to CH3COOH. The added acid is all consumed" along with an equal amount of acetate ion. We keep track of the amounts of acetic acid and acetate ion when a strong acid (or base) is added by writing the starting amounts above the equation and the final amounts (after the added substance has been consumed) below the equation: '

As long as the amount of strong acid added to the buffer does not exceed the amount of conjugate base originally present, all the added acid will be consumed and converted to weak acid,

Upon addition of H +:

"

~

1.0 mol

0.1 mol

1.0 mol

CH3COO- (aq) + H+(aq)

After H + has been consumed:

, CH3COOH(aq) 1.1 mol

o mol

0.9 mol

We can use the resulting amounts of acetic acid and acetate ion to construct a new equilibrium table: CH3COOH(aq)

+:.=::z:'

H+(aq)

+ CH 3COO-(aq)

Initial concentration (M):

1.1

0

0.9

Change in concentration (M):

-x

+x x

+x

Equilibrium concentration (M):

1.1 - x

0.9

+x

We can solve for pH as we have done before, assuming that x is small enough to be neglected,

SECTION 17.2

1. 8 XlO

-5

=

(x)(0.9

+ x)

1.1 - x

Buffer Solutions

685

(x)(0.9) = -- -

1.1

x = 2.2 X 10- 5 M Thus, when equilibrium is reestablished, [H+] = 2.2 X 10- 5 M and pH = 4.66 ...~. ~h~~g~' ~f"~'~iy """ 0.08 pH unit. In the determination of the pH of a buffer such as the one just described, we always neglect the small amount of weak acid that ionizes (x) because ionization is suppressed by the presence of a common ion. Similarly, we ignore the hydrolysis of the acetate ion because of the presence of acetic acid. This enables us to derive an expression for determining the pH of a buffer. We begin with the equilibrium expression

Had we added 0.1 0 mole of He I to 1 L of pUlE water, the pH would have gone from 7.00 to 1.00!

Rearranging to solve for [H+] gives

Taking the negative logarithm of both sides, we obtain =

[HA] -log Ka - log [A ]

=

[A- ] -log Ka + log [HA]

+

- log [H ] or - log [H+] Thus,

pH

= pKa +

[A - ] log [HA]

Equation 17.1

where pKa = - log Ka

Equation 17.2

Equation 17.1 is known as the Henderson-Hasselbalch equation. Its more general for III is

pH = pKa

+

[conjugate base] log [weak acid]

Equation 17.3

In the case of our acetic acid (1.0 M) and sodium acetate (1.0 M) buffer, the concentrations of weak acid and conjugate base are equal. When this is true, the log term in the Henderson-Hasselbalch equation is zero and the pH is numerically equal to the pKa. In the case of an acetic acid-acetate ion buffer, pKa = -log 1.8 X 10- 5 = 4.74. After the addition of 0.10 mole of HCl, we determined that the concentrations of acetic acid and acetate ion were 1.1 M and 0.9 M, respectively. Using these concentrations in the HendersonHasselbalch equation gives [CH 3COO-] pH = 4.74 + log [CH COOH] .

3

=

4.74

+ 1og 0.9 M 1.1 M

= 4.74 + ( - 0.087) = 4.65 The small difference between this pH and the 4.66 calculated using an equilibrium table is due to differences in rounding. Figure 17.1 illustrates how a buffer solution resists drastic changes in pH .

•

Media Player/MPEG Animation : Figure 17.1, Buffer Solutions, pp. 686-687.

Figure 17.1

•

When we add 0.001 mol of strong acid, it is completely consumed by the acetate ion in the buffer. Before reaction: 0.001 mol 0.010 mol 0.010 mol H +(aq)

After reaction: 0.100 MCH 3COOH 0.100 MCH3COO[CH 3COO - ] = 4.74 + log [CH COOH] = 4.74 3

+ CH3COO-(aq)

0 mol

0.009 mol

• CH3COOH(aq) 0.011 mol

Water pH = 7.00

2S.0 ·c

aO The buffer solution is 0.100 M in acetic acid and 0.100 M in sodium acetate. 100 mL of this buffer contains (0.100 mollL)(O.lO L) = 0.010 mol each acetic acid and acetate ion.

When we add 0.001 mol of strong base, it is completely consumed by the acetic acid in the buffer. Before reaction: 0.001 mol 0.010 mol 0.010 mol OH-(aq)

After reaction:

0 mol

+ CH 3COOH(aq)

• H20 (0

0.009 mol

+ CH 3COO - (aq) 0.011 mol

N

7 •

686

-

We can calculate th€ new pH using the Henderson-Hasselbalch equation: 0.009 pH = 4.74 + log 0.011 = 4.65 There is nothing in pure water to consume strong acid. Therefore, its pH drops drastically.

= -1 0.001 mol = 200 H P og O.lDL .

We can calculate the new pH using the Henderson-Hasselbalch equation: 0.011 pH= 4.74 + log 0.009 =4.83 -

.

Ther€ is nothing in pure water to Gonsume strong base. Therefore, its pH rises drastically. 0.001 mol pOH = -log 0.10 L = 2.00, pH = 12.00

~--

What's the point? A buffer contains both a weak acid and its conjugate base. * Small amounts of strong acid or strong base are consumed by the buffer components, thereby preventing drastic pH changes. Pure water does not contain species that can consume acid or base. Even a very small addition of either causes a large change in pH.

*A buffer could also be prepared using a weak base and its conjugate acid. 687

688

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

Sample Problem 17.2 shows how the Henderson-Hasselbalch equation is used to determine the pH of a buffer after the addition of a strong base.

Sample Problem 17.2 Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (Assume that the addition does not change the volume of the solution.) Strategy Added base will react with the acetic acid component of the buffer, converting OH- to CH 3COO- : The volume of the buffer is 1 L in this example, so the number of moles of a substance is equal to the molar concentration. In cases where the buffer volume is something other than 1 L, however, we can still use molar amounts in the Henderson-Hasselbalch equation because the volume would cancel in the top and bottom of the log term.

•

• •

•••• ••• • • ••• •• • ••• ••

•

• • •

•

• • • •

Write the starting amount of each species above the equation and the final amount of each species below the equation. Use the final amounts as concentrations in Equation 17.1. Setup Upon addition of OH- :

1.00 mol CH3COOH(aq)

Think About It Always do a "reality check" on a calculated pH. Although a buffer does minimize the effect of added base, the pH does increase. If you find that you've calculated a lower pH after the addition of a base, check for errors like mixing up the weak acid and conjugate base concentrations or losing track cif a minus sign.

After OH- has been consumed:

0.10 mol

1.00 mol

+ OW(aq) -----... H 2 0(/) + CH3COO - (aq)

0.90 mol

omol

1.10 mol

Solution H = 4.74 + I 0a 1.10 M P 0 0.90M =

474 .

+ Iog 0.90 1.10 M M

=

483 .

Thus, the pH of the buffer after addition of 0.10 mole of NaOH is 4.83.

Practice Problem A Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.0 Min sodium acetate after the addition of 0.25 mole of NaOH. Practice Problem B Calculate the pH of 1 L of a buffer that is 1.0 M in acetic acid and 1.5 Min sodium acetate after the addition of 0.20 mole of HCI.

Preparing a Buffer Solution with a Specific pH A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added. If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity. Therefore, we consider a solution a buffer, and can use Equation 17.1 to calculate its pH, only if the following condition is met: 10 >

Weak Acid

Ka

pKa

HF HN0 2 HCOOH C6 HsCOOH CH 3COOH

7.1 x 10-4 4.5 X 10- 4 1.7 x 10- 4 6.5 x lO- s 1.8 X lO- s 4.9 x 10- 10 1.3 x 10- 10

3.15 3.35 3.77 4.19 4.74 9.31 9.89

HCN C6 HsOH

[conjugate base] [weak acid]

>0.1

Consequently, the log term in Equation 17.1 can only have values from -1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pKa of the weak acid it contains. This is known as the range of the buffer, where pH = pKa + 1. This enables us to select the appropriate conjugate pair to prepare a buffer with a specific, desired pH. First, we choose a weak acid whose pKa is close to the desired pH. Next, we substitute the pH and pKa values into Equation 17.1 to obtain the necessary ratio of [conjugate base]/[ weak acid]. This ratio can then be converted to molar quantities for the preparation of the buffer. Sample Problem 17.3 demonstrates this procedure.

Sample Problem 17.3 . Select an appropriate weak acid from the table in the margin, and describe how you would prepare a buffer with a pH of 9.50.

SECTION 17.2

Buffer Solutions

Strategy Select an acid with a pKa within one pH unit of 9.50. Use the pKa of the acid and Equation

17.1 to calculate the necessary ratio of [conjugate base]/[ weak acid]. Select concentrations of the buffer components that yield the calculated ratio. Setup Two of the acids listed in Table 16.6 have pKa values in the desired range: hydrocyanic acid

(HCN, pKa = 9.31) and phenol (C 6 H sOH, pKa = 9.89). Solution Plugging the values for phenol into Equation 17.1 gives

Think About It There is an

Therefore, the ratio of [C 6 HsO- ] to [C 6H sOH] must be 0.41 to 1. One way to achieve this would be to dissolve 0.41 mole of C 6HsONa and 1.00 mole of C 6HsOH in 1 L of water.

Practice Problem A Select an appropriate acid from Table 16.6, and describe how you would

prepare a buffer with pH = 4.5. Practice Problem B What range of pH values could be achieved with a buffer consisting of nitrous

acid (HN02 ) and sodium nitrite (N0 2 )7

Bringing Chemistry to Life Maintaining the pH of Blood There are about 5 L of blood in the average adult. Circulating blood keeps cells alive by providing them with oxygen and nutrients and by removing carbon dioxide and other waste materials. The efficiency of this enolmously complex system relies on buffers. The two primary components of blood are blood plasma and red blood cells, or erythrocytes. Blood plasma contains many compounds, including proteins, metal ions, and inorganic phosphates. The erythrocytes contain hemoglobin molecules, as well as the enzyme carbonic anhydrase, which catalyzes both the formation and the decomposition of carbonic acid (H2C0 3 ) :

The substances inside the erythrocytes are protected from extracellular fluid (blood plasma) by a semipermeable cell membrane that allows only certain molecules to diffuse through it. The pH of blood plasma is maintained at about 7.40 by several buffer systems, the most important of which is the HCO;- IH2C0 3 system. In the erythrocyte, where the pH is 7.25, the principal buffer systems are HCO;- IH 2C0 3 and hemoglobin. The hemoglobin molecule is a complex protein molecule (molar mass of 65,000 g) that contains a number of ionizable protons. As a very rough approximation, we can treat it as a monoprotic acid in the form HHb:

where HHb represents the hemoglobin molecule and Hb - is the conjugate base of HHb. Oxyhemoglobin (HHb0 2), fOlmed by the combination of oxygen with hemoglobin, is a stronger acid than HHb:

Carbon dioxide produced by metabolic processes diffuses into the erythrocyte, where it is rapidly converted to H 2C0 3 by carbonic anhydrase:

The ionization of the carbonic acid,

infinite number of combinations of [conjugate base] and [weak acid] that will give the necessary ratio. Note that this pH could also be achieved using HCN and a cyanide salt. For most purposes, it is best to use the least toxic compounds available.

689

690

CHAPTER 17

Acid-Base Equi libria and Solubility Equilibria

has two important consequences. First, the bicarbonate ion diffuses out of the erythrocyte and is carried by the blood plasma to the lungs. This is the major mechanism for removing carbon dioxide. Second, the H+ ions shift the equilibrium in favor of the un-ionized oxyhemoglobin molecules:

Because HHb0 2 releases oxygen more readily than does its conjugate base (Hb0 2 ), the formation of the acid promotes the following reaction from left to right: HHbOzeaq)

+:.==:!:'

HHb(aq)

+ 0 2(aq)

The O2 molecules diffuse out of the erythrocyte and are taken up by other cells to carry out metabolism. When the venous blood returns to the lungs, the preceding processes are reversed. The bicarbonate ions now diffuse into the erythrocyte, where they react with hemoglobin to form carbonic acid: HHb(aq)

+ HC0 3 (aq) +:.==:!:' Hb - (aq) + H 2C0 3(aq)

Most of the acid is then converted to CO 2 by carbonic anhydrase:

The carbon dioxide diffuses to the lungs and is eventually exhaled. The formation of the Hb - ions (due to the reaction between HHb and HC0 3") also favors the uptake of oxygen at the lungs, Hb -(aq)

+ O?(aq) :;::::.===z:' Hb0 2 (aq)

because Hb- has a greater affinity for oxygen than does HHb. ' When the arterial blood flows back to the body tissues, the entire cycle is repeated.

Checkpoint 17.2 17.2.1

Buffer Solutions

Which of the following combinations can be used to prepare a buffer?

17.2.3

c) CH3 COOHlOH-

Consider 1 L of a buffer that is 0.85 M in formic acid (HCOOH) and 1.4 M in sodium formate (HCOONa). Calculate the pH after the addition of 0.15 mol HCI. (Assume the addition causes no volume change.)

d) HN0 2IN0 2

a) 4.11

e) HN0 3IN0 3

b) 3.99

a) HCIICIb) HFIF-

17.2.2

c) 3.87

What is the pH of a buffer that is 0.76 M in HF and 0.98 Min NaF?

d) 10.13

e) 10.01

a) 3.26 b) 3.04 c) 3.15 d) 10.85 e) 10.74

17.2.4

Consider 1 L of a buffer that is 1.5 M in hydrocyanic acid (HCN) and 1.2 M in sodium cyanide (NaCN). Calculate the pH after the addition of 0.25 mol NaOH. (Assume the addition causes no volume change.) a) 9.21

b) 9.37 c) 9.04 d) 4.63 e) 4 .96

Acid-Base Titrations In Section 4.6 we introduced acid-base titrations as a form of chemical analysis. Having discussed buffer solutions, we can now look in more detail at the quantitative aspects of acid-base titrations. We will consider three types of reactions: (1) titrations involving a strong acid and a strong base,

SECTION 17.3

Acid-Base Titrations

69'

Fig u re 17.2

. u " ~

A pH meter is used monitor an acid-base titration.

I

[0

!

(2) titrations involving a weak acid 'and a strong base, and (3) titrations involving a strong acid and a weak base. Titrations involving a weak acid and a weak base are complicated by the hydrolysis of both the cation and the anion of the salt formed. These titrations will not be discussed here. Figure, 17.2 shows the experimental setup for monitoring the pH over the course of an acid-base titration.

Strong Acid- Strong Base Titrations The reaction between the strong acid. HCl and the strong base NaOH can be represented by NaOH(aq)

I

+ HCI(aq) ---. NaCI(aq) + H 20(l)

Multimedia

Acids and Bases- titration of He l with NaOH .

or by the net ionic equation,

,

Consider the addition of a 0.100 M NaOH solution (from a buret) to a vessel containing 25.0 mL of 0.100 M HCI. For convenience, we will use only three significant figures for volume and con-. . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . centration and two significant figures for pH. Figure 17.3 shows the titration curve the plot of pH ....... . ........ . ... x-.-.................... . . ...... . ..... . .... . ........................................... . as a function of titrant volume added. : Before the addition of NaOH begins, the pH of the acid is given by -log(0,100), or 1.00. : When NaOH is added, the pH of the solution increases slowly at first. Near the equivalence point, ..... . the pH begins to rise steeply, and at the equivalence point, when equimolar amounts of acid and : base have reacted, the curve rises almost vertically. In a strong acid-strong base titration, both the . · hydrogen ion and hydroxide ion concentrations are very small at the equivalence point (roughly ····· 1 X 10- 7 M); consequently; the addition of a single drop of the base causes a large increase in [OH- ] and a steep rise in the pH of the solution. Beyond the equivalence point, the pH again increases slowly with the continued addition of NaOH.

Recall that only the digits to the right of the decimal point are significant in a pH value.

•

. ~">"

The titrant is the solution that is added from the buret.

•

Figure 17.3

Titrationcurve(pH as a function of volume titrant added) of a strong acid-strong base titration. A 0.100 M NaOH solution, the titrant, is added from a buret to 25.0 mL of a 0.100 M Hel solution in an Erlenmeyer flask.

14 13

- --

F-

12 11 10 9 8 -

:r:

0,

7

Recall that for an acid and base that combine in a 1:1 ratio, the equivalence point is where equal molar amounts of acid and base have beer combined [ ~~ Sec tio n 4.5 ].

Equivalence point

---------------

6 5 4 3 2

-~

J __ I I _

~--

I -I-

1

I

0

20

30

Volume of NaOH added (mL)

40

50

692

CHAPTER 17

Acid-Base Equilibria and Solubility Equ ilibria

It is possible to calculate the pH of the solution at every stage of titration. Here are three sample calculations.

These calculations could also be done using moles, but using millimoles simplifies the calculations. Remember that millimoles = M x mL [ ~ Section 4.5] .

1. Consider the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCI: The total .............. vciiume' of"ihe ·sciiutlori.·ls·3S".() ·fiC.·The· number· of"millirnoles of NaOH in 10.0 mL is 10.0 mL X 0.100 mIllOI NaOH 1 mL

=

1.00 mIllol

The number of millimoles of HCI originally present in 25.0 mL of solution is 25.0 mL X 0.100 ~l HCI

= 2.50 mIllol

Thus, the amount of HClleft after partial neutralization is 2.50 - 1.00, or 1.50 mIllol. Next, we determine the resulting concentration of H +. We have 1.50 mIllol in 35.0 mL: 1.50 mIllol HCI = 0.0429 M 35.0 mL Thus [H+] = 0.0429 M, and the pH of the solution is pH = - log (0.04289) = 1.37 2. Consider the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCI: This is a straightforward calculation, because it involves a complete neutralization reaction and neither ion in the salt (NaCl) undergoes hydrolysis [ ~~ Section 16.10] . At the equivalence point, [H+] = [OH-] = 1.00 X 10- 7 M and the pH of the solution is 7.000. 3. Consider the addition of 35.0 mL of 0.100 M NaOH to 25 .0 mL of 0.l00 M HCI: The total volume of the solution is now 60.0 mL. The number of rnillimoles of NaOH added is 35.0 mL X 0.100 ~: NaOH = 3.50 mIllol There are 2.50 mIllol of HCl in 25.0 mL of solution. After complete neutralization of HCI, 2.50 mIllol of NaOH have been consumed, and the number of rnillimoles of NaOH remaining is 3.5 - 2.5 or 1.00 mrnol. The concentration of NaOH in 60.0 mL of solution is 1.00 mIllol NaOH 60.0 mL

= 00167 M .

Thus [OH-] = 0.0167 M and pOH = - log (0.0167) = 1.78. The pH of the solution is 14.00 - 1.78 or 12.22. Table 17.1 lists the data at nine different points during a strong acid - strong base titration along with the calculated pH at each point.

Volume OHadded (mL)

OHadded (mmol)

H+ remaining (mmol)

Total volume (mL)

0

0

2.5

25 .0

0.100

LOOO

5.0

0.50

2.0

30.0

0.0667

1.176

10.0

1.0

1.5

35.0

0.0429

1.364

15.0

1.5

1.0

40.0

0.0250

1.602

20.0

2.0

0.5

45.0

0.0111

1.955

25.0

2.5

0

50.0

1.00 X 10- 7

7.000

Volume OHadded (mL)

OHadded (mmol)

Excess OH- (mmol)

Total volume (mL)

[OH - ] (moI / L)

pOH

pH

30.0

3.0

0.5

55.0

0.0091

2.04

11.96

35 .0

3.5

1.0

60.0

0.0167

1.78

12.223

[H+] (moI / L)

pH

SECTION 17.3

Acid-Base Titratio ns

693

Weak Acid-Strong Base Titrations Consider the neutralization reaction between acetic acid (a weak acid) and sodium hydroxide (a strong base):

+ NaOH(aq) -

CH3COOH(aq)

-+. CH3COONa(aq)

+ H 20(l)

This equation can be simplified to

The acetate ion that results from this neutralization undergoes hydrolysis follows:

[ ~~

Section 16.10] as

At the equivalence point, therefore, when we only have sodium acetate in solution, the pH will be greater than 7 as a result of the OH- formed by hydrolysis of the acetate ion. The curve for titration of 25.0 mL of 0.1 M acetic acid with 0.10 M sodium hydroxide is shown in Figure 17.4. Note how the shape of the curve differs from the one in Figure 17.3. Compared to the curve for titration of a strong acid with a strong base, the curve for titration of a weak acid with a strong base has a higher initial pH, a more gradual change in pH as base is added, and a shorter vertical region near the equivalence point. Again, it is possible to calculate the pH at every stage of the titration. Here are four sample calculations. 1. Prior to the addition of any base, the pH is determined by the ionization of acetic acid. We use its concentration (0.10 M) and its K a (1.8 X 10- 5) to calculate the H + concentration using an equilibrium table: CH3COOH(aq) :;::: , ~. H +(aq)

0.10

o

o

-x

+x

+x

0.10 - x

x

x

Initial concentration (M) : Change in concentration (M): Equilibrium concentration (M):

2 X

0.10 - x We can neglect x on the bottom of the equation 2

0~1O = 1.8

+ CH 3COO- (aq)

X

10-

[ ~~

= 1.8

X

10- 5

Section 16.5] . Solving for x,

5

x 2 = (1.8 X 10- 5)(0.10)

=

1.8

X

10- 6

x = ~1.8 X 10- 6 = 1.34 X 10- 3 M

gives [H+]

=

1.34

X

10- 3 M and pH = 2.87. Figure 17.4

Titration curve of a weak acid-strong base titration. A 0.100 M NaOH solution is added from a buret to 25.0 mL of a 0.100 M CH3COOH solution in an Erlenmeyer flask. Because of the hydrolysis of the salt formed, the pH at the equivalence point is greater than 7.

1413 12 -

11 10 9-

:a

- - - - - - - - - - - - - - - - - Equivalence 8point

7-

.?:

6-

543 2-

I I I I I I I I I

-'

I-

0 0

, 10

I

,

I

1

20

30

40

50

Volume of NaOH added (mL)

694

CHAPTER 17

Acid-Base Equil ibria and Solubi lity Equi li bria

2. After the first addition of base, some of the acetic acid has been converted to acetate ion via the reaction CH3COOH(aq)

+ OH-(aq) -

-+.

+ H 20 (I)

CH 3COO - (aq)

With significant amounts of both acetic acid and acetate ion in solution, we now treat the solution as a buffer and use the Henderson-Hasselbalch equation to calculate the pH. After the addition of 10.0 mL of base, the solution contains 1.5 mmol of acetic acid . . . . . . . . . ... . . . . . . . . and 1.0 mmol of acetate ion (see Table 17.2).

Remember that we can use mol or mmol amounts in place of concentrations in the Henderson-Hasselbalch equation.

+ I0",a

H = 474 p .

1.0 mmol = 456 1.5 mmol .

Each of the points between the beginning of the titration and the equivalence point can be calculated in this way. 3. At the equivalence point, all the acetic acid has been neutralized and we are left with acetate ion in solution. (There is also sodium ion, which does not undergo hydrolysis and therefore does not impact the pH of the solution.) At this point, pH is determined by the concentration and the Kb of acetate ion, The equivalence point occurs when 25.0 mL of base has been added, making the total volume 50.0 mL. The 2.5 mmol of acetic acid (see Table 17.2) has all been converted to acetate ion. Therefore, the concentration of acetate .

,

lOn IS

[CH COO-] = 2,5 mmol = 0050 M 3 50.0 mL . As we did at the beginning of the titration, we construct an equilibrium table: CH3COO-(aq) + H 2O(l) •

0.050

0

0

Change in concentration (M) :

-x

+x

+x

Equilibrium concentration (M):

0.050 - x

x

x

Initial concentration (M) :

Ka XKb=Kw [ ~.

Section 16.7].

.............. 'T he Kb for acetate ion is 5.6 X 10- 10 .

x2 _ - 0.050 _ x - 5.6

K - [OH- ][CH3COOH] _

[CH3COO]

b -

Volume OH added (mL) · . . . ...... .. . . . . . . . .. . '"

Prior to the addition of any base, and at the equivalence point, this is an equilibrium problem that is solved using a concentration, an ionization constant, and an equilibrium table.

• OH-(aq) + CH3COOH(aq)

o

OH- added (mmol) ... . . .. -. . . . . . .

•

X

- 10

10

•

remaining pH .. ..... ....... -. . .. .. . . . . . . . . . . .. . . .. . ....... • • • • • • 0.0 2,87* 2,5 o . . . . . .. ..... ......... . . . .. . . ... .,. . . . . . .. . .... ...... . . . . . . . . .. . . . . . . . . . 0.50 2.0 0.50 4.14 • •• •

•

10.0

1.0

1.5

1.0

4,56

15.0

1.5

1.0

1.5

4,92

using the Henderson-Hasselbalch equation:

20.0

2.0

0.5

2.0

5.34

I [conjugate base] . pH = pKa + og ~-----=o---,-;:--' [weak acid]

25,0

2.5

0,0

2.5

8.72t [OH-] (moIlL) 0.0091

Prior to the equivalence point, pH is determined

· . ..

•

After the equivalence point, the titration curve of a weak acid is identical to that of a strong acid.

••

'

•

•

Volume OH added (mL)

OH- added (mmol)

Excess OH- (mmol)

Total volume (mL)

30,0

3.0

0.5

55.0

• • •• •

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • ••

35,0 *[CH 3COOH]

••••••••••••••••••••••

3.5 = 0.10 M,

1.0

60.0

Ka = 1.8 x 10- '.

t[CH 3 COO- ] = 0.050 M, Kb = 5.6 X 10- 10 •

••••••••••• ••••••••••

!

• ••••• • • •

•

• • • • •

0.Ql7

pOH

pH

2.04 11.96 • ••

••• •••••

1.78 12.22

SECTION 17.3

As before, we can neglect x in the denominator of the equation. Solving for x, 2

x = 56 X 10- 10 0.050 . x2

= (5.6

X

10- 10)(0.050)

x = ~2.78 X 10

II =

= 2.8

X

10- 11

5.3 X 10- 6 M

gives [OH- ] = 5.3 X 10- 6 M, pOH = 5.28, and pH = 8.72. 4. After the equivalence point, the curve for titration of a weak acid with a strong base is identical to the curve for titration of a strong acid with a strong base. Because all the acetic acid has been consumed, there is nothing in solution to consume the additional added OH- , and the pH levels off between 12 and 13. Table 17.2 lists the data for the titration of 25.0 mL of 0.10 M acetic acid with 0.10 M NaOH. Sample Problem 17.4 shows how to calculate the pH for the titration of a weak acid with a strong base.

I Sample Problem 17.4 Calculate the pH in the titration of 50.0 rnL of 0.120 M acetic acid by 0.240 M sodium hydroxide after the addition of (a) 10.0 rnL of base, (b) 25.0 rnL of base, and (c) 35.0 rnL of base.

Strategy The reaction between acetic acid and sodium hydroxide is

Prior to the equivalence point [part (a)], the solution contains both acetic acid and acetate ion, making the solution a buffer. We can solve part (a) using Equation 17.1, the Henderson-Hasselbalch equation. At the equivalence point [part (b)], all the acetic acid has been neutralized and we have only acetate ion in solution. We must determine the concentration of acetate ion and solve part (b) as an equilibrium problem, using the Kb for acetate ion. After the equivalence point [part (c)], all the acetic acid has been neutralized and there is nothing to consume the additional added base. We must determine the concentration of excess hydroxide ion in the solution and solve for pH using Equations 16.4 and 16.6.

Setup Remember that M can be defined as either molfL or mmoVrnL [ ~ Section 4.5] . For this type of problem, it simplifies the calculations to use millimoles rather than moles. Ka for acetic acid is 1.8 X 10- 5, so pKa = 4.74. Kb for acetate ion is 5.6 X 10- 10 (a) The solution originally contains (0.120 mmoVrnL)(50.0 rnL) = 6.00 mmol of acetic acid. A 1O.0-rnL amount of base contains (0.240 mmoVrnL)(1O.0 rnL) = 2.40 mmol of base. After the addition of 10.0 rnL of base, 2.40 mmol of OH- has neutralized 2.40 mmol of acetic acid, leaving 3.60 mmol of acetic acid and 2.40 mmol acetate ion in solution: Upon addition of OH- :

6.00 mmol

2.40 mmol

CH 3COOH(aq)

After OH- has been consumed:

3.60 mmol

0 mmol

+ OH- (aq) :;::.==::::!:' H 20 (l) + CH3COO - (aq) Ommol

2.40 mmol

(b) After the addition of 25.0 rnL of base, the titration is at the equivalence point. We calculate the pH using the concentration and the Kb of acetate ion. (c) After the addition of 35.0 rnL of base, the titration is past the equivalence point and we solve for pH by determining the concentration of excess hydroxide ion.

Solution (a) pH = pKa

+ log 2.40 = 3.60

4.74 - 0.18 = 4.56

(b) At the equivalence point, we have 6.0 mmol of acetate ion in the total volume. We determine the total volume by calculating what volume of 0.24 M base contains 6.0 mmol: (volume)(0.240 mmoVrnL)

=

6.00 mmol

6.00 mmol volume = 0.240 mmoVrnL = 25.0 rnL

(Continued)

Acid-Base Titrations

695

696

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

Therefore, the equivalence point occurs when 25.0 mL of base has been added, making the total volume 50.0 mL + 25.0 mL = 75.0 mL. The concentration of acetate ion at the equivalence point is therefore Think About It For each point in a titration, decide first what species are in solution and what type of problem it is. If the solution contains only a weak acid (or weak base), as is the case before any titrant is added, or if it contains only a conjugate base (or conjugate acid), as is the case at the equivalence point, when pH is determined by salt hydrolysis, it is an equilibrium problem that requires a concentration, an ionization constant, and an equilibrium table. If the solution contains comparable concentrations of both members of a conjugate pair, which is the case at points prior to the equivalence point, it is a buffer problem and is solved using the Henderson-Hasselbalch equation. If the solution contains excess titrant, either a strong base or strong acid, it is simply a pH problem requiring only a concentration.

6.00 mmol CH 3COO- = 0.0800 M 75.0mL We can construct an equilibrium table using this concentration and solve for pH using the ionization constant for CH 3COO- (Kb = 5.6 X 10- 10):

Initial concentration (M): Change in concentration (M): Equilibrium concentration (M):

CH3COO-(qq)

+ H 20(I) :;::.=::!:" OW(aq) + CH3COOH(aq)

0.0800

o

o

-x

+x

+x

0.0800 - x

x

x

Using the equilibrium expression and assuming that x is small enough to be neglected, [CH3COOH] [OW] (x) (x) x2 10 Kb = [CH 3COO-] = 0.0800 _ x = 0.0800 = 5.6 X 10x = ~4.48

X

10- 11 = 6.7

X

10- 6 M

According to the equilibrium table, x = [OH-], so [OH- ] = 6.7 X 10- 6 M. At equilibrium, therefore, pOH = -log (6.7 X 10- 6) = 5.17 and pH = 14.00 - 5.17 = 8.83. (c) After the equivalence point, we must determine the concentration of excess base and calculate pOH and pH using Equations 16.4 and 16.6. A 35.0-mL amount of the base contains (0.240 mmol/mL)(35.0 mL) = 8.40 mmol of OH-. After neutralizing the 6.00 mmol of acetic acid originally present in the solution, this leaves 8.40 - 6.00 = 2.40 mmol of excess OH-. The total volume is 50.0 + 35.0 = 85.0 mL. Therefore, [OH- ] = 2.40 mmo1/85.0 mL = 0.0280 M, pOH = -log (0.0280) = 1.553, and pH = 14.000 - 1.553 = 12.447. In summary, (a) pH = 4.56, (b) pH = 8.83, and (c) pH = 12.447.

Practice Problem A For the titration of 10.0 mL of 0.15 M acetic acid with 0.10 M sodium hydroxide, determine the pH when (a) 10.0 mL of base has been added, (b) 15.0 mL of base has been added, and (c) 20.0 mL of base has been added. Practice Problem B For the titration of 25.0 mT , of 0,20 M hydrofluoric acid with 0.20 M sodium hydroxide, determine the volume of base added when pH is (a) 2,85, (b) 3,15, and (c) 11.89,

Strong Acid-Weak Base Titrations Consider the titration of HCl, a strong acid, with NH 3, a weak: base:

or simply

The pH at the equivalence point is less than 7 because the ammonium ion acts as a weak: Bn'lnsted acid:

or simply

Because of the volatility of an aqueous ammonia solution, it is more convenient to use hydrochloric acid as the titrant (i.e" to add HCl solution from the buret), Figure 17.5 shows the titration curve for this experiment. Analogous to the titration of a weak: acid with a strong base, the initial pH is determined by the concentration and the Kb of ammonia:

SECTION 17.3

Figure 17.5

14

13 12 11 10 9 8

:r:0-

Acid-Base Titrations

7 6

Equivalence - - - - - - - - - -- -point

- - - - - - - - ----- - - -

5 4 3 2 1

•

0 0

I I I I I I I I I

20

10

40

30

Volume HCI added (mL)

pH

0.0 5.0 10.0 15.0 20.0 22.0 24.0 25.0 26.0 28.0 30.0 35.0 40.0 45 .0 50.0

11.13 9.86 9.44 9.08 8.66 8.39 7.88 5.28 2.70 2.22 2.00 1.70 1.52 1.40 1.30

50

Volume of HCI added (mL)

Consider the titration of 25 .0 mL of 0.10 M NH3 with 0.10 M HC!. We calculate the initial pH by constructing an equilibrium table and solving for x: NH3(aq) + H 20(l) Initial concentration (M) :

0.10

0

0

-x

+x

+x

x

x

Change in concentration (M):

-_.-

0.10 - x

Equilibrium concentration (M):

===' NH1(aq) + OH-(aq)

+=.

Using the equilibrium expression and assuming that x is small enough to be neglected, [NH1] [OH-] Kb = [NH3]

(x) (x)

0.10 - x

2

=

X

0.10

= 1.8

X

10- 5

x 2 = l.8 X 10- 6 x = ~l.8 X 10- 6 = 1.3 X 10- 3

pH

=

11.11

The pH at the equivalence point is calculated using the concentration and Ka of the conjugate base of NH 3, the NH1 ion, and an equilibrium table. Sample Problem 17.5 shows how this is done.

Sample Problem 17.5 Calculate the pH at the equivalence point when 25.0 mL of 0.100 M NH3 is titrated with 0.100 M HC!.

Strategy The reaction between NH3 and HCl is NH 3(aq)

+ H +(aq) - _. NHt (aq)

At the equivalence point, all the NH3 has been converted to NHt . Therefore, we must determine the concentration of NHt at the equivalence point and use the Ka for NHt to solve for pH using an equilibrium table.

Setup The solution originally contains (0.100 mmollrnL)(25.0 rnL) = 2.50 mmol NHt . At the equivalence point, 2.50 mmol of HCI has been added. The volume of 0.100 M HCl that contains 2.50 mmol is (volume)(0.100 mmolirnL) = 2.50 mmol vo Iume

=

2.50 mmol 0.100 mmol/mL

= 25 .0 rnL (Continued)

69-

Titration curve of a strong acid-weak base titration. A 0.100 M HCI solution is added from a buret to 25.0 rnL of a 0.100 M NH3 solution in an Erlenmeyer flask. As a result of salt hydrolysis, the pH at the equivalence point is lower than 7.

698

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

It takes 25.0 mL of titrant to reach the equivalence point, so the total solution volume is 25.0 + 25.0 = 50.0 mL. At the equivalence point, all the NH3 originally present has been converted to NH1. The concentration of NH 1 is (2.50 mmol)/(50.0 mL) = 0.0500 M. We must use this concentration as the starting concentration of ammonium ion in our equilibrium table. Solution NH 1(aq)

0.0500

o

o

-x

+x

+x

0.0500 - x

x

x

Initial concentration (M): Think About It In the titration of a weak base with a strong acid, the species in solution at the equivalence point is the conjugate acid. Therefore, we should expect an acidic pH. Once all the NH3 has been converted to NH1, there is no longer anything in the solution to consume added acid. Thus, the pH after the equivalence point depends on the number of rnillimoles of H + added and not consumed divided by the new total volume.

Change in concentration (M): Equilibrium concentration (M):

+ H 20(l) +.==" NH 3 (aq) + H30(aq)

The equilibrium expression is 2 NH K = [ 3][H+] = (x ) (x ) _ x = x = 5.6 X 10- 10 a [NH1] 0.0500 0.0500

x2 =2.8 X IO- 11

x = ~2.8

X

10

II

= 5.3

X

10- 6 M

[H+] = x = 5.3 X 10- 6 M. At equilibrium, therefore, pH = -log (5.3 X 10- 6) = 5.28.

Practice Problem A Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.10 M methylamine (see Table 16.7) with 0.20 M HCI. Practice Problem B Calculate the pH at the equivalence point in the titration of 35 mL of 0.12 M NH3 with 0.16 M RN0 3.

Acid-Base Indicators The equivalence point is the point at which the acid has been neutralized completely by the added base. The equiValence point in a titration can be determined by monitoring the pH over the course of the titration, or it can be determined using an acid-base indicator. An acid-base indicator is usually a weak organic acid or base for which the ionized and un-ionized forms are different colors. Consider a weak organic acid that we will refer to as HIn. To be an effective acid-base indicator, HIn and its conjugate base, In -, must have distinctly different colors. In solution, the acid ionizes to a small extent:

The end point is where the color changes. The equivalence point is where neutralization is complete. Experimentally, we use the end point to estimate the equivalence point.

In a sufficiently acidic medium, the ionization of HIn is suppressed according to Le Chatelier's principle, and the preceding equilibrium shifts to the left. In this case, the color of the solution will be that of HIn. In a basic medium, on the other hand, the equilibrium shifts to the right and the color of the solution will be that of the conjugate base, In- . .......... .... 'the' end point of a titration is the point at which the color of the indicator changes. Not all indicators change color at the same pH, however, so the choice of indicator for a particular titration depends on the strength of the acid (and the base) used in the titration. To use the end point to determine the equivalence point of a titration, we must select an appropriate indicator. The end point of an indicator does not occur at a specific pH; rather, there is a range of pH over which the color change occurs. In practice, we select an indicator whose color change occurs over a pH range that coincides with the steepest part of the titration curve. Consider the information in Figure 17.6, which shows the titration curves for hydrochloric acid and acetic acid each being titrated with sodium hydroxide. Either of the indicators shown can be used for the titration of a strong acid with a strong base because both end points coincide with the steepest part of the HCl-NaOH titration curve. However, methyl red changes from red to yellow over the pH range of 4.2 to 6.3. This end point occurs significantly before the equivalence point in the titration of acetic acid, which occurs at about pH 8.7. Therefore, methyl red is not a suitable indicator for use in the titration of acetic acid with sodium hydroxide. Phenolphthalein, on the other hand, is a suitable indicator for the CH3COOH-NaOH titration. Many acid-base indicators are plant pigments. For example, boiling red cabbage in water extracts pigments that exhibit a variety of colors at different pH values (Figure 17.7). Table 17.3 lists a number of indicators commonly used in acid-base titrations. The choice of indicator for a particular titration depends on the strength of the acid and base to be titrated.

SECTION 17.3

14 I

12 -

10

Titration curve of a strong acid with a strong base (blue) and titration curve of a weak acid with a strong base (red). The indicator phenolphthalein can be used to determine the equivalence point of either titration. Methyl red can be used for the strong acid- strong base titration but cannot be used for the weak acid- strong base titration because its color change does not coincide with the steepest part of the curve.

! r,

/

11 -

-

~,

I

Phenolphthalein (8.3 - 10) I

9,

8-

:a.

--

-

7

546

3

Methyl red (4.2 - 6.3)

/'

-

r

2-

1

•

o 0

699

Figure 17.6

,

13 -

Acid-Base Titrations

.

10

20

io

3'0

'.,~



17.6.2

Barium nitrate is added slowly to a solution that is 0.10 M in SO ~- ions and 0.10 M in F- ions. Calculate the concentration of Ba2+ ions (in mol/L) required to initiate the precipitation of BaS04 without precipitating BaF1 . a) 1.7 X 10- 6 M b) 1.1 X 10- 9 M

d) AgI

c) 1.7 X 10- 4 M

e) Ag2 S04

d) 1.7 X 10- 5 M e) 1.1 X 10- 8 M

•

716

CHAPTER 17

Acid-Base Equilibria and Solubility Equ i libria

Applying What You've Learned Most toothpastes contain fluoride, which helps to reduce tooth decay. The F- ions in toothpaste replace some of the OH- ions during the remineralization process:

SCa2+(aq)

•

+ 3PO ~ - (aq) + F-(aq)

+-

•

CaS(P04)3F(S)

Because F- is a weaker base than OH - , the modified enamel, called fluoroapatite, is more resistant to the acid produced by bacteria .

Problems:

(a) Calculate the molar solubility of hydroxyapatite given that its [ ~~ Sample Problem 17 .7]

K sp is

2 X 10- 59 .

(b) Calculate the K sp of fluoroapatite given that its molar solubility is 7 X 10- 8 M. [ ~~ Sample Problem 17.7] (c) Calculate the molar solubility of fluoroapatite in an aqueous solution in which the concentration of fluoride ion is 0.10 M. [ ~~ Sample Problem 17 .7] Cd) Calculate the molar solubility of hydroxyapatite in a buffered aqueous solution with pH = 4.0. [ ~~ Sample Problem 17.7]

KEY EQUATIONS

-

717

CHAPTER SUMMARY =~~---

Section 17.1 o

o

a titration is the point at which the color of the indicator changes. It is used to estimate the equivalence point of a titration.

The presence of a common ion suppresses the ioni zation of a weak acid or weak base. This is known as the common ion effect.

o

A common ion is added to a solution in the form of a salt.

Section 17.2 o

o

o

o

o

Section 17.4

A solution that contains significant concentrations of both members of a conjugate acid-base pair (weak acid -conjugate base or weak base - conjugate acid) is a buffer solution or simply a buffer.

o

Buffer solutions resist pH change upon addition of small amounts of strong acid or strong base. Buffers are important to biological systems .

o

The pH of a buffer can be calculated using an equilibrium table or with the Henderson-Hasselbalch equation.

o

The pKa of a weak acid is - log Ka. When the weak acid and conjugate base concentrations in a buffer solution are equal, pH = pKa.

o

o

o

o

o

The solubility product constant (Ksp) is the equilibrium constant that indicates to what extent a slightly soluble ionic compound dissociates m water.

Ksp can be used to determine molar solubility or solubility in gIL, and • VIce versa. Ksp can also be used to predict whether or not a precipitate will form when two solutions are mixed.

Section 17.5

We can prepare a buffer with a specific pH by choosing a weak acid with a pKa close to the desired pH.

o

Solubility is affected by common ions, pH, and complex ion formation. Theformation constant (Kr) indicates to what extent complex ions form.

o

A salt that dissoc iates to give a strong conjugate base such as fluoride ion will be more soluble in acidic solution than in pure water.

o

A salt that dis sociates to give hydroxide ion will be more soluble at lower pH and less soluble at higher pH.

o

The solubility of an ionic compound increases when the formation of a complex ion consumes one of the products of dissociation.

Section 17.3 o

The indicator used for a particular titration should exhibit a color change in the pH range corresponding to the steep region of the titration curve.

The titration curve of a strong acid - strong base titration has a long, steep region near the equivalence point. Titration curves for weak acid-strong base or weak base-strong acid titrations have a significantly shorter steep region. The pH at the equivalence point of a strong acid-strong base titration is 7.00.

Section 17.6

The pH at the equivalence point of a weak acid-strong base titration is above 7.00. The pH at the equivalence point of a weak base - strong acid titration is below 7.00.

o

Ions can be separated using fractional precipitation.

o

Fractional precipitation schemes can be designed based on K,p values.

o

Acid-base indicators are usually weak organic acids that exhibit two different colors depending on the pH of the solution. The end point of

Groups of cations can be identified through the use of selective precipitation. This is the basis of qualitative analysis.

KEyWORDS Buffer, 683

End point, 698

Common ion effect, 683

Formation constant (Kr), 710

Complex ion, 710

Fractional precipitation , 713

Henderson-Hasselbalch equation, 685 Molar solubility, 701

KEY EQUATIONS +

[A - ] 10" = ...,..-:b [HA]

17. I

PH

17.2

pKa = -log Ka

17.3

=

pKa

pH = pKa +

[conjugate base] log [weak acid]

Qualitative analysis, 714 Solubility, 701 Solubility product constant (Ksp) , 70 1

718

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

QUESTIONS AND PROBLEMS Section 17.1: The Common Ion Effect

17.17

Calculate the pH of the 0.20 M NHi O.20 M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCI to 65.0 mL of the buffer?

17.18

Calculate the pH of 1.00 L of the buffer 1.00 M CH3COONa/l.00 M CH3COOH before and after the addition of (a) 0.080 mol NaOH and (b) 0.12 mol HC!. (Assume that there is no change in volume.)

17.19

A diprotic acid, H 2 A, has the following ionization constants: Ka = 1.1 X 10- 3 and Ka = 2.5 X 10- 6. To make up a buffer 1 2 solution of pH 5.80, which combination would you choose: NaHA!H 2A or Na2A1NaHA?

17.20

A student is asked to prepare a buffer solution at pH 8.60, using one of the following weak acids: HA (Ka = 2.7 X 10- 3), HB (Ka = 4.4 X 10- 6), HC (Ka = 2.6 X 10- 9). Which acid should the student choose? Why?

17.21

The following diagrams contain one or more of the compounds: HzA, NaHA, and NazA, where HzA is a weak diprotic acid. (1 ) Which of the solutions can act as buffer solutions? (2) \Vhich solution is the most effective buffer solution? Water molecules and Na + ions have been omitted for clarity.

Review Questions 17.1

Use Le Chatelier's principle to explain how the common ion effect affects the pH of a weak acid solution.

17.2

Describe the effect on pH Gincrease, decrease, or no change) that results from each of the following additions: (a) potassium acetate to an acetic acid solution, (b) ammonium nitrate to an ammonia solution, (c) sodium formate (HCOONa) to a fonnic acid (HCOOH) solution, (d) potassium chloride to a hydrochloric acid solution, (e) barium iodide to a hydroiodic acid solution.

17.3

Define pKa for a weak acid. What is the relationship between the value of the pKa and the strength of the acid?

17.4

The pK;s of two monoprotic acids HA and HE are 5.9 and 8.1, respectively. Which of the two is the stronger acid?

Problems

17.5

Determine the pH of (a) a 0.40 M CH 3COOH solution and (b) a solution that is 0.40 M CH 3COOH and 0.20 M CH3COONa.

17.6

Determine the pH of (a) a 0.20 M NH3 solution, and (b) a solution that is 0.20 M NH3 and 0.30 M NH4Cl.

= HA-

Section 17.2: Buffer Solutions

Review Questions 17.7

17.8

What is a buffer solution? What must a solution contain in order to be a buffer? Using only a pH meter, water, and a graduated cylinder, how would you distinguish between an acid solution and a buffer solution at the same pH?

17.22

Problems

17.9

17.10

Which of the following solutions can act as a buffer: (a) KCVHCI, (b) KHS0 41H 2S0 4 , (c) Na2HP04INaHzP04, (d) KN0 2IHN02 ?

-

(c)

(b)

(a)

(d)

The following diagrams represent solutions containing a weak acid HA (pKa = 5.0) and its sodium salt NaA. (I).Which solution has the lowest pH? Which has the highest pH? (2) How many different species are present after the addition of two H + ions to solution (a)? (3) How many different species are present after the addition of two OH- ions to solution (b)? =HA

= A-

Which of the following solutions can act as a buffer: (a) KCN/HCN, (b) Na2S041NaHS04' (c) NHi NH4 N0 3 , (d) NaIIHI? •

17.11

17.12

17.13

Calculate the pH of the buffer system made up of 0.15 M NH3/0.35 M NH4 Cl. Calculate the pH of the following two buffer solutions: (a) 2.0 M CH3COONa/2.0 M CH3COOH, (b) 0.20 M CH 3COONa/0.20 M CH 3COOH. Which is the more effective buffer? Why? The pH of a bicarbonate-carbonic acid buffer is 8.00. Calculate the ratio of the concentration of carbonic acid (H2C0 3) to that of the bicarbonate ion (HC03").

17.14

What is the pH of the buffer 0.10 M Na2HPOJ O.15 M KH ZP04 ?

17.15

The pH of a sodium acetate-acetic acid buffer is 4.50. Calculate the ratio [CH3COO- ]I[CH3COOH].

17.16

The pH of blood plasma is 7.40. Assuming the principal buffer system is HC0 3"1H2C0 3 , calculate the ratio [HC03"]/[H2 C0 3]. Is this buffer more effective against an added acid or an added base?

(a)

(b)

(c)

(d)

Section 17.3: Acid-Base Titrations

Review Questions 17.23

Briefly describe what happens in an acid-base titration.

17.24

Sketch titration curves for the following acid-base titrations: (a) HCI versus NaOH, (b) HCI versus CH 3NH2, (c) CH 3COOH versus NaOH. In each case, the base is added to the acid in an Erlenmeyer flask. Your graphs should show the pH on the y axis and the volume of base added on the x axis.

17.25

Explain how an acid-base indicator works in a titration. What are the criteria for choosing an indicator for a particular acid-base titration?

17.26

The amount of indicator used in an acid-base titration must be small. Why?

•

719

QUESTIONS AND PROBLEMS

Problems

17.27

A 0.2688-g sample of a monoprotic acid neutralizes 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid.

17.28

A S.OO-g quantity of a diprotic acid was dissolved in water and made up to exactly 2S0 mL. Calculate the molar mass of the acid if 2S.0 mL of this solution required 11.1 mL of 1.00 M KOH for neutralization. Assume that both protons of the acid were titrated.

17.29

17.30

17.31

In a titration experiment, 20.4 mL of 0.883 M HCOOH neutralizes 19.3 mL of Ba(OH)z. What is the concentration of the Ba(OHh solution? A 0.1276-g sample of an unknown.monoprotic acid was dissolved in 2S.0 mL of water and titrated with a 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be S.87. What is the Ka of the unknown acid? A solution is made by mixing exactly SOO mL of 0.167 M NaOH with exactly SOO mL of 0.100 M CH 3COOH. Calculate the equilibrium concentrations ofH+, CH3COOH, CH 3COO- , OH-, and Na +.

17.33

Calculate the pH at the equivalence point for the following titration: 0.20 M HCI versus 0.20 M methylamine (CH3NH 2).

17.35

17.36

17.37

17.38

17.39

17.40

17.41

=B

• •

In a titration experiment, 12.S mL of O.SOO M H 2S0 4 neutrali zes SO.O mL of NaOH. What is the concentration of the NaOH solution?

17.32

17.34

the solution that corresponds to (1) the initial stage before the addition of HCl, (2) halfway to the equivalence point, (3) the equivalence point, (4) beyond the equivalence point. Is the pH greater than, less than, or equal to 7 at the equivalence point? Water and Cl- ions have been omitted for clarity.

"'"

(c)

(d)

The following diagrams represent solutions at various stages in the titration of a weak acid HA with NaOH. Identify the solution that corresponds to (1) the initial stage before the addition of NaOH, (2) halfway to the equivalence point, (3) the equivalence point, (4) beyond the equivalence point. Is the pH greater than, less than, or equal to 7 at the equivalence point? Water and Na + ions have been omitted for clarity. =

=HA

(a)

OH-

(b)

(c)

(d)

Review Questions

A 2S.0-mL solution of 0.100 M CH 3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) S.O mL, (c) 10.0 mL, (d) 12.S mL, (e) IS.0 mL. A 10.0-mL solution of 0.300 M NH3 is titrated with a 0.100 M HCI solution. Calculate the pH after the following additions of the HCl solution: (a) 0.0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 30.0 mL, (e) 40.0 mL. Referring to Table 17.3, specify which indicator or indicators you would use for the following titrations: (a) HCOOH versus NaOH, (b) HCl versus KOH, (c) HN0 3 versus CH 3NH2 . A student carried out an acid-base titration by adding NaOH solution from a buret to an Erlenmeyer flask containing an HCl solution and using phenolphthalein as the indicator. At the equivalence point, she observed a faint reddish-pink COIOf. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened? 6

The ionization constant Ka of an indicator HIn is 1.0 X 10- The color of the nonionized form is red and that of the ionized form is yellow. What is the color of this indicator in a solution whose pH is 4.00? The Ka of a certain indicator is 2.0 X 10- 6 The color of HIn is green and that of In- is red. A few drops of the indicator are added to an HCI solution, which is then titrated against an NaOH solution. At what pH will the indicator change color? The following diagrams represent solutions at various stages in the titration of a weak base B (such as NH 3) with HC!. Identify

/

17.42

(b)

Section 17.4: Solubility Equilibria

Calculate the pH at the equivalence point for the following titration: 0.10 M HCOOH versus 0.10 M NaOH.

/

(a)

•

17.43

Use BaS04 to distinguish between the terms solubility and solubility product.

17.44

Why do we usually not quote the K sp values for soluble ionic compounds?

17.4S

Write balanced equations and solubilIty product expressions for the solubility equilibria of the following compounds: (a) CuBr, (b) ZnC 2 0 4, (c) Ag2Cr04, (d) Hg 2C12, (e) AuCI 3 , (f) Mn3(P04h.

17.46

Write the solubility product expression for the ionic compound AxBy.

17.47

How can we predict whether a precipitate will form when two solutions are mixed?

17.48

Silver chloride has a larger K sp than silver carbonate (see Table 17.4). Does this mean that AgCI also has a larger molar solubility than Ag2 C0 3?

Problems

17.49

Calculate the concentration of ions in the following saturated solutions: (a) W] inAgI sol~tion with [Ag+] = 9.1 X 10- 9 M, (b) [A]3+ ] in Al(OH)3 solution with [OH-] = 2.9 X 10- 9 M.

17 .SO

From the solubility data given, calculate the solubility products for the following compounds: (a) SrF2, 7.3 X 10- 2 gIL, (b) Ag3P0 4, 6.7 X 10- 3 gIL.

17.51

The molar solubility of MnC0 3 is 4.2 X 10- 6 M. What is this compound?

17.S2

The solubility of an ionic compound MX (molar mass = 346 g) is 4.63 X 10- 3 gIL. What is K sp for this compound?

K sp

for

720

CHAPTER 17

Acid-Base Equilibria and Solubility Equilibria

The solubility of an ionic compound M 2X 3 (molar mass = 288 g) is 3.6 X 10- 17 gIL. What is Ksp for this compound?

17.71

Calculate whether or not a precipitate will form if 2.00 mL of 0.60 M NH3 is added to 1.0 L of 1.0 X 10- 3 M FeS04'

17.54

Using data from Table 17.4, calculate the molar solubility ofCaFl .

17.72

17.55

What is the pH of a saturated zinc hydroxide solution?

If 2.50 g of CUS04 are dissolved in 9.0 X 102 mL of 0.30 M NH 3 , what are the concentrations of Cu2+, Cu(NH3)~ + ' and NH3 at eq uilibrium?

17.56

The pH of a saturated solution of a metal hydroxide MOH is 9.68. Calculate the Ksp for this compound.

17.73

17.57

If 20.0 mL of 0.10 MBa(N0 3)l is added to 50.0 mL ofO.IOM Na2C03, will BaC03 precl.pitate?

Calculate the concentrations of Cd 2 +, Cd(CN)~- , and CN- at equilibrium when 0.50 g of Cd(N0 3)2 dissolves in 5.0 X 102 mL of 0.50 M NaCN.

17.74

A volume of75 mL of 0.060 MNaF is mixed with 25 mL of 0.15 M Sr(N03)2' Calculate the concentrations in the final solution of NO) , Na+, S?+, and F - . (Ksp for SrF2 = 2.0 X 10- 10 .)

If NaOH is added to 0.010 M AlH, which will be the predominant species at equilibrium: AI(OH)3 or AI(OH)4 ? The 33 pH of the solution is 14.00. [Kf for Al(OH)4 = 2.0 X 10 ]

17.75

Calculate the molar solubility of AgI in a 1.0 M NH3 solution.

17.76

Both Ag + and Zn2+ form complex ions with NH3. Write balanced equations for the reactions. However, Zn(OHh is soluble in 6 M NaOH, and AgOH is not. Explain.

17.77

Explain, with balanced ionic equations, why (a) CuI2 dissolves in ammonia solution, (b) AgBl' dissolves in NaCN solution, and (c) HgCll dissolves in KCl solution.

17.53

•

17.58

Section 17.5: Factors Affecting Solubility

Review Questions 17.59

How does the corrunon ion effect influence solubility equilibria? Use Le Chiltelier's principle to explain the decrease in solubility of CaC0 3 in an Nal C0 3 solution.

17.60

The molar solubility of AgCl in 6.5 X 10- 3 M AgN0 3 is 2.5 X 10- 8 M. In deriving Ksp from these data, which of the following assumptions are reasonable? (a) Ksp is the same as solubility. (b) Ksp of ~gtl is }he same in 6.5 X 10- 3 M AgN0 3 as in pure water. (c) Solubility of AgCl is independent of the concentration of AgN0 3. (d) [Ag +] in solution does not change significantly upon the addition of AgCl to 6.5 X 10- 3 M AgN0 3. (e) [Ag +] in solution after the addition of AgCl to 6.5 X 10- 3 M AgN0 3 is the same as it would be in pure water.

17.61

Give an example to illustrate the general effect of complex ion formation on solubility.

Section 17.6: Separation of Ions Using Differences in Solubility

Review Questions 17.78

Outline the general procedure of qualitative analysis.

17.79

Give two examples of metal ions in each group (1 through 5) in the qualitative analysis scheme.

Problems 17.80

Solid NaI is slowly added to a solution that is 0.010 M in Cu + and 0.010 M in Ag +. (a) Which compound will begin to precipitate first? (b) Calculate [Ag +] when CuI just begins to precipitate. (c) What percent of Ag + remains in solution at this point?

17.81

Find the approximate pH range suitable for the separation of Fe H and Zn2+ ions by precipitation of Fe(OH)3 from a solution that is initially 0.010 M in both Fe H and Zn H

17.82

In a group I analysis, a student obtained a precipitate containing both AgCI and PbCI 2 . Suggest one reagent that would enable the student to separate AgCl(s) from PbCI 2 (s).

17.83

In a group 1 analysis, a student adds HCI acid to the unknown solution to make [Cl ~ ] = 0.15 M. Some PbCl 2 precipitates. Calculate the concentration of Pb 2 + remaining in solution.

17.84

Both KCl and NH4Cl are white solids. Suggest one reagent that would enable you to distinguish between these two compounds.

17.85

Describe a simple test that would allow you to distinguish between AgN0 3 (s) and Cu(N03lz(s) .

Problems 17.62

17.63

How many grams of CaC0 3 will dissolve in 3.0 X 102 mL of 0.050 M Ca(N0 3 h? 6 The solubility product ofPbBr2 is 8.9 X 10- . Determine the , molar solubility in (a) pure water, (b) 0.20 M KEr solution, and (c) 0.20 M Pb(N0 3)2 solution.

17.64

Calculate the molar solubility of AgCl in a 1.00-L solution containing 10.0 g of dissolved CaCl l .

17.65

Calculate the molar solubility of BaS04 in (a) water and (b) a solution containing 1.0 M SO~- ions.

17.66

Which of the following ionic compounds will be more soluble in acid solution than in water: (a) BaS04, (b) PbCl l , (c) Fe(OHh (d) CaC0 3?

17.67

Which of the following will be more soluble in acid solution than in pure water: (a) CuI, (b) Ag l S0 4 , (c) Zn(OH)l, (d) BaC 20 4, (e) Ca3(P04)2?

Additional Problems 17.68

Compare the molar solubility of Mg(OH)l in water and in a solution buffered at a pH of 9.0.

17.69

Calculate the molar solubility of Fe(OH)2 in a solution buffered at (a) a pH of 8.00 and (b) a pH of 10.00.

17.70

17.86

The buffer range is defined by the equation pH = pKa + 1. Calculate the range of the ratio [conjugate base]/ [acid] that corresponds to this equation.

17.87

The pKa of the indicator methyl orange is 3.46. Over what pH range does this indicator change from 90 percent HIn to 90 percent In- ?

17.88

Sketch the titration curve of a weak acid with a strong base like the one shown in Figure 17.4. On your graph, indicate the volume of base used at the equivalence point and also at the half-equivalence

11

The solubility product of Mg(OH)l is 1.2 X 10- • What minimum OH- concentration must be attained (for example, by adding NaOH) to decrease the Mg concentration in a solution of Mg(N0 3)2 to less than 1.0 X 10- 10 M?

-:2~

QUESTIONS AND PROBLEMS

point, that is, the point at which half of the acid has been neutralized. Show how you can measure the pH of the solution at the half-equivalence point. Using Equation 17.3, explain how you can determine the pKa of the acid by thi s procedure.

17.89

17.90

A 200-mL volume of NaOH solution was added to 400 mL of a 2.00 M HNO z solution. The pH of the mixed solution was 1.50 units greater than that of the original acid solution. Calculate the molarity of the NaOH solution.

plotted the mass of the precipitate versus the volume of the KJ solution added and obtained the following graph. Explain the shape of the graph. v ,-------------------------, 0), even for substances in their standard states. (Remember that the standard enthalpy of formation, D..H'f, for elements in their standard states is arbitrarily set to zero, and for compounds it may be either positive or negative [ ~~ Secti on 5 .6] .) Referring to Table 18.2, we can identify several important trends:

•••••••••••••••••••.•••• • ••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •

• The standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in either the liquid or solid phase (examples are water, bromine, and iodine). • For two different substances in the same phase, the substance with the more complex structure has the higher entropy (for example, methane versus ethane). • When there are allotropic forms of an element, the more highly ordered form has the smaller entropy (for example, diamond versus graphite). • For monatomic species, the heavier substance has the greater entropy (for example, helium versus neon).

Entropy Changes in a System As we said earlier, entropy often is described as a measure of randomness or disorder. Although this can be a useful description, it should be used with caution and not taken too literally. It is generally preferable to view the change in entropy of a system in terms of the change in the number of microstates of the system. Nevertheless, we can use the concept of disorder to make some qualitative predictions about the entropy changes that accompany certain processes. Consider the processes represented in Figure 18.2 that lead to an increase in entropy of the system:

Figure JB.2(a). In a solid, the particles (atoms, molecules, or ions) are confined to fixed positions [ ~~

Figure 18.2{a) Sliquid

>

>

Melting:

S solid '

Figure 18.2{b) svapor

Section 12.3] and the number of microstates is small. Upon melting, the particles can

Sliquid.

Vaporization:

---

SECTION 18.2

Figure 18.2(c) S aqueous

>

S pure

Entropy

73

Dissolving:

o

•

Figure 18.2(d)

Heating:

Shigher temperature> Slower temperature '

I I

+

Figure 18.2(e) of gaseous moles.

+

occupy many more positions as they move away from the lattice points. Consequently, the number of microstates increases because there are now many more ways to arrange the particles. • disorder" phase transition results in an increase in entropy because Therefore, this "order the number of microstates has increased. Figure JB.2(b). Vaporization will also lead to an increase in the entropy of the system. The increase will be considerably greater than that for melting, however, because molecules in the gas pbase occupy much more space and therefore there are far more microstates than in the liquid phase. Figure JB.2( c). The solution process usually leads to an increase in entropy. When a sugar crystal dissolves in water, the highly ordered structure of the solid and patt of the ordered structure of the water are disrupted. Thus, the number of microstates in the solution is greater than the number in tbe pure solute and the number in the pure solvent combined. When an ionic solid such as NaCI

•

Increase in number

,

,

I· I

732

CHAPTER 18

Entropy, Free Energy, and Equilibrium

Figure 18.3

(a) Vibrational motion in a water molecule. The atoms are displaced, as shown by the arrows, and then reverse their directions to complete a cycle of vibration. (b) Rotational motion of a water molecule about an axis through the oxygen atom. The molecule can also vibrate and rotate in other ways.

/

\

(a) Vibrations

/

/

\

/

•

/

(b)

Rotation

dissolves in water, there are two contributions to the entropy increase namely, the solution process (the mixing of solute with solvent) and the dissociation of the compound into ions: NaCI(s)

H 20.

Na +(aq)

+ CI - (aq)

A greater number of particles leads to a greater number of microstates. However, we must also consider hydration, which actually causes water molecules to become more ordered around the ions. This part of the process decreases entropy because it reduces the number of microstates of the solvent molecules. (For small, highly charged ions such as AI3+ and Fe3+, the decrease in entropy due to hydration can outweigh the increase in entropy due to mixing and dissociation, so the entropy change for the overall process can actually be negative.) Figure JB.2 (d). Heating always increases the entropy of a system. The molecules of a system possess several types of energy, including translational (the motion through space of the whole molecule), rotational (rotation about a specific axis), and vibrational (alternating shortening and lengthening of bonds within the molecule). The rotations and vibrations of a water molecule are shown in Figure 18.3. As the temperature is increased, the energies associated with all types of molecular motion increase. This increase in the number of possible energies associated with each molecule is an increase in the number of microstates-and therefore an increase in the entropy. Figure JB.2(e). A reaction that results in an increase in the number of moles of gas always increases the entropy of a system. Sample Problem 18.1 lets you practice the qualitative prediction of entropy change for a process. •

Sample Problem 18.1 For each process, determine the sign of I1S for the system: (a) decomposition of CaC0 3 (s) to give CaO(s) and CO 2(g), (b) heating brorrtine vapor from 45°C to 80°C, (c) condensation of water vapor on a cold surface, (d) reaction of NH 3 (g) and HCl(g) to give NH4 Cl(s), and (e) dissolution of sugar in water. . Think About It Although it

is the change in the number of microstates that determines the change in entropy, you can visualize changes in entropy in terms of ordeJ: If a process involves matter becoming less ordered, the sign of I1S is positive. If a process involves matter becorrting more ordered, I1S is negative.

Strategy Consider the change in number of microstates (the number of possible positions that

each particle can occupy) in each case. An increase in the number of microstates corresponds to an increase in entropy and therefore a positive I1S. Setup Increases in entropy generally accompany solid-to-liquid, liquid-lo-gas, and solid-to-gas

transitions; the dissolving of one substance in another; a temperature increase; and reactions that increase the net number of moles of gas. Solution I1S is (a) positi ve, (b) positive, (c) negative, (d) negative, and (e) positive.

SECTION 18.3

The Second and Third Laws of Thermodynamics

73~

Practice Problem For each of the following processes, determine the sign of t:.S: (a) crystallization of sucrose from a supersaturated solution, (b) cooling water vapor from ISO°C to 110°C, (c) sublimation of dry ice.

Checkpoint 18.2 18.2.1

Entropy

For which of the following physical processes is t:.S negative? (Select all that apply.)

18.2.2

For which of the following chemical reactions is t:.S negative? (Select all that apply.)

a) Freezing ethanol

a) 20 3 (g)

b) Evaporating water

b) 4Fe(s)

c) Mixing carbon tetrachloride and benzene

c) 2H2 0 2 (aq)

d) Heating water

d) 2Li(s)

e) Condensing bromine vapor

e) 2NH 3(g)

• 30 2 (g)

+ 302 (g)

•

2F~03(s)

• 2H20(l) + 0 2(g)

+ 2H 20(l)

• 2LiOH(aq)

• Nig)

+ Hig)

+ 3H2 (g)

The Second and Third Laws of Thermodynamics Recall that the universe is made up of two parts: the system and the surroundings [ ~~ Section 5.1] . The system typically is the part of the universe we are investigating (e.g., the reactants and products in a chemical reaction). The surroundings are everything else. Both the system and the surroundings undergo a change in entropy in a chemical or physical process. The second law of thermodynamics relates the entropy change of the universe overall to the type of process. In this context, there are two types of processes: spontaneous processes, which we defined in Section . . . . . . . .. ... . . ... . . . .. ...,.... ...... .... ... ...... . 18.1, and equilibrium processes. An equilibrium process is one that can be made to occur by the addition or removal of energy-but that does not happen on its own. An example of an equilibrium process is the melting of ice at O°e. (Remember that at ODC, the normal melting point, ice and liquid water are in equilibrium with each other [ ~~ Section 12.6] .) According to the second law of thermodynamics, the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Mathematically, the second law of thermodynamics is expressed as follows: For a spontaneous process:

Equation 18.5

For an equilibrium process:

Equation 18.6

The term equilibrium process is different from the term chemical equilibrium. An equilibrium process is one that we cause to occur by adding or removing energy from a system that is at equilibrium.

For a spontaneous process, the second law says that the entropy change for the universe is greater than zero (Equation 18.5). Because ilSuniv consists of two tenlls, ilSsys and ilSsum it is possible for the entropy change for the system (or the surroundings) to be negative just as long as the sum of the two terms is positive. In the case of an equilibrium process, such as a phase change, the terms ilS sys and ilSsurr must be equal in magnitude and opposite in sign.

Entropy Changes in the System Consider a system represented by the following equation: aA

+ bB - _. cC + dD

Just as the enthalpy change of a reaction is the difference between the en thai pies of the products and reactants (Equation 5.12), the entropy change is the difference between the entropies of the products and reactants: Equation 18.7 Or, using 2, to represent summation and In and n to represent the stoichiometric coefficients of the reactants and products, respectively, Equation 18.7 can be generalized as follows: Equation 18.8

'

734

CHAPTER 18

Entropy, Free Energy, and Equilibrium

The reaction generally is the system. Therefore, t.S~n is t.S~ys .

The standard entropy values of a large number of substances have been measured in 11K· mol. tiie ' s't'lliidarci"entrcipy ' 'li' reaction (ilS~n)' we look up the standard entropies of the products and reactants and use Equation 18.7. Sample Problem 18.2 demonstrates this approach.

...... To ·caiCi.iiate·

di.ange·for

.

.

SampleProble l11.J8.2 '.' . From the standard entropy values in Appendix 2, calculate the standard entropy changes for the following reactions at 25°C:

• •

Recall that here per mole means per mole of reaction as written [ H~ Section 5 .6].

•

•

••

•

•

• • • •

(b) N 2 (g)

·

(c) H 2(g)

•

•

·

Think About It The results

are consistent with the fact that production of gases causes an increase in entropy. In part (a), the increase in moles of gas gives a large positive value for LlS:'xn' In part (b), the decrease in moles of gas gives a large negative value for LlS:'x n- In part (c), where there is no change in the number of moles of gas in the reaction, LlS ~xn is positive but is also fairly small. In general, in cases where there is no net change in the number of moles of gas in a reaction, we cannot predict whether LlS :'x n will be positive or negative-but we can predict that it will be a relatively small number. This makes sense given that gases invariably have greater entropy than liquids and solids.

(a) CaC0 3 (s)

:

• CaO(s)

+ 3H2(g) + CI 2 (g)

+ CO 2 (g)

• 2NH 3(g) • 2HCl(g)

Strategy Look up standard entropy values and use Equation 18.7 to calculate LlS~xn ' Just as we did

:· . . . . . when . . . . . . . .we . . . .calculated . . . . . . . . . . . . .standard . . . . . . . . . . .enthalpies . . . . . . . . . . ' .' .of . . .reaction, we consider stoichiometric coefficients to be dimensionless-giving LlS:'x n units of J/K . mol. Setup From Appendix 2, SO[CaC0 3 (s)) = 92.9 J/K . mol, SO[CaO(s)] = 39.8 JIK . mol,

SO[COzeg) ] = 213.6 JIK . mol, SO[N 2(g)] = 191.5 J/K . mol, SO[H2(g) ] = 131.0 J/K . mol, SO[NH3(g) ] = 193.0 JIK . mol, SO[CI 2(g)] = 223.0 JIK . mol, and SO[HCl(g)] = 187.0 JIK . mol. Solution

(a) M:'xn = [SO(CaO)

+ SO(C0 2)]

= [(39.8 JIK . mol)

-

[SO(CaC0 3) ]

+ (213.6 J/K

. mol)] - (92.9 J/K . mol)

= 160.5 JIK . mol

(b) M :'x n = [2S0(NH3) ]

-

[SO(N2)

+ 3S0(H2)]

= (2)( 193 .0 JIK . mol) - [(191.5 J/K . mol)

+ (3)( 131.0 J/K . mol)]

= - 198.5 JIK· mol

(c) M:'xn = [2S0(HCl)] - [SO(H2 )

+ SO(CI2)]

= (2)(187 .0 J/K . mol) - [(131. 0 J/K . mol)

+ (223. 0 JIK . mol)]

= 20.0 JIK . mol

• • • •

For reactions involving only liquids and solids, predicting the sign of t.S" is more difficult, but in many such cases an increase in the total number of molecules and/or ions is accompanied by an increase in entropy.

Practice Problem Calculate the standard entropy change for the following reactions at 25 °C. Predict

first whether each one will be positive, negative, or too close to call.

(a) 2C0zCg) - - . . 2CO(g) (b) 30 2(g)

+ 0 2(g)

• 20 3(g)

•

Entropy Changes in the Surroundings Next we discuss how ilS ~urr is calculated. When an exothermic process takes place in the system, the heat transferred to the surroundings increases the motion of the molecules in the surroundings. Consequently, there is an increase in the number of microstates and the entropy of the surroundings increases . Conversely, an endothermic process in the system absorbs heat from the surroundings and so decreases the entropy of the surroundings by slowing molecular motion. Remember that for constant-pressure processes, the heat released or absorbed, q, is equal to the enthalpy change of the system, ilHsys [ ~~ Section 5 .3] . The change in entropy for the surroundings, ilSsum is directly proportional to ilHsys:

The minus sign indicates that a negative enthalpy change in the system (an exothermic process) corresponds to a positive entropy change in the surroundings. For an endothermic process, the enthalpy change in the system is a positive number and corresponds to a negative entropy change in the surroundings.

SECTION 18.3

The Second and Third Laws of Thermodynamics

735

In addition to being directly proportional to b.Hsys, b.Ssurr is inversely propOltional to temperature:

Combining the two expressions gives Equation 18.9 With Equations 18.8 and 18.9, we can calculate the entropy change in both the system and surroundings for a chemical reaction, and we can determine whether the reaction is spontaneous. Consider the synthesis of ammonia at 25°C: b.H~xn

From Sample Problem 18.2(b), we have ( - 92.6 kJ/mol) into Equation 18.9, we get

b.S ~ys

= -92.6 kJ/mol

-198.5 J/K . mol, and substituting

b.H ~ys

-(-92.6 X 1000) J/mol _ . uSsurr 298 K - 311 J/K mol A

_

The entropy change for the universe is

= -199 J/K· =

mol + 311 J/K· mol

112 J/K . mol

Because b.S ~niv is positive, the reaction will be spontaneous at 25 °C. Keep in mind, though , that just because a reaction is spontaneous does not mean that it will occur at an observable rate. The synthesis of ammonia is, in fact, extremely slow at room temperature. Thermodynamics can tell us whether or not a reaction will occur spontaneously under specific conditions, but it does not tell us how fast it will occur.

Third Law of Thermodynamics Finally, we consider the third law of thermodynamics briefly in connection with the determination of standard entropy. So far we have related entropy to microstates the greater the number of microstates a system possesses, the larger is the entropy of the system. Consider a perfect crystalline substance at absolute zero (0 K). Under these conditions, there is essentially no molecular motion and the number of microstates (W) is 1 (there is only one way to arrange the atoms or molecules to form a perfect crystal). From Equation 18.2, we write S = kin W =klnl=O According to the third law of thermodynamics, the entropy of a perfect crystalline substance is zero at absolute zero. As temperature increases, molecular motion increases, causing an increase in the number of microstates. Thus, the entropy of any substance at any temperature above 0 K is greater than zero. If the crystalline substance is impure or imperfect in any way, then its entropy is greater than zero even at 0 K because without perfect order there is more than one microstate. The important point about the third law of thermodynamics is that it enables us to determine of a pure crysthe absolute entropies of substances. Starting with the knowledge that the entropy . . . .. ...... . . . . .. ., .... . . talline substance is zero at 0 K, we can measure the increase in entropy of the substance when it is heated. The change in entropy of a substance, b.S, is the difference between the final and initial entropy values: b.S

= Sf -

Si

where Si is zero if the substance starts at 0 K. Therefore, the measured change in entropy is equal to the absolute entropy at the new temperature.

f

Although the complete details of these measurements are beyond the scope of this book, entropy changes are determined in part by measuring the heat capacity of a substance [ ~~ Section 5.4] as a function of absolute temperature.

736

CHAPTER 18

Entropy, Free Energy, and Equilibrium

Figure 18.4 Entropy increases in a substance as temperature increases from absolute zero.

Boiling

•

(~Svap)

Gas

Liquid

Melting Solid

(~Sfus)

Temperature (K)

The entropy values arrived at in this way are called absolute entropies because they are true values-unlike standard enthalpies of formation, which are derived using an arbitrary reference. Because the tabulated values are determined at I atm, we usually refer to absolute entropies as standard entropies, So. Figure 18.4 shows the increase in entropy of a substance as temperature increases from absolute zero. At 0 K , it has a zero entropy value (assuming that it is a perfect crystalline substance). As it is heated, its entropy increases gradually at first because of greater molecular motion within the crystal. At the melting point, there is a large increase in entropy as the solid is transformed into the liquid. Further heating increases the entropy of the liquid again due to increased molecular motion. At the boiling point, there is a large increase in entropy as a result of the liquid-to-vapor transition. Beyond that temperature, the entropy of the gas continues to increase with increasing temperature.

Checkpoint 18.3 18.3.1

The Second and Third Laws of Thermodynamics

Using data from Appendix 2, calculate t:.so (in 11K · mol) for the following reaction: 2NO(g)

18.3.2

Using data from Appendix 2, calculate t:.so (in 11K· mol) for the following reaction:

+ 0 2(g) - - - +. 2NOig)

CH4 (g)

+ 20ig) ---+. COig) + 2H20(I)

a) 145.3 11K· mol

a) 107.7 11K· mol

b) -145.3IIK·mol

b) -107.7IIK· mol

c) 59.7 J/K . mol

c) 2.6 11K· mol

d) -59.7IIK· mol

d) 242.8 11K· mol

e) -421.2IIK· mol

e) -242.8 11K· mol

Gibbs Free Energy According to the second law of thermodynamics, D.Suniv > 0 for a spontaneous process. We are usually concerned with and usually measure, however, the properties of the system rather than those of the surroundings or those of the universe overall. Therefore, it is convenient to have a thermodynamic function that enables us to determine whether or not a process is spontaneous by considering the system alone. We begin with Equation 18.5. For a spontaneous process, D.Suniv

= D.Ssys + D.Ssurr > 0

SECTION 18.4

Gibbs Free Energy

737

Substituting - fUisy/T for ~SSUIT> we write ~Suniv = ~Ssys

+-

~Hsys

T

>

0

Multiplying both sides of the equation by T gives T~Suniv = T~Ssys - ~Hsys

>0

Now we have an equation that expresses the second law of thermodynamics (and predicts whether or not a process is spontaneous) in terms of only the system. We no longer need to consider the surroundings. For convenience, we can rearrange the preceding equation, multiply through by -1, and replace the> sign with a < sign:

- T~Suniv = ~Hsys - T~Ssys < 0 According to this equation, a process carried out at constant pressure and temperature is spontaneous if the changes in enthalpy and entropy of the system are such that fUi sys - T~Ssys is less than zero. To express the spontaneity of a,process more directly, we introduce another thermodynamic function called the Gibbs! free energy (G), or simply free energy. G

= H- TS

Equation 18.10

Each of the terms in Equation 18.10 pertains to the system. G has units of energy just as Hand TS do. Furthermore, like enthalpy and entropy, free energy is a state function. The change in free energy, ~G, of a system for a process that occurs at constant temperature is ~G= ~H- T~S

Equation 18.11

Equation 18.11 enables us to predict the spontaneity of a process using the change in enthalpy, the change in entropy, and the absolute temperature. At constant temperature and pressure, for processes that are spontaneous as written (in the forward direction), ~G is negative. For processes that are.......... spontaneous in the reverse direction, ~G is positive. For are not spontaneous as written but that .. .., ................................. ..........,.......................... In this context, free energy is the energy systems at equilibrium, ~G is zero.

•

~G
0

The reaction is nonspontaneous in the forward direction (and spontaneous in the reverse direction).

•

~G

=0

available to do work. Thus, if a particular process is accompanied by a release of usable energy (i.e., if LiG is negative), this fact alone guarantees that it is spontaneous, and there is no need to consider what happens to the rest of the universe .

The system is at equilibrium.

Often we can predict the sign of ~G for a process if we know the signs of ~H and ~S. Table 18.3 shows how we can use Equation 18.11 to make such predictions. Based on the information in Table 18.3, you may wonder what constitutes a " low" or a "high" temperature. For the example given in the table, O°C is the temperature that divides high from low. Water freezes spontaneously at temperatures below O°C, and ice melts spontaneously at temperatures above O°C. At O°C, a system of ice and water is at equilibrium. The temperature that

When

aH Is

And

as

Is

aG Will Be

And the Process Is

Example

Negative

Positive

Negative

Always spontaneous

2H 20 2 (aq) --+. 2H20(l)

Positive

Negative

Positive

Always nonspontaneous

30z(s)

•

+ °z(g)

• 20 3(g)

Negative

Negative

Negative when T~S < fUi Positive when T~S > ~H

Spontaneous at low T Nonspontaneous at high T

Hz0(l)

Positive

Positive

Negative when T~S > fUi Positive when T~S < fUi

Spontaneous at high T Nonspontaneous at low T

2HgO(s) --+. 2Hg(l) + 0 z(g)

1. Josiah Willard Gibbs (1839-1903). American physicist. One of the founders of thermodynamics. Gibbs was a modest

and private individual who spent almost all his professional life at Yale University. Because he published most of his work in obscure journals, Gibbs never gained the eminence that his contemporary and admirer James Maxwell did. Even today, very few people outside of chemistry and physics have ever heard of Gibbs.

• HzO(s) (freezing of water)

738

CHAPTER 18

Entropy, Free Energy, and Equilibrium

divides "high" from "low" depends, though, on the individual reaction. To determine that temperature, we must set t1G equal to 0 in Equation 18.11 (i.e., the equilibrium condition):

0= t1H - Tt1S Rearranging to solve for T yields

T= t1H t1S

•

The temperature that divides high from low for a particular reaction can now be calculated if the values of t1H and t1S are known. Sample Problem 18 .3 demonstrates the use of this approach .

Sample Problem 18.3 According to Table IS.3, a reaction will be spontaneous only at high temperatures if both tlH and tlS are positive. For a reaction in which tlH = 199.5 kJ/mol and tlS = 476 J/K . mol, determine the temperature (in 0c) above which the reaction is spontaneous. Think About It Spontaneity is

favored by a release of energy (tlH bei ng negative) and by an increase in entropy (tlS being positive). When both quantities are positive, as in thi s case, only the entropy change favors spontaneity. For an endothermic process such as this, which requires the input of heat, it should make sense that adding more heat by increasing the temperature will shift the equilibrium to the right, thus making it "more spontaneous."

Strategy The temperature that divides high from low is the temperature at which tlH = TtlS

(tlG = 0) . Therefore, we use Equation IS.11, substituting 0 for tlG and solving for T to determine temperature in kelvins; we then convert to degrees Celsius. Setup

tlS = ( 476 J ) ( 1 kJ ) = 0.476 kJ/K . mol mol, K 1000 J Solution

T = D.H = 199.5 kllmol = 419 K tlS 0.476 kllK . mol = (419 - 273) = 146°C

Practice Problem A reaction will be spontaneous only at low temperatures if both tlH and tlS are

negative. For a reaction in which tlH = - 3S0. 1 kJ/mol and tlS = -95.001IK· mol, determine the temperature (in 0c) below which the reaction is spontaneous .

Standard Free-Energy Changes The introduction of the term

~GO enables us to

write Equation 18.11 as

~GO = M ? - T/:,S

...... The"sia'ndardfree~eneiijY

'of'riiaction '(t1G~Xll )

is the free energy change for a reaction when it occurs under standard-state conditions-that is, when reactants in their standard states are converted to products in their standard states. The conventions used by chemists to define the standard states of pure substances and solutions are • Gases

1 atm pressure

• Liquids • Solids

Pure liquid

• Elements • Solutions

The most stable allotropic form at 1 atm and 25 °C

To calculate

t1G ~xn '

Pure solid 1 molar concentration we start with the general equation

aA

+ bB - _ . cC + dD

The standard free-energy change for this reaction is given by Equation 18.12

t1G ~xn =

[ct1G r(C)

+ dt1G r(D)) - [at1G r(A) + bt1G r(B)) •

Equation 18.12 can be generalized as follows: Equation 18.13

t1G ~xn

= lnt1G r (products)

- lmt1G r (reactants)

SECTION 18.4

•

Gibbs Free Energy

739

where m and n are stoichiometric coefficients. The term !1GJ is the standard free energy of formation of a compound that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its constituent elements, each in its standard state. For the combustion of graphite,

the standard free-energy change (from Equation 18.13) is !1G~xn =

[!1G HC0 2)] -

[!1G ~ (C,

graphite)

+ !1G ~ (02)]

As with standard enthalpy of formation, the standard free energy of formation of any element (in its most stable allotropic form at 1 atm) is defined as zero. Thus, !1GJ?(C, graphite) = 0

and

Therefore, the standard free-energy change for the reaction in this case is equal to the standard free energy of formation of CO 2: !1G ~xn =

!1Gf(C02 )

Appendix 2 lists the values of !1GJ? at 25 °C for a number of compounds. Sample Problem 18.4 demonstrates the calculation of standard free-energy changes.

Sample Problem 18.4 Calculate the standard free-energy changes for the following reactions at 25°C: (a) CH4 (g)

+ 20z(g)

(b) 2MgO(s)

• COz(g)

+ 2H 20 (I)

• 2Mg(s) + Oz(g)

Strategy Look up the !:J.G'f values for the reactants and products in each equation, and use Equation 18.13 to solve for !:J.G ~xn' Setup From Appendix 2, we have the following values: !:J.Gf[CH4 (g)] = -50.8 kllmol, !:J.Gf[COz(g)] = -394.4 kllmol, !:J.Gf[H20(l)] = -237.2 kJ/mol, and !:J.Gf[MgO(s)] = - 569.6 kllmo!. All the other substances are elements in their standard states and have, by definition, !:J.G'f = O. Solution (a) !:J.G ~n = (!:J.Gf[C0 2(g)] + 2!:J.Gf[H20(I)]) - (!:J.Gf[CHig)] + 2!:J.Gf[02(g)]) = [( -394.4 kllmol) + (2)( -237.2 kJ/mol)] - [( -50.8 kllmol ) + (2)(0 kllmol )] = - 818.0 kll mol

(b) !:J.G ~xn = (2!:J.G'f[Mg(s)] = [(2)(0 kllmol)

+ !:J.Gf[02(g)]) - (2!:J.Gf[MgO (s)]) + (0 kllmol)] - [(2)(-569.6 kJ/mol)]

= 1139 kJ/mol

Think About It Note that, like standard enthalpies of formation (DJIf), standard free energies of formation (!:J.G7) depend on the state of matter. Using water as an example, !:J.Gf[H 20(l)] = -237.2 kll mol and !:J.Gf[H 20(g)] = -228.6 kJ/mo!. Always double-check to make sure you have selected the right value from the table.

Practice Problem Calculate the standard free-energy changes for the following reactions at 25°C:

+ Br2(l) ---+. 2HBr(g) (b) 2C 2H 6 (g) + 70 2(g) • 4COzeg) + (a) H 2(g)

6H 20(I)

Using L1G and L1G to Solve Problems O

, It is the sign of !1G, the free-energy change, not the sign of !1Go, the standard free-energy change, . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

that indicates whether or not a process will occur spontaneously under a given set of conditions. What the sign of !1Go tells us is the same thing that the magnitude of the equilibrium constant (K) tells us [ ~~ Section 15.2] . A negative !1Go value corresponds to a large K value (products favored at equilibrium), whereas a positive !1Go value corresponds to a small K value (reactants favored at equilibrium). Like equilibrium constants, !1Go values change with temperature. One of the uses of Equation 18.11 is to determine the temperature at which a particular equilibrium will begin to favor a desired product. For example, calcium oxide (CaO), also called quicklime, is an extremely valuable inorganic substance with a variety of industrial uses , including water treatment and

The sign of !:J.GOdoes indicate whether or not a process is spontaneous when all reactants ana products are in their standard states, but this is very seldom the case.

740

CHAPTER 18

Entropy, Free Energy, and Equilibrium

pollution control. It is prepared by heating limestone (CaC0 3), which decomposes at a high temperature:

•

The reaction is reversible, and under the right conditions, CaO and CO 2 readily recombine to form CaC0 3 again. To prevent this from happening in the industrial preparation, the system is never maintained at equilibrium; rather, CO 2 is constantly removed as it forms, shifting the equilibrium from left to right, thus promoting the formation of calcium oxide. An important piece of information for the chemist responsible for maximizing CaO production is the temperature at which the decomposition equilibrium of CaC0 3 begins to favor products . We can make a reliable estimate of that temperature as follows. First we calculate b..Ho and b..So for the reaction at 2S oC, using the data in Appendix 2. To determine b..Ho, we apply Equation S.1 9:

b..W

=

[b..Ht'(CaO) + b..Ht'(C0 2 )]

= [(-63S.6 kJ/mol) + (= 177.8

-

[b..Ht'(CaC0 3 )]

393.S kJ/mol )] - (-1206.9 kJ/mol)

kJ/mol

Next we apply Equation 18.8 to find b..So:

b..So = [SO(CaO) + SOCO?)] - SO(CaC0 3)

= [(39.8 JIK . mol) + (213.6 J/K . mol)] - (92.9 J/K . mol) = 160.S J/K . mol From Equation 18.11, we can write

and we obtain . . .. ... . . . . . . . . . . . ... . . . . Be carefu l with units in problems of this type. 5" values are·tabulated using joules, whereas Ll.H\' values are tabulated using kilojoules.

... .. ... ....... ......... ... ... ...... . .... . b..Go = (177 .8 kJ/mol) - (298 K)(0.160S kJ/K . mol)

. . . ..

= 130.0 kJ/mol Because b..Go is a large positive number, the reaction does not favor product formation at 2S oC (298 K). And, because b..Ho and b..So are both positive, we know that b..Go will be negative (product formation will be favored) at high temperatures. We can determine what constitutes a high temperature for this reaction by calculating the temperature at which b..Go is zero.

or

(177 .8 kJ/mol)(1000 JlkJ) 0.160S kJ/K . mol •

= 1108 K (83S°C) At temperatures higher than 83S oC, b..Go becomes negative, indicating that the reaction would then favor the formation of CaO and CO? At 840°C (1113 K), for example,

b..Go = b..Ho - Tb..So = 177.8

kJ/mol - (1113 K)(0.160S kJ/K . mol)

( l~okg J

= -0,8 kJ/mol At still higher temperatures, b..Go becomes increasingly negative, thus favoring product formation even more. Note that in this example we used the b..Ho and b..So values at 2S oC to calculate changes to b..Go at much higher temperatures. Because both b..Ho and b..So actually change with temperature, this approach does not give us a truly accurate value for b..Go, but it does give us a reasonably good estimate. Equation 18.11 can also be used to calculate the change in entropy that accompanies a phase change. At the temperature at which a phase change occurs (i.e., the melting point or boiling point of a substance), the system is at equilibrium (b..G = 0). Therefore, Equation 18.11 becomes

SECTION 18.4

Gibbs Free Energy

0= tlH - TtlS

or tlS = tlH T Consider the ice-water equilibrium. For the ice-to-water transition, tlH is the molar heat of fusion (see Table 12.8) and T is the melting point. The entropy change is therefore tlSice --+. water

6010 Ilmol = 273 K = 22.0 IlK · mol

Thus, when 1 mole of ice melts at ODC, there is an increase in entropy of 22.0 IlK· mol. The increase in entropy is consistent with the increase in microstates from solid to liquid. Conversely, for the water-to-ice transition, the decrease in entropy is given by tlS •

water

•

. = -6010 Ilmol = -220 IlK · I Ice 273 K . mo

The same approach can be applied to the water-to-steam transition. In this case, tlH is the heat of vaporization and T is the boiling point of water. Sample Problem 18.5 examines the phase transitions in benzene.

Sample Problem 18.S The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol , respectively. Calculate the entropy changes for the solid-to-liquid and liquid-to-vapor transitions for benzene. At 1 atm press ure, benzene melts at 5.5°C and boils at 80. 1DC. Strategy The solid-liquid transition at the melting point and the liquid- vapor transition at the boiling

point are equilibrium processes. Therefore, because D.G is zero at equ ilibrium, in each case we can use Equation 18.11, substituting 0 for D.G and solving for D.S, to determine the entropy change associated with the process. Setup The melting point of benzene is 5.5

+ 273.15

= 278.7 K and the boiling point is 80.1

+

273.15 = 353.3 K. Solution

D.S

fu s

=

D.Hru s Tmelting

10.9 kJ/mol 278.7 K

= 0.0391 kJ/K . molar

D.Hvap

D.Syap = -

--'-

T boiling

Think About It For the same

39.1 J/K . mol

31.0 kJ/mol 353.3 K

= 0.0877 kJ/K . mol

or

87.7 JIK . mol

substance, D.Svap is always significantly larger than D.S fus ' The change in number of microstates is always bigger in a liquid-to-gas transition than in a solid-to-liquid transition.

Practice Problem The molar heats of fu sion and vaporization of argon are 1.3 and 6.3 kJ/mol ,

respectively, and argon's melting point and boiling poin t are -190°C and -186°C, respectively. Calculate the entropy changes for the fu sion and vaporization of argon .

Checkpoint 18.4

Gibbs Free Energy •

18.4.1

A reaction for which D.H and D.S are both negative is a) nonspontaneous at all temperatures b) spontaneous at all temperatures c) spontaneous at high temperatures d) spontaneo us at low temperatures e) at equilibrium

18.4.2

At what temperature (i n 0c) does a reaction go from being nonspontaneous to spontaneous if it has D.H = 17 1 kllmol and D.S = 161 J/K . mol? a) 270°C b) 670°C c) 1100°C d) 790°C e) 28°C

741

742

CHAPTER 18

Entropy, Free Energy, and Equilibrium 18.4.3

Using data from Appendix 2, calculate LlGo (in kJ/mol) at 25°C for the reaction:

CH4 (g) + 20 2 (g)

•

• CO 2(g)

18.4.4

Calculate LlSvap (in J/K . mol) for the vaporization of bromine: Br2(l) - _ . Br2(g)

+ 2H2 0(l)

a) -580.8 kJ/mol

LlHvap = 31 kJ/mol, and the boiling point of bromine is 59°C.

b) 580.8 kJ/mol

a) 9311K'mol

c) -572.0 kJ/mol

b) 0.53 11K· mol

d) 572.0 kJ/mol

c) 11 11K· mol

e) -818.0 kJ/mol

d) l.0 X 104 J/K . mol e) 2. 1 X 102 11K . mol

Free Energy and Chemical Equilibrium • • •

Even for a reaction that starts with all reactants and products in their standard states, as soon as the reaction begins, the concentrations of all species change and standard-state conditions no longer exist.

•

•

Reactants and products in a chemical reaction are almost always in something other than their standard states-that is, solutions usually have concentrations other than 1 M and gases usually have pressures other than 1 atm. To determine whether or not a reaction is spontaneous, therefore, we must take into account the actual concentrations and/or pressures of the species involved. And although we can determine t::..Go from tabulated values, we need to know t::..G to determine spontaneity.

Relationship Between LlG and LlGO The relationship between t::..G and t::..Go, which is derived from thermodynamics, is

t::..G

Equation 18.14

= t::..Go + RTln Q 3

The Q used in Equation 18.1 4 can be either Q, (for reactions that take place in solution) or Qp (for reactions that take place in the gas phase).

where R is the gas constant (8.314 11K· mol or 8.314 X 10- kllK . mol), T is the absolute temis the reaction quotient [ ~~ Secti on 15.2] . Thus, t::..G depends on two terms: t::..Go and RT In Q. For a given reaction at temperature T, the value of t::..Go is fixed but that of RT In Q can vary because Q varies according to the composition of the reaction mixture. Consider the following equilibrium:

..... ,pe'ratiiie'at whIch'the'reac"ii"on 'takes'pi ace:and'Q

Using Equation 18.13 and information from Appendix 2, we find that t::..Go for this reaction at 25 °C is 2.60 kllmol. The value of t::..G, however, depends on the pressures of all three gaseous species. If we start with a mixture of gases in which P H2 = 2.0 atm, PIz = 2.0 atm, and PHI = 3.0 atm, the reaction quotient, Qp, is (P HI )2

Qp

=

(PHz)(PIz )

9.0 4.0

(3.0)2 (2.0)(2.0)

= 2.25 Using this value in Equation 18.14 gives

t::..G = 2.60 kl

+

mol = 4.3

3

8.314 X 10- kl (298 K)(ln 2.25) K'mol

kllmol

Because t::..G is positive, we conclude that, starting with these concentrations, the forward reaction will not occur spontaneously as written. Instead, the reverse reaction will occur spontaneously and the system will reach equilibrium by consuming part of the HI initially present and producing more H 2 and I 2 . If, on the other hand, we start with a mixture of gases in which PH2 = 2.0 atm, PI2 = 2.0 atm, and PHI = 1.0 atm, the reaction quotient, Qp, is (P HI )2

Qp = (PHz)(PIz)

= 0.25

(1.0)2 -

(2.0)(2.0)

1 4

-

SECTION 18.5

Free Energy and Chemical Equilibrium

743

Using this value in Equation 18.14 gives AG

= 2.6 kJ + 8.314 mol

3

X 10- kJ (298 K)(ln 0.25)

K . mol

= - 0.8 kJ/mol

With a negative value for AG, the reaction will be spontaneous as written in the forward direction. In this case, the system will achieve equilibrium by consuming some of the H2 and 12 to produce more HI. Sample Problem 18.6 uses AGo and the reaction quotient to determine in which direction a reaction is spontaneous . •

Sample Problem 18.6 The equilibrium constant, Kp , for the reaction

is 0.1l3 at 298 K, which corresponds to a standard free-energy change of 5.4 kllmo!. In a certain experiment, the initial pressures are P N 0 = 0.453 atm and PNO = 0.122 atm. Calculate 6.G for the 2 4 2 reaction at these pressures, and predict the direction in which the reaction will proceed spontaneously to establish equilibrium. Strategy Use the partial pressures of N 2 0 4 and N02 to calculate the reaction quotient Qp, and then

use Equation 18.14 to calculate 6.G. Setup The reaction quotient expression is

Qp =

(PNO /

=

P N20 4

(0. 122l

0.453

= 0.0329

Solution

6.G = 6.Go

+ RT In Qp 3

kl + (8 .3 14 X 10- kJ (298 K)(In 0.0329) mol K· mol

= 5.4 = 5.4

kllmol - 8.46 kllmol

= -3 .1

Think About It Remember, a

kJ/mol

Because 6.G is negative, the reaction proceeds spontaneously from left to right to reach equilibrium.

-

reaction with a positive 6.Go value can be spontaneous if the starting concentrations of reactants and products are such that Q < K.

Practice Problem A 6.Go for the reaction

HzCg)

+ 12(g) :;:::,~. 2HI(g)

is 2.60 kllmol at 25°C. Calculate 6.G, and predict the direction in which the reaction is spontaneous if the starting concentrations are P H2 = 3.5 atm, P I2 = 1.5 atm, and PHI = 1.75 atm. Practice Problem B What is the minimum partial pressure of 12 required for the precedi ng reaction

to be spontaneous in the forward direction at 25°C if the partial pressures of H2 and HI are 3.5 and 1.75 atm, respectively?

•

Relationship Between By definition , AG

~Go

and K

= 0 and Q = K at equilibrium, where K is the equilibrium constant. Thus,

AG = AGO+ RTln Q (Equation 18.14) becomes

0 = AGO+ RTlnK

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . In this equation and the one that follows, K IS for reactions that take place in solution and Ko for reactions that take place in the gas phase.

or

AGO= - RTln K

Equation 18.15

_

744 •

CHAPTER 18

Entropy, Free Energy, and Equilibrium

, Multimedia

Chemical Equilibrium

.,.

"..

equilibrium .

Multimedia

Chemical Kinetics- interactive reaction coordinate diagram.

• The sign of 6 GO tells us the same thing that the magnitude of K tells us. The sign of 6G tells us the same thing as the comparison of Q and K values [ ~~ Section 15.4] .

(a) !:lGo < O. At equilibrium, there is a significant conversion of reactants to products. (b) !:lGo > O. At equilibrium, reactants are favored over products. In both cases, the net reaction toward equilibrium is from left to right (reactants to products) if Q < K and right to left (products to reactants) if Q > K. At equilibrium, Q = K.

According to Equation 18.15, then, the larger K is, the more negative !1Go is. For chemists, Equation 18.15 is one of the most important equations in thermodynamics because it enables us to find the equilibrium constant of a reaction if we know the change in standard free energy, and vice versa. It is significant that Equation 18.15 relates the equilibrium constant to the standard freeenergy change, !1Go, rather than to the actual free-energy change, !1G. The actual free-energy change of the system varies as the reaction progresses and becomes zero at equilibrium. On the other hand, !1Go, like K, is a constant for a particular reaction at a given temperature. Figure 18.5 shows plots of the free energy of a reacting system versus the extent of the reaction for two reactions. Table 18.4 summarizes the relationship between the magnitude of an equilibrium constant and the sign of the corresponding !1Go. Remember this important distinction: It is the sign of !1G and not that of !1Go that determines the direction of reaction spontaneity. The sign of !1Go only the..relative amounts products and reactants when equilibrium is reached, not the directells .us ... . . .. . .of .. .. .. tion the reaction must go in order to reach equilibrium.

Figure 18.5

GO(reactants) - , . -~ - - - - - - - - - - - -- - - - - -- - - - - -- - - - - - - - - --

t:.GOis negative

---- --------- --------------- ------- +- G°(products) I I I I I I I

Equilibrium position (products favored)

I

I

~~----------------v~---------------t~-----v~----_/

Q K 0 t:.G > 0

f-----------------Extent of reaction ----------------+-.1

(a)

------------------------------------t:.G Ois positive

GO(reactants) -.,. ~~- -- - - - - - - - - - - - - - - - - -

- - - - - - - - - - - --

I

I I I I I I

I I

Equilibrium position (reactants favored)

v---t

v---------------/

Q < K Q=K t:.G < 0 t:.G = 0

Q>K t:.G > 0

f----------------- Extent of reaction - - - - - - - - - - - - - 1

(b)

GO (products)

SECTION 18.5

Free Energy and Chemical Equilibrium

K

In K

>1

Positive

Negative

Products are favored.

o

o

Neither products nor reactants are favored.

Negative

Positive

Reactants are favored.

=

1



0

= D.Ssys + D.Ssurr = 0 + dSO(D)] - [aSO(A) + bSO(B)]

LnSO(products) - LmSO(reactants) -D.Hsys

18.9

D.Ssurr =

18.10

G = H- TS

18.11

D.G = D.H - TD.S

18.12

D.G;'.n = [cD.Gr(C)

18.13

D.G~xn =

18.14

D.G = D.Go + RTln

18.15

D.Go =

T

+ dD.G)?(D)] - [aD.G)?(A) + MG)?(B)]

LnD.G)?(products) - LmD.G)?(reactants)

-

RTln K

Q

QUESTIONS AND PROBLEMS

QUESTIONS AND PROBLEMS =========================-~~-

•

Section 18.1: Spontaneous Processes

(c) 2Na(s) (d) N 2(g)

Review Questions 18.1

Explain what is meant by a spontaneous process. Give two examples each of spontaneous and nonspontaneous processes.

18.2

Which of the following processes are spontaneous and which are nonspontaneous: (a) di ssolving table salt (NaCI) in hot soup, (b) climbing Mt. Everest, (c) spreading fragrance in a room by removing the cap from a perfume bottle, (d) separating helium and neon from a mixture of the gases?

18.3

Which of the following processes are spontaneous and which are nonspontaneous at a given temperature? 2

(a) NaN0 3 (s)

H 0. NaN0 3(aq)

saturated soln

(b) NaN0 3(s)

H20. NaN0 3(aq) 0 H2 • NaN0 3(aq)

unsaturated soln

(c) NaN0 3(s)

18.11

• 2N(g)

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. • PCls(s) (a) PCI 3 (l) + CI2 (g) (b) 2HgO(s) • 2Hg(l) + 02(g) (c) H 2(g) • 2H(g) (d) U(s) + 3F2 (g) • UF6(s)

Section 18.3: The Second and Third Laws of Thermodynamics

Review Questions 18.12

State the second law of thermodynamics in words, and expr -mathematically.

18.13

State the third law of thermodynamics in words, and expl ai n iusefulness in calculating entropy values.

supersaturated soln

Section 18.2: Entropy

+ 2H20(l) -_I 2NaOH(aq) + H 2(g)

j-

Review Questions Problems 18.4

Define entropy. What are the units of entropy? .

18.5

Define microstate. What is the relationship between microstates and entropy?

18.6

How does the entropy of a system change for each of the following processes? (a) A solid melts. (b) A liquid freezes. (c) A liquid boils. (d) A vapor is converted to a solid. (e) A vapor condenses to a liquid. (f) A solid sublimes. (g) A solid dissolves in water.

18.7

18.14

For each pair of substances listed here, choose the one having th~ larger standard entropy value at 25°C. The same molar amoun is used in the comparison. Explain the basis for your choice. (a) Li(s) or Li(l), (b) C 2H sOH(l) or CH 30CH 3 (l) (Hint: Which molecule can hydrogen-bond?), (c) Ar(g) or Xe(g), (d) CO(g ) or CO 2(g), (e) 02(g) or 03(g), (f) N0 2(g) or N 20ig).

18.15

Arrange the following substances (1 mole each) in order of increasing entropy at 2s oe: (a) Ne(g), (b) S02(g), (c) Na(s), (d) NaCl(s), (e) H 2(g) . Give the reasons for your arrangement.

18.16

Using the data in Appendix 2, calculate the standard entropy changes for the following reactions at 2ye: (a) S(rhombic) + 0 2(g) • S02(g) (b) MgC0 3 (s) • MgO(s) + COzeg)

How does the entropy of a system change for each of the following processes? (a) Bromine liquid vaporizes. (b) Water freezes to form ice. (c) Naphthalene, the key component of mothballs, sublimes . (d) Sugar crystals form from a supersaturated solution. (e) A block of lead melts. (f) Iodine vapor condenses to form solid iodine. (g) Carbon tetrachloride dissolves in liquid benzene.

(c) 2C 2H 6(g)

18.17

+ 70 2(g)

• 4C02(g)

+ 6H20 (I)

U sing the data in Appendix 2, calculate the standard entropy changes for the following reactions at 2YC: (a) H 2(g) + CuO(s) -_I Cu (s) + H 20(g) (b) 2Al(s) + 3ZnO(s) • AI2 0 3(s) + 3Zn(s) (c) CH4 (g)

18.18

Problems

+ 20zeg)

• COzeg)

+ 2H20(l)

According to the second law of thermodynamics, the entropy of an irreversible process in an isolated system must always increase. On the other hand, it is well known that the entropy of li ving systems remains small. (For example, the synthesis of highly complex protein molecules from individual amino acids is a process that leads to a decrease in entropy.) Is the second law invalid for livi ng systems? Explain.

18.8

Referring to the setup in Figure 18.1 , calculate the probability of all the molecules ending up in the same flask if the number is (a) 6, (b) 60, (c) 600.

18.9

Referring to the setup in Figure 18.1, calculate the probability of all the molecules ending up in the same flask if the number is (a) 4, (b) 10, (c) 200.

Section 18.4: Gibbs Free Energy

Predict whether the entropy change is positive or negative for each of the following reactions. Give reasons for your predictions.

18.19

Definefree energy. What are its units?

18.20

Why is it more convenient to predict the direction of a reaction in terms of I1Gsys instead of I1Suni v? Under what conditions can I1G sys be used to predict the spontaneity of a reaction?

18.10

(a) 2KCl0 4 (s) (b) H 20 (g)

-_I 2KCl0 3(s) + 0 2(g) • H 20 (l)

Review Questions

752

18.21

CHAPTER 18

Entropy, Free Energy, and Equilibrium

From the following combinations of !:::Ji and !::,.S, predict if a process will be spontaneous at a high or low temperature: (a) both !::,.H and !::,.S are negative, (b) !:::Ji is negative and !::,.S is positive, (c) both !::"H and !::,.S are positive, (d) !::,.H is positive and !::,.S is negative.

18.31

Given that the standard Gibbs free energies of formation of N0 2 and NO ) are -34.6 and - 110.5 kllmol, respectively, calculate the amount of Gibbs free energy released when 1 mole of NO z is oxidized to 1 mole of NO) .

Calculate!::"Go for the following reactions at 25°C: (a) Nz(g) + 02(g) • 2NO(g) (b) HzO(I) • H zO(g) • (c) 2C 2HzCg) + 50 z(g) • 4CO z(g)

Section 18.5: Free Energy and Chemical Equilibrium

+ 2HzO(I)

(Hint: Look up the standard free energies of formation of the reactants and products in Appendix 2.) 18.23

+ O 2 - - - . . 2NO)

2N0 2

Problems 18.22

Certain bacteria in the soil obtain the necessary energy for growth by oxidizing nitrites to nitrates:

Calculate!::"Go for the following reactions at 25 °C:

Review Questions 18.32

Explain the difference between !::"G and !::"Go.

18.33

Explain why Equation 18.15 is of great importance in chemistry.

18.34

Fill in the missing entries in the following table:

(a) 2Mg(s) + 02(g) ---.. 2MgO(s) (b) 2S0zCg) + 02(g) • 2S0 3(g) (c) 2C 2H 6(g) + 70 z(g ) • 4C0 2(g) + 6H zO(I)

K < 1

!::"Go

InK

0

(See Appendix 2 for thelIDodynamic data.) 18.24

18.25

From the values of !::"H and !::,.S, predict which of the following reactions would be spontaneous at 25°C: reaction A: !::,.H = 10.5 kllmol, !::,.S = 30 11K . mol; reaction B: !:::Ji = 1.8 kll mol , !::,.S = - 113 11K · mol. If either of the reactions is nonspontaneous at 25 °C, at what temperature might it become spontaneous ? Find the temperatures at which reactions with the following !::"H and !::"S values would become spontaneous: (a) !::,.H = - 126 kll mol , !::,.S = 84 11K . mol; (b) !::,.H = -11.7 kllmol, !::,.s = -105 11K· mol.

Product s are fa vare d

Problems 18.35

Calculate Kp for the following reaction at 25 °C: !::"Go = 2.60 kJ/mol

18.36

For the autoionization of water at 25 °C,

Kw is 1.0 18.37

X

10- 14 What is !::"Go for the process?

Consider the following reaction at 25 °C: • Fe2+ (aq)

Fe(OH)z(s), 18.26

18.27

18.28

The molar heats affusion and vaporization of ethanol are 7.61 and 26.0 kllmol, respectively. Calculate the molar entropy changes for the solid-liquid and liquid-vapor transitions for ethanol. At 1 atm pressure, ethanol melts at - 117.3 °C and boils at 78.3°C. The molar heats of fusion and vaporization of mercury are 23.4 and 59.0 kllmol , respectively. Calculate the molar entropy changes for the solid-liquid and liquid-vapor transitions for mercury. At 1 atm pressure, mercury melts at - 38.9°C and boils at 357°C.

18.29

18.30

As an approximation, we can assume that proteins exist either in the native (physiologically functioning) state or the denatured state. The standard molar enthalpy and entropy of the denaturation of a certain protein are 512 kllmol and 1.60 kllK . mol, respectively. Comment on the signs and magnitudes of these quantities, and calculate the temperature at which the denaturation becomes spontaneous.

10- 14

Calculate!::"Go and K p for the following equilibrium reaction at 25 °C:

18.39

(a) Calculate!::"Go and Kp for the following equilibrium reaction at 25 °C. The !::"Gf values are 0 for Clz(g), -286 kll mol for PCI 3(g), and -325 kllmol for PCls(g). PCls(g) '

• PCI 3(g)

+ CI 2(g)

(b) Calculate!::"G for the reaction if the partial press ures of the initial mixture are PPC1 5 = 0.0029 atm, PPC1, = 0.27 atm, and PC!, = 0.40 atm. 18.40

The equilibrium constant (Kp) for the reaction H 2(g)

+ COzCg) :;:.==' H zO (g) + CO(g)

is 4.40 at 2000 K. (a) Calculate!::"GO for the reaction. (b) Calculate!::"G for the reaction when the partial pressures are PH,_ = 0.25 atm, Peo2 = 0.78 atm, PH 20 = 0.66 atm , and P eo = 1.20 atm.

Use the values listed in Appendix 2 to calculate!::"Go for the following alcohol fermentation: .

X

18.38

, P2

At 25°C, we have !::,.Ho = 17 kJ/mol and !::"So = 65 11K . mol. Is the dimerization favored at this temperature? Comment on the effect of lowering the temperature. Does your result explain why some enzymes lose their activities under cold conditions?

+ 20H- (aq)

Calculate!::"Go for the reaction. Ksp for Fe(OHh is 1.6

Consider the formation of a dimeric protein: 2P

Result at equilibriuill

18.41

Consider the decomposition of calcium carbonate: CaC0 3 (s) :;:.=

=' CaO(s) + CO 2(g)

Calculate the pressure in atm of COz in an equilibrium process (a) at 25 °C and (b) at 800°C. Assume that !::,.W = 177.8 kllmol and !::"So = 160.5 11K· mol for the temperature range. 18.42

The equilibrium constant Kp for the reaction CO(g)

+

= =' COCI2(g)

Clz(g) :;:,

is 5.62 X 103s at 25°C. Calculate!::,.Gf for COCl 2 at 25°C.

QUESTIONS AND PROBLEMS

18.43

At 25 °C, !J.G o for the process H 2 0(I)

18.54

Give a detailed example of each of the following, with an explanation: (a) a thermodynamically spontaneous process, (b) a process that would violate the first law of thermodynamics, (c) a process that would violate the second law of thermodynamics, (d) an irreversible process, (e) an equilibrium process.

18.55

Predict the signs of !J.H, !J.S, and !J.G of the system for the following processes at 1 atm: (a) ammonia melts at -60°C, (b) ammonia melts at -77.7°C, (c) ammonia melts at - 100°e. (The normal melting point of ammonia is -77.7°e.)

18.56

Consider the following facts: Water freezes spontaneously at -5°C and 1 atm, and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

18.57

Ammonium nitrate (NH4 N0 3 ) dissolves spontaneously and endotherrnically in water. What can you deduce about the sign of !J.S for the solution process?

18.58

Calculate the equilibrium pressure of CO 2 due to the decomposition of barium carbonate (BaC0 3) at 25 °e.

18.59

(a) Trouton's rule states that the ratio of the molar heat of vaporization of a liquid (!J.Hvap) to its boiling point in kelvins is approximately 90 J/K . mol. Use the following data to show that this is the case and explain why Trouton's rule holds true:

+.=::t:, H 20(g)

is 8.6 kJ/mol. Calculate the vapor pressure of water at this temperature. 18.44

Calculate !J.G o for the process C(diamond) +.=::t:, C(graphite) Is the formation of graphite from diamond favored at 25°C? If so, why is it that diamonds do not become graphite on standing?

Section 18.6: Thermodynamics in Living Systems

Review Questions 18.45

18.46

What is a coupled reaction? What is its importance in biological reactions? What is the role of ATP in biological reactions?

Problems 18.47

18.48

Referring to the metabolic process involving glucose on page 747, calculate the maximum number of moles of ATP that can be synthesized from ADP from the breakdown of 1 mole of glucose. In the metabolism of glucose, the first step is the conversion of glucose to glucose 6-phosphate: glucose

+ H 3P0 4 ---+, glucose 6-phosphate + H 20

Benzene Hexane Mercury Toluene

o

!J.G = 13.4 kJ/mol Because !J.G o is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction, and estimate the equilibrium constant for the coupled process.

Additional Problems 18.49

Explain the following nursery rhyme in terms of the second law of thermodynamics.

Referring to Problem 18.59, explain why the ratio is considerably smaller than 90 J/K . mol for liquid HF.

18.61

Carbon monoxide (CO) and nitric oxide (NO) are polluting gases contained in automobile exhaust. Under suitable conditions, these gases can be made to react to form nitrogen (N 2 ) and the less harmful carbon dioxide (C0 2 ) . (a) Write an equation for this reaction. (b) Identify the oxidizing and reducing agents. (c) Calculate the Kp for the reaction at 25°e. (d) Under normal atmospheric conditions, the partial pressures are P N2 = 0.80 atm, P eo, = 3.0 X 10- 4 atm, P eo = 5.0 X 10- 5 atm, and P NO = 5.0 X 10- 7 atm. Calculate Qp, and predict the direction toward which the reaction will proceed. (e) Will rai sing the temperature favor the formation of N2 and CO 2?

18.62

For reactions can"ied out under standard-state conditions, Equation 18.11 takes the form !J.G o = !J.Ho - T!J.S o . (a) Assuming !J.Ho and !J.S o are independent of temperature, derive the equation

at 25 °C for the following conditions: [H+] [H +] [H +] [H+]

= 1.0 X 10- 7 M, [OH-] = 1.0 X 10- 7 M = 1.0 X 10- 3 M, [OW] = 1.0 X 10- 4 M = 1.0 X 10- 12 M, [OH-] = 2.0 X 10- 8 M

= 3.5 M , [OW] = 4.8

X

10- 4 M

18.51

Which of the following thermodynamic function s are associated only with the first law of thermodynamics: S, E, G, and H ?

18.52

A student placed 1 g of each of three compounds A, B, and C in a container and found that after 1 week no change had occun"ed. Offer some possible explanations for the fact that no reactions took place. Assume that A, B, and C are totally miscible liquids.

18.53

The enthalpy change in the denaturation of a certain protein is 125 kJ/mo!. If the entropy change is 397 J/K " mol , calculate the minimum temperature at which the protein would denature spontaneously.

!J.Hvap(kJ/mol) 31.0 30.8 59.0 35.2

18.60

Calculate !J.G for the reaction .

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

Tbp COc) 80.1 68.7 357 110.6

(b) Use the values in Table 12.6 to calculate the same ratio for ethanol and water. Explain why Trouton's rule does not apply to these two substances as well as it does to other liquids.

Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King's horses and all the King's men Couldn't put Humpty together again. 18.50

753

where K I and K 2 are the equilibrium constants at TI and T z, respectively. (b) Given that at 25°C Kc is 4.63 X 10- 3 for the reaction

!J.W = 58.0 kJ/mol calculate the equilibrium constant at 65 °C.

754

CHAPTER 18

Entropy, Free Energy, and Equilibrium

18.63

The K,p of AgCl is given in Table 17.4. What is its value at 6Q°C? [Hint: You need the result of Problem 18.62(a) and the data in Appendix 2 to calculate LVio.J

18.64

Under what conditions does a substance have a standard entropy of zero? Can an element or a compound ever have a negative standard entropy?

18.65

18.76

Ni(s)

==' CO(g) +

Hlg)

18.77

Prom the data in Appendix 2, estimate the temperature at which the reaction begins to favor the formation of products. 18.66

+ Cnaq)

18.67

18.68

Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see page S09). Based on this, what can you deduce about the signs of !:1S and LVi?

18.69

Consider the thermal decomposition of CaC0 3:

18.78

A certain reaction is spontaneous at 72°e. If the enthalpy change for the reaction is 19 kJ/mol , what is the minimum value of !:1S (in J/K . mol) for the reaction ?

18.71

Predict whether the entropy change is positive or negative for each of these reactions: (a) Zn(s) + 2HCI(aq) • ' ZnCI 2(aq) + H 2(g) (b) O(g) + O(g) . ' 02(g) (c) NH4N0 3(s) • ' N 20(g) + 2H20(g) (d) 2H2 0 2 (l). ' 2HzO(l) + 0 2(g)

18.72

The reaction NH 3 (g) + HCI(g) , NH4 Cl(s) proceeds spontaneously at 2S oC even though there is a decrease in disorder in the system (gases are converted to a solid). Explain.

18.73

Use the following data to determine the normal boiling point, in kelvins, of mercury. What assumptions must you make in order to do the calculation?

Hg(l): !:1H'f

==' Ni(s) + ~02(g)

18.79

Comment on the statement: "Just talking about entropy increases its value in the universe."

18.80

Por a reaction with a negative !:1Go value, which of the following statements is fal se? (a) The equilibrium constant K is greater than one. (b) The reaction is spontaneous when all the reactants and products are in their standard states. (c) The reaction is always exothermic.

18.81

Consider the reaction N 2(g) + 0 2(g) :;::.:=::::!' 2NO(g)

Given that !:1Go for the reaction at 2SoC is 173.4 kJ/mol, (a) calculate the standard free energy of formation of NO and (b) calculate Kp of the reaction. (c) One of the starting substances in smog formation is NO. Assuming that the temperature in a running automobile engine is 1100°C, estimate Kp for the given reaction. (d) As farmers know, lightning helps to produce a better crop. Why ? 18.82

==' Cu(s) + W 2(g)

So = 174.7 J/K . mol 18.74

The molar heat of vaporization of ethanol is 39.3 kJ/mol , and the boiling point of ethanol is 78.3°e. Calculate!:1S for the vaporization of O.SO mole of ethanol.

18.75

A certain reaction is known to have a !:1Go value of -122 kJ/moi. Will the reaction necessarily occur if the reactants are mixed together?

!:1Go = 127.2 kJ/mol

However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process, and calculate the equilibrium constant for the coupled reaction. 18.83

Consider the decomposition of magnesium carbonate:

==' MgO(s) + COig)

MgC0 3(s) ••

Calculate the temperature at which the decomposition begins to favor products. Assume that both !:1Ho and !:1So are independent of temperature.

= 0 (by definition)

= 60.78 kJ/mol

Heating copper(II) oxide at 400°C does not produce any appreciable amount of Cu: CuO(s) ••

So = 77.4 JIK . mol

Hg(g): !:1H'f

Calculate the pressure of O 2 (in atm) over a sample of NiO at 2S oC if !:1Go = 212 kJ/mol for the reaction NiO(s) ••

The equilibrium vapor pressures of CO 2 are 22.6 rnmHg at 700°C and 1829 mmHg at 9S0°C. Calculate the standard enthalpy of the reaction. [Hint: See Problem 18.62(a).J 18.70

' HBr(g)

Account for the differences in !:1Go and Kp obtained for parts (a) and (b).

==' HCI (aq) + P-(aq)

The pH of gastric juice is about 1.00 and that of blood plasma is 7.40. Calculate the Gibbs free energy required to secrete a mole of H + ions from blood plasma to the stomach at 37°e.

Ni(COMg)

==' 2HBr(g)

••

(a) Predict whether K will be greater or smaller than 1. (b) Does !:1So or !:1Ho make a greater contribution to !:1GO? (c) Is !:1Ho likely to be positive or negative?

+=.:=::::!'

Calculate!:1Go and Kp for the following processes at 2SoC: (a) Hig) + Bri l) •• (b) !H2(g) + ~Br2(l).

Consider the following BriZlnsted acid-base reaction at 2SoC: HF(aq)

+ 4CO(g)

Given that the standard free energies of formation of CO(g) and Ni(CO)ig) are -137.3 and -S87.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 80°e. Assume that !:1G'f is temperature independent.

Water gas, a mixture of H2 and CO, is a fuel made by reacting steam with red-hot coke (a by-product of coal distillation): H 20 (g) +, C(s) ••

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a gas and can therefore be separated from solid impurities:

18.84

(a) Over the years, there have been numerous claims about " perpetual motion machines," machines that will produce useful work with no input of energy. Explain why the first law of thermodynamics prohibits the possibility of such a machine existing. (b) Another kind of machine, sometimes called a "perpetual motion of the second kind," operates as follows. Suppose an ocean liner sails by scooping up water from the ocean and then extracting heat from the water, converting the heat to electric power to run the ship, and dumping the water

QUESTIONS AND PROBLEMS

back into the ocean. This process does not violate the first law of thermodynamics, for no energy is created-energy from the ocean is just converted to electric energy. Show that the second law of thermodynamics prohibits the existence of such a machine.

18.85

Calculate the I1Go for the overall reaction if this process is coupled to the conversion of sulfur to sulfur dioxide and I1Gr(CU2S) = -86.1 kl/mo!.

18.93

The activity series in Table 4.6 shows that reaction (a) is spontaneous whereas reaction (b) is nonspontaneous at 25 °C: (a) Fe(s) + 2H+ (b) Cu(s) + 2H+

• Fe2+ (aq) + Hig) • Cu 2+(aq) + Hig)

Use the data in Appendix 2 to calculate the equilibrium constant for these reactions and hence confirm that the activity series is correct. 18.86

K +(15 mM)

18.94

2

is '(. 1 X 10 1M • s at 25 °C. What is the rate constant for the reverse reaction at the same temperature? 18.87

The following reaction was described as the cause of sulfur deposits formed at volcanic sites.

It may also be used to remove 502 from powerplant stack gases. (a) Identify the type of redox reaction it is. (b) Calculate the equilibrium constant (Kp) at 25 °C, and comment on whether this method is feasible for removing 502, (c) Would this procedure become more effective or less effective at a higher temperature? 18.88

Describe two ways that you could determine I1Go of a reaction.

18.89

The following reaction represents the removal of ozone in the stratosphere:

18.91

18.92

When a native protein in solution is heated to a high enough temperature, its polypeptide chain will unfold to become the denatured protein. The temperature at which a large portion of the protein unfolds is called the melting temperature. The melting temperature of a certain protein is found to be 46°C, and the enthalpy of denaturation is 382 kl/mo!. Estimate the entropy of denaturation, assuming that the denaturation is a two-state process; tnat is, native protein • denatured protein. The single polypeptide protein chain has 122 amino acids. Calculate the entropy of denaturation per amino acid. Comment on your result. A 74.6-g ice cube floats in the Arctic Sea. The pressure and temperature of the system and surroundings are at 1 atrn and O°C, respectively. Calculate I1Ssys , I1Ssurn and I1Sun iv for the melting of the ice cube. What can you conclude about the nature of the process from the value of I1Suniv ? (The molar heat of fu sion of water is 6.01 kJ/mo!.) Comment on the feasibility of extracting copper from its ore chalcocite (CU2S) by heati ng: CU2S(S)

----+. 2Cu(s) + S(s)

Large quantities of hydrogen are needed for the synthesis of ammonia. One preparation of hydrogen involves the reaction between carbon monoxide and steam at 300°C in the presence of a copper-zinc catalyst:

Calculate the equilibrium constant (Kp) for the reaction and the temperature at which the reaction favors the formation of CO and H 2 0. Will a larger Kp be attained at the same temperature if a more efficient catalyst is used?

18.95

Calculate the equilibrium constant (Kp) for this reaction. In view of the magnitude of the equilibrium constant, explain why this reaction is not considered a major cause of ozone depletion in the absence of human-made pollutants such as the nitrogen oxides and CFCs. Assume the temperature of the stratosphere is - 30°C and I1G? is temperature independent. 18.90

• K +(400 mM)

In this calculation, the I1Go term can be set to zero. What is the justification for this step?

The rate constant for the elementary reaction 9

Active transport is the process in which a substance is transferred from a region of lower concentration to one of higher concentration. This is a nonspontaneous process and must be coupled to a spontaneous process, such as the hydrolysis of ATP. The concentrations of K + ions in the blood plasma and in ner 'e cells are 15 mM and 400 mM, respectively (1 mM = 1 X 10- 3 M). Use Equation 18.14 to calculate I1G for the process at the physiological temperature of 37°C:

Consider two carboxylic acids (acids that contain the -COOH group): CH 3COOH (acetic acid, Ka = 1.8 X 10- 5 ) and CH2CICOOH (chloroacetic acid, Ka = 1.4 X 10- 3). (a) Calculate D.Go for the ionization of these acids at 25 °C. (b) From the equation I1Go = I1Ho -TI1So, we see that the contributions to the D.Go term are an enthalpy term (D.HO) and a temperature times entropy term (TD.SO). These contributions are listed here for the two acids: CH 3COOH CH2CICOOH

l1Ho (kJ/mol) -0.57 -4.7

nso(kJ/mol) - 27.6 - 21.l

Which is the dominant term in determining the value of D.Go (and hence Ka of the acid)? (c) What processes contribute to I1HO? (Consider the ionization of the acids as a Br0nsted acid-base reaction.) (d) Explain why the TI1So term is more negative for CH 3COOH. 18.96

Many hydrocarbons exist as structural isomers, which are compounds that have the same molecular formula but different structures. For example, both butane and isobutane have the same molecular formula of C 4H IO (see Problem 12.20 on page 497). Calculate the mole percent of these molecules in an equilibrium mixture at 25°C, given that the standard free energy of formation of butane is - 15.9 kJ/mol and that of isobutane is - 18.0 kl/mo!. Does your result support the notion that straightchain hydrocarbons (that is, hydrocarbons in which the C atoms are joined along a line) are less stable than branch-chain hydrocarbons?

•

18.97

One of the steps in the extraction of iron from its ore (FeO) is the reduction of iron (II) oxide by carbon monoxide at 900°C: FeO(s)

+ CO(g) +.=~. Fe(s) + CO2(g)

If CO is allowed to react with an excess of FeO, calculate the mole fractions of CO and CO 2 at equilibrium. State any assumptions.

756

CHAPTER 18

18.98

Entropy, Free Energy, and Equilibrium

Derive the equation t::.G = RTln (QIK)

where Q is the reaction quotient, and describe how you would use it to predict the spontaneity of a reaction.

18.99

The sublimation of carbon dioxide at -78°C is given by t::.Hsub = 25.? kJ/mol

Calculate t::.Ssub when 84.8 g of CO 2 sublimes at this temperature. 18.100 Entropy has sometimes been described as "time's arrow" because it is the property that determines the forward direction of time. Explain.

18.101 Referring to Figure 18.1 , we see that the probability of finding all 100 molecules in the same flask is 8 X 10- 31 . Assuming that the age of the universe is 13 billion years, calculate the time in seconds during which tbis event can be observed. 18.102 A student looked up the t::.G't, t::.H't, and t::.so values fo r CO 2 in Appendix 2. Plugging these values into Equation 18.11, the student found that t::.G't *- t::.H't - Tt::.So at 298 K. What is wrong with thi s approach ?

18.108 Give a detailed example of each of the following , with an explanation: (a) a thermodynamically spontaneous process, (b) a process that would violate the first law of thermodynamics, (c) a process that wo uld violate the second law of thermodynamics, (d) an irreversible process, (e) an equilibrium process.

18.109 At 0 K, the entropy of carbon monoxide crystal is not zero but has a value of 4.2 J/K . mol, called the residual entropy. According to the third law of thelIDodynamics, this means that the crystal does not have a perfect arrangement of the CO molecules. (a) What would be the residual entropy if the arrangement were totally random? (b) Comment on the difference between the result in part (a) and 4.2 J/K . mol. (Hint: Assume that each CO molecule has two choices for Olientation, and use Equation 18.2 to calculate the residual entropy.) 18.11 0 Comment on the COlTectness of the analogy sometimes used to relate a student's dormitory room becoming untidy to an increase in entropy.

18.111 The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and 269.2 11K· mol, respectively. Calculate t::.Ho, t::.So, and t::.Go for the given process at 25°C. Comment on your answers.

18.103 Consider the following reaction at 298 K: t::.Ho = - 571.6 kJ/mol

Calculate t::.Ssys, t::.SSUIT> and t::.Suniv for the reacti on.

18.112 The following diagram shows the variation of the equilibrium constant with temperature for the reaction

18.104 The concentration of glucose inside a cell is 0.12 mM and that outside a cell is 12.3 mM. Calculate the Gibbs free-energy change for the transport of 3 moles of glucose into the cell at 37°C.

Calculate t::.Go, t::.Ho, and t::.So for the reaction at 872 K. (Hint: See Problem 18.62.)

18.105 Which of the following are not state fu nctions: S, H, q, w, T? ~

18.106 Which of the following is not accompanied by an increase in the entropy of the system: (a) mixing of two gases at the same temperature and pressure, (b) mixing of ethanol and water, (c) dischargin g a battery, (d) expansion of a gas followed by compression to its original temperature, pressure, and volume?

1 1

K? = 0.0480 K J = 1.80

X

1

1 1 1

: T2 1

1 1 1

18.107 Hydrogenation reactions (for example, the process of converting C=C bonds to C-C bonds in the food industry) are facilitated by the use of a transition metal catalyst, such as Ni or Pt. The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of t::.H, t::.S, and t::.G when hydrogen gas is adsorbed onto the surface of Ni metal.

=

1173 K

T

J

=

872 K

1 I I J

liT

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES

•

In chemistry, the standard state for a solution is 1 M. This means that each solute concentration expressed in molarity is divided by 1 M. In biological systems, however, we define the standard state for the H+ ions to be 1 X 10- 7 M. Consequently, the change in the standard Gibbs free energy according to these two conventions will be different for reactions involving the uptake or release of H + ions, depending on which convention is used. We will therefore replace t::.Go with t::.G0 where the prime denotes that it is the standard Gibbs free-energy change for a biological process . 1

,

1.

What is most likely the reason for the standard state for H+ being defined as 1 X 10- 7 M in biological systems? a) It is an arbitrary convention. b) Physiological conditions reduce an concentrations by a factor of 107

10- 4

•

c) Physiological conditions increase all concentrations by a factor of 107 d) Bi ological systems have a pH of about 7. 2.

Consider the reaction A

+ B - _ . C + xH +

where x is a stoichiometric coefficient. Using the equation t::.G = t::.Go + RTln Q I

determine the relationship between t::.GO and t::.Go, keeping in mind that t::.G is the same for a process regardless of which convention is used.

a) t::.GOI = t::.Go + RTln (1 X 107 ) b) tGO = tGo + RTln (1 X 10- 7 ), I

ANSWERS TO IN-CHAPTER MATERIALS

c) !1G01 = !1Go + e -7/RT d) !1G01 = !1Go + RT X e- 7

3.

1

Determine the relationship between !1G0 and !1Go for the reverse process:

C a) !1G01 b) !1G01 c) !1G01 d) !1G01 4.

+ RT In (1 X 107 t !1Go + RTln (1 X 10- 7) !1Go + e71RT !1Go + RT X e7

= !1Go = =

=

!1GOis - 21.8 kllmol at 298 K. Calculate !1G01 for this proce . and calculate!1G using either the chemical or the biological convention when [NADH] = 1.5 X 10- 2 M, [H +] = 3.0 X 10- 5 M, [NAD] = 4.6 X 10- 3 M, and PH, = 0.010 atm. a) b) c) d)

+ xH+ - _ . A + B

18.1 kllmol, 50.3 kllmol 21.8 kllmol, 29.6 kllmol 18.1 kllmol, 10.3 kllmol 21.8 kllmol, 10.3 kllmol

NAD + and NADH are the oxidized and reduced forms of nicotinamide adenine dinucleotide, two key compounds in the metabolic pathways. For the oxidation of NADH NADH

+ H+

• NAD +

7:> 1

+ H2

ANSWERS TO IN-CHAPTER MATERIALS Answers to Practice Problems

Answers to Checkpoints

18.1 (a) Negative, (b) negative, (c) positive. 18.2 (a) 173.6 11K· mol , (b) -139.8 11K· mol, (c) 215.3 11K· mol. 18.3 3728°C. 18.4 (a) - 106 kllmol, (b) - 2935 kllmol. 18.5 !1Sfu s = 16 11K' mol , !1Svap = 72 11K· mol. 18.6A 1.3 kllmol , reverse reaction is spontaneous. 18.6B 2.5 atm. 18.7 2 X 10- 57 18.8 32.9 kllmol.

18.2.1 a, e. 18.2.2 b. 18.3.1 b. 18.3.2 e. 18.4.1 d. 18.4.2 d. 18.4.3 e. 18.4.4 a. 18.5.1 b. 18.5.2 d. 18.5.3 b. 18.5.4 d.

Answers to Applying What You've learned a) Positive. b) 1061.4 11K· mol. c) I 114.9 kllmol. d) 1075.4°C.

,

•

ectroc emlstr •

19.1

Balancing Redox Reactions

19.2 19.3

Galvanic Cells Standard Reduction Potentials

19.4

Spontaneity of Redox Reactions Under Standard-State Conditions

19.5

Spontaneity of Redox Reactions Under Conditions Other Than Standard State

•

•

19.6 •

• • •

19.7 • •

• •

19.8

""', e.

The Nernst Equation Concentration Cells

Batteries Dry Cells and Alkaline Batteries Lead Storage Batteries Lithium-Ion Batteries Fuel Cells

1

,

•

. .,

Electrolysis Electrolysis of Molten Sodium Chloride Electrolysis of Water Electrolysis of an Aqueous Sodium Chloride Solution Quantitative Applications of Electrolysis

Corrosion

•

--

••
2Ag +(aq) + Fe(s)

:;:,

Practice Problem B Calculate the equilibrium constant for the following reaction at 25 °C: 2Cr(s)

Checkpoint 19.4 19.4 .1

+ 3Mn2+ (aq) :;:.= =' 2Cr3+(aq) + 3Mn(s)

Spontaneity of Redox Reactions Under Standard-State Conditions

Calculate K at 25°C for the following reaction: Fe2+(aq)

+ Ni(s) •

>

Fe(s)

19.4 .2

+ Ni2+(aq)

Calculate I1Go at 25°C for the following reaction: 3Cu 2+(aq)

+ Cr(s) •

a) 6 X 10- 4

a) 2.6 X 102 kJ/mol

b) 4 X 10- 7

b) - 2.6 X 102 kJ/mol

c) 3 X 106

c) 1 X 1045 kJ/mol

d) 1 X 10- 13

d) -86 kJ/mol

e) 2 X 103

e) 86 kJ/mol

Spontaneity of Redox Reactions Under Conditions Other Than Standard State So far we have focused on redox reactions in which the reactants and products are in th~ir standard states. Standard-state conditions, though, are difficult to come by and usually impossible to maintain. Just as there is an equation that relates IlG to IlGo [ ~~ Section 18.5 , Equation 18 .14] , there is an equation that relates E to EO. We will now derive this equation.

The Nernst Equation Consider a redox reaction of the type aA

+ bB - -> cC + dD

•

From Equation 18.14,

IlG

=

IlGO+ RTln Q

Because IlG = - nFE and IlGo = - nFEo, the equation can be expressed as -nFE = - nFEo + RTln Q

Dividing the equation through by - nF, we get E = EO - RT In Q nF

Equation 19.6

>

3Cu+(aq)

+ Cr3+ (a q)

I

/'

774

CHAPTER 19

Electrochemistry where Q is the reaction quotient [ ~~ Section 15.2] . Equation 19.6 is known as the NernsP eqllation. At 298 K, Equation 19.6 can be rewritten as E = EO - 0.0257 V In Q n

or using the base-l 0 logarithm of Q as E = E O - 0.0592 V Io cr Q n 0

Equation 19.7

• During the operation of a galvanic cell, electrons flow from the anode to the cathode, resulting in product formation and a decrease in reactant concentration. Thus, Q increases, which means that E decreases. Eventually, the cell reaches equilibrium. At equilibrium, there is no net transfer of electrons, so E = 0 and Q = K, where K is the equilibrium constant. The Nemst equation enables us to calculate E as a function of reactant and product concentrations in a redox reaction. For example, for the cell pictured in Figure 19.1, Zn(s)

+ Cu 2+(aq)

• Zn2 +(aq)

+ Cu(s)

The Nemst equation for this cell at 25 °C can be written as E

•

= 1.l0 V _ 0.0592 V Io cr [Zn2+ ] 2

0

[Cu2+]

If the ratio [Zn2+]/[Cu 2 +] is less than 1, log ([Zn?+]/ [Cu2+]) is a negative number, making the second term on the right-hand side of the preceding equation a positive quantity. Under this condition, E is greater than the standard potential EO. If the ratio is greater than 1, E is smaller than EO. Sample Problem 19.6 shows how to use the Nemst equation.

Sample Problem 19.6 Predict whether the following reaction will occur spontaneously as written at 298 K: Co(s)

+ Fe2+(aq) ----+. C0 2+(aq) + Fe(s)

assuming [C0 2 +] = 0. 15 M and [Fe2+] = 0.68 M. Strategy Use EOvalues from Table 19.1 to determine EOfor the reaction, and use Equation 19.7 to calculate E. If E is positive, the reaction will occur spontaneously. Setup From Table 19.1 , Cathode (reduction): Fe 2+(aq)

+ 2e - ----+. Fe(s) Co(s)' C0 2 +(aq) + 2e -

Anode (oxidation):

EO - EO cell cathode

-

EO anode

=

-0.44 V - (-0.28 V)

=

- 0.16 V

•

The reaction quotient, Q, for the reaction is [Co2+]/[Fe2+]. Therefore, Q = (0.15/0.68) = 0.22. Solution From Equation 19.7,

E = EO- 0.0592 V log Q n

=

-0.16 V - 0.0592 V log 0.22 2

=

-0. 14 V

The negative E value indicates that the reaction is not spontaneous as written under the conditions described.

2. Walter Hermann Nernst (1864-1941). German chem ist and physicist. Nernst's work was mainly on electrolyte solutions and thermodynamics. He also invented an electric piano. Nernst was awarded the Nobel Prize in Chemistry in 1920 for hi s contribution to thermo ~

SECTION 19.5

Spontaneity of Redox Reactions Under Conditions Other Than Standard State

77':>

Think About It For this reaction to be spontaneous as written, the ratio of [Fe2+] to [Co2+] would

have to be enormous. We can determine the required ratio by first setting E equal to zero:

oV

= - 016 V - 0.0592 V log

.

2

Q

(0.16 V)(2) - 0.0592 V = log Q

log Q = - 5.4

Q = 10- 54 =

0+ [Co- ] = 4 X

10- 6

[Fe2 +] For E to be positive, therefore, the ratio of [Fe 2 +] to [Co2+], the reciprocal of Q, would have to be greater than 3 X 105 to 1. 2

Practice Problem A Will the following reaction occur spontaneously at 298 Kif [Fe +] = 0.60 M and [Cd2+] = 0.010 M?

Cd(s)

+ Fe2 +(aq) - _ . Cd2+ (aq) + Fe(s) 2

Practice Problem B What concentration of Sn + would be necessary to make the following reaction

spontaneous at 25°C if [Pb2+]

=

1.0 M?

Pb(s)

+ Sn2+ (aq)

• Pb2+ (aq)

+ Sn(s)

Concentration Cells Because electrode potential depends on ion concentrations, it is possible to construct a galvanic cell from two half-cells composed of the same material but differing in ion concentrations. Such a cell is called a concentration cell. Consider a galvanic cell consisting of a zinc electrode in 0.10 M zinc sulfate in one compartment and a zinc electrode in l.0 M zinc sulfate in the other compartment (Figure 19.5). According to Le Chatelier's principle, the tendency for the reduction . Zn2+(aq)

+ 2e-

• Zn(s)

Figure 19.5

A concentration cell. Oxidation occurs in the container with the lower Zn2 + concentration. Reduction occurs in the container with the higher Zn2+ concentration .

..)

2e -lost

2e - gained

.j

per Zn2+ ion reduced

+

Anode (-) Zn Salt bridge

Cathode (+ )

N;:a~+::;=.~N~a~+-======-::s-::o~~=--"""'" _

Zn

•

o

o Zn2+ (0.10 M)

.02 o Zn2 + (1.0 M)

o

776

CHAPTER 19

Electrochemistry

to occur increases with increasing concentration of Zn2+ ions. Therefore, reduction should occur in the more concentrated compartment and oxidati on should take place on the more dilute side. The cell diagram is Zn(s) I Zn 2+(aq)(O .lO M) II Zn2+ (aq)(1.0 M) I ( Zn(s) and the half-cell reactions are Zn(s) - _ . Zn2+ (aq)(O .lOM) + 2e-

Oxidation:

,

+ 2e -

R eduction: Zn?+ (aq) ( l.O M)

• Zn(s)

Zn 2+(aq)( 1.0 M) - _ . Zn 2+(aq)(0 .10 M)

Overall:

The cell potential is

E

=

EO _ 0.0592 V log _ [Z _ _n_2_ +]_d_ilu_te_

2

?+

[Zn- ]concentrated

where the subscripts "dilute" and "concentrated" refer to the 0.10 M and 1.0 M concentrations, respectively. The EOfor this cell is zero (because the same electrode-ion combination is used in both half-cells; i.e. , E ~.thode = E ~node), so E

=0 -

0.0592 V loa 0.10 2 b 1.0 =

0.030 V

The cell potential for a concentration cell is typically small and decreases continually during the operation of the cell as the concentrations in the two compartments approach each other. When the concentrations of the ions in the two compartments are equal, E becomes zero and no further change occurs.

Bringing Chemistry to Life Biological Concentration Cells A biological cell can be compared to a concentration cell for the purpose of calculating its membrane potential. Membrane potential is the electric potential that exists across the membrane of various kinds of cells, including muscle cells and nerve cells. This potential is responsible for the propagation of nerve impulses and heartbeat. A membrane potential is established whenever there are unequal concentrations of the same type of ion in the interior and exterior of a cell and the membrane is permeable to the ion. For example, the concentrations of N a + ions in the interior and 2 1 exterior of a nerve cell are 1.5 X 10- M and 1.5 X 10- M, respectively. Treating the situation as a concentration cell and applying the Nernst equation for a single ion, we can write ECG strip

ENa+

- EO -

N.+ -

0.0592 V I [Na + ] ex og ---:-1 [Na +]in

= - (0.0592 V) log 1.5

1

X 101.5 X 10- 2

= 0.059 V

or

59 mV

where the subscripts "ex" and "in" denote "exterior" and "interior," respectively. We have set EN. = 0 because the same type of ion is involved. Thus, an electric potential of 59 m V exists across the membrane due to the unequal concentrations of Na + ions. When a nerve cell is stimulated, there is a large change in the membrane permeability, decreasing the membrane potential temporarily to about 34 m V. This sudden change in the potential is called the action potential. Once created, the action potential propagates along the nerve fiber until it reaches either a synaptic junction (the connection between nerve cells) or a neuromuscular junction (the connection between a nerve cell and a muscle cell). In the muscle cell of the heart, an action potential is generated during the heartbeat. This potential produces enough current to be detected by electrodes placed on the chest. The amplified signals can be recorded either on a moving chart or displayed on an oscilloscope. The record, called an electrocardiogram (ECG, also known as an EKG, where K is from the German word kardio for heart), is a valuable tool in the diagnosis of heart disease. Patient with ECG electrodes

SECTION 19.6

Checkpoint 19.5 19.5.1

n7

Batteries

Spontaneity of Redox Reactions Under Conditions Other Than Standard State 19.5.2

Calculate E at 25 °C for a galvanic cell based on the reaction,

Calculate the cell potential at 25°C of a galvanic cell consisting of an Ag electrode in 0.15 M AgN0 3 and an Ag electrode in l.0 M AgN0 3 .

+ Cu2+(aq) • Zn2+(aq) + Cu(s), in which [Zn2+] = 0.55 M and [Cu2+] = l.02 M. Zn(s)

a) O.OV

a) 1.10 V

b) 0.049 V

b) 1.11 V

c) - 0.049 V

c) l.09 V

d) 0.024 V

d) 0.0118 V

e) - 0.024 V

e) l.03 V

Batteries A battery is a galvanic cell, or a series of connected galvanic cells, that can be used as a portable, self-contained source of direct electric current. In this section we examine several types of batteries.

Dry Cells and Alkaline Batteries The most common batteries, dry cells and alkaline batteries, are those used in flashlights, toys, and certain portable electronics such as CD players. The two are similar in appearance, but differ in the spontaneous chemical reaction responsible for producing a voltage. Although the reactions that take place in these batteries are somewhat complex, the reactions shown here approximate the overall processes. A dry cell, so named because it has no fluid component, consists of a zinc container (the anode) in contact with manganese dioxide and an electrolyte (Figure 19.6). The electrolyte consists of ammonium chloride and zinc chloride in water, to which starch is added to thicken the solution to a paste so that it is less likely to leak. A carbon rod, immersed in the electrolyte in the center of the cell, serves as the cathode. The cell reactions are:

+ ;---Paper spacer '---Moist paste of ZnC]2 and NH4 Cl

Zn(s) -_I Zn2+(aq) + 2e -

Anode: Cathode: Overall:

2NHt(aq) Zn(s)

+ 2Mn02(S) + 2e-

+ 2NHt(aq) + 2Mn02(s)

• Mn203(S) • Zn2+(aq)

+ 2NH3(aq) + H 20(l)

---1'-f'---Layer of Mn02

+ Mn203(S) + 2NH3(aq) + H 2 0(l)

---1-+':---

The voltage produced by a dry cell is about 1.5 V. An alkaline battery is also based on the reduction of manganese dioxide and the oxidation of zinc. However, the reactions take place in a basic medium, hence the name alkaline battery. The anode consists of powdered zinc suspended in a gel, which is in contact with a concentrated solution of KOH. The cathode is a mixture of manganese dioxide and graphite. The anode and cathode are separated by a porous barrier (Figure 19.7).

Cathode:

2Mn02(S)

Zn(s)

Anode: Overall:

+ 2H20(l) + 2e - -

Zn(s)

Figure 19.6

Interior view of the type of dry cell used in flashlights and other small devices.

-+. 2MnO(OH)(s) + 20H -(aq)

+ 20H- (aq) --+. Zn(OH)z(s) + 2e-

+ 2Mn02(S) + 2H20(l) -

~Positive

-+. Zn(OH)z(s) + 2MnO(OH)(s) + 20H- (aq)

,

,..----Cathode

Lead Storage Batteries

---Porous

The lead storage battery commonly used in automobiles consists of six identical cells joined together in series. Each cell has a lead anode and a cathode made of lead dioxide (Pb0 2) packed on a metal plate (Figure 19.8). Both the cathode and the anode are immersed in an aqueous solution of sulfuric acid, which acts as the electrolyte. The cell reactions are

Cathode:

Pb(s) Pb0 2(s)

Overall: Pb(s)

cap

---Outer sleeve

Alkaline batteries are more expensive than dry cells and offer superior performance and shelf life.

Anode:

Graphite cathode anode

+ SO~-(aq) -

....... PbS0 4 (s)

---Anode

Negati ve cap

+ 2e -

+ 4H+(aq) + SO~ - (aq) + 2e- - _ I PbSOis) + 2H20(l)

+ Pb02(s) + 4H+(aq) + 2S0~-(aq)

barrier

• 2PbS0 4 (s)

+ 2H20(l) •

Figure 19.7 alkaline battery.

Interior view of an

778

CHAPTER 19

Electrochemistry

Figure 19.8

Interior view of a lead storage battery. Under normal operating conditions, the concentration of the sulfuric acid solution is about 38 percent by mass.

Anode

I -

Removable cap

/

Cathode

,

- - - H2 S04 electrolyte ---::C---....,--Negative plates (lead grills filled with spongy lead)

•

- ----:l'---'--

Positi ve plates (lead grills filled with Pb02)

Under normal operating conditions, each cell produces 2 V. A total of 12 V from the six cells is used to power the ignition circuit of the automobile and its other electric systems. The lead storage battery can deliver large amounts of current for a short time, such as the time it takes to start the engme. Unlike dry cells and alkaline batteries, the lead storage battery is rechargeable. Recharging the battery means reversing the normal electrochemical reaction by applying an external voltage at the cathode and the anode. (This kind of process is called electrolysis, which we discuss in Section 19.7.) •

Lithium-Ion Batteries Sometimes called "the battery of the future," lithium-ion batteries have several advantages over other battery types. The overall reaction that takes place in the lithium-ion battery is

Anode: Cathode: Overall: . ... . .. .

Lithium has the largest negative reduction potential of all the metals, making it a powerful reducing agent.

Li(s) ----+. Li+ Li +

+ Co02 + e Li(s)

+ Co02

+ e-

----0-' LiCo0 2(s) •

LiCo0 2 (s)

.. The overall cell potential is 3.4 V, which is relatively large. Lithium is also the lightest metal only 6.941 g of Li (its molar mass) are needed to produce 1 mole of electrons. Furthermore, a lithiumion battery can be recharged hundreds of times. These qualities make lithium batteries suitable for use in portable devices such as cell phones, digital cameras, and laptop computers. .

Fuel Cells Fossil fuels are a major source of energy, but the conversion of fossil fuel into electric energy is a highly inefficient process. Consider the combustion of methane:

Strictly speaking, a fuel cell is not a battery because it is not self-contained.

To generate electricity, heat produced by the reaction is first used to convert water to steam, which then drives a turbine, which then drives a generator. A significant fraction of the energy released in the form of heat is lost to the surroundings at each step (even the most efficient power plant converts only about 40 percent of the original chemical energy into electricity). Because combustion reactions are redox reactions, it is more desirable to carry them out directly by electrochemithereby greatly increasing the efficiency of power production. This objective can be . . . cal ...means, . ... . . . . .... . . . . . . . . . accomplished by a device known as afuel cell, a galvanic cell that requires a continuous supply of reactants to keep functioning. In its simplest form, a hydrogen-oxygen fuel cell consists of an electrolyte solution, such as a potassium hydroxide solution, and two inert electrodes. Hydrogen and oxygen gases are bubbled

SECTION 19.6

Anode

---+-- Porous carbon electrode -~""'-

containing Ni and NiO

o

o

Hot KOH solution ---\-•

Motor

Oxidation 2H2(g)

+ 40H-(aq)

Reduction OzCg) + 2H20(l) + 4e-

• 4H20(l) + 4e -

• 40H - (aq)

through the anode and cathode compartments (Figure 19.9), where the following reactions take place: Anode:

2H2(g)

Cathode: 0 2(g)

+ 40H-(aq) --+. 4H20(l) + 4e-

+ 2H20(l) + 4e-

Overall:

2H 2(g)

• 40H - (aq)

+ 0 2(g)

• 2H20(l)

Using EOvalues from Table 19.1, the standard cell potential is calculated as follows: o E cell

-

EO cathode

-

EO anode

= 0.40 V - (-0.83 V) = 1.23 V

Thus, the cell reaction is spontaneous under standard-state conditions. Note that the reaction is the same as the hydrogen combustion reaction, but the oxidation and reduction are carried out separately at the anode and the cathode. Like platinum in the standard hydrogen electrode, the electrodes serve two purposes. They serve as electrical conductors, and they provide the necessary surfaces for the initial decomposition of the molecules into atomic species, which must take place before electrons can be transferred. Electrodes that serve this particular purpose are called "electrocatalysts." Metals such as platinum, nickel, and rhodium also make good electrocatalysts. In addition to the Hz-0 2 system, a number of other fuel cells have been developed. Among these is the propane-oxygen fuel cell. The corresponding half-cell reactions are Anode: Cathode: 50 2(g) Overall:

C3HS(g)

+ 6H20(l) --+. 3C02(g) + 20H +(aq) + 20e-

+ 20H+(aq) + 20e C 3H s(g)

+ 50 2(g)

779

Figure 19.9 A hydrogen-oxygen fuel cell. The Ni and NiO embedded in the porous carbon electrodes are catalysts.

Cathode

Porous carbon eie(;tro,dl;'---I containing Ni

Batteries

) lOH 20(l) ) 3COzCg)

+ 4H 20(I)

The overall reaction is identical to the burning of propane in oxygen. Unlike batteries, fuel cells do not store chemical energy. Reactants must be constantly resupplied as they are consumed, and products must be removed as they form. However, properly designed fuel cells may be as much as 70 percent efficient, which is about twice as efficient as an internal combustion engine. In addition, fuel-cell generators are free of the noise, vibration, heat transfer, thermal pollution, and other problems normally associated with conventional power plants. Nevertheless, fuel cells are not yet in widespread use. One major problem is the expense of electrocatalysts that can function efficiently for long periods of time without contamination. One notable application of fuel cells is their use in space vehicles. Hydrogen-oxygen fuel cells provide electric power (and drinking water!) for space flight.

•

780

._ .

[ .~

CHAPTER 19

Electrochemistry

~.

Electrolysis .... .....

Multimedia

Electrochemistry-the electrolysis of potassium iodide.

Cell Type

Chemical Reaction

Electric Energy

Galvanic Electrolytic

Spontaneous Nonspontaneous

Producea Consumed

In Section 19.6, we mentioned that lead storage batteries are rechargeable and that recharging means reversing the electrochemical processes by which the battery ordinarily operates through the application of an external voltage. This process, the use of electric energy to drive a nonspontaneous chemical reaction , is called electrolysis. An electrolytic cell is one used to carry out . . . . . . . . . .. ............................. .... . .... ......... . . . . . . . .. ............. .. electrolysis. The same principles apply to the processes in both galvanic and electrolytic cells. In this section, we discuss three examples of electrolysis based on those principles. We then examine some of the quantitative aspects of electrolysis. .

Electrolysis of Molten Sodium Chloride In its molten (melted) state, sodium chloride, an ionic compound, can be electrolyzed to separate it into its constituent elements, sodium and chlorine. Figure 19.1 O(a) is a diagram of a Downs cell, which is used for the large-scale electrolysis of NaCl. In molten NaCl, the cations and anions are the Na + and Cl - ions, respectively. Figure 19.10(b) is a simplified diagram showing the reactions that occur at the electrodes. The electrolytic cell contain s a pair of electrodes connected to the battery. The battery serves to push electrons in the direction they would not flow spontaneously. The electrode toward which the electrons are pushed is the cathode, where reduction takes place. The electrode away from which electrons are drawn is the anode, where oxidation takes place. The reactions at the electrodes are

Anode (oxidation): Cathode (reduction): Overall: We ca n only estimate the value of E~ell because the val ues in Table 19.1 refer to the species in aqueous solution.

. ..

2Cl-(l)

---+.

CI 2(g)

2Na+(l) + 2e-

• 2Na(l)

+ 2Cl- (l)

• 2Na(l)

2Na+(l)

+ 2e+ Clig)

This process is a major industrial source of pure sodium metal and chlorine gas. ........... .. . . . . . .. .. ........... ................. .. .................................. ....... .

Using data from Table 19.1 , we estimate E ~ell to be -4 V for this process. The negative standard reduction potential indicates that for the process to occur as written, a minimum of approximately 4 V must be supplied by the battery to drive the reaction in the desired direction. In practice, an even higher voltage is required because of inefficiencies in the electrolytic process and because of overvoltage, a phenomenon we will discuss later in this section.

Electrolysis of Water Under ordinary atmospheric conditions (1 atm and 25 °C), water will not spontaneously decompose to form hydrogen and oxygen gas because the standard free-energy change for the reaction is a large positive quantity: !1Go = 474.4 kJ/mol

Figure 19.10

(a) A practical arrangement called a Downs cell for the electrolysis of molten NaCI (m.p. = 801 °C). The sodium metal formed at the cathodes is in the liquid state. Because liquid sodium metal is lighter than molten NaCl, the sodium floats to the surface, as shown, and is collected. Chlorine gas forms at the anode and is collected at the top. (b) A simplified diagram showing the electrode reactions during the electrolysis of molten NaCI. The battery is needed to drive the nonspontaneous reaction.

NaCI Molten NaCl

!

Cl2 gas

i

Anode

\

.

Cathode

Liquid Na

\

o

Molten NaCl-+Iron cathode

Iron

cathode

Carbon anode 2Cl (~

Oxidation • CI2 (g)

+

Reduction 2e- 2Na+ + 2e• 2Na(/)

(b)

SECTION 19.7

Anode

Electrolysis

78 1

Figure 19.11

Apparatus for small-scale electrolysis of water. The volume of hydrogen gas generated at the cathode is twice that of oxygen gas generated at the anode .

Cathode

•

•

Dilute H 2S0 4 -+- !-solution

Battery

Oxidation 2H20 (l) _. 0 2(g)

Reduction

+ 4H +(aq) + 4e -

4H + (aq)

+ 4e-

• 2Hig)

\

However, this reaction can be made to occur in an electrolytic cell like the one shown in Figure 19.11. This cell consists of a pair of electrodes made of a nonreactive metal, such as platinum, immersed in water. When the electrodes are connected to the battery, nothing happens because there are not enough ions in pure water' to'carr)': illiidiof 'an 'electrIC curreiii.· the ·rea'CilClll. occur's ' '" readily in a 0.1 M H 2S04 solution , however, because there is a sufficient ion concentration to conduct electricity. Immediately, gas bubbles begin to appear at both electrodes. The processes at the electrodes are Anode: Cathode:

2H 20(l) - _ I 0 2(g) 4H+(aq )

Overall:

+ 4e-

Remember that at 2 SoC pure water has only a very low concentration of ions [ H~

[H+] = [OW) = 1 X 10- 7 M

+ 4H+(aq) + 4e-

• 2H2 (g)

2H 20 ( l ) ' 0 2(g) + 2H2(g)

Note that there is no net consumption of the acid.

Electrolysis of an Aqueous Sodium Chloride Solution An aqueous sodium chloride solution is the most complicated of the three examples of electrolysis considered here because NaCI(aq) contains several species that could be oxidized and reduced. The reductions that might occur at the cathode are

+ 2e-

- _ I H2(g )

2H 2 0(l) + 2e -

• H ig)

2H +(aq)

EO= O.OOV

+ 20H- (aq)

EO= - 0. 83 V

or

EO = - 2.71 V We can rule out the reduction of Na + ion because of the large negative EOvalue. Under standardstate conditions, the reduction of H + is more apt to occur than the reduction of H 20. However, in a solution of NaCI, the H + concentration is very low, making the reduction of H 20 the more probable reaction at the cathode. The oxidation reactions that might occur at the anode are

or

Section 16.2] :

•

782

CHAPTER 19

Electrochemistry

Referring to Table 19.1, we find CI 2(g) 0 2(g)

•

+ 2e-

+ 4H+(aq) + 4e -

• 2H20(l)

EO= 1.36 V EO= 1.23 V

The standard reduction potentials of the two reactions are not very different, but the values do suggest that the oxidation of H 20 should occur more readily. However, by experiment we find that the gas produced at the anode is C1 2, not O 2, In the study of electrolytic processes, we sometimes find that the voltage required for a reaction is considerably higher than the electrode potentials would indicate. The overvoltage is the difference between the calculated voltage and the actual voltage required to cause electrolysis. The overvoltage for O2 formation is quite high. Under normal operating conditions, therefore, Cl2 gas forms at the anode instead of O2, Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are Anode (oxidation):

2CI- (aq) - _. CI 2(g)

. . ..

..

....

+ 2e-

+ 2e • H 2(g) + 20H- (aq) ------~~----------~~------~~------

Cathode (reduction) : Keep in mind that in the electrolysis of aqueous solutions, the water itself may be oxidized and/or reduced .

- _. 2CI- (aq)

2H20(l)

2H20(l)

Overall:

+ 2CI-(aq) --+. HzCg) + ClzCg) + 20H-(aq)

As the overall reaction shows, the concentration of the Cl- ions decreases during electrolysis and that of the OH- ions increases. Therefore, in addition to H2 and C1 2, the useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of the electrolysis. Electrolysis has many important applications in industry, mainly in the extraction and purification of metals. We discuss some of these applications in Chapter 23.

Quantitative Applications of Electrolysis

Current (amperes) and time (seconds)

t

The quantitative treatment of electrolysis was developed primarily by Faraday. He observed that the mass of product formed (or reactant consumed) at an electrode was proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance being produced (or consumed). In the electrolysis of molten NaCl, for example, the cathode reaction tells us that one Na atom is produced when one Na + ion accepts an electron from the electrode. To reduce 1 mole of Na + ions, we must supply an Avogadro's number (6.02 X 1023 ) of electrons to the cathode. On the other hand, stoichiometry tells us that it takes 2 moles of electrons to reduce 1 mole of Mg2+ ions and 3 moles of electrons to reduce 1 mole of A1 3+ ions:

+ e• Na Mg2+ + 2e- --+. Mg Na+

Multiply current a nd time

AI3+ Charge in coulombs

t

Divide by the Farad ay constant

Number of moles of electrons

t

Use mole ratio in half-ce II reaction

Moles of substance reduced or oxidized

t

Use mol ar mass or ideal g as equation

Grams or liters of product

+ 3e-

• Al

In an electrolysis experiment, we generally measure the current (in amperes) that passes through an electrolytic cell in a given period of time. The relationship between charge (in coulombs) and the current is lC=IA X ls That is, a coulomb is the quantity of electric charge passing any point in the circuit in 1 s when the current is 1 A. Therefore, if we know the current (in amperes) and how long it is applied (in seconds), we can calculate the charge (in coulombs). Knowing the charge enables us to determine the number of moles of electrons. And knowing the number of moles of electrons allows us to use stoichiometry to determine the number of moles of product. Figure 19.12 shows the steps involved in calculating the quantities of substances produced in electrolysis. To illustrate this approach, consider an electrolytic cell in which molten CaCl 2 is separated into its constituent elements, Ca and C1 2. Suppose a current of 0.452 A is passed through the cell for 1.50 h. How much product will be formed at each electrode? The first step is to determine which species will be oxidized at the anode and which species will be reduced at the cathode. Here the choice is straightforward because we have only Ca2+ and CI- ions. The cell reactions are 2Cl - (l) - _ . CI 2(g)

Anode (oxidation) :

Figure 19.12

Steps involved in calculating amounts of substances reduced or oxidized in e1ecrolysis.

Cathode (reduction) : Overall:

+ 2e-

• Ca(l)

+ 2CI- (l)

• Ca(l)

Ca2+ (l) c i+(l)

+ 2e + CI 2(g)

SECTION 19.7

Electrolysis

The quantities of calcium metal and chlorine gas formed depend on the number of electrons that pass through the electrolytic cell, which in turn depends on charge, or cunent X time:

3600 s coulombs = 0.452 A X 1.50 h X 1 h X

1 C = 2.441 X 10 3 C 1A . s

Because 1 mol e- = 96,500 C and 2 mol e - are required to reduce 1 mole of Ca2 + ions , the mass of Ca metal formed at the cathode is calculated as follows:

1 mol e(2.44,1 X 10 C) 96,500 C 3

grams Ca =

\

1 mol Ca 2 mol e

40.08 g Ca -

- l -c-1 rna C a

= 0.507 g Ca

The anode reaction indicates that 1 mole of chlorine is produced per 2 mol e- of electricity. Hence, the mass of chlorine gas formed is 3 1 mol e1 mol CI 2 70.90 g Cl 2 --l-C-l- = 0.897 g Cl 2 grams Cl 2 = (2.441 X 10 C) 96,500 C 2 mol e

1 rna

2

Sample Problem 19.7 applies this approach to electrolysis in an aqueous solution.

,

A current of 1.26 A is passed through an electrolytic cell containing a dilute sulfuric acid solution for 7.44 h. Write the half-cell reactions, and calculate the volume of gases generated at STP. Strategy As shown in Figure 19.12, we can use current and time to determine charge. We can then

convert charge to moles of electrons, and use the balanced half-reactions to determine how many moles of product forms at each electrode. Finally, we can convert moles to volume. Setup The half-cell reactions for the electrolysis of water are

Anode:

+ 4H+(aq) + 4e -

2H 20(l) - _ . OzCg)

Cathode: 4H+ ~aq)

+ 4e-

• 2H2 (g)

+ 2H 2(g )

2H 20 ( I ) . 0 2(g)

Overall:

Remember that STP for gases means 273 K and 1 atm. Solution

coulombs = (1.26 A)(7.44 h) ( 3600 s ) ( 1 C ) = 3.375 X 104 C lh 1A· s At the anode: moles O2

_ -

(3.375

X

1 mol e - ) ( 1 mol O2 = 0.OS74 mol O 2 10 C) 96,500 C 4 mol e 4

(

The volume of 0.OS74 mol O2 at STP is given by

V= nRT p (0.OS74 mol)(0.OS206 L . atmlK . mol)(273.15) 6 = = 1.9 L 0 0 ( 1 atm Similarly, for hydrogen we write F

)=

1 mol e - 1 mol H2, moles Ho (3.375 X 10 C) ( )( ' 96,500 C 2 mol e 4

0.175 mol H2

The volume of 0.175 mol H2 at STP is given by

V= nRT

•

p

•

_ (0.175 mol)(0.OS206 L . atmlK· mol )(273.15 ) _ 3 92 - . L Ho 1 atm -

Practice Problem A A constant current of 0.912 A is passed through an electrolytic cell containing

molten MgCl 2 for IS h. What mass of Mg is produced? Practice Problem B A constant current is passed through an electrolytic cell containing molten

MgCl 2 for 12 h. If 4.S3 L of C12 (at STP) is produced at the anode, what is the current in amperes?

,

Think About It The volume of

H2 is twice that of O 2 (see Figure 19.11), which is what we would e pect based on Avogadro's law (at the same temperature and pressure, volume is directly proportional to the number of moles of gas: V cc n) [I~~ Section 11.2] .

783

784

CHAPTER 19

Electrochemistry

Checkpoint 19.7 19.7.1

•

Electrolysis 19.7.2

In the electrolysis of molten CaCh, a current of 1.1 2 A is passed through the cell for 3 .0 h . What is the mass of Ca produced at the cathode?

How long will a current of 0.995 A need to be passed through water (containing H 2 S04) for 5.00 L of O 2 to be produced at STP?

a) 2 .51 g

a) 6.0 h

b) 1.26 g

b) 8.2 h

c) 5.02 g

c) 1.5 h

d) 10.0 g

d) 3.0 h 5

e) 2 .42 X 10 g

•

e) 24.0 h

Corrosion .

The tenn corrosion generally refers to the deterioration of a metal by an electrochemical process. There are many examples of corrosion, including rust on iron, tarnish on silver, and the green layer that fonns on copper and brass. In this section we discuss the processes involved in corrosion and some of the measures taken to prevent it. The fOlmation of rust on iron requires oxygen and water. Although the reactions involved are quite complex and not completely understood, the main steps are believed to be as follows. A region of the metal's surface serves as the anode, where the following oxidation occurs: Fe(s)

• Fe2+ (aq) + 2e -

The electrons given up by iron reduce atmospheric oxygen to water at the cathode, which is another region of the same metal's surface:

The overall redox reaction is

With data from Table 19.1, the standard potential for this process can be calculated as follows: E cell O -- E cathode O

-

E anode O

= l.23 V - (-0.44 V) =

l.67 V

Note that this reaction occurs in an acidic medium; the H + ions are supplied in part by the reaction of atmospheric carbon dioxide with water to form the weak acid, carbonic acid (H 2C0 3). The Fe2+ ions formed at the anode are further oxidized by oxygen as follows':

o

M edia Player/ M PEG Content Eledrochemistry- The produdion of aluminum.

From Table 19. 1:

+ 2e- - _ . Fe(s) EO = - 0.44 V • AI(s) EO = - 1.66 V AI3+(aq) + 3e-

Fe 2 +(aq)

AI, with the more negative reduction potential, is less likely to be reduced (more likely to be oxidized) than Fe.

•

This hydrated fonn of iron(ill) oxide is known as rust. The amount of water associated with the iron (III) oxide varies, so we represent the fOlmula as Fez0 3 . xHzO. Figure 19.13 shows the mechanism of rust fonnation. The electric circuit is completed by the migration of electrons and ions; this is why rusting occurs so rapidly in saltwater. In cold climates, salts (NaCl or CaCI 2 ) spread on roadways to melt ice and snow are a major cause of rust formation on automobiles. Other metals also undergo oxidation. Aluminum, for example, which is used to make air. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ................................................... .................,........ ....... .. .. planes, beverage cans, and aluminum foil, has a much greater tendency to oxidize than does iron. Unlike the corrosion of iron, though, corrosion of aluminum produces an insoluble layer of protective coating (Al z0 3) that prevents the underlying metal from additional corrosion. Coinage metals such as copper and silver also corrode, but much more slowly than either iron or aluminum: Cu(s) - _ . Cu2+ (aq) Ag(s)

• Ag + (aq)

+ 2e-

+ e-

SECTION

Rust

Iron

Fe(s) ---+. Fe2+ (aq) Fe 2 +(aq)

• Fe3+ (aq)

+ 2e+ e-

•

In ordinary atmospheric exposure, copper forms a layer of copper carbonate (CuC0 3), a green substance referred to as patina, that protects the metal underneath from further corrosion. Likewise, silverware that comes into contact with foodstuffs develops a layer of silver sulfide (Ag?S). A number of methods have been devised to protect metals from corrosion. Most of these methods are aimed at preventing rust formation. The most obvious approach is to coat the metal surface with paint to prevent exposure to the substances necessary for corrosion. If the paint is scratched or otherwise damaged, however, thus exposing even the smallest area of bare metal, rust will form under the paint layer. The surface of iron metal can be made inactive by a process called passivation. A thin oxide layer is formed when the metal is treated with a strong oxidizing agent such as concentrated nitric acid. A solution of sodium chromate is often added to cooling systems and radiators to prevent rust formation. The tendency for iron to oxidize is greatly reduced when it is alloyed with certain other metals. For example, in stainless steel, an alloy of iron and chromium, a layer of chromium oxide forms that protects the iron from corrosion. An iron container can be covered with a layer of another metal such as tin or zinc. A "tin" can is made by applying a thin layer of tin over iron. Rust formation is prevented as long as the tin layer remains intact. However, once the surface has been breached by a scratch or a dent, rusting occurs rapidly. If we look up the standard reduction potentials in Table 19.1, we find that when tin and iron are in contact with each other, tin, with its greater reduction potential, acts as a cathode. Iron acts as an anode and is therefore oxidized. Sn2 +(aq)

Fe2+(aq)

+ 2e- --+. Sn(s) + 2e • Fe(s)

. -~

.

-

+ 2e-

- -..... Zn(s)

EO= - 0.14 V EO= -0.44 V

EO= -0.76 V

so zinc is more easily oxidized than iron. Like aluminum, zinc oxidizes to form a protective coating. Even when the zinc layer is compromised, though, and the underlying iron is exposed, zinc is still the more easily oxidized of the two metals and will act as the anode. Iron will be the cathode, thereby remaining reduced. Galvanization is one example of cathodic protection, the process by which a metal is protected by being made the cathode in what amounts to a galvanic cell. Another example is the use of zinc or magnesium bars to protect underground storage tanks and ships. When a steel tank or hull is connected to a more easily oxidized metal, corrosion of the steel is prevented.

-

Multimedia

Electrochemistry---chrome plating.

Zinc-plating, or galvanization, protects iron from corrosion by a different mechanism. According to Table 19.1,

Zn2+ (aq)

785

Figure 19.13 The electrochemical process involved in rust formation. The H + ions are supplied by H 2C0 3 , which forms when CO 2 from air dissolved in water.

Air Water

19.8 Corrosion

•

786

CHAPTER 19

Electrochemistry

Applying What You've Learned

•

The pain caused by having aluminum foil contact a dental filling results from the filling being made the cathode in an electrochemical cell. Another type of discomfort can result from the filling being made the anode in an electrochemical cell. This occurs when the filling touches a metal with a greater reduction potential than the components of the amalgam , such as gold. When an amalgam filling comes into contact with a gold inlay, the tin in the filling (the most easily oxidized of the major amalgam components) is oxidized creating an unpleasant metallic taste in the mouth. A simplified, unbalanced equation for the redox reaction that takes place is

•

Gold inlay I

",,) + 4H +(aq) + 4e -

• 2H20(l)

_. SnLT filling

Gold inlay touching amalgam dental filling I

Problems:

a)

Balance (in acidic medium) the equation for the oxidation of tin from an amalgam filling. [ ~. Sample Problem 19.1]

b)

Calculate the standard cell potential for the reaction in part (a). [ ~. Sample Problem 19.2]

c)

Which of the components of dental amalgam (mercury, silver, tin, copper, or zinc) would be oxidized when a filling is brought into contact with lead? [ ~. Sample Problem 19.3]

d)

Calculate the standard free-energy change for the reaction in part (a). [ ~. Sample Problem 19.4]

e)

Calculate the equilibrium constant at 25°C for the reaction in part (a). [ ~. Sample Problem 19 .5]

I

KEY WORDS

787

CHAPTER SUMMARY •

Section 19.1 o

o

o

Redox reactions are those in which oxidation numbers change. Halfreactions are the separated oxidation and reduction reactions that make up the overall redox reaction. Redox equations can be balanced via the half-reaction method, which allows for the addition of H 20 to balance 0, H + to balance H, and OH- for reactions taking place in basic solution.

Section 19.6 Batteries are portable, self-contained sources of electric energy consisting of galvanic cells o

Section 19.2 o

o

o

o

The electrode at which reduction occurs is called the cathode; the electrode at which oxidation occurs is called the anode.

o

o

Half-cell potentials are measured relative to the standard hydrogen electrode (SHE), the half-reaction for which has an arbitrarily defined standard reduction potential of zero.

o

o

E ~ell

is related to the standard free-energy change (C:..GO) and to the equilibrium constant, K. A positive E ~ell corresponds to a negative (C:..GO) value and a large K value.

o

o

Ecell under other than standard-state conditions is determined from E ~ell and the reaction quotient, Q, using the Nerizst equation.

We can calculate the amount of a substance produced in electrolysis if we know the current applied to the cell and the length of time for which it is applied.

Corrosion is the undesirable oxidation of metals. Corrosion can be prevented by coating the metal surface with paint, a less easily oxidized metal, or a more easily oxidized metal such as •

ZlllC. o

Section 19.5

Electrolysis is used to recharge lead storage batteries, separate compounds into their constituent elements, and separate and purify metals.

Section 19.8 o

Section 19.4

The voltage that must actually be supplied to drive a nonspontaneous redox reaction is greater than the calculated amount because of

overvoltage.

Section 19.3

o

Electrolysis is the use of electric energy to drive a nonspontaneous redox reaction. An electrochemical cell used for this purpose is called an electrolytic cell.

The difference in electric potential between the cathode and the anode is the cell potential (E cell )'

We use standard reduction potentials (EO) to calculate the standard cell voltage or standard cell potential (E ~eID.

Fuel cells are not really batteries but also supply electric energy via a

Section 19.7

o

o

or series of galvanic cells.

spontaneous redox reaction. Reactants must be supplied constantly for a fuel cell to operate.

An electrochemical cell in which a spontaneous chemical reaction generates a flow of electrons through a wire is called a galvanic cell. Half-reactions in a galvanic cell take place in separate compartments called half-cells. Half-cells contain electrodes in solutions and are connected via an external wire and by a salt bridge.

A concentration cell has the same type of electrode and the sa me ion in solution (at different concentrations) in the anode and C8 ihode compartments.

The use of a more easily oxidized metal is known as cathodic protection, wherein the metal being protected is made the cathode in a galvanic cell. Galvanization is the cathodic protection of iron or steel . . USlllg zmc.

I(EyWORDS Anode, 763

Corrosion, 784

Galvanic cell, 763

Overvoltage, 782

Battery, 777

Electrode, 763

Galvanization, 785

Salt bridge, 763

Cathode, 763

Electrolysis, 780

Half-cell, 763

Cell potential (Ecell)' 764

Electrolytic cell, 780

Half-reaction, 760

Standard hydrogen electrode (SHE), 765

Concentration cell, 775

Fuel cell, 778

Nemst equation, 774

• Standard reduction potential (EO), 765

788

CHAPTER 19

Electrochemistry

KEY EQUATIONS

•

19.1

E cell O -- E cathode O

19.2

I::..G

19.3

I::..Go =

-nFE ~ell

19.4

E OII

=

RT In K nF

19 . 5

o E cell

=

0.0592 V I n

19.6

E

=

EO- RT In Q nF

19.7

E

=

EO _ 0.0592 V log Q

=

ce

-

E anode O

-nFEceli

°b

ct

K

n

QUESTIONS AND PROBLEMS ,

Section 19.1: Balancing Redox Equations

Section 19.3: Standard Reduction Potentials

Problems

Review Questions

19.1

19.8

Discuss the spontaneity of an electrochemical reaction in terms of its standard emf (E~cll) '

19.9

After operating a Daniell cell (see Figure 19.1) for a few minutes, a student notices that the cell emf begins to drop. Why?

Balance the following redox equations by the half-reaction method: (a) (b) (c) (d) (e)

19.2

H 20 2 + Fe2+ • Fe3+ + H 20 (in acidic solution) • Cu2+ + NO + H 20 (in acidic solution) Cu + HN0 3 CN- + Mn04 • CNO - + Mn02 (in basic solution) Br2 • BrO} + Br - (in basic solution) S20 ~- + 12 • 1- + S40 ~ - (in acidic sol ution)

Problems 19.10

Calculate the standard emf of a cell that uses the Mg/Mg2 + and Cu/Cu 2+ half-cell reactions at 25 °C. Write the equation for the cell reaction that occurs under standard-state conditions.

19.11

Calculate the standard emf of a cell that uses Ag/Ag + and half-cell reactions. Write the cell reaction that occurs AllAI3+ . under standard-state conditions.

19.1?

Predict whether Fe3+ can oxidize 1- to 12 under standard-state conditions.

19.13

Which of the following reagents can oxidize H 20 to 0 2(g) under standard-state conditions: H +(aq) , Cl-(aq), CI 2(g), Cu2+ (aq), Pb2+ (aq), Mn0 4 (aq) (i n acid)?

19.14

Consider the following half-reactions:

Balance the following redox equations by the half-reaction method: (a) Mn2+ + H 20 2 - -.... Mn02 + H 20 (i n basic solution) (b) Bi(OH)3 + SnOi• SnO ~ - + Bi (in basic solution) (c) Cr20 ~ - + C20 ~ • Cr3+ + CO 2 (in acidic solution) (d) CIO} + Cl• Cl2 + CI0 2 (in acidic solution) • Bi3+ + Mn04 (in acidic solution) (e) Mn2 + + BiO}

Section 19.2: Galvanic Cells

Review Questions 19.3

19.4

\I

Define the following terms: anode, cathode, cell voltage, electromotive jorce, standard reduction potential. Describe the basic features of a gal vanic cell. Why are the two components of the cell separated from each other?

19.5

What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge?

19.6

What is a cell diagram? Write the cell diagram for a galvanic cell consisting of an Al electrode placed in a I M Al(N0 3)3 solution and an Ag electrode placed in aIM AgN0 3 solution.

19.7

What is the difference between the half-reactions discussed in redox processes in Chapter 4 and the half-cell reactions discussed in Section 19.2?

Mn04 (aq) + .8H +(aq) NO } (aq)

+ 5e -

+ 4H+(aq) + 3e -

- _ . Mn2+ (aq) • NO(g)

+ 4H 20(I)

+ 2H20(l)

Predict whether NO} ions will oxidize Mn 2+ to Mn04 under standard-state conditions.

19.15

Predict whether the following reactions would occur spontaneously in aqueous solution at 25 °C. Assume that the initial concentrations of dissolved species are all 1.0 M. (a) Ca(s) + Cd2+(aq) (b) 2Br-(aq) + Sn2+ (aq) (c) 2Ag(s) + Ni2 +(aq) (d) Cu +(aq) + Fe3+ (aq)

• Ca2+ (aq) + Cd(s) • Br2(l) + Sn(s) • 2Ag+(aq) + Ni(s) • Cu2+(aq) + Fe2+ (aq)

•

QUESTIONS AND PROBLEMS 19 .16

19.17

Which species in each pair is a better oxidizing agent under standard-state conditions: (a) Br2 or Au H , (b) H2 or Ag +, (c) Cd2+ or Cr H , (d) O 2 in acidic media or O 2 in basic media? Which species in each pair is a better reducing agent under standard-state conditions: (a) Na or Li, (b) H2 or 12> (c) Fe 2+ or Ag, (d) Br- or Co2+ ?

Problems

19.29

What is the potential of a cell made up of Z n/Zn2+ and ClLCU:half-ce lls at 25 °C if [Zn2+ ] = 0.25 M and [Cu 2 +] = 0.1 5 ,\f'?

19.30

Calculate EO, E, and I1G for the following cell reactions. (a) Mg(s) + Sn 2 +(aq ) . • Mg2+(aq) + Sn(s) [Mg2+] = 0.045 M, [Sn2+] = 0.035 M (b) 3Zn(s) + 2CrH (aq) . • 3Zn2+ (aq) + 2Cr(s) [CrH ] = 0.010 M, [Zn2+] = 0.0085 M

Section 19.4: Spontaneity of Redox Reactions Under Standard-State Conditions Review Questions 19.18

19.19

19.31

Use the information in Table 2.1, and calculate the Faraday constant. Write the equations rel ating I1Go and Kto the standard emf of a cell. Define all the terms.

789

Calculate the standard potential of the cell consisting of the ZnlZn 2 + half-cell and the SHE. What will the emf of the cell be if [Zn2+] = 0.45 M, PH = 2.0 atm, and [H+] = 1.8 M? 2

19.32

What is the emf of a cell consisting of a Pb2+/Pb half-cell and a PtlH+1H2 half-cell if [Pb2+] = 0.10 M, [H] = 0.050 M, and PH = 1.0 atm? 2

19.20

Compare the ease of measuring the equilibrium constant electrochemically with that by chemical means [see Equation (1 8.15)] .

19.33

Refening to the anangement in Figure 19.1, calculate the [Cu2+]/[Zn 2+] ratio at which the following reaction is spontaneous at 25 °C:

Problems

19.21

Cu(s)

What is the equilibrium constant for the fo llowing reaction at 25 °C?

19.34

The equilibrium constant for the reacti on

Mg(s) I Mg2+(0.24 M)

19.35

What is a battery? Describe several types of batteries.

19.36

Explain the differences between a primary galvanic cell-one that is not rechargeable-and a storage cell (for example, the lead storage battery), which is rechargeable.

19.37

Discuss the advantages and di sadvantages of fuel cells over conventional power plants in produci ng electricity.

+

Calculate I1Go and Kc fo r the following reactions at 25 °C: (a) Mg(s) + Pb2+(aq) • • Mg2+ (aq) + Pb(s) (b) 0 2(g) + 4H+(aq) + 4Fe2+(aq) . ' 2H 2 0 (l) (c) 2AI(s) + 3lz(s) • ' 2AIH (aq) + 6r-(aq)

II Mg2 +(0.53 M) I Mg(s)

Review Questions

Use the standard reduction potentials to find the equilibrium constant for each of the following reactions at 2YC: (a) Brz(l) + 21 -(aq) • ' 2Br- (aq) + 12 (s) 4 (b) 2Ce +(aq) + 2Cqaq) • • CI 2(g) + 2Ce 3+(aq) (c) 5Fe 2+(aq) + Mn04(aq) + 8H +(aq). ' Mn2+(aq) 4H 2 0 + 5FeH (aq)

19.24

+ Zn(s)

Section 19.6: Batteries

Sr(s) + Mg2+ (aq). ' Sr2+(aq) + Mg(s) is 2.69 X 10 12 at 25°C. Calculate EOfor a cell made up of Sr1Sr2+ and Mg/Mg2+ half-cells.

19.23

• Cu2+ (aq)

Calculate the emf of the following concentration cell:

Mg(s) + Zn2+(aq ) -.:.=:!:' Mg2+ (aq) + Zn(s) 19.22

+ Zn2+ (aq)

Problems 19.38

The hydrogen-oxygen fuel cell is described in Section 19.6. (a) What volume of H ig), stored at 25°C at a pressure of 155 atm , wo uld be needed to run an electric motor drawing a cunent of 8.5 A for 3.0 h? (b) What volume (in liters) of air at 25°C and 1.00 atm will have to pass into the cell per minute to run the motor? Assume that air is 20 percent O 2 by volume and that all the Oz is consu med in the cel!. The other components of air do not affect th e fuel-cell reaction s. Assume ideal gas behavior.

19.39

Calculate the standard emf of the pro pane fuel cell discussed on page 779 at 2YC, given that I1G 'f for propane is - 23.5 kJ/mo!.

+ 4Fe H (aq)

19.25

Under standard-state conditions, what spontaneo us reaction will . aqueous so Iuli.on among the IOns ' Ce4 + , C e'H , F e3+ , an d occur 1n Fe2+? Calculate I1Go and Kc for the reaction.

19. 26

Given that EO= 0.52 V for the reduction Cu +(aq) + e • Cu(s), calculate EO, I1Go, and K for the following reaction at 25 °C:

Section 19.7: Electrolysis Section 19.5: Spontaneity of Redox Reactions Under Conditions Other Than Standard State

Review Questions

Review Questions

19.40 What is the difference be een a galvanic cell (such as a Daniell . cell) and an electrolytic cell?

19. 27

19.41

What is Faraday's contribution to quantitative electrolysis?

19.42

Define the term over voltage. How does overvoltage affect electrolytic processes?

19.28

Write the Nernst equation, and explain all the terms. Write the Nernst equati on for the following processes at some I temperature T (a) Mg(s) + Sn2+(aq). (b) 2Cr(s) + 3Pb2+ (aq) •

+ Sn(s) 2CrH(aq) + 3Pb(s)

• Mg2+(aq) •

790

CHAPTER 19

Electrochemistry

Problems 19.43

19.55

The passage of a current of 0.750 A for 25.0 min deposited 0.369 g of copper from a CUS04 solution. From this information, calculate the molar mass of copper.

19.56

A quantity of 0.300 g of copper was deposited from a CUS04 solution by passing a current of 3.00 A through the solution for 304 s. Calculate the value of the Faraday constant.

19.57

In a certain electrolysis experiment, 1.44 g of Ag were deposited

The half-reaction at an electrode is Mg2+ (molten) + 2e- --~. Mg(s) Calculate the number of grams of magnesium that can be produced by supplying 1.00 F to the electrode.

19.44

. 19.45

19.46

19.47

Consider the electrolysis of molten barium chloride (BaCI2 ) . (a) Write the half-reactions. (b) How many grams of barium metal can be produced by supplying 0.50 A for 30 min?

• Considering only the cost of electricity, would it be cheaper to produce a ton of sodium or a ton of aluminum by electrolysis? If the cost of electricity to produce magnesium by the electrolysis of molten magnesium chloride is $155 per ton of metal, what is the cost (in dollars) of the electricity necessary to produce (a) 10.0 tons of aluminum, (b) 30.0 tons of sodium, and (c) 50.0 tons of calcium?

One of the half-reactions for the electrolysis of water is 2H20(l)

• 0 2(g) + 4H+(aq) + 4e -

If 0.076 L of O2 is collected at 25 °C and 755 mmHg, how many faradays of electricity had to pass through the solution ?

19.48

19.49

19.50

in one cell (containing an aqueous AgN0 3 solution), while 0.120 g of an unknown metal X was deposited in another cell (containing an aqueous XCl3 solution) in series with the AgN0 3 cell. Calculate the molar mass of X.

How many faradays of electricity are required to produce (a) 0.84 L of O2 at exactly 1 atm and 25°C from aqueous H 2S0 4 solution, (b) 1.50 L of Cl2 at 750 mmHg and 20°C from molten NaCl, and (c) 6.0 g of Sn from molten SnCI 2?

19.58

If 0.845 L of H2 is collected at 25°C and 782 mmHg, how many faradays of electricity had to pass through the solution?

Section 19.8: Corrosion

Review Questions 19.59

Steel hardware, including nuts and bolts, is often coated with a thin plating of cadmium. Explain the function of the cadmium layer.

19.60

"Galvanized iron" is steel sheet that has been coated with zinc; "tin" cans are made of steel sheet coated with tin. Discuss the functions of these coatings and the electrochemistry of the corrosion reactions that occur if an electrolyte contacts the scratched surface of a galvanized iron sheet or a tin can.

19.61

Tarnished silver contains Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte solution, such as NaCI. Explain the electrochemical principle for this procedure. [The standard reduction potential for • 2Ag(s) + S2 -(aq) is the half-cell reaction Ag 2S(s) + 2e-0.7 1 V.]

19.62

How does the tendency of iron to rust depend on the pH of the solution?

Calculate the amounts of Cu and Br2 produced in 1.0 h at inert electrodes in a solution of CuBr2 by a current of 4 .50 A.

In the electrolysis of an aqueous AgN0 3 solution, 0.67 g of Ag is deposited after a certain period of time. (a) Write the half-reaction for the reduction of Ag +. (b) What is the probable oxidation half-reaction? (c) Calculate the quantity of electricity used (i n coulombs).

19.51

19.52

19.53

19.54

A steady current was passed through molten CoS04 until 2.35 g of metallic cobalt was produced. Calculate the number of coulombs of electricity used. A constant electric current flows for 3.75 h through two electrolytic cells connected in series. One contains a solution of AgN0 3 and the second a solution of CuCI 2. During this time, 2.00 g of silver is deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing (in amperes)? What is the hourly production rate of chlorine gas (in kg) from an electrolytic cell using aqueous NaCI electrolyte and carrying a current of 1.500 X 10 3 A? The anode efficiency for the oxidation of Cl- is 93.0 percent.

Additional Problems 19.63

For each of the following redox reactions, (i) write the halfreactions, (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions: • H +(aq) + Ni(s) (b) Mn04 (aq) + Cl - (aq) • Mn2+ (aq) + CI 2(g) (in acid

(a) H 2(g)

+ Ne+(aq)

solution) (c) Cr(s) + Zn2+(aq)

--+.

CrH(aq)

+ Zn(s)

19.64

The oxidation of 25.0 mL of a solution containing Fe2+ requires 26.0 mL of 0.0250 M K2Cr207 in acidic solution. Balance the following equation, and calculate the molar concentration of Fe2 +:

19.65

The S02 present in air is mainly responsible for the phenomenon of acid rain. The concentration of S02 can be determined by titrating against a standard permanganate solution as follows:

Chromium plating is applied by electrolysis to objects suspended in a dichromate solution, according to the following (unbalanced) half-reaction:

How long (in hours) would it take to apply a chromium plating 1.0 X 10- 2 mm thick to a car bumper with a surface area of 0.25 m 2 in an electrolytic cell carrying a current of 25.0 A? (The density of chromium is 7.1 9 g/cm 3 .)

One of the half-reactions for the electrolysis of water is

5S02 + 2Mn04 +2H 20

• 5S0~ - + 2Mn2+ + 4H+

Calculate the number of grams of S02 in a sample of air if 7.37 mL of 0.00800 M KMn04 solution is required for the titration .

QUESTIONS AND PROBLEMS

19.66

19.67

A sample of iron ore weighing 0.2792 g was dissolved in an excess of a dilute acid solution. All the iron was first converted to Fe(II) ions. The solution then required 23.30 mL of 0.0194 M KMn04 for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

19.75

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation:

Calcium oxalate (CaC 20 4) is insoluble in water. This property has been used to determine the amount of Ca2+ ions in blood. The calcium oxalate isolated from blood is dissolved in acid and titrated against a standardized KMn04 solution as described in Problem 19.68. In one test it is found that the calcium oxalate isolated from a 1O.0-mL sample of blood requires 24.2 mL of 9.56 X 10- 4 M KMn04 for titration. Calculate the number of milligrams of calcium per milliliter of blood.

19.70

Complete the following table. State whether the cell reaction is spontaneous, nonspontaneous, or at equilibrium.

E

D.G

19.78

Describe what you would observe at the anode and the cathode. (Hint: Molecular iodine is only slightly soluble in water, but in the presence of r ions, it forms the brown color of 1:3 ions. See Problem 13.114.)

> 0 =0

19.72

19.73

19.74

An aqueous KI solution to which a few drops of phenolphthalein have been added is electrolyzed using an apparatus like the one shown here:

Cell Reaction

> 0

19.71

II soln B I Hg(l)

where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained 2.63 g mercury(I) nitrate per liter. If the measured emf of such a cell is 0.0289 Vat 18°C, what can you deduce about the nature of the mercury(I) ions?

• Mn2+ + CO 2

19.69

• ZnMn204(S)

For a number of years, it was not clear whether mercury(I) ions existed in solution as Hg + or as Hg~ + . To distinguish between these two possibilities, we could set up the following system:

Hg(l) I soln A

Oxalic acid (H2C20 4) is present in many plants and vegetables. (a) Balance the following equation in acid solution:

(b) If a 1.00-g sample of plant matter requires 24.0 mL of 0.0100 M KMn04 solution to reach the equivalence point, what is the percent by mass of H2C 20 4 in the sample?

+ Zn2+ (aq) + 2e-

If a Leclanche cell produces a current of 0.0050 A, calculate how many hours this current supply will last if there is initially 4.0 g of Mn02 present in the cell. Assume that there is an excess of Zn 2+ ions.

(a) Balance this equation. (b) If 36.44 mL of a 0.01652 M KMn04 solution is required to completely oxidize 25.00 mL of an H 20 2 solution, calculate the molarity of the H 20 2 solution.

Mn04 + C20~-

II Cu2+ (1.2 M) I Cu(s)

The cathode reaction in the Leclanche cell is given by 2Mn02(S)

19.77

19.68

Calculate the emf of the following concentration cell at 25 °C: Cu(s) I Cu2+ (0.080 M)

19.76

791

From the following information, calculate the solubility product of AgBr: Ag+(aq)

+ e- - _ . Ag(s)

eo =

AgBr(s)

+ e-

EO= 0.07V

• Ag(s) + Br - (aq)

0.80V

Consider a galvanic cell composed of the SHE and a half-cell • Ag(s). (a) Calculate the using the reaction Ag +(aq) + e standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell potential when [H+] in the hydrogen electrode is changed to (i) 1.0 X 10- 2 M and (ii) 1.0 X 10- 5 M, all other reagents being held at standard-state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter. A galvanic cell consists of a silver electrode in contact with 346 mL of 0.100 M AgN0 3 solution and a magnesium electrode in contact with 288 mL of 0.100 M Mg(N0 3h solution. (a) Calculate E for the cell at 25°C. (b) A current is drawn from the cell until 1.20 g of silver has been deposited at the silver electrode. Calculate E for the cell at this stage of operation. Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of NaCI but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

19.79

A piece of magnesium metal weighing 1.56 g is placed in 100.0 mL of 0.100 M AgN0 3 at 25 °C. Calculate [Mg2+] and [Ag +] in solution at equilibrium. What is the mass of the magnesium left? The volume remains constant.

19.80

Describe an experiment that would enable you to detennine which is the cathode and which is the anode in a galvanic cell using copper and zinc electrodes. •

19.81

An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g after 1.52 X 10 3 s. (a) What is the gas produced at the cathode, and what is its volume at STP? (b) Given that the charge of an electron is 1.6022 X 10- 19 C, calculate Avogadro's number. Assume that copper is oxidized to Cu2+ ions.

19.82

In a certain electrolysis experiment involving AI3+ ions, 60.2 g of AI is recovered when a current of 0.352 A is used. How many minutes did the electrolysis last?

19.83

Consider the oxidation of ammonia:

(a) Calculate the D.Go for the reaction. (b) If this reaction were used in a fuel cell, what would the standard cell potential be?

•

792

CHAPTER 19

Electrochemistry

19.84

When an aq ueous solution contairting gold(III) salt is electrolyzed, metallic gold is deposited at the cathode and oxygen gas is generated at the anode. (a) If 9.26 g of Au is deposited at the cathode, calculate the volume (in liters) of O 2 generated at 23 °C and 747 mmHg. (b) What is the current used if the electrolyti c process took 2.00 h?

19.94

Gold will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. However, the metal does dissolve in a mixture of the acids (one part HN0 3 and three parts HCI by volume), called aqua regia. (a) Write a balanced equation for this reaction. (Hint: Among the products are HAuCl4 and NO z.) (b) What is the function of HCl?

19.85

In an electrolysis experiment, a student passes the same quantity of electricity through two electrolytic cells, one containing a sil ver salt and the other a gold salt. Over a certain period of time, the student finds that 2.64 & of Ag and 1.61 g of Au are deposited at the cathodes. What is the oxidation state of gold in the gold salt?

19.95

Explain why most useful galvanic cells give voltages of no more than 1.5 to 2.5 V What are the prospects for developing practical galvanic cells with voltages of 5 V or more?

19.96

A silver rod and a SHE are dipped into a saturated aqueous solution of si lver oxalate (Ag 2C 20 4), at 25 °C. The measured potential difference between the rod and the SHE is 0.589 V, the rod being positive. Calculate the solubility product constant for silver oxalate.

19.97

Zinc is an amphoteric metal; that is, it reacts with both acids and bases. The standard reduction potential is -1.36 V for the reaction

19.86

19.87

People livi ng in cold-climate countries where there is plenty of snow are advised not to heat their garages in the winter. What is the electrochemical basis for this recommendation ? Given that 2Hg2+(aq)

Hg~+(aq)

+ 2e+ 2e-

------. Hg ~ + (aq) • 2Hg(l)

EO= 0.92 V EO= 0 .85 V

Calculate the formation constant (Kr) for the reaction

calculate GOand K for the following process at 25°C: Hg ~ + (aq)

19.88

•

• Hg2+(aq)

Use the data in Table 19.1 to determine whether or not hydrogen peroxide will undergo disproportionation in an acid medium: • 2H zO + Oz· 2H zO z

Fluorine (F2) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containin g potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process. (b) What is the purpose of KF? (c) Calculate the volume of F2 (in liters) collected at 24.0°C and 1.2 atm after electrolyzing the solution for 15 h at a current of 502 A.

19.99

The magnitudes (but not the signs) of the stand ard reducti on potentials of two metals X and Y are

19.90

Industrially, copper is purified by electrolysis . The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a CUS04 solution. During electrolysis, copper at the anode enters the solution as Cu2+ while Cu2+ ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrol ytic process . (b) Suppose the anode was co ntaminated with Zn and Ag. Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain 1.00 kg of Cu at a current of 18.9 A ?

19.93

• Zn(OH) ~ -(aq)

19.98

A 300-mL solution of NaCl was electrolyzed for 6.00 min. If the pH of the final solution was 12.24, calculate the average current used.

19.92

+ 40W(aq)'

(The precedi ng reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced. )

19.89

19.91

Z n2+ (aq)

+ Hg(l)

An aqueous solution of a platinum salt is electrolyzed at a current of 2.50 A for 2.00 h. As a result, 9.09 g of metallic Pt is formed at the cathode. Calculate the charge on the Pt ions in this solution. Consider a galvanic cell co nsisting of a magnesium electrode in contact with 1.0 M Mg(N0 3)2 and a cadmium electrode in contact with 1.0 M Cd(N0 3h Calculate EOfor the cell, and draw a diagram showing th e cathode, anode, and direction of electron flow. A current of 6.00 A passes through an electrolytic cell containing dilute sulfuric acid for 3.40 h. If the volume of O 2 gas generated at the anode is 4.26 L (at STP), calculate the charge (in co ulombs) on an electron.

y 2+

+ 2e -

_ _~. Y

X2+ + 2e - ---+. X

I EO1= 0.34 V I EOI = 0.25 V

where the II notation denotes that onl y the magnitude (but not the sign) of the EOvalue is shown. When the half-cells of X and Y are connected, electrons flow from X to Y. When X is connected to a SHE, electrons flow from X to SHE. (a) Are the EOvalues of the half-reactions positive or negative? (b) What is the standard emf of a cell made up of X and Y? 19.100 A galvanic cell is constructed as follows. One half-cell consists of a platinum wire immersed in a solution containing 1.0 M Sn 2 + H and 1.0 M Sn ; the other half-cell has a thallium rod immersed in a solution of 1.0 M TI + (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at 25°C? (c) What is the cell voltage if the TI + concentration is increased 10-fo1d? (EOTtm = -0.34 V)

19.101 Given the standard reduction potential for Au 3 + in Table 19.1 and Au + (aq)

+ e-

• Au(s)

EO= 1.69 V

answer the fo llowing questions. (a) Why does gold not tarnish in air? (b) Will the following disproportionation occur spon taneously? 3Au+(aq)

• Au3+ (aq)

+ 2Au(s)

(c) Predict the reaction between gold and fluorine gas. 19.102 The ingestion of a very small quantity of mercury is not considered too harmful. Would this statement still hold if the gastric juice in your stomach were mostly nitric acid instead of hydrochloric acid?

QUESTIONS AND PROBLEMS

19.103 When 25.0 mL of a solution containing both Fe2 + and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMn04 (in dilute sulfuric acid), all the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all the Fe3 + ions to Fe2+ ions. Finally, 40.0 mL of the same KMn04 solution is added to the solution in order to oxidize the Fe2+ ions to Fe3+ . Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution. 19.104 Consider the Daniell cell in Figure 19.1. When viewed externally, the anode appears negative and the cathode positive (electrons are flowing from the anode to the cathode). Yet in solution anions are moving toward the anode, which means that it must appear positive to the anions. Because the anode cannot simultaneously be negative and positive, give an explanation for this apparently contradictory situation .

19.105 Use the data in Table 19.1 to show that the decomposition of H20i (a disproportionation reaction) is spontaneous at 25 °C:

19.106 The concentration of sulfuric acid in the lead-storage battery of an automobile over a period of time has decreased from 38.0 percent by mass (density = 1.29 g/mL) to 26.0 percent by mass (1.19 g/mL). Assume the volume of the acid remains constant at 724 mL. (a) Calculate the total charge in coulombs supplied by the battery. (b) How long (in hours) will it take to recharge the battery back to the original sulfuric acid concentration using a current of 22.4 A?

793

19.113 The zinc-air battery shows much promise for electric cars because it is lightweight and rechargeable:

·r?1Air cathode 1
ZnO(s). (a) Write the half-reactions at the zinc-air electrodes, and calculate the standard emf of the battery at 25 °C. (b) Calculate the emf under actual operating conditions when the partial pressure of oxygen is 0.21 atm. (c) What is the energy density (measured as the energy in kilojoules that can be obtained from I kg of the metal) of the zinc electrode? (d) If a CUlTent of 2.1 X 105 A is to be drawn from a zinc-air battery system, what volume of air (in liters) would need to be supplied to the battery every second? Ass ume that the temperature is 25 °C and the partial pressure of oxygen is 0.21 atm. 19.114 Calculate EOfor the reactions of mercury with (a) 1 M HCI and (b) I M HN0 3 . Which acid will oxidize Hg to Hg~ + under standard-state conditions? Can you identify which pictured test tube contains HN0 3 and Hg and which contains HCI and Hg?

19.107 Consider a Daniell cell operating under non-standard-state conditions. Suppose that the cell's reaction is mUltiplied by 2. What effect does this have on each of the following quantities in the Nernst equation: (a) E, (b) EO, (c) Q, (d) In Q, (e) n? 19.108 A spoon was silver-plated electrolytically in an AgN0 3 solution . (a) Sketch a diagram for the process. (b) If 0.884 g of Ag was deposited on the spoon at a constant cunent of 18.5 rnA , how long (in min) did the electrolysis take?

19.109 Comment on whether F2 will become a stronger oxidizing agent with increasing H+ concentration.

19.115 Because all alkali metals react with water, it is not possible to measure the standard reduction potentials of these metals directly as in the case of, say, zinc. An indirect method is to consider the following hypothetical reaction: Li +(aq)

19.110 In recent years there has been much interest in electric cars. List some advantages and disadvantages of electric cars compared to automobiles with internal combustion engines.

19.111 Calculate the pressure of H2 (in atm) required to maintain equilibrium with respect to the following reaction at 25 °C, Pb(s) given that [Pb2+] 1.60.

+ 2H +(aq). =

>

Pb2+ (aq )

+ Hi

g)

0 .035 M and the solution is buffered at pH

19.112 A piece of magnesium ribbon and a copper wire are partially immersed in a 0.1 M HCI solution in a beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at both the Mg and Cu surfaces. (a) Write equations representing the reactions occuning at the metals. (b) What visual evidence would you seek to show that Cu is not oxidized to Cu2+ ? (c) At some stage, NaOH solution is added to the beaker to neutralize the HCI acid. Upon further addition of NaOH, a white precipitate form s. What is it?

+

m ig )

>

Li(s)

+ H +(aq)

Using the appropriate equation presented in this chapter and the thermodynamic data in Appendix 2, calculate EOfor Li+(aq) +e > Li(s) at 298 K. Use 96,485 .338 C/mol e - for the Faraday constant. Compare your result with that listed in Table 19.1. 19.116 A gal vanic cell using Mg/Mg2+ and Cu/Cu2+ half-cells operates under standard-state conditions at 25 °C, and each compartment has a volume of 218 mL. The cell delivers 0.22 A for 31.6 h. (a) How many grams of Cu are deposited ? (b) What is the [Cu2+ ] remaining?

19.117 Given the follow ing standard reduction potentials, calculate the ion-product, Kw, for water at 25°C:

•

+ 2e - ---+> H 2 (g) 2H 20 (l ) + 2e• H 2(g ) + 20W(aq)

2H+(aq)

EO= O.OOV EO= -0.83 V

19.118 Compare the pros and cons of a fuel cell, such as the hydrogenoxygen fuel cell, and a coal-fired power station for generating electricity.

794

CHAPTER 19

Electrochemistry

19.119 Lead storage batteries are rated by ampere-hours, that is, the number of amperes they can deliver in an hour. (a) Show that 1 Ah = 3600 C. (b) The lead anodes of a certain lead-storage battery have a total mass of 406 g. Calculate the maximum theoretical capacity of the battery in ampere-hours. Explain why in practice we can never extract this much energy from the battery. (Hint: Assume all the lead will be used up in the electrochemical reaction, and refer to the electrode reactions on page 777.) (c) Calculate E ~ell and AGOfor the battery. 19.120 Use Equations 18.11 and 19.3 to calculate the emf values of the Daniell cell at 25 °C and 80°C. Comment on your results. What assumptions are used in the derivation? (H int: You need the thermodynamic data in Appendix 2.) 19.121 A construction company is installing an iron culvert (a long cylindrical tube) that is 40.0 m long with a radius of 0.900 m. To prevent corrosion, the culvert must be galvanized. This process is carried out by first passing an iron sheet of appropriate dimensions through an electrolytic cell containing Zn2+ ions, using graphite as the anode and the iron sheet as the cathode. If the voltage is 3.26 V, what is the cost of electricity for depositing a layer 0.200 mm thick if the efficiency of the process is 95 percent? The electricity rate is $0.12 per kilowatt hour (kWh), where 1 W = 1 l is and the density of Zn is 7.1 4 g/cm3 19.122 A 9.00 X 102 mL amount of 0.200 M MgI2 solution was electrolyzed. As a result, hydrogen gas was generated at the cathode and iodine was formed at the anode. The volume of hydrogen collected at 26°C and 779 mmHg was 1.22 X 103 mL. (a) Calculate the charge in coulombs consumed in the process. (b) How long (in min) did the electrolysis last if a current of 7.55 A was used ? (c) A white precipitate was formed in the process. What was it, and what was its mass in grams? Assume the volume of the solution was constant.

19.123 Based on the following standard reduction potentials, Fe 2 +(aq) + 2e -

• Fe(s)

E'l = -0.44 V E~

= 0.77 V

calculate the standard reduction potential for the half-reaction FeH (aq)

+ 3e -

• Fe(s)

E '3 =?

19.124 To remove the tarnish (Ag2S) on a silver spoon, a student carried out the following steps. First, she placed the spoon in a large pan filled with water so the spoon was totally immersed. Next, she added a few tablespoonfuls of baking soda (sodium bicarbonate), which readily dissolved. Finally, she placed some aluminum foil at the bottom of the pan in contact with the spoon and then heated the solution to about 80°C. After a few minutes, the spoon was removed and rinsed with cold water. The tarnish was gone, and the spoon regained its original shiny appearance. (a) Describe with equations the electrochemical basis for the procedure. (b) Adding N aCI instead of N aHC0 3 would also work because both compounds are strong electrolytes. What is the added advantage of using NaHC0 3? (Hint: Consider the pH of the solution.) (c) What is the purpose of heating the solution? (d) Some commercial tarnish removers contain a fluid (or paste) that is a dilute HCI solution. Rubbing the spoon with the fluid will also remove the tarnish. Name two disadvantages of using this procedure compared to the one described previously. 19.125 Calculate the equilibrium constant for the following reaction at 298 K: Zn(s)

+ Cu2+(aq) +.==' Zn2+(aq) + Cu(s)

19.126 Cytochrome-c is a protein involved in biological electron transfer processes. The redox half-reaction is shown by the reduction of the Fe 3+ ion to the Fe2+ ion: EO = 0.254 V cyt c(Fe 3+) + e - --+. cyt c(Fe2+) Calculate the number of moles of cyt c(FeH ) formed from cyt c(Fe2+) with the Gibbs free energy derived from the oxidation of I mole of glucose.

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES A galvanic cell is constructed by immersing a piece of copper wire in 25 .0 mL of a 0.20 M CUS04 solution and a zinc strip in 25.0 mL of a 0 .20 M ZnS04 solution. Cu2+ ions react with aqueous NH3 to form the complex ion Cu(NH3)~+: Cu2+ (aq)

1.

+

4NH 3(aq)

U si ng the eq uati on

calculate the emf of the cell at 25 °C. a) 0.0 V b) 1.10 V c) 0.90 V d) 1.30 V

What would happen if a small amount of concentrated NH3 solution were added to the CUS04 solution? a) Nothing. b) Emf would increase. c) Emf would decrease. d) Not enough information to determine.

What would happen if a small amount of concentrated NH3 solution were added to the ZnS04 solution? a) Nothing. b) Emf would increase. c) Emf would decrease. d) Not enough information to determine.

--+. Cu(NH3)~+

E = EO_ 0.0592 V lou Q n b

2.

3.

4.

In a separate experiment, 25.0 mL of 3.00 M NH3 is added to the CUS04 solution . If the emf of the cell is 0.68 V at equilibrium, calculate the formation constant (K r) of Cu(NH3) ~+ ' a) 9.4 X 1022 b) 1.1 X 10- 23 c) 1.5 X 10- 14 d) 1.5 X 10 14

ANSWERS TO IN-CHAPTER MATERIALS

795

ANSWERS TO IN-CHAPTER MATERIALS Answers to Practice Problems

Answers to Checkpoints

19.1A Mn04 + 5Fe2+ + 8H+ • Mn2+ + 5Fe3+ + 4H20. 19.1B 2Mn04 + H 2 0 + 3CN• 2Mn02 + 20H- + 3CNO-. 19.2A Cd + Pb2+ • Cd2+ + Pb, E ~ell = 0.27 V. 19.2B 2Al + 3Cu2+ • 2AI3+ + 3Cu, E ~ell = 2.00 V. 19.3A (a) No reaction, • H2 + Pb H 19.3B (a) No reaction, (b) 2H+ + (b) 2H+ + Pb Zn • H2 + Zn2+. 19.4A - 411 kJ/mol. 19.4B 23.2 kJ/mol. 19.5A 1 X 10- 42 . 19.5B 3 X 10- 45 . 19.6A Yes, the reaction is spontaneous. 19.6B 2.2 M. 19.7A 7.44 g Mg. 19.7B 0.96 A.

19.1.1 a, c, e. 19.1.2 c. 19.3.1 d. 19.3.2 c. 19.3.3 b. 19.3.4 e. 19.4.1 b. 19.4.2 b. 19.5.1 b. 19.5.2 b. 19.7.1 a. 19.7.2 e.

Answers to Applying What You've Learned a) 2Sn + O 2 + 4H + • 2Sn2+ + 2H 20 . b) 1.37 V. c) Tin and zinc. d) - 5.3 X 102 kJ/ mol. e) 4 X 1092

•

lie ear

20.1 20.2 • •

20.3 • •

2004 20.5 20.6 20.7 •

•

20.8

Nuclei and Nuclear Reactions

emlstr

,

Nuclear Stability

,', •

Patterns of Nuclear Stability Nuclear Binding Energy

Natural Radioactivity

•

•

-

Kinetics of Radioactive Decay Dating Based on Radioactive Decay

•

Nuclear Transmutation Nuclear Fission Nuclear Fusion

'.

Uses of Isotopes

,

Chemical Analysis Isotopes in Medicine

Biological Effects of Radiation

•

•

•

•

•

Nuclear Chemistry in the Treatment of Cancer Brain tumors are some of the most difficult cancers to treat because the site ofthe malignant growth makes surgical excision difficult or impossible. Likewise, conventional radiation therapy using X rays or 'Y rays from outside the skull is usually not effective. An ingenious approach to this problem is boron neutron capture therapy (BNCT). This technique involves first administering a boron-lO compound that is selectively taken up by tumor cells and then applying a beam of low-energy neutrons to the tumor site. lOB captures a neutron to produce lIB, which disintegrates via the following nuclear reaction: l~ B

+ bn-_. j Li + ia

The highly energetic particles produced by this reaction destroy the tumor cells in which the lOB is concentrated. Because the particles are confined to just a few micrometers, they preferentially destroy tumor cells without damaging neighboring normal cells. BNCT is a highly promising treatment and is an active area of research. One of the major goals of the research is to develop suitable compounds to deliver lOB to the desired site. For such a compound to be effective, it must meet several criteria. It must have a high affinity for tumor cells, be able to pass through membrane barriers to reach the tumor site, and have minimal toxic effects on the human body. Thi s is one example of how nuclear chemistry is impOitant in the treatment of cancer.

In Th is Chapter, You Will Learn

some of the fundamentals of nuclear chemistry and how nuclear reactions are

important to living systems and to society.

Before you begin, you should review •

Radioactivity

[~~

Section 2 .2]

•

Atomic number, mass number, and isotopes

•

First-orderkinetics [~

[~~

Section 2.3]

Section 14.3]

Media Playerl MPEG Content Chapter in Review

Th e CT scan shows a brain tumor that might be difficult or imp ossible to treat by conventional surgical methods. Nuclear medicine, including boron neutron capture therapy (BNCT), enables doctors to treat cancers of this type. 79 7

798

CHAPTER 20

Nuclear Chemistry

Nuclei and Nuclear Reactions

•

With the exception of hydrogen cl B ), all nuclei contain protons and neutrons. Some nuclei are unstable and undergo radioactive decay, emitting particles and/or electromagnetic radiation [ ~~ Section 2.2]. Spontaneous emission of particles or electromagnetic radiation is known as radioactivity. All elements having an atomic number greater than 83 are unstable and are therefore radioactive. Polonium-210 e~~ po), for example, decays spontaneously to Pb by emitting an a particle. Another type of nuclear process, known as nuclear transmutation, results from the bombardment of nuclei by neutrons, protons, or other nuclei. An example of a nuclear transmutation is the conversion of atmospheric IjN to I ~ C and IB, which results when the nitrogen isotope is bombarded by neutrons (from the sun). In some cases, heavier elements are synthesized from lighter elements. This type of transmutation occurs naturally in outer space, but it can also be achieved artificially, as we will see in Section 20.4. Radioactive decay and nuclear transmutation are nuclear reactions, which differ significantly from ordinary chemical reactions. Table 20.1 summarizes the differences. To discuss nuclear reactions in any depth, we must understand how to write and balance nuclear equations. Writing a nuclear equation differs somewhat from writing equations for chemical reactions. In addition to writing the symbols for the various chemical elements, we must also explicitly indicate the number of subatomic particles in every species involved in the reaction. The symbols for subatomic particles are as follows:

°

lB or lp

bn

_ ~e or - ~f3

+ I e or + 0f3 I

proton

neutron

electron

positron

iaor i Be CI.

particle

In accordance with the notation introduced in Section 2.3 , the superscript in each case denotes the mass number (the total number of neutrons and protons present) and the subscript is the atomic number (the number of protons). Thus, the "atomic number" of a proton is 1, because there is one proton present, and the "mass number" is also 1, because there is one proton but no neutrons present. On the other hand, the mass number of a neutron is 1, but its atomic number is zero, because there are no protons present. For the electron, the mass number is zero (there are neither protons nor neutrons present), but the atomic number is - 1, because the electron possesses a unit negative charge. The symbol represents an electron in or from an atomic orbital. The symbol _?f3, on the other hand, represents an electron that, although physically identical to any other electron, comes from a nucleus (in a decay process in which a neutron is converted to a proton and an electron) and not from an atomic orbital. The positron has the same mass as the electron, but bears a charge of + 1. The a particle has two protons and two neutrons, so its atomic number is 2 and its mass . .. ... . number is 4. In balancing any nuclear equation, we must balance the total of all atomic numbers and the total of all mass numbers for the products and reactants. If we know the atomic numbers and mass numbers of all but one of the species in a nuclear equation, we can identify the unknown species by applying these rules, as shown in Sample Problem 20.1.

_?e

An CI. particle is identical to a helium-4 nucleus and can be represented either as ju or j He.

Chemical Reactions

Nuclear Reactions

1. Atoms are rearranged by the breaking and fOIlning of chemical bonds.

1. Elements are converted to other elements (or isotopes).

2. Only electrons in atomic or molecular orbitals are involved in the reaction.

2. Protons, neutrons, electrons, and other subatomic particles such as a particles may be involved.

3. Reactions are accompanied by the absorption or release of relatively small amounts of energy.

3. Reactions are accompanied by the absorption or release of tremendous amounts of energy.

4. Rates of reaction are influenced by temperature, pressure, concentration, and catalysts.

4. Rates of reaction normally are not affected by temperature, pressure, or catalysts.

SECTION 20.2

Nuclear Stability

Sample Problem 20.1 . Identify the missing species X in each of the following nuclear equations: (a) 2~~ PO

• 2~~ Pb

(b) ~~ Sr

+ -?f3 • I~O + +?f3

(c)

X

+X

"X

Strategy Determine the mass number for the unknown species, X, by summing the mass numbers on both sides of the equation: ~

reactant mass numbers = ~ product mass numbers

Similarly, determine the atomic number for the unknown species: ~

reactant atomic numbers = ~ product atomic numbers

Use the mass number and atomic number to determine the identity of the unknown species. Setup (a) 212 = (208 + mass number of X); mass number of X = 4. 84 = (82 X); atomic number of X = 2.

+ atomic number of

(b) 90 = (mass number of X atomic number of X = 39.

+ 0) ; mass number of X = 90. 38 = [atomic number of X + (-1 )];

(c) Mass number of X = (18 number of X = 9.

+ 0); mass number of X = 18. Atomic number of X = (8 + 1); atomic Think About It The rules of summation we apply to balance nuclear equations can be thought of as the conservation of mass number and the conservation of atomic number.

ia: 2~~ PO - _I 2~~ Pb + ia

Solution (a) X =

(b) X = ~~ Y: ~~ Sr

" ~~ Y

+ -?f3

(c) X = I ~F: I~ F

• I~ O

+

+?f3

Practice Problem A Identify X in each of the following nuclear equations:

(a) ~~ As

•X

+ -?f3

(b)jH + iHe

·X

(c) T6~ Fm - -" T 6ZFm

+X

Practice Problem B Identify X in each of the following nuclear equations: (a) X

-?f3

+

(b) 2~~ U (c) X

• 2~PU "X

+ i He

" li N +

_?e

Checkpoint 20.1 20.1.1

Nuclei and Nuclear Reactions

Identify the species X in the following nuclear equation: 2~~ Rn

• X +

20.1.2

Identify the species X in the following nuclear equation:

I~O

ia

a) 2~~ PO

a) I~ F

b) 2~~ Ra

b) I~ O

c) 2~~ PO

c) I~F

d) 2~~PO

d) li N

e) 2~~ Ra

e) li N

• X

+ -?f3

Nuclear Stability The nucleus occupies a very small portion of the total volume of an atom, but it contains most of the atom's mass because both the protons and the neutrons reside there. In studying the stability of the atomic nucleus, it is helpful to know something about its density, because it tells us how tightly the particles are packed together. As a sample calculation, let us assume that a nucleus has

•

799

800

. CHAPTER 20

Nuclear Chemistry

a radius of 5 X 10- 3 pm and a mass of 1 X 10- 22 g. These figures conespond roughly to a nucleus containing 30 protons and 30 neutrons, Density is mass/volume, and we can calculate the volume from the known radius (the volume of a sphere is ~'ITr3, where r is the radius of the sphere). First 3 we convert the picometer units to centimeters. Then, we calculate the density in g/cm : r = (5 X 10- 3 pm) 1

. density =

•

-~~

1 X 10 -- g

mass volume

=2

X

\1~:2 m (1~0 ~m ) = 5 X 10-

= - - -3-~ 'ITr

13

cm

1 X 10- 22 g ~ 'IT(5 X 10-

13

cm)3

10 14 g/cm 3

This is an exceedingly high density. The highest density known for an element is 22.6 g/cm3, for 12 osmium (Os). Thus, the average atomic nucleus is roughly 9 X 10 (or 9 trillion) times as dense as the densest element known! The enormously high density of the nucleus means that some very strong force is needed to hold the particles together so tightly. From Coulomb's law we know that like charges repel and unlike charges attract one another. We would thus expect the protons to repel one another strongly, particularly when we consider how close they must be to each other. This indeed is so. However, in addition to the repulsion, there are also short-range attractions between proton and proton, proton and neutron , and neutron and neutron. The stability of any nucleus is determined by the difference between coulombic repulsion and the shOlt-range attraction. If repulsion outweighs attraction, the nucleus disintegrates, emitting particles and/or radiation, If attractive forces prevail, the nucleus is stable.

Patterns of Nuclear Stability The principal factor that determines whether a nucleus is stable is the neutron-to-proton ratio (nip). For stable atoms of elements having low atomic number «20), the nip value is close to 1. As the atomic number increases, the neutron-to-proton ratios of the stable nuclei also increase. This deviation at higher atomic numbers arises because more neutrons are needed to counteract the strong repulsion among the protons and stabilize the nucleus, The following rules are useful in gauging whether or not a particular nucleus is expected to be stable:

.. The significance of these numbers for nuclear stability is similar to the numbers of electrons associated with the very stable noble gases (i.e., 2, 10, 18, 36, 54, and 86 electrons).

1. There are more stable nuclei containing 2, 8, 20, 50, 82, or 126 protons or neutrons than there are containing other numbers of protons or neutrons. For example, there are 10 stable isoof tin (Sn) .with the atomic number 50 and isotopes of antimony (Sb) with topes . . . . .. .. . .. . . .. . . only ... .2 stable . . . . . ... . . .. ... . the atomic number 51. The numbers 2,8,20, 50, 82, and 126 are called magic numbers. ... 2. There are many more stable nuclei with even numbers of both protons and neutrons than

with odd numbers of these particles (Table 20.2), 3. All isotopes of the elements with atomic numbers higher than 83 are radioactive.

.... Of the two stable isotopes of antimony mentioned in rule 1, both have even numbers of neutrons: '~lSb and '~~Sb.

4. All isotopes of technetium (Tc, Z = 43) and promethium (Pm, Z = 61) are radioactive, Figure 20.1 shows a plot of the number of neutrons versus the number of protons in various isotopes. The stable nuclei are located in an area of the graph known as the belt of stability. Most radioactive nuclei lie outside this belt. Above the belt of stability, the nuclei have higher neutronto-proton ratios than those within the belt (for the same number of protons). To lower this ratio (and hence move down toward the belt of stability), these nuclei undergo the following process, called {3-particle emission:

Protons

Neutrons

Number of Stable Isotopes

Odd

Odd

4

Odd

Even

50

Even

Odd

53

Even

Even

164

SECTION 20.2

Nuclear Stability

Figure 20.1

Plot of neutron \ 'erslb protons for various stable isotopes. represented by dots. The straight line represents the points at which the neutron-to-proton ratio is 1. The haded area represents the belt of stability.

(nIp ratio 1.5: 1) 2~~Pb------,.c..:

120

100 Belt of stability

'" 80

g

•

~

•

.

• • • •. ...

...• ..

..... •

'- B

'"k

•

"

E

~

CH4 15%

CFCs 24%

Global climate change is the subject of the Academy Award-winning 2006 documentary, An Inconvenient Truth, presented by former vice president AI Gore.

-0.2

-0.4

-0 .6 1-------.-------.------.-------r~~~_r~~~

Figure 21 .19

Contribution to global warming by various greenhouse gases. The concentrations of CFCs and methane are much lower than that of carbon dioxide. However, because they can absorb IR radiation much more effectively than CO 2, they make a significant contribution to the overall warming effect.

0

1880

1900

1920

1940

1960

1980

2000

Year

take into account before concluding that global warming is inevitable and irreversible. For example, the ash from volcanic eruptions diffuses upward and can stay in the atmosphere for years. By reflecting incoming sunlight, volcanic ash can cause a cooling effect. Furthermore, the warming effect of CFCs in the troposphere is offset by its action in the stratosphere. Because ozone is a polar poly atomic molecule, it is also an effective greenhouse gas. A decrease in ozone brought about by CFCs actually produces a noticeable drop in temperature. To combat the greenhouse effect, we must lower carbon dioxide emissions. This can be done by improving energy efficiency in automobiles and in household heating and lighting, and by developing nonfossil fuel energy sources, such as photovoltaic cells. Nuclear energy is a viable alternative, but its use is highly controversial due to the difficulty of disposing of radioactive waste and the fact that nuclear power stations are more prone to accidents than conventional power stations (see Chapter 20). The proposed phasing out of CFCs, the most potent greenhouse gas, will help to slow down the warming trend. The recovery of methane gas generated at landfills and the reduction of natural gas leakages are other steps we could take to control CO 2 emissions. Finally, the preservation of the Amazon jungle, tropical forests in Southeast Asia, and other large forests is vital to maintaining the steady-state concentration of CO 2 in the atmosphere. Converting forests to crops................................... and grassland for cattle may do irreparable damage to the delicate ecosystem . . . . .farmland . . . . . . . . . . . . for . .. ...,.. . and permanently alter the climate pattern on Earth.

Which of the following gases qualify as a greenhouse gas: CO, NO, N0 2, C12, H 2 , Ne? Strategy To behave as a greenhouse gas, either the molecule must possess a dipole moment or some

of its vibrational motions must generate a temporary dipole moment. Setup The necessary conditions immediately rule out homonuclear diatomic molecules and atomic •

speCIes. Think About It CO 2, the best-

known greenhouse gas, is nonpolar. It is only necessary for at least one of a molecule's vibrational modes to induce a temporary dipole for it to act as a greenhouse gas.

Solution Only CO, NO, and N0 2, which are all polar molecules, qualify as greenhouse gases.

Both Cl 2 and H2 are homonuclear diatomic molecules, and Ne is atomic. These three species are all IR-inactive.

Practice Problem Which of the following is a more effective greenhouse gas: CO or H 2 0?

SECTION 21.6

The Greenhouse Effect

Checkpoint 21.5 21.5.1

Acid Rain

Which of the following can act as a greenhouse gas? (Select all that apply.)

The greenhouse effect is

21.5.2

a) CH4

a) caused by depletion of stratospheric ozone.

b) N2

b) entirely the result of human activity.

c) Rn

c) a natural phenomenon that has been enhanced by human activity.

d) 0 3

d) the absorption of the sun's energy by molecules and atoms in the upper atmosphere.

e) Xe

e) responsible for the aurora borealis and the aurora australis . •

Acid Rain Every year acid rain causes hundreds of millions of dollars' worth of damage to stone buildings and statues throughout the world. The term stone leprosy is used by some environmental chemists to describe the corrosion of stone by acid rain (Figure 21.20). Acid rain is also toxic to vegetation and aquatic life. Many well-documented cases show dramatically how acid rain has destroyed agricultural and forest lands and killed aquatic organisms. Precipitation in the northeastern United States has an average pH of about 4.3 (Figure 21.21). Because atmospheric CO 2 in equilibrium with rainwater would not be expected to result in a pH less than 5.5, sulfur dioxide (S02) and, to a lesser extent, nitrogen oxides from auto emissions are believed to be responsible for the high acidity of rainwater. Acidic oxides, such as SOb react with water to give the corresponding acids. There are several sources of atmospheric S02' Nature itself contributes much S02 in the form of volcanic eruptions. Also, many metals exist combined with sulfur in nature. Extracting the metals often entails smelting, or roasting, the ores that is, heating the metal sulfide in air to form the metal oxide and S02' For example, 2ZnS(s)

+ 30 2(g) --+. 2ZnO(s) + 2S0 2(g)

The metal oxide can be reduced more easily than the sulfide (by a more reactive metal or in some cases by carbon) to the free metal. Although smelting is a major source of SOb the burning of fossil fuels in industry, in power plants, and in homes accounts for most of the S02 emitted to the atmosphere (Figure 21.22). The

5.3 5.3 5.1 4.9

4.5

4.7

4.3 4.3

Figure 21.20

The effect of acid rain on a marble statue. The photos were taken just a few decades apart.

,;'

Multimedia

Organic and Biochemistry-oil refining processes. 5.3

5.1

5.1

4.7 4.9 5.1

Figure 21.21

Mean precipitation pH in the United States in 1994. Most S02 comes from the midwestern states. Prevailing winds carry the acid droplets formed over the Northeast. Nitrogen oxides also contribute to acid rain formation.

844

CHAPTER 21

Environmental Chemistry

Sulfur dioxide an~ other air pollutants being released into the atmosphere from a coal-burning power plant.

Figure 21.22

sulfur content of coal ranges from 0.5 to 5 percent by mass, depending on the source of the coal. The sulfur content of other fossil fuels is similarly variable. Oil from the Middle East, for instance, is low in sulfur, whereas that from Venezuela has a high sulfur content. To a lesser extent, the nitrogen-containing compounds in oil and coal are converted to nitrogen oxides, which can also acidify rainwater. All in all, some 50 million to 60 million tons of S02 are released into the atmosphere each year! In the troposphere, SO? is almost all oxidized to H 2S0 4 in the form of aerosol, which ends up in wet precipitation or acid rain. The mechanism for the conversion of S02 to H 2 S0 4 is quite complex and not fully understood. The reaction is believed to be initiated by the hydroxyl radical (OH): •

The HOSO? radical is further oxidized to S03:

The sulfur trioxide formed would then rapidly react with water to form sulfuric acid:

S02 can also be oxidized to S03 and then converted to H 2S0 4 on particles by heterogeneous catalysis. Eventually, the acid rain can corrode limestone and marble (CaC0 3). A typical reaction is

Sulfur dioxide can also attack calcium carbonate directly:

There are two ways to minimize the effects of S02 pollution. The most direct approach is to remove sulfur from fossil fuels before combustion, but this is technologically difficult to accomplish. A cheaper but less efficient way is to remove S02 as it is formed. For example, in one process powdered limestone is injected into the power plant boiler or furnace along with the coal (Figure 21.23). At high temperatures, the following decomposition occurs: CaC03 (s) --.. CaO(s) limestone

+ CO 2(g)

quicklime

The quicklime reacts with S02 to form calcium sulfite and some calcium sulfate:

+ S02(g) - - . . CaS03(S) + 2S0ig) + 0 2(g) • 2CaS04(s) CaO(s)

2CaO(s)

Figure 21.23

Common procedure for removing S02 from burning fossil fuel. Powdered limestone decomposes into CaO, which reacts with S02 to form CaS03' The remaining S02 is combined with an aqueous suspension

Mostly CO 2 and air --,.

Smokestack - -

S

of CaO to form CaS03'

+ 0 2 - - ' S02

CaC0 3 CaO

•

+ S02

CaO

Purification chamber

+ CO 2

• CaS03

nnll

Aqueous suspension of CaO Furnace

Coal

SECTION 21.7

Photochemical Smog

845

To remove any remaining S02, an aqueous suspension of quicklime is injected into a purification chamber prior to the gases' escape through the smokestack. Quicklime is also added to lakes and soils in a process caJled liming to reduce their acidity (Figure 21.24). Installing a sulfuric acid plant near a metal ore refining site is also an effective way to cut SOz emission because the S02 produced by roasting metal sulfides can be captured for use in the synthesis of sulfuric acid. This is a very sensible way to turn what is a pollutant in one process into a starting material for another process!

Figure 21 .24

Spreading calcium oxide (CaO) over acidified soil. This process is called limjng.

Photochemical Smog The word smog was coined to describe the combination of smoke and fog that shrouded London during the 1950s. The primary cause of this noxious cloud was sulfur dioxide. Today, however, photochemical smog, which is formed by the reactions of automobile exhaust in the presence of sunlight, is much more common. Automobile exhaust consists mainly of NO, CO, and various unburned hydrocarbons. These gases are called primary pollutants because they set in motion a series of photochemical reactions that produce secondary pollutants. It is the secondary pollutants chiefly N02 and 0 3 that are responsible for the buildup of smog. Nitric oxide is the product of the reaction between atmospheric nitrogen and oxygen at high temperatures inside an automobile engine:

Once released into the atmosphere, nitric oxide is oxidized to nitrogen dioxide:

Sunlight causes the photochemical decomposition of N0 2 (at a wavelength shorter than 400 nm) into NO and 0:

NOig)

+ hv --+. NO(g) + O(g)

Atomic oxygen is a highly reactive species that can initiate a number of important reactions, one of which is the formation of ozone:

where M is some inert substance such as N2. Ozone attacks the C=C linkage in rubber:

R \ /

R

C=C

/ \

R

R

+ R

03

•

\

/0., /

C/

1\

R H2 0

'c

j\

ROO

R

R

\

•

/

C=O

R

where R represents groups of C and H atoms. In smog-ridden areas, this reaction can cause automobile tires to crack. Similar reactions are also damaging to lung tissues and other biological substances. Ozone can be formed also by a series of very complex reactions involving unburned hydrocarbons, nitrogen oxides, and oxygen. One of the products of these reactions is peroxyacetyl nitrate (PAN): CH3-C-0-0-N02

II

o PAN is a powerful lachrymator, or tear producer, and causes breathing difficulties. Figure 21.25 shows typical variations with time of primary and secondary pollutants. Initially, the concentration of N0 2 is quite low. As soon as solar radiation penetrates the atmosphere, though, more N0 2 is formed from NO and O 2 . The concentration of ozone remains fairly constant at a low level in the early morning hours. As the concentration of unburned hydrocarbons and aldehydes increases in the air, the concentrations of N0 2 and 0 3 also rise rapidly. The actual amounts depend on the location, traffic, and weather conditions, but their presence is always accompanied by haze (Figure 21.26). The oxidation of hydrocarbons produces'var~ous organic intermediates, such as alcohols and carboxylic acids, which are all less volati~ than the hydrocarbons themselves. These substances eventually condense into small droplets of llq~. The dispersion of these droplets in air, called an aerosol, scatters sunlight and reduces visibility. This interaction also makes the air look hazy.

•

846

CHAPTER 21

Environmental Chemistry

Figure 21 .25

Typical variations with time in concentration of air pollutants on a smoggy day. Hydrocarbons

NO;.2_

• NO

4

8

6 A.M.

10

12 Noon

2

4

6 P.M.

As the mechanism of photochemical smog formation has become better understood, major efforts have been made to reduce the buildup of primary pollutants. Most automobiles now are equipped with catalytic converters designed to oxidize CO and unburned hydrocarbons to CO 2 and H 20 and to reduce NO and N0 2 to N2 and O2 [ ~~ Section 14.6] . More efficient automobile engines and better public transportation systems would also help to decrease air pollution in urban areas. A recent technological innovation to combat photochemical smog is to coat automobile radiators and air conditioner compressors with a platinum catalyst. So equipped, a running car can purify the air that flows under the hood by converting ozone and carbon monoxide to oxygen and carbon dioxide:

Figure 21.26 big city.

A smoggy day in a

In a city like Los Angeles, where the number of miles driven in one day equals nearly 300 million, this approach would significantly improve the air quality and reduce the "high-ozone level" warnings frequently issued to its residents.

Indoor Pollution Difficult as it is to avoid air pollution outdoors, it is no easier to avoid pollution indoors. The air quality in homes and in the workplace is affected by human activities, by construction materials, and by other factors in our immediate environment. The common indoor pollutants are radon, carbon monoxide, carbon dioxide, and formaldehyde.

The Risk from Radon In a highly publicized case about 35 years ago, an employee reporting for work at a nuclear power plant in Pennsylvania set off the plant's radiation monitor. Astonishingly, the source of his contamination turned out not to be the plant, but radon in his home! A lot has been said and written about the potential dangers of radon as an air pollutant. Just what is radon? Where does it come from? And how does it affect our health? Radon is a member of Group 8A (the noble gases). It is an intermediate product of the radioactive decay of uranium-238 (see Figure 20.3). All isotopes of radon are radioactive, but radon-222 is the most hazardous because it has the longest half-life 3.8 days. Radon, which accounts for slightly over half the background radioactivity on Earth, is generated mostly from the phosphate minerals of uranium (Figure 21.27). Since the 1970s, high levels of radon have been detected in homes built on reclaimed land above uranium mill tailing deposits. The colorless, odorless, and tasteless radon gas enters a building through tiny cracks in the basement floor. It is slightly soluble in water, so it can be spread in different media. Radon-222 is an a-emitter. When it decays, it produces radioactive polonium-214 and polonium-218, which can build up to high levels in an enclosed space. These solid radioactive particles can adhere to airborne dust and smoke, which are inhaled into the lungs and deposited in the respiratory tract. Over a long period of time, the a particles emitted by polonium and its decay products, which are also radioactive, can cause lung cancer.

SECTION 21.8

Indoor Pollution

Figure 21.27

Other human-made radiation 3%

Nuclear testing 2%

Sources of background radiation.

Radon 54%

--~~~?/~----:--I Other natural radiation 27%

Medical procedures 14%

What can be done to combat radon pollution indoors? The first step is to measure the radon level in the basement with a reliable test kit. Short-term and long-term kits are available (Figure 21.28). The short-term tests use activated charcoal to collect the decay products of radon over a period. of several days. The container is sent to a laboratory where a technician measures the radioactivity (I' rays) from radon-decay products lead-214 and bismuth-214. Knowing the length of exposure, the lab technician back-calculates to determine radon concentration. The long-term test kits use a piece of special polymer film on which an Q' particle will leave a "track." After several months' exposure, the film is etched with a sodium hydroxide solution and the number of tracks counted. Knowing the length of exposure enables the technician to calculate the radon concentration. If the radon level is unacceptably high, then the house must be regularly ventilated. This precaution is particularly important in recently built houses, which are well insulated. A more effective way to prevent radon pollution is to reroute the gas before it gets into the house (e.g., by installing a ventilation duct to draw air from beneath the basement floor to the outside) . Currently there is considerable controversy regarding the health effects of radon. The first detailed studies of the effects of radon on human health were carried out in the 1950s when it was recognized that uranium miners suffered from an abnormally high incidence of lung cancer. Some scientists have challenged the validity of these studies because the miners were also smokers. It seems quite likely that there is a synergistic effect between radon and smoking on the development of lung cancer. Radon decay products will adhere not only to tobacco tar deposits in the lungs, but also to the solid particles in cigarette smoke, which can be inhaled by smokers and nonsmokers. More systematic studies are needed to evaluate the environmental impact of radon. In the meantime, the Environmental Protection Agency (EPA) has recommended remedial action where the radioactivity level due to radon exceeds 4 picocuries (pCi) per liter of air. (A curie COfresponds to 3.70 X 10 10 disintegrations of radioactive nuclei per second; a picocurie is a trillionth of a curie, Of 3.70 X 10- 2 disintegrations per second.)

Fig u re 21.28

Home radon

detector.

•

:Sample Problem 21.3 The half-life of Rn-222 is 3.8 days. Starting with 1.0 g of Rn-222, how much will be left after 10 half-lives? Strategy All radioactive decays obey first-order kinetics, making the half-life independent of the initial concentration. Setup Because the question involves an integral number of half-lives, we can deduce the amount of Rn-222 remaining without using Equation 14.3. Solution After one half-life, the amount of Rn left is 0.5 X 1.0 g, or 0.5 g. After two half-lives, only 0.25 g of Rn remains. Generalizing the fraction of the isotope left after n half-lives as (1/2)", where n = 10, we write

quantity of Rn-222 left

=

1.0 g X (1/2)10

=

9.8 X 10- 4 g

6

Practice Problem The concentration of Rn-222 in the basement of a house is 1.8 X 10- mol/L. Assume the air remains static, and calculate the concentration of the radon after 2.4 days.

Think About It An alternative solution is to calculate the firstorder rate constant from the halflife and use Equation 14.3: In

N

t

=

-kt

No where N is the mass of Rn-222. Try this nd verify that your answers are the same. (Since most of the kinetics problems we encounter do not involve an integral number of half-lives, we generally use Equation 14.3 as the first approach to solving them.)

847

848

CHAPTER 21

Environmental Chemistry

Carbon Dioxide and Carbon Monoxide

•

Both carbon dioxide (C0 2 ) and carbon monoxide (CO) are products of combustion. In the presence of an abundant supply of oxygen, CO 2 is formed; in a limited supply of oxygen, both CO and CO 2 are formed. The indoor sources of these gases are gas cooking ranges, woodstoves, space heaters, tobacco smoke, human respiration, and exhaust fumes from cars (in garages). Carbon dioxide is not a toxic gas, but it does have an asphyxiating effect. In airtight buildings, the concentration of CO 2 can reach as high as 2000 ppm by volume (compared with 3 ppm outdoors). Workers exposed to high concentrations of CO 2 in skyscrapers and other sealed environments become fatigued more easily and have difficulty concentrating. Adequate ventilation is the solution to CO 2 pollution. Like CO?, CO is a colorless and odorless gas, but it differs from CO 2 in that it is highly poisonous. The toxicity of CO lies in its unusual ability to bind very strongly to hemoglobin, the oxygen carrier in blood. Both O? and CO bind to the Fe(II) ion in hemoglobin, but the affinity of hemoglobin for CO is about 200 times greater than it is for O2 , Hemoglobin molecules with tightly bound CO (called carboxyhemoglobin) cannot carry the oxygen needed for metabolic processes. At a concentration of 70 ppm, CO can cause drowsiness and headache; at higher concentrations, death may result when about half the hemoglobin molecules become complexed with CO. The best first-aid response to CO poisoning is to remove the victim immediately to an area with a plentiful oxygen supply or to give mouth-to-mouth resuscitation.

Formaldehyde Formaldehyde (CH?O) is a rather disagreeable-smelling liquid used as a preservative for laboratory specimens. Industrially, formaldehyde resins are used as bonding agents in building and furniture construction materials such as plywood and particle board. In addition, urea-formaldehyde insulation foams are used to fill wall cavities. The resins and foams slowly break down to release free formaldehyde, especially under acid and humid conditions. Low concentrations of formaldehyde in the air can cause drowsiness, nausea, headaches, and other respiratory ailments. Laboratory tests show that breathing high concentrations of formaldehyde can induce cancers in animals, but whether it has a similar effect in humans is unclear. The safe standard of formaldehyde in indoor air has been set at 0.1 ppm by volume. Because formaldehyde is a reducing agent, devices have been constructed to remove it by means of a redox reaction. Indoor air is circulated through an air purifier containing an oxidant such as A120 i KMn04, which converts formaldehyde to the less harmful and less volatile formic acid (RCOOH). Proper ventilation is the best way to remove formaldehyde. However, care should be taken not to remove the air from a room too quickly without replenishment, because a reduced pressure would cause the formaldehyde resins to decompose faster, resulting in the release of more formaldehyde.

Checkpoint 21.8 21 .8.1

Indoor Pollution

The risk posed by radon gas in homes is

21 .8.2

a) respiratory distress

What mass of 1.0-g sample of radon remains after 30 days? (The half-life of radon is 3.8 days.)

b) nausea

a) 0.030 g

c) dizziness

b) 0.0040 g

d) coma

c) 0.99 g

e) lung cancer

d) 0.97 g e) 0.0010 g

•

APPLYING WHAT YOU'VE LEARNED

Applying What You've Learned The research of Paul Crutzen, the third recipient of the Nobel Prize for Chemistry in 1995, involved the effect of nitric oxide (NO) on the destruction of stratospheric ozone. Unlike CFCs, which may take 50 to 100 years to diffuse into the upper atmosphere, nitric oxide is introduced directly to the stratosphere in the exhaust of high-altitude aircraft. Early in the 1970s, the United States considered construction of a large fleet of supersonic transport airplanes (SSTs), similar to the Concorde. Environmentalists argued, based in part on the work of Paul Crutzen, that to do so would significantly endanger the ozone layer. Problems:

a)

The bond enthalpy of NO is 630.6 kJ/mol. Determine the maximum wavelength of . light required to break the bond in an NO molecule. [ ~~ Sample Problem 21.1]

b)

Can nitric oxide act as a greenhouse gas? Explain.

[ ~~

Sample Problem 21.2]

•

849

850

CHAPTER 21

Environmental Chemistry

CHAPTER SUMMARY Section 21.1

Section 21.5

•

Earth's atmosphere is made up mainly of nitrogen and oxygen, plus a number of other trace gases. Molecular nitrogen in the atmosphere is incorporated into other compounds via nitrogen fixation.

•

•

The regions of the atmosphere, from Earth's surface outward, are the troposphere, the stratosphere, the mesosphere, the thermosphere, and the ionosphere.

•

The chemical processes that go on in the atmosphere are influenced by solar radiation, volcanic erupti on, and human activities .

Section 21.6 •

Section 21.2 •

In the outer regions of the atmosphere, the bombardment of molecules and atoms by solar particles gives rise to the aurora borealis in the Northern Hemisphere and the aurora australis in the Southern Hemisphere. The glow on a space shuttle is caused by excitation of molecules adsorbed on the shuttle's surface.

Carbon dioxide's ability to absorb infrared radiation enables it to trap some of the outgoing heat from Earth, warming its surface-a phenomenon known as the greenhouse effect. Other gases such as the CFCs and methane also contribute to the greenhouse effect. Global warming refers to the result of the enhanced greenhouse effect caused by human activities.

Sulfur dioxide, and to a lesser extent nitrogen oxides, generated mainly from the burning of fo ssil fuels and from the roasting of metal sulfi des, causes acid rain.

Section 21.7 •

Photochemical smog is formed by the photochemical reaction of automobile exhaust in the presence of sunlight. It is a complex reaction involving nitrogen oxides, ozone, and hydrocarbons.

Section 21.3 •

Ozone in the stratosphere absorbs harmful UV radiation in the 200- to 300-nm range and protects life underneath. For many years, chlorofluorocarbons have been destroying the ozone layer.

Section 21.4 •

Section 21.8 •

Indoor air pollution is caused by radon, a radioactive gas formed during uranium decay; carbon monoxide and carbon dioxide, products of combustion; and formaldehyde, a volatile organic substance released from resins used in construction materials.

Volcanic eruptions can lead to air pollution, deplete ozone in the stratosphere, and affect climate.

I(EY WORDS Greenhouse effect, 838

Mesosphere, 833

Photochemical smog, 845

Thelmosphere, 833

Ionosphere, 833

Nitrogen fixation, 831

Stratosphere, 833

Troposphere, 832

UESTIONS AND PROBLEMS •

Section 21.1: Earth's Atmosphere

Problems

Review Questions

21.5

Referring to Table 21.1, calculate the mole fraction of CO 2 and its concentration in parts per million by volume.

21.6

Calculate the partial pressure of CO2 (in atm) in dry air when the atmospheric pressure is 754 mmHg.

21.7

Describe the processes that result in the warming of the stratosphere.

21.8

Calculate the mass (in kg) of nitrogen, oxygen, and carbon dioxide gases in the atmosphere. Assume that the total mass of air in the atmosphere is 5.25 X 102 1 g.

21.1 21.2

21.3

21.4

Describe the regions of Earth's atmosphere. Briefly outline the main processes of the nitrogen and oxygen cycles. Explain why, for maximum performance, supersonic airplanes need to fly at a high altitude (in the stratosphere). Jupiter's atmosphere consists mainly of hydrogen (90 percent) and helium (9 percent). How does this mixture of gases contrast with the composition of Earth's atmosphere? Why does the composition differ?

QUESTIONS AND PROBLEMS

85 1

Section 21.2: Phenomena in the Outer Layers of the Atmosphere

21.24

Why are CFCs not decomposed by UV radiation in the troposphere?

Review Questions

21.25

The average bond enthalpies of the C-Cl and C-F bonds are 340 and 485 kJ/mol, respectively. Based on this information, explain why the C-CI bond in a CFC molecule is preferentially broken by solar radiation at 250 nm.

21.26

Like CFCs, certain bromine-containing compounds such as CF3Br can also participate in the destruction of ozone by a similar mechanism starting with the Br atom:

21.9

What process gives rise to the aurora borealis and aurora australis?

21.10

Why can astronauts not release oxygen atoms to test the mechanism of shuttle glow?

Problems

21.11

The highly reactive OH radical (a species with an unpaired electron) is believed to be involved in some atmospheric processes. Table 8.6 lists the bond enthalpy for the oxygento-hydrogen bond in OH as 460 kJ/mol. What is the longest wavelength (in nm) of radiation that can bring about the following reaction? •

OH(g) -

21.12

.... O(g)

The green color observed in the aurora borealis is produced by the emission of a photon by an electronically excited oxygen atom at 558 nm. Calculate the energy difference between the two levels involved in the emission process.

Review Questions

21.14

21.15

21.16

Briefly describe the absorption of solar radiation in the stratosphere by O 2 and 0 3 molecules. Explain the processes that have a warming effect on the stratosphere. List the properties of CFCs, and name four major uses of these compounds. How do CFCs and nitrogen oxides destroy ozone in the stratosphere?

21.17

What causes the polar ozone holes?

21.18

How do volcanic eruptions contribute to ozone destruction?

21.19

Describe ways to curb the destruction of ozone in the stratosphere.

21.20

21.27

Draw Lewis structures for chlorine nitrate (CION0 2 ) and chlorine monoxide (ClO).

21.28

Draw Lewis structures for HCFC-123 (CF 3CHCI 2 ) and CF 3CFH 2 .

+ H(g)

Section 21.3: Depletion of Ozone in the Stratosphere

21.13

Given that the average C-Br bond energy is 276 kJ/mol, estimate the longest wavelength required to break this bond. Will this compound be decomposed in the troposphere only or in both the troposphere and stratosphere?

Section 21.4: Volcanoes

Review Questions 21.29

What are the effects of volcanic eruptions on climate?

21.30

Classify the reaction between H 2S and S02 that leads to the formation of sulfur at the site of a volcanic eruption.

Section 21.5: The Greenhouse Effect

Review Questions 21.31

What is the greenhouse effect? What is the criterion for classifying a gas as a greenhouse gas?

21.32

Why is more emphasis placed on the role of carbon dioxide in the greenhouse effect than on that of water?

21.33

Describe three human activities that generate carbon dioxide. List two major mechanisms for the uptake of carbon dioxide.

21.34

Deforestation contributes to the greenhouse effect in two ways. What are they?

21.35

How does an increase in world population enhance the greenhouse effect?

Discuss the effectiveness of some of the CFC substitutes. .

Problems

21.21

21.22

21.23

Given that the quantity of ozone in the stratosphere is equivalent to a 3.0-mm-thick layer of ozone on Earth at STP, calculate the number of ozone molecules in the stratosphere and their mass in kilograms. (Hint: The radius of Earth is 6371 km and the surface area of a sphere is 47Tr2, where r is the radius. ) Referring to the answer in Problem 21.21, and assuming that the level of ozone in the stratosphere has already fallen 6.0 percent, calculate the number of kilograms of ozone that would have to be manufactured on a daily basis so that we could restore the ozone to the original level in 100 years. If ozone is made according to the process 30 2 (g) • 20 3 (g), how many kilojoules of energy would be required? Both Freon-ll and Freon-12 are made by the reaction of carbon tetrachloride (CC14 ) with hydrogen fluoride. Write equations for these reactions.

21.36

Is ozone a greenhouse gas? If so, sketch three ways an ozone molecule can vibrate.

21.37

What effects do CFCs and their substitutes have on Earth's temperature?

21.38

Why are CFCs more effective greenhouse gases than methane and carbon dioxide?

Problems

21.39

•

The annual production of zinc sulfide (ZnS) is 4.0 X 104 tons. Estimate the number of tons of S02 produced by roasting it to extract zinc metal.

852

21.40

CHAPTER 21

Environmental Chemistry

Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the thermal decomposition of calcium carbonate: CaC0 3(s) - _ . CaO(s)

+ CO 2 (g)

Calculate the yearly release of CO 2 (in kg) to the atmosphere if the annual production of CaO in the United States is 1.7 X 1010 kg.

21.41

21.42

The molar heat capacity of a diatomic molecule is 29.1 JIK . mol. Assuming the atmosphere cbntains only nitrogen gas and there is no heat loss, calculate the total heat intake (in kJ) if the atmosphere warms up by 3°C during the next 50 years. Given that there are 1.8 X 1020 moles of diatomic molecules present, how many kilograms of ice (at the North and South Poles) will this quantity of heat melt at O°C? (The molar heat of fusion of ice is 6.01 kJ/mol.) As mentioned in the chapter, spraying the stratosphere with hydrocarbons such as ethane and propane should eliminate CI atoms . What is the drawback of this procedure if used on a large scale for an extended period of time?

21.54

The safety limits of ozone and carbon monoxide are 120 ppb by vol ume and 9 ppm by volume, respectively. Why does ozone have a lower limit?

21.55

Suggest ways to minimi ze the formation of photochemical smog.

21.56

In which region of the atmosphere is ozone beneficial? In which region is it detrimental ?

Problems 21.57

2NO(g)

21.58

Review Questions Name the gas that is largel y responsible for the acid rain phenomenon.

21.44

List three detrimental effects of acid rain.

21.45

Briefly discuss two industrial processes that lead to acid rain.

21.46

Discuss ways to curb acid rain.

21.47

Water and sulfur dioxide are both polar molecules, and their geometry is similar. Why is S02 not considered a major greenhouse gas?

21.48

Describe the removal of S02 by CaO (to form CaS03) in terms of a Lewis acid-base reaction.

Problems 21.49

21.50

+

OzCg) - _ . 2N02(g)

is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the conditions described in part (b), the half-life of the reaction has been estimated to be 6.4 X 103 min. What would the half-life be if the initial concentration of NO were 10 ppm?

Section 21.6: Acid Rain

21.43

Assume that the formation of nitrogen dioxide:

The gas-phase decomposition of peroxyacetyl nitrate (PAN) obeys first-order kinetics: with a rate constant of 4.9 X 10- 4 s - I Calculate the rate of decomposition (in Mis) if the concentration of PAN is 0.55 ppm by volume. Assume STP conditions.

21.59

On a smoggy day in a certain city, the ozone concentration was 0.42 ppm by volume. Calculate the partial pressure of ozone (in atm) and the number of ozone molecules per liter of air if the temperature and press ure were 20.0°C and 748 mmHg, respectively.

21.60

Which of the following settings is the most suitable for photochemical smog fonnation: (a) Gobi desert at noon in June, (b) New York City at 1 P.M. in July, (c) Boston at noon in January? Explain your choice.

Section 21.8: Indoor Pollution Review Questions

An electric power station annually burns 3.1 X 107 kg of coal containing 2.4 percent sulfur by mass. Calculate the volume of S02 emitted at STP. The concentration of S02 in the troposphere over a certain region is 0.16 ppm by volume. The gas dissolves in rainwater as follows: S02(g)

+ H 20 (l)

.

• H +(aq)

+ HS0 3"(aq)

Given that the equilibrium constant for the preceding reaction is 2 1.3 X 10- , calculate the pH of the rainwater. Assume that the reaction does not affect the partial pressure of S02'

Section 21.7: Photochemical Smog

21.61

List the major indoor pollutants and their sources.

21.62

What is the best way to deal with indoor pollution?

21.63

Why is it dangerous to idle a car's engine in a poorly ventilated place, such as the garage?

21.64

Describe the properties that make radon an indoor pollutant. Would radon be more hazardous if 222Rn had a longer half-life?

Problems 21.65

A concentration of 8.00 X 102 ppm by volume of CO is considered lethal to humans. Calculate the minimum mass of CO (in grams) that would become a lethal concentration in a closed room 17.6 m long, 8.80 m wide, and 2.64 m high. The temperature and pressure are 20.0°C and 756 mmHg, respectively.

21.66

A volume of 5.0 L of polluted air at 18.0°C and 747 mmHg is passed through lime water [an aqueous suspension of Ca(OH)z], so that all the carbon dioxide present is precipitated as CaC0 3 . If the mass of the CaC0 3 precipitate is 0.026 g, calculate the percentage by volume of CO 2 in the air sample.

Review Questions 21.51

What is photochemical smog? List the factors that favor the formation of photochemical smog.

21.52

What are primary and secondary pollutants?

21.53

Identify the gas that is respon sible for the brown color of photochemical smog.

QUESTIONS AND PROBLEMS

Additional Problems

21.75

21.67

Briefly describe the harmful effects of the following substan ces : 0 3 , S02, N0 2, CO, CH 3COOON0 2 (PAN), Rn.

21.68

The equilibrium constant (Kp) for the reaction

Although the hydroxyl radical (OH) is present only in a trace amount in the troposphere, it plays a central role in its chemi ~ because it is a strong ox idi zing agent and can react with many pollutants as well as some CFC sub stitutes. (a) The hydrox yl radical is formed by the fo llowing reactions:

03 is 4.0 X 10- 3 1 at 25 °C and 2.6 X 10- 6 at 1l00°C, the temperature of a running car's engine. Is this an endoth ermic or exothermic reaction ? 21.69

+ Hb0 2(aq)

=::t. OzCg) + HbCO(aq)

::;:.

where Hb0 2 and HbCO are oxygenated hemoglobin and carboxyhemoglobin , respectively. (b) The composition of a • breath of air inhaled by a person smoking a cigarette is 1.9 X 10- 6 mollL CO and 8.6 X 10- 3 mollL O 2, Calcul ate the ratio of [HbCO] to [Hb0 2], given that Kc is 212 at 37°C. 21.70

21.71

21.72

A glass of water initiall y at pH 7.0 is exposed to dry air at sea level at 20°C. Calculate the pH of the water when equilibrium is reached between atmospheric CO 2 and COz di ssolved in the water, given that Henry's law constant for COz at 20°C is 0.032 11l01lL . atm. (Hint: Assume no loss of water due to evaporation, and use Table 2 1.1 to calculate the pal1ial pressure of CO 2, Your answer should cOlTespond roughly to the pH of rainwater.)

21.73

A 14-m by lO-m by 3.0-m basement had a high radon content. On the day the basement was sealed off from its surroundings so that no exchange of air could take place, the partial pressure 2')2 - - 6 'J?? of - Rn was 1.2 X 10 mmHg. Calculate the number of ---Rn isotopes (tli2 = 3.8 days) at the beginning and end of 31 days. Assume STP conditi ons.

21.74

Ozone in the troposp here is formed by the following steps : NO z - - - . NO

0+ O2 ---.03

+0

(I) (2)

The first step is initiated by the absorpti on of visible li ght (N0 2 is a brown gas) . Calcul ate the longest wavelength required for step 1 at 25 °C. (H int: You need to first calculate flH and hence flE fo r step 1. Next, determine the wavelength for decomposing NO z from flE.)

320 !UTI.

0 * + O2

21.76

The equilibrium constant (K p) for the reaction 2CO(g) + 90 Oz(g). • 2CO z(g) is 1.4 X 10 at 25 °C. Given this enormous value, why doesn't CO convert totally to CO 2 in the troposphere?

21.77

A person was found dead of carbon monoxide poisoning in a well-insul ated cabin. Investigation showed that he had used a blackened bucket to heat water on a butane burner. The burner was found to functio n properly with no leakage. Explain, with an appropriate equation, the cause of his death .

2 1.78

The carbon dioxide level in th e atmosphere today is often compared with that in preindustrial days. Explain how scientists use tree rings and ai r trapped in polar ice to arrive at the . co mpanson.

21.79

What is ironic about the following cartoon?

21.80

Calculate the standard enthalpy of formation (flH 'f) of CIO from the follow ing bond energies: Clz: 242.7 kllmol; Oz: 498.7 kllmol ; CIO: 206 kllmo!.

21.81

Methyl bromide (CH 3Br, b.p. = 3.6°C) is used as a soil fumigant to control insects and weeds.' It is also a marine by-product. Photodissociation of the C - Br bond produces Br atoms that can react with ozone similar to Cl, except more effectively. Do you expect CH3Br to be photolyzed in the troposphere? The bond enthalpy of the C- Br bond is about 293 kllmo!.

Instead of monitoring carbon dioxide, suggest another gas that scientists could study to substantiate the fact that COz concentration is steadil y increasing in the atm osphere. In 1991 it was discovered that nitrous oxide (NzO) is produced in the synthesis of nylon. This compound, which is released into the atmosphere, contributes both to the depletion of ozone in the stratosphere and to the greenhouse effect. (a) Write equati ons representing the reactions between N 20 and oxygen atoms in the stratosphere to produce nitric oxide (NO), whi ch is then oxidi zed by ozone to form nitrogen dioxide. (b) Is NzO a more effective greenhouse gas than carbon dioxide? Explain. (c) One of the intermediates in nyl on manufacture is adipic acid [HOOC(CHz)4COOHl About 2.2 X 109 kg of adipic acid is consumed every year. It is estim ated th at for every mole of adipic acid produced, I mole of NzO is generated. What is th e maximum number of moles of 0 3 that can be destroyed as a result of this process per year?

,,=

where 0 * denotes an electroni cally excited atom. (a) Explain why the concentration of OH is so small even though the concentrations of 0 3 and H 20 are quite large in the troposphere. (b) What property makes OH a strong oxidizi ng agent? (c) The reaction between OH and N0 2 contributes to acid rain. Write an equation for thi s process. (d) The hydroxyl radical can oxidize SOz to H 2S0 4 , The first step is the formation of a neutral HS0 3 species, followed by its reacti on with Oz and HzO to form HzSO, and the hydroperoxyl radical (HO z). Write eq uati ons for these processes.

As stated in the chapter, carbon monoxide has a much hi gher affinity for hemoglobin than oxygen does. (a) Write the equilibrium constant expression (Ke) for the foll owing process: CO(g)

8:;,3

854

CHAPTER 21

Environmental Chemistry

21.82

The effective incoming solar radiation per unit area on Earth is 342 W /m2 Of this radiation, 6.7 W /m2 is absorbed by CO 2 at 14,993 nm in the atmosphere. How many photons at this wavelength are absorbed per second in 1 m 2 by CO 2? (1 W = 1 J/s)

21.83

As stated in the chapter, about 50 million tons of sulfur dioxide are released into the atmosphere every year. (a) If 20 percent of the S02 is eventually converted to H 2S0 4 , calculate the number of 1000-lb marble statues ~he resulting acid rain can damage. As an estimate, assume that the acid rain only destroys the surface layer of each statue, which is made up of 5 percent of its total mass . (b) What is the other undesirable result of the acid rain damage?

21.84

21.85

How are past temperatures determined from ice cores obtained from the Arctic or Antarctica? (Hint: Look up the stable isotopes of hydrogen and oxygen . How does energy required for vaporization depend on the masses of H 20 molecules containing different isotopes? How would you determine the age of an ice core?)

21.86

The balance between S02 and S03 is important in understanding acid rain formation in the troposphere. From the following information at 25°C S (s) + 0 2(g) :;:::.~. SOig) 2S(s) + 30 2(g).

• 2S0 3(g)

KI = 4.2 X 1052

K2 = 9.8

X

10

128

calculate the equilibrium constant for the reaction

Peroxyacetyl nitrate (PAN) undergoes thermal decomposition as follow s:

The rate constant is 3 .0 X 10- 4 S -I at 25°C. At the boundary between the troposphere and stratosphere, where the temperature is about -40°C, the rate constant is reduced to 2.6 X 10- 7 S- l (a) Calculate the activation energy for the decomposition of PAN. (b) What is the half-life of the reaction (in min) at 25 °C ?

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES Indoor air pollutants, including radon gas, can pose significant health risks. Radon is a radioactive gas found in the soils and rocks of Earth's crust. It finds its way into homes through cracks in the foundation or basement, and sometimes through the water supply. Radon is invisible and odorless, and becomes a health hazard when it is allowed to build up inside the home. The isotope of radon commonly found in homes is radon-222, a product of the uranium-238 decay chain, formed by the C\' decay of radium-226. Radon-222, in turn, decays by C\' emission. 1.

Select the equation that correctly represents the radium-226. a) 226Ra - -.... 226Rn + C\' b) 226Ra • 222Rn + C\' c) 226 Ra + C\' • 226Rn d) 226 Ra + C\' • 222Rn

C\'

2.

How many neutron s does a radium-226 nucleus contain? a) 226 b) 88 c) 113 d) 138

3.

What is the product when radon-222 decays by

C\'

emission?

a) Radon-218 b) Radium-218 c) Polonium-218 s) Polonium-226

decay of

4.

What would the product be if radon-222 decayed by a) Francium-222 b) Radon-223 c) Radon-221 d) Astatine-222

f3 emission?

•

ANSWERS TO IN-CHAPTER MATERIALS

ANSWERS

To

IN-CHAPTER MATERIALS

Answers to Practice Problems 21.11120 nm. 21.2 H 20. 21.3 1.2

x

10- 6 mollL.

Answers to Checkpoints

Answers to Applying What You've Learned a) 190 nm. b) Yes, because it is a polar molecule and therefore is IR-active.

21.2.1 a. 21.2.2 d. 21.5.1 a, d. 21.5.2 c. 21.8.1 e. 21.8.2 b.

•

•

855

•

•

. . . . . oor Ination •

22.1

Coordination Compounds

• • •

Properties of Transition Metals Ligands Nomenclature of Coordination Compounds

22.2

Structure of Coordination Compounds

22.3

Bonding in Coordination Compounds: Crystal Field Theory

• • • •

Crystal Field Splitting in Octahedral Complexes Color Magnetic Properties Tetrahedral and SquarePlanar Complexes

22.4

Reactions of Coordination Compounds

22.5

Applications of Coordination Compounds

•

emIstr

Coordination Chemistry in the Treatment of Lead Poisoning

WHITE LEAD

ABS OLUTE LY

PURE

Although the Consumer Product Safety Commission (CPSC) banned the residential use of lead-based paint in 1978, millions of children remain at risk for exposure to lead from deteriorating paint in older homes. Lead poisoning is especially harmful to children under the age of 5 years because it interferes with growth and development and it has been shown to lower IQ. Symptoms of chronic exposure to lead include diminished appetite, nausea, malaise, and convulsions. Blood lead level (BLL), expressed as micrograms per deciliter (J..Lg/dL) , is used to monitor the effect of chronic exposure. A BLL < 10 J..Lg/dL is considered normal; a BLL > 45 J..Lg/dL requires medical and environmental intervention. At high levels (> 70 J..Lg/dL), lead can cause seizures, coma, and death. Treatment for lead poisoning involves chelation therapy, in which a chelating agent is administered orally, intravenously, or intramuscularly. Chelating agents form strong coordinate-covalent bonds to metal ions, forming stable, water-soluble complex ions that are easily removed from the body via the urine. One of the drugs commonly used for this purpose is dimercaptosuccinic acid (DMSA), marketed under the name Chemet. Chelation therapy relies on coordination chemistry.

o

Covers And Lasts longer 1 Than "OLD DUTCH " Process While Le:.~ __ J

SH OH

HO SH

0

Dimercaptosuccinic acid (DMSA)

In This Chapter You, Will Learn

about the properties of coordination compounds and how, through the use of chelates, coordination chemistry is used to solve a variety of medical and other societal problems.

Before you begin. you should review

=,

$"'. '11

[ ~~

•

Lewis acids and bases

•

The shapes of d orbitals

Section 16.12]

[ ~~

Section 6.7]

Media Player/

MPEG Content Chapter in Review

•

Lead paint, known for its brightness and durability, was commonly used to paint homes, fences, and interior walls. Its use in homes was banned in 1978 because of the health risks associated with exp osure to lead. 857

858

CHAPTER 22

Coordination Chemistry

Coordination Compounds

·. . . A complex ion is one in which a metal cation is covalently bou nd to one or more molecules 0r ions [ ~~ Section 17 .5].

Coordination compounds contain coordinate covalent bonds [~~ Sectio n 8,8 ] formed by the reactions of metal ions with groups of anions or polar molecules. The metal ion in these kinds of reactions acts as a Lewis acid, accepting electrons, whereas the anions or polar molecules act as Lewis bases, donating pairs of electrons to form bonds to the metal ion. Thus, a coordinate covalent bond is a covalent bond in which one of the atoms donates both of the electrons that constitute .the'ballet" Often ci"coordlnatlon co·ri.i.po·uncl corisis'ts "Of a' complex ion and one or more counter ions. In writing formu las for such coordination compounds, we use square brackets to separate the complex ion from the counter ion. ..

.

Th is compound consists of the complex We can use the term coordination complex to refer to a compound, such as Fe(CO)s, or to a complex ion.

.. ion PtCI ~-

and two K - counter ions.

· . . . . .. .......... ... . Some coordination compounds, such as Fe(CO)s, do not contain complex ions. Most but not all of the metals in coordination compounds are transition metals. Our understanding of the nature of coordination compounds stems from the classic work of Alfred Werner, I who prepared and characterized many coordination compounds. In 1893, at the age of26 , Werner proposed what is now commonly referred to as Werner's coordination theory. Nineteenth-century chemists were puzzled by a certain class of reactions that seemed to violate valence theory. For example, the valences of the elements in cobalt(III) chloride and those in ammonia seem to be completely satisfied, and yet these two substances react to form a stable compound having the formula CoCl3 • 6NH 3 . To explain this behavior, Werner postulated that most elements exhibit two types of valence: primary valence and secondary valence. In modern terminology, primary valence corresponds to the oxidation number and secondary valence to the coordination number of the element. In CoCl 3 • 6NH 3 , according to Werner, cobalt has a primary valence of 3 and a secondary valence of 6. Today we use the formula [Co(NH 3)6]CI 3 to indicate that the ammonia molecules and the cobalt atom form a complex ion ; the chloride ions are not part of the complex but are counter ions, held to the complex ion by Coulombic attraction.

Properties of Transition Metals The Group 2B metals- Zn, Cd, and Hg--do not fit either of these criteria. As a result, they are d-block metals, but they are not actually transition metals.

'-. -,, '

Multimedia

Chemical Reactions-oxidation states of vanadium .

Figure 22.1

The transition metals

(shown in green). Note that although the Group 2B elements (Zn, Cd, Hg) are described as transition metals b y some chemists, neither the metals nor their ions possess incompletely filled d subshells.

metals are those that either have incompletely filled d subshells or form ions with · . . . . . .Transition . . .. . .... . .. ...... ... .. .. ... . .. incompletely filled d subshells (Figure 22.1). Incompletely filled d subshells give rise to several notable properties, including distinctive colors, the formation of paramagnetic compounds, catalytic activity, and the tendency to form complex ions. The most common transition metals are scandium through copper, which occupy the fourth row of the periodic table. Table 22.1 lists the electron configurations and some of the properties of these metals. Most of the transition metals exhibit a close-packed structure in which each atom has a coordination number of 12. FurtherlIlore, these elements have relatively small atomic radii. The combined effect of closest packing and small atomic size results in strong metallic bonds. Therefore,

IA 1 GJ 2A H ? 0 4 Li Be 11 12 Na Mg 19 20 K Ca 37 38 Rb Sr 55 56 Cs Ba 87 88 Fr Ra

-

Is

3B 4B 5B 6B 7B 8B~ 1B 2B 3 4 5 6 78 91011 12 2t 22 23 24 25 26 27 28 29 30 Sc Ti V Cr Mn Fe Co Ni Cu Zn 42 43 44 45 46 47 48 39 40 41 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd 57 72 73 74 75 76 77 78 79 80 La Hf Ta W Re Os Ir Pt Au Hg 89 104 105 106 107 108 109 110 I 1I 112 Ac Rf Db Sg Bh Hs Mt Ds Rg

3A 13 5 B 13 Ai 31 Ga 49 In 81 T1 113

4A SA 6A 14 15 16 6 7 8 N 0 C 14 15 16 Si P S 32 33 34 Ge As Se 50 51 52 Sn Sb Te 82 83 84 Pb Bi Po 114 lIS 116

8A 18 7A 2 17 He 9 10 F Ne 17 18 Cl Ar 35 36 Br Kr 53 54 I Xe 85 86 At Rn (ll7) 118

1. Alfred Werner (1866- 1919). Swiss chemist. Werner started as an organic chemist but did his most notable work in

coordination chemistry. For his theory of coordination compounds, Werner was awarded the Nobel Prize in Chemistry in 1913.

SECTION 22.1

Sc

Ti

Coordination Compounds

859

The trends in radii, electronegativity, and . . . . . . ionization energy for the first row of transition metals are somewhat different from those for main group elements. This is largely the effect of shielding of the 4s electrons by the 3d electrons.

Cr Mn Fe Co Ni Cu . . . . . . . . . . . . . .. . . . . . . . . . . . . .. ......... .... .. ... .. ... ... ... ... ... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V

~

Electron configuration 4i3d1 4i3d2 M M 2+

-

M3+

[Ar]

Electronegativity 1.3

4i3d3

2 3d

3

4i3dS

4i3d 6

4S23d7

4i3d 8

4S13d 1O

4

3d S

3d 6

3d7

3d 8

3d 9

4

S

6

7

3d

8

4s 13d S 3d

3i

3d 2 3d

3

3d

3d

l.5

l.6

l.6

l.5

l.8

l.9

l.9

l.9

3d

3d

3d

The electron configu rations of the first-row transition metals and their ions were discussed in Section 6.9 and Section 7.5, respectively.

Ionization energy (kJ/mol) First

631

658

650

652

717

759

760

736

745

Second

1235

1309

1413

1591

1509

1561

1645

1751

1958 .

Third

2389

2650

2828

2986

3250

2956

3231

3393

3578

162

147

134

130

135

126

125

124

128

90

88

85

80

77

75

69

72

77

74

64

66

60

64

Radius (pm) M M 2+ M 3+

81

1A

V

Transition Metals Cr Mn Fe

Co

Ni

Cu

2B Zn

147

134

130

135

126

125

124

128

138

1539

1668

1900

1875

1245

1536

1495

1453

1083

419.5

1440

2730

3260

3450

2665

2150

2900

2730

2595

906

4.51

3.0

4.51

6.1

7.19

7.43

3000 7.86

8.9

8.9

8.96

7.14

K

2A Ca

Sc

Ti

Atomic radius (pm)

235

197

162

Melting point (0C)

63.7

838

Boiling point COC)

760

Density (g/cm 3)

0.86

transition metals have higher densities , higher melting points and boiling points, and higher heats of fusion and vaporization than the main group and Group 2B metals (Table 22.2). Transition metals exhibit variable oxidation states in their compounds. Figure 22.2 shows the oxidation states of the first row of transition metals. Note that all these metals can exhibit the oxidation state + 3 and nearly all can exhibit the oxidation state + 2. Of these two, the + 2 oxidation state is somewhat more common for the heavier elements. The highest oxidation state for a 5 transition metal is + 7, exhibited by manganese (4i 3d ). Transition metals exhibit their highest oxidation states in compounds that contain highly such as oxygen and . . . . .electronegative . . . . . . . . .. , . . . . . . . , . . .elements . . . . . . . . . . . . . . . . . . .. . . . .. . . . . . . . . . . . . . . . . . . . . fluorine- for example, V 20 S, er03, and Mn20 7'

Figure 22.2

Oxidation states of the first-row transition metals. The most

+7

stabl~

+6 +6 + 6

oxidation numbers are shown in red. The zero oxidation state is encountered in some compounds, such

+5 + 5 + 5 + 5

as Ni(CO)4 and Fe(CO)s.

+4 + 4 +4 +4 + 4 +4 +3 + 3 + 3 +3 +3 +3 +3 +3 +3 + 2 + 2 + 2 +2 +2 +2 +2 +2 +1 Se

Ti

V

The oxidation state of 0 in each of these compounds is -2, making those of V, Cr, and Mn +5, +6, and + 7, respectively [ ~~ Section 4.4]

Cr Mn Fe Co Ni Cu

860

CHAPTER 22

Coordination Chemistry

Structure

Name Monodentate

•

..

Ammonia

H-N-H I H

Carbon monoxide

.·C=O· .

Chloride ion

:cr-

Cyanide ion

[:C=N:]-

Thiocyanate ion

[:~-C=Nr

••

••

••

H-O-H ••

Water Bidentate

••

Ethylenediamine

••

H2N-CH2-CH2-NH2

:0

b: 2-

\\

•

Oxalate ion

.I

II

C- C

\ .

:0. • •

.0: • •

Polydentate

Ethylenediaminetetraacetate ion (EDTA)

Ligands The molecules or ions that surround the metal in a complex ion are called ligands (Table 22.3). The formation of covalent bonds between ligands and a metal can be thought of as a Lewis acidbase reaction. (Recall that a Lewis base is a species that donates a pair of electrons [ ~~ Section 16.12] .) In order to be a ligand, a molecule or ion must have at least one unshared pair of valence electrons, as these examples illustrate: ••

:CF••

.·C=O· .

Therefore, ligands play the role of Lewis bases. The transition metal, on the other hand, acts as a Lewis acid, accepting (and sharing) pairs of electrons from the Lewis bases. The atom in a ligand that is bound directly to the metal atom is known as the donor atom. For example, nitrogen is the donor atom in the [Cu(NH3)4]2+ complex ion:

H

?+

I

H-N-H

H

H

I

I

H- N-Cu-N-H

I

I

H

H

H-N-H

I

H

SECTION 22.1

Coordination Compounds

861

Figure 22.3

(a) Structure of a metal-ethylenediamjne complex cation, such as [Co(enh]2+. Each ethylenediamine molecule provides two N donor atoms and is therefore a bidentate ligand. (b) Simplified structure of the same complex cation.

(a)

(b)

Figure 22.4

(a) EDTA complex of lead. The complex bears a net charge of 2 - because each 0 donor atom has one negative charge and the lead ion carries two positive charges. Only the lone pairs that participate in bonding are shown. Note the octahedral geometry about the Pb2+ ion. (b) Molecular model of the Pb 2 + -EDTA complex. The light green sphere is the Pb2+ ion.

-+--CH?

'\

0:

:N-CH 2

0:

CH 2

CH

o~/

2

C

II

o (a)

(b)

The coordination number in a coordination compound refers to the number of donor atoms surrounding the central metal atom in a complex ion. The coordination number of Cu2+ in [Cu(NH 3)4]2 + is 4. The most common coordination numbers are 4 and 6, although coordination numbers of 2 and 5 are also known. Depending on the number of donor atoms a ligand possesses, it is classified as mono dentate (1 donor atom), bidentate (2 donor atoms), or polydentate (> 2 donor atoms). Table 22.3 lists some common ligands. Figure 22.3 shows how ethylenediamine, sometimes abbreviated "en," forms two bonds to a metal atom. Bidentate and polydentate ligands are also called chelating agents because of their ability to hold the metal atom like a claw (from the Greek ehele, meaning "claw"). One example is EDTA (Figure 22.4), a polydentate ligand used to treat metal poisoning. Six donor atoms enable EDTA to form a very stable complex ion with lead. This stable complex enables the body to remove lead from the blood. The oxidation state of a transition metal in a complex ion is determined using the known charges of the ligands and the known overall charge of the complex ion. In the complex ion [PtCI 6 for example, each chloride ion ligand has an oxidation number of -' Foi: 't he ovei:ail ' charge of the ion to be -2, the Pt must have an oxidation number of +4 . Sample Problem 22.1 shows how to determine transition metal oxidation states in coordination compounds.

f -,

1:

Remember that the oXidation number of a monoatomic ion is equal to the charge [ ~~ Section 4.4] .

•

Sample Problem 22.1 Determjne the oxidation state of the central metal atom in each of the following compounds: (a) [Ru(NH 3)s(H20)]CI 2 , (b) [Cr(NH 3)6](N0 3) 3, and (c) Fe(CO)s.

I I

Strategy Identify the components of each compound, and use known oxidation states and charges to determine the oxidation state of the metal. ( Continued)

862

CHAPTER 22

Coordination Chemistry

Setup (a) [Ru(NH3)s(H2 0)]CI 2 consists of a complex ion (the part of the formula enclosed in square

brackets) and two Cl- counter ions. Because the overall charge on the compound is zero, the complex ion is [Ru(NH3)s(H2 0)f +. There are six ligands: five ammonia molecules and one water molecule. Each molecule has a zero charge (i.e., each ligand is neutral), so the charge on the metal is equal to the overall charge on the complex ion. (b) [Cr(NH 3)6](N0 3)3 consists of a complex ion and three N0 3- ions, making the complex ion [Cr(NH3)6]3+ Each of the six ammonia molecule ligands is neutral (i.e., each has a zero charge), making the charge on the metal equal to the overall charge on the complex ion. (c) Fe(CO)s does not contain a complex ion. The ligands are CO molecules, which have a zero charge, so the central metal also has a zero charge.

• Think About It To solve a

Solution (a) + 2

problem like this, you must be able to recognize the common polyatomic ions and you must know their charges.

(b) +3 (c) 0

Practice Problem Give oxidation numbers for the metals in (a) K[Au(OH)4] and (b) ~[Fe(CN)6]'

Nomenclature of Coordination Compounds Now that we have discussed the various types of ligands and the oxidation numbers of metals, our next step is to learn how to name coordination compounds. The rules for naming ionic coordination compounds are as follow s: 1. The cation is named before the anion, as in other ionic compounds. The rule holds regardless of whether the complex ion bears a net positive or a net negative charge. In the compounds K 2 [Fe(CN)6] and [Co(NH 3)4]CI, for example, we name the K + and [Co(NH 3)4] + cations first, respectively.

2. Within a complex ion, the ligands are named first, in alphabetical order, and the metal ion is named last.

3. The names of anionic ligands end with the letter 0, whereas neutral ligands are usually called by the names of the molecules. The exceptions are H 2 0 (aquo), CO (carbonyl), and NH3 (ammine). Table 22.4 lists some common ligands and their nomenclature.

4. When two or more of the same ligand are present, use Greek prefixes di-, tri-, tetra-, penta-, and hexa- to specify their number. Thus, the ligands in the cation [Co(NH3)4Cl2t are "tet. . . . . . . . . . . . raamminedichloro." (Note that prefixes are not used for the purpose of alphabetizing the ,

When the name of a ligand already contains a Greek prefix, a different set of prefixes is used to denote the number of the ligand: 2 bis 3 tris 4 tetrakis Two ethylenediamine ligands, for example, would be specified by bis(ethylenediamine).

ligands.)

Ligand

Name of Ligand in Coordination Compound

Bromide, (Br-)

Bromo

Chloride, (CI-)

Chi oro

Cyanide, (CN-)

Cyano

Hydroxide, (OH-)

Hydroxo

Oxide, (0 2 -)

Oxo

Carbonate, (CO~ - )

Carbonato

Nitrite, (NO 3)

Nitro

Oxalate, (C20~-)

Oxalato

Ammonia, (NH3)

Ammine

Carbon monoxide, (CO)

Carbonyl

Water, (H 20)

Aquo

Ethylenediamine

Ethylenediamine

Ethy lenediarninetetraacetate

Ethylenediaminetetraacetate

•

SECTION 22.1

Metal

,

Coordination Compounds

863

Name of Metal in Anionic Complex

Aluminum

Aluminate

Chromium

Chromate

Cobalt

Cobaltate

Copper

Cuprate

Gold

Aurate

Iron

Ferrate

Lead

Plumbate

Manganese

Manganate

Molybdenum

Molybdate

Nickel

Nickelate

Silver

Argentate

Tin

Stannate

Tungsten

Tungstate

Zinc

Zincate

5. The oxidation number of the metal is indicated in Roman numerals immediately following the name of the metal. For example, the Roman numeral III is used to indicate the + 3 oxida- . . . . . tion state of chromium in [Cr(NH3)4Cl2t, which is called tetraamminedichlorochromium(III) •

lOn.

6. If the complex is an anion, its name ends in -ate. In

~[Fe(CN)6], for example, the anion

Note that there is no space between the compound name and the parenthetical Roman numeral.

[Fe(CN)6t- is called hexacyanoferrate(II) ion. Note that the Roman numeral indicating the oxidation state of the metal follows the suffix -ate. Table 22.5 lists the names of anions containing metal atoms. Sample Problems 22.2 and 22.3 apply these rules to the nomenclature of coordination compounds.

Write the names of the following coordination compounds: (a) [Co(NH3)4CI2lCl and (b) K3[Fe(CNM Strategy For each compound, name the cation first and the anion second. Refer to Tables

22.4 and

22.5 for the names of ligands and anions containing metal atoms. Setup (a) The cation is a complex ion containing four ammonia molecules and two chloride ions.

The counter ion is chloride (Cl- ), so the charge on the complex cation is state of cobalt + 3.

+ 1, making the oxidation

(b) The cation is K+, and the anion is a complex ion containing six cyanide ions. The charge on the complex ion is - 3, making the oxidation state of iron + 3. Solution (a) Tetraamminedichlorocobalt(III) chloride

(b) Potassium hexacyanoferrate(III)

Practice Problem Give the con-ect name for (a) [Cr(H20 )4Cl2lCI and (b) [Cr(en)3lCI3.

Sample Problem 22.3 •

Write formulas for the following compounds: (a) pentaamminechlorocobalt(III) chloride and (b) dichlorobis( ethylenediamine )platinum(IV) nitrate.

(Continued)

Think About It When the anion is

a complex ion, its name must end in -ate, followed by the metal's oxidation state in Roman numerals. Also, do not use prefixes to denote numbers of counter ions.

864

CHAPTER 22

Coordination Chemistry

Strategy If you can't remember them yet, refer to Tables 22.4 and

22.5 for the names of ligands

and anions containing metal atoms. Setup (a) There are six ligands: five NH3 molecules and one Cl- ion. Tb.e oxidation state of cobalt

is +3, making the overall charge on the complex ion +2. Therefore, there are two chloride ions as counter ions.

Think About It Although ligands

are alphabetized in a compound's name, they do not necessarily appear in alphabetical order in the compound's formula.

(b) There are four ligands: two bidentate ethylenediamines and two Cl- ions. The oxidation state of platinum is +4, making the overall charge on the complex ion +2. Therefore, there are two nitrate ions as counter ions. Solution (a) [Co(NH 3)sCl]CI 2

(b) [Pt(en)zCI2](N03)2

Practice Problem Write the formulas for (a) tris(ethylenediamine)cobalt(lII) sulfate and (b) sodium

hexanitrocobaitate(III).

---------------------------------------------------------~

Checkpoint 22.1 22 .1.1

22.1.2

Coordination Compounds

Select the correct name for the compound [Cu(NH 3)4]CI2 .

22 .1.3

Select the correct formula for pentaaminenitrocobalt(III).

a) Copperteu'aammine dichloride

a) [Co(NH 3)sN0 3]3+

b) Tetraamminecopper(II) chloride

b) [Co(NH 3 )sN0 3 f

c) Tetraaminedichlorocuprate(II)

c) Co(NH3)sN03

d) Dichlorotetraaminecopper(II)

d) [Co(NH 3)s](N03)

e) Tetraaminedichlorocopper(II)

e) [Co(NH3)S] (N0 3)2

Select the correct name for the compound K 3[FeF6].

22.1.4

+

a) Tripotassiumironhexaft uoride

Select the correct formula for tetraaq uodichlorochromi urn (III) chloride.

b) Hexaftuorotripotassiumferrate(III)

a) [Cr(H 2O)4 CI2]CI3

c) Hexaftuoroiron(III) potassium

b) [Cr(H 2O)4CI2]CI2

d) Potassi um hexaftuoroferrate(lII)

c) [Cr(H2O)4CI 2]Cl

e) Potassium ironhexaftuorate

d) [Cr(H2O)4]CI3 e) [Cr(H2O)4]Clz

Structure of Coordination Compounds The geometry of a coordination compound often plays a significant role in determining its properties. Figure 22.5 shows four different geometric arrangements for metal atoms with monodentate ligands. In these diagrams we See that structure and the coordination number of the metal relate to each other as follow s :

Coordination Number 2

Structure Linear Tetrahedral or square planar Octahedral

4 6 Figure 22.5

Common geometries of complex ions. In each case M is a metal and L is a monodentate ligand.

L

L

M

L

M

L - - ; -------L

L L

~M /

L

/~L

L Linear

Tetrahedral

Square planar

/

L

-M-

/

L

L Octahedral

L •

SECTION 22.2

Structure of Coordination Compounds

In studying the geometry of coordination compounds, we sometimes find that there is more than one way to an'ange the ligands around the central atom. Such compounds in which ligands are arranged differently, known as stereoisomers, have distinctly different physical and chemical properties. Coordination compounds may exhibit two types of stereoisomerism: geometric and optical. Geometric isomers are stereoisomers that cannot be interconverted without breaking chemical bonds. Geometric isomers come in pairs. We use the terms cis and trans to distinguish one ' geometric isomer of a compound from the other. Cis means that two palticular atoms (or groups of atoms) are adjacent to each other, and trans means that the atoms (or groups of atoms) are on opposite sides in the structural formula. The cis and trans isomers of coordination compounds generally have quite different colors, melting points, dipole moments, and chemical reactivities. Figure 22.6 shows the cis and trans isomers of diamrninedichloroplatinum(II). Note that although the types of bonds are the same in both isomers (two Pt-N and two Pt-Cl bonds), the spatial arrangements are different. Another example is the tetraamminedichlorocobalt(III) ion, shown in Figure 22.7. Optical isomers are nonsuperimposable mirror images. (Superimposable means that if one structure is laid over the other, the positions of all the atoms will match.) Like geometric isomers, optical isomers come in pairs. However, the optical isomers of a compound have identical physical and chemical properties, such as melting point, boiling point, dipole moment, and chemical reactivity toward molecules that are not themselves optical isomers. Optical isomers differ from each other, though, in their interactions with plane-polarized light, as we will see. The structural relationship between two optical isomers is analogous to the relationship between your left and right hands. If you place your left hand in front of a mirror, the image you see will look like your right hand (Figure 22.8). Your left hand and right hand are mirror images of each other. They are nonsuperimposable, however, because when you place your left hand over your right hand (with both palms facing down), they do not match. This is why a right-handed glove will not fit comfortably on your left hand. Figure 22.9 shows the cis and trans isomers of dichlorobis(ethylenediamine)cobalt(III) ion and the mirror image of each. Careful examination reveals that the trans isomer and its mirror • ••

•••

•••

••••

a

4-1

o

•

.0 ..... > ..... ......

g

Metal

Reduction Process

Lithium, sodium, magnesium, calcium

Electrolytic reduction of the molten chloride

Aluminum

Electrolytic reduction of anhydrous oxide (in molten cryolite)

Chromium, manganese, titanium, vanadium, iron, zinc

of the metal oxide with a more Reduction . electropositive metal, or reduction with coke and carbon monoxide

Mercury, silver, platinum, copper, gold

These metals occur in the free (uncombined) state, or they can be obtained by roasting their sulfides

01)

c:

.~

'" ~

~

~

o

high temperatures. The reduction in these procedures may be accomplished either chemically or electrolytically.

Chemical Reduction A more electropositive metal can be used as a reducing agent to separate a less electropositive metal from its compound at high temperatures: V 2 0 S(s) TiCli g)

CrZ0 3(S)

+ SCa(l) ---. 2V(l) + SCaO(s) + 2Mg(l) • Ti(s) + 2MgCl z(l) + 2AI(s)

3Mn30 4(S) + 8AI(s)

• 2Cr(l)

+ AI2 0 3(s)

• 9Mn(l) + 4AI20 3(s)

In some cases, even molecular hydrogen can be used as a reducing agent, as in the preparation of tungsten (used as filaments in lightbulbs) from tungsten(VI) oxide:

Electrolytic Reduction Electrolytic reduction is suitable for very electropositive metals, such as sodium, magnesium, and aluminum. The process is usually carried out on the anhydrous molten oxide or halide of the metal: 2MO(l) - _ . 2M (at cathode) + O 2 (at anode) 2MCl(l)

• 2M (at cathode)

+ C12 (at anode)

We will describe the specific procedures later in this section.

The Metallurgy of Iron _ _ Multimedia Chemical Reactions- smelting of iron.

Iron exists in Earth's crust in many different minerals, such as iron pyrite (FeS2), siderite (FeC0 3), hematite (Fe20 3), and magnetite (Fe30 4, often represented as FeO . Fe20 3)' Of these, hematite and magnetite are particularly suitable for the extraction of iron. The metallurgical processing of iron involves the chemical reduction of the minerals by carbon (in the form of coke) in a blast furnace (Figure 23.3). The concentrated iron ore, limestone (CaC0 3), and coke are introduced into the furnace from the top. A blast of hot air is forced up the furnace from the bottom-hence, the name blast furna ce. The oxygen gas reacts with the carbon in the coke to form mostly carbon monoxide and some carbon dioxide. These reactions are highly exothermic, and as the hot CO and CO 2 gases rise, they react with the iron oxides in different temperature zones as shown in Figure 23.3. The key steps in the extraction of iron are 3Fez0 3(s) + CO(g) - _ . 2Fe30 4(S) + COz(g)

+ CO(g) FeO (s) + CO(g)

Fe30 i s)

• 3FeO(s) • Fe(l)

+ CO 2(g)

+ CO 2(g)

SECTION 23.2

.,..-CO,C02

3Fe203

+ co

•

3Fe304

CaC0 3

• CaO

+ CO

• 3FeO

Fe304

-

+ CO2

+ CO 2

+ CO2

-+ CO2 FeO + CO C

'"

50 .g'"

1200°C

'" ~

lS00°C

..c:

-

• 2CO • Fe

+ CO 2

Iron melts Molten slag forms 2C

0

+ 02

• 2CO

"0

.~

2000°C

0

CI)

,

Hot air blast

Hot air blast

The limestone decomposes in the furnace as follows:

The calcium oxide then reacts with the impurities in the iron, which are mostly sand (Si0 2 ) and aluminum oxide (AI 20 3): CaO(s) + Si02 (s) --+. CaSi03 (l) CaO(s)

+ AI20 3(s)

885

Figure 23.3 A blast furnace. Iron ore, limestone, and coke are introduced at the top of the furnace. Iron is obtained from the ore by reduction with carbon.

Charge (ore, limestone, coke)

"0

Metallurgical Processes

• Ca(AI0 2)z(1)

The mixture of calcium silicate and calcium aluminate that remains molten at the furnace temperature is known as slag. By the time the ore works its way down to the bottom of the furnace, most of it has already been reduced to iron. The temperature of the lower part of the furnace is above the melting point of impure iron, and so the molten iron at the lower level can be run off to a receiver. The slag, because it is less dense, forms the top layer above the molten iron and can be run off at that level, as shown in Figure 23.3. Iron extracted in this way contains many impurities and is called pig iron; it may contain up to 5 percent carbon and some silicon, phosphorus, manganese, and sulfur. Some of the impurities stem from the silicate and phosphate minerals, while carbon and sulfur come from coke. Pig iron ., is granular and brittle. It has a relatively low melting point (about 1180°C), so it can be cast in various forms; for this reason it is also called cast iron.

Steelmaking Steel manufacturing is one of the most important metal industries. In the United States, the annual consumption of steel is well above 100 million tons. Steel is an iron alloy that contains from 0.03 to 1.4 percent carbon plus various amounts of other elements. The wide range of useful mechanical properties associated with steel is primarily a function of the chemical composition and heat treatment of a particular type of steel. Whereas the production of iron is basically a reduction process (converting iron oxides to metallic iron), the conversion of iron to steel is essentially an oxidation process in which the unwanted impurities are removed from the iron by reaction with oxygen gas. One of several methods used in steelmaking is the basic oxygen process. Because of its ease of operation and

•

886

CHAPTER 23

Metallurgy and the Chemistry of Metals

Figure 23 .4 The basic oxygen process of steelmaking. The capacity of a typical vessel is 100 tons of cast iron. ____-

CaO or Si0 2

•

Horizontal position

Molten steel

-------- Molten steel + slag Vertical position I

the relatively short time (about 20 minutes) required for each large-scale (hundreds of tons) conversion, the basic oxygen process is by far the most common means of producing steel today. Figure 23.4 shows the basic oxygen process. Molten iron from the blast furnace is poured into an upright cylindrical vessel. Pressurized oxygen gas is introduced via a water-cooled tube above the molten metal. Under these conditions, manganese, phosphorus , and silicon, as well as excess carbon, react with oxygen to form oxides. These oxides are then reacted with the appropriate fluxes (e.g., CaO or Si02) to form slag. The type of flux chosen depends on the composition of the iron. If the main impurities are silicon and phosphorus, a basic flux such as CaO is added to the iron:

,

Si02(s) P4 0

lO (l)

+ CaO(s) -

+ 6CaO(s)

...... CaSi0 3(l) • 2Ca3(P04hCZ)

On the other hand, if manganese is the main impurity, then an acidic flux such as Si0 2 is needed to form the slag:

The molten steel is sampled at intervals. When the desired blend of carbon has been reached, the vessel is rotated to a horizontal position so that the tapped off (Figure 23.5). The properties of steel depend not only on its chemical composition treatment. At high temperatures, iron and carbon in steel combine to form called cementite: 3Fe(s)

Figure 23.5

Steelmaking.

+ C(s) -

and other impurities molten steel can be but also on the heat iron carbide (Fe3C),

...... Fe3C(S)

The forward reaction is endothermic, so the formation of cementite is favored at high temperatures. When steel containing cementite is cooled slowly, the preceding equilibrium shifts to the left and the carbon separates as small particles of graphite, which give the steel a grey color. (Very slow decomposition of cementite also takes place at room temperature.) If the steel is cooled rapidly, equilibrium is not attained and the carbon remains largely in the form of cementite (Fe3C)' Steel containing cementite is light in color, and it is harder and more brittle than that containing graphite. Heating the steel to some appropriate temperature for a short time and then cooling it rapidly in order to give it the desired mechanical properties is known as "tempering." In this way, the ratio of carbon present as graphite and as cementite can be varied within rather wide limits. Table 23.3 lists the composition, properties, and uses of various types of steel.

SECTION 23.2

Metallurgical Processes

887

Composition (Percent by Mass)* Type

C

Mn

P

S

Si

Plain

1.35

1.65

0.04

0.05

0.06

High-strength

0.25

1.65

0.04

0.05

0.15-0.9

0.4-1.0

0.3-1.3

Stainless

0.03-1.2

1.0-10

0.04-0.06

0.03

1-3

1-22

4.0-27

*A

Ni

Cr

Others

Uses

Cu (0.2-0.6)

Sheet products, tools

Cu (0.01-0.08)

Construction, steam turbines Kitchen utensils, razor blades

single number indicates the maximum amount of the substance present.

Purification of Metals Metals prepared by reduction usually need further treatment to remove impurities. The extent of purification depends on how the metal will be used. Three common purification procedures are distillation, electrolysis, and zone refining. Distillation

Metals that have low boiling points, such as mercury, magnesium, and zinc, can be separated from other metals by fractional distillation. One well-known method of fractional distillation is the Moni process for the purification of nickel. Carbon monoxide gas is passed over the impure nickel metal at about 70°C to form the volatile tetracarbonylnickel (b.p. 43 °C), a highly toxic substance, which is separated from the less volatile impurities by distillation: Ni(s) + 4CO(g) -

..... Ni(CO)ig)

Pure metallic nickel is recovered from Ni(CO)4 by heating the gas at 200°C: Ni(CO)ig) -

..... Ni(s)

+ 4CO(g)

The carbon monoxide that is released is recycled back into the process. Electrolysis

Electrolysis is another important purification technique. The copper metal obtained by roasting copper sulfide usually contains impurities such as zinc, iron, silver, and gold. The more electropositive metals are removed by an electrolysis process in which the impure copper acts as the anode and pure copper acts as the cathode in a sulfuric acid solution containing Cu2+ ions (Figure 23.6). The reactions are

Cu(s)' Cu2+(aq)

Anode (oxidation): Cathode (reduction): Impure copper anode

Cu 2 +(aq)

+ 2e -

+ 2e-

• Cu(s)

Pure copper cathode

Figure 23.6

Electrolytic purification of copper.

I

•

1. Ludwig Mond (1839-1909). British chemist of German origin. Mond made many important contributions to industrial ::" mistry. His method for purifying nickel by converting it to the volatile Ni(CO)4 compound has been described as having ::;" "en "wings" to the metal.

I

888

CHAPTER 23

Metallurgy and the Chemistry of Metals

Reactive metals in the copper anode, such as iron and zinc, are also oxidized at the anode and enter the solution as Fe2+ and Zn2+ ions. They are not reduced at the cathode, however. The less electropositive metals, such as gold and silver, are not oxidized at the anode. Eventually, as the copper anode dissolves, these metals fall to the bottom of the cell. Thus, the net result of this electrolysis process is the transfer of copper from the anode to the cathode. Copper prepared this way has a purity greater than 99.5 percent (Figure 23.7).

Zone Refining

Figure 23.7 Copper cathodes used in the electrorefining process.

Another often-used method of obtaining extremely pure metals is zone refining. In this process, a metal rod containing a few impurities is drawn through an electric heating coil that melts the metal (Figure 23.8). Most impurities dissolve in the molten metal. As the metal rod emerges from the heating coil, it cools and the pure metal crystallizes, leaving the impurities in the molten metal portion that is still in the heating coil. (This is analogous to the freezing of seawater, in which the solid that separates is mostly pure solvent-water. In zone refining, the liquid metal acts as the solvent and the impurities act as the solutes.) When the molten zone carrying the impurities, now at increased concentration, reaches the end of the rod, it is allowed to cool and is then cut off. Repeating this procedure a number of times results in metal with a purity greater than 99.99 percent.

Band Theory of Conductivity To gain a better understanding of the conductivity properties of metals, we must also apply our knowledge of quantum mechanics. The model we will use to study metallic bonding is band theory, so called because it states that delocalized electrons move freely through "bands" formed by overlapping molecular orbitals. We will also apply band theory to certain elements that are semiconductors.

Conductors Metals are characterized by high electrical conductivity. Consider magnesium, for example. The electron configuration of Mg is [Ne]3i, so each atom has two valence electrons in the 3s orbital. In a metallic crystal, the atoms are packed closely together, so the energy levels of each magnesium atom are affected by the immediate neighbors of the atom as a result of orbital overlaps. According to molecular orbital theory [ ~~ Section 9 .6] , the interaction between two atomic orbitals leads to the fOImation of a bonding and an antibonding molecular orbital. Because the number of atoms in even a small piece of magnesium is enormously large (on the order of 1020 atoms), the number of molecular orbitals they form is also very large. These molecular orbitals are so closely spaced on the energy scale that they are more appropriately described as a "band" (Figure 23.9). The closely spaced filled energy levels make up the valence band. The upper half of the energy

Figure 23.8 Zone-refining technique for purifying metals. Top to bottom: An impure metal rod is moved slowly through a heating coil. As the metal rod moves forward, the impurities dissolve in the molten portion of the metal while pure metal crystallizes out in front of the molten zone.

Heating coil ,

, ....... .:

,-

,

'JVvV

'

"

,

-,

,-,

.. . -

l

..... -. .. , ", .,. . ,.. ..- . " ,'.,'.. -,

.. .....' ' . - j ' . '. . , •...• ,'* •

. :' .. s:.... 'h

•

Metal rod

..

~

•.•• .•

l ' ,. •

*

",

e

.'~'

'

. . " -, ..... .. ' ",. " .. . .,... .... ., .. ... ' ..

. .. , " , '. , .

'~

,-'

jiiA1t~g;?.. "

. , . ..•. =

+.",;t;"

~

.'~'

'....

SECTION 23.3

Band Th eory of Co nd uct ivit y

Figure 23.9

Conduction band

/ I

Valence band

",

/

",

/

'I.

./

./ 'I.

2p

2s

Is

r--

-

f-

12+

12 +

Mg

Mg

-

r-

r--

12

+

Mg

-

+

12+

Mg

Mg

12

Formation of conduction bands in magnesium . The electrons in the Is, 2s, and 2p orbitals are localized on each Mg atom . However, the 3s and 3p orbitals overlap to form delocalized molecular orbitals. Electrons in these orbitals can travel throughout the metal, and this accounts for the electrical conductivity of the metal.

•

Figure 23.10

Comparison of the energy gaps between the valence band and the conduction band in a metal, a semiconductor, and an insulator. In a metal the energy gap is virtually nonexistent; in a semiconductor the energy gap is small; and in an insulator the energy gap is very large, thus making the promotion of an electron from the valence band to the conduction band difficult.

levels corresponds to the empty, delocalized molecular orbitals formed by the overlap of the 3p orbitals. This set of closely spaced empty levels is called the conduction band. We can imagine a metallic crystal as an array of positive ions immersed in a sea of delocalized valence electrons. The great cohesive force resulting from the delocalization is partly responsible for the strength noted in most metals. Because the valence band and the conduction band are adjacent to each other, the amount of energy needed to promote a valence electron to the conduction band is negligible. There the electron can travel freely through the metal, because the conduction band is void of electrons. This freedom of movement explains why metals are good conductors that is , they are capable of conducting an electric current. Why don't substances like wood and glass conduct electricity as metals do ? Figure 23.10 provides an answer to this question. Basically, the electrical conductivity of a solid depends on the spacing of the energy bands and the extent to which they are occupied. In magnesium and other metals, the valence bands are adjacent to the conduction bands, so these metals readily act as conductors. In wood and glass, on the other hand, the gap between the valence band and the conduction band is considerably greater than it is in a metal. Consequently, much more energy is needed to excite an electron into the conduction band. Lacking this energy, electrons cannot move freely. Therefore, glass and wood are insulators, ineffective conductors of electricity.

Semiconductors Semiconductors are elements that normally are not conductors, but will conduct electricity at elevated temperatures or when combined with a small amount of certain other elements. The Group 4A elements silicon and germanium are especially suited for thi s purpose. The use of semiconductors in transistors and solar cells, to name two applications, has revolutionized the electronic industry in recent decades, leading to the increased miniaturization of electronic equipment. The energy gap between the filled and empty bands of these solids is much smaller than it is for insulators (see Figure 23.10). If the energy needed to excite electrons fro m the valence band into the conduction band is provided, the solid becomes a conductor. This behavior is opposite that of the metals. A metal's ability to conduct electricity decreases with increasing temperature, because the enhanced vibration of atoms at higher temperatures tends to disrupt the flow of electrons. The ability of a semiconductor to conduct electricity can also be enhanced by adding small amounts of certain impurities to the element, a process called doping. Consider what happens,

889

,

890

CHAPTER 23

Metallurgy and the Chemistry of Metals

Figure 23.11

(a) Silicon crystal doped with phosphorus. (b) Silicon crystal doped with boron. Note the formation of a negative center in (a) and a positive center in (b) .

•

(b)

(a)

• for example, when a trace amount of boron or phosphorus is added to solid silicon. (Only about five out of every million Si atoms are replaced by B or P atoms.) The structure of solid silicon is similar to that of diamond; that is, each Si atom is covalently bonded to four other Si atoms. Phosphorus ([Ne]3/3p 3) has one more valence electron than silicon ([Ne ]3 s23p 2), so there is a valence electron left over after four of them are used to form covalent bonds with silicon (Figure 23.11). This extra electron can be removed from the phosphorus atom by applying a voltage across the solid. The free electron can move through the structure and function as a conduction electron. Impurities of this type are known as donor impurities, because they provide conduction electrons. Solids containing donor impurities are called n-type semiconductors, where n stands for negative (the charge of the "extra" electron). The opposite effect occurs if boron is added to silicon. A boron atom has three valence electrons Oi2/2pl), one less than silicon. Thus, for every boron atom in the silicon crystal, there is a single vacancy in a bonding orbital. It is possible, though, to excite a valence electron from a nearby Si into this vacant orbital. A vacancy created at that Si atom can then be filled by an electron from a neighboring Si atom, and so on. In this manner, electrons can move through the crystal in one direction while the vacancies, or "positive holes," move in the opposite direction, and the solid becomes an electrical conductor. Impurities that are electron deficient are called acceptor impurities. Semiconductors that contain acceptor impuritie ~ are called p-type semiconductors, where p stands for positive. In both the p-type and n-type semiconductors, the energy gap between the valence band and the conduction band is effectively reduced, so only a small amount of energy is needed to excite the electrons. Typically, the conductivity of a semiconductor is increased by a factor of 100,000 or so by the presence of impurity atoms. The growth of the semiconductor industry since the early 1960s has been truly remarkable. Today semiconductors are essential components of nearly all electronic equipment, ranging from radios and television sets to pocket calculators and computers. One of the main advantages of solid-state devices over vacuum-tube electronics is that the former can be made on a single "chip" of silicon no larger than the cross section of a pencil eraser. Consequently, much more equipment can be packed into a small volume a point of particular importance in space travel, as well as in handheld calculators and microprocessors (computers-on-a-chip).

Periodic Trends in Metallic Properties Metals are lustrous in appearance, solid at room temperature (with the exception of mercury), good conductors of heat and electricity, malleable (can be hammered fiat), and ductile (can be drawn into wires). Figure 23.12 shows the positions of the representative metals and the Group

Figure 23.12

Main group metals (green) and Group 2B metals (blue) according to their positions in the periodic table.

lA

8A 18

1

2A H 2

3A 4A 5A 6A 7A 13 14 15 16 17 He

Li Be

B C N 0

3B 4B 5B 6B 7B1s8B~ IB 2B Na Mg 3 4 5 6 7 8 9 10 11 12 AI Si K

P

F Ne

S Cl Ar

Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I

Xe

Cs Ba La Hi Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg

SECTION 23.5

The Alkali M etals

2B metals in the periodic table. (The transition metals are discussed in Chapter 22.) As we saw in Chapter 8, the electronegativity of elements increases from left to right across a period and from bottom to top in a group [ ~~ Section 8.4, Figure 8.6] . The metallic character of metals increases in just the opposite directions-that is, from right to left across a period and from top to bottom in a group. Because metals generally have low electronegativities, they tend to form cations and almost always have positive oxidation numbers in their compounds. However, beryllium and magnesium in Group 2A and the metals in Group 3A and beyond also form covalent compounds. In Sections 23.5 through 23.7 we will study the chemistry of selected metals from Group lA (the alkali metals), Group 2A (the alkaline earth metals), and Group 3A (aluminum).

The Alkali Metals As a group, the alkali metals (the Group 1A elements) are the most electropositive (or the least electronegative) elements known. They exhibit many similar properties, some of which are listed in Table 23.4.. Based on their electron configurations, we expect the oxidation number of these elements in their compounds to be + 1 because the cations would be isoelectronic with the preceding noble gases. This is indeed the case. The alkali metals have low melting points and are soft enough to be sliced with a knife. These metals all possess a body-centered crystal structure with low packing efficiency. This accounts for their low densities among metals. In fact, lithium is the lightest metal known. Because of their great chemical reactivity, the alkali metals never occur naturally in elemental form; instead, they are found combined with halide, sulfate, carbonate, and silicate ions. In this section we will describe the chemistry of two members of Group lA sodium and potassium. The chemistry of lithium, rubidium, and cesium is less important; all isotopes of francium, the last member of the group, are radioactive. Sodium and potassium are about equally abundant in nature. They occur in silicate minerals such as albite (NaAISi30g) and orthoclase (KAISi30g). Over long periods of time (on a geologic scale), silicate minerals are slowly decomposed by wind and rain, and their sodium and potassium ions are converted to more soluble compounds. Eventually rain leaches these compounds out of the soil and carries them to the sea. Yet when we look at the composition of seawater, we find that the concentration ratio of sodium to potassium is about 28 to 1. The reason for this uneven distribution is that potassium is essential to plant growth, while sodium is not. Thus, plants take up many of the potassium ions along the way, while sodium ions are free to move on to the sea. Other minerals that contain sodium or potassium are halite (NaCl), shown in Figure 23.13, Chile saltpeter (NaN0 3), and sylvite (KCI). Sodium chloride is also obtained from rock salt. Metallic sodium is most conveniently obtained from molten sodium chloride by electroly-' sis in the Downs cell (review Figure 19.10). The melting point of sodium chloride is rather high (801 °C), and much energy is needed to keep large amounts of the substance molten. Adding a suitable substance, such as CaCl 2 , lowers the melting point to about 600°C a more convenient temperature for the electrolysis process.

Li

Na

K

Rb

Cs

Valence electron configuration

2S1

3s 1

4S1

5s 1

6s 1

Density (g/cm 3)

0.534

0.97

0.86

1.53

1.87

Melting point (OC)

179

97.6

63

39

28

Boiling point (0C)

1317

892

770

688

678

Atomic radius (pm)

155

190

235

248

267

Ionic radius (pm)*

60

95

133

148

169

Ionization energy (kJ/mol)

520

496

419

403

375

Electronegativity

1.0

0.9

0.8

0.8

0.7

-3.05

-2.71

-2.93

-2.93

-2.92

Standard reduction potential (V)t

* Refers to the cation M +, where M denotes an alkali metal atom. t The ha lf-reaction is M +(aq)

+ e-

• M(s) .

Figure 23.13

,

Halite (NaCt).

891

892

CHAPTER 23

Metallurgy and the Chemistry of Metals

Metallic potassium cannot be easily prepared by the electrolysis of molten KCl because it is too soluble in the molten KCl to float to the top of the cell for collection. Moreover, it vaporizes readily at the operating temperatures, creating hazardous conditions. Instead, it is usually obtained by the distillation of molten KCl in the presence of sodium vapor at 892°C. The reaction that takes place at this temperature is Na(g)

, .

••

'

••

>':

Multimedia

Periodic Table properties of the alkali and alkaline earth metals.

+ KCl(l) --+. NaCl(l) + K(g)

This reaction may seem strange given that potassium is a stronger reducing agent than sodium (see Table 23.4). Potassium has a lower boiling point (770°C) than sodium (892°C), however, so it is more volatile at 892°C and distills off more easily. According to Le Chfltelier's principle, constantly removing the potassium vapor drives the reaction to the right, ensuring metallic potassium is recovered. Sodium and potassium are both extremely reactive, but potassium is the more reactive ofthe two. Both react with water to form the corresponding hydroxides. In a limited supply of oxygen, sodium bums to form sodium oxide (Na?O). In the presence of excess oxygen, however, sodium forms the pale-yellow peroxide:

Sodium peroxide reacts with water to give an alkaline solution and hydrogen peroxide:

Like sodium, potassium forms the peroxide. In addition, potassium also forms the superoxide when it bums in air: K(s)

+ 0 2(g) --+. K0 2 (s)

When potassium superoxide reacts with water, oxygen gas is evolved:

This reaction is utilized in breathing equipment (Figure 23.14). Exhaled air contains both moisture and carbon dioxide. The moisture reacts with K0 2 in the apparatus to generate oxygen gas as shown in the preceding reaction. Furthermore, K0 2 also reacts with exhaled CO 2 , which produces more oxygen gas:

Thus, a person using the apparatus can continue to breathe oxygen without being exposed to toxic fumes outside. Sodium and potassium metals dissolve in liquid ammonia to produce beautiful blue solutions: Na NH3.Na+ + e-

.

Both the cation and the electron exist in the solvated form, and the solvated electrons are responsible for the characteristic blue color of such solutions. Metal-ammonia solutions are powerful reducing agents (because they contain free electrons); they are useful in synthesizing both organic and inorganic compounds. It was discovered that the hitherto unknown alkali metal anions, M-, are also formed in such solutions. This means that an ammonia solution of an alkali metal contains ion pairs such as Na +Na- and K+K- ! (In each case, the metal cation exists as a complex ion with crown ether, an organic compound with a high affinity for cations.) In fact, these "salts" are so stable that they can be isolated in crystalline form. This finding is of considerable theoretical interest, because it shows clearly that the alkali metals can have an oxidation number of -1 , although - 1 is not found in ordinary compounds. Sodium and potassium are essential elements of living matter. Sodium ions and potassium ions are present in intracellular and extracellular fluids, and they are essential for osmotic balance and enzyme functions. We now describe the preparations and uses of several of the important compounds of sodium and potassium.

Figure 23.14

Self-contained breathing apparatus.

Sodium Chloride The source, properties, and uses of sodium chloride were discussed in Chapter 7.

SECTION 23.6

The Alkaline Earth Metals

Sodium Carbonate Sodium carbonate (called soda ash) is used in all kinds of industrial processes, including water treatment and the manufacture of soaps, detergents, medicines, and food additives. Today about half of all Na2C03 produced is used in the glass industry. Sodium carbonate ranks eleventh among the chemicals produced in the United States. For many years, Na2C03 was produced by the Solvay2 process, in which ammonia is first dissolved in a saturated solution of sodium chloride. Bubbling carbon dioxide into the solution precipitates sodium bicarbonate as follows:

Sodium bicarbonate is then separated from the solution and heated to give sodium carbonate:

However, the rising cost of ammonia and the pollution problem resulting from the by-products have prompted chemists to look for other sources of sodium carbonate. One is the mineral trona [Nas(C0 3MHC0 3 ) • 2H 20], large deposits of which have been found in Wyoming. When trona is crushed and h.,ated, it decomposes as follows:

The sodium carbonate obtained this way is dissolved in water, the solution is filtered to remove the insoluble impurities, and the sodium carbonate is crystallized as Na2C03 . lOH 2 0. Finally, the hydrate is heated to give pure, anhydrous sodium carbonate.

Sodium Hydroxide and Potassium Hydroxide The properties of sodium hydroxide and potassium hydroxide are very similar. These hydroxides are prepared by the electrolysis of aqueous NaCI and KCI solutions; both hydroxides are strong bases and very soluble in water. Sodium hydroxide is used in the manufacture of soap and many organic and inorganic compounds. Potassium hydroxide is used as an electrolyte in some storage batteries, and aqueous potassium hydroxide is used to remove carbon dioxide and sulfur dioxide from air.

Sodium Nitrate and Potassium Nitrate Large deposits of sodium nitrate (Chile saltpeter) are found in Chile. It decomposes with the evolution of oxygen at about 500°C:

Potassium nitrate (saltpeter) is prepared beginning with the "reaction" KCI(aq)

+ NaN0 3 (aq) --+. KN0 3(aq) + NaCI(aq)

This process is carried out just below 100°C. Because KN0 3 is the least soluble salt at room remperature, it is separated from the solution by fractional crystallization. Like NaN0 3, KN0 3 decomposes when heated. Gunpowder consists of potassium nitrate, wood charcoal, and sulfur in the approximate proportions of 6: 1: 1 by mass. When gunpowder is heated, the reaction is

The sudden formation of hot nitrogen and carbon dioxide gases causes an explosion.

The Alkaline Earth Metals The alkaline earth metals are somewhat less electropositive and less reactive than the alkali metals. Except for the first member of the family, beryllium, which resembles aluminum (a Group 3A metal) in some respects, the alkaline earth metals have similar chemical properties. Because their ~.1-- ions attain the stable electron configuration ofthe preceding noble gas , the oxidation number 0.- alkaline earth metals in the combined form is almost always +2. Table 23.5 lists some common ~

::.. F::nest Solvay (1838-1922). Belgian chemist. Solvay's main contribution to industrial chemistry was the development of !he process for the production of sodium carbonate that now bears his name.

•

893

894

CHAPTER 23

Metal lurgy and the Chemistry of Metals

Be

•

Sr

Mg

Ca

5s

2

6i

Ba

Valence electron configuration

2i

3i

4i

Density (g/cm3)

1.86

1.74

1.55

2.6

3.5

Melting point (DC)

1280

650

838

770

714

Boiling point (DC)

2770

1107

1484

1380

1640

Atomic radius (pm)

112

160

197

215

222

Ionic radius (pm)*

31

65

99

113

135

First ionization energy (kJ/mol)

899

738

590

548

502

Second ionization energy (kJ/mol)

1757

1450

1145

1058

958

Electronegativity

1.5

1.2

1.0

1.0

0.9

Standard reduction potential (V) t

-1.85

-2.37

-2.87

-2.89

-2.90

* Refers to the cation M 2 +, w here M denotes an alkali earth metal atom . t The half-reaction is M 2 +(aq)

+ 2e -

• M(s).

properties of these metals. Radium is not included in the table because all radium isotopes are radioactive and it is difficult and expensive to study the chemistry of this Group 2A element.

Magnesium Magnesium is the sixth most plentiful element in Earth's crust (about 2.5 percent by mass). Among the principal magnesium ores are brucite [Mg(OH)z], dolomite (CaC0 3 • MgC0 3) (Figure 23.l5), and epsomite (MgS04 . 7HzO). Seawater is a good source of magnesium there are about 1.3 g of magnesium in each kilogram of seawater. As is the case with most alkali and alkaline earth metals, metallic magnesium is obtained by electrolysis, in this case from its molten chloride, MgCl 2 (obtained from seawater). The chemistry of magnesium is intermediate between that of beryllium and the heavier Group 2A elements. Magnesium does not react with cold water but does react slowly with steam: Mg(s)

+ HzO(g) -

-+.

MgO(s)

+ HzCg)

It burns brilliantly in air to produce magnesium oxide and magnesium nitride: 2Mg(s)

+ Oz(g) -

3Mg(s)

+ Nz(g)

-+.

2MgO(s)

• Mg 3N 2(s)

This property makes magnesium (in the form of thin ribbons or fibers) useful in flash photography and flares. Magnesium oxide reacts very slowly with water to form magnesium hydroxide, a white solid suspension called milk of magnesia, which is used to treat acid indigestion: MgO(s)

+ HzO(I) -

-+.

Mg(OHh(s)

Magnesium is a typical alkaline earth metal in that its hydroxide is a strong base. (The only alkaline earth hydroxide that is not a strong base is Be(OHh which is amphoteric.) The major uses of magnesium are in lightweight structural alloys, for cathodic protection; in organic synthesis; and in batteries. Magnesium is essential to plant and animal life, and Mg2+ ions are not toxic. It is estimated that the average adult ingests about 0.3 g of magnesium ions daily. Magnesium plays several important biological roles. It is present, for instance, in intracellular and extracellular fluids, and magnesium ions are essential for the proper functioning of a number of enzymes. Magnesium is also present in the green plant pigment chlorophyll, which plays an important part in photosynthesis.

Calcium Figure 23.15 MgC0 3) ·

Dolomite (CaC0 3

•

Earth 's crust contains about 3.4 percent calcium by mass. Calcium occurs in limestone, calcite, chalk, and marble as CaC03 ; in dolomite as CaC0 3 • MgC0 3 (see Figure 23.l5); in gypsum as

SECTION 23.7

Aluminum

895

CaS04 . 2H 20; and in fluorite as CaF2 (Figure 23.16). Metallic calcium is best prepared by the electrolysis of molten calcium chloride (CaCI2). As we read down Group 2A from beryllium to barium, metallic properties increase. Unlike beryllium and magnesium, calcium (like strontium and barium) reacts with cold water to yield the corresponding hydroxide, although the rate of reaction is much slower than those involving the alkali metals: Ca(s)

+ 2H20(l) -

...... Ca(OHMaq)

+ H 2(g)

Calcium hydroxide [Ca(OH)zJ is commonly known as slaked lime or hydrated lime. Lime (CaO), which is also referred to as quicklime, is one of the oldest materials known to humankind. Quicklime is produced by the thermal decomposition of calcium carbonate:

Figure 23.16

Fluorite (CaF 2 ) .

Figure 23 .17

Corundum (A12 0 3).

whereas slaked lime is produced by the reaction between quicklime and water: CaO(s)

+ H 20 (I) -

...... Ca(OHMaq)

Quicklime is used in metallurgy (see Section 23.2) and in the removal of S02 when fossil fuel is burned. Slaked lime is used in water treatment. For many years, farmers have used lime to lower the acidity of the soil for their crops (a process called liming) . Nowadays lime is also applied to lakes affected by acid rain. Metallic calcium has rather limited uses. It serves mainly as an alloying agent for metals like aluminum and copper and in the preparation of beryllium metal from its compounds. It is also used as a dehydrating agent for organic solvents. Calcium is an essential element in living matter. It is the major component of bones and teeth; the calcium ion is present in a complex phosphate salt called hydroxyapatite [Ca5(P04)30HJ. A characteristic function of Ca2+ ions in living systems is the activation of a variety of metabolic processes, including a vital role in heart action, blood clotting, muscle contraction, and nerve impulse transmission.

Aluminum Aluminum is the most abundant metal and the third most plentiful element in Earth's crust (7.5 percent by mass). The elemental fOlln does not occur in nature; instead, its principal ore is bauxite (A1 20 3 . 2H 20). Other minerals containing aluminum are orthoclase (KA1Si 30 8), beryl (Be3AI2Si6018), cryolite (Na3AlF6), and corundum (A1 20 3) (Figure 23.17). Aluminum is usually prepared from bauxite, which is frequently contaminated with silica (Si0 2), iron oxides, and titanium(IV) oxide. The ore is first heated in sodium hydroxide solution to convelt the silica into soluble silicates: Si02(s)

+ 20H - (aq) -

...... SiO j - (aq)

+ H 20 (l)

At the same time, aluminum oxide is converted to the aluminate ion (AI0 2 ):

Iron oxide and titanium oxide are unaffected by this treatment and are filtered off. Next, the solution is treated with acid to precipitate the insoluble aluminum hydroxide:

After filtration , the aluminum hydroxide is heated to obtain aluminum oxide:

Anhydrous aluminum oxide, or corundum, is reduced to aluminum by the Ha1l 3 process. Figure 23.18 shows a Hall electrolytic cell, which contains a series of carbon anodes. The cathode is also made of carbon and constitutes the lining inside the cell. The key to the Hall process is the use of

3. Charles Martin Hall (1863-1914). American inventor. While Hall was an undergraduate at Oberlin College. he became interested in finding an inexpensive way to extract aluminum. Shortly after graduation, when he was only 22 years old, Hall succeeded in obtaining aluminum from aluminum oxide in a backyard woodshed. Amazingly, the same discovery was made at almost the same moment in France by Paul Heroult, another 22-year-old inventor working in a similar makeshift laboratory.

896

CHAPTER 23

Metallurgy and the Chemistry of Metals

Figure 23.18

Electrolytic production of aluminum based on the Hall process.

Carbon anodes _

'

"-

...,,-

• cathode

Molten

Al2 0 3 in molten cryolite

aluminum

cryolite (Na3AIF6; m.p. lOOO°C) as the solvent for aluminum oxide (m.p. 2045 °C). The mixture is electrolyzed to produce aluminum and oxygen gas: 3[20

Anode (oxidation): Cathode (reduction):

4[AI3+

Overall:

•

2

- ----.

+ 3e -

2Al2 0 3

0 2(g)

+ 4e - ]

• AI(l)] •

4AI(l) + 30 2 (g)

Oxygen gas reacts with the carbon anodes (at elevated temperatures) to form carbon monoxide, which escapes as a gas. The liquid aluminum metal (m.p. 660.2°C) sinks to the bottom of the vessel, from which it can be drained from time to time during the procedure. Aluminum is one of the most versatile metals known. It has a low density (2.7 g/cm3) and high tensile strength (i.e. , it can be stretched or drawn out). Aluminum is malleable, it can be rolled into thin foils, and it is an excellent electrical conductor. Its conductivity is about 65 percent that of copper. However, because aluminum is cheaper and lighter than copper, it is widely used in highvoltage transmission lines. Although aluminum's chief use is in aircraft construction, the pure metal itself is too soft and weak to withstand much strain. Its mechanical properties are greatly improved by alloying it with small amounts of metals such as copper, magnesium, and manganese, as well as silicon. Aluminum is not used by living systems and is generally considered to be nontoxic. As we read across the periodic table from left to right in a given period, metallic properties gradually decrease. Thus, although aluminum is considered an active metal, it does not react with water as do sodium and calcium. Aluminum reacts with hydrochloric acid and with strong bases as follows: 2AI(s) 2AI(s)

+ 6HCI(aq) ---+. 2AICI 3(aq) + 3Hig)

+ 2NaOH(aq) + 2H2 0(/)

• 2NaAI0 2 (aq)

•

+ 3H2 (g)

Aluminum readily forms the oxide Al 20 3 when exposed to air:

A tenacious film of this oxide protects metallic aluminum from further corrosion and accounts for some of the unexpected inertness of aluminum. Aluminum oxide has a very large exothermic enthalpy of formation (!1H'f = -1670 kJ/mol). This property makes aluminum suitable for use in solid propellants for rockets such as those used for some space shuttles. When a mixture of aluminum and ammonium perchlorate (NH4 CI0 4 ) is ignited, aluminum is oxidized to A12 0 3, and the heat liberated in the reaction causes the gases that are formed to expand with great force. This action lifts the rocket. The great affinity of aluminum for oxygen is illustrated nicely by the reaction of aluminum powder with a variety of metal oxides, particularly the transition metal oxides, to produce the corresponding metals. A typical reaction is

!1Ho = -852 kJ/mol

SECTION 23.7

Aluminum

897

Figure 23.19

The temperature of a thermite reaction can reach 3000°C.

which can result in temperatures approaching 3000°C. This transformation, which is used in the welding of steel and iron, is called the thermite reaction (Figure 23.19). Aluminum chloride exists as a dimer:

Promotion of electron

Each of the bridging chlorine atoms forms a normal covalent bond and a coordinate covalent bond (each indicated by an arrow) with two aluminum atoms. Each aluminum atom is assumed to be Sp 3-hybridized, so the vacant hybrid orbital can accept a lone pair from the chlorine atom (Figure 23.20). Aluminum chloride undergoes hydrolysis as follows:

si

AICI 3(s)

+ 3H20(l) --+. AI(OHMs) + 3HCI(aq)

Aluminum hydroxide, like Be(OHh, is amphoteric: AI(OHMs)

+ 3H+(aq)

• AI3+ (aq)

+ 3H20(I)

AI(OHMs)

+ OH- (aq)

• AI(OH)4 (aq)

In contrast to the boron hydrides, which are a well-defined series of compounds, aluminum hydride is a polymer in which each aluminum atom is surrounded octahedrally by bridging hydrogen atoms (Figure 23.21). When an aqueous mixture of aluminum sulfate and potassium sulfate is evaporated slowly, crystals of KAI(S04h . 12H20 are formed. Similar crystals can be formed by substituting Na + or NHt for K +, and Cr3+ or Fe3+ for AI 3 +. These compounds are called alums, and they have the general formula M +: K+, Na+, NH4+ M3+ : AI3+ , Cr3+ , Fe3 + Alums are examples of double salts

that is, salts that contain two different cations.

11 1 1

Ground state

Sp3_

hybridized state

3s

3p

IT]

11 11 1 1

3s

3p

1111111 I

Figure 23.20

I

sp3 orbitals

The Sp3

hybridization of an Al atom in A1 2Cl 6 . Each Al atom has one vacant sp3 hybrid orbital that can accept a lone pair from the bridging CI atom.

Figure 23.21

Structure of aluminum hydride. Note that this compound is a polymer. Each Al atom is surrounded in an octahedral arrangement by six bridging H atoms.

898

CHAPTER 23

Metallurgy and the Chemistry of Metals

Applying What You've Learned

•

Most health problems related to copper are the result of errors in copper metabolism. However, although it is rare, copper deficiency can result from a diet that is poor in copper. Symptoms of dietary copper deficiency include anemia (a deficiency of red blood cells) and neutropenia (a deficiency of a particular type of white blood cell). The fact that copper is essential to human health was first demonstrated with a group of children in Peru. One patient's ordeal was detailed by Cordano and Graham in the journal Pediatrics in 1966. During her first few years of life, the patient was hospitalized several times with anemia, neutropenia, osteoporosis, and multiple fractures. At age 6, over a period of 3 months, she received 20 blood transfusions for her severe anemia, which had not responded to treatment. When Dr. Cordano became aware of the patient's history, he initiated treatment with copper supplementation. The patient never required another transfusion and after 6 months on copper supplements, at age 7, she walked for the first time in her life.

Writing Prompt: Research the subject on the Web, and write a SOO-word essay on the causes, diagnosis, and treatment of dietary copper deficiency. Include a specific case study.

•

l

QUESTIONS AND PROBLEMS

899

CHAPTER SUMMARY Section 23.1 •

Depending on their reactivities, metals exist in nature in either the free or combined state. (More reactive metals are found combined with other elements.) Most metals are found in minerals. Minerals with high metal content are called ores.

p-type semiconductors.

Section 23.4

Section 23.2 • Metallurgy involves recovering metal from ores. The three stages of metal recovery are preparation, separation, and purification. An alloy is a solid mixture of one or more metals, sometimes also containing one or more nonmetals. An amalgam is a mixture of mercury and one or more other metals. •

with a small amount of certain other elements. Semiconductors in which an electron-rich impurity is added to enhance conduction are known as n-type semiconductors. Semiconductors in which an electron-poor impurity is added to enhance conduction are known as

The methods' commonly used for purifying metals are distillation, electrolysis, and zone refining. Pyrometallurgy refers to metallurgical processes carried out at high temperatures.

•

Metals typically are good conductors and are malleable and ductile. Metallic character increases from top to bottom in a group and decreases from left to right across a period.

Section 23.5 •

The alkali metals are the most reactive of all the metallic elements. They have an oxidation state of + I in their compounds. Under special conditions, some of them can form anions with an oxidation state of

-1.

Section 23.3 •

•

Metallic bonds can be thought of as the force between positive ions immersed in a sea of electrons. In terms of band theory, the atomic orbitals merge to form energy bands. A substance is a conductor when electrons can be readily promoted to the conduction band, where they are free to move through the substance. In an insulator, the energy gap between the valence band and the conduction band is so large that electrons cannot be promoted into the conduction band.

• Semiconductors are substances that normally are not conductors but will conduct electricity at elevated temperatures or when combined

Section 23.6 •

The alkaline earth metals are somewhat less reactive than the alkali metals. They almost always have an oxidation number of +2 in their compounds. The properties of the alkaline earth elements become increasingly metallic from top to bottom in their group.

Section 23.7 •

Aluminum ordinarily does not react with water due to a protective coating of aluminum oxide; its hydroxide is amphoteric. The Hall process is used to reduce aluminum oxide to aluminum.

KEyWORDS Alloy, 883

Insulator, 889

Mineral, 882

p-type semiconductor, 890

Amalgam, 883

Hall process, 895

n-type semiconductor, 890

Pyrometallurgy, 883

Band theory, 888

Metallurgy, 883

Ore, 882

Semiconductor, 889

Conductor, 889

QUESTIONS AND PROBLEMS Section 23.1: Occurrence of Metals

23.4

Review Questions 23 .1

Define the terms mineral and ore.

23 .2

List three metals that are usually found in an uncombined state in nature and three metals that are always found in a combined state in nature.

23 .3

Write chemical formulas for the following minerals: (a) calcite, (b) dolomite, (c) fluorite, (d) halite, (e) corundum, (f) magnetite, (g) beryl, (h) galena, (i) epsomite, (j) anhydrite.

Name the following minerals: (a) MgC0 3, (b) Na3AIF6, (c) A1 20 3, (d) Ag2S; (e) HgS, (f) ZnS, (g) SrS04, (h) PbC0 3, (i) Mn02, (j) Ti0 2· •

Section 23.2: Metallurgical Processes

Review Questions 23.5

Define the terms metallurgy, alloy, and amalgam.

23.6

Describe the main steps involved in the preparation of an ore.

23.7

What does roasting mean in metallurgy? Why is roasting a major source of air pollution and acid rain?

900

CHAPTER 23

Metallurgy and the Chemistry of Metals

State whether silicon would form n-type or p-type semiconductors with the following elements: Ga, Sb, AI, As.

23.8

Describe with examples the chemical and electrolytic reduction processes used in the production of metals.

23.24

23.9

Describe the main steps used to purify metals.

Section 23.4: Periodic Trends in Metallic Properties

23.10

Describe the extraction of iron in a blast furnace.

Review Questions

23.11

Briefly discuss the steelmaking process.

23 .25

Discuss the general properties of metals.

23.12

Briefly describe the zone refining process.

23 .26

Use periodic trends in ionization energy and electronegativity to show how the metallic character changes within a group.

23.27

Use periodic trends in ionization energy and electronegativity to show how the metallic character changes across a period.

•

Problems 23.13

In the Mond process for the purificati on of nickel, CO is passed over metallic nickel to give Ni(CO)4: Ni(s)

+ 4CO(g) +.==' Ni(COMg)

Given that the standard free energies of formation of CO (g) and Ni(COMg) are - 137.3 and - 587.4 kJ/mol, respectively, calculate the equilibrium constant of the reaction at 80°C. (Assume i1 G f to be independent of temperature.) 23.14

23.15

Copper is purified by electrolysis (see Figure 23.6). A 5.00-kg anode is used in a cell where the current is 37.8 A. How long (in hours) must the current run to dissolve this anode and electroplate it onto the cathode? Consider the electrolytic procedure for purifying copper described in Figure 23.6. Suppose that a sample of copper contains the following impurities: Fe, Ag, Zn, Au, Co, Pt, and Pb. Which of the metals will be oxidized and dissolved in solution and which will be unaffected and simply form the sludge that accumulates at the bottom of the cell?

23.16

How would you obtain zinc from sphalerite (ZnS)?

23.17

A certain mine produces 2.0 X 10 8 kg of copper from chalcopyrite (CuFeS2) each year. The ore contains only 0.80 percent Cu by mass. (a) If the density of the ore is 2.8 g/cm 3 , calculate the volume (in cm3) of ore removed each year. (b) Calculate the mass (in kg) of SOl produced by roasting (assume chalcopyrite to be the only source of sulfur).

23.18

Starting with rutile (TiOl ), explain how you would obtain pure titanium metal. (Hint: First convert TiO z to TiCI 4. Next, reduce TiCl4 with Mg. Look up physical properties of TiCI 4, Mg, and MgCI l in a chemistry handbook.)

23.19

Which of the following compounds would require electrolysis to yield the free metals: Agl S, CaCI2 , NaCI, Fez0 3, AI1 0 3 , TiCI4?

23.20

Although iron is only about two-thirds as abundant as aluminum in Earth's crust, mass for mass it costs only about one-quarter as much to produce. Why?

Section 23.5: The Alkali Metals

Review Questions 23 .28

How is sodium prepared commercially?

23.29

Why is potassium usually not prepared electrolytically from one of its salts?

23 .30

Describe the uses of the following compounds: NaCl, Na2C03, NaOH, KOH, K0 2.

23.3 1

Under what conditions do sodium and potassium form Na - and K- ions?

Problems 23 .32

Complete and balance the following equations:

+ H10(1) (b) NaH(s) + H 2 0(l) (c) Na(s) + Oz(g) (d) K(s) + 0 2(g) (a) K(s)

•

•

• •

23.33

Write a balanced equation for each of the following reactions: (a) sodium reacts with water, (b) an aqueous solution of NaOH reacts with COlo (c) solid Nal C0 3 reacts with an HCl solution, (d) solid NaHC0 3 reacts with an HCl solution, (e) solid NaHC0 3 is heated, (f) solid N al C0 3 is heated.

23.34

Sodium hydride (NaH) can be used as a drying agent for many organic sol vents. Explain how it works.

23.35

Calculate the volume of COl at 1O.0°C and 746 mmHg pressure obtained by treating 25.0 g of Na2C03 with an excess of hydrochloric acid.

Section 23.6: The Alkaline Earth Metals

Review Questions 23 .36

List the common ores of magnesium and calcium.

23.37

How are magnesium and calcium obtained commercially?

Section 23.3: Band Theory of Conductivity

Review Questions

Problems 23.21

23.22

23.23

Define the following terms: conductor, insulator, semiconducting elements, donor impurities, acceptor impurities, n-type semiconductors, p-type semiconductors. Briefly discuss the nature of bonding in metals, insulators, and semiconducting elements. Describe the general characteristics of n-type and p-type semiconductors.

23.38

From the thennodynamic data in Appendix 2, calculate the i1Ho values for the following decompositions: (a) MgC0 3 (s) - _ . MgO(s) + CO 2 (g) (b) CaC0 3 (s) • CaO(s) + COi g) Which of the two compounds is more easily decomposed by heat?

QUESTIONS AND PROBLEMS

23.39

Starting with magnesium and concentrated nitric acid, describe how you would prepare magnesium oxide. [Hint: First convert Mg to Mg(N0 3 )2' Next, MgO can be obtained by heating Mg(N0 3)2']

23.40

Describe two ways of preparing magnesium chloride.

23.41

The second ionization energy of magnesium is only about twice as great as the first, but the third ionization energy is 10 times as great. Why does it take so much more energy to remove the third electron?

23.42

List the sulfates of the Group 2A metals in order of increasing solubility in water. Explain the trend. (Hint: You need to consult a chemistry handbook.)

23.43

Helium contains the same number of electrons in its outer shell as do the alkaline earth metals. Explain why helium is inert whereas the Gn:>up 2A metals are not.

23.44

When exposed to air, calcium first forms calcium oxide, which is then converted to calcium hydroxide, and finally to calcium carbonate. Write a balanced equation for each step.

23.54

Describe some of the properties of aluminum that make it one of the most versatile metals known.

23.55

The pressure of gaseous Al2Ci 6 increases more rapidly with temperature than predicted by the ideal gas equation even though Al2Cl 6 behaves like an" ideal gas. Explain.

23.56

Starting with aluminum, describe with balanced equations how you would prepare (a) A1 2 C1 6 , (b) A1 20 3, (c) AI 2(S04)3, (d) NH4AI(S04h . 12H20.

23.57

Explain the change in bonding when Al 2Ci6 di ssociates to form AlCl 3 in the gas phase.

Additional Problems 23.58

In steelmaking, nonmetallic impurities such as P, S, and Si are removed as the corresponding oxides. The inside of the furnace is usually lined with CaC0 3 and MgC0 3, which decompose at high temperatures to yield CaO and MgO. How do CaO and MgO help in the removal of the nonmetallic oxides?

23.59

When 1.164 g of a certain metal sulfide was roasted in air, 0.972 g of the metal oxide was formed. If the oxidation number of the metal is + 2, calculate the molar mass of the metal.

23.60

An early view of metallic bonding assumed that bonding in metals consisted of localized, shared electron-pair bonds between metal atoms. What evidence would help you to argue against this viewpoint?

23.61

Referring to Figure 23.6, would you expect H 20 and H to be reduced at the cathode and H 20 oxidized at the anode?

23.62

A 0.450-g sample of steel contains manganese as an impurity. The sample is dissolved in acidic solution and the manganese is oxidi zed to the permanganate ion MnO 4. The MnO 4 ion is reduced to Mn2+ by reacting with 50.0 mL of 0.0800 M FeS04 solution. The excess Fe 2 + ions are then oxidized to Fe 3+ by 22.4 mL of 0.0100 M K2Cr207' Calculate the percent by mass of manganese in the sample.

23.63

Given that LlGf?(Fe20 3) = -741.0 kl/mol and that LlGf?(AI20 3) = - 1576.4 kl/mol, calculate LlGo for the following reactions at 25 °C:

23.45 . Write chemical formulas for (a) quicklime and (b) slaked lime.

Section 23 .7: Aluminum Review Questions 23.46 23.47

Describe the Hall process for preparing aluminum. What action renders aluminum inert?

Problems 23.48

Before Hall invented his electrolytic process, aluminum was produced by the reduction of its chloride with an active metal. Which metals would you use for the production of al uminum in that way?

23.49

With the Hall process, how many hours will it take to deposit 664 g of Al at a current of 32.6 A ?

23.50

23.51

Aluminum forms the complex ions AlCi 4 and AIF ~ - . Describe the shapes of these ions. AlCI ~ - does not form. Why? (Hint: Consider the relative sizes of AI3+ , F-, and Cl - ions.) The overall reaction for the electrolytic production of aluminum by means of the Hall process may be represented as AI 20 3(s)

73.52

23.53

(a) 2Fe20 3(S) - _ . 4Fe(s) • 4AI(s) (b) 2AI 20 3(s)

In basic solution, aluminum metal is a strong reducing agent and is oxidized to Al0 2 . Give balanced equations for the reaction of Al in basic solution with the following: (a) NaN0 3, to give ammonia; (b) water, to give hydrogen; (c) Na2Sn03, to give metallic tin. Write a balanced equation for the thermal decomposition of aluminum nitrate to form aluminum oxide, nitrogen dioxide, and oxygen gas.

+ 30 2(g) + 30 2(g)

23.64

Use compounds of aluminum as examples to explain what is meant by amphoterism.

23.65

When an inert atmosphere is needed for a metallurgical process, nitrogen is frequently used. However, in the reduction of TiCl 4 by magnesium, helium is used. Explain why nitrogen is not suitable for this process.

23.66

It has been shown that Na2 species form in the vapor phase.

+ 3C(s) - _ . 2AI(l) + 3CO(g)

At 1000°C, the standard free-energy change for this process is 594 kl/mol. (a) Calculate the minimum voltage required to produce 1 mole of aluminum at this temperature. (b) If the actual voltage applied is exactly three times the ideal value, calculate the energy required to produce 1.00 kg of the metal.

901

Describe the formation of the "Eiisodium molecule" in terms of a molecular orbital energy level diagram. Would you expect the alkaline earth metals to exhibit a similar property?

23.67

Explain each of the following statements: (a) An aqueous solution of AlCl 3 is acidic. (b) Al(OH)3 is soluble in NaOH solution but not in NH3 solution.

902

Metallurgy and the Chemistry of Metals

CHAPTER 23

23.68

Write balanced equations for the following reactions: (a) the heating of aluminum carbonate, (b) the reaction between AICI 3 and K, (c) the reaction between solutions of Na2C03 and Ca(OH)2'

23.69

Write a balanced equation for the reaction between calcium oxide and dilute HCI solution.

23.70

23.75

Lithium and magnesium exhibit a diagonal relationship in some chemical properties. How does lithium resemble magnesium in its reaction with oxygen and nitrogen? Consult a handbook of chemistry and compare the solubilities of carbonates, fluorides , and phosphates of these metals.

23.76

To prevent the formation of oxides, peroxides, and superoxides, alkali metals are sometimes stored in an inert atmosphere. Which of the foll owing gases should not be used for lithium: Ne, Al", N z, Kr?Why?

23.77

Which of the following metals is not found in the free state in nature: Ag, Cu, Zn, Au, Pt?

23.78

After heating, a metal surface (such as that of a cooking pan or skillet) develops a color pattern like an oil slick on water. Explain.

23.79

A sample of 10.00 g of sodium reacts with oxygen to form 13.83 g of sodium oxide (Na20) and sodium peroxide (Na202)' Calculate the percent composition of the mixture.

23.80

The electrical conductance of copper and metal decreases with temperature, but that of a CUS04 solution increases with temperature. Explain.

23.81

As stated in the chapter, potassium superoxide (KO z) is a useful source of oxygen employed in breathing equipment. Calculate the pressure at which oxygen gas stored at 20°C would have the same density as the oxygen gas provided by KO z. The density of KO z at 20°C is 2.15 g/cm 3

23.82

Chemical tests of four metals A, B, C, and D show the following results: (a) Only Band C react with 0.5 M HCI to give H2 gas. (b) When B is added to a solution containing the ions of the other metals, metallic A, C, and D are formed. (c) A reacts with 6 M HN0 3, but D does not. Arrange the metals in the increasing order as reducing agents. Suggest four metals that fit these descriptions.

What is wrong with the following procedure for obtaining magnesium?

•

MgC0 3(s) - _ . MgO(s) MgO(s)

+ CO(g)

• Mg(s)

+ CO 2(g)

+ COz(g)

23.71

Explain why most metals have a flickering appearance.

23.72

Predict the chemical properties of francium, the last member of Group lA.

23.73

Describe a medicinal or health-related application for each of the following compounds: NaP, Li 2C0 3, Mg(OH)2, CaC0 3, BaS04'

23.74

The following are two reaction schemes involving magnesium. Scheme I: When magnesium burns in oxygen, a white solid (A) is formed. A dissolves in 1 M HCI to give a colorless solution (B). Upon addition of Na2C03 to B, a white precipitate is formed (C). On heating, C decomposes to D and a colorless gas is generated (E). When E is passed through limewater [an aqueous suspension of Ca(OH)2], a white precipitate appears (F). Scheme II: Magnesium reacts with 1 M H 2S04 to produce a colorless solution (G). Treating G with an excess of NaOH produces a white precipitate (H). H dissolves in 1 M HN0 3 to form a colorless solution. When the solution is slowly evaporated, a white solid (I) appears. On heating I, a brown gas is given off. Identify A - I, and write equations representing the reactions involved.

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: VERBAL REASONING

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: VERBAL REASONING Copper is essential for a wide range of biochemical processes, which are vital for human health. However, copper is also potentially toxic. Under physiological conditions, copper exists in two different oxidation states : Cu + (cuprous) and Cu2+ (cupric). It is the one-electron shift back and forth between these two oxidation states that makes copper essential to the function of cuproenzymes. However, this electron transfer also contributes to the potential toxicity of copper in the body. Cycling between Cu + and Cu2+ can generate hydroxyl radicals (·OH), highly reactive oxygen species that can damage biological molecules such as DNA and proteins. Recently, intracellular proteins have been discovered that appear to prevent the toxic effects of copper ions. These so-called copper chaperones escort copper ions directly to the cuproenzymes that need them, keeping the cells free of "unchaperoned" copper ions, which would otherwise be involved in the formation of hydroxyl radicals. Menkes disease results from an inability to absorb copper in the intestine. Lack of availability of copper resul ts in reduced activity of cuproenzymes. Symptoms of Menkes disease include mental retardation and abnormalities in connective tiss ue. Victims of Menkes disease are almost always male, and they typically do not survive beyond the age of 3. In recent years, an experimental treatment for Menkes disease has been reported. The treatment involves injection of a copper compound (copper histadine) to supply copper to the body 's cells. One report details the effects of this treatment on four patients who exceeded the life expectancy of Menkes disease sufferers by as much as 20 years with regular injections of copper histadine. Although the mental retardation was largely mitigated by the treatment, abnormalities in connective tissue were not and so although the patients survived, they were severely disabled by their symptoms.

1.

According to the passage, the potential toxicity of copper in the human body results from a) copper ions damaging DNA and proteins. b) increased activity of cuproenzymes. c) the formation of hydroxyl radicals . d) excess copper accumulating in the brain.

2.

What is the function of a copper chaperone? a) To convert copper atoms into copper ions. b) To convert Cu + ions to Cu 2 + ions. c) To convert Cu2+ ions to Cu+ ions. d) To escort copper ions to the cuproenzymes that require them.

3.

According to the passage, what is the cause of Menkes disease? a) Abnonnalities in connective ti ssue b) Failure to absorb copper in the intestine c) Excess copper in the cells d) Faulty cuproenzymes

4.

According to the passage, victims of Menkes disease a) are typically male. b) typicall y live up to 20 years. c) can be cured with copper histidine injections. d) all the above.

•

903

•

ements oun s

onmeta Ie •

elf

24.1

General Properties of Nonmetals

24.2

•

Hydrogen

• • • •

24.3 24.4 • •

24.5

Binary Hydrides Isotopes of Hydrogen Hydrogenation The Hydrogen Economy

Carbon Nitrogen and Phosphorus Nitrogen Phosphorus

Oxygen and Sulfur

• •

24.6 •

• •

Oxygen Sulfur

The Halogens Preparation and General Properties of the Halogens Compounds of the Halogens Uses of the Halogens

OP

· R.

Toxicity of Arsenic(lll) Compounds Arsenic has always been the poison of choice for murder mysteries. In the 1960s, samples of Napoleon's hair were analyzed and found to contain a high level of arsenic. It was naturally concluded that he was intentionally poisoned. Studies in recent years, however, have shown that Napoleon's exposure to arsenic may have been environmental. The wallpaper in his drawing room was found to contain the green pigment copper arsenite (CuHAs0 4 ). The mold growing on the papers could have converted the compound to the volatile, toxic trimethyl arsine [(CH3)3As], which Napoleon then ingested. Unlike the first two elements in Group SA, nitrogen and phosphorus, arsenic is not an essential element in the human body. Moreover, elemental arsenic itself is not all that harmful. The commonly used poison is actually arsenic(III) oxide (AS 2 0 3), a white compound that dissolves in water, has no taste, and if administered over a period of time, is hard to detect. The toxicity of As(III) inorganic compounds lies in their ability to bind to the sulfhydryl group (-SH) of proteins and enzymes, thus impairing their normal functions. Arsenite also inhibits enzyme activities in the mitochondria and uncouples oxidative phosphorylation. This results in a decrease in ATP production and an increase in harmful reactive oxygen species (ROS) such as hydrogen peroxide and superoxide ion. The accumulation of ROS can lead to DNA damage and initiate carcinogenic processes. Arsenic is a metalloid. The metalloids and the nonmetallic elements and their compounds exhibit chemistry that varies considerably.

In This Chapter, You Will Learn

some of the properties of metalloids and nonmetals and which compounds they

form.

Before you begin, you should review •

General trends in chemical properties

[~

Section 7.7]

,

Media Player/ MPEG Content Chapter in Review

Arsenic, the poison of many murder mysteries, can cause severe abdominal pain, intense thirst, vomiting, convulsions, and ultimately death. Despite its toxicity, it has been used historically in medicine, cosmetics, and pigments. 905

906

CHAPTER 24

Nonmetallic Elements and The ir Compounds

General Properties of Nonmetals

•

Properties of nonmetals are more varied than those of metals. Hydrogen, oxygen, nitrogen, fluorine, chlOline, and the noble gases are all gases in the elemental state, whereas only bromine is a liquid. All the remaining nonmetals are solids at room temperature. Unlike metal s, nonmetals are poor conductors of heat and electricity, and when they form compounds, nonmetals can exhibit either positive or negative oxidation numbers. A small group of elements, called metalloids, have properties characteristic of both metals and nonmetals. The metalloids boron, silicon , germanium, and arsenic are semiconducting elements (see Section 20.3). Nonmetals are more electronegative than metals [ ~~ Section 8.4] . The electronegativity of elements increases from left to right across any period and from bottom to top in any group in the periodic table (see Figure 8.6). With the exception of hydrogen, the nonmetals are concentrated in the upper right-hand corner ofthe periodic table (Figure 24.1). Compounds formed by a combination of metals with nonmetals tend to be ionic, having a metallic cation and a nonmetallic anion. In this chapter we will discuss the chemistry of a number of common and important nonmetallic elements-namely, hydrogen; carbon (Group 4A); nitrogen and phosphorus (Group SA); oxygen and sulfur (Group 6A); and the halogens: fluorine, chlorine, bromine, and iodine (Group 7 A).

Hydrogen Hydrogen is the simplest element known its most common atomic form contains only one proton and one electron. The atomic form of hydrogen exists only at very high temperatures, however. Normally, elemental hydrogen is a diatomic molecule, the product of an exothermic reaction between H atoms:

H(g) + H(g) ---. HzCg)

Remember that the halogens also form anions that are isoelectronic with the noble gases [ ~~ Section 7.5].

Figure 24.1 Main group nonmetallic elements (in blue) and metalloids (in orange).

6..HO= -436.4 kJlmol

Molecular hydrogen is a colorless, odorless, and nonpoisonous gas. At 1 atm, liquid hydrogen has a boiling point of -2S2.9°C (20.3 K). Hydrogen is the most abundant element in the universe, accounting for about 70 percent of the universe's total mass. It is the tenth most abundant element in Earth's crust, where it is found in combination with other elements. Unlike Jupiter and Saturn, Earth does not have a strong enough gravitational pull to retain the lightweight Hz molecules, so hydrogen is not found in our atmosphere. The ground-state electron configuration of H is Is'. It resembles the alkali metals (Group IA) in that it can be oxidized to the H + ion, which exists in aqueous solutions in the hydrated form. On the other hand, hydrogen resembles the') halogens (Group 7A) in that it forms the hydride . . . . . . . . . . . . . . . . .. .. . . . . . . .. . . . . ion (H - ), which is isoelectronic with helium (ls-) . Hydrogen is found in a large number of covalent compounds. It also has the unique capacity, when bonded to small, electronegative atoms, for hydrogen-bond formation [ ~~ Section 12.1] . •

IA 1

1

8A 18 2

3A 13 5

4A 14 6

SA 15 7

6A 16 8

7A 17 9

He

3

2A 2 4

Li

Be

B

C

N

0

F

Ne

11

12

H

2B 12 30

13

14

15

16

17

18

Al

Si

P

S

CI

Ar

31

32

33

34

35

36

19

20

3B 3 21

K

Ca

Sc

Ti

V

Cr Mn · Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

37

38

39

40

41

41

Rb

Sr

Y

Zr

Nb Mo

55

56

57

72

73

Cs

Ba

La

Hf

87

88

89

Fr

Ra

Ac

Na Mg

4B 4

10

22

5B 5 23

6B 6 24

7B ~8B~ IB 7 8 9 10 11 25 26 27 28 29 43

44

45

46

47

48

49

50

51

52

53

54

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

74

75

76

77

78

79

80

81

82

83

84

85

86

Ta

W

Re

Os

Ir

Pt

Au Hg

TI

Pb

Bi

Po

At

Rn

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

-

-

-

-

-

-

SECTION 24.2

Figure 24.2

907

Apparatus for the laboratory preparation of hydrogen gas. The gas is collected over water.

Hel

-

-

Water Z--Zn

Hydrogen gas plays an important role in industrial processes. About 95 percent of the hydrogen produced is produced at or near the plant where it is used for industrial processes, such as the synthesis of ammonia. The large-scale industrial preparation is the reaction between propane (from natural gas and also as a product of oil refineries) and steam in the presence of a catalyst at 900°C:

In another process, steam is passed over a bed of red-hot coke:

The mixture of carbon monoxide and hydrogen gas produced in this reaction is commonly known as water gas. Because both CO and H2 bum in air, water gas was used as a fuel for many years. But because CO is poisonous, water gas has been replaced by natural gases, such as methane and propane. Small quantities of hydrogen gas can be prepared conveniently in the laboratory by combining zinc with dilute hydrochloric acid (Figure 24.2): Zn(s)

Hydrogen

+ 2HCl(aq) --+. ZnCliaq) + H 2(g)

Hydrogen gas can also be produced by the reaction between an alkali metal or an alkaline earth metal (Ca or Ba) and water (see Section 7.7), but these reactions are too violent to be suitable for the laboratory preparation of hydrogen gas. Very pure hydrogen gas can be obtained by the electrolysis of water, but this method consumes too much energy to be practical on a large scale.

Binary Hydrides Binary hydrides are compounds containing hydrogen and another element, either a metal or a nonmetal. Depending on the structure and properties, these hydrides are broadly divided into three types: (1) ionic hydrides, (2) covalent hydrides, and (3) interstitial hydrides. Ionic hydrides are formed when molecular hydrogen combines directly with any alkali metal or with the alkaline earth metals Ca, Sr, or Ba: 2Li(s)

+ H 2(g) --+. 2LiH(s)

Ca(s)

+ Hig) --+. CaH2(s)

All ionic hydrides are solids that have the high melting points characteristic of ionic compounds. The anion in these compounds is the hydride ion (H~), which is a very strong Br0nsted base. It readily accepts a proton from a proton donor such as water:

,

908

CHAPTER 24

Nonmetallic Elements and Their Compounds

D D D

Figure 24.3

Binary hydrides of the representative elements. In cases in which hydrogen forms more than one compound with the same element, only the formula of the simplest hydride is shown. The properties of many of the transition metal hydrides are not well characterized.

•

IA 1

2A 2

Discrete molecular units SA 18

Polymeric structure·, covalent compound 3A l3

Ionic compound

LiH BeH2

NaH McrHo 3B b

_

3

4B 4

5B

6B

)

6

-

7B 7

Is

8B 9

~

IB II

2B 12

4A 14

SA 15

6A 16

7A 17

B2H6 CH4 NH3 H2O

HF

A1H3 SiH4 PH3 H2S

HCI

KH CaH2

GaH3 GeH4 AsH3 H2Se HBr

RbH SrH2

InH3 SnH4 SbH 3 H2Te

CsH BaH2

T1H3 PbH4 BiH3

HI

Because of their high reactivity with water, ionic hydrides are frequently used to remove traces of water from organic solvents. In covalent hydrides, the hydrogen atom is covalently bonded to the atom of another element. There are two types of covalent hydrides those containing discrete molecular units, such as CH4 and NH 3 , and those having complex polymeric structures, such as (BeH 2)x and (AIH3\ where x is a very large number. Figure 24.3 shows the binary ionic and covalent hydrides of the main group elements. The physical and chemical properties of these compounds change from ionic to covalent across a given period. Consider, for example, the hydrides of the second-period elements: LiH, BeH2, B?H 6 , CH4 , NH 3, H 20, and HF. LiH is an ionic compound with a high melting point (680°C). The structure of BeH2 (in the solid state) is polymeric; it is a covalent compound. The molecules B2H6 and CH4 are nonpolar. In contrast, NH 3, H 20 , and HF are all polar molecules in which the hydrogen atom is the positive end of the polar bond. Of this group of hydrides (NH 3 , H 20, and HF), only HF is acidic in water. As we move down any group in Figure 24.3, the compounds change from covalent to ionic. In Group 2A, for example, BeH2 and MgH2 are covalent, but CaH 2, SrH2> and BaH2 are ionic. Molecular hydrogen forms a number of hydrides with transition metals. In some of these compounds, the ratio of hydrogen atoms to metal atoms is not a constant. Such compounds are called interstitial hydrides. Depending on conditions, for example, the formula for titanium hydride can vary between TiH1.8 and TiH2. Many of the interstitial hydrides have metallic properties such as electrical conductivity. It is known , however, that hydrogen is definitely bonded to the metal in these compounds,.although the exact nature of the bonding is often unclear. Molecular hydrogen interacts in a unique way with palladium (Pd). Hydrogen gas is readily adsorbed onto the surface of the palladium metal, where it dissociates into atomic hydrogen. The H atoms then "dissolve" into the metal. On heating and under the pressure of H2 gas on one side of the metal, these atoms diffuse through the metal and recombine to form molecular hydrogen, which emerges as the gas from the other side. Because no other gas behaves in this way with palladium, this process has been used to separate hydrogen gas from other gases on a small scale.

Isotopes of Hydrogen Hydrogen has three isotopes: lH (hydrogen), ~H (deuterium, symbol D), and fH (tritium, symbol T). The natural abundances of the stable hydrogen isotopes are hydrogen, 99.985 percent; and deuterium, 0.015 percent. Tritium is a radioactive isotope with a half-life of about 12.5 years. Table 24.l compares some of the common properties of H 20 with those of D 20. Deuterium oxide, or heavy water as it is commonly called, is used in some nuclear reactors as a coolant and a moderator of nuclear reactions [ ~~ Section 20.5] . D 20 can be separated from H 20 by fractional distillation because H?O boils at a lower temperature, as Table 24.1 shows. Another technique for

SECTION 24.2

Property

H2O

D2 0

Molar mass (g/mol)

18.02

20.03

Melting point COc)

0

3.8

Boiling point COc)

100

101.4

Density at 4°C (g/cm

3

1.000

)

1.108

separating D 20 is the electrolysis of water. Because H2 gas is formed about eight times as fast as D2 during electrolysis, the water remaining in the electrolytic cell becomes progressively enriched with D 20. Interestingly, the Dead Sea, which for thousands of years has entrapped water that has no outlet other than through evaporation, has a higher [D )O] / [H 20] ratio than water found elsewhere. Although DzO resembles H 20 chemically in most respects, it is still a toxic substance. The reason is that deuterium is heavier than hydrogen, so its compounds often react more slowly than those of the lighter isotope. Drinking D 20 instead of H 20 on a regular basis could prove fatal because of the slower rate of transfer of D compared with that of H in the acid-base reactions involved in enzyme catalysis. This kinetic isotope effect is also manifest in acid ionization constants. For example, the ionization constant of acetic acid

•

Ka = 1.8 X 10- 5

is about three times as large as that of deuterated acetic acid, in which the ionizable hydrogen atom is replaced by a deuterium atom:

Ka = 6

X 10- 6

Hyd rogenation Hydrogenation is the addition of hydrogen to compounds containing multiple bonds, usually C= C and C C bonds. A simple example of hydrogenation is the conversion of ethylene to ethane:

H

\ /

/

•

C=C

H

\

H H

H H

Ethylene

Hydrogen

I I H-C-C-H I I

H H

Ethane

This reaction is quite slow under normal conditions, but the rate can be greatly increased by the presence of a catalyst such as nickel or platinum. Hydrogenation is an extremely important process in the food industry. Vegetable oils have considerable nutritional value, but some oils must be hydrogenated before we can use them because of their unsavory flavor and their inappropriate molecular structures (i.e. , there are too many C=C bonds present). Upon exposure to air, these polyunsaturated molecules (i.e. , molecules with many C=C bonds) undergo oxidation to yield unpleasant-tasting products (oil that has oxidized is said to be rancid). In the hydrogenation process, a small amount of nickel (about 0.1 percent by mass) is added to the oil and the mixture is exposed to hydrogen gas at high temperature and pressure. Afterward, the nickel is removed by filtration. Hydrogenation reduces the number of double bonds in the molecule but does not completely eliminate them. If all the double bonds are eliminated, the oil becomes hard and brittle. Under controlled conditions, suitable cooking oils and margarine may be prepared by the hydrogenation of vegetable oils extracted from cottonseed, corn, and soybeans.

The Hydrogen Economy The world's fossil fuel reserves are being depleted at an alarmingly fast rate. Faced with this dilemma, scientists have made intensive efforts in recent years to develop a method of obtaining hydrogen gas as an alternative energy source. Hydrogen gas could replace gasoline to power automobiles (after considerable modification of the engine) or be used with oxygen gas in fuel cells to

,

909

910

CHAPTER 24

Nonmetallic Elements and Their Compounds

generate electricity (see page 779). One major advantage of using hydrogen gas in these ways is that the reactions are essentially free of pollutants; the end product formed in a hydrogen-powered engine or in a fuel cell would be water, just as in the burning of hydrogen gas in air:

•

•

•

Despite these attractive features, though, the success of a hydrogen economy would depend on how cheaply we could produce hydrogen gas and how easily we could store it. Although electrolysis of water consumes too much energy for large-scale application, if scientists can devise a more practical method of "splitting" water molecules, we could obtain vast amounts of hydrogen from seawater. One approach that is currently in the early stages of development would use solar energy. In this scheme, a catalyst (a complex molecule containing one or more transition metal atoms, such as ruthenium) absorbs a photon from solar radiation and becomes energetically excited. , In its excited state, the catalyst is capable of reducing water to molecular hydrogen . Some of the interstitial hydrides we have discussed would make suitable storage compounds for hydrogen. The reactions that form these hydrides are usually reversible, so hydrogen gas can be obtained simply by reducing the pressure of the hydrogen gas above the metal. The advantages of using interstitial hydrides are as follows: (1 ) many metals have a high capacity to take up hydrogen gas sometimes up to three times as many hydrogen atoms as there are metal atoms; and (2) because these hydrides are solids, they can be stored and transported more easily than gases or liquids.

Carbon Although it constitutes only about 0.09 percent by mass of Earth's crust, carbon is an essential element of living matter. It is found free in the form of diamond and graphite, and it is also a component of natural gas, petroleum, and coal. (Coal is a natural dark-brown to black solid used as a fuel; it is formed from fossilized plants and consists of amorphous carbon with various organic and some inorganic compounds.) Carbon combines with oxygen to form carbon dioxide in the atmosphere and occurs as carbonate in limestone and chalk. Diamond and graphite are allotropes of carbon. Figure 24.4 shows the phase diagram of carbon. Although graphite is the stable form of carbon at 1 atm and 25 °C, owners of diamond jewelry need not be alarmed, because the rate of the spontaneous process C(diamond) - _ . C(graphite) -~

Diamond 04 -

/

/"

Liquid

Graphite Vapor

3300 t ( 0 C)

Figure 24.4

Phase diagram of carbon. Note that under atmospheric conditions, graphite is the stable form of carbon.

!1GO= -2.87 kJ/mol

is extremely slow. In fact, millions of years may pass before a diamond turns to graphite. Synthetic diamond can be prepared from graphite by applying very high pressures and temperatures. Figure 24.5 shows a synthetic diamond and its starting material, graphite. Synthetic diamonds generally lack the optical properties of natural diamonds. They are useful, however, as abrasives and in cutting concrete and many other hard substances, including metals and alloys. Graphite is used as a lubricant and as the "lead" in pencils. Carbon has the unique ability to form long chains (consisting of more than 50 C atoms) and stable rings with five or six members. This phenomenon is called catenation, the linking of like atoms. Carbon's versatility is responsible for the millions of organic compounds (made up of carbon and hydrogen and other elements such as oxygen, nitrogen, and the halogens) found on Earth [ ~~ Chapter 10] . Carbon combines with metals to form ionic compounds called carbides, such as CaC 2 and Be2C, in which carbon is in the form of C ~- or C 4 - ions. These ions are strong Br¢nsted bases and react with water as follows: C ~- (aq)

+ 2H2 0(l) --.. 20H- (aq) + C2H 2(g)

C 4 -(aq) + 4H20(I)

• 40H-(aq) + CH4 (g)

Carbon also forms a covalent compound with silicon. Silicon carbide (SiC) is called carborundum and is prepared as follows: Si0 2(s)

-

.'

Figure 24.5

A synthetic diamond and the starting material-graphite.

+ 3C(s) --.. SiC(s) + 2CO(g)

Carborundum is also formed by heating silicon with carbon at 1500°C. Carborundum is almost as hard as diamond, and it has the diamond structure; that is, each carbon atom is bonded tetrahedrally to four Si atoms, and vice versa. It is used mainly for cutting, grinding, and polishing metals and glasses.

SECTION 24.4

Nitrogen and Phosphorus

Another important class of carbon compounds, the cyanides, contain the anion group :C - N: - . Cyanide ions are extremely toxic because they bind almost irreversibly to the Fe(III) ion in cytochrome oxidase, a key enzyme in metabolic processes. Hydrogen cyanide, which has the aroma of bitter almonds, is even more dangerous because of its volatility (b.p. 26°C). A few tenths of 1 percent by volume of HCN in air can cause death within minutes. Hydrogen cyanide can be prepared by treating sodium cyanide or potassium cyanide with acid: NaCN(s)

+ HCl(aq) --+. NaCI(aq) + HCN(aq)

Because HCN (in solution it is called hydrocyanic acid) is a very weak acid (Ka = 4.9 X 10- 10), most of the HCN produced reaction form and ....................... leaves the solution as . .. . . . . .. ... ...in. . .this .. ... . . . ... is. . .in . . . the .. . . .nonionized . . . . . . . . .. .. ............... ...... hydrogen cyanide gas. For this reason, acids should never be mixed with metal cyanides in the laboratory without proper ventilation. Cyanide ions are used to extract gold and silver. Although these metals are usually found in the uncombined state in nature, in other metal ores they may be present in relatively small concentrations and are more difficult to extract. In a typical process, the crushed ore is treated with an aqueous cyanide solution in the presence of air to dissolve the gold by forming the soluble complex ion [Au(CNh]-: 4Au(s)

+

8CN-(aq)

Hydrogen cyanide (HeN) is the gas used in gas-chamber executions.

+ 02(g) + 2H20(I) --+. 4[Au(CN)2] - (aq) + 40H-(aq)

The complex ion [Au(CN)2] - (along with some cation, such as Na +) is separated from other insoluble materials by filtration and treated with an electropositive metal such as zinc to recover the gold: Zn(s) + 2[Au(CN)2r(aq) --+. [Zn(CN)4f-(aq) + 2Au(s) Figure 24.6 shows an aerial view of a "cyanide pond" used for the extraction of gold. Of the several oxides of carbon, the most important are carbon monoxide (CO) and carbon dioxide (C0 2), Carbon monoxide is a colorless, odorless gas formed by the incomplete combustion of carbon or carbon-containing compounds: 2C(s)

+ 0 2(g) --+. 2CO(g)

Carbon monoxide is used in metallurgical processes for extracting nickel in organic synthesis and in the production of hydrocarbon fuels with hydrogen. Industrially, it is prepared by passing steam over heated coke. Carbon monoxide burns readily in oxygen to form carbon dioxide: 2CO(g)

+ 02(g) --+. 2C0 2(g)

t1W

= - 566 kJ/mol

Carbon monoxide is not an acidic oxide (it differs from carbon dioxide in that regard), and it is only slightly soluble in water. Carbon dioxide is a colorless and odorless gas. Unlike carbon monoxide, CO 2 is nontoxicalthough it is a simple asphyxiant. It is an acidic oxide. Carbon dioxide is used in beverages, in fire extinguishers, and in the manufacture of baking soda (NaHC0 3) and soda ash (Na2C03)' Solid carbon dioxide, called dry ice, is used as a refrigerant.

Nitrogen and Phosphorus Nitrogen About 78 percent of air by volume is nitrogen. The most important mineral sources of nitrogen are saltpeter (KN0 3) and Chile saltpeter (NaN0 3 ). Nitrogen is an essential element of life because it is a component of proteins and nucleic acids. Molecular nitrogen is obtained by the fractional distillation of air (the boiling points of liquid nitrogen and liquid oxygen are - 196°C and -183°C, respectively). In the laboratory, very pure nitrogen gas can be prepared by the thennal decomposition of ammonium nitrite:

The N2 molecule contains a triple bond and is very stable with respect to dissociation into atomic species. However, nitrogen forms a large number of compounds with hydrogen and oxygen in which the oxidation number of nitrogen varies from -3 to + 5 (Table 24.2). Most nitrogen compounds are covalent; when heated with certain metals, however, nitrogen forms ionic nitrides containing the N 3 - ion:

Figure 24.6

A cyanide pond for extracting gold from metal are.

911

912

CHAPTER 24

Nonmetallic Elements and Their Compounds

Oxidation Number -3

Formula

Compound

••

H-N- H I H

Ammonia

-2

Structure

••

••

I

I

H-N-N-H

Hydrazine

H

• -1

o

••

Hydroxylamine

H-N-O-H I H

Nitrogen (for reference)

:N-N:

+1

Nitrous oxide

+2

Nitric oxide

+3

••

:N-N-O: •• NO HN0 2

Nitrous acid

+4

N0 2

Nitrogen dioxide

+5

H

HN0 3

Nitric acid

:N=O: •

•

••

••

•

••

:O=NO - H • •• •

:O-N=O: •• • ••

•

:O=N-O-H

·

I

..

• • '0' ••

The nitride ion is a very strong Br0nsted base and reacts with water to produce ammonia and hydroxide ions:

Ammonia is one of the best-known nitrogen compounds. It is prepared industrially from nitrogen and hydrogen by the Haber process. It can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide: NH 4 Cl(aq)

+ NaOH(aq) --+. NaCI(aq) + HzO(l) + NH3(g)

Ammonia is a colorless gas (b.p. - 33.4 DC) with an irritating odor. Most of the ammonia produced annually in the United States is used in fertilizers. Liquid ammonia, like water, undergoes autoionization:

or simply

where NH ? is called the amide ion. Both the H + and NH z ions are solvated with NH3 molecules an example of an ion-dipole interaction [ ~~ Section 12.1] . At 50 DC, the ion product [H+][NH 2] is about 1 X 10- 33, considerably smaller than the 1 X 10- 14 for water at 25 DC. Nevertheless, liquid ammonia is a suitable solvent for many electrolytes, especially when a more basic medium is required or if the solutes react with water. Another important hydride of nitrogen is hydrazine:

H

\ ..

.. /

H

N-N

/

H

\

H

E ach N atom is sp3-hybridized. Hydrazine is a colorless liquid that smells like ammonia. It melts at 2 DC and boils at 114 DC. Hydrazine is a base that can be protonated to give the NzH ~ and NzH ~ + ions. A reducing agent, it can reduce Fe3+ to Fe z+, Mn04 to Mnz+, and I? to 1-. Its reaction with oxygen is highly exothermic:

D..HD= - 666.6 kJ/mol

SECTION 24.4

Nitrogen and Phosphorus

913

Hydrazine and its derivative methylhydrazine [N?H3(CH3)], together with the oxidizer dinitrogen tetroxide (N20 4 ), are used as rocket fuels. Hydrazine also plays a role in polymer synthesis and in the manufacture of pesticides. There are many nitrogen oxides, but the three particularly important ones are nitrous oxide, nitric oxide, and nitrogen dioxide. Nitrous oxide (N 20) is a colorless gas with a pleasing odor and sweet taste. It is prepared by heating ammonium nitrate to about 270°C:

Nitrous oxide resembles molecular oxygen in that it SUppOlts combustion. It does so because it decomposes when heated to form molecular nitrogen and molecular oxygen:

It is chiefly used as an anesthetic in dental procedures and other minor surgery. Nitrous oxide is

also called "laughing gas" because a person inhaling the gas becomes somewhat giddy. No satisfactory explanation has yet been proposed for this unusual physiological response. Nitric oxide (NO) is a colorless gas. The reaction of N2 and O2 in the atmosphere, N 2 (g)

+ 0 2(g)

-.::.~. 2NO(g)

6.GO= 173.4 kllmol

is a form of nitrogen fixation. The equilibrium constant for the preceding reaction is very small at room temperature: Kp is only 4.0 X 10- 31 at 25 °C, so very little NO will form at that temperature. However, the equilibrium constant increases rapidly with temperature, such as in a running auto engine. An appreciable amount of nitric oxide is formed in the atmosphere by the action of lightning. In the laboratory, the gas can be prepared by the reduction of dilute nitric acid with copper: 3Cu(s) + 8HN0 3(aq) ---.. 3Cu(N0 3Maq) + 4H 20(l) + 2NO(g) . . . . . .. . ...... . ... . ... . . . . . . . . . . . . . .. . ... ... .. ..... . . . The nitric oxide molecule is paramagnetic, containing one unpaired electron. It can be represented by the following resonance structures: ~

:N=O: • •

+.---.

The NO molecule does not obey the octet rule

[ 11~ Section 8.8] .

. .+ :N=O. • •

Unlike nitrous oxide and nitric oxide, nitrogen dioxide is a highly toxic, yellow-brown gas with a choking odor. In the laboratory, nitrogen dioxide is prepared by the action of concentrated nitric acid on copper (Figure 24.7):

Nitrogen dioxide is paramagnetic, and it has a strong tendency to dimerize to dinitrogen tetroxide, which is diamagnetic: •

This reaction occurs in both the gas phase and the liquid phase. Nitrogen dioxide is an acidic oxide; it reacts rapidly with cold water to form both nitrous acid (HN0 2) and nitric acid (HN0 3):

1

•

•

4Zn(s) + lOH+(aq) + N0 3 (aq ) - _ . 4Zn2+(aq) + NH t (aq ) + 3H?O(l)

• •

This is a disproportionation reaction in which the oxidation number of nitrogen changes from +4 (in N0 2) to + 3 (in HN0 2) and +5 (in HN0 3). This reaction is quite different from that between CO 2 and H20, in which only one acid (carbonic acid) is formed. Nitric acid is one of the most important inorganic acids. It is a liquid (b.p. 82.6°C), but it does not exist as a pure liquid because it decomposes spontaneously to some extent as follows:

The concentrated nitric acid used in the laboratory is 68 percent HN0 3 by mass (density 1.42 g/cm\ which corresponds to 15.7 M. Nitric acid is a powerful oxidizing agent. The oxidation number of N in HN0 3 is +5. The most common reduction products of nitric acid are N0 2 (oxidation number of N = + 4), NO (oxidation number of N = + 2), and NH t (oxidation number of N = - 3). Nitric acid can oxidize metals both above and below hydrogen in the activity series. Copper, for example, is oxidized by concentrated nitric acid. In the presence of a strong reducing agent, such as zinc metal, nitric acid can be reduced all the way to the ammonium ion:

•

•

J... •

.'

,

Figure 24.7

The production of N0 2 gas when copper reacts with concentrated nitric acid.

914

CHAPTER 24

Nonmetallic Elements and Their Compounds

Concentrated nitric acid does not oxidize gold. However, when the acid is added to concentrated hydrochloric acid in a 1:3 ratio by volume (one part HN0 3 to three parts HC1), the resulting solution, called aqua regia, can oxidize gold as follows: Au(s)

+ 3HN0 3 (aq) + 4HCl(aq) - - .. HAuCI4 (aq) + 3H20(I) + 3NOig)

The oxidation of Au is promoted by the complexing ability of the Cl - ion (to form the AuCl 4 ion). Concentrated nitric acid also oxidizes a number of nonmetals to their corresponding oxoacids:

P4 (s)

+ 20HN0 3(aq) - - . . 4H 3POiaq) + 20N0 2(g) + 4H20(l)

S(s)

+ 6HN0 3(aq)

• H 2S0 4(aq)

+ 6N0 2(g) + 2H20(I)

Nitric acid is used in the manufacture of fertilizers, dyes , drugs, and explosives.

Phosphorus Like nitrogen, phosphorus is a member of the Group SA family, and in some respects the chemistry of phosphorus resembles that of nitrogen. Phosphorus occurs most commonly in nature as phosphate rocks, which are mostly calcium phosphate [Ca3(P04)2] and fiuoroapatite [Cas(P04)3F] (Figure 24.8). Elemental phosphorus can be obtained by heating calcium phosphate with coke and silica sand:

Figure 24.8

Fluoroapatite

[CaS(P04 )3 F j.

•

There are several allotropic forms of phosphorus, but only white phosphorus and red phosphorus are of importance. White phosphorus consists of discrete tetrahedral P 4 molecules (Figure 24.9). A solid (m.p. 44.2°C), white phosphorus is insoluble in water but quite soluble in carbon disulfide (CS?) and organic solvents such as chloroform (CHCI 3). White phosphorus is a highly toxic substance. It bursts into flames spontaneously when exposed to air; hence, it is used in incendiary bombs and grenades:

The high reactivity of white phosphorus is attributed to structural strain: the P-P bonds are compressed in the tetrahedral P 4 molecule. White phosphorus was once used in matches, but because of its toxicity it has been replaced by tetraphosphorus trisulfide (P4S 3). When heated in the absence of air, white phosphorus is slowly converted to red phosphorus at about 300°C: nPi white phosphorus) ---+. (P4 )ired phosphorus) Red phosphorus has a polymeric structure (see Figure 24.9) and is more stable and less volatile than white phosphorus. The most important hydride of phosphorus is phosphine (PH 3), a colorless, very poisonous gas formed by heating white phosphorus in concentrated sodium hydroxide:

•

Figure 24.9

The structures of white and red phosphorus. Red phosphorus is believed to have a chain structure, as shown.

White phosphorus

Red phosphorus

SECTION 24.4

Nitrogen and Phosphorus

915

Phosphine is moderately soluble in water and more soluble in carbon disulfide and organic solvents. Its aqueous solution is neutral, unlike that of ammonia. In liquid ammonia, phosphine dissolves to give NH1PH 2 . Phosphine is a strong reducing agent; it reduces many metal salts to the corresponding metals. The gas burns in air:

Phosphorus forms binary compounds with halogens-namely, the trihalides (PX 3) and the pentahalides (PX s), where X denotes a halogen atom. In contrast, nitrogen can form only trihalides (NX 3). Unlike nitrogen, phosphorus has a 3d subshell, which can be used for valence-shell expansion. We can explain the bonding in PCIs by assuming that phosphorus undergoes sp 3d hybridization of its 3s, 3p, and 3d orbitals [ ~~ Section 9.4] . The five sp3d hybrid orbitals also account for the trigonal bipyramidal geometry of the PCls molecule (see Table 9.4). Phosphorus trichloride is prepared by heating white phosphorus in chlorine:

A colorless liquid (b.p. 76°C), PCl 3 is hydrolyzed according to the following equation:

In the presence of an excess of chlorine gas, PCl 3 is converted to phosphorus pentachloride, which is a light-yellow solid: PCI 3(1)

+ Clig) --+. PCIs(s)

X-ray studies indicate that solid phosphorus pentachloride exists as [PCI1] [PCI 6 ], in which the pCl1 ion has a tetrahedral geometry and the PCl 6 ion has an octahedral geometry. In the gas phase, PCIs (which has trigonal bipyramidal geometry) is in equilibrium with PCl 3 and C1 2 :

Phosphorus pentachloride reacts with water as follows:

The two important oxides of phosphorus are tetraphosphorus hexaoxide (P40 6) and tetraphosphorus decaoxide (P40 lO), shown in Figure 24.10. The oxides are obtained by burning white . phosphorus in limited and excess amounts of oxygen gas, respectively: Pis) Pis)

+ 30ig) --+. P40 6(S) + 50 2(g) • P40 lO (S)

Both oxides are acidic; that is, they are converted to acids in water. The compound P 4 0 flocculent powder (m.p. 420°C) that has a great affinity for water:

lO

is a white

For this reason, it is often used for drying gases and for removing water from solvents. There are many oxoacids containing phosphorus. Some examples are phosphorous acid (H 3P0 3), phosphoric acid (H 3P04), hypophosphorous acid (H 3P0 2), and triphosphOlic acid (H SP 30 IO) (the structures of which are shown in Figure 24.11). Phosphoric acid, also called orthophosphoric acid, is a weak triprotic acid. It is prepared industrially by the reaction of calcium phosphate with sulfuric acid:

Phosphorus Oxygen

Fig ure 24.10

The structures of R40 6 and P40 1O' Note the tetrahedral arrangement of the P atoms in P4° 10.

916

CHAPTER 24

Nonmetallic Elements and Their Compounds •

Figure 24.11

•

'0'

'0'

Structures of some common phosphorus-containing oxoacids.

. . II H-O-P-H •• I

. . II .. H-O-P-O-H .. I ..

H

H

H ypophosphorous acid (H 3P0 2l

'0'

'0'

. . II .. H-O-P . . I - O.. H

..

II

..

II - ..

II

..

"

[

"

1

1

"

:0: I

I

Pregnant women are advised not to consume large quantities of soda because of the phosphate content.

.t)o

H-O-P-O-P-O-P-O-H

'0' • •

"

:0: I

H

H

•

'0'

'0' • • I H

H

Phosphoric acid (H 3P0 4l

In the pure form, phosphoric acid is a colorless solid (m.p. 42.2°C). The phosphoric acid we use in the laboratory is usually an 82% H 3P0 4 solution (by mass). Phosphoric acid and phosphates have many commercial applications in detergents, fertilizers, flame retardants, and toothpastes, and as buffers in carbonated beverages. Like nitrogen, phosphorus is an element that is essential to life. It constitutes only about 1 percent by mass of the human body, but it is a very important 1 percent. About 23 percent of the human skeleton is mineral matter. The phosphorus content of this mineral matter, calcium phosphate [Ca3(P04)2], is 20 percent. Our teeth are basically Ca3(P04)2 and CaS(P04)30H. Phosphates are also important components of the genetic materials deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) .

Oxygen and Sulfur Oxygen Oxygen is by far the most abundant element in Earth 's crust, constituting about 46 percent of its mass. In addition, the atmosphere contains about 21 percent molecular oxygen by volume (23 percent by mass) . Like nitrogen, oxygen in the free state is a diatomic molecule (0 2), In the laboratory, oxygen gas can be obtained by heating potassium chlorate: 2KCI0 3 (s) --+. 2KCl(s)

+ 30 2 (g)

The reaction is usually catalyzed by manganese(IV) dioxide (Mn02)' Pure oxygen gas can be prepared by electrolyzing water (page 781 ). Industrially, oxygen gas is prepared by the fractional distillation of liquefied air. Oxygen gas is colorless and odorless. Oxygen is a building block of practically all biomolecules, accounting for about a fourth of the atoms in living matter. Molecular oxygen is the essential oxidant in the metabolic breakdown of food molecules. Without it, a human being cannot survive for more than a few minutes. Although oxygen has two allotropes, O 2 and 0 3, when we speak of molecular oxygen, we normally mean O? Ozone (0 3 ) is less stable than O 2 , The O 2 molecule is paramagnetic because it contains two unpaired electrons (see Section 10.7). A strong oxidizing agent, molecular oxygen is one of the most widely used industrial chemicals. Its main uses are in the steel industry and in sewage treatment. Oxygen is also used as a bleaching agent for pulp and paper, in medicine to ease breathing difficulties, in oxyacetylene torches, and as an oxidizing agent in many inorganic and organic reactions. Oxygen forms three types of oxides: the normal oxide (or simply the oxide), which contains the 2 0 - ion; the peroxide, which contains the O ~ - ion; and the superoxide, which contains the O 2 ion: ••

•

'0"•.2•

'0"•'0"•.2•

'0'0'• • • •• ••

Oxide

Peroxide

Superoxide

••

••

••

The ions are all strong Br0nsted bases and react with water as follows: Oxide:

+ H?O(l) - - + . 20H- (aq)

20 ~ - (aq)

+ 2H?0(1)

• 0 2(g)

Superoxide: 40 2 (aq)

+ 2H20(l)

• 30?(g)

Perox ide:

The reaction of 0 processes.

0 2- (aq)

2

-

+ 40H- (aq)

+ 40H - (aq)

with water is a hydrolysis reaction, but those involving O ~- and O 2 are redox

SECTION 24.5

Oxygen and Sulfur

The nature of bonding in oxides changes across any period in the periodic table. Oxides of elements on the left side of the periodic table, such as those of the alkali metals and alkaline earth metals, are generally ionic solids with high melting points. Oxides of the metalloids and of the metallic elements toward the middle of the periodic table are also solids, but they have much less ionic character. Oxides of nonmetals are covalent compounds that generally exist as liquids or gases at room temperature. The acidic character of the oxides increases from left to right. Consider the oxides of the third-period elements: MgO •

J

\,

Acidic

,

1

"

Amphoteric

Basic

The basicity of the oxides increases as we move down a paIticular group. MgO does not react with water, for example, but reacts with acid as follows:

On the other hand, BaO, which is more basic, undergoes hydrolysis to yield the corresponding hydroxid~:

BaO(s)

+ H 20(l) --+. Ba(OHMaq)

The best-known peroxide is hydrogen peroxide (H 2 0 ?). It is a colorless, syrupy liquid (m.p. -O.9°C), prepared in the laboratory by the action of cold dilute sulfuric acid on barium peroxide octahydrate:

The structure of hydrogen peroxide is shown in Figure 24.12. Using the VSEPR method, we see that the H -0 and 0-0 bonds are bent about each oxygen atom in a configuration similar to the structure of water. The lone-pair-bonding-pair repUlsion is greater in H 20 2 than in H 20, so the H-O-O angle is only 97° (compared with 104.5° for H-O-H in H 2 0) . Hydrogen peroxide is a polar molecule (f.L = 2.16 D). Hydrogen peroxide readily decomposes when heated or exposed to sunlight or even in the presence of dust particles or certain metals, including iron and copper: t).H O

= -196.4 kJ/mol

This is a disproportionation reaction. The oxidation number of oxygen changes from - 1 to - 2 and O. Hydrogen peroxide is miscible with water in all proportions due to its ability to hydrogenbond with water. Dilute hydrogen peroxide solutions (3 percent by mass), available in drugstores , are used as mild antiseptics; more concentrated H 2 0 2 solutions are employed, as bleaching agents for textiles, fur, and hair. The high heat of decomposition of hydrogen peroxide also makes it a suitable component in rocket fuel. Hydrogen peroxide is a strong oxidizing agent; it can oxidize Fe2+ ions to Fe 3 + ions in an acidic solution: H 20 2(aq)

+ 2Fe2+ (aq) + 2H+(aq) -_I 2Fe 3+(aq) + 2H 20(l)

It also oxidizes SO~- ions to SO~- ions:

In addition, hydrogen peroxide can act as a reducing agent toward substances that are stronger oxidizing agents than itself. For example, hydrogen peroxide reduces silver oxide to metallic silver,

•Figure 24.12 H 20 2 ·

The structure of

c- j

918

CHAPTER 24

Nonmetallic Elements and Their Compounds

Fig ure 24.13

The preparation of 0 3 from O2 by electrical discharge. The outside of the outer tube and the inside of the inner tube are coated with metal foils that are connected to a highvoltage source. (The metal foil on the inside of the inner tube is not shown.) During the electrical discharge, O2 gas is passed through the tube. The 0 3 gal' formed exits from the upper right-hand tube, along with some umeacted O2 gas.

,.--..:-_ lute~

tube

Metal foil on outer tube

.... 0 3 plus some unreacted 02

0,

Inner tube

High-voltage source

and permanganate (Mn0 4) to manganese(II) in an acidic solution,

If we want to determine hydrogen peroxide concentration, this reaction can be carried out as a redox titration, using a standard permanganate solution. There are relatively few known superoxides (i.e., compounds containing the O 2 ion). In general, only the most reactive alkali metals (K, Rb, and Cs) form superoxides. Both the peroxide ion and the superoxide ion are by-products of metabolism. Because these ions are highly reactive, they can inflict great damage on living cells. Fortunately, our bodies are equipped with the enzymes catalase, peroxidase, and superoxide dismutase, which convert these toxic substances to water and molecular oxygen. Ozone is a rather toxic, light-blue gas (b.p. -111.3°C). Its pungent odor is noticeable around sources of significant electrical discharges (such as a subway train). Ozone can be prepared frommolecular oxygen, either photochemically or by subjecting O2 to an electrical discharge (Figure 24.13):

D..GO= 326.8 kJ/mol Because the standard free energy of formation of ozone is a large positive quantity (D..G'f 163.4 kJ/mol), ozone is less stable than molecular oxygen. The ozone molecule has a bent structure in which the bond angle is 116.5°: ••

••

o ./ ~ :0. 0: • •

••

•

•

/P :01" :0: ••

•

•

Ozone is mainly used to purify drinking water; to deodorize air and sewage gases; and to bleach waxes, oils, and textiles. Ozone is a very powerful oxidizing agent-its oxidizing power is exceeded only by that of molecular fluorine (see Table 19.1). For example, ozone can oxidize sulfides of many metals to the corresponding sulfates:

Ozone oxidizes all the common metals except gold and platinum. In fact, a convenient test for ozone is based on its action on mercury. When exposed to ozone, mercury loses its metallic luster and sticks to glass tubing (instead of flowing freely through it). This behavior is attributed to the change in surface tension caused by the formation of mercury(II) oxide: 0 3(g)

+ 3Hg(l) --+. 3HgO(s)

The beneficial effect of ozone in the stratosphere and its undesirable action in smog formation were discussed in Chapter 21.

Sulfur Although sulfur is not a very abundant element (it constitutes only about 0.06 percent of Earth's crust by mass), it is readily available because it occurs commonly in nature in the elemental form.

SECTION 24.5

The largest known reserves of sulfur are found in sedimentary deposits. In addition, sulfur occurs widely in gypsum (CaS04 . 2H 20) and various sulfide minerals such as pyrite (FeS2) (Figure 24.14). Sulfur is also present in natural gas as H 2S, S02, and other sulfur-containing compounds. Sulfur is extracted from underground deposits by the Frasch l process, shown in Figure 24.15. In this process, superheated water (liquid water heated to about 160°C under high pressure to prevent it from boiling) is pumped down the outermost pipe to melt the sulfur. Next, compressed air is forced down the innermost pipe. Liquid sulfur mixed with air forms an emulsion that is less dense than water and therefore rises to the surface as it is forced up the middle pipe. Sulfur produced in this manner, which amounts to about 10 million tons per year, has a purity of about 99.5 percent. There are several allotropic forms of sulfur, the most important being the rhombic and monoclinic forms. Rhombic sulfur is thermodynamically the most stable form; it has a puckered Ss ring structure: • •

Oxygen and Sulfur

Figure 24.14

919

Pyrite (FeS2)'

• •

'S'

'S'

. / '-.... . /

.S. o.

.S.

'5" ..

. / '-....

.S. • •

•

.S• •.

.

" '~"

.-.

~

.

Multimedia

Nonmetallic elements- the extraction of

sulfur.

It is a yellow, tasteless, and odorless solid (m.p. 112°C) that is insoluble in water but soluble in carbon disulfide. When heated, it is slowly converted to monoclinic sulfur (m.p. 119°C), which also consists of the Sg units. When liquid sulfur is heated above 150°C, the rings begin to break up, and the entangling of the sulfur chains results in a sharp increase in the liquid's viscosity. Further heating tends to rupture the chains, so the viscosity decreases. Like nitrogen, sulfur shows a wide variety of oxidation numbers in its compounds (Table 24.3). The best-known hydrogen compound of sulfur is hydrogen sulfide, which is prepared by the action of an acid on a sulfide; for example,

Compressed air -----.-

Figure 24.15

.....f

I---'=--+ Sulfur

Superheated water

-......,,110-~

The Frasch process. Three concentric pipes are inserted into a hole drilled down to the sulfur deposit. Superheated water is forced down the outer pipe into the sulfur, causing it to melt. Molten sulfur is then forced up the middle pipe by compressed air.

,

Molten sulfur

1. Hennan Frasch (1 851- 1914). German chemical engineer. Besides inventing the process for obtaining pure sulfur, Frasch developed methods for refining petroleum.

920

CHAPTER 24

Nonmetallic Elements and Their Compounds

Oxidation Number

-2

,

Compound

Structure

Formula

Hydrogen sulfide

o

Sulfur (for reference)

·s.

• •

'S'

.............

/'

'S'

••

+1

.. .. /'"Cl: .. .. ,./ s-s .. ..

Disulfur dichloride

:CI ••

+2

• •

Sulfur dichloride

/

.'

S

".Cl.. '

.Cl. o.

"

••

+4

S -0 0-

Sulfur dioxide

:0'/ "0: .'

'.

'0' +6

Sulfur tlioxide

.

/

:0. • •

II S

".0:. • •

Nowadays, hydrogen sulfide used in qualitative analysis is prepared by the hydrolysis of thioacetamide:

o

S

II

+ 2H20 + H +

CH3-C

\

NH2 Thioacetamide

-_.

II CH3-C \

+

H 2S

+

NHt

O-H

Acetic acid

Hydrogen sulfide is a colorless gas (b.p. -60.2°C) that smells like rotten eggs. (The odor of rotten eggs actually does come from hydrogen sulfide, which is formed by the bacterial decomposition of sulfur-containing proteins.) Hydrogen sulfide is a highly toxic substance that, like hydrogen • cyanide, attacks respiratory enzymes. It is a very weak diprotic acid (see Table 15.5). In basic solution, H 2S is a reducing agent. For example, it is oxidized by perrnanganate to elemental sulfur: 3H 2S(aq)

+ 2Mn04 (aq) --+. 3S(s) + 2Mn02(S) + 2H 20(l) + 20H- (aq) •

Sulfur has two important oxides: sulfur dioxide (S02) and sulfur trioxide (S03)' Sulfur dioxide is formed when sulfur burns in air:

In the laboratory, it can be prepared by the action of an acid on a sulfite; for example,

or by the action of concentrated sulfuric acid on copper:

Sulfur dioxide (b.p. -10°C) is a pungent, colorless gas that is quite toxic. An acidic oxide, it reacts with water as follows:

Sulfur dioxide is slowly oxidized to sulfur trioxide, but the reaction rate can be greatly enhanced by a platinum or vanadium oxide catalyst:

'.

SECTION 24.6

Sulfur trioxide dissolves in water to form sulfuric acid:

The contributing role of sulfur dioxide to acid rain is discussed on page 844. Sulfuric acid is the world's most important industrial chemical. It is prepared industrially by first burning sulfur in air:

Next is the key step of converting sulfur dioxide to sulfur trioxide:

Vanadium(V) oxide (V 205) is the catalyst used for the second step. Because the sulfur dioxide and oxygen molecules react in contact with the surface of solid V?OS, the process is referred to as the contact process. Sulfuric acid is a diprotic acid. It is a colorless, viscous liquid (m.p. 10.4 QC) . The concentrated sulfuric acid we use in the laboratory is 98 percent H2S04 by mass (density = 1.84 g/cm\ which corresponds to a concentration of 18 M. The oxidizing strength of sulfuric acid depends on its temperature and concentration. A cold, dilute sulfuric acid solution reacts with metals above hydrogen in the activity series, thereby liberating molecular hydrogen in a displacement reaction:

This is a typical reaction of an active metal with an acid. The strength of sulfuric acid as an oxidizing agent is greatly enhanced when it is both hot and concentrated. In such a solution, the oxidizing agent is actually the sulfate ion rather than the hydrated proton, H +(aq) . Thus, copper reacts with concentrated sulfuric acid as follows:

Depending on the nature of the reducing agents, the sulfate ion may be further reduced to elemental sulfur or the sulfide ion. For example, the reduction of H 2 S04 by HI yields H 2 S and 12 :

Concentrated sulfuric acid oxidizes nonmetals. For example, it oxidizes carbon to. carbon dioxide and sulfur to sulfur dioxide: C(s)

+ 2H 2S0 4 (aq) --+. CO 2(g) + 2S0 2 (g) + 2H 20(l)

S(s)

+ 2H 2S0 4(aq)

• 3S0 2 (g)

+ 2H 20(t) Q

Carbon disulfide, a colorless, flammable liquid (b.p. 46 C), is formed by heating carbon and sulfur to a high temperature: C(s) + 2S(l) --+. CS 2 (l) It is only slightly soluble in water. Carbon disulfide is a good solvent for sulfur, phosphorus, iodine, and nonpolar substances such as waxes and rubber. Another interesting compound of sulfur is sulfur hexafluoride (SF6)' which is prepared by heating sulfur in an atmosphere of fluorine: Set)

+ 3Fig) --+. SF6(g)

Sulfur hexafluoride is a nontoxic, colorless gas (b.p. 63.8 C). It is the most inert of all sulfur compounds; it resists attack even by molten KOH. The structure and bonding of SF6 were discussed in Chapters 8 and 9. Q

•

The Halogens The halogens fluorine, chlorine, bromine, and iodine-are reactive nonmetals. Table 24.4 lists some of the properties of these elements. Although all halogens are highly reactive and toxic, the magnitude of reactivity and toxicity generally decreases from fluorine to iodine. The chemistry of fluorine differs from that of the rest of the halogens in the following ways: 1. Fluorine is the most reactive of all the halogens. The difference in reactivity between fluorine and chlorine is greater than that between chlorine and bromi ne. Table 24.4 shows that

The Halogens

-.

922

CHAPTER 24

Nonmetallic Elements and Their Compounds

F

(I

Br

Valence electron configuration

2i2l

3i3p s

4i4p 5

5s 5l

Melting point (0C)

-223

-102

-7

114

Boiling point (0C)

-187

-35

59

183

Appearance*

Pale-yellow gas

Yellow~green

gas

Red-brown liquid

Dark-violet vapor Dark metalliclooking solid

Atomic radius (pm)

72

99

114

133

Ionic radius (pm)-'-

136

181

195

216

Ionization energy (kJ/mol)

1680

1251

1139

1003

Electronegativity

4.0

3.0

2.8

2.5

Standard reduction potential (V)*

2.87

1.36

1.07

0.53

Bond enthalpy (kJ/mol)*

150.6

242.7

192.5

151.0

Property

•

1 2

* These va lues and descriptions apply to the diatomic species X" where X represents a halogen atom. The halfreaction is X2 (g) + 2e• 2X-(aq). t Refers to the anion X- .

the F-F bond is considerably weaker than the Cl-Cl bond. The weak bond in F2 can be explained in terms of the lone pairs on the F atoms: ••

••

:F-F: •• •• The small size of the F atoms (see Table 24.4) allows a close approach of the three lone pairs on each of the F atoms , resulting in a greater repulsion than that found in C1 2 , which consists of larger atoms. 2. Hydrogen fluoride (HF) has a relatively high boiling point (19.5 °C) as a result of strong intermolecular hydrogen bonding, whereas all other hydrogen halides have much lower boiling points. 3. Hydrofluoric acid is a weak acid, whereas all other hydrohalic acids (HCl, HBr, and HI) are strong acids. 4. Fluorine reacts with cold sodium hydroxide solution to produce oxygen difluoride as follows: 2F2(g)

+ 2NaOH(aq) - - + . 2NaF(aq) + H 20(l) + OF2(g)

The same reaction with chlorine or bromine, on the other hand, produces a halide and a hypohalite: Xig)

+ 2NaOH(aq) - - + . NaX(aq) + NaXO(aq) + H 20(l)

where X stands for Cl or Br. Iodine does not react under the same conditions. 5. Silver fluoride (AgF) is soluble. All other silver halides (AgCl, AgBr, and AgI) are insoluble. The element astatine also belongs to the Group 7 A family. However, all isotopes of astatine are radioactive; its longest-lived isotope is astatine-21O, which has a half-life of 8.3 h. As a result, it is both difficult and expensive to study astatine in the laboratory. The halogens form a very large number of compounds. In the elemental state, they form diatomic molecules (X 2). In nature, however, because of their high reactivity, halogens are always found combined with other elements. Chlorine, bromine, and iodine occur as halides in seawater, and fluorine occurs in the minerals fluorite (CaF2 ) and cryolite (Na3AlF6)'

SECTION 24.6

The Halogens

923

Preparation and General Properties of the Halogens Because fluorine and chlorine are strong oxidizing agents, they must be prepared by electrolysis rather than by chemical oxidation of the fluoride and chloride ions. Electrolysis does not work for aqueous solutions of fluorides, however, because fluorine is a stronger oxidizing agent than oxygen. From Table 19.1. we find that

+ 2e -

--+. 2F- (aq)

EO = 2.87 V

+ 4H+(aq) + 4e -

• 2H20(I)

EO = 1.23 V

F2(g) 02(g)

IfF2 were formed by the electrolysis of an aqueous fluoride solution, it would immediately oxidize water to oxygen. For this reason, fluorine is prepared by electrolyzing liquid hydrogen fluoride containing potassium fluoride to increase its conductivity, at about 70°C (Figure 24.16): 2F- --+. Fig)

Anode (oxidation): Cathode (reduction): 2H+

+ 2e-

Overall reaction:

2HF(l) --+. H 2(g)

+ 2e-

---+~ Hi g )

+ F2(g)

Chlorine gas (CI 2) is prepared industrially by the electrolysis of molten NaCI or by the chlor-alkali process, the electrolysis of a concentrated aqueous NaCI solution (called brine). (Chlor denotes chlorine, and alkali denotes an alkali metal, such as sodium.) Two of the common cells used in the chlor-alkali process are the mercury cell and the diaphragm cell. In both cells , the overall reaction is 2NaCI(aq)

+ 2H20(l)

electrolysi~ 2NaOH(aq)

+ H2(g ) + Cli

g)

As you can see, this reaction yields two useful by-products, NaOH and H 2. The cells are designed to separate the molecular chlorine from the sodium hydroxide solution and the molecular hydrogen to prevent side reactions such as 2NaOH(aq)

+ CI2(g) --+. NaOCI(aq) + NaCI(aq) + H 20(l)

and Hig) + C1 2(g) --+. 2HCI(g)

These reactions must be prevented because they consume the desired products and can be dangerous because a mixture of H2 and Cl 2 is explosive. Figure 24.17 shows the mercury cell used in the chlor-alkali process. The cathode is a liquid mercury pool at the bottom of the cell, and the anode is made of either graphite or titanium coated

Figure 24.16

Fz gas

1, H2 gas (

. 11

,Carbon anode

•

Hz gas

•

Diaphragm to prevent mixing of Hz and Fz gases Steel cathode

Liquid HF

Electrolytic cell for the preparation of fluorine gas. Note that because Hz and F z form an explosive mixture, these gases must be separated from each other.

,

924

CHAPTER 24

Nonmetallic Elements and Their Compounds

Figu re 24.17

Mercury cell used in the chlor-alkai process. The cathode contains mercury. The sodium-mercury amalgam is treated with water outside the cell to produce sodium hydroxide and hydrogen gas.

•

Graphite anode

I •

Brine

•

•

•Brine

,tr--Hg cathode

• Hg plus NalHg

with platinum. Brine is continuously passed through the cell as shown in the diagram. The electrode reactions are

Anode (oxidation): Cathode (reduction) :

2CI - ( a q ) ' Clig)

2Na(aq)

Overall reaction:

Figu re 24.18

The industrial manufacture of chlorine gas.

+ 2e-

H g(l).

+ 2e-

2NaHg

2NaCI(aq). 2NafHg

+ Clig)

where NafHg denotes the formation of sodium amalgam. The chlorine gas generated this way is very pure. The sodium amalgam does not react with the brine solution but decomposes as follows when treated with pure water outside the cell: 2NafHg

+ 2H 20(l) ---. 2NaOH(aq) + Hig) + 2Hg(l)

The by-products are sodium hydroxide and hydrogen gas. Although the mercury is cycled back into the cell for reuse, some of it is always discharged with waste solutions into the environment, resulting in mercury pollution. This is a major drawback of the mercury cell. Figure 24.18 shows the industrial manufacture of chlorine gas. The half-cell reactions in a diaphragm cell are shown in Figure 24.19. The asbestos diaphragm is permeable to the ions but not to the hydrogen and chlorine gases and so prevents the gases from mixing. During electrolysis, a positive pressure is applied on the anode side of the compartment to prevent the migration of the OH- ions from the cathode compartment. Periodically, fresh brine solution is added to the cell and the sodium hydroxide solution is run off as shown. The diaphragm cell presents no pollution problems. Its main disadvantage is that the sodium hydroxide solution is contaminated with unreacted sodium chloride.

Figure 24.19

Diaphragm cell used in the chlor-alkali process.

Anode

Cathode

Asbestos

:;:::::.~J=~r,diaphragm •

1

-- '

Brine

.------1-- • NaOH solution Battery

Reduction

Oxidation - _ . Cl , (g)

+ 2e -

SECTION 24.6

In the laboratory, chlorine, bromine, and iodine can be prepared by heating the alkali halides (NaCI, KBr, or KI) in concentrated sulfuric acid in the presence of manganese(IV) oxide. A representative reaction is Mn02(s)

+ 2H2SOiaq) + 2NaCI(aq) - _ . MnS04(aq) + Na2S0iaq) + 2H20(l) + Cli

g)

Compounds of the Halogens Most of the halides can be categorized as either ionic or covalent. The fluorides and chlorides of many metallic elements, especially those belonging to the alkali metal and alkaline earth metal (except beryllium) families, are ionic compounds. Most of the halides of nonmetals such as sulfur and phosphorus are covalent compounds. The oxidation numbers of the halogens can vary from -1 to + 7. The only exception is fluorine. Because it is the most electronegative element, fluorine can have only two oxidation numbers, 0 (as in F 2 ) and - 1, in all its compounds. The hydrogen halides, an important class of halogen compounds, can be formed by the direct combination of the elements: H2(g)

+ Xi

g) :;:::.=::!:' 2HX(g)

where X denotes a halogen atom. These reactions (especially the ones involving F2 and C1 2 ) can occur with explosive violence. Industrially, hydrogen chloride is produced as a by-product in the manufacture of chlorinated hydrocarbons:

In the laboratory, hydrogen fluoride and hydrogen chloride can be prepared by combining the metal halides with concentrated sulfuric acid: CaP2 (s) 2NaCl(s)

+ H 2 SOiaq) --+. 2HF(g) + CaS04(s) + H 2SOiaq) • 2HCl(g) + Na2S0i aq)

Hydrogen bromide and hydrogen iodide cannot be prepared this way because they are oxidized to elemental bromine and iodine. For example, the reaction between NaBr and H 2S04 is

Instead, hydrogen bromide is prepared by first reacting bromine with phosphorus to form phosphorus tribromide:

Next, PBr3 is treated with water to yield HBr:

Hydrogen iodide can be prepared in a similar manner. HF is so highly reactive that it attacks silica and silicates: 6HF(aq)

+ Si02(g) --+. H 2SiF6 (aq) + 2H2 0(l)

This property makes HF suitable for etching glass and is the reason that hydrogen fluoride must be kept in plastic or inert metal (e.g., Pt) containers. Hydrogen fluoride is used in the manufacture of Freons (see Chapter 17); for example, CClil) CFC1 3 (g)

+ HF(g) --+. CFCI 3(g) + HCl(g) + HF(g)

• CF2C1 2(g)

+ HCl(g)

It is also important in the production of aluminum. Hydrogen chloride is used in the preparation of hydrochloric acid, inorganic chlorides, and in various metallurgical processes. Hydrogen bromide and hydrogen iodide do not have any major industrial uses. Aqueous solutions of hydrogen halides are acidic. The strength of the acids increases as follows:

HF < < HCl < HBr < HI The halogens also form a series of oxoacids with the following general formulas: HXO Hypohalous acid

HX0 2 Halous acid

HX0 3 Halic acid

HX0 4 Perhalic acid

•

The Halogens

925

926

CHAPTER 24

Nonmetallic Elements and Their Compounds

(for example, HCIO, hypochlorous acid; HCI0 2, chlorous acid; HCl0 3, chloric acid; and HCl0 4, perchloric acid). Chlorous acid (HCI0 2) is the only known halous acid. All the halogens except fluorine fOIm halic and perhalic acids. The Lewis structures of the chlorine oxoacids are

..

H:O:Cl: •• ••

H:O:Cl:O: •• •• ••

H:O:Cl:O: •• •• • • • '0' .. •

• • '0' •• •• H:O:Cl:O: •• •• •• • '0' .. •

Hypochlorous acid

Chlorous acid

Chloric acid

Perchloric acid

••

•

••

••

••

••

••

••

••

••

For a given halogen, the acid strength decreases from perhalic acid to hypohalous acid [ ~~ Section 16.9] . Table 24.5 lists some of the halogen compounds. Periodic acid (HI04) does not appear because this compound cannot be isolated in the pure form. Instead the formula H 5I0 6 is often used to represent periodic acid.

Uses of the Halogens Applications of the halogens and their compounds are widespread in industry, health care, and other areas. One such application is fluoridation, the practice of adding small quantities of fluorides (about 1 ppm by mass) such as NaP to drinking water to reduce dental caries. One of the most important inorganic fluorides is uranium hexafluoride (UF6), which is essential to the gaseous diffusion process for separating isotopes of uranium (U-235 and U-238). Industrially, fluorine is used to produce polytetrafluoroethylene, a polymer better known as Teflon:

where n is a large number. Teflon is used in electrical insulators, high-temperature plastics, cooking utensils, and so on. Chlorine plays an important biological role in the human body, where the chloride ion is the principal anion in intracellular and extracellular fluids. Chlorine is widely used as an industrial bleaching agent for paper and textiles. Ordinary household laundry bleach contains the active ingredient sodium hypochlorite (about 5 percent by mass), which is prepared by combining chlorine gas with a cold solution of sodium hydroxide: C12 (g) + 2NaOH(aq) -

-+. NaCl(aq) + NaCIO(aq) + H 20(l)

Chlorine is also used to purify water and disinfect swimming pools. When chlorine dissolves in water, it undergoes the following reaction: Cl2(g)

+ H 20(l) -

-+. HCl(aq)

+ HCIO(aq)

It is thought that the CIO- ions destroy bacteria by oxidizing life-sustaining compounds within them. Chlorinated methanes, such as carbon tetrachloride and chloroform, are useful organic solvents. Large quantities of chlorine are used to produce insecticides, such as DDT. However, in

Compound

F

CI

Br

I

Hydrogen halide

HF (-1)

HCl ( - 1)

HBr (- 1)

HI (-1)

Oxides

OF2 (-1)

C1 20 (+1)

Br20(+I)

120 5 (+5)

Cl0 2 (+4)

Br02 (+4)

C120? (+7) Oxoacids

HFO (-1)

HCIO (+1)

HBrO(+1)

HIO(+I)

HCl0 3 (+5)

HBr03 (+5)

HI0 3 (+5)

HCI04 (+7)

HBr04 (+7)

H 5I06 (+7)

HCI0 2 (+3)

* The number in pa rentheses indicates the oxidation number of the halogen.

SECTION 24.6

view of the damage they inflict on the environment, the use of many of these compounds is either totally banned or greatly restricted in the United States. Chlorine is also used to produce polymers such as poly(vinyl chloride). So far as we know, bromine compounds occur naturally only in some marine organisms. Seawater is about 1 X 10- 3 M Br-, so it is the main source of bromine. Bromine is used to prepare ethylene dibromide (BrCH 2 CH 2 Br), which is used as an insecticide and as a scavenger for lead (i.e., to combine with lead) in gasoline to keep lead deposits from clogging engines. Studies have shown that ethylene dibromide is a very potent carcinogen. Bromine combines directly with silver to form silver bromide (AgBr), which is used in photographic films. Iodine is not used as widely as the other halogens. A 50 percent (by mass) alcohol solution of iodine, known as tincture of iodine, is used medicinally as an antiseptic. Iodine is an essential constituent of the thyroid hormone thyroxine: I

HO

I

Figure 24.20 Agi particles.

I

IP

H I CH 2 -C-C I \ NH 2 0H

0

The Halogens

I

Iodine deficiency in the diet may result in enlargement of the thyroid gland (known as goiter). Iodized table salt sold in the United States usually contains 0.01 percent Kl or Nal, which is more than sufficient to satisfy the 1 mg of iodine per week required for the formation of thyroxine in the human body. Silver iodide (AgI) is a pale-yellow solid that darkens when exposed to light. In this respect, it is similar to silver bromide. Silver iodide is sometimes used in cloud seeding, a process for inducing rainfall on a small scale (Figure 24.20). The advantage of using silver iodide is that enormous numbers of nuclei (i.e., small particles of silver iodide on which ice crystals can form) become available. About 1015 nuclei are produced from 1 g of AgI by vaporizing an acetone solution of silver iodide in a hot flame. The nuclei are then dispersed into the clouds from an airplane.

,

927

Cloud seeding using

928

CHAPTER 24

Nonmetallic Elements and Their Compounds

Applying What You've Learned

•

King George III of Britain (1738-1820) suffered from periodic physical and mental illness througbout his adult life. Several of the episodes were severe enough to render the king temporarily unable to rule. Although most of the king's symptoms are now attributed to porphyria, a hereditary metabolic disorder, a 2004 analysis of samples of the king's hair revealed high levels of arsenic. One possibility that has been raised regarding the source of the arsenic is the antimony used in the king's treatment for chronic illness. Antimony is an element known since ancient times and used in cosmetics (stibnite, the most common antimony ore was used as eye liner during biblical times) and in medicine.

Writing Prompt: Research arsenic and antimony, and write a SOO-word essay describing their history and explaining how medicinal antimony might have been the source of the arsenic found in the hair samples from Britain's King George III.

•

•

QUESTIONS AND PROBLEMS

929

CHAPTER SUMMARY Section 24.1

•

•

Properties of nonmetals vary more than properties of metals.

•

onmetal elements may be solid, liquid, or gaseous and may exhibit both positive and negative oxidation numbers in their compounds.

Section 24.2 • H ydrogen atoms contain one proton and one electron. They are the simplest atoms.

• Hydrogen combines with many metals and nonmetals to form

Phosphorus forms oxides and halides with oxidation numbers of + 3 and + 5 and several oxoacids. The phosphates are the most important phosphorus compounds.

Section 24.5 •

Elemental oxygen (02 ) is paramagnetic and contains two unpaired electrons.

•

Oxygen forms ozone (0 3), oxides (0 2 superoxides (0 2 ),

),

peroxides (O~-), and

• The most abundant element in Earth's crust, oxygen is essential for

hydrides; some hydrides are ionic and some are covalent.

life on Earth.

• There are three isotopes of hydrogen: lH, iH (deuterium), and ~H (tritium).

• Sulfur is obtained from Earth's crust as a molten liquid via the Frasch process.

• "Heavy water" (D 20) contains deuterium.

• Sulfur exists in a number of allotropic forms and has a variety of oxidation numbers in its compounds.

Section 24.3

•

• The important inorganic compounds of carbon are the carbides; the cyanides, most of which are extremely toxic; carbon monoxide, also toxic and a major air pollutant; the carbonates and bicarbonates; and carbon dioxide, an end product of metabolism and a component of the global carbon cycle.

Section 24.6 •

The halogens are toxic and reactive elements that are found only in compounds with other elements.

•

Fluorine and chlorine are strong oxidizing agents and are prepared by electrolysis.

•

With the exception of fluorine, the halogens may have both negative and positive oxidation states.

•

Fluorine's oxidation state in its compounds is always -1.

Section 24.4 •

Elemental nitrogen (N2) contains a triple bond and is very stable.

•

Compounds in which nitrogen has oxidation numbers from - 3 to are formed between nitrogen and hydrogen and/or oxygen atoms.

•

Ammonia (NH3 ) is widely used in fertilizers.

•

White phosphorus (P4 ) is highly toxic, very reactive, and flammable; the polymeric red phosphorus [(P4 )nl is more stable.

+5

Sulfuric acid is the cornerstone of the chemical industry. It is produced from sulfur via sulfur dioxide and sulfur trioxide by means of the contact process.

QUESTIONS AND PROBLEMS Section 24.1: General Properties of Nonmetals

Section 24.2: Hydrogen

Review Questions

Review Questions

_4.1

2-t2

_4.3

_·t4

Without referring to Figure 24.1, state whether each of the following elements are metals, metalloids, or nonmetals: (a) Cs, (b) Ge, (c) I, (d) Kr, (e) W, (f) Ga, (g) Te, (h) Bi.

24.5

Explain why hydrogen has a unique position in the periodic table.

24.6

Describe two laboratory and two industrial preparations for hydrogen.

List two chemical and two physical properties that distinguish a metal from a nonmetal.

24.7

Hydrogen exhibits three types of bonding in its compounds. Describe each type of bonding with an example.

Make a list of physical and chemical properties of chlorine (CI2 ) and magnesium. Comment on their differences with reference to the fact that one is a metal and the other is a nonmetal. Carbon is usually classified as a nonmetal. However, the graphite used in "lead" pencils conducts electricity. Look at a pencil, and list two nonmetallic properties of graphite.

,

24.8

What are interstitial hydrides?

24.9

Give the name of (a) an ionic hydride and (b) a covalent hydride. In each case describe the preparation and give the structure of the compound.

24.10

Describe what is meant by the "hydrogen economy."

930

CHAPTER 24

Nonmetallic Elements and Their Compounds

Problems 24.11

24.12

Elements number 17 and 20 fOim compounds with hydrogen. Write the formulas for these two compounds, and compare their chemical behavior in water.

Magnesium chloride is di ssolved in a solution containing sodium bicarbonate. On heating, a white precipitate is fonned. Explain what causes the precipitation.

24.30

A few drops of concentrated ammonia solution added to a calcium bicarbonate solution cause a white precipitate to form. Write a balanced equation for the reaction.

24.31

Sodium hydroxide is hygroscopic-that is, it absorbs moisture when exposed to the atmosphere. A student placed a pellet of NaOH on a watch glass. A few days later, she noticed that the pellet was covered with a white solid. What is the identity of this solid? (Hint: Air contains CO 2,)

24.32

A piece of red-hot magnesium ribbon will continue to burn in an atmosphere of CO 2 even though CO 2 does not support combustion. Explain.

24.33

Is carbon monoxide isoelectronic with nitrogen (N 2)?

Give an example of hydrogen as (a) an oxidizing agent and (b) a reducing agent.

24.13

Compare the physical and chemical properties of the hydrides of each of the following elemerats: Na, Ca, C, N, 0, Cl.

24.14

Suggest a physical method that would enable you to separate hydrogen gas from neon gas.

24.15

Write a balanced equation to show the reaction between CaH2 and H 20. How many grams of CaH2 are needed to produce 26.4 L of H2 gas at 20°C and 746 mmHg?

24.16

How many kilograms of water must be processed to obtain 2.0 L of D 2 at 25°C and 0.90 atm pressure? Assume that deuterium abundance is 0.015 percent and that recovery is 80 percent.

24.17

24.29

Predict the outcome of the following reactions:

Section 24.4: Nitrogen and Phosphorus

Review Questions

(a) CuO(s) + H 2(g) - _ . (b) Na20(s) + H2(g) - _ .

24.34

Describe a laboratory and an industrial preparation of nitrogen gas.

Starting with H 2, describe how you would prepare (a) HCl, (b) NH 3, and (c) LiOH.

24.35

What is meant by nitrogen fixation? Describe a process for fixation of nitrogen on an industrial scale.

24.36

Describe an industrial preparation of phosphorus.

Review Questions

24.37

Why is the P 4 molecule unstable?

24.19

Give an example of a carbide and a cyanide.

Problems

24.20

How are cyanide ions used in metallurgy?

24.38

24.21

Briefly discuss the preparation and properties of carbon monoxide and carbon dioxide.

Nitrogen can be obtained by (a) passing ammonia over red-hot copper(II) oxide and (b) heating ammonium dichromate [one of the products is Cr(Ill) oxide]. Write a balanced equation for each preparation.

24.22

What is coal?

24.39

24.23

Describe two chemical differences between CO and CO 2,

Write balanced equations for the preparation of sodium nitrite by (a) heating sodium nitrate and (b) heating sodium nitrate with carbon.

24.24

Describe the reaction between CO 2 and OH- in term s of a Lewis acid-base reaction.

24.40

Sodium amide (NaN Hz) reacts with water to produce sodium hydroxide and ammonia. Describe this reaction as a Br~;nsted acid-base reaction.

24.41

Write a balanced equation for the formation of urea, [(NH2)2CO], from carbon dioxide and ammonia. Should the reaction be run at • a high or low pressure to maximize the yield?

24.42

Some farmers feel that lightning helps produce a better crop. What is the scientific basis for this belief?

24.43

Explain why nitric acid can be reduced but not oxidized.

24.44

At 620 K, the vapor density of ammonium chloride relative to hydrogen (H 2) under the same conditions of temperature and pressure is 14.5, although, according to its formula mass, it should have a vapor density of 26.8. How would you account for this discrepancy?

24.45

Write a balanced equation for each of the following processes: (a) On heating, ammonium nitrate produces nitrous oxide. (b) On heating, potassium nitrate produces potassium nitrite and oxygen gas. (c) On heating, lead nitrate produces lead(II) oxide, nitrogen dioxide (N0 2), and oxygen gas.

24.18

Section 24.3: Carbon

Problems 24.25

Draw a Lewis structure for the C ~ - ion.

24.26

Balance the following equations: (a) Be2C(S) (b) CaC 2 (s)

24.27

24.28

+ H 20(I) - _ . + H 20(I) - _ .

Unlike CaC0 3, Na2C03 does not readily yield CO 2 when heated. On the other hand, NaHC0 3 undergoes thermal decomposition to produce CO 2 and Na2C03' (a) Write a balanced equation for the reaction. (b) How would you test for the CO 2 evolved? [Hint: Treat the gas with limewater, an aqueous solution of Ca(OH)z.] Two solutions are labeled A and B. Solution A contains Na2C03, and solution B contains NaHC0 3. Describe how you would distinguish between the two solutions if you were provided with an MgCI2 solution. (Hint: You need to know the solubilities of MgC0 3 and MgHC0 3.)

QUESTIONS AND PROBLEMS

24.46

24.47

24.48

24.49

Explain why, under normal conditions, the reaction of zinc with nitric acid does not produce hydrogen.

Problems

24.63

Potassium nitrite can be produced by heating a mixture of potassium nitrate and carbon. Write a balanced equation for this reaction. Calculate the theoretical yield of KN0 2 produced by heating 57.0 g of KN0 3 with an excess of carbon. 24.64

Consider the reaction

24.65

24.52

24.53

24.54

24.55

+ 50 2(g) - _ . 4NO(g) + 6H20(l)

24.66

Oxygen forms double bonds in O 2, but sulfur forms single bonds in S8' Explain .

When 1.645 g of white phosphorus is dissolved in 75.5 g of CS 2, the solution boils at 46.709°C, whereas pure CS 2 boils at 46.300°C. The molal boiling-point elevation constant for CS 2 is 2.34°Clm. Calculate the molar mass of white phosphorus, and give the molecular formula.

24.67

What are the oxidation numbers of 0 and F in HFO?

24.68

In 2004 about 48 million tons of sulfuric acid was produced in the United States. Calculate the amount of sulfur (in grams and moles) used to produce that amount of sulfuric acid.

Explain why two N atoms can form a double bond or a triple bond, whereas two P atoms normally can form only a single bond.

24.69

Sulfuric acid is a dehydrating agent. Write balanced equations for the reactions between sulfuric acid and the following substances: (a) HCOOH, (b) H 3P0 4, (c) RN0 3, (d) HCI0 3. (Hint: Sulfuric acid is not decomposed by the dehydrating action.)

Dinitrogen pentoxide is a product of the reaction between P40 lO and RN03 . Write a balanced equation for this reaction. Calculate the theoretical yield of N 20 S if 79.4 g of P4010 is combined with an excess of HN0 3 . (Hint: One of the products is HP0 3 .)

24.70

Calculate the amount of CaC0 3 (in grams) that would be required to react with 50.6 g of S02 emitted by a power plant.

Starting with elemental phosphorus (P4), show how you would prepare phosphoric acid.

24.71

SF6 exists, but OF6 does not. Explain.

24.72

Explain why SCI6, SBr6, and SI6 cannot be prepared.

24.73

The bad smell of water containing hydrogen sulfide can be removed by the action of chlorine. The reaction is

What is the hybridization of phosphorus in the phosphonium ion

Explain why (a) NH3 is more basic than PH3, (b) NH3 has a higher boiling point than PH3 , (c) PCls exists but NCls does not, and (d) N2 is more inert than P 4.

H 2S(aq)

Review Questions Describe one industrial and one laboratory preparation of O 2,

+ CI2(aq) - _ . 2HCl(aq) + S(s)

If the hydrogen sulfide content of contaminated water is 22 ppm by mass, calculate the amount of Cl2 (in grams) required to remove all the H2S from 2.0 X 102 gal of water. (1 gallon = 3.785 L.)

Section 24.5: Oxygen and Sulfur

24.57

Hydrogen peroxide is unstable and decomposes readily:

This reaction is accelerated by light, heat, or a catalyst. (a) Explain why hydrogen peroxide sold in drugstores comes in dark bottles. (b) The concentrations of aqueous hydrogen peroxide solutions are normally expressed as percent by mass. In the decomposition of hydrogen peroxide, how many liters of oxygen gas can be produced at STP from 15.0 g of a 7.50% hydrogen peroxide solution?

From the data in Appendix 2, calculate tlHo for the synthesis of NO (which is the first step in the manufacture of nitric acid) at 25°C:

(PH~)?

24.56

22 .

+ 0 2(g) :;::.=:!:' 2NO(g)

•

24.51

Draw molecular orbital energy level diagrams for O 2, O 2 , and

0

•

4NHig)

+ 03(g) - _ . NOz(g) + 02(g)

From the data in Appendix 2, calculate tlGo, K p , and Kc for the reaction at 25°C.

Given that the tlGo for the reaction at 298 K is 173.4 kJ/mol, calculate (a) the standard free energy of formation of NO, (b) Kp for the reaction, and (c) Kc for the reaction. 24.50

One of the steps involved in the depletion of ozone in the stratosphere by nitric oxide may be represented as NO(g)

Predict the geometry of nitrous oxide (N20), by the VSEPR method, and draw resonance structures for the molecule. (Hint: The atoms are arranged as NNO.)

N 2(g)

931

24.74

Compare the physical and chemical properties of H 20 and H2S.

24.75

Concentrated sulfuric acid reacts with sodium iodide to produce molecular iodine, hydrogen sulfide, and sodillm hydrogen sulfate. Write a balanced equation for the reaction. Describe two reactions in,which sulfuric acid acts as an oxidizing agent.

24.58

Give an account of the various kinds of oxides that exist, and illustrate each type by two examples.

24.59

Hydrogen peroxide can be prepared by treating barium peroxide with sulfuric acid. Write a balanced equation for this reaction.

24.76

24.60

Describe the Frasch process for obtaining sulfur.

Section 24.6: The Halogens

24.61

Describe the contact process for the production of sulfuric acid.

Review Questions

24.62

How is hydrogen sulfide generated in the laboratory?

24.77

Describe an industrial method for preparing each of the halogens.

24.78

Name the major uses of the halogens.

932

CHAPTER 24

Nonmetallic Elements and Their Compounds

Problems 24.79

Sulfuric acid is a weaker acid than hydrochloric acid. Yet hydrogen chloride is evolved when concentrated sulfuric acid is added to sodi um chloride. Ex plain.

24.81

Use the VSEPR method to predict the geometries of the following species: (a) 13 , (b) SiCI4 , (c) PFs, (d) SF4 .

24.82

Metal chlorides can be prepa:ed in a number of ways: (a) direct combination of metal and molecular chlorine, (b) reaction between metal and hydrochloric acid, (c) acid-base neutrali zation, (d) metal carbonate treated with hydrochloric acid, (e) precipitation reaction . Give an example for each type of preparation.

24.84

24.85

A 37S-gal tank is filled with water containing 167 g of bromine in the form of Br - ions. How many liters of Cl 2 gas at 1.00 atm and 20°C will be required to oxidize all the bromide to molecular bromine?

What volume of bromine (Br2) vapor measured at 100°C and 700 mmHg pressure would be obtained if 2.00 L of dry chlorine (Cl z), measured at 15°C and 760 mmHg, was absorbed by a potassium bromide solution? Aqueous copper(II) sulfate solution is blue. When aqueous potassium fluoride is added to the CUS04 sol ution, a green precipitate is formed. If aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what happens in each case.

24.87

Iodine pentoxide (1z0s) is sometimes used to remove carbon monoxide from the air by forming carbon dioxide and iodine. Write a balanced equation for this reaction, and identify species that are oxidized and reduced . Show that chlorine, bromine, and iodine are very much alike by giving an account of their behavior (a) with hydrogen, (b) in producing silver salts, (c) as oxidizing agents, and (d) with sodium hydroxide. (e) In what respects is fluorine not a typical halogen element?

Additional Problems 24.89

24.93

What is the change in oxidation number for the following reaction?

24.94

Starting with deuterium oxide (DzO), describe how you would prepare (a) NaOD, (b) DCl, (c) ND 3 , (d) C 2Dz, (e) CD4 , and (f) D 2 S04 ·

24.95

Solid PCIs exists as [pCl r] [PCI 6 ]. Draw Lewis structures for these ions. Describe the hybridization state of the P atoms.

24.96

Consider the Frasch process. (a) How is it possible to heat water well above 100°C without turning it into steam? (b) Why is water sent down the outermost pipe? (c) Why would excavating a mine and digging for sulfur be a dangerou s procedure for obtaining the element?

24.97

As we saw in Section 23.2, the reduction of iron oxides is accomplished by using carbon monoxide as a reducing agent. Starting with coke in a blast furnace, the following equilibrium plays a key role in the extraction of iron:

Hydrogen fluoride can be prepared by the action of sulfuric acid on sodium fluoride. Explain why hydrogen bromide cannot be prepared by the action of the same acid on sodium bromide.

24.86

24.88

Describe the bonding in the C~ - ion in terms of the molecular orbital theory.

Draw structures for (a) (HF)2 and (b) HF 2 .

24.80

24.83

24.92

Write a balanced equation for each of the following reactions: (a) Heating phosphorous acid yields phosphoric acid and phosphine (PH 3). (b) Lithium carbide reacts with hydrochloric acid to give lithium chloride and methane. (c) Bubbling HI gas through an aqueous solution of HN0 2 yields molecular iodine and nitric oxide. (d) Hydrogen sulfide is oxidized by chlorine to give HCI and SCI 2 .

24.90

Both N 2 0 and O 2 support combustion. Suggest one physical and one chemical test to distinguish between the two gases.

24.91

(a) Which of the following compounds has the greatest ionic character: PCls, SiCI 4 , CCI4 , BCI 3 ? (b) Which of the following ions has the smallest ionic radius: F -, C 4 - , N 3 -, 0 2-? (c) Which of the following atoms has the highest ionization energy: F, Cl, Br, I? (d) Which of the following oxides is most acidic: H 2 0, SiOl> CO 2 ?

C(s)

+ CO 2(g) - - +. 2CO(g)

Use the data in Appendix 2 to calculate the equilibrium constant at 25°C and 100°C. Assume !1Ho and !1So to be independent of temperature. 24.98

Lubricants used in watches usually consist of long-chain hydrocarbons. Oxidation by air forms solid polymers that eventually destroy the effectiveness of the lubricants. It is believed that one of the initial steps in the oxidation is removal of a hydrogen atom (hydrogen abstraction). By replacing the hydrogen atoms at reactive sites with deuterium atoms, it is possible to substantially slow down the overall oxidation rate. Why? (Hint: Consider the kinetic isotope effect.)

24.99

How are lightbulbs frosted? (Hint: Consider the action of hydrofluoric acid on glass, which is made of silicon dioxide.)

24.100 Life evolves to adapt to its environment. In this respect, explain why life most frequently needs oxygen for survival, rather than the more abundant nitrogen. 24.101 A I O.O-g sample of white phosphorus was burned in an excess of oxygen. The product was dissolved in enough water to make 500 mL of solution. Calculate the pH of the solution at 25°C. .

24.102 Predict the physical and chemical properties of astatine, a radioactive element and the last member of Group 7 A. 24.103 Assuming ideal behavior, calculate the density of gaseous HF at its normal boiling point (19.S°C). The experimentally measured density under the same conditions is 3.10 giL. Account for the discrepancy between your calculated value and the experimental result. 24.104 Ammonium nitrate is the most important nitrogen-containing fertilizer in the world. Given only air and water as starting materials and any equipment and catalyst at your disposal, describe how you would prepare ammonium nitrate. State conditions under which you can increase the yield in each step.

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: VERBAL REASONING

933

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: VERBAL REASONING Iodi ne deficiency results in a condition known as goiter, which is characterized by an enlarged thyroid, often appearing as a large, bulbous protrusion on the neck. Although it is relatively rare today, goiter was once common in regions where the soil and food supply are iodine poor, including the northern half of the contiguous United States. In 1918, 30 percent of the men registering for the World War I draft in Michigan were found to have ignificant goiter symptoms. Many of the young men were so ill as to be disqualified from enlisting in the Army. In 1922, concerned about the high incidence of goiter in his home state, David Murray Cowie, a professor of pediatrics at the University of ~ 1ichigan, began a campaign for the addition of a small amount of sodium iodide or potassium iodide to the table salt used by all Americans. Dr. Cowie was actually modeling a successful public health program that had all but eliminated iodine deficiency disorders (IDD) in Switzerland. In 1924 the Morton Salt Company began distributing iodized salt nationwide.

1.

The main point of the passage is that a) goiter was once common but is now rare because of iodized salt. b) goiter is caused by iodine deficiency. c) David Murray Cowie was a pediatrician in Michigan. d) men suffering from goiter were disqualified from serving in the Army.

2.

The reason dietary iodine is added to table salt is most likely that a) sodium iodide closely resembles sodium chloride. b) citizens of all socioeconomic strata use table salt. c) sodium iodide is not water soluble. d) elemental iodine is toxic.

3.

The most likely reason it took 2 years to have iodized salt made available nationwide is that a) shipping took longer without interstate highways. b) the Morton Salt Company didn't exist before 1924. c) public health officials had to be convinced that iodization of salt was safe and effective. d) Michigan was not yet part of the United States.

4.

According to the passage, goiter a) can be caused by a low-salt diet. b) can be cured by the addition of iodide to the diet. c) is potentially debilitating. d) was a more serious public health problem in Switzerland than in the United States.

,

ern •

25.1 • •

25.2 •

•

25.3 25.4 • • •

25.5 •

25.6 25.7

Polymers Addition Polymers Condensation Polymers

Ceramics and Composite Materials Ceramics Composite Materials

Liquid Crystals Biomedical Materials Dental Implants Soft Tissue Materials Artificial]oints

Nanotechnology Graphite, Buckyballs, and Nanotubes

Semiconductors Superconductors

•

aterla s

Modern Materials Chemistry and the 2005 Nobel Prize

•

The 2005 Nobel Prize in Chemistry was awarded to Yves Chauvin, Robert H. Grubbs, and Richard R. Schrock for the development of the metathesis method in organic synmolecules are broken, and thesis. In the metathesis method, double bonds in organic . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ....... . then new double bonds are formed. In olefin metathesis, a carbon-carbon double bond is broken in each of two olefins. The subsequent fOIIl1ation of two new carbon-carbon double bonds results in products in which the compounds have exchanged substituent groups with each other:

R

\

C=C

/

/

\

H

R'

R

\

+

H

C=C

/

H

/

\

R

R'

•

H

\

C=C

/

H

/

\

R

R'

+

H

\

C=C

/

H

/

\

.

.

Metathesis is a word meaning" an exchange of position, " and olefin is an old-fashioned word for an alkene, a molecule with a carbon-carbon double bond.

R'

H

Note that each reactant molecule has one R group and one R' group, whereas each product molecule has two identical groups (i.e., two R groups or two R' groups). This process requires a metal catalyst and is believed to occur by the formation of a four-membered nng: •

R

\

C=C

/

R'

R R'

/ l -\ H

"-... H

H~ \ )

/

R

/

C=C

\

H

•

I I H-C-C-H 1:/(';1 H-C-C-H I I R R'

•

R'

The metathesis method can be applied to many industrial chemical processes, including those that potentially can produce pharmaceuticals for the treatment of such conditions as bacterial and viral infections, cancer, and Alzheimer's disease.

In This Chapter, You Will Learn

about some of the chemistry involved in the development of modem materials.

Before you begin, you should review •

Functional groups

[ I~~

Section 10.2]

•

Organic polymers

[ ~~

Section 10.6]

,

Media Player/ MPEG Content J- Chapter in Review

Yves Chauvin (France), Robert H. Grubbs (USA), and Richard R. Schrock (USA) shared the 2005 Nobel Prize in Chemistry for their work on the metathesis method of organic synthesis. 935

936

CHAPTER 25

Modern Materials

Polymers Many biological molecules, such as DNA, starch, and proteins, have very large molecular masses. These molecules are polymers because they are made up of many smaller parts linked together [ ~~ Section 10.6] . The prefix poly comes from a Greek root meaning "many," and the small molecules that make up the individual building blocks of polymers are called monomers. Many natural polymers, such as silk and rubber, now have synthetic analogues that are manufactured on large scales by the chemical industry. Polymers are used in many medical applications, such as flexible wound dressings, surgical implants, and prosthetics. Some of these polymers are thermoplastic, which means that they can be melted and reshaped, or heated and bent. Others are thermosetting, which means that their shape is determined as part of the chemical process that fOllIled the polymer. Thermosetting polymers cannot be reshaped easily and are not easily recycled, whereas thermoplastic polymers can be melted down and cast into new shapes in different products.

Addition Polymers Bottles that can be recycled are made of thermoplastic polymers.

The simplest type of polymerization reaction involves the bonding of monomer molecules by movement of electrons from a multiple bond into new single bonds between molecules. This type of polymerization is called addition polymerization [ ~~ Section 10.6]. A molecule with a carbon-carbon double bond is called an alkene, and a molecule with a carbon-carbon triple bond . ....... ................. . . .. . . . .. . . . . .. . . - . . . . . . . . . . . .. is called an alkyne. The synthesis of polyethylene from ethylene (C ZH4 ), illustrates an addition polymerization. The arrows with half-pointed heads show movement of a single electron. The of "splitting open" the .double process .. . . . .. ...,......... ............... .. ... . bond is initiated by a molecule with a single 0-0 bond that can be broken in half by heating. The free radical initiator molecule attaches to one of the ethylene carbon atoms and breaks the pi bond between the two carbon atoms in ethylene. Figure 25.1 shows the scheme of addition polymerization with respect to the double bonds opening up to form single bonds between monomer units to form polyethylene from ethylene. The structure of polyethylene can be represented as -fCH z-CHz7n, where n is the number of ethylene monomer molecules that reacted (which could be hundreds, thousands, or more). The end of the chain contains a carbon atom with an unpaired electron (i.e., -CHzCHz'), though, so it must be closed by bonding with another atom. This might be a hydrogen atom (e.g., -CHzCHz-H), but it might also be part of the initiator molecule (e.g., -CHzCHz-OR). Although the residue of the initiator molecule has very different chemical properties than the main polymer chain, it will not significantly affect the polymer's properties because the polyethylene chain is so long. Although the structure of a polyethylene chain is drawn in Figure 25.1 as a straight line, the bond angles around a central carbon atom with four bonds are approximately 109.5° each [ ~~ Section 9.1] . Thus, the "straight" chain really zigzags, instead. The polymer chain bends, too, because there is free rotation about carbon-carbon single bonds. These molecular-level features manifest themselves in such macroscopic properties as flexibility. It is possible, depending on the reaction conditions, for branches to form off the main polyethylene chain during polymerization. Polymer chains with many branches cannot pack together as efficiently as "straight" or unbranched chains, so branched-chain polyethylene is more flexible and lower in density than straight-chain polyethylene. The difference between branched and unbranched polyethylene is shown schematically in Figure 25.2. Polyethylene consisting primarily of unbranched chains is known as high-density polyethylene (HDPE), whereas polyethylene that consists primarily of branched chains is known as lowdensity polyethylene (LDPE). HDPE is used in plastic bottles and containers where it is important to maintain shape, whereas LDPE is used in plastic food bags and wraps. Natural rubber fOlIlled from latex is an addition polymer. The monomer unit is isoprene, a molecule with two C=C double bonds. In the reaction that forms polyisoprene, a pair of electrons ,

The systematic name of C2 H4 is ethene, but the common name ethylene is still widely used.

Recall that species with an unpaired electron are called free radicals [~ Section 8.8] .

Multimedia Organic and Biochemistry- natural and synthetic polymers.

~C=C~~C=C~~C=C~ - _ . H

Figure 25.1

/

\

/

HH

\

/

HH

\

H

~~~~~~ -C-C-C-C-C-CI11111

HHHHHH

Addition polymerization to form polyethylene from ethylene. Each curved, single-headed arrow represents the movement of one electron.

SECTION 2S.1

Polymers

937

Figure 25.2

,

t •

Linear polyethylene is used to make plastic milk bottles. Plastic bags and wrap like Glad Cling Wrap are made from branched polyethylene.

3

--

A molecule of linear polyethylene, or HOPE

A molecule of branched polyethylene, or LDPE

•

moves from one double bond to form a new double bond between the middle two carbon atoms, and the other double bond opens up to form a carbon-carbon single bond to the next monomer unit. H 2C

CH2

\\

II

I

\

C-C

H3C H Isoprene

-H2C

CH 2 H 2C-J-CH 2 H 2C+--CH2 H 2C

CH 2 -

H n

cis-Polyisoprene Natural rubber is not very durable it only retains its useful properties in a relatively narrow l temperature range, To address this problem, Charles Goodyear developed industrial vulcanization in 1839, a process in which sulfur is used to create linkages between individual polymer chains in the rubber. By heating the rubber with sulfur (S8) under carefully controlled conditions, some of the carbon-hydrogen bonds are broken and replaced with carbon-sulfur bonds. The sulfur atoms are bonded to one or more additional sulfur atoms and then to another polymer chain through another sulfur-carbon bond (Figure 25.3). These linkages, called cross-links, make the rubber much stronger and more rigid than natural polyisoprene. Rubber is an elastomer, a material that can stretch or bend and then return to its original shape as long as the limits of its elasticity are not exceeded. The cross-linking of the polyisoprene chains plays an important role in maintaining rubber's elastomer properties. . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . Goodyear, the current supplier of tires to NASCAR teams, varies the composition of the rubber in its tires for each race depending on the expected tire wear and banking of the track. The left- and right-side tires can even be of different compositions depending on the track layout and

1. Charles Goodyear (1800-1860). American chemi·st. Goodyear was the first person to reali ze the potential of natural rubber. His vulcanization process made the use of rubber practical, contributing greatly to the development of the automobile industry.

Each tire used in the NASCAR race series costs approximately $400. A single race car can go through as many as 32 tires in one SOO-mile race, depending on track conditions. There are dozens of races throughout the year, and even more tires are used up during practices and prerace qualifying. It takes a lot of tires to compete in NASCAR!

938

CHAPTER 25

Figure 25.3

Modern Materials ,"

Vulcanization.

"-

CH3 S

I

CH 3

I

I

-t-C-C-C C-C-C-C=C-C-C-+H2 H2 I H Hz H2 H H2 H2 S-----. S-----. CH3 SI CH3

+

•

I

•

=l=C-C-C-C C-C-C-C=C-C-+ H H2 H2 I H H2 H2 H Hz S-----. S ..........

n

Sulfur

cis-Polyisoprene

I

CH 3

CH 3

I

I

-+-C=C-C-C-C H H2 H2 I

S

I

CH3

I

C-C - C- C=C-+H H2 H2 H

/S ,"

n

Cross-linked pol yisoprene

This figure shows the structure of a general "vinyl" compound. Styrene and vinyl chloride are vinyl compounds in which R is a phenyl group and a chlorine atom, respectively.

H

R

f=< H H Traditional record albums are made of PVC mixed with carbon .

conditions. Softer tire compounds offer more grip but wear out faster, whereas harder compounds last longer but grip less . Many other molecules containing carbon-carbon double bonds can form addition polymers, Styrene [Figure 25.4(a)] is an analogue of ethylene in which a hydrogen atom has been replaced with a C6HS (a phenyl) group. The polymerization of styrene forms polystyrene (Figure 25.5). As or some other gas blown into the solid), polystyrene is used in disposable cups . . . . . . .a. .foam . . . . . . . .(with . . . . . . . .air . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . and plates as well as Styrofoam insulation and craft materials, Vinyl chloride [Figure 25.4(b)], an analogue of ethylene in which one of the hydrogen atoms has been replaced by a chlorine atom, is another monomer that can be used to form a polymer with many common uses. Poly(vinyl chlo,... ,.fIde)' '(PYC)' ls"used 'iii' and 'fiiii'rigs ;' 'aiid' the' "vinyl" records that were the predecessor to musical recordings on cassettes and CDs.

PIpes'

some'iiiastlC 'wraps:

(b)

(a)

Figure 25.4

(c)

(a) Styrene, (b) vinyl chloride, and (c) tetrafluoroethylene (the monomer of Teflon).

H

I C I

H

H

H

I C I

I C I

I C I

C HC-7 'CH

II

HC"" /CH C H

Structure of polystyrene.

H

C H HC-7 'CH

I

Figure 25.5

Ciasslc'

I n

II

HC"" /CH C H

SECTION 25.1

Polymer

Structure

Monomer

939

Uses

Polyethylene

Plastic bags and wraps, bottles, toys, Tyvek wrap

H

H

H

I

I

I

C

HC~ ""'CH

I

H

II

HC

II

\

HC~ .......CH

C H

Poly( viny 1 chloride)

I -+-C I

H

H

IC-+I

Cl

\

H

H

I C--+I

CN

C

Ii

n

H H

I -t-C I

Disposable cups and plates, insulation, packing materials

~ CH /

C-CH H

n

H H

•

H

t/ =c!\ H H C-C

~----C-----C-+---

Polystyrene

Polyacrylonitrile

Polymers

\

/

C

H

n

H

/

C

\:1 /

Pipes and fittings, plastic wraps, records

H

C

\

Fibers for carpeting and clothing (Orlon)

CN

Poly butadiene

Synthetic rubber

n

Polytetrafluoroethylene, more commonly known as the trademarked brand name Teflon, is formed from the addition polymerization of tetrafluoroethylene. Tetrafluoroethylene, shown in Figure 25.4(c), is an analogue of ethylene in which fluorine atoms have replaced all four of the hydrogen atoms. Teflon is a very good electrical insulator, so it is commonly used to coat wires. It is probably best known, though, as a nonstick substance used to coat bakeware, frying pans, and pots. It is also used in films that can be inserted into threaded joints between metal pipes to make it easier to unscrew the connection when necessary. Because Teflon is chemically inert, it is not possible to cross-link the chains like an elastomer. The structures of some addition polymers, including the structures of their respective monomers and what they are typically used for, are summarized in Table 25.1. The diversity of polymers increases greatly when different monomers are used in the same polymer chain. Copolymers are polymers made from two or more different monomers [ ~~ Section 10.6]. Different combinations of the two or more monomers give rise to different types of copolymers. These different copolymers are summarized in Table 25.2.

,

Copolymer Type

Representation*

Random

... -A-A-B-A-B-B-B-A-A-B-A-B-A-B-A-B-B-A- ...

Alternating

... -A-B -A- B-A-B-A-B-A-B-A-B-A-B-A-B- ...

Block

, .. -A-A-A-A-A-A-B-B-B-B-B-B-A-A-A-A-A-A- ...

Graft

... -A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A- ,.. I

B-B-B-B-B-B-B-B-B ... *A and B represent different monomers (e.g., ethylene and styrene). ,

940

CHAPTER 25

Modern Materials

Sample Problem 2S. 1 lets you practice deducing the structure of an addition polymer based on the structure of its monomer.

Sample Problem 25.1 . .-. An Australian banknote issued in 1988 is printed on polypropylene for increased durability and security. •

Polypropylene is used in applications that range from indoor-outdoor carpeting to soda bottles to bank notes. It is produced by the addition polymerization of propylene, H 2C=CH-CH3 . Draw the structure of polypropylene, showing at least two repeating monomer units, and write its general formula. Strategy Addition polymers form via a free-radical reaction in which the pi electrons in the carbon-

~ 10

--

carbon double bond of a monomer molecule are used to form carbon-carbon single bonds to other monomer molecules. Draw the structural form ula of propylene such that the double bond can be "opened up" to form single bonds between consecutive monomer units. Setup The structural formu la of propylene is

H

H

f=
--C n

Write the structure of the two monomers that form Kevlar.

Strategy Identify the condensation linkages, split them apart, and then add H to one of the resulting • exposed bonds and OH to the other.

-

SECTION 25.1

Polymers

943

o

I

Setup Kevlar contains an amide (linkage between C and N). We split the C-N bond, adding

an H to the exposed N and an OH to the exposed C to produce an amine and a carboxylic acid, respectively. Solution To determine what the monomers are, first remove the C - N bond:

o -NH---
--NH

Next, add H 20 as H to the Nand OH to the C:

o I >--C--

o

I

--NH-

HO-C-

Finally, add the components of another water molecule to the open ends of the molecules-H to N and OH to C, as before:

•

Think About It To determine

o I

o

II

HO-C-------C-OH

Practice Problem Nylon 6, like nylon 6,6, is a condensation polymer (a polyamide). Unlike nylon

6,6, however, nylon 6 is not a copolymer because it is synthesized from a single monomer called 6-aminohexanoic acid:

o i

Draw the structure of at least three repeating units of ny Ion 6.

..

~,----------------------------------------------------~---

Bringing Chemistry to life Electrically Conducting Polymers You have seen so far that it is possible to polymerize organic compounds with carbon-carbon double bonds (alkenes). The resulting addition polymers (e.g., polyethylene) have carbon-carbon trisingle bonds. It is also possible, though, to polymerize organic compounds with carbon-carbon . . . . . . . . . . . . .. . . . pIe bonds. The simplest of these compounds (known collectively as alkynes) is acetylene (C 2H 2) , which is commonly used in welding torches. The polymerization of acetylene is very similar to the addition polymerization of ethylene. That is, a free-radical initiator molecule attaches to one of the carbon atoms and breaks one of the pi bonds between the two carbon atoms in acetylene. The resulting polymer, polyacetylene, thus retains a carbon-carbon double bond: H H-C

C-H

H

I

H

I

I

H-C-C-H H-C-C-H - _ . -C=C-C=C-C=C-

~~~~~~

I

H

I

H

I

H

The generalized form of the reaction can be represented as follows: H

I

n H-C=C-H ----+. -+C=C+

I H

The systematic name of acetylene is ethyne, but the common name acetylene is far more commonly used, w hich is somewhat unfortunate since the -ene ending makes it sound like it is an alkene with a double bond instead of an alkyne with a triple bond I

n

The structure of polyacetylene contains alternating carbon-carbon single and carboncarbon double bonds. Recall from Chapter 9 that the carbon atoms in a C=C double bond are

,

944

CHAPTER 25

Modern Materials

•

Sp2 hybridized and that it is the leftover p orbital on each carbon atom that overlaps to form the pi bond. Each carbon atom in polyacetylene is identical (structurally and electronically), so there is extended overlap of the p orbitals throughout the polymer chain. The p electrons are delocalized!that is, they can move throughout the network of overlapping orbitals that extend the length of the polymer chain. By being able to move electrons from one end of the polymer chain to the other, poly acetylene can conduct electricity much like a wire. That is, polyacetylene is a plastic that conducts electricity! For comparison, consider polyethylene and Teflon. Both are polyalkenes in which the polymer chain contains only carbon-carbon single bonds. Since all the electrons are localized in sigma bonds between the carbon atoms, there are no delocalized electrons available to carry charge throughout the polymer chain. The presence of electrons in delocalized orbitals, such as those in polyacetylene, is the key to electrical conductivity. Sample Problem 25.3 shows how to determine the structure of an electrically conducting polymer.

Sample Problem 25.3 Propyne (HC=CCH 3) can be used to form an electrically conducting polymer. Draw the structure of polypropyne, showing at least three repeating units, and write the general formula for the polymer. Strategy Addition polymers, whether they are syn thesized from alkenes or alkynes, form via

a free-radi cal reaction in which one pair of pi electrons in the carbon-carbon multiple bond of a monomer molecule is used to form carbon-carbon single bonds to other monomer molecules. Draw the structural formula of propyne such that the triple bond can be "opened up" to form single bonds between consecutive monomer units. Setup The structural formula of propyne can be determined from the formula given and by analogy

to acetylene. The only difference between propyne and acetylene is the CH3 group in place of one of the hydrogen atoms: H-C-C-CH 3 By drawing three adjacent propyne molecules, 'we can rearrange the bonds to show three repeating units:

Solution The structure of polypropylene showing three monomer units is

CH3

Think About It The repeating

polymer unit in the general structure resembles the original monomer except that there is a carbon-carbon double bond in place of a carbon-carbon triple bond. If each of the carbon-carbon single bonds at the ends of the repeating unit in parentheses were split open, then one electron from each bond could be combined to re-form the second pi bond in the carboncarbon triple bond of the monomer.

CH3

CH 3

I

I

I

-C=C-C=C-C=C-

I

H The repeating unit is -CH=C(CH 3 ) -

,

I

H

I

H

so the general structure is

CH3 I -+-C=C-+I H n

Practice Problem Draw the structures of poly(l-butyne) and poly(l-butene), showing three or more

repeating units. How are they different? How are they similar? Which of them, if any, do you think will conduct electricity?

H-C

C

CH2CH3

I-butyne

H2C

CH CH2CH3 I-butene

SECTION 25.2

Checkpoint 25.1 25.1 .1

Ceramics and Composite Materials

945

Polymers

Classify the following copolymer: - B-B-B-B-B-A-A-A-A-A-B-B-B-B - B-A-A-A-A-A(Select all that apply.) a) Block b) Random c) Graft d) Alternating e) All the above

25.1.2

What feature is common to molecules that can undergo polymerization? a) Fluorine b) Hydrogen bonds c) Sulfur d) Multiple bonds e) Lone pairs

· · · · 25.2 .

Ceramics and Composite Materials

Ceramics The use of ceramics in the form of pottery dates back to antiquity. Along with common ceramic substances like brick and cement, modern advanced ceramics are found in electronic devices and on the exterior of spaceships. All these ceramics are polymeric inorganic compounds that share the properties of hardness, strength, and high melting points. 'Cer--C-OH

/CH 2,----/ CH2,----CH2 CH2 CH 3

Liquid Crystals

Which of the following is a good analogy for anisotropy? (Select all that apply. )

25 .3.2

What characteristics make a molecule likely to exhibit liquid crystal properties? (Select all that apply. )

a) Glass of water

a) Long narrow shape

b) Bucket of ice cubes

b) Flexibility

c) Box of spaghetti

c) Rigidity

d) Box of rice

d) Low molar mass

e) Box of macaroni

e) High molar mass

Biomedical Materials Many modern materials are finding uses in medical applications . Replacement joints, dental implants, and artificial organs all contain modern polymers, composites, and ceramics. To function successfully in a biomedical application, though, a material must first be compatible with the living system. The human body very easily recognizes foreign substances and attacks them in an effort to rid them from the body. Thus, a biomaterial must be designed with enough similarity to the body's own systems that the body will accept the material as its own. Additionally, if the polymer, ceramic, or composite contains leftover chemicals from its manufacture, these contaminants may lead to adverse reactions with the body over the lifetime of the medical implant. It is important, therefore, that the substance be as pure as possible. The physical properties of a biomedical material are important, too, because the longer the material lasts, the less often the medical device has to be replaced during the lifetime of the patient. · Most biomedical materials must possess great strength and flexibility to perform in the body. For example, the materials in artificial joints and heart valves must be able to flex many times without breaking. Materials used in dental implants must show great hardness, moreover, so as not to crack during biting and chewing. It takes years of research to develop successful biomedical materials that meet these needs.

•

949

950

CHAPTER 25

Modern Materials

Dental Implants

•

Dental implant materials have been used for many years. The oldest dental fillings were made of various materials including lead (which fell out of favor due to its softness the danger of lead poisoning was as yet unknown), tin, platinum, gold, and aluminum. The remains of Confederate soldiers have been shown to include fillings made of a lead-tungsten mixture, probably from shotgun pellets; tin-iron; a mercury amalgam; gold; and even radioactive thorium. (The dentist probably thought he was using tin, which is similar in appearance.) Many modern fillings are made of dental amalgam [ ~~ Chapter 19] , a solution of several metals in mercury. Modern dental amalgam consists of 50 percent mercury and 50 percent of an alloy powder that usually contains (in order of abundance) silver, tin, copper, and zinc. These metals tend to expand slightly with age, causing fissures and cracks in the tooth that may require further intervention (e.g., crowns, root canals, or tooth replacement). Some people consider amalgam fillings to be unsightly, too. Dental composite materials are now used that have several advantages over traditional amalgams. The composites can be made in a wide range of colors, for example, to match the color of the other teeth. The existing healthy portion of the tooth can be etched with acid, moreover, to create pores into which the composite material can bond. With traditional dental amalgam, the dentist must create indentations in the healthy tooth to hold the amalgam in place. Destroying a portion of the healthy tooth is undesirable and can be avoided with the use of the composite. The dental composite consists of a matrix (made from a methacrylate resin) and a silica filler. The composite material is applied to the cavity in a putty-like consistency and then is dried and polymerized by a photochemical reaction initiated by light of a particular wavelength. Because the light does not penetrate very far into the composite, the thickness of the applied composite is critical. If the layer of composite is too thick, some of the composite will remain soft. A properly constructed dental composite filling will last more than 10 years. Over time, though, composite fillings tend to shrink, leaving breaches that can cause leakage, a situation that must be addressed to prevent further tooth decay. Porcelain ceramic fillings and crowns are very common, but the ceramic has two disadvantages: it is very hard and can wear on the opposing teeth, and it is brittle and may crack if subjected to great force. Most dentists, therefore, do not recommend porcelain ceramic crowns and fillings for the molar teeth, which do the bulk of the grinding work during chewing. A material used in dentistry must have properties that maximize both patient comfort and the lifetime of the implant. Dental fillings and tooth replacements must be resistant to acids, for example, because many foods (such as citrus fruits and soft drinks) contain acids directly. Any food containing carbohydrates can produce acids, though, if traces are left in the mouth because the bacteria that reside there consume carbohydrates, producing acids in the process. These acids eat away at the natural hydroxyapatite in teeth, creating caries (cavities). Materials like dental amalgam, gold, and dental composite resist attack by acids in the mouth. A dental material should also have low thermal conductivity; that is, it should conduct heat poorly. This is especially important in applications where the implant material is in contact with the nerve inside the tooth. If the implant material conducts heat well, then whenever hot or cold foods come in contact with the implant, the hot or cold will be transmitted through the material to the nerve, causing pain. Metals are good thermal conductors, so dental amalgams must not touch nerves. Finally, dental materials should resist wear (so they last a long time) and they should resist expansion and contraction as the temperature fluctuates. The close spacing and precision fits involved in dental fillings and implants would result in discomfort and possibly failure of the implant if the material expanded or contracted appreciably with changes in temperature.

Soft Tissue Materials Burn victims who have lost large amounts of skin do not have enough cells to grow new skin, so a synthetic substitute must be used. The most promising artificial skin material is based on a polymer of lactic acid and glycolic acid. Both of these compounds contain an alcohol group (-OH) and a carboxy group (-COOH), so they can form a polyester copolymer in a condensation reaction that mirrors the formation of Dacron polyester (Section 25.1):

o II

0

II

HOCH2C~,-O_ H _H ---,t--OCHC-OH

I

CH3 glycolic acid

0

lactic acid

0

II

--+.

II

--+-OCH2C-OCHC-+-

I

CH3

n

Biomedical Materials

SECTION 25.4

This copolymer forms the structural mesh that supports the growth of skin tissue cells from a source other than the patient. Once the skin cells grow on the structural mesh, the artificial skin is applied to..the patient, and the copolymer mesh eventually disappears as the ester linkages are . . .. .. .............................................. ......... ... ............. .. ..... ...... . .. . .. .. ... . hydrolyzed. During this treatment, the patients must take drugs that will suppress their immune systems, which ordinarily would attack these substances as foreign to prevent their bodies from rejecting the new skin. Sutures, commonly known as stitches, are now made of the same lactic acid-glycolic acid copolymer as the structural mesh of artificial skin. Before, sutures were made of materials that did not degrade so they eventually had to be removed again. This type of suture is still used in cases where the degrading suture is deemed a risk or there might be a need to reopen the incision at a later time. Biodegradable sutures are routinely used during operations on animals, too, which decreases the number of trips to the veterinarian, thereby reducing the stress on the pet and its owner. When portions of blood vessels need to be replaced or circumvented in the body, it may or may not be possible to obtain a suitable vascular graft from another part of the patient's body. Dacron polyester (Section 25.1) can be woven into a tubular shape for this purpose, and pores can . . be worked into the material so that the graft can integrate itself into the tissue around it, as well as to allow capillaries to connect. Unfortunately, the body 's immune system recognizes this as a foreign substance and blood platelets stick to the walls of the Dacron graft, causing clotting. Artificial hearts and heart valves are potential solutions to the shortage of donor organs. Artificial hearts have not yet been perfected to the point that they can work on a permanent basis, but valve replacement is a common procedure that can successfully prolong a patient's life. The most common replacement heart valve is fixed in position by surrounding the valve with lactic acid-glycolic acid copolymer, the same polyester that is used in skin grafts and biodegradable sutures. The polymer is woven into a mesh ring that allows the body tissue to grow into the ring and hold it in place. Researchers today are using electrospinning to place nanofibers directly onto wounds. (Nanofibers are products of nanotechnology; see Section 25.5.) Electrospinning uses a highvoltage electric field to create a jet of polymer solution that transforms into a spiraling dry fiber only nanometers in diameter. The wound dressings applied by direct electrospinning have high rates of moisture transmission and bacterial resistance. Enzymes can be added to the solution containing the nanofiber polymer, and these enzymes can be released into the skin after the dressing is applied. Proteins can also be cross-linked onto the nanofiber, making possible many applications of the nanofiber wound dressing.

Hydrolysis is essentially the opposite of a condensation reaction.

Artificial Joints Joint replacements, such as artificial knees, elbows, and hips, require major surgery but have become relatively commonplace because suitable materials have become available for not only the bone replacement but also the contact surfaces. In knee replacement, for example, the goal is usually to replace worn-out cartilage surfaces (which may have resulted from arthritis or injury). It is possible to replace shattered bones that cannot heal back to their original state as well. The materials originally used in joint replacements were ivory, metals, and glass, but they are fast being replaced by polymeric plastics and ceramics. Polymethyl methacrylate (PMMA) , which was the subject of Practice Problem 25.lB in Section 25.1 , and polyethylene have been used in many total joint replacements. However, these joints tend to fail over time when they are implanted in younger, more active patients, so efforts are under way to improve the wear quality of the PMMA and polyethylene parts. Although metallurgical research provided new substances that were much stronger and less likely to break, including titanium-based alloys, total joint replacements were predicted to last no more than 20 years in a patient before needing to be replaced again. Ultrahigh-molecularweight polyethylene, treated with radiation to initiate cross-linking of the polymer chains and stored under inert nitrogen gas to stop free-radical damage to the chain from the radiation , is now commonly used in many replacement joints, with the expectation of improved wear. It remains to be seen whether ultrahigh-molecular-weight polyethylene will prolong the life of these devices, since many of the joints have been in service in patients for less than a decade. Other surfaces have also been researched to reduce wear. Metal-on-metal bearing surfaces suffered early setbacks due to the manufacturing tolerances of the cobalt-based alloy used. As the bearing surfaces wore, they produced metal sludge, and high friction between components caused loosening of the joint. Modern metal-on-metal bearing surfaces have improved as the manufacturing has become more consistent and produced components lower in friction.

,

95 1

952

CHAPTER 25

Modern Materials

Ceramic-on-ceramic bearing surfaces have also been developed. Some ceramics have been prone to cracking, but many have shown great promise for long-lasting use. When a ceramic component in a replacement joint does crack, however, it can be disastrous because the entire joint may need to be replaced and further healthy bone may need to be sacrificed in order to install the new replacement joint. Despite the progress made so far on developing safe, durable, and useful artificial joints, much work remains to be done. One long-term issue that has yet to be researched extensively, for example, is the effect of the release of metal cations from the joint into the body over the lifetime of the joint replacement. This may become a more urgent health issue as the lifetimes of the artificial joints increases and the ages at which patients get the replacements decreases.

Nanotechnology Atoms and molecules have sizes on the order of a tenth of a nanometer (where 1 nm = 1O-9m) [ ~~ Section 2.2] . Nanotechnology is the development and study of such small-scale materials and objects. An atomic force microscope [Figure 2S.13(a)] is one way to image the surface of a material at the atomic or molecular level. A sharp-pointed stylus is moved over the surface of the substance under study, and the intermolecular forces between the material and the stylus force the stylus up and down. The stylus is mounted on a probe that reflects a laser beam into a detector. From the differences in deflection of the laser beam, the electronics attached to the probe can create an image of the surface of the object [Figure 2S.13(b)]. The scanning tunneling microscope (STM) works on a similar principle but only for samples that conduct electricity. The STM measures the peak and valley heights of the sample from the flow of electric current. The tip of the STM must move up and down to keep the current constant, and the up and down movements are translated into an image of the atomic and molecular structure of the substance [Figure 2S.13 (c)].

Graphite, Buckyballs, and Nanotubes Recall that allotropes are different forms of the same element [ ~~ Section 2.6] .

..... . Carbon 'exists' as 'seve'ral allotropes, one of which is graphite. Graphite consists of sheets of carbon atoms that are all sp2-hybridized [ ~~ Section 9.4] :

Intermolecular forces [ ~~ Chapter 11] hold the sheets together. Because of the delocalization of the electrons in the extended network of p orbitals perpendicular to the graphite sheet, graphite

(a)

Figure 25.13

(b)

(c)

(a) Atomic force microscope (AFM). (b) AFM image of a yeast cell. (c) STM image of iron atoms arranged to display the Chinese characters for atom on a copper surface.

SECTION 25.5

Nanotechnology

9:>3

is an electrical conductor within its sheets, but not between them. Electrical conductivity is much more common for metals than it is for nonmetals like graphite. Graphite and diamond were long thought to be the only two allotropes of carbon. In 1985, though, a compound consisting of 60 carbon atoms was isolated from graphite rods that had been vaporized with an intense pulse of laser light. The structure of the compound (C 60) was identical to that of a soccer ball, with 12 pentagonal and 20 hexagonal faces and carbon atoms at the comers of each face: /

/.,

This pattern also resembles an architectural structure called a geodesic dome, so the C60 molecule was named buckminsteifullerene in honor of R. Buckminster Fuller, the designer of the geodesic dome. The nickname for C60 is the buckyball. Since the discovery of CGO , elongated and elliptical cages of 70 and 80 carbon atoms have also been discovered. These molecules, calledfullerenes as a group, have been proposed to exist in such exotic places as stars and interstellar media and have been observed in such mundane places as deposits of chimney soot. Imagine wrapping a segment of a single graphite sheet (correctly called a graphene) around on itself. The resulting cylinder made' o{carboii ' ittomtdias it diameter' on the s'caie 'of iiiuiometers, ...... so it is called a carbon nanotube. Two types of nanotube structures are possible, as shown in Figure 25.14. The differences in arrangement of hexagons give the two kinds of nanotubes different electrical conductivities. Carbon nanotubes of different diameters can nest inside each other, producing multiplewalled nanotubes. Most nanotubes are single-walled, though, and they can be as narrow as 1 nm in diameter. These nanotubes are remarkably strong-stronger, in fact, than a comparably sized piece of steel! Because nanotubes consist only of carbon atoms, they can come into close contact, thereby allowing intermolecular forces to hold large numbers of tubes together. This, in turn, makes it possible to align groups of tubes, potentially forming long strands and fibers. Lightweight fibers and strands with tremendous strength have many applications in such things as auto parts, medical implants, and sports equipment. Besides carbon, other elements have been fashioned into nanotubes, too. Molybdenum sulfide nanotubes have been synthesized, for example, with openings between atoms that are just the right size to permit some gas molecules to pass through, but not others. This may have potential for the nanoscale storage of gases for fuel cells. Many universities have developed large-scale facilities for nanotechnology research in collaboration with industrial partners. Since the lower limits of size have been reached in the production of silicon chip-based electronic circuits, nanotechnology may provide the pathway to circuits that are much smaller than have ever been imagined.

An analogy is to take a piece of paper and ro ~ so that yo.u have a tube.

Figure 25.14

•

Carbon nanotubes.

954

CHAPTER 25

Modern Materials

Figure 25.15

Energy bands in metals (conductors), semiconductors, and insulators. Conduction band No energy gap

Valence band

k

Conduction band

~~

Large energy gap

Small energy gap

---i11 ~ k

Valence band

• Conductor

Conduction band

Semiconductor

Valence band Insulator

Semiconductors Recall from Section 9.6 that bonds between atoms can be described as resulting from the combination of atomic orbitals to form molecular orbitals. In a bulk sample of metal containing many, many atoms, there are many, many molecular orbitals. These molecular orbitals are so close in energy that instead of forming individual bonding and antibonding orbitals for each pair of atomic orbitals that combine, they form a band of bonding levels and a band of anti bonding levels. The energies of the bonding and antibonding bands depend on the energies of the atomic orbitals that combined to form them in the first place. As a result, the band structure of a bulk sample depends on the original atoms' energy levels. The band energies and the gaps (or lack of gaps) between the bonding band (called the valence band) and the antibonding band (called the conduction band) make it possible to classify a substance as an electrical conductor, a semiconductor, or an electrical insulator based on the behavior of the electrons in the bands (Figure 25.15). The valence band is filled with the valence (bonding) electrons, whereas the conduction band is empty or only partially filled with electrons. It is the conduction band that allows electrons to move between atoms. The gap between the valence band and the conduction band can vary from nothing (for an electrical conductor) to a small amount (for a semiconductor) to a very large amount (for an electrical insulator). Thus, the size of the band gap determines the conduction behavior of the material. Metals have no band gap, so they are good conductors of electricity. The valence band and conduction band actually overlap or are one and the same in metallic conductors, so there is no energy balTier to the movement of electrons from one atom to another in a metal. Semiconductors have a band gap, but it is relatively small, so there is limited movement of electrons from the valence band to the conduction band. Nonconductors (electrical insulators) have large band gaps, so it is nearly impossible to promote electrons from the valence band to the conduction band. Silicon, germanium, and carbon in the form of graphite are the only elemental semicon. . .. .. .. . . .... .. . . . . . ... . . . . . . . .. .. ..... . .. .. ...... ... . . . . . . . .. . . .. . . . . .. .. . The group number for a main group element ductors at room temperature. All three of these elements are in Group 4A and have four valence corresponds to the number of valence electrons electrons. (Tin and lead are also in Group 4A, but they are metals, not semiconductors.) Other thatthe atom has [ ~~ Section 8.5]. semiconductors consist of combinations of elements whose valence electron count totals 8. For example, gallium (Group 3A) and phosphorus (Group SA) form a semiconductor because gallium contributes three valence electrons and phosphorus contributes five, giving a total of eight valence electrons. Semiconductors have also been formed between Group 2B elements (particularly Zn and Cd) and Group 6A elements. Sample Problem 25.5 lets you practice identifying combinations of elements that can exhibit semiconductor properties. ~

Sample State whether each of the following combinations of elements could form a semiconductor: (a) GaSe, (b) In-P, (c) Cd-Teo

Strategy Count the valence electrons in each type of atom. If they total eight for the two elements, then the combination will probabl y form a semiconductor.

SECTION 25.6

Semiconductors

---

c .:;-::;

Setup Locate each element in the periodic table. All of them are main group elements, so we can use

their group numbers to directly determine the number of valence electrons each element contributes: (a) Ga is in Group 3A, and Se is in Group 6A. (b) In is in Group 3A, and P is in Group SA. (c) Cd is in Group 2B, and Te is in Group 6A. Solution (a) Ga has three valence electrons, and Se has six. This gives a total of nine, which is too

many to form a semiconductor. (b) In (three valence electrons) and P (five valence electrons) combine for a total of eight, so In-P should be a semiconductor. (c) Cd (two valence electrons) and Te (six valence electrons) also combine for a total of eight, so CdTe should be a semiconductor, too.

Think About It Semiconductors

consisting of Group 3 and Group S elements are used in the contact layers of light -emitting diodes (LEDs).

Practice Problem Name an element in each case that could be combined with each of the following

elements to form a semiconductor: (a) 0, (b) Sb, (c) Zn.

The conductivity of a semiconductor can be enhanced greatly by doping, the addition of very small quantities of an element with one more or one fewer valence electron than the natural semiconductor. For the purpose of simplicity, we will consider a pure silicon semiconductor. Silicon is a Group 4A element, so it has four valence electrons per atom. A small (parts per million) amount of phosphorus (Group SA, five valence electrons) can be added, thus doping the silicon with phosphorus. Since each phosphorus atom has an "extra" electron relative to the pure semiconductor, these extra electrons must reside in the conduction band, where they increase its conductivity. This type of doped semiconductor is called an n-type semiconductor because the semiconductivity has been enhanced by the addition of negative particles, the extra electrons. It is also possible to dope a semiconductor with an element that has fewer valence electrons than the semiconductor itself. For example, a silicon semiconductor could be doped with small amounts (again, parts per million) of gallium (Group 3A, three valence electrons). Now, instead of an "extra" electron, there is a "hole" (cine less electron) in every place that a gallium atom has replaced a silicon atom. These holes are effectively positive charges (because each is the absence of an electron), so this type of material is called a p-type semiconductor. A p-type semiconductor has increased conductivity because the holes (which are in the valence band) move through the silicon rather than electrons. The energy required to move an electron from the valence band into a hole (also in the valence band) is considerably less than the energy needed to promote an electron from the valence band to the conduction band of a semiconductor (Figure 2S .16). Thus, the movement of the holes results from the movement of charge, and conductivity is the movement of charge. A combination of p-type and n-type semiconductors can be used to create a solar battery in which the holes and extra electrons in the semiconductors are at equilibrium in the dark. At equilibrium, the positive holes in the n-type layer have offset the movement of electrons into the p-type layer. Light falling on the solar cell causes the electrons that have entered the p-type layer to go back to the n-type layer, and other electrons to pass through a wire from the n-type layer to the p-type layer. This movement of electrons constitutes an electric cunent. Diodes, which are electronic devices that restrict the flow of electrons in a circuit to one direction, work in essentially the opposite direction from solar cells. A light-emitting diode (LED) consists of n-type and p-type semiconductor layers placed in contact. When a small voltage is applied, the "extra" electrons from the n-type side combine with the holes in the p-type side, thus Figure 25.16

The effects of doping on the conduction of semiconducto . CB

CB

CB

VB

VB

VB

Intrinsic

n-Type

p-Type

956

CHAPTER 25

Modern Materials

emitting energy (light) whose wavelength depends on the band gap. The band gap is different for different semiconductors (e.g. , Ga-As versus Ga-P), so the specific combination of semiconductor materials can be chosen to make the band gap correspond to a desired color of light. LEDs are finding increasing use in applications previously dominated by traditional incandescent lightbulbs such as emergency exit signs, car taillights, and traffic signals (Figure 25.17).

Checkpoint 25.6 •

Figure 25.17 LED arrays in traffic signals. Many communities have replaced the red and green signals with LED anays that are much more energy efficient than the incandescent bulbs used previously. The yellow lights are being replaced more slowly, however, because LED arrays are expensive and the short duration of yellow lights would not necessaril y allow an overall cost saving when the energy efficiency is factored in.

Allex Muller and Georg Bednorz were awarded the 1987 Nobel Prize in Physics "for their important breakthrough in the discovery of superconductivity in ceramic materials. "

25.6.1

Semiconductors

a) have a large band gap.

Elements that are semiconductors at room temperature are

b) have a small band gap.

a) members of Group 2A.

c) have no band gap.

b) members of Group 3A.

d) do not conduct electricity.

c) members of Group 4A.

e) conduct electricity only when combined with other elements.

d) combinations of elements from Group 3A and Group SA.

25.6.2

Semiconductors are substances that

e) combinations of elements from Group 2A and Group 6A.

Superconductors No ordinary electrical conductor is perfect. Even metals have some resistance to the flow of electrons, which wastes energy in the form of heat. Superconductors have no resistance to the flow of electrons and thus could be very useful for the transmission of electricity over the long distances between power plants and cities and towns. The first superconductor was discovered in 1911 by the Dutch physicist H . Kamerlingh-Onnes. He received the 1913 Nobel Prize in Physics for showing that mercury was a superconductor at 4 K , the temperature of liquid helium. Liquid helium is very expensive, however, so there were few feasible applications of a mercury superconductor. Since..that have been discovered including a lanthanum-, barium-, . . . . .. ... .....time, more. . .superconductors . .. . . . copper- , and oxygen-containing ceramic compound, and a series of copper oxides ("cuprates"), all of which become superconducting below 77 K, the temperature of liquid nitrogen (a much more common and less expensive refrigerant than liquid helium). The temperature below which an element, compound, or material becomes superconducting is called the superconducting transition temperature, Te. The higher the Te, the more useful the superconductor. By 1987, superconductivity was observed in another ceramic, yttrium barium copper oxide (YBCO) (YBa2Cu30 7) ' Commonly called a " 1-2-3 compound" because of the 1:2:3 ratio of yttrium to barium to copper, YBCO has a Te of 93 K (above the temperature of liquid nitrogen). Although 93 K is still cold, it qualifies as quite warm in the world of superconductors, so YBCO is considered to be the first high-temperature superconductor. YBCO was first prepared by combining the three metal carbonates at high temperature (1000 to 1300 K). The copper ions are present in a mixture of + 2 and + 3 oxidation states: . . . .. ... .. . ... . .. . . . . . . . . . . . . .. .' 4BaC0 3 + Y 2(C0 3)3 + 4CuC0 3 + CU2(C0 3h • 2YBa2Cu30 7 + 14 CO 2 -

The decomposition of a metal carbonate to the metal oxide and carbon dioxide gas is a very common reaction.

YBCO must be sintered carefully to obtain the optimal superconductivity. Superconducting wires were first made from Bi2Sr2CaCu20 g, known as BSCCO-2212, a high-temperature superconductor with Te = 95 K. It can be prepared in wire forms because it has layers of bismuth and oxygen atoms that YBCO does not have. BSCCO-2212 is so named because its formula consists of two bismuth, two strontium, one calcium, and two copper atoms. Other BSCCO compounds exist, too, and they all have a general formula of Bi2Sr2Ca1/Cul/+ 1021l+6. BSCCO-2223, for example, which has the formula Bi2Sr?Ca2Cu30 1O (where n = 2), is a superconductor with Te = 107 K. Superconductors at or below Te exhibit the Meissner effect, the exclusion of magnetic fields. In Figure 25.18, for example, a magnet is shown levitating above a YBCO pellet, a superconductor at or below - 180°C, its Te. Magnetic levitation is being researched for use in trains, because the amount of friction between the train and the tracks would be drastically reduced in a magnetically levitated ("maglev") train versus one running on ordinary tracks and wheels. The engineering challenge to a

SECTION 25.7

Superconductors

957

Figure 25.18

The Meissner effect. A magnet levitates above a YBCO pellet at the temperature of liquid nitrogen (- 196°C).

maglev train is that the superconductor must be kept at or below its Tc for the magnetic levitation to occur, and the highest temperatures obtained for superconducting substances are approximately 140 K-warm on the superconductivity scale but still very cold by ordinary standards. Superconductivity in metals can be explained satisfactorily by BCS theory, first proposed by John Bardeen, Leon. . . Neil Robert 1957. They received the . . . . . . .Cooper, . . . . . . . . . . . .and . . . . . .John . . . .. .... ..... Schrieffer . . . . . . . . . . . . in . . . . . . . .. ... ...................... .... . Nobel Prize in Physics in 1972 for their work. BCS theory treats superconductivity using quantum mechanical effects, proposing that. .electrons opposite spin...... can...pair due to fundamental ... . .. ... . .. . . . . ..with .................... ............................... attractive forces between the electrons. At temperatures below Tc, the paired electrons resist energetic interference from other atoms and experience no resistance to flow. Superconductivity in ceramics has yet to be satisfactorily explained. ,

Many a Nobel Prize has been won by scientists who research and develop modern materials!

You might be skeptical that two negatively charged electrons would attract each other, but there is evidence that all subatomic particles do exert some attractive forces on each other.

•

,

958

CHAPTER 25

\

Modern Materials

I

Applying What You've Learned Now that you know more about polymers, we can apply some of the concepts you've learned to ring-opening metathesis polymerization (ROMP), a polymerization reaction that utilizes the metathesis reaction introduced at the beginning of the chapter. The general mechanism for ROMP is follows: I

R

•

M

\-.

/

I

H 2C

~

M

HC"\ CH

R

R

•

I

CH2

M

\~CH I I

H2 C

~CH ~CH

•

I

I

The pi bond from the carbon-carbon double bond in the ring of the monomer reacts with the metal-carbon double bond in the transition metal catalyst (M=CHR) to first form a metallacyclobutane complex. The metallacyclobutane complex then breaks up to form a new carbon-carbon double bond and a new metal-carbon double bond. The new carboncarbon double bond is the first in the polymer chain, and the new metal-carbon double bond is available to react with additional monomers and thus grow the chain. As long as the metal from the catalyst is attached to the polymer via this metal-carbon double bond, the polymer is said to be "living" because it will continue to grow as more monomer is added. The process we will consider is the polymelization of two cyclic alkenes (cyclobutene and 3-chlorocyclobutene) using ROMP with an appropriate transition metal catalyst (abbreviated as M=CHR):

I

CH

I

C(CH-CH2 Cyclobutene

3-Chlorocyclobutene

Problems:

a)

What is the formula of the polymer formed by polymerizing first fifty cyclobutene molecules and then seventy 3-chlorocyclobutene molecules? [ ~. Sample Problem 25.1]

b)

What type of polymer (see Table 25.2) is this?

c)

Would this polymer conduct electricity?

[ ~.

I

H 2C- CH 2

CH2

HC

h

[ ~.

Sample Problem 25.2]

Sample Problem 25.3]

,

•

!

CHAPTER SUMMARY

959

CHAPTER SUMMARY Section 25.1

Section 25.3

• Polymers are made up of many repeating units, called monomers, that

o

are linked together by covalent bonds.

in the liquid phase. Molecules containing structurally rigid regions (e.g., rings and multiple bonds) and molecules that are longer in one dimension than in another seem to form the best liquid crystals. Although ordinary liquids are isotropic, liquid crystals are anisotropic, meaning that their apparent properties depend on the direction from which we view them.

• Thermoplastic polymers can be melted and reshaped, whereas thermosetting polymers assume their final shape as part of the chemical reaction that forms them in the first place.

• Addition polymers are formed via a radical reaction in which a pi bond in an alkene or alkyne is opened up to form new bonds to adjacent monomers. •

Liquid crystals are substances that maintain an ordered arrangement

o

Polymer chains can be cross-linked to provide added durability and flexibility.

• Elastomers, such as natural rubber, can stretch and bend and then return to their original shape.

Liquid crystals may be nematic, smectic, or cholesteric, depending on the type of ordering the molecules can adopt. Nematic liquid crystals contain order in only one dimension (parallel molecules), smectic liquid crystals contain order in two dimensions (parallel molecules in parallel layers), and cholesteric liquid crystals contain layers that are rotated with respect to one another. Within each cholesteric layer, the molecules are parallel.

• Copolymers contain more than one type of monomer. Copolymers can be random, alternating, block, or graft, depending on the sequence of the different monomers and the way that the monomers are attached in the polymer.

Section 25.4 o

• Tacticity describes the relative geometric arrangements of the groups attached to the chiral carbon atoms in vinyl polymers. In isotactic polymers, all the groups attached to the chiral carbon atoms have identical arrangements. In syndiotactic polymers, they alternate positions along the polymer chain. In atactic polymers, they have random arrangements.

o

• Condensation polymers form when two different groups on monomers react, thus producing a covalent bond between monomers and a small molecule (usually water) that is expelled. The two reacting groups may be on the same monomer (as in the amino acids that form polypeptides and proteins), or they may be on two different monomers (as in hexamethylenediarnine and adipic acid, which form nylon 6,6). Most condensation polymers are really copolymers. •

•

Amines and organic acids form condensation polymers known as polyamides. Alcohols and organic acids form condensation polymers known as polyesters. Amino acids form polyamide polymers known as polypeptides. Long, straight-chain polypeptides are known as proteins.

•

o

o

•

The sol-gel process is used to prepare ceramics for structural applications because it produces particles of nearly uniform size that are much more likely to produce a solid ceramic without gaps or cracks.

•

o

Composite materials contain several components each of which contributes properties to the overall material. Modern composites include reinforcing fibers and a surrounding matrix. The reinforcing fibers may be made of polymers, metals, or ceramics.

Pure carbon exists as diamond, graphite,fuUerenes, and carbon Electrical conductors (e.g., metals) have no gap between their valence band and conduction band, so electrons are easily promoted from the valence to the conduction band. Semiconductors, which conduct electricity more than nonconductors (insulators) but less than metallic conductors, have a small gap. The gap in nonconductors is too large for electrons to be promoted from the valence to the conduction band.

Semiconductors generally have a total of eight valence electrons per formula unit. Thus, they typically form between two Group 4A elements, one Group 3A element and one Group 5A element, or one Group 2B element and one Group 6A element.

•

Semiconductors can be doped with an element with fewer valence electrons to produce a p-type semiconductor, or with an element with more valence electrons to produce an n-type semiconductor.

•

o o

Nanotechnology is the study and development of objects with sizes on

nanotubes.

• Ceramics are polymeric inorganic compounds that are hard and strong Ceramics are sintered to bond the particles to each other and to close any gaps that may exist in the material.

Artificial joint replacements can be made of metal, ceramics, and organic polymers.

the order of a nanometer (one billionth of a meter).

Section 25.2

•

Biodegradable sutures and artificial skin are made of synthetic polyesters that support the growth of tissue around the polymer and then hydrolyze into the body.

Section 25.6

DNA and RNA are condensation polymers consisting of sugars, organic bases, and phosphate ions.

and have high melting points.

Dental amalgam consists of several metals dissolved in an alloy with mercury. Dental composite materials consist of a methacrylate polymer in a silica matrix. The composite can be varied in color to match existing teeth, making the repair more aesthetically pleasing than metallic amalgam.

p- Type and n-type semiconductors can be used together in solar cells and diodes.

960

CHAPTER 25

Modern Materials

Section 25.7 •

Superconductors offer no resistance to the flow of electrons. The earliest known superconductors only exhibited the behavior at liquid helium temperature (4 K), but modern high-temperature superconductors conduct at temperatures as high as 140 K.

KEyWORDS

•

The highest temperature at which a superconductor exhibits superconductivity is called the superconducting transition temperature (TJ.

•

Superconducting substances exhibit the Meissner effect, the exclusion of magnetic fields in the substance's volume. This can be demonstrated by levitating a magnet above a superconductor.

•

-.--~~-

Addition polymerization, 936

Diode, 955

Nanotechnology, 952

Anisotropic, 947

Doping, 955

Nematic, 947

Superconducting transition temperature (Tc), 956

Atactic, 941

Elastomer, 937

Nonconductor, 954

Superconductor, 956

Carbon nanotube, 953

Fullerene, 953

n- Type semiconductor, 955

Suture, 951

Ceramic, 945

Polyamide, 942

Syndiotactic, 941

Cholesteric, 947

High-temperature superconductor, 956

Polyester, 942

Tacticity, 941

Composite material, 946

Isotactic, 941

Polymer, 936

Thermoplastic, 936

Condensation polymer, 941

Isotropic, 947

p -Type semiconductor, 955

Thermosetting, 936

Conduction band, 954

Liquid crystal, 947

Semiconductor, 954

Valence band, 954

Copolymer, 939

Meissner effect, 956

Sintering, 945

Cross-link, 937

Monomer, 936

Smectic, 947

QUESTIONS AND PROBLEMS Section 25.1: Polymers

25.6

Review Questions 25.1

25.2

Draw the structure of an alternating copolymer of these two amino acids, showing at least two repeating units:

o II

Bakelite, the first commercially produced polymer, contains monomer units of phenol and formaldehyde. If an item made of Bakelite were broken, it could not be melted down and re-formed. Is Bakelite a thermoplastic or thermosetting polymer? Can Bakelite be recycled? How are SBS rubber [Figure 25.6(a)] and nylon 6,6 similar? How are they different?

H 2 N-CH-C-OH

I

CH 2

I

SH

o II

H2 N-CH-C-OH

I

CH-OH

I

CH 3

Section 25.2: Ceramics and Composite Materials

Problems

Review Questions

25.3

What type of monomer unit is needed to make an electrically conducting organic polymer?

25.7

What types of atoms and bonds are present in the polymer "backbone" in a ceramic containing a transition metal?

25.4

Draw the structure of the addition polymer formed by I-butene:

25.8

Describe two natural types of composite materials and the substances that they contain.

H ~C"",,- /CH 3

H2C

C

H2

Problems 25.9

What steps would be used to form a ceramic material from scandium and ethanol (CH 3CH20H) using the sol-gel process?

25.10

Amorphous silica (Si0 2) can be formed in uniform spheres from a sol-gel process using methanol (CH 30H) solvent. List the chemical reactions involved in this process .

25.11

Bakelite, described in Problem 25.1, is manufactured using materials like cellulose paper or glass fibers that are compressed and heated among the monomer units. Is Bakelite a simple polymer? If not, what classification best fits Bakelite?

Show at least two repeating units. 25.5

Draw the structure of the alternating copolymer of these two compounds, showing at least two repeating units.

H-C-C-H

QUESTIONS AND PROBLEMS

Se-ction 25.3: liquid Crystals :::E .

-

Section 25.5: Nanotechnology

'e N Questions

-- - ..,. --'- -

961

Review Questions

Is a norn1al liquid isotropic or anisotropic? How is an anisotropic material different from an isotropic material?

25.22

How does an STM measure the peak and valley heights along the surface of a sample?

Describe how nematic, smectic, and cholesteric liquid crystals are similar and different.

25.23

Name four allotropic forms of carbon.

Problems :::- DIem s -~

- -.1. 1

•

25.24

How does an AFM differ from an STM? Why can graphite possibly be investigated with either instrument whereas diamond can only be investigated with an AFM?

25.25

What is the hybridization of the carbon atoms in a single-walled nanotube?

25.26

What structural or bonding aspect of graphite and carbon nanotubes allows their use as semiconductors, whereas diamond is an electrical insulator?

25.27

What type of intermolecular forces holds the sheets of carbon atoms together in graphite? What type of intermolecular forces hold carbon nanotubes together in strands and fibers?

\ ould each of these molecules be likely to form a liquid crystal? If not, why not? (a)

o o o

Section 25.6: Semiconductors Review Questions

25.1-

Would long-chain hydrocarbons or addition polymers of alkene molec ules form liquid crystals? Why or why not?

:5.16

Wou ld an ionic compound form a liquid crystal? Why or why not?

Secti on 25.4: Biomedical Materials

25.28

How do the band gaps differ among conductors, semiconductors, and insulators?

25.29

The only elemental semiconductors (Si, Ge, and graphite) are members of Group 4A. Why are Sn and Pb, which are also in Group 4A, not classified as semiconductors?

Problems

25.30

State whether each combination of elements would be expected to produce a semiconductor: (a) Ga and Sb, (b) As and N, (c) B and P, (d) Zn and Sb, (e) Cd and S.

25.31

What type of semiconductor would be formed (a) if Si were doped with Sb and (b) If Ge were doped with B?

25.32

Describe the movement of electrons in a solar battery that uses p-type and n-type semiconductors when light shines on the solar cell.

eview Questions

:5.1 -;

::":.1

Name three major concerns that must be addressed when propos ing any material to be used in medical applications inside the body. arne a necessary chemical property and a necessary physical property of dental materials.

OOrob l ems

Section 25.7: Superconductors

....5.19

Review Questions

Is the polymer commonly used in artificial ski n material (shown here) an addition polymer or a condensation polymer? Is it a block, alternating, graft, or random copo lymer?

o 0 II " -+OCH2 C-OCHC-+I CH 3

25.33

Why is YBCO called a "high-temperature" semiconductor when it exhibits superconductivity only below -l80°C, which is very cold?

25.34

How might the Meissner effect be used in a real-world application? What are the engineering challenges to this application with the current superconductors?

11

•

::":_0

.:5_1

What are some advantages and disadvantages of the use of ceramics in replacement joint bearing surfaces relative to metals? How do amalgam and composite tooth fillings differ in the ways they fai lover time?

Problems

25.35

What is the molecular formula of BSCCO-220 1?

25.36

Superconductivity at room temperature is sometimes a subject of science fiction or fantasy. For example, in the movie Terminator 2: Judgment Day, a cyborg has a CPU that contains a roomtemperature superconductor. How might a computer with such a CPU differ in design from current computers?

CHAPTER 25

962

25.37

Modern Materials

What are the electron configurations of the copper ions in YBCO? Are both ions expected oxidation states for copper?

25.43

Fluoride ion is commonly used in drinking water supplies and in toothpaste (in the form of sodium fluoride) to prevent tooth decay. The fluoride ions replace hydroxide ions in the enamel mineral hydroxyapatite [Ca5(P04)30H, the same mineral as in bone] , forming fluoroapatite [Ca5(P04)3F]. Why does replacing the hydroxide ion with fluoride ion prevent tooth decay? How does this relate to the properties of materials used in artificial fillings?

25.44

Would the metal alkoxide in the sol-gel process be acidic or basic?

25.45

Would the compound shown form a liquid crystal? Why or why not?

25.46

The semiconductor band gap energy determines the color of an LED. Knowing what you learned in Chapter 6 about the energies of photons of different colors of light, which semiconductor would have the larger band gap: the one in a green exit sign 's LEDs, or the one in a red traffic signal?

Additional Problems 25.38

25.39

Draw representations of block copolymers and graft copolymers using the symbols A and B for two types of monomer units. Are the categories exclusive? Using a third monomer type, see if yo u can draw a representation of a polymer that is both a block and a graft copolymer.

•

Draw the structures of the two monomer units in this polymer:

NH

o II

CHz

NH-C

0 CH z

8

25.40

II

C-+4 n

Is the polymer in Problem 25.39 a condensation polymer or an addition polymer? Is it a polyester, a polyamide, or something else?

25.41

What types of bonding (covalent, ionic, network, metallic) are present in a plastic polymer? In a ceramic material?

25.42

Draw representations of isotactic, syndiotactic, and atactic polyacrylonitrile (for the monomer unit, see Table 25 .1).

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS: PHYSICAL AND BIOLOGICAL SCIENCES These questions are not based on a passage. l.

Which of the following would form an n-type semiconductor: (i) doping Ge with As, (ii) doping Si with B, (iii) doping graphite with P? a) i, ii, and iii b) i and iii c) i and ii d) iii only

2.

3.

What is unusual about the copper in YBCO? a) It exists as a mixture of + I and +2 oxidation states. b) It exists as a mixture of + I and 0 oxidation states. c) It exists as a mixture of + 1 and + 3 oxidation states. d) It exists as a mixture of +2 and +3 oxidation states.

A semiconductor can be produced by the combination of two elements with a total of how many valence electrons? a) 6 b) 8 c) 12 d)16

4.

Metals are good conductors of electricity because they a) have no band gap. b) have a large band gap. c) have a small band gap. d) combine with nonmetals to form semiconductors.

ANSWERS TO IN-CHAPTER MATERIALS

963

-"-~~SWERS TO IN-CHAPTER MATERIALS J"\.J

swers to Practice Problems H Cl H Cl

I

I

I

I

I

I

I

I

H

:S.H - C-C-C-CH Cl

.25.1B

\

/

H

H Cl

/

CH 3

C=C

\

COCH 3

0 0 0 ~.1

II

II

II

- NHCH2(CH2hCH2CNHCH2(CH2hCH2CNHCH2(CH2)3CH2C-

The two structures are similar in that they both have ethyl groups on alternating carbons. They are different in that poly(1-butyne) is a conjugated system and poly(l-butene) is not. The first structure is more likely to be an electrically conducting polymer. 25.4 The top molecule would be more likely to exhibit liquid crystal behavior because it is long, thin, and has conjugated rings that limit free rotation. 25.5 (a) Elements with two valence electrons, e.g., Be, Sr, Cd, (b) elements with three valence electrons, e.g., B, Ga, In, (c) elements with six valence electrons, e.g., 0 , Se, Te.

Answers to Checkpoints 25.1.1 a. 25.1.2 d. 25.3.1 c. 25.3.2 a, c. 25.6.1 b. 25.6.2 c.

•

•

MATHEMATICAL OPERATIONS

Scientific Notation Chemists often deal with numbers that are either extremely large or extremely small. For example, in I g of the element hydrogen there are roughly 602,200,000,000,000,000,000,000 hydrogen atoms. Each hydrogen atom has a mass of only 0.00000000000000000000000166 g These numbers are cumbersome to handle, and it is easy to make mistakes when using them in arithmetic computations. Consider the following multiplication: 0.0000000056 X 0.00000000048 = 0.000000000000000002688 It would be easy for us to miss one zero or add one more zero after the decimal point. Consequently, when working with very large and very small numbers, we use a system called scientific notation. Regardless of their magnitude, all numbers can be expressed in the form

N X IOn where N is a number between 1 and 10 and n, the exponent, is a positive or negative integer (whole number). Any number expressed in this way is said to be written in scientific notation. Suppose that we are given a certain number and asked to express it in scientific notation. Basically, this assignment calls for us to find n. We count the number of places that the decimal point must be moved to give the number N (which is between 1 and 10). If the decimal point has to be moved to the left, then n is a positive integer; if it has to be moved to the right, n is a negative integer. The following examples illustrate the use of scientific notation: 1. Express 568.762 in scientific notation: 568.762 = 5.68762 X 10

2

Note that the decimal point is moved to the left by two places and n = 2.

,

2. Express 0.00000772 in scientific notation: 0.00000772

= 7.72

X 10-

6

Here the decimal point is moved to the right by six places and n = -6. Keep in mind the following two points. First, n = 0 is used for numbers that are not expressed in scientific notation. For example, 74.6 X 10° (n = 0) is equivalent to 74.6. Second, the usual practice is to omit the superscript when n = 1. Thus the scientific notation for 1 74.6 is 7.46 X 10 and not 7.46 X 10 Next, we consider how scientific notation is handled in arithmetic operations. A-1

A-2

APPENDIX 1

Mathematical Operations

Addition and Subtraction To add or subtract using scientific notation, we first write each quantity say NI and N 2 -with the same exponent n. Then we combine NI and N 2; the exponents remain the same. Consider the following examples:

+ (2.1 104) + (3.9

(7.4 X 10 3) (4.31 X

X 10

3

)

X 10 3

= 9.5

X 103) = (4.31 X 104)

+ (0.39

X 104)

4 = 4.70 X 10

• (2.22 X 10- 2) - (4.10 X 10- 3)

= (2.22

X 10- 2) - (0.41 X 10- 2)

1.81 X 10- 2

=

Multiplication and Division To multiply numbers expressed in scientific notation, we multiply NI and N2 in the usual way, but add the exponents together. To divide using scientific notation, we divide NI and N2 as usual and subtract the exponents. The following examples show how these operations are performed: (8 .0 X 104) X (5.0 X 102 ) = (8.0 X 5.0)( 104 + ?)

(4.0 X 10- 5) X (7 .0 X 103)

=

40 X 106

=

7 4.0 X 10

=

(4.0 X 7.0)(10 - 5+ 3)

= 28 =

7

6.9 X 10 3.0 X 10- 5

4

8.5 X 10 5.0 X 109

X 10- 2

2.8 X 10- 1

= 6.9

X 107-(-5)

3.0 12

= 2.3

X 10

= 8.5

X 104 - 9

5.0 =

1.7 X 10- 5

Basic Trigonometry B

a

AL---------~b~------~ C

In the triangle shown, A , B , and C are angles (C = 90°) and a, b, and c are side lengths. To calculate unknown angles or sides, we use the following relationships :

a? + b? = c2 . A

SIn

= -a

c

b cos A = -

c

a

tan A = -

b

LOGARITHMS

A-3

Logarithms Common Logarithms The concept of the logarithms is an extension of the concept of exponents, which is discussed on page A-I. The common, or base-lO, logarithm of any number is the power to which 10 must be raised to equal the number. The following examples illustrate this relationship:

Exponent

Logarithm log log log log log

1= 0 10 = 1 100 = 2 10- 1 = -1 10- 2 = -2

10° = 1 101 = 10 1O? = 100 10- 1 = 0.1 10- 2 = 0.01

In each case the logarithm of the number can be obtained by inspection. Because the logarithms of numbers are exponents, they have the same properties as exponents. Thus, we have

Logarithm

Exponent

log AB = log A + log B A log B = log A - log B

lOA X loB

•

= lOA+B

lOA = lO A- B lOB

Furthermore, log An = n log A. Now suppose we want to find the common logarithm of 6.7 X 10- 4 . On most electronic calculators, the number is entered first and then the log key is pressed. This operation gives us log 6.7 X 10- 4

= - 3.17

Note that there are as many digits after the decimal point as there are significant figures in the original number. The original number has two significant figures, and the "17" in -3.l7 tells us that the log has two significant figures. The "3" in - 3.17 serves only to locate the decimal point in the number 6.7 X 10- 4 . Other examples are

Number

Common Logarithm

62 0.872 1.0 X 10- 7

1.79

•

-0.0595 -7.00

Sometimes (as in the case of pH calculations) it is necessary to obtain the number whose logarithm is known. This procedure is known as taking the antilogarithm; it is simply the reverse of taking the logarithm of a number. Suppose in a certain calculation we have pH = 1.46 and are asked to calculate [H +]. From the definition of pH (pH = -log [H+]) we can write [H+]

=

10 - 1.46

Many calculators have a key labeled log- lor INV log to obtain antilogs. Other calculators have or yXkey (where x corresponds to -1.46 in our example and y is 10 for base-l0 logarithm). a Therefore, we find that [H+] = 0.035 M.

lax

•

Natural Logarithms Logarithms taken to the base e instead of 10 are known as natural logarithms (denoted by In or loge); e is equal to 2.7183. The relationship between common logarithms and natural logarithms is as follows:

=1 In 10 = 2.303

log 10

10

= 10

e 2 .303 = 10

Thus, In x

1

= 2.303 log x

A-4

APPENDIX 1

Mathematical Operations

To find the natural logarithm of 2.27, say, we first enter the number on the electronic calculator and then press the In key to get In 2.27 = 0.820

If no In key is provided, we can proceed as follows: 2.303 log 2.27 = 2.303 X 0.356 = 0.820 •

Sometimes we may be given the natural logarithm and asked to find the number it represents. For example, In x=59.7 On many calculators, we simply enter the number and press the e key:

e59 7 = 8 X 1025

The Quadratic Equation A quadratic equation takes the form ~

aX- + bx + c

0

=

If coefficients a, b, and c are known, then x is given by

- b : :': : -Vb2

x =

-

4ac

2a

Suppose we have the following quadratic equation: ~

2x-

+ 5x -

12 = 0

Solving for x, we write x=

-5 : :': : >/ (5)2 - 4(2)(- 12) 2(2) - 5 : :': : --125

•

+ 96

4

Therefore, x=

-5

+ 11

3 2

4

and

-5 - 11 x= - -- - = -4 4

Successive Approximation In the determination of hydrogen ion concentration in a weak acid solution, use of the quadratic equation can sometimes be avoided using a method known as successive approximation. Consider the example of a 0.0150 M solution of hydrofluoric acid, (HF). The Ka for HF is 7.10 X 10- 4 . To determine the hydrogen ion concentration in this solution , we construct an equilibrium table and enter the initial concentrations, the expected change in concentrations, and the equilibrium concentrations of all species: HF(aq) Initial concentration (M):

====' H +(aq) + F-(aq)

+=.

0.0150

o

o

-----~---~-----

Change in concentration (M):

+x +x -----~----+----

Equilibrium concentration (M) :

0.0150 - x

- x

x

x

SUCCESSIVE APPROXIMATION

Using the rule that x can be neglected if the initial acid concentration divided by the Ka is greater than 100, we find that in this case x cannot be neglected (0.0150/7.1 X 10- 4 = 21). Successive approximation involves first neglecting x with respect to initial acid concentration _ _

x2

_

_ =

0.0150 - x

X

2

= 7.10

X

10- 4

0.0150

and solving for x:

x2 x

=

(0.0150)(7.10 X 10- 4 ) = 1.07 X 10- 5

= ~1.07

X

10

5

= 0.00326 M

We then solve for x again, this time using the calculated value of x on the bottom of the fraction: 2

2

x 0.0150 - x

0.0150

~ 0.00326 = 7.10 X

4

10-

x 2 = (0.0150 - 0.00326)(7.10 X 10- 4) = 8.33 X 10- 6 x = ~8.33 X 10- 6 = 0.00289 M

Note that the calculated value of x decreased from 0.00326 to 0.00289. We now use the new calculated value of x on the bottom of the fraction and solve for x again: 2

0.0150

0.0150 - x

~ 0.00289 = 7.10 X

2

x = (0.0150 - 0.00289)(7.10 x

=

~8.60

X 10- 6 = 0.00293

X

4

1010-

4

)

= 8.60

X

10- 6

M

This time the value of x increased slightly. We use the new calculated value and solve for x again. 2

x2 0.0150 - x

0.0150

~ 0.00293

4

=

7.10 X 10-

2

x = (0.0150 - 0.00293)(7.10 X 10-

x

=

~8.57

4

)

=

8.57 X 10- 6

X 10- 6 = 0.00293 M

Thi s time we find that the answer is still 0.00293 M, so there is no need to repeat the process. In general, we apply the method of successive approximation until the value of x obtained does not differ from the value obtained in the previous step. The value of x determined using successive approximation is the same value we would get if we were to use the quadratic equation.

,

A-5

• •

THERMODYNAMIC DATA AT

1 ATM AND

25°C~:-

INORGANIC SUBSTANCES

Substance Ag(s)

I1H'f (kJ / mol)

aG'f (kJ/mol)

0

0

42.7

105.9

77 .1

73.9

AgCI(s)

- 127.0

-109 .7

96.1

AgB r(s)

-99.5

- 95 .9

107.1

Agl(s)

-62.4

-66.3

114.2

-1 23.1

-32.2

140.9

0

28.3

Ag +(aq)

AgN0 3(s) AI(s)

0

AI3+(aq)

- 524.7

- 481.2

-313 .38

AI 20 3(S)

- 1669.8

- 1576.4

50.99

As(s)

AsOl- (aq) AsH 3(g) H3As0 4(s) Au(s) AU 203(S)

0 -870 .3

0 - 635.97

-144.77

-900.4 0

0

47.7

80.8

163.2

125.5

- 35.2

AuCI 3(s)

-118.4 0

0

B20 3(S)

- 1263.6

H3B0 3(s)

- 1087.9

- 963.16

H3B0 3(aq)

- 1067.8

- 963.3

Ba(s)

35.15

171.5

AuCI (s)

B(s)

0

- 1184 .1

6.5 54.0 89.58 159.8

0

66.9

Ba 2 +(aq)

- 538.4

- 560.66

12. 55

BaO(s)

-558.2

-52 8.4

70 .3

BaC 12(5)

-860.1

- 810.66

125.5

BaS04(s)

-1464.4

- 1353.1

132.2

BaC0 3(s)

-12 18.8

- 1138.9

112 .1

Be(s) BeO(s)

0 -610.9

Br 2(1)

0

Br(g)

30.7

0

9.5

-581.58

14.1

0

152.3

3.14

245.13

*The thermodynamic quantities of ions are based on the reference states that 6.Hf[ H +(aq) ] and SO[H +(aq)] = 0.

A-6

S° (J/K . mol)

= 0, 6.Gf[H+(aq)] = 0,

INORGANIC SUBSTANCES

Substance

I1Hf (kJlmol)

I1Gf (kJlmol)

S° (J/K . mol)

Br-(aq)

- 120.9

-102.8

80 .?

HBr(g)

- 36 .2

-53.2

198.48

C(graphite)

0

0

5.69

C(diamond)

1.90

2.87

2.4

CO(9)

- 110.5

- 137.3

197.9

CO z(9)

- 393.5

-394.4

213.6

COz(aq)

- 412.9

-386 .2

12 1.3

CO~-(aq)

- 676 .3

- 528.1

-53.1

HC0 3(aq)

-691.1

- 587.1

HZC0 3(aq)

-699.7

- 623.2

187.4

C5 z(9)

11 5.3

65 .1

237.8

C5 z(l)

87.3

63.6

151.0

HCN(aq)

105.4

11 2. 1

128.9

CW(aq)

151.0

165.69

117.99

94.98

(NHzhCO(s)

- 333 .19

-197.15

104.6

(N HzhCO(aq)

- 319.2

-203.84

173.85

Ca(s)

0

0

41.6

Ca(9)

179 .3

145 .5

154.8

Ca2+(aq)

-542 .96

-553.0

- 55 .2

CaO(s)

- 635.6

-604.2

39.8

Ca(OHh(s)

- 986.6

- 896.8

83.4

- 1214.6

- 1161.9

68.87

CaFz(s) CaClz(s)

- 794.96

- 750.19

113.8

Ca50 4 (s)

-1432.69

- 1320.3

106.69

CaC0 3 (s)

-1206.9

- 1128.8

92.9

Cd(s)

Cd 2+ (aq)

0

0

51.46

- 72.38

- 77.7

-6109

CdO(s)

-254.6

- 225 .06

54.8

CdC lz(s)

-389.1

- 342.59

118.4

Cd50 4 (s)

-926.17

-820.2

137.2

Cl z(9)

0

0

223 .0

CI(9)

121 .7

105. 7

165 .2

CI - (aq)

- 167.2

- 131.2

56 .5

HCI(g)

- 92.3

- 95.27

0

0

- 67.36

-51.46

155.2

- 213.38

43.9

0

23.77

Co(s) Coz +(aq)

CoO(s) Cr(s)

-239.3 0

Cr2+(aq)

-138.9

CrZ03(S)

- 11 28.4

C r O~ - (aq)

CrzO? - (aq)

C5(S)

- 863 .1 6 -1460 .6 0

-1046 .8 -706.26 -1257.29 0

187.0 28.45

81.1 7 38.49 213.8 82.8

(Continued)

•

A-?

A-8

APPENDIX 2

Thermod ynam ic Data at 1 atm and 25 °(

Substance C5(g)

76 .50

IlGf (kJ/mo l) 49 .53

S° (J/K . mol) 175 .6

Cs +(aq)

-247.69

-282 .0

133 .05

CsCI(s)

-442.8

-414. 4

10 1.2

Cu(s)

•

Il Hf (kJ/mol)

0

0

33 .3

Cu +(aq)

51.88

50.2

- 26.4

Cu 2+(aq)

64 .39

64.98

-99.6

CuO(s)

- 155 .2

-127 .2

43.5

CU20(S)

- 166.69

-146 .36

100.8

CuC I(s)

-134.7

-118 .8

91 .6

CuC I2(s)

-205.85

CuS(s) CUS04(S)

7

7

-48.5

-49 .0

66.5

- 769 .86

-661.9

113.39

F2(g)

0

0

203 .34

F(g)

80.0

61.9

158 .7

naq)

-329.1

-276.48

-9 .6

HF(g)

-271.6

- 270.7

173 .5 27.2

0

0

Fe 2+(aq)

-87.86

-84. 9

-1 13. 39

Fe3+(aq)

- 47.7

- 10.5

-293 .3

FeO(s)

-272.0

-2 55. 2

60.8

Fe203(S)

-822.2

-741.0

90.0

Fe(OHh(s)

- 568.19

-483.55

79 .5

Fe(OHh(s)

-824.25

7

7

H(g)

218.2

203.2

114.6

H2(g)

0

0

13 1. 0

H+(aq)

0

0

0

OW(aq)

- 229.94

- 157.30

- 10.5

H2 O(g)

-241.8

- 228.6

188.7

H2 O(l)

-285.8

-237.2

69 .9

H2 0 2 (1)

-187.6

- 118 .1

7

Fe(s)

Hg(l)

0

Hg 2+(aq) HgO(s)

77 .4

- 164.38 -90.7

HgCI 2 (s)

- 230.1

Hg 2 CI 2 (s)

-264.9

HgS(s)

0

- 58.16

- 58.5

720

- 210.66

196.2

- 48 .8

77.8

HgS0 4 (s)

- 704.17

Hg 2 S0 4 (S )

-741 .99

- 623 .92

200 .75

12(s)

0

0

116. 7

l(g)

106.8

70. 21

180. 67

I- (aq)

-55.9

-51.67

109. 37

HI(g)

25.9

1.30

K(s)

0

0

206.3 63 .6

INORGANIC SUBSTANCES

Sub stance

A.Hf (kJ/mol)

A.Gf (kJ / mol) - 282.28

S° (J/ K . mol)

iCaq)

- 251.2

mH(s)

- 425.85

CI(s)

-435.87

- 408.3

82.68

KCI 0 3(s)

-391.20

- 289.9

142.97

KCI0 4 (s)

- 433.46

-304.18

151 .0

KBr(s)

- 392.17

- 379.2

96.4

KI(s)

-327.65

-322.29

104.35

KN 0 3(s)

-492.7

-393.1

132.9

102.5

Li(s)

0

0

28.0

Li(g)

159.3

126.6

138.8

Li +(aq)

- 278.46

- 293.8

14.2

LiC I(s)

- 408.3

- 384.0

59.30

Li 2O(s)

-595.8

LiOH(s)

- 487.2

")

-443.9

")

50.2

Mg(s)

0

0

Mg(g)

150

115

Mg2+(aq)

- 461.96

-456.0

-117 .99

MgO(s)

-601.8

- 569.6

26.78

Mg(O Hh(s)

- 92 4 .66

- 833.75

63.1

MgCI 2(s)

-64 1.8

- 592 .3

89.5

MgS0 4 (s)

- 1278.2

-1 173.6

91.6

MgC0 3 (s)

-1112.9

-1029.3

65.69

Mn(s)

0

0

32.5 148.55

31.76

Mn2+(aq)

-218.8

-223.4

-83.68

Mn02(S)

- 520.9

-466.1

53 .1

0

0

191.5

N3"(aq)

245.18

")

")

NH3(g)

- 46.3

- 16.6

193.0

NH!(aq)

-132.80

- 79.5

112.8

NH4 CI(s)

-3 15.39

-203.89

NH3(aq)

- 80.3

-26.5

11 1.3

N2(g)

94.56

N2H4 (1)

50.4

NO(g)

90.4

86.7

210.6

N0 2(g)

33.8 5

51 .8

240.46

N20 4 (g)

9.66

98 .29

304.3

N2°(g)

81. 56

103.6

219.99

HN02(aq)

- 118.8

-53.6

HN0 3(1)

- 173.2

- 79.9

155.6

N0 3"(aq)

-206.57

-110.5

146.4

Na(s)

0

0

Na(g)

107 .7

77.3

•

51.05 153.7

Na+(aq)

-239 .66

- 261.87

60.25

Na20(S)

- 41 5.9

- 376.56

72.8

(Continued)

A -9

A-l0

APPENDIX 2

The rmodynamic Data at 1 atm and 25°C

Substance NaCI(s)

- 410.9

Na l(s)

- 288 .0

Na2504(S)

-1384.49

NaN0 3(s)

-466.68

Na2C03(S) J

tl.Hf (kJlmol)

NaHC03(s) Ni(s)

Ni2+(aq)

tl.Gf (kJlmol)

S° (JIK . mol)

-384.0

72.38

- 1266.8

149.49

-365.89

116.3

-1047.67

135.98

-947.68

- 851.86

102.09

0

0

30.1

- 64.0

- 46.4

- 1130.9

-159.4

NiO(s)

- 244.35

-216.3

38.58

Ni(OHhCs)

- 538.06

-453.1

79.5

O(g)

249.4

230.1

160.95

0 2(g)

0

0

205.0

0 3(aq)

- 12.09

16.3

0 3(g)

142.2

163.4

237.6

0

0

44.0

-18.4

13.8

29.3

P(white) P(red) pot(aq)

-128407

P4O lO(S)

- 3012.48

PH 3(g)

9.25

- 1025.59

18. 2

110.88

- 217.57

210.0

HPOl - (aq)

- 1298.7

-1094.1

- 35.98

H2P0 4 (aq)

- 1302.48

-1135.1

89.1

Pb(s)

0

Pb 2+(aq)

1.6

0

64.89

- 24.3

21.3

PbO(s)

- 217.86

-188.49

69.45

Pb0 2(s)

- 276.65

-218.99

76.57

PbCI 2(s)

- 359.2

- 313.97

136.4

-94.3

-92.68

91.2

Pb5(s) Pb50 4(s) Pt(s)

PtCll-(aq)

- 918.4 0 - 516.3

-811.2 0 -384.5

147.28 41.84 175.7

Rb(s)

0

0

Rb(g)

85.8

55.8

Rb+(aq)

- 246.4

-282.2

124.27

RbBr(s)

- 389.2

-378.1

108.3

RbCI(s)

- 435 .35

-407.8

Rbl(s)

- 328

-326

5(rhombic)

0

0

31.88

5(monociinic)

0.30

0.10

32.5 5

50(g)

5.01

69.45 . 170.0

95.90 118.0

-1 9.9

221.8

50 2(g)

-296.4

-300.4

248. 5

50 3(g)

- 395.2

-370.4

256.2

50 ~- (aq)

- 624 .2 5

-497.06

43.5

50 l-(aq)

- 907 .5

-741.99

17.1 5

ORGANIC SUBSTANCES

Il H¥ (kJ/mol)

Substance

Il G¥ (kJlmol)

S° (J / K . mol)

- 20.15

-33.0

205.64

H50 3 (aq)

-627.98

-527.3

132.38

H50 4(aq)

-885.75

-752.87

126.86

H2 50 4(1)

-8 11.3

7

7

-1096.2

7

7

H2 S(g)

5F 6(g) 5i(s)

0

5i0 2(s)

0

-859.3

5r(s)

18.70

-805.0

0

41.84 54.39

0

5r 2+(aq)

-545.5

- 557.3

-39.33

5rCI 2 (s)

-828.4

-781.15

117.1 5

5r50 4 (s)

- 1444.74

- 1334.28

121.75

5rC0 3 (s)

-1218.38

-1137.6

97.07

Zn(s)

0

0

41.6

Zn 2 +(aq)

- 152.4

- 147.2

ZnO(s)

- 348.0

- 318.2

43.9

ZnCI 2 (s)

- 415.89

-369.26

108.37

Zn5(s)

- 202.9

-198.3

57.7

ZnS04(s)

- 978.6

- 871.6

124.7

- 106.48

ORGANIC SUBSTANCES Il H¥ (kJ / mol)

S° (JIK . mol)

Substance

Formula

Acetic acid(l)

CH 3 COOH

-484.2

- 389.45

159.8

Acetaldehyde(g)

CH 3 CHO

- 166.35

-139.08

264.2

Acetone(l)

CH 3 COCH 3

-246.8

-153.55

198.7

Acetylene(g)

C2 H2

209.2

200.8

Benzene(l)

C6 H6

124.5

172.8

Butane(g)

C4H10

-124.7

- 15.7

310.0

Ethanol(l)

C2 HsOH

- 276.98

-174.18

161.0

Ethane(g)

C2 H6

-84.7

-32.89

229.5

Et hylene(g)

C2 H4

52.3

68.1

219.5

Formic acid(l)

HCOOH

-409.2

- 346.0

129.0

Glucose(s)

C6 H12 0 6

- 1274.5

- 910.56

212.1

Methane(g)

CH 4

-50.8

186.2

Methanol(l)

CH 30H

- 238.7

-166 .3

126.8

Propane(g)

C3 HS

-103.9

- 23.5

269.9

5ucrose(s)

C12 H22 0 11

-2221.7

- 1544.3

360.2

226.6 49.04

-74.85

Il G¥ (kJlmol)

•

A-l l

A absolute temperature scale. A scale based on -273.15 °C (absolute zero) being the lowest point. (11.2) absolute zero. Theoretically the lowest obtainable temperature: - 273.15°C orO K. (11.2) accuracy. The closeness of a measurement to the true or accepted value. (1.5) acid. See Arrhenius acid, Br¢nsted acid, and Lewis acid. acid ionization constant (K.). The equilibrium constant that indicates to what extent a weak acid ionizes. (16.5) actinide series. Series of elements that has partially filled Sf and/or 6d subshells. (6.9) activated complex. A transient species that forms when molecules collide in an effective collision. (14.4) activation energy (E.). The minimum amount of energy to begin a chemical reaction. (14.4) activity series. A list of metals arranged from top to bottom in order of decreasing ease of oxidation. (4.4) actual yield. The amount of product actually obtained from a reaction. (3.7) addition polymer. A large molecule that form s when small molecules known as monomers join together. (10.6) adhesion. The attractions between unlike molecules. (12.2) alcohol. A compound consisting of an alkyl group and the functional group -OH. (10.2) aldehyde. A compound containing a hydrogen atom bonded to a carbonyl group. (10.2) aliphatic. Descri bes organic molecules that do not contain the benzene ring. (10.1) alkali metal. An element from Group lA, with the exception of H (i.e., Li, Na, K, Rb, Cs, and Fr). (2.4) alkaline earth metal. An element from Group 2A (Be, Mg, Ca, Sr, Ba, and Ra). (2.4) alkane. Hydrocarbons having the general formula C"H 2,, +2, where 11 = 1, 2, ... (2.6) alkyl group. A portion of a molecule that resembles an alkane. (10.2) allotrope. One of two or more distinct forms of an element. (2.6) alpha particle. A helium ion with a positive charge of + 2. (2.2)

alpha ray. See alpha particle. amalgam. A substance made by combining mercury with one or more other metals. (Chapter 19, Opening Essay; 23.2) amide. An organic molecule that contains an amide group. (10.2) amine. An organic molecule that contains an amino group. (10.2) amino acid. A compound that contains both an amino group and a carboxy group. (10.2) amorphous solid. A solid that lacks a regular three-dimensional arrangement of atoms. (12.5) amphoteric. Describes an oxide that displays both acidic and basic properties. (7.7) amplitude. The vertical distance from the midline of a wave to the top of the peak or the bottom of the trough. (6.1 ) angular momentum quantum number (f). Describes the shape of the atomic orbital. (6.6) anion. An ion with a negative charge. (2.7) anisotropic. Dependent upon the axis of measurement. (25.3) anode. The electrode at which oxidation occurs. (19.2) antibonding molecular orbital. A molecular orbital that is higher in energy than the atomic orbitals that combined to produce it. (9.6) aromatic. Describes organic compounds that are related to benzene or that contain one or more benzene rings. (10.1) Arrhenius acid. Substance that increases H + concentration when added to water. (2.6, 4.3) Arrhenius base. Substance that increases OHconcentration when added to water. (4.1, 4.3) Arrhenius equation. An equation that gives the dependence of the rate constant of a reaction . on temperature: k = Ae- E/ RT . (14.4) atactic. Describes polymers in which the substituents are oriented randomly along the polymer chain. (25.2) atom. The basic unit of an element that can enter into chemical combination. (2.2) atomic mass. The mass of the atom given in atomic mass units (amu). (2.5) atomic mass unit (amu). A mass exactly equal to one-twelfth the mass of one carbon-12 atom. (2.5)

atomic number (Z). The number of protons in the nucleus of each atom of an element. (2.3) atomic orbital. The wave function of an electron in an atom. (6.5) atomic radius. Metallic: One-half the distance between the nuclei in the two adjacent atoms of the same element in a metal. Covalent: One-half the distance between the nuclei of the two identical atoms in a diatomic molecule. (7.4) atomic weight. The average atomic mass. (2.5) Autbau principle. The process by which the periodic table can be built up by successively adding one proton to the nucleus and one electron to the appropriate atomic orbital. (6.8) autoionization of water. Ionization of water molecules to give H + and OH- ions. (16.2) Avogadro's law. The volume of a sample of gas (V) is directly proportional to the number of moles (n) in the sample at constant temperature and pressure: Vex n. (11.2) Avogadro's number (NA ). The number of atoms in exactly 12 g of carbon-12: 6.022 X 1023. (3.4) axial. Describes the two bonds that form an axis perpendicular to the trigonal plane. (9.1)

B band theory. A theory wherein atomic orbitals merge to form energy bands. (23.3) barometer. An instrument used to measure atmospheric pressure. (11.1) base. See Arrhenius base, Br¢nsted base, and Lewis base. base ionization constant (Kb ). The equmbrium constant that indicates to what extent a weak base ionizes. , (16.6) battery. A portable, self-contained source of electric energy consisting of galvanic cells or a series of galvanic cells. (19.6) beta particle. An electron. (2.2) beta ray. See beta particle. bimolecular. Describes a reaction in which two reactant molecules collide. (14.5) binary compound. A substance that consists of just two different elements. (2.6) blackbody radiation. The electromagnetic radiation emitted from a heated solid. (6.2) G-1

G-2

GLOSSARY

body-centered cubic cell. A unit cell with one atom at the center of the cube and one atom at each of the eight comers. (12.3) boiling point. The temperature at which vapor pressure equals atmospheric pressure. (12.6) bond angle. The angle between two adjacent A - B bonds. (9.1 ) bond enthalpy. The enthalpy change associated with breaking a particular bond in 1 mole of gaseous molecules. (8 .9) • bond order. A number based on the number of electrons in bonding and antibonding molecular orbitals that indicates, qualitatively, how stable a bond is. (9.6) bonding molecular orbital. A molecular orbital that is lower in energy than the atomic orbitals that combined to produce it. (9 .6) Born-Haber cycle. The cycle that relates the lattice energy of an ionic compound to quantities that can be measured. (8 .2) Boyle's law. The pressure of a fixed amount of gas at a constant temperature is inversely proportional to the volume of the gas: p ex IIV. (11.2) Bragg equation. An equation relating the wavelength of X-rays, the angle of diffraction. and the spacing between atoms in a lattice. (12.3) breeder reactor. A nuclear reactor that produces more fi ssionable material than it consumes. (20.5) Brjjnsted acid. A substance that donates a proton (H+). (4.3, 16.1 ) Brjjnsted base. A substance that accepts a proton (H +) . (4.3, 16.1) buffer. A solution that contains significant concentrations of both members of a conjugate pair (weak acid/conjugate base or weak base/conjugate acid). (17.2)

c calorimetry. The measurement of heat changes . (SA)

capillary action. The movement of liquid up a narrow tube as the result of adhesive forces. (12.2) carbocation. A species in which one of the carbons is surrounded by only six electrons. (10.5) carbon nanotube. A tube made of carbon atoms with dimensions on the order of nanometers. (25.5) carboxylic acid. An organic acid that contains a carboxy group. (10.2) catalyst. A substance that increases the rate of a chemical reaction without itself being consumed. (14.6) catenation. The formation of long carbon chains. (10.1 ) cathode. The electrode at which reduction occurs. (19.2) cation. An ion with a positive charge. (2.7)

cell potential (E cell ) ' The difference in electric potential between the cathode and the anode. (19.?) ceramics. Pol ymeric inorganic compounds that share the properties of hardness, strength, and high melting points . (25.2) chalcogens. Elements in Group 6A (0 , S, Se, Te, and Po). (? A) Charles and Gay-Lussac's law. See Charles's law. Charles's law. The volume of a fixed amount of gas ( \I) maintained at constant pressure is directl y proportional to absolute temperaulre (T): Vex T. (11.2 ) chelating agent. A polydentate ligand that forms complex ions with metal ions in solution. (22 .1 ) chemical change. A process in which one or more substances are changed into one or more new substances. (I A) chemical energy. Energy stored within the structural units (molecules or polyatomic ions) of chemical substances. (5.1 ) chemical equation. Chemical symbols used to represent a chemical reaction. (3 .3) chemical formula. Chemical symbols and numerical subscripts used to denote the composition of the substance. (2.6) chemical property. Any property of a substance that cannot be studied without converting the substance into some other substance. (1 A) chemistry. The study of matter and the changes it undergoes. (1.1 ) chiral. Describes molecules with nonsuperimposable mirror images. (lOA ) cholesteric. Describes molecules that are parallel to each other within each layer, but where each layer is rotated with respect to the layers above and below it. (25 .3) cis. Describes the isomer in which two substituents both lie on the same side of a double bond. ( lOA) Clausius-Clapeyron equation. A linear relationship that exists between the natural log of vapor press ure and the reciprocal of absolute temperature. (12.2) closed system. A system that can exchange energy (but not mass) with the surroundings. (5.2) cohesion. The attraction between like molecules. (12.2) colligative properties. Properties that depend on the number of solute particles in solution but do not depend on the nature of the solute particles. (13.5) collision theory. The reaction rate is directly prop0!1ional to the number of molecular collisions per second. (14A) colloid. A di spersion of particles of one substance throughout another substance. (13.7) combination reaction. A reaction in which two or more reactants combine to form a single product. (3.7)

combustion analysis. An experimental determination of an empirical formula by a reaction with oxygen to produce carbon dioxide and water. (3.5) combustion reaction. A reaction in which a substance bums in the presence of oxygen. (3 .7) common ion effect. The presence of a common ion suppresses the ionization of a weak acid or weak base. (17.1) composite material. A material made from two or more substances with different properties that remain separate in the bulk material. (25.2) compound. A substance composed of atoms of two or more elements chemicalJy united in fixed prop0!1ions. (1.2) concentration cell. A cell that has the same type of electrode and the same jon in solution (at different concentrations) in the anode and cathode compartments. (19.5) condensation. The phase transition from gas to liquid. (12.2) condensation polymer. A large molecule that forms when small molecules undergo condensation reactions. (10.6) condensation reaction. An elimination reaction in which two or more molecules become connected with the elimination of a small molecule, often water. (10.6) condensed structural formula. Shows the same information as a structural formula, but in a condensed form. (10.3) conduction band. The antibonding band. (25.6) conductor. A substance through which electrons move freely. (23.3) conjugate acid. The cation that remains when a Brpnsted base accepts a proton. (16.1) conjugate base. The anion that remains when a Brpnsted acid donates a proton. (16.1) conjugate pair. The combination of a Brpnsted acid and its conjugate base (or the combination of a Brpnsted base and its conjugate acid). (16.1) constitutional isomers. Compounds with the same chemical formula but different structures. conversion factor. A fraction in which the same quantity is expressed one way in the numerator and another way in the denominator. (1 .6) coordinate covalent bond. A covalent bond in which one of the atoms donates both electrons. (8.8) coordination compound. A compound that contains coordinate covalent bonds between a metal ion (often a transition metal ion) and two or more polar molecules or ions. (22.1) coordination number. The number of atoms sun'ounding an atom in a crystal lattice. (12.3) The number of donor atoms surrounding a metal in a complex. (22. 1) copolymer. A polymer made of two or more different monomers. (10.6) corrosion. The undesirable oxidation of metals. (19.8)

GLOSSARY

Coulomb's law. The force (F) between two charged objects COl and Q2) is directly proportional to the product of the two charges and inversely proportional to the distance (d) between the objects squared. (7.3) covalent bond. A shared pair of electrons. (8.3) covalent bonding. Two atoms sharing a pair of electrons. (8.3) covalent radius. Half the distance between adjacent, identical nuclei in a molecule. (7.4) critical mass. The minimum amount of fissionable material required to sustain a reaction. (20.5) critical pressure (Pc). The minimum pressure to at must be applied to liquefy a substance at its critical temperature. (12.6) critical temperature (Tc). The temperature at which the gas phase cannot be liquefied, no matter how great the applied pressure. (12.6) cross-link. A bond that forms between a functional group off the backbone of the polymer chain that interacts with another functional group of a second polymer strand creating a new covalent bond and causing the polymer to be stronger and more rigid. (25.1) crystal field splitting (11). The difference in energy between the lower and higher d-orbital energy levels. (22.3) crystalline solid. A solid that possesses rigid and long-range order; its atoms, molecules, or ions occupy specific positions. (12.3)

D Dalton's law of partial pressures. The total pressure exerted by a gas mixture is the sum of the partial pressures exerted by each component of the mixture. (11.5) dative bond. A covalent bond in which one of the atoms donates both electrons. (8.8) de Broglie wavelength. A wavelength calculated using the following equation: A. = hlmu. (6.4) decomposition reaction. A reaction in which one reactant forms two or more products. (3.7) degenerate. Having an equal energy. (6.8) delocalized. Spread out over the molecule or part of the molecule, rather than confined between two specific atoms . (9.7) density. The ratio of mass to volume. (1.3) deposition. The phase change from gas to solid. (12.6) dextrorotatory. The term used to describe the enantiomer that rotates the plane-polarized light to the right. (22.2) diamagnetic. A species without unpaired electrons that is weakly repelled by magnetic fields. (9.6) diatomic molecule. A molecule that contains two atoms. (2.6) diffusion. The mixing of gases. (11. 6) dilution. The process of preparing a less concentrated solution from a more concentrated one. (4.5)

dimensional analysis. The use of conversion factors in problem solving. (1.6) diode. An electronic device that restricts the flow of electrons in a circuit to one direction. (25.6) dipole moment (f1-). A quantitative measure of the polarity of a bond. (8.4) dipole-dipole interactions. Attractive forces that act between polar molecules. (12.1) diprotic acid. An acid with two ionizable protons. (4.3) dispersion forces. See London dispersion forces. displacement. An atom or ion in a compound is replaced by an atom of another element. (4.4) disproportionation. When an element undergoes both oxidation and reduction in the same reaction. (4.4) dissociation. The process by which an ionic compound, upon dissolution, breaks apart into its constituent ions. (4.1) donor atom. The atom that bears the unshared pair of electrons. (22.1) doping. The addition of very small quantities of an element with one more or one fewer valence electron than the natural semiconductor. (25.6) double bond. A multiple bond in which the atoms share two pairs of electrons. (8.3) dynamic equilibrium. Occurs when a forward process and reverse process are occurring at the same rate. (12.2)

E effective collision. A collision that results in a reaction. (14.4) effective nuclear charge (Zeff). The actual magnitude of positive charge that is "experienced" by an electron in the atom. (7.3) effusion. The escape of a gas from a container into a vacuum. (11.6) elastomer. A material that can stretch or bend and then return to its original shape as long as the limits of its elasticity are not exceeded. (25.1 ) electricity. The movement of electrons. (4.1) electrode. A piece of conducting metal in an electrochemical cell at which either oxidation or reduction takes place. (19 .2) electrolysis. The use of electric energy to drive a nonspontaneous redox reaction. (19.7) electrolyte. A substance that dissolves in water to yield a solution that conducts electricity. (4.1) electrolytic cell. An electrochemical cell used for electrolysis. (19.7) electromagnetic spectrum. Consists of radio waves, microwave radiation, infrared radiation, visible light, ultraviolet radiation, X rays, and gamma rays. (6.1) electromagnetic wave. A wave that has an electric field component and a magnetic field component. (6.1)

G-3

electron. A negatively charged subatomic particle found outside the nucleus of all atoms. (2.2) electron affinity (EA). The energy released (the negative of the enthalpy change, 11H) when an atom in the gas phase accepts an electron. (7.4) electron configuration. The distribution of electrons in the atomic orbitals of an atom. (6.8) electron density. The probability that an electron will be found in a particular region of an atom. (6.5) electron domain. A lone pair or a bond, regardless of whether the bond is single, double, or triple. (9.1) electron-domain geometry. The arrangement of electron domains (bonds and lone pairs) around a central atom. (9.1) electron spin quantum number (m s ). The fourth quantum number that differentiates two electrons in the same orbital. (6.6) electro negativity. The ability of an atom in a compound to draw electrons to itself. (8.4) electrophile. A region of positive or partial positive charge. (10.5) electrophiJic addition. An addition reaction that begins when an electrophile approaches a region of electron density. (10.5) electrostatic energy. The potential energy that results from the interaction of charged particles. (5.1) element. A substance that cannot be separated into simpler substances by chemical means. (1.2) elementary reaction. A reaction that occurs in a single collision of the reactant molecules. (14.5) elimination reaction. A reaction in which a double bond forms and a molecule such as water is removed. (10.5) emission spectrum. The light emitted, either as a continuum or in discrete lines, by a substance in an excited electronic state. (6.3) empirical formula. The chemical formula that conveys with the smallest possible whole numbers the ratio of combination of elements in a compound. (2.6) enantiomers. Molecules that are mirror images of each other but cannot be superimposed. (10.4,22.2) end point. The point at which the color of the indicator changes. (17.3) endothermic process. A process that absorbs heat. (5. ) energy. The capacity to do work or transfer heat. (5.1) enthalpy (n). A thermodynamic quantity defined by the equation H = V + pv. The change in enthalpy, I1H, is equal to the heat exchanged between the system and surroundings at constant pressure qp. (5.3) enthalpy of reaction (!1Hrxn). The difference between the enthalpy of the products and the enthalpy of the reactants. (5.3)

G-4

GLOSSARY

entropy (S). A thermodynamic state function often described as a measure of disorder. (18.2) enzymes. Biological catalysts. (14.6) equatorial. The three bonds that are arranged in a trigonal plane in a trigonal bipyramidal geometry. (9.1) equilibrium. A state in which forward and reverse processes are occurring at the same • rate. (15.1) equilibrium constant (K). A number equal to the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants, with each concentration raised to the power of its stoichiometric coefficient. (15.2 )

equilibrium expression. The quotient of product concentrations and reactant concentrations, each raised to the power of its stoichiometric coefficient. (15.2) equilibrium process. A process that can be made to occur by the addition or removal of energy but does not happen on its own. (18.3) equilibrium vapor pressure. The pressure exerted by the molecules that have escaped to the gas phase, once the pressure has stopped increasing. (12.2) Equivalence point. The point in a titration where the reaction (e.g., neutralization) is complete. (4.6) ester. An organic molecule containing a -COOR group. (10.2) evaporation. The phase change from liquid to solid at a temperature below the boiling point. (12.2) excess reactant. The reactant present in a greater amount than necessary to react with all the limiting reactant. (3 .7) excited state. A state that is higher in energy than the ground state. (6.3) exothermic process. A process that gives off heat. (5.1) extensive property. A property that depends on the amount of matter involved. (1.4)

F face-centered cubic cell. A cubic unit cell with one atom on each of the six faces and one atom at each of the eight corners. (12.3) family. The elements in a vertical column of the periodic table. (2.4) first law of thermodynamics. Energy can be converted from one form to another, but cannot be created or destroyed. (5 .2) first-order reaction. A reaction whose rate depends on the reactant concentration raised to the first power. (14.3) formation constant (Kr). The equilibrium constant that indicates to what extent complex-ion formation reactions occur. (17 .5) formula mass. The mass of a formula unit. (3. 1) formula weight. See formula mass.

fractional precipitation. The separati on of a mixture based upon the components' solubilities. (17 .6) free energy. The energy available to do work. (18.4) free radical. A molecule with an odd number of electrons. (8.8) freezing point. The temperature at which a liquid undergoes the phase transition to solid. (12 .6) frequency (v) . The number of waves that pass through a par1icular point in 1 s. (6.1) fuel cell. A voltaic cell in which reactants must be continually supplied . (19.6) fullerenes. Molecules with elongated and elliptical cages of 70 and 80 car'bon atoms. (25.5 )

functional group. The part of a molecule characterized by a special arrangement of atoms that is largely responsible for the chemical behavior of the parent molecule. (2.6) fusion. The phase transiti on from solid to liquid (melting) (12.6) or the combination of light nuclei to form heavier nuclei. (20.5)

G galvanic cell. An electrochemical cell in which a spontaneous chemical reaction generates a flow of electrons. (19 .2) galvanization. The cathodic protection of iron or steel using zinc. (19. 8) gamma rays. High-energy radiation. (2.2) gas constant (R). The proportionality constant that appears in the ideal gas equation. (11.3) gas laws. Equations that relate the volume of a gas sample to its other parameters: temperature (T ), press ure (P ), and number of moles (n). (11.2) geometric isomers. Molecules that contain the same atoms and bonds arranged differently in space. (10.4, 22.?) Gibbs free energy. The energy available to do work. (18.4) glass. Commonly refers to an optically transparent fu sion product of inorganic materials that has cooled to a rigid state without crystallizing. (12.5 ) Graham's law. The rates of diffusion and effusion are inversely proportional to the square root of the molar mass of the gas. ( 11.6) gravimetric analysis. An analytical technique based on the measurement of mass . (4.6) greenhouse effect. Describes the trapping of heat near Earth's surface by gases in the atmosphere, particularly carbon dioxide. (2 1.5 )

ground state. The lowest energy state of an atom. (6.3) group. The elements in a vertical column of the periodic table. (2.4)

H half-cell. One compartment of an electrochemical cell containing an electrode immersed in a solution. (19.2) half-life (t112)' The time required for the reactant concentration to drop to half its original value. (14.3) half-reaction method. Balancing an oxidationreduction equation by separating the oxidation and the reduction, balancing them separately, and adding them back together. (4.4) half-reaction. The separated oxidation and reduction reactions that make up the overall redox reaction. (19. I ) halogens. The elements in Group 7 A (F, Cl, Br, I, and At). (2.4) heat. The transfer of thermal energy between two bodies that are at different temperatures. (5.1 )

heat capacity (C). The amount of heat required to raise the temperature of an object by 1°C. (5.4) Heisenberg uncertainty principle. It is impossible to know simultaneously both the momentum (p) (defined as mass times velocity, m X u) and the position (x) of a particle with certainty. (6.5) Henderson-Hasselbalch equation. pH = pKa + log ([conjugate base]/ [weak acid]). (17.2) Henry's law. The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution. (13.4) Henry's law constant (k). The proportionality constant that is specific to the gas-solvent combination and varies with temperature. (13.4) Hess's law. The change in enthalpy that occurs when reactants are converted to products in a reaction is the same whether the reaction takes place in one step or in a series of steps. (5.5) heteroatom. Any atom in an organic molecule other than carbon or hydrogen. (10.3) heterogeneous catalysis. A catalysis in which the reactants and the catalyst are in different phases. (14.6) heterogeneous mixture. A mixture in which the composition varies. (1.2) heteronuclear. Containing two or more different elements. (2.6) high-temperature superconductor. A material that shows no resistance to the flow of electrons at an unusually high temperature. (25.7) homogeneous catalysis. A catalysis in which the reactants and the catalyst are in the same phase. (14.6) homogeneous mixture. A mixture in which the composition is uniform. Also called a solution. (1.2) homonuclear. Containing atoms of only one element. (2.6) Hund's rule. The most stable arrangement of electrons in orbitals of equal energy is the

GLOSSARY

one in which the number of electrons with the same spin is maximized. (6.8) hybridization. The mixing of atomic orbitals. (9.4) hydrate. A compound with a specific number of water molecules within its solid structure. (2.7) hydration. The process by which water molecules surround solute particles in an aqueous solution. (4.2) hydrocarbon. A compound containing only carbo n and hydrogen. (2.6) hy drogen bonding. A special type of dipoledipole interaction that occurs only in molecules that contain H bonded to a small, highly electronegative atom, such as N , 0 , or F(12. 1) hydronium ion. A hydrated proton (H 30 +). (4.3) h ydrophilic. Water-loving. (13.7) hydrophobic. Water-fearing. (13.7) h ypertonic. Describes a solution with a higher concentration of dissolved substances than plasma. (13.5 ) h ypothesis. A tentative explanation for a set of observations. (1.1) hypotonic. Describes a solution that has a lower concentration of dissolved substances than plasma. (13.5)

I ideal gas. A hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas eq uation . (11.3) ideal gas equation. An equation that describes the relationship among the four variables P, V, n, and T. ( 11.3 ) ideal solution. A sol ution that obeys Raoult 's law. (13.5) indicator. A sub stance that has a distinctly different color in acidic and basic media. (4.6) initial rate. The instantaneous rate at the beginning of a reaction. (14.2) inorganic compounds. Compounds that do not contain carbon or that are derived from nonliving sources . (2 .6) instantaneous dipole. A fleeting nonuniform distribution of electron density in a molecule without a permanent dipole. (12.1) instantaneous rate. The reaction rate at a specific time. (14.1) insulator. A substance that does not co nduct electricity. (23 .3) integrated rate law. In ([AJ/[AJ o) = - kt. (14.3) intensive property. A property that does not depend on the amount of matter involved. (1 .4) intermediate. A chemical species that is produced in one step of a reaction mechanism and consumed in a subsequent step. (14.5)

intermolecular forces. The attractive forces that hold particles together in the condensed phases. (12.1) ion pair. Ions in sol ution that are held together by electrostatic forces. (13.5) ion-dipole interactions. Coulombic attractions between ions and polar molecules. (12. 1) ionic bonding. An electrostatic attraction that holds oppositely charged ions together in an ionic compound. (8.2) ionic radius. The radius of a cation or an anion. (7 .6) ionizable hydrogen atom. A hydrogen atom that can be lost as a hydrogen ion, (H +). (2.6) ionization. The process by which a molecular' compound forms ions when it di ssolves. (4. 1) ionization energy (IE). The minimum energy required to remove an electron from an atom in the gas phase. (7.4) ionosphere. See thermosphere. isoelectronic. Describes two or more species with identical electron configurations. (7. 5) isoelectronic series. A series of two or more species that have identical electron configurations but different nuclear charges. (7.6) isolated system. A system that can exchange neither energy nor mass with the surroundings. (5.2) isomerization reaction. A reaction in which one isomer is converted to another. (10.5) isotactic. Polymers in which all the substituents (i.e., the R groups) are in the same relative orientation (i.e. , on the same side of the polymer chain). (25.1) isotonic. Equal in concentration and osmotic pressure. (13.5) isotope. Atoms that have the same atomic number (Z) but different mass numbers (A) . (2.3) isotropic. Independent of the axis of measurement. (25 .3)

J joule. SI unit of energy, I kg . m 2/s2 . (5. 1)

K Kekule structure. A structure similar to a Lewis structure in which the lone pairs may not be shown. (10.3) kelvin. The SI base unit of temperature. (1.3) Kelvin temperature scale. A temperature scale offset from the Celsius scale by 273.15. One kelvin (1 K) is equal in magnitude to one degree Celsius (1 °C) . (1 1.2) ketone. An organic co mpound consisti ng of two R groups bonded to a carbonyl. (1 0.2) kinetic energy. The energy that results from motion. (5.1) kinetic molecular theory. A theory that explains how the molecular' nature of gases gives rise to their macroscopic properties . (11.6)

G-S

L Lanthanide (rare earth) series. A series of 14 elements that have incompletely filled 4f subshells or that readily give ri se to cations that have incompletely filled 4f subshells. (6.9) lattice. A three-dimensional array of cations and anion s. (2.6) lattice energy. The amount of energy required to convert a mole of ionic solid to its constituent ions in the gas phase. (8.2) lattice points. Positions occupied by atoms, ions, or molecules in a unit cell. (12.3) lattice structure. The arrangement of the particles in a crystalline solid. (12.3) law. A concise verbal or mathematical statement of a reliable relationship between phenomena. ( Ll ) law of conservation of energy. First law of thermodynamics stating that energy can be neither created nor destroyed. (5.1) law of conservation of mass. An alternate statement of the first law of thermodynamics stating that matter can be neither created nor destroyed. (2 .1 ) law of definite proportions. Different samples of a given compound always contain the same elements in the same mass ratio. (2.1) law of mass action. For a reversible reaction at eq uilibrium and a constant temperature, the reaction qu otient, Q, has a constant value, K (the eq uilibrium constant). (15.2) law of multiple proportions. Different compounds made up of the same elements differ in the number of atoms of each kind that combine. (2.1 ) LeOChatelier's principle. When a stress is applied to a system at equilibrium, the system will respond by shifting in the direction that minimizes the effect of the stress . (15.5) levorotatory. The term used to describe the enantiomer that rotates the plane-polarized light to the left. (22.2) Lewis acid. A species that can accept a pair of electrons . (16 .1 2) Lewis base. A species that can donate a pair of electrons. (16.12) Lewis dot symbol. An elemental sy mbol surrounded by dots, where each dot represents a valence electron . (8.1) Lewis structure. A representation of covalent bonding in which shared electron pairs are shown either as dashes or as pairs of dots between two atoms, and lone pairs are shown as pairs of dots on individual atoms . (8 .4) Lewis theory of bonding. A chemical bond involves atoms sharing electrons . (8.3) ligand. A molecule or anion that can form coordi nate bonds to a metal to form a coordinati on complex. (22.1) limiting reactant. The reactant that is completely consumed and determines the , amount of product formed . (3.7)

G-6

GLOSSARY

line spectra. The emission or absorption of light only at discrete wavelengths. (6.3) liquid crystal. A substance that exhibits properties of both a liquid, such as the ability to flow and to take on the shape of a container, and those of a crystal, such as a regular arrangement of particles in a lattice. (25 .3) London dispersion forces. Attractive forces that act between all molecules, including nonpolar molecules, resulting frqm the formation of instantaneous dipoles and induced dipoles. (12.1) lone pair. A pair of valence electrons that are not involved in covalent bond formation. (S.3)

M magnetic quantum number (m e). Describes the orientation of an orbital in space. (6.6) main group elements. Elements in the sand p blocks of the periodic table. (7 .2) manometer. A device used to measure the pressure of gases relative to atmospheric pressure. (1Ll) mass. A measure of the amount of matter in an object or sample. (1.3) mass defect. The difference between the actual mass of a nucleus and the mass calculated by summing the masses of the individual nucleons. (20.2) mass number (A). The number of neutrons and protons present in the nucleus of an atom of an element. (2.3 ) matter. Anything that occupies space and has mass. ( Ll) Meissner effect. The exclusion of magnetic fields. (25.7) melting point. The temperature at which solid and liquid phases are in equilibrium. (12.6) mesosphere. The region located above the stratosphere in which the concentration of ozone and other gases is low and temperature decreases again with increasing altitude. (21.1) metallic radius. Half the distance between the nuclei of two adjacent, identical metal atoms. (7.4) metalloid. Elements with properties intermediate between metals and nonmetals. (7.4) metallurgy. The preparation, separation, and purification of metals. (23.2) microstate. A specific microscopic configuration of a system. (1S.2) mineral. A naturally occurring substance with a characteristic chemical composition and specific physical properties. (23 .1 ) mixture. A combination of two or more substances in which the substances retain their distinct identities. (1.2) moderator. A material that limits the speed of liberated neutrons but does not itself undergo fission when bombarded with neutrons. (20.5) molality (m). The number of moles of solute dissolved in 1 kg (1000 g) of solvent. (13.3)

molar concentration. See molarity (M) . molar heat of fusion (MIrlli). The energy, usually expressed in kJ/mol, required to melt 1 mole of a solid. (12.6) molar heat of sublimation (MIsub ) ' The energy, usually expressed in kilojoules, required to sublime 1 mole of a solid. (12.6) molar heat of vaporization (MI"ap)' The amount of heat required to vaporize a mole of substance at its boiling point. (12.6) molar mass. The mass in grams of 1 mole of the substance. (3.4) molar solubility. The number of moles of solute in one liter of saturated solution (mol/I.,). (17.4) molarity (M). The number of moles of solute per liter of solution. (4.5) mole (mol). The amount of a substance that contains as many elementary entities (atoms, molecules, formula units, etc.) as there are atoms in exactly 0.012 kg (12 g) of carbon12. (3.4) mole fraction CxJ The number of moles of a component divided by the total number of moles in a mixture. (1 l.5 ) molecular equation. A chemical equation written with all compounds represented by their chemical formulas. (4.2) molecular formula. A chemical formula that gives the number of atoms of each element in a molecule. (2.6) molecular geometry. The arrangement of bonded atoms. (9 .1 ) molecular mass. The sum of the atomic masses (in amu) of the atoms that make up a molecule. (3. 1) molecular orbital theory. A theory that describes the orbitals in a molecule as bonding and anti bonding combinations of atomic orbitals. (9 .6) molecular orbital. An orbital that results from the interaction of the atomic orbitals of the bonding atoms. (9.6) molecular weight. Average molecular mass. (3 .1 ) molecularity. The number of molecules involved in a specific step in a reaction mechanism. (14.5) molecule. A combination of two or more atoms in a specific arrangement held together by chemical bonds. (2.6) monatomic ion. An ion that contains only one atom. (2.7) monomer. A small molecule that can be linked in large numbers to form a large molecule (polymer). (10.6) monoprotic acid. An acid with one ionizable proton. (4.3) multiple bond. A chemical bond in which two atoms share two or more pairs of electrons. (S .3)

N nanotechnology. The development and study of extremely small-scale materials and objects. (25 .5)

nematic. Describes an arrangement in which molecules are all aligned parallel to one another but with no organization into layers or rows. (25.3) Nernst equation. An equation relating the emf of a galvanic cell with the standard emf and the concentrations of reactants and products. (19.5) neutralization reaction. A reaction between an acid and a base. (4.3) neutron. An electrically neutral subatomic particle with a mass slightly greater than that of a proton . (2.2) Newton (N). The SI unit of force. (11.1) nitrogen fixation. The conversion of molecular nitrogen into nitrogen compounds. (21.1) noble gas core. A representation in an electron configuration that shows in brackets the most recently completed noble gas . (6.9) noble gases. Elements in Group SA (He, Ne, Ar, Kr, Xe, and Rn). (2.4) node. A collection of points at which electron density in an atom is zero. (6.4) nonconductor. A substance that does not conduct electricity. (25.6) nonelectrolyte. A substance that dissolves in water to yield a solution that does not conduct electricity. (4.1 ) nonpolar. Having a uniform distribution of electron density. (S.4) nonspontaneous process. A process that does not occur under a specified set of conditions. (1S.1) nonvolatile. Having no measurable vapor pressure. (13.5) n-type semiconductor. Semiconductors in which an electron-rich impurity is added to enhance conduction. (23.3, 25.6) nuclear binding energy. The energy required to separate the nucleons in a nucleus. (20.2) nuclear chain reaction. A self-sustaining reaction sequence of fission reactions. . (20.5) nuclear fission. The splitting of a large nucleus into smaller nuclei and one or more neutrons. (20.5) nuclear fusion. The combination of two light nuclei to form one heavier nucleus. (20.6) nuclear transmutation. The conversion of one nucleus to another. (20.1) nucleophile. A region of negative or partial negative charge. (10.5) nucleophilic addition. An addition reaction that begins when a nucleophile donates a pair of electrons to an electron-deficient atom. (10.5) nucleus. The central core of the atom that contains the protons and neutrons. (2.2)

o octet rule. Atoms will lose, gain, or share electrons in order to achieve a noble gas electron configuration. (S.3) open system. A system that can exchange mass and energy with its surroundings. (5.2)

GLOSSARY

optical isomers. Nonsuperimposable mirror images. (10.4, 22.2) ore. A mineral deposit concentrated enough to allow economical recovery of a desired metal. (23.1) organic compounds. Compounds containing carbon and hydrogen, sometimes in combination with other elements such as oxygen, nitrogen, sulfur, and the halogens. (2.6) osmosis. The selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated one. (13.5) osmotic pressure (rr). The pressure required to stop osmosis. (13 .5) overvoltage. The difference between the electrode potential and the actual voltage required to cause electrolysis. (19.7) oxidation. Loss of electrons. (4.4) oxidation number. See oxidation state. oxidation state. The charge an atom would have if electrons were transferred completely. (4.4) oxidation-reduction reaction. See redox reaction. oxidizing agent. A species that accepts electrons. (4.4) oxoacid. An acid consisting of one or more ionizable protons and an oxoanion. (2.7) oxoanion. A polyatomic anion that contains one or more oxygen atoms bonded to a central atom. (2.7)

p paramagnetic. A species with unpaired electrons that are attracted by magnetic fields. (9.6) partial pressure (Pi)' The pressure exerted by a component in a gas mixture. (11.5) pascal (Pa). The SI unit of pressure. (11.1) Pauli exclusion principle. No two electrons in an atom can have the same four quantum numbers. (6.8) peptide bond. The bond that forms between amino acids. (10.6) percent by mass. The ratio of the mass of an individual component to the total mass, multiplied by 100 percent. (13.3) percent composition by mass. The percent of the total mass contributed by each element in a compound. (3.2) percent dissociation. The percentage of dissol ved molecules (or formula units, in the case of an ionic compound) that separate into ions in solution. (13.6) percent yield. The ratio of actual yield to theoretical yield, multiplied by 100 percent. (3.7) period. A horizontal row of the periodic table. (2.4) periodic table. A chart in which elements having similar chemical and physical properties are grouped together. (2.4)

pH. A scale used to measure acidity. pH = -log [H+]. (16.3) phase change. When a substance goes from one phase to another phase. (12.6) phase diagram. Summarizes the conditions (temperature and pressure) at which a substance exists as a solid, a liquid, or a gas. (12.7) photoelectric effect. A phenomenon in which electrons are ejected from the surface of a metal exposed to light of at least a certain minimum frequency. (6.2) photon. A particle of light. (6.2) physical change. A process in which the state of matter changes but the identity of the matter does not change. (1.4) physical property. A property that can be observed and measured without changing the identity of a substance. (1.4) pi ('IT) bond. Bonds that form from the interaction of parallel p orbitals. (9.5) pOH. A scale used to measure basicity. pOH = -log [OW]. (16.3) polar. Having a nonuniform electron density. (8.4) polar covalent bond. Bonds in which electrons are unequally shared. (8.4) polarized. A molecule in which a dipole moment has been induced. (12.1) polyamide. Polymers in which the monomers are connected by amide linkages. (25.1) polyatomic. Molecules containing more than two atoms. (2.6) polyester. Polymers in which the monomers are connected by ester linkages. (25 .1 ) polymer. Molecular compounds, either natural or synthetic, that are made up of many repeating units called monomers. (10.6) polypeptide. Short chains of amino acids. (10.6) polyprotic acid. An acid with more than two ionizable protons. (2.7, 4.3) positron. A subatomic particle with the same mass as an electron, but with a positive charge. (20.1) potential energy. The energy possessed by an object by virtue of its position. (5 .1 ) precipitate. An insoluble solid product that separates from a solution. (4.2) precipitation reaction. A chemical reaction in which a precipitate forms . (4.2) precision. The closeness of agreement of two or more measurements of the same quantity. (1.5)

pressure. The force applied per unit area. (11.1) principal quantum number (n). Designates the size of the orbital. (6.6) product. A substance that forms in a chemical reaction. (3 .3) protein. A polymer of amino acids. (10.6) proton. A positively charged particle in the nucleus of an atom. (2.2) p-type semiconductor. A semiconductor in which an electron-poor impurity is added to enhance conduction. (23.3, 25.6) pyrometallurgy. Metallurgic processes carried out at high temperatures. (23.2)

G-,

qualitative analysis. The determination of the types of ions present in a solution. (17.6) qualitative property. A property of a system that can be determined by general observation. (1.4) quantitative property. A property of a system that can be measured and expressed with a number. (1.4) quantum. The smallest quantity of energy thar can be emitted (or absorbed) in the form of electromagnetic radiation. (6.2) quantum numbers. Numbers required to describe the arrangement of electrons in an atom. (6.6)

R racemic mixture. An equimolar mixture of enantiomers that does not rotate the plane of plane-polarized light. (10.4,22.2) racemization. The conversion of a single enantiomer to a racemic mixture of both enantiomers. (10.5) radiation. The emission and transmission of energy through space in the form of wave . (2.2) radical. A chemical species with an odd number of electrons. (8 .8, 20.8) radioactive decay series. A sequence of nuclear reactions that ultimately result in the formation of a stable isotope. (20.3) radioactivity. The spontaneous emission of particles or radiation from unstable nuclei. (2.2,20.1) Raoult's law. The partial pressure of a soh'en' over a solution, Pi, is given by the vapor pressure of the pure solvent, p o, times the mole fraction of the solvent in the solution. Xi ' (13 .5) rate constant (k). The proportionality constan' in a rate law. (14.1) rate law. An equation relating the rate of reaction to the concentrations of reactants. (14.2) rate of reaction. The change in concentratio 0; reactants or products per unit time. (14.1 rate-determining step. The slowest step in 2 reaction mechanism. (14.5) reactant. A substance that is consumed in a chemical reaction. (3 .3) reaction order. The sum of the powers to whi::~ all reactant concentrations appearing in the • rate law are raised. (14.2) reaction quotient (Q). A fraction with prodn . concentrations in the numerator and concentrations in the denominator, each raised to its stoichiometric coefficient. (1.:--= redox reaction. A chemical reaction in wtllcb. electrons are transferred from one rea laL; il another. (4.4) reducing agent. A species that can donate electrons. (4.4)

G-8

GLOSSARY

reduction. A gain of electrons. (4.4) resonance structures. Two or more equally valid Lewis structures for a single molecule that differ only in the positions of electrons. (8.7) reversible process. A process in which the products can react to form reactants. (15.1 ) root-mean-square speed (JLrms)' The molecular speed that is inversely proportional to the molecular mass. (1 1.6)

5

•

salt. An ionic compound made up of the cation from a base and the anion from an acid. (4.3) salt bridge. An inverted U tube containing an inert electrolyte solution, such as KCI or NH4 N0 3 , that maintains electrical neutrality in an electrochemical cell. (19.2) salt hydrolysis. The reaction of a salt's constituent ions with water to produce either hydroxide ions or hydronium ions. (16.10) saturated solution. A solution that contains the maximum amount of a sol ute that will dissolve in a solvent at a specific temperature. (13.1) scientific method. A systematic approach to experimentation. (1.1) second law of thermodynamics. The entropy of the universe increases in a spo ntaneous process and remains unchanged in an equilibrium process. (18.2) second-order reaction. A reaction whose rate depends on the concentration of one reactant raised to the second power or on the product of the concentrations of two different reactants each raised to the first power. (14.3 ) semiconductor. A substance that normally does not conduct electricity but that will conduct at elevated temperatures or when combined with a small amount of certain other elements. (23.3,25.6) semipermeable membrane. A membrane that allows the passage of solvent molecules but blocks the passage of solute molecules. (13.5) shielding. The partial obstruction of nuclear charge by core electrons. (7.3) SI units. International System of Units . A system of units based on metric units. (1.3) sigma (0") bond. A bond in which the shared electron density is concentrated directly along the internuclear axis. (9.5) significant figures. Meaningful digits in a measured or calculated value. ( 1.5) simple cubic cell. The basic repeating unit in which there is one atom at each of the eight corners in a cube. (12.3) single bond. A pair of electrons shared by two atoms. (8.3) sintering. A method used to form objects by heating a finely divided substance. (25.2) skeletal structure. A structure in which straight lines represent carbon-carbon bonds . (10.3) smectic. Containing molecules ordered in two dimensions. The molecules are aligned parallel to one other and are further arranged

in layers that are parallel to one another. (25.3) solubility. The maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. (4.2,13.1,17.4) solubility product constant (Ksp). The equilibrium constant that indicates to what extent a slightly soluble ionic compound dissolves in water. (17.4) solute. The dissolved substance in a solution . (4.1) solution. A homogeneous mixture consisting of a solvent and one or more solutes. (4.1) solvent. A substance in a sol ution that is present in the largest amount. (4.1) specific heat (s) . The amount of heat required to raise the temperature of 1 g of a substance by 1°e. (5.4) spectrochemical series. A list of ligands arranged in increasing order of their abilities to split the d orbital energy levels. (22.3) spontaneous process. A process that occurs under a specified set of conditions. (18.1) standard atmospheric pressure. The pressure that would support a column of mercury exactly 760 mm high at O°e. (11.1) standard enthalpy of formation (!:::JIf). The heat change that results when I mole of a compound is formed from its constituent elements in their standard states . (5.6) standard enthalpy of reaction (!:::JI~xn)' The enthalpy of a reaction carried out under standard conditions. (5.6) standard entropy (S O). The absolute entropy of a substance at 1 atm. (18 .2) standard free energy of formation (i1G The free-energy change that occurs when 1 mole of the compound forms from its constituent elements in their standard states. (18.4) standard free energy of reaction (i1G~x"]])' The free-energy change for a reaction when it occurs under standard-state conditions. (18.4) standard hydrogen electrode (SHE). A half-cell based on the half-reaction 2H+(I M) + 2e • H2 (l atm), which has an arbitrarily defined standard reduction potential of zero. (19.3) standard reduction potential (E red ) . The potential associated with a reduction half-reaction at an electrode when the ion concentration is 1 M and the gas pressure is 1 atm. (19.3) standard solution. A solution of precisely known concentration. (4.6) standard temperature and pressure (STP). O°C and 1 atm. (11.3) state function. Properties that are determined by the state of the system, independent of how the state was achieved. (5.2) state of a system. The values of all relevant macroscopic properties, such as composition, energy, temperature, pressure, and volume. (5.2) stereoisomers. Molecules that contain identical bonds but differ in the orientation of those bonds in space. (10.4, n.2)

n.

stoichiometric amounts. Quantities of reactants in the same relative amounts as those represented in the balanced chemical equation. (3 .6) stoichiometric coefficients. The numeric values written to the left of each species in a chemical equation to balance the equation. (3.3) stratosphere. The region of the atmosphere located above the troposphere and consisting of nitrogen, oxygen, and ozone. (2 1.1) strong conjugate acid. The conjugate acid of a weak base. Acts as a weak Br\2lnsted acid in water. (16.7) strong conjugate base. The conjugate base of a weak acid. Acts as a weak Br\2lnsted base in water. (16.7) strong electrolyte. An electrolyte that ionizes or dissociates completely. (4.1 ) structural formula. A chemical formula that shows the general arrangement of atoms within the molecule. (2.6) structural isomer. Molecules that have the same chemical formula but different arrangements of atoms. (9.2) sublimation. The phase change from solid to gas. (12.6) substance. Matter with a definite (constant) composition and distinct properties. (1.2) substituent. A group other than - H bonded to the carbons of an organic molecule. (10.2) substitution reaction. One group is replaced by another group by electrophilic or nucleophilic attack. (10.5) superconducting transition temperature (Tc). The temperature below which an element, compound, or material becomes superconducting. (25.7) superconductor. A substance with no resistance to the flow of electrons. (25.7) supercooling. A phenomenon in which a liquid can be temporarily cooled to below its freezing point. (12.6) supercritical fluid. A fluid at a temperature and pressure that exceed Tc and Pc. (12 .6) supersaturated solution. A solution that contains more dissolved solute than is present in a saturated solution. (13.1) surface tension. The amount of energy required to stretch or increase the surface of a liquid by a unit area. (12.2) surroundings. The part of the universe not included in the system. (5 .1 ) syndiotactic. Polymers in which the substituents alternate positions along the polymer chain. (25 .1 ) system. The specific part of the universe that is of interest to us. (5 .1 )

T tacticity. Descri bes the relative arrangements of chiral carbon atoms within a polymer. (25.1) termolecular. lnvol ving three reactant molecules. (14.5)

GLOSSARY

theoretical yield. The maximum amount of product that can be obtained from a reaction. (3.7) theory. A unifying principle that explains a body of experimental observations and the laws that are based on them. (1.1 ) thermal energy. The energy associated with the random motion of atoms and molecules. (S.l) thermochemical equation. A chemical equation that includes the enthalpy change. (S.3) thermochemistry. The study of the heat associated with chemical reactions and physical processes. (S .l ) thermodynamics. The scientific study of the interconversion of heat and other kinds of energy. (S .l ) thermonuclear reaction. Generally refers to a fusion reaction. (20.6) thermoplastic. Polymers that can be melted and reshaped. (2S.1) thermosetting. Polymers that assume their final shape as part of the chemical reaction that forms them. (2S.1 ) thermosphere. The outermost layer of the atmosphere. (21.1 ) third law of thermodynamics. The entropy of a pure crystalline solid is zero at absolute zero (0 K). (1S.3) titration. The gradual addition of a solution of known concentration to another solution of unknown concentration until the chemical reaction between the two solutions is complete. (4.6) tracers. Radioacti ve isotopes that are used to trace the path of the atoms of an element in a chemical or biological process . (20.7) trans. The isomer in which two substituents lie on opposite sides of a double bond. (10.4) transition metals. The elements in Group 1B and Groups 3B - SB . (2.4) transition state. See activated complex.

transuranium elements. Elements with atomic numbers greater than 92, created by bombarding other elements with accelerated neutrons, protons, alpha particles, or other nuclei. (20.4) triple bond. A multiple bond in which the atoms share three pairs of electrons. (S.3) triple point. The point at which all three phase boundary lines meet. (12.7) triprotic acid. An acid molecule with three ionizable protons. (4.3) troposphere. The layer of the atmosphere closest to Earth's surface. (21.1 ) tyndall effect. The scattering of visible light by colloidal particles. (13.7)

u unimolecular. Describes a reacti on involving one reactant molecule. (14 .S) unit cell. The basic repeating structural unit of a crystalline solid. ( 12.3) unsaturated solution. A solution that contains less solute than it has the capacity to dissolve. (13.1 )

v valence band. The bonding band. (2S .6) valence bond theory. Atoms share electrons when an atomic orbital on one atom overlaps with an atomic orbital on the other. (9.3) valence electrons. The outermost e lectrons of an atom. (7 .2) valence· shell electron-pair repulsion (VSEPR). A model that accounts for electron pairs in the valence shell of an atom repelling one another. (9.1 ) van der Waal's equation. An equation relating the volume of a real gas to the other parameters, P, T, and 11. (11.7)

G-9

van der Waal's forces. The attractive forces that hold particles together in the condensed phases that include dipole-dipole interaction (including hydrogen bonding) and dispersion forces. (12. 1) van ' t Hoff factor (i). The ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved in solution . (13.S) vaporization. The phase change from liquid to gas at the boiling point. (12.6) viscosity. A measure of a fluid 's resistance to flow. (12.2)

wavelength (A). The distance between identical points on successive waves. (6.1) weak acid. An acid that ionizes only partially. ( 16.S ) weak base. A base that ionizes only partially. ( 16.6) weak conjugate acid. A conjugate acid of a strong base. Does not react with water. (16.7) weak conjugate base. A conjugate base of a strong acid. Does not react with water. (16.7) weak electrolyte. A compound that produces ions upon di ssolving but exi sts in solution predominantly as molecules that are not ionized. (4. 1)

x X-ray diffraction. A method of using X rays to bombard a crystalline sample to determine the structure of the crystal. ( 12.3)

z zeroth-order reaction. A constant rate, independent of reactant concentration. (14.3)

,

To Chapter 1 •

1.5 (a) Law. (b) Theory. (c) Hypothesis. 1.7 (a) C and 0. (b) F and H. (c) Nand H. (d) 0. 1.13 (a) K. (b) Sn. (c) Cr. (d) B. (e) Ba. (f) Pu. (g) S. (h) Ar. (i) Hg. 1.15 (a) Homogeneous mixture. (b) Element. (c) Compound. (d) Homogeneous mixture. (e) Heterogeneous mixture. (f) Homogeneous mixture. (g) Heterogeneous mixture. 1.17 (a) Element. (b) Compound. (c) Compound. (d) Element. 1.233.12 g/mL. 1.25 (a) 35 °C. (b) - 11 0C. (c) 39°C. (d) 1011 0C. (e) -459.67"F. 1.272.52 mL. 1.29 (a) 386 K. (b) 3.10 X 102 K. (c) 6.30 X 102 K. 1.35 (a) Quantitative. (b) Qualitative. (c) Qualitative. (d) Qualitative. (e) Qualitative. 1.37 (a) Physical. (b) Chemical. (c) Physical. (d) Chemical. (e) Physical. 1.3999.9 g, 20°C, 11.35 g/cm 3 . 1.45 (a) 0.0152. (b) 0.0000000778. 1.47 (a) 1.8 X 10- 2 (b) 1.14 X 10 10. (c) -5 X 104. (d) 1.3 X 103 .1.49 (a) 1. (b) 3. (c) 3. (d) 4. (e) 3. (f) 1. (g) lor 2.1.51 (a) 1.28. (b) 3.18 X 10- 3 mg. (c) 8.14 X 107 dm. 1.53 Z most accurate, Y least accurate; X most precise, Y least precise. 1.55 (a) 1.10 X 108 mg. (b) 6.83 X 10- 5 m 3. (c) 7.2 X 103 L. (d) 6.24 X 10- 8 lb. 1.573.1557 X 107 s. 1.59 (a) 81 in/so (b) 1.2 X 102 mlmin. (c) 7.4 krnIh. 1.6188 krnIh. 1.633.7 X 10- 3 g Pb. 1.65 (a) 1.85 X 10- 7 m. (b) 1.4 X 10 17 S. (c) 7.12 X 10- 5 m 3. (d) 8.86 X 104 L.1.67 6.25 X 10- 4 g/cm 3. 1.690.88 S. 1.71 (a) Chemical. (b) Chemical. (c) Physical. (d) Physical. (e) Chemical. 1.734.75 X 107 tons. 1.75 (a) 8.08 X 104 g. (b) 1.4 X 10- 6 g. (c) 39.9 g.1.77 31.35 cm 3. 1.79 10.50 g/cm 3.1.8111.4 g/cm 3 1.83 %Error (OF) = 0.1 %, %Error (0C) = 0.3 %. 1.85 - 40°C = - 40°F. 1.875 X 102 mL/breath. 1.894.8 X 10 19 kg NaCI, 5.3 X 10 16 tons NaCI. 1.91 The density of the crucible is equal to the density of pure platinum. 1.93 (a) 75.0 g. (b) A troy oz (31.103 g) is heavier than an oz (28.35 g). 1.95 (a) 0.5 %. (b) 3.1 %.1.97 20919 s = 3 min 29.19 s.1.991.77 X 106 g Cu. 1.101 6.0 X 10 12 g Au, $1.3 X 10 14 1.1037.3 X '1021 kg Si. 1.1059.5 X 1010 kg CO2, 1.107 (a) Homogeneous. (b) Heterogeneous. 1.109 1.1 X 102 yr. 1.111 2.3 X 104 kg NaF/yr, 99% NaF wasted. 1.113 4.0 X 10- 19 gIL. 1.1157.20 g/cm3 0.853 cm. 1.117 Element. Most compounds would decompose on heating, making them easy to identify. 1.119 Gently heat the liquid to see if any solid remains after the liquid evaporates; collect the vapor and then compare the densities of the condensed liquid with the original liquid. 1.121 The glass bottle would break because water expands when it freezes.

Chapter 2 2.90.12 mi. 2.15 145. 2.17 I~N: protons = 7, electrons = 7, neutrons = 8; f~S: protons = 16, electrons = 16, neutrons = 17; g~Cu: protons = 29, electrons = 29, neutrons = 34; ~~Sr: protons = 38, electrons = 38, neutrons = 46; I ~~a: protons = 56, electrons = 56, neutrons = 74; I ~~W: protons = 74, electrons = 74, neutrons = 112; 2~6Hg: protons = 80, electrons = 80, neutrons = 122.2.19 (a) I ~~. (b) 2~6Hg. (c) j~Se. (d) 2§~PU. 2.21 (a) 19. (b) 34. (c) 75. (d) 192.2.29 Metallic character (a) increases as we move down a periodic group and (b) decreases as we move from left to right across the periodic table. 2.31 Na and K, N and P, F and Cl. 2.33 Iron: Fe, period 4,

ODD-NuMBERED PROBLEMS

upper-left square of 8B Group; Iodine: I, period 5, Group 7 A; Sodium: Na; period 3, Group 1A; Phosphorus: P, period 3, Group 5A; Magnesium: period 3, Group 2A. 2.39 207.2 amu. 2.41 Lithium-6 = 7.5%, lithium-7 = 92.5 %.2.435.1 X 1024 amu. 2.53 (a) Polyatomic, elemental form, not a compound. (b) Polyatomic, compound. (c) Diatomic, compound. 2.55 Elements: N 2, S8, H 2; Compounds: NH 3, NO, CO, CO 2, S02' 2.57 (a) CN. (b) CH. (c) C 9H 20. (d) P20 S ' (e) BH 3 . 2.59 C 3H 7N0 2. 2.61 (a) Nitrogen trichloride. (b) Iodine heptafluoride. (c) Tetraphosphorus hexoxide. (d) Disulfur dichloride. 2.63 (a) Nitrogen trifluoride. (b) Phosphorus pentabromide. (c) Sulfur dichloride. 2.69 Na +: 11 protons, 10 electrons; Ca2+ : 20 protons, 18 electrons; AIH : 13 protons, 10 electrons; Fe2+ : 26 protons, 24 electrons; r: 53 protons, 54 electrons; F - : 9 protons, 10 electrons; S2-: 16 protons, 18 electrons; 0 2- : 8 protons, 10 electrons; N 3 - : 7 protons, 10 electrons. 2.71 (a) Na20. (b) FeS. (c) CO 2(S04)3' (d) BaF2. 2.73 Ionic: LiF, BaCI2, KCI; Molecular: SiCI4 , B 2H 6, C 2 H 4. 2.75 (a) Potassium dihydrogen phosphate. (b) Potassium hydrogen phosphate. (c) Hydrogen bromide. (d) Hydrobromic acid. (e) Lithium carbonate. (f) Potassium dichromate. (g) Ammonium nitrite. (h) Hydrogen iodate (in water, iodic acid). (i) Phosphorus pentafluoride. U) Tetraphosphorus hexoxide. (k) Cadmium iodide. (1) Strontium sulfate. (m) Aluminum hydroxide. 2.77 (a) RbN0 2. (b) K 2S. (c) NaHS. (d) Mg 3(P0 4)2' (e) CaHP0 4· (f) KH2P04 · (g) IF7 · (h) (NH4)2S04' (i) AgC10 4· (j) BCI3. 2.79 (a) Mg(N0 3)z. (b) Al 20 3 . (c) LiH. (d) Na2S, 2.81 Acid: produces H+; Base: produces OH- ; Oxoacid: an acid with oxygen; Oxoanion: anion of an oxoacid; Hydrate: crystalline solids containing water molecules. 2.83 Changing the electrical charge of an atom usually has a major effect on its chemical properties. The two electrically neutral carbon isotopes should have nearly identical chemical properties. 2.85 r-. 2.87 NaCI is an ionic compound; it doesn't consist of molecules. 2.89 (a) Molecule and compound. (b) Element and molecule. (c) Element. (d) Molecule and compound. (e) Element. (f) Element and molecule. (g) Element and molecule. (h) Molecule and compound. (i) Compound but not molecule. (j) Element. (k) Element and molecule. (1) Compound but not molecule. 2.91 It establishes a standard mass unit which permits the measurement of masses of all other isotopes relative to carbon-12. 2.93 19B, protons = 5, neutrons = 6, electrons = 5, charge = 0; ~~Fe2+ , protons = 26, neutrons = 28, electrons = 24, charge = +2; f~p3- , protons = 15, neutrons = 16, electrons = 18, charge = -3; l~gAu, protons = 79, neutrons = 117, electrons = 79, charge = 0; 2~~Rn, protons = 86, neutrons = 136, electrons = 86, charge = 0.2.95 (a) U +. (b) S2-. (c) I- . (d) N 3 - . (e) AIH . (f) Cs +. (g) Mg2+ . 2.97 Group 7 A, binary: HF, hydrofluoric acid; HCI, hydrochloric acid; RBr, hydrobromic acid; HI, hydroiodic acid. Group 7A, oxoacid: HCI04 , perchloric acid; HCI0 3, chloric acid; HCI0 2, chlorous acid; HCIO, hypochlorous acid, bromic acid; bromous acid; hypobromous acid; periodic acid; iodic acid; hypoiodous acid. Examples of oxoacids containing other Group A-block elements: H 3B03 , boric acid; H 2C0 3, carbonic acid; RN0 3 , nitric acid; H 3P0 4, phosphoric acid; H 2S04, sulfuric acid. Binary acids formed from Group A-block elements other than 7 A: H 2S, hydrosulfuric acid. 2.99 iHe, protons = 2, neutrons = 2, neutrons/ protons = 1.00; ~gNe, protons = 10, neutrons = 10, neutrons/protons = AP-1

AP-2

ANSWERS TO ODD-NUMBERED PROBLEMS

1.00; i~Ar, protons = 18, neutrons = 22, neutrons/protons = 1.22; ~Kr, 1'1 protons = 36, neutrons = 48, neutrons/protons = 1.33 ; 54Xe, protons = 54, neutrons = 78, neutrons/protons = 1.44. The neutron:proton ratio increases with increasing atomic number. 2.101 Cu, Ag, and Au are fairly chemically unreactive. This makes them especially suitable for making coins and jewelry that you want to last a very long time. 2.103 MgO and SrO. 2.105 (a) Berkelium (Berkeley, CA); Europium (Europe); Francium (France); Scandium (Scandinavi a); Ytterbium (Ytterby, Sweden); Yttrium (Ytterby, Sweden). (b) Einsteinium (Albert Einstein); Fermium (Enrico Fermi); Curium (Marie and Pierre Curie); Mendelevium (Dmitri Mendeleev); Lawrencium (Ernest Lawrence); Meitnerium (Lise Meitner). (c) Arsenic, Cesium , Chlorine, Chromium, Iodine. 2.107 The mass of fluorine reacting with hydrogen and deuterium would be the same. The ratio of F atoms to hydrogen (or deuterium) atoms is 1: 1 in both compounds. This does not violate the law of definite proportions . When the law of definite proportions was formulated, scientists did not know of the existence of isotopes. 2.109 (a) Br. (b) Rn. (c) Se. (d) Rb. (e) Pb. 2.111 Ml +, HCO 3, Mg(HC0 3)2, magnesium bicarbonate; Sr2+, Cl-, SrCI 2 , strontium chlOlide; Fe3+ , NO 2, Fe(N0 2 h iron (III) nitrite; Mn2+, CIO 3, Mn(CI0 3h, manganese(II) chlorate; Sn 4+, Br -, SnBr 4, tin(N) brOinide; Hg~ + , 1- , Hg212 , mercury(I) iodide; Cu +, CO~ - , CU2C03, copper(I) carbonate; Li +, N 3- , Li3N, lithium nitride; AI3+, S2- , A12 S3, aluminum sulfide. 2.113 1.908 X lO- s g. In principle, the law is violated, but by a negligible amount. HI

2.115

C~:

HH HHH I I I I I H-C-H ; C 2H6 : H-C-C-H ; C3Hs: H-C-C-C-H' I I I I I I ' H H H H H H H

I

H H H H H H-C-H H IIII I I I C4H IO : H-C-C-C-C-H , H-C C--C-H; I I I I I I I HHHH H H H

H H

H

H

I

II

H

H

II

I

H H-C-H H

H

I

I

II

111'1

I

II'

H

HH

C sHI2: H-C-C-C-C-C-H H-C--C--C-C-H

11

HHHHH

H

H

I

H H-C-H H I I I H-C C--C-H

I

I

I

H H-C-H H

I

iodide reacts to form hydrogen and iodine. 3.23 (a) 2NP5 • 2N20 4 • 2KN0 2 + O2 , (c) NH4N0 3 + O2, (b) 2KN0 3 • N 20 + 2H20. (d) ~N02 • N2 + 2H20. (e) 2NaHC0 3 • NaZC0 3 + H20 + CO 2, (f) P40 lO + 6H20 • 4H 3P04. (g) 2HCI + CaC03 • CaCI 2 + H20 + CO2, (h) 2Al + 3H2S04 • Al2 (S04) 3 + 3H2. (i) COz + 2KOH • CO2 + 2H20. (k) BezC + - _ . K2C0 3 + H20; (j) C~ + 202 • 2Be(0H)2 + CH4 · (1) 3Cu + 8HN03 4H20 • 3Cu(N03)z + • H2S04 + 6N0 2 + 2H20. (n) 2NH3 2NO + 4H20. (m) S + 6HN0 3 + 3CuO • 3Cu + N2 + 3H20. 3.25 (d) 3A + 2B • 2C + D. 3.315.8 X 10 3 light-years. 3.33 9.96 X 1O- 1s mol Co. 3.35 3.01 X 103 g Au. 3.37 1.244 X 10- 22 g = 1 As atom, 9.746 X 10- 23 g = 1 Ni atom. 3.392.98 X 1022 Cu atoms. 3.41 Two Pb atoms have more mass than 5.1 22 22 X 10-23 mol He. 3.43 409 g/mol. 3.45 3.01 X 10 C, 6.02 X 10 H, 3.01 X 1022 O. 3.47 2.1 X 109 molecules. 3.49 (a) 80.56% C, 7.51 % H, 11.93% O. (b) 2.11 X 1021 molecules. 3.51 C 2 H 3N O s. 3.53 39.3 g S. 3.555.97 g F. ,3.57 (a) CH20. (b) KCN. 3.59 C SH lO N40 Z' 3.611.25 g As0 3, 6.12 X 1021 molecules. 3.65 C lO H20 0. 3.67 CH20. 3.69 Empirical formula C3H2Cl, molecular formula C6 H4 C1 2. 3.71 C 3H70 2NS. 3.75 1.01 mol C1 2. 3.77 2.0 X 10 1 mol CO 2, 3.79 (a) 2NaHC0 3 • Na 2 CO, + CO 2 + H20. (b) 78 .3 g NaHC0 3. 3.81 255.9 g, 0.324 L. 3.83 0.294 mol KCN. 3.852.0 X 10 1g N 20. 3.8718.0 g O 2, 3.93 6 moles ofNH3 are formed and 1 mole of H2 is left. 3.95 0 3 is the limiting reactant, 0.709 g N0 2 produced. 0.0069 mol NO remains. 3.97 HCl is the limiting reactant, 23.4 g Cl2 produced. 3.99 (a) 7.05 g O 2, (b) 92.9 %. 3.101 3.48 3 X 10 g hexane. 3.103 8.55 g S2C12, 76.6%. 3.105 (a) Combustion. (b) Combination. (c) Decomposition. 3.107 Diagram b. 3.109 C 3H2CIFsO, 184.50 g/mol. 3.111 C1 20 7 . 3.113 (a) 0.212 mol O. (b) 0.424 mol O. 3.11539.6% NO. 3.117 30.20% C, 5.069% H, 44.57 % C1, 20.16% S. 3.119700 g. 3.121 (a) Zn(s) + H 2S0 4(aq) • ZnS04(aq) + H2(g). (b) 64.2%. (c) Assume Zn is the limiting reactant and none of the impurities reacts with the acid to produce hydrogen. 3.123 1.7 X 10 15 g CO 2 per year. 3.125 0.98 g chlorophyll, 0.16 g more than the oxygen. 3.127 (a) 6.532 X 104 g/mol hemoglobin (hg). (b) 7.6 X 102 g hemoglobin. 3.129 (a) 4.24 X 1022 K+ ions and 4.24 X 1022 Br- ions. (b) 4.58 X 1022 Na + ions and 2.29 X 1022 sol- ions. (c) 4.34 X 1022 Ca2+ ions and 2.89 X 1022 PO~- ions. 3.131 (a) 0.307 g sal. (b) 0.410 g sal. (c) 12.08 g aspirin, Yield = 90.2%.3.133 (a) C3H7NO. (b) C6H14N202' 3.135 1 g = 6.022 X 1023 amu. 3.13716.00 amu. 3.139 (e) 0.50 mol Cl2 contains 35.45 g Cl. 3.141 PtCl2 and PtCI 4 . 3.143 (a) X = MnOz, Y = Mn30 4' (b) 3Mn02 • Mn30 4 + O 2, 3.145 6.1 X 105 tons. 3.147 Mg 3N 2, magnesium nitride. 3.149 PbC gH20 . 3.151 (a) 4.3 X 1022 atoms. (b) 160 pm. 3.15328.97 g/mol. 3.155 3.1 X 1023 3.157 (a) C 3Hg + 3H20 • 3CO + 7H2. (b) 909 kg H 2. 3.159 (a) $0.47lkg. (b) 0.631 kg K20. 3.161 BaBr2' 3.163 32.l7% NaCl, 20.09% Na2S04, 47.75 % NaN0 3.

H

2.117 (a) Yes. (b) Acetylene: any fOllliula with C:H = 1:1 (CH, C2H 2, etc.); Ethane: any formula with C:H = 1:3 (CH 3, C 2H 6, etc.). 2.119 Manganese, Mn. 2.121 HCI0 3, chloric acid; HN0 2, nitrous acid; HCN, hydrocyanic acid; H2S0 4 , sulfuric acid.

Chapter 3 3.3 (a) 50.49 amu. (b) 92.01 amu. (c) 64.07 amu. (d) 84.16 amu. (e) 34.02 amu . (f) 342.3 amu. (g) 17.03 amu. 3.5 (a) 16.04 amu. (b) 46.01 amu. (c) 80.07 amu. (d) 78 .11 amu. (e) 149.9 amu. (f) 174.27 amu. (g) 310.18 amu. 3.978.77 % Sn, 21.23 % o. 3.11 (d) Ammonia, NH 3, 82.25% N. 3.13 39.90% Ca, 18.50% P, 41.41 % 0 , 0.20% H. 3.19 (a) KOH + H 3P04 • K3P04 + H 20. (b) Zn + AgCl • ZnCl2 + Ag. (c) NaHC0 3 • Na2C03 + H 20 + CO 2 , (d) NH4N0 2 • N2 + H 20 , (e) CO 2 + KOH • K 2C0 3 + H20. 3.21 (a) Potassium and water react to form potassium hydroxide and hydrogen. (b) Bar'ium hydroxide and hydrochloric acid react to form barium chloride and water. (c) Copper and nitric acid react to form copper(II) nitrate, nitrogen monoxide and water. (d) Aluminum and sulfuric acid react to form aluminum sulfate and hydrogen. (e) Hydrogen

Chapter 4 4.7 Diagram (c). 4.9 (a) Strong electrolyte. (b) Non-electrolyte. (c) Weak electrolyte. (d) Strong electrolyte. 4.11 (a) Non-conducting. (b) Conducting. (c) Conducting. 4.13 Since HCI dissolved in water conducts electricity, HC1(aq) must exist as H +(aq) and Cl-(aq). Since HCl dissolved in benzene solvent does not conduct electricity, the HCI molecules in benzene solvent exist as un-ionized molecules. 4.17 Diagram (c) . 4.19 (a) Insoluble. (b) Insoluble. (c) Soluble. (d) Soluble. 4.21 (a) 2Ag+(aq) + 2N0 :i(aq)+ 2Na+ + SO~ - (aq) - _ . AgSOis) + 2Na+ + 2NO :i(aq); 2Ag+(aq) + SO~-(aq) • Ag 2S0 4 (s) . (b) Ba2+(aq) + 2Cqaq) + Zn2 +(aq) + SO~ - (aq) • BaSOis) + Zn2 +(aq) + 2C1- (aq); Ba2+(aq) + SO~-(aq) • BaS04(s). (c) 2NH !(aq) + CO~-(aq) + Ca2+(aq) + 2Cl -(aq) • CaC0 3(s) + 2NH !(aq) + 2Cl-(aq); CO~-(aq) + Ca2+(aq) • CaC0 3(s). 4.23 (a) No precipitate form s. (b) Ba2+(aq) + SO~-(aq) • BaS04(s). 4.31 (a) Br0nsted base. (b) Br0nsted base. (c) Br0nsted acid. (d) Br0nsted acid and Br0nsted base. 4.33 (a) CH 3COOH(aq) + KOH(aq) • CH 3COOK (aq) + H 20 (I); ionic: CH 3COOH(aq) + K +(aq) + OH -(aq) • K +(aq) + CH3COO -(aq) + H 2 0(l) ; net: CH 3COOH(aq) + OH-(aq)

ANSWERS TO ODD-NUMBERED PROBLEMS

AP-3

----+. CH 3COO - (aq) + H 20(l). (b) H 2C0 3(aq) + 2NaOH(aq) • _-_:CO,(aq) + 2H 20(I), ionic: H 2C0 3(aq)+ 2Na+(aq) + 20H -(aq)

CaS04(s) + 2HF(g); 2NaCI(s) + H 2S04 (aq) • Na2S0iaq) + 2HCI(g) . (b) The sulfuric acid would oxidize the Br - and r ions to Br2 and

---+. 71 a+(aq) + CO~ - (aq) + 2H 20(I); net: H 2C0 3(aq) + 20H - (aq)

12. (c) PBr3(1) + 3H20(I) • 3HBr(g) + H 3POiaq). 4.131 (1) Pis) + 50 2(g) • P 4 0 IO (S) (redox), P40 IO (S ) + 6H20(I) • 4H3POil) (acid-base), (2) CaS(P0 4)3F(s) + 5H 2S0 4 (aq) • 3H 3P04(aq) + HF(aq) + 5CaS04(s) (acid-base and precipitation). 4.133 (a) 4K0 2(s) + 2COz(g) • 30 2(g) + 2K2C0 3(S). (b) O~ - (-1),0"2 (-1/2). (c) 34.4 L air. 4.135 No, the oxidation number on all atoms is zero. 4.137 (a) Acid H30 +, base OH+ • +

---+ . CO~- (aq) + 2H20(l). (c) 2HN0 3 (aq) + Ba(OH )z(aq) • 3 a -:\03)z(aq) + 2H 20(l), ionic : 2H+(aq) + 2N0 "3 (aq) + Ba2+(aq) +

::OH -(aq)

• Ba2+(aq) + 2N0 "3 (aq) + 2H2 0(l); net: 2H+(aq) +

• 2H zO(l). 4.41 (a) 2Sr • 2Sr2+ + 4e- , Sr is the reducing agent; O 2 + 4e• 20 2- , O 2 is the oxidizing agent. (b) _Li • 2Li + + 2e - , Li is the reducing agent; H2 + 2e • 2H- , _OH-(aq)

H _ i the ox idizing agent. (c) 2Cs • 2Cs + + 2e - , Cs is the reducing • 2Br - , Br2 is the oxidizing agent. (d) 3Mg • agent; Br2 + 2e 3~[g2+ + 6e -, Mg is the reducing agent; N2 + 6e • 2N 3- , N2 is the oxidizing agent. 4.43 H 2S (-2), S2- ( - 2), HS - (-2) < S8 (0) < S02 (.4) < S03 (+6), H 2S0 4 (+6). 4.45 (a) +1. (b) + 7 . (c) - 4. (d) - 1. (e) - 2 . (f) +6. (g) +6. (h) + 3. (i) + 4. (j) O. (k) +5. (I) - 1/2. (m) +5 . (n) -.-3. 4.47 (a) + 1. (b) - l. (c) +3 . (d) + 3. (e) +4. (f) +6. (g) +2. (h) +4. (i) + 2. (j) + 3. (k) +5.4.49 If nitric acid is a strong oxidizing agent and zi nc is a strong reducing agent, then zinc metal will probably reduce nitric acid when the two react; that is, N wi ll gain electrons and the oxidation number of N must decrease. Since the oxidation number of nitrogen in nitric acid is + 5 (verify!), then the nitrogen-containing product must have a smaller oxidation number for nitrogen. The o nly compound in the list that doesn't have a nitrogen oxidation number less than +5 is N 20 S , (what is the oxidation number of N in N 20 S ?). This is never a product of the reduction of nitric acid. 4.51 O 2 is an oxidizin g agent. In S03, sulfur is already in its maximum oxid ation state (+6), so S03 cannot be oxidized any further. 4.53 (a) Decomposition. (b) Displacement. (c) D ecomposition. (d) Combination. 4.59232 g KI. 4.61 6.00 X 10- 3 mol YlgCI2. 4.63 (a) 1.16 M. (b) 0.608 M. (c) 1.78 M. 4.65 (a) 136 mL. (b) 62.2 mL. (c) 47 mL. 4.67 Dilute 323 mL of the 2.00 M H CI to a volume of 1.00 L. 4.69 Dilute 3.00 mL of the 4.00 M HN0 3 to a volu me of 60.0 mL. 4.71 1.41 M. 4.730.00666 M. 4.810.215 g AgC!. 4.83 0.165 g aCl;Ag+(aq) + Cqaq) • AgCI(s). 4.85 0.1106 M. 4.87 (a) 42.78 mL. (b) 158.5 mL. (c) 79.23 mL. 4.89 (a) Redox. (b) Precipitation. (c) Acid-base. (d) Combination. (e) Redox. (f) Redox. (g) Precipitation. (h) Redox. (i) Redox. (j) Redox. 4.91 (d) 0.20 M Mg(N0 3)2 (greatest concentrati on of ions). 4.93774 mL. 4.95 (a) Weak electrolyte. (b) Strong electrolyte. (c) Strong electrolyte. (d) Nonelectrolyte. 4.97 (a) C 2 H sONH 2 molecules. (b) K + and F - ions. (c) NH ;; and NO 3 ions. (d) C 3H 70H molecules. 4.991146 g/mo!. 4.1011.28 M. 4.10343.4 g BaS04. 4.105 (a) React any soluble magnesium salt with a soluble hydroxide. S ince Mg(OH)2 is insoluble in water, it will precipitate and can be filtered out of the solution. (b) 0.78 L. 4.1071.72 M. 4.109 (1) Electrolysis-to ascertain presence of Hand 0, (2) Reactivity with alkali metal-water produces H 2(g) and a basic solution, (3) Metal oxide solubility-if the liquid is water, a basic solution should form . 4.111 Diagram (a) showing Ag +(aq) and N0 3 (aq). The reaction is AgOH(s) + HN0 3(aq) • AgN0 3(aq) + H 20(l). 4.113 (a) Check with litmus paper, test reactivity with carbonate, or mix with NaOH(aq) and demonstrate neutralization (use an acid-base indicator). (b) Titrate a known quantity of acid with standard NaOH(aq) solution. (c) Visually compare the conductivity of an acid solutio n with that of a sodium chloride solution of the same molarity. 4.115 (a) The complete reaction is Pb2+ (aq) + 2N0 3" (aq ) + Na2S0iaq) ---+. PbSOis) + 2NaN0 3(aq) . The net ionic reaction is Pb 2 +(aq ) • PbS0 4(s) . (b) 6.34 X 10- s M. 4.117 (a) HI(aq) + + SOl - (aq) KOH(aq)

• KI(aq) + H 20(I), evaporate to dryness. (b) 2H1(aq) +

K 2C0 3(aq) • 2KI(aq) + CO 2(g) + H 20(l), evaporate to dryness. 4.119 (a) React any soluble magnesium salt with a soluble hydroxide, filter the precipitate. (b) React any soluble si lver salt with any soluble iodide salt, filter the precipitate. (c) React any soluble barium salt with any soluble phosphate salt, filter the precipitate. 4.121 (a) Add Na2S04. (b) Add KOH. (c) Add AgN0 3. (d) Add Ca(N0 3)2. (e) Add Mg(N0 3)2. 4.123 (1) SO ~- (aq)

+ H 20 2(aq) • SO~ - (aq) + H 20(l), (2) SOr(aq) + Ba2+(aq) • BaSOis). 4.125 C l20 (+ I), CI 20 3 (+3), CI02 (+4), C I20 6 ( + 6), Cl 20 7 • (+7).4.1274.99 grains. 4.129 (a) CaF2(s) + H 2S04(aq)

°

(b) Acid NH t ; base NH2 :

+

•

+

4.139 When a solid dissolves in solution, the volume of the solution usuall y changes. 4.141 (a) The precipitate CaS04 formed over Ca preventing the Ca from reacting with the sulfuric acid. (b) Aluminum is protected by a tenacious oxide layer w ith the composition A1 20 3 . (c) These metals react more readily with water: 2Na(s) + 2H zO(l) • 2NaOH(aq) + H 2(g) . (d) Below Fe and above H. (e) Any metal above AI: Li , K, Ba, Ca, Na, Mg. 4.143 (a) (1) Cu(s) + 4HN0 3(aq) • Cu(N0 3)z(aq) + 2N0 2(g) + 2H20(l) (redox), (2) Cu(N0 3)z(aq) + 2NaOH(aq)

• CU(OH)2(S) + 2NaN0 3(aq) (precipitation) , (3) • CuO(s) + H 2 0 (g ) (decomposition), (4) CuO(s) +

Cu(OH)z(s) H 2S0 4 (aq) • CuSOi aq) + H 20(I) (acid-base), (5) CuS04(aq) + Zn(s) • Cu(s) + ZnS04(aq) (redox), (6) Zn(s) + 2HCI(aq) • ZnCI 2 (aq) + H2(g) (redox). (b) (I ) 194 g Cu(N0 3)z, (2) 101 g CU(OH)2, (3) 82.1 gCuO, (4) 165 gCUS04, (5) 65.6g Cu . (c) All of the reaction steps are clean and almost quantitative; therefore, the recovery yield should be high. 4.145 [Na +] = 0 .5295 M, [NO 3] = 0.4298 M, [OW] = 0.09968 M, [Mg2+] = 0 M.

Chapter 5 5.13 48 J. 5.15 577 J, work done on the system. 5.23 (a) w = 0 (b) w = - 9 .5 J. (c) w = -18 J. 5.254.51 kJ/g . 5.274.80 X 102 kJ. 5.29 For processes occurring in soluti on (cellular fluids), there is usually almost no volume change, so Pfl V = 0 and flU + Pfl V = flU (see Equation 5.10). 5.35 728 kJ. 5.37 50.7°C. 5.39 26.3 °C. 5.41 47.8°C/day, 4.1 kg water/day. 5.45 0.30 kJ/mo!. 5.47 -238.7 kJ/mo!. 5.53 CH4(g), H(g). 5.55 flHf[H 20(I)]. 5.57 H 2 0(I) has a more negative flH F than H z0 2(l). 5.59 177.8 kJ/mo!. 5.61 (a) - 571.6 kJ/mo!. (b) -2599 kJ/mo!. 5.63 (a) -724 kJ/mo!. (b) - 1.37 X 103 kJ/mo!. (c) -2.01 X 103 kllmo!. 5.65 -3924 kJ/mo!. 5.67 -3.43 X 104 kJ. 5.69 In a chemical reaction the same elements and the same numbers of atoms are always on both sides of the equation . This provides a consistent reference which allows the energy change in the reaction to be interpreted in terms of the chemical or physical changes that have occun-ed. In a nuclear reaction the same elements are not always on both sides of the equation and no common reference point exists . 5.71 - 44.35 kJ/mo!. 5.73 -175.3 kJ. 5.75 (a) Note that the reverse reaction is combustion of glucose, so flH ~n = -flH ~omb. Burn one mole of glucose in a bomb calorimeter and measure the heat evol ved. (b) 1.1 X 10 19 kJ. 5.77 -350.7 kJ/mo!. 5.790.492 J/g· °C. 5.81 The first reaction is exothermic, the second reaction is endothermic. The heat in the first reaction is used in the second reaction. 5.83 446 mol H 2. 5.850.167 mol ethane. 5.875.60 kJ/m.o!. 5.89 (a). 5.91 (a) 0 J. (b) - 9.1 J. 5.93 The more closely packed, the greater the mass of food. Heat capacity depends on both the mass and specific heat: C = ms. The heat capacity of the food is greater than the heat capacity of air; hence, the cold in the freezer will be retained longer. Tea and coffee are mostly water; whereas soup might contain vegetables and meat. Water has a higher heat capacity than the other ingredients in soup; therefore, coffee and tea retain heat longer than soup. 5.95 - 1.84 X 10 3 kJ. 5.97 3.0 X 109 bombs. 5.99 (a) 2LiOH(aq) + CO 2 (g) • Li 2C0 3(aq ) + H 20 (I). (b) l.l kg COl> 2 6 1.2 kg LiOH. 5.101 - 5.2 X 10 kJ. 5.103 -3.60 X 10 kJ/mol Zn. 5.105 We ass ume that when the car is stopped, its kinetic energy is completely converted into heat (friction of the brakes and friction between the tires

AP-4

ANSWERS TO ODD-NUMBERED PROBLEMS 2

and the road). Thus, q = (1/2)mu , and the amount of heat generated must be proportional to the braking distance, d: d IX q or d IX u 2 Therefore, as u increases to 2u, d increases to (2U)2 = 4u2 which is proportional to 4d. 5.107 Water has a larger specific heat than air. Thus, cold damp air can extract more heat from the body than cold dry air. B y the same token, hot humid air can deliver more heat to the body. 5.1095.8 X 102 m. 5.111 The metals in the table are AI, Au, Cu, Fe, and Hg. Dulong and Petit's law does not apply to mercury; it is not a solid at room temperature and 1 atm. 5.113 LlH'i(C 2H2) = 226.6 kJ/mol, LlHf(C6H6) = 49.0 kJ/mol, LlH~xn = -630.8 kJ/moi. 5.115 4.1 cellts. 5.117 (a) CaC2(s) + H 20(l) • CaO(s) + C2 H 2 (g). (b) -1.51 X 106 J. 5.119 Glucose: (a) 31 kJ. (b) 15 m.; Sucrose: (a) 33 kJ. (b) 16 m. 5.121 U = -5 151 kJ/mol, w = O. 5.123 -564.2 kJ/moi. 5.125 96.21 %. 5.1270.61 kJ

Chapter 6 6.5 (a) 3.5 X 103 nm. (b) 5.30 X 10 14 Hz. 6.77.0 X 102 s. 6.9 Microwave, 3.26 X 107 nm. 6.15 2.82 X 10- 19 J/photon. 6.17 (a) 4.6 X 10 7 nm, not in the visible region. (b) 4.3 X 10- 24 J/photon. (c) ?6 J/mol. 6.19 1.29 15 X 10- J/photon. 6.21 Infrared photons have insufficient energy to cause the chemical changes. 6.23 9.29 X 10- 13 m, 3 .23 X 1020 S-l 6.29 Analyze the emitted light by passing it through a prism. 6.31 Excited atoms of the chemical elements emit the same characteristic frequencie s or lines in a terrestrial laboratory, in the sun, or in a star many light-years distant from Earth. 6.33 3.027 X 10- 19 J. 6.35 1.60 X 10 14 Hz, 1.88 X 103 nm, Infrared. 6.375.6.410.565 nm. 6.43 9.96 X 10- 32 cm. 6.51 Llu = 26 . 4.38 X 10- m (far smaller than can be measured). 6.55 .e = 0 (s ), me = 0; = 1 (P), me = - 1,0, + 1. 6.57 = 0 (s), me = O;.e = 1 (P), me = - l , 0, + 1; .e = 2 (d), me = - 2, - l , 0, +1, +2;.e = 3 (j), me = -3, - 2, -1,0, + 1, +2, +3 . 6.63 (a) n = 2,.e = 1, me = - 1,0 or 1. (b) n = 3,.e = 0, me = O. (c) n = 5, .e = 2, me = -2, -1 , 0, 1 or 2.6.65 A 2s orbital is larger than a Is orbital. Both have the same spherical shape. The Is orbital is lower in energy than the 2s. 6.67 In H, energy depends only on n, but for all other atoms, energy depends on n and .e. 6.69 (a) 2s. (b) 3p. (c) equal. (d) equal. (e) 5s. 6.75 In the nth shell, there are n2 orbitals and 2n2 electrons. 6.77 (a) 3. (b) 6. (c) O. 6.79 AI: too few 2p, too many 3p electrons ; B: too many 2p electrons; F: too many 2p electrons . 6.81 Is electrons: (n, .e, me, ms) = (1,0,0, + 112) (1, 0, 0, 112 ); 2s electrons: (2,0,0, +112), (2,0,0, -1 12); 2p electrons: (2, 1, -1, + 112), (2, l, -1, - 112), (2, 1,0, + 112), (2, 1,0, -112), (2, 1, + 1, + 1/2), (2, 1, + 1, -112); 3s electrons: (3, 0, 0, + 112), (3, 0, 0, -112). The element is Mg. 6.91 [Kr]5s 2 4d5 . 6.93 Ge: [Ar]4s23d l 04/; Fe: [Ar]4i3~; Zn: [Ar]4s 23d lO ; Ni: [Ar]4i3d 8 ; W: [Xe]6i4fI4 5~; Tl: [Xe]6s 24/ 45d106pl 6.95 Part (b) is correct in the view of contemporary quantum theory. Bohr's explanation of emission and absorption line spectra appears to have universal validity. Parts (a) and (c) are artifacts of Bohr's early planetary model of the hydrogen atom and are not considered to be valid today. 6.97 (a) 4. (b) 6. (c) 10. (d) 1. (e) 2. 6.99 Since the electrons are charged particles, the metal surface becomes positively charged as more electrons are lost. After a long enough period of time, the positive surface charge becomes large enough to start attracting the ejected electrons back toward the metal with the result that the kinetic energy of the departing electrons becomes smaller. 6.101 (a) 18 1.20 X 10 photons/pulse. (b) 3.76 X 108 W. 6.103419 nm. 6.1053.0 X 10 19 photons/s. 6.107 He+: A (nm) = 164,121,109,103 (UV). H: A (nm)

e

e

°-

= 657,487, 434,411 (visible). 6.1091-. = 1-. + 1-.. 6.111 (a) He,

Al

A3 (b) N, li2i2p 3 (c) Na, IS22s22l3s 1, (d) As, [Ar]4i3d 104p 3 (e) Cl, [Ne]3i3 p 5 6.113 (a)

H H HH1

(c)[Ar]

H H H1 1 1 4s 2

A2

Is2.

(b)[Ne]

H 11 1

3d7

6.1150.596 m (microwave/radio wave). 6.117 (a) False. (b) False. (c) True. (d) False. (e) True. 6.1194.9 X 1023 photons, 34 molecules/photon. 6.1212.2 X 105 J. 6.123 Only (b) and (d) are allowed. 6.125 17.4 pm.

6.1270.382 pm, 7 .86 X 1020 Hz. 6.129 1.96 X 10- 17 J; 7.85 X 10- 17 J; 10.6 nm. 6.131 3.87 X 105 mls. 6.133 (a) We note that the maximum solar radiation centers around 500 nm. Thus, over billions of years, organisms have adjusted their development to capture energy at or near this wavelength . The two most notable cases are photosynthesis and vision. (b) Astronomers record blackbody radiation curves from stars and compare them with those obtained from objects at different temperatures in the laboratory. Because the shape of the curve and the wavelength corresponding to the maximum depend on the temperature of an object, astronomers can reliably determine the temperature at the surface of a star from the closest matching curve and wavelength.

Chapter 7 7.17 S, [Ne ]3hp4 7.19 (a) and (d), (b) and (e), (c) and (f). 7.21 (a) Na, 1 (1A) . (b) P, 15 (5A). (c) Ar, 18 (8A) . (d) Ni, 10 (8B). 7.25 (a) (J" = 2 and Zeff = + 4. (b) 2s, Zeff = +3.22; 2p, Zeff = +3. 14. The values are higher than part (a) because the 2s electrons and 2p electrons actually do shield each other somewhat. 7.33 Cl < P < Al < Mg < Na. 7.35 Fluorine. 7.37 Left to right: S, Se, Ca, K. 7.39 The atomic radius is largely determined by how strongly the outer-shell electrons are held by the nucleus. The larger the effective nuclear charge, the more strongly the electrons are held and the smaller the atomic radiu s. For the second period, the atomic radius of Li is largest because the 2s electron is well shielded by the filled I s shell. The effective nuclear charge that the outermost electrons • feel increases across the pel10d as a result of incomplete shielding by electrons in the same shell. Consequently, the orbital containing the electrons is compressed and the atomic radius decreases. 7.41 K < Ca < P < F < Ne. 7.43 The Group 3A elements (such as AI) all have a single electron in the outermost p subshell, which is well shielded from the nuclear charge by the inner electrons and the ni electrons. Therefore, less energy is needed to remove a single p electron than to remove a paired s electron from the same principal energy level (such 1 as for Mg). 7.45 1i 2s22p 6 • 2080 kJ/mol, 1i2i2l3s • 496 kJ/moi. 7.47 8.40 X 106 kJ/moi. 7.49 Cl. 7.51 Alkali metals have a 1 valence electron configuration of ns so they can accept another electron in the ns orbital. On the other hand, alkaline earth metals have a valence electron configuration of ni. Alkaline earth metals have little tendency to accept another electron, as it would have to go into a higher energy p orbital. 7.57 Fe. 7.59 (a) [Ne]. (b) [Ne]. (c) [Ar). (d) [Ar). (e) [Ar]. (f) [Ar]3d 6 . (g) [Ar]3d 9. (h) [Ar]3d lO . 7.61 (a) Cr3+ . (b) Sc3+ . (c) Rh3+ . (d) Ir 3+. 7.63 Be2+ and He; N 3- and F - ; Fe2+ and Co3+ ; S2- and Ar. 7.67 (a) Cl. (b) Na +. (c) 0 2-. (d) AIH (e) Au3+ . 7.69 The Cu + ion is larger than Cu2+ because it has one more electron. 7.71 The binding of a cation to an anion results from electrostatic attraction. As the +2 cation gets smaller (from Ba2+ to Mg2+), the distance between the opposite charges decreases and the electrostatic attraction increases. 7.75 -199.4°C. 7.77 Since ionization energies decrease going down a column in the periodic table, francium should have the lowest first ionization energy of all the alkali metals. As a result, Fr should be the most reactive of all the Group 1A elements toward water and oxygen. The reaction with oxygen would probably be similar to that of K, Rb , or Cs. 7.79 The Group 1B elements are much less reactive than the Group 1A elements. The IB elements are more stable because they have much higher ionization energies resulting from incomplete shielding of the nuclear charge by the inner d electrons. 1 The ns electron of a Group 1A element is shielded from the nucleus more effectively by the completely filled noble gas core. Consequently, the outer s electrons of 1B elements are more strongly attracted by the nucleus. 7.81 (a) Li 20(s) + H 20 (l) • 2LiOH(aq). (b) CaO(s) + H 2 0(!) • Ca(OHMaq). (c) S03(g) + H 20(l) • H 2 SOiaq). 7.83 BaO. 7.85 (a) Br. (b) N. (c) Rb. (d) Mg. 7.87 0 2 - < F- < Na+ < Mg2+ . 7.89 M = K, X = Br. 7.91 0 + and N; S2- and Ar; N 3 - and Ne; As3+ and Zn; Cs + and Xe. 7.93 (a) and (d). 7.95 Fluorine is a yellow-green gas that attacks glass; chlorine is a pale yellow gas; bromine is a fuming red liquid ; iodine is a dark, metallic-looking solid. 7.97 F. 7.99 H - . 7.101 Li20 , lithium oxide, basic; BeO, beryllium oxide, amphoteric; B 20 3 diboron trioxide, acidic; COl> carbon dioxide, acidic; N 20 5, dinitrogen

ANSWERS TO ODD-NUMBERED PROBLEMS

pentoxide, acidic. 7.103 Hydrogen has just one valence electron but it is also just one electron short of a noble gas configuration. Thus, it can behave like a member of Group lA or like a member of Group 7 A. 7.105 0.66.7.10777.5%.7.1096.94 X 10- 19 J/electron; if there are no other electrons with lower kinetic energy, then this is the electron from the valence shel!. 7.111 X must belong to Group 4A; it is probably Sn or Pb because it is not a very reactive metal (it is certainly not reactive like an alkali metal). Y is a nonmetal since it does not conduct electricity. Since it is a light yellow solid, it is probably phosphorus (Group 5A). Z is an alkali metal since it reacts with air to form a basic oxide or peroxide. • IE2 and IE9 • IEw) represents the 7.113 Each break (IEI transition to another shell (n = 1 • 2 and n = 2 • 3).

(EN = 3.2).8.39 CI-CI < Br-CI < Si -C < Cs-F. 8.41 (a) Nonpolar. (b) Polar covalent. (c) Ionic. (d) Polar covalent. H H 8.43 I I •• •• •• •• •• •• •• (a) :F-O-F: (b) :F-N=N-F: (c) H-Si-Si-H •• •• •• •• ••

I I H H H H I 1+ (f) H-C-N-H I I H H

H :0:

I

••

(d) :O-H ..

II

..

(e) H-C-C-O:

I

..

:Cl: •• 8.45 (a)

.. (b) H-Se-H · ..

:p-'ci-' p: .. I ..

..

(c) H-N-H

I

:F:

:O-H ••

••

• • '0'

.. II .. (d) :CI- P-CI: .. I ..

IE IEs IE9 7

AP-5

(e)

:CI: ..

H H I I .. H-C-C-Br: I I .. H H

••

••

••

(f) :CI-N-CI: •• I •• :Cl: ..

..

H-N-H

I

(g) H-C-H

I

I

I

I

I

I

I

I

I

H

I

••

1234567891011

S=C=N •• ••

Electron Ionized 6

7.115 Considering electron configurations Fe2+ [ArJ3d • Fe 3+[ArJ3d s and Mn2 +[Arl3d s • Mn 3+[Arl3ct. A half-filled shell has extra stability. In oxidizing Fe2+ the product is a d S-half-filled shel!. In oxidizing Mn2+, a d S-half-filled shell electron is being lost, which requires more energy. 7.117 Only K 2Ti0 4 (Ti = +6), is unlikely to exist because of the very high ionization energies required to remove the fifth and sixth electron from Ti. 7.119 (a) 2KCI0 3(s) • 2KCI(s) + • 2NH3(g) (industrial), NH4CI(s) + • NH3(g) + NaCI(aq) + H 2 0(l). (c) CaC0 3(s) •

.. (b) :S-C=N: ..

.. + .. 8.47 (a) O=N=O •• ••

I

II

.. .. + .. (d) :F-CI-F: •• •• ••

••

(c) :~-V

..

(b) H-C-C-O-H

I

..

H

:0:

:0: /; ....-.... H-C\ :0: .. -

/

• CaClz(aq) + CaO(s) + CO 2(g) (industrial), CaC0 3(s) + 2HCI(aq) H 20(l) + CO 2(g). (d) Zn(s) + H ZS04(aq) • ZnS04(aq) + Hig). (e) Same as (c), (first equation). 7.121 Examine a solution of Na2S04, which is colorless. This shows that the SO ~- ion is colorless. Thus the blue color is due to Cu 2 +(aq). 7.123 Zeff increases from left to right across the table, so electrons are held more tightly. (This explains the electron affinity values of C and 0.) Nitrogen has a zero value of electron affinity because of the stability of the half-filled 2p subshell (that is, N has little tendency to accept another electron). 7.125 Once an atom gains an electron forming a negative ion, adding additional electrons is typically an unfavorable process due to electron-electron repulsions. 2nd and 3rd electron affinities do not occur spontaneously and are therefore difficult to measure. 7.127 (a) It was determined that the periodic table was based on atomic number, not atomic mass. (b) Argon: 39.95 amu; Potassium: 39.10 amu. 4 I0 i 7.129 Element 119: [Rn]7i5/ 6d 7l8s . 7.1312A. 7.133 (a) SiH4, GeH4, SnH4, PbH 4. (b) RbH should be more ionic than NaB. (c) Ra(s) + 2H zO(l) • Ra(OHMaq) + H 2(g). (d) Be. 7.135 See sections 7.1 and 7.2.7.137 (c) Carbon. 7.139 Li: Z eff = l.26, Zeffln = 0.630; Na: Z eff = l.84, Zeffln = 0.613; K: Z eff = 2.26, Zeffln = 0.565. The Zeffln values are fairly constant, meaning that the screening per shell is about the same.

••

8.49 (a) Neither oxygen atom has a complete octet, and the left-most hydrogen atom shows two bonds (4 electrons). Hydrogen can hold only two electrons in its valence shell. H :0:

30 2(g). (b) N 2(g) + 3H2(g) NaOH(aq)

••

8.53 (a) H-C~ :0:

..

'0;• • H :0: H +/ '\+/ '\ +---.. C-N ••--+. (b) C=N • \ / /" ~ '0' • .. • H :0: H

H

:0:

\-

+ /;

C-N /" \

H

:O=..

8.55 .. H-N

"_ N:'

+ N

"_ + ) H-N-N==N:' ..

+ + .. 2• H-N=N-N: ..

~:57....

+ .. 2) :O-C=N: ••-+. :O=C-N: .. ..

O=C=N' .. ..

..

..

8.59

..

H-N-H I

/,IN--..c./ H-C II \ .. C

/N~

"

C

..

"'N"

N-:::-

I C

H-N-H I

C

;,N--..C-:::-

....-

"

H

H

"N'

I II .... H-C'I \ .. C C N --- '::::oN'/ "H / .. H

••

••

8.65 BeCl2 is "electron deficient", :CI-Be-CI: •• ••

+.. 2 .. + CI=Be=CI •• ••

• Chapter 8 8.19 (a) Decreases the ionic bond energy. (b) Triples the ionic bond energy. (c) Increases the bond energy by a factor of 4. (d) Increases the bond energy by a factor of 2. 8.21 .. .. 2(b) 2K· + 'S' (a) Na' + :F' • 2K+ :S: • Na+ :p: .. .. .. ..

..

•

(c) Ba •

.. + ·0· ..

•

Ba

2+ .. 2(d) AI· :0: .. • •

+

..

·N· •

•

AI3+ :N? -

..

8.23 860 kJ/mo!. 8.33 (a) BF3, boron trifluoride, covalent. (b) KEr, potassium bromide, ionic. 8.37 C-H (EN = 0.4) < Br-H (EN = 0.7) F-H (EN = l.9) < Li-CI (EN = 2.0) < Na-CI (EN = 2.1) < K-F


l (Ul p/ (7T2p/(7T2P)ZC7T;p/ (no unpaired electrons, like the Lewis diagram). 9.107 (a) Although the 0 atoms are Sp 3 hybridized, they are locked in a planar structure by the benzene rings. The molecule is symmetrical and therefore does not possess a dipole moment.

Kekule H3C

HO

OH

I

0 O~

+

(b) 24 u-bonds and 6 7T-bonds. 9.109 (a) :C=O: The electronegativity difference between 0 and C suggests that electron density should concentrate on the 0 atom, but assigning formal charges places a negative charge on the C atom. Therefore, we expect CO to have a small dipole moment. (b) CO is isoelectronic with Nz, bond order 3. This agrees with the triple bond in the Lewis structure. (c) Since C has a negative formal charge, it is more likely to form bonds with Fe 2 + . (OC - Fe2+ rather than CO-Fe 2+).

•

o

Hl

-

I

CH 3

.'

.'

-

10.23 (a) DMF is:

I

./'C ~ *~C"'--- /CH l o C C

o

II

. I .. ' C=O H 2 CH I .. C I 3 H 2C/ "'---C....--C~

Hl CH3 CH2 /C"'--- I /CH /CH_ / H C C '''''CH C 1I Hl

A P-7

OH Skeletal

0

o

(c) Isoamyl acetate is:

II CH37HCH2CH20CCH3

(CH3)?CH(CH2h0 2CCH3 Condensed

CH 3 Kekule

o

Chapter 10

o

10.7 (a) Amine. (b) Aldehyde. (c) Ketone. (d) Carboxy lic acid. (e) Alcohol. 10.9 (a) 3-ethyl-2,4,4-trimethylhexane. Cb) 6,6dimethyl-2-heptanol. (c) 4-chlorohexanal. 10.11 3,S-dimethyloctane. 10.13 (a) (CH3)3CCH2CH(CH3h (b) HO(CH2)2CH(CH3)2' (c) CH 3(CH 2)4C(0)NH2. Cd) CI 3CCHO. 10.15 A = Carbonyl (Ketone) B = Carboxy (Caroboxylic acid) C = Hydroxy (Alcohol)

OH C AO

CH 3

Skeletal

10.25 (a)

OH

o II

(b) (CH3)3CCH27HCH2CH or (CH3)3CCH2CH(Br)CH2CHO (c)

Br

n 10.27 (a) CH 3-C=N:

.--+-----, OH B

CH3

I

H

\ Primary . 10.17 N ammo / H

~

j

I

• + ?'C "'" :0 CHo

(b) .0--C , • :0 + CH 3 I

P C Carboxy

+ "+-.-~. CH 3- C=N: CH 3

I

H

'oH

CH 3

I

(c) _ 0 / C""

:..0 /

" CH 2 U

0

H

+.-_.

, CH 3

I

"",C, _ :0/ ' CH? ••• • -

••

CH 3

Br

I

(c) CH3CHCH27HCH3

C6HS

I

CH 3

(0: ! 10.29 (a) H- C

c\..:

I

(d) CH3CH2CH7HCHCH2CH2CH3 CH 3

(b)

b~

CH / " ' CH 3

+ 2

••

0:

+-.-~.

+.-_.

Ii

H-C

\: ..

0 CH3/ " '~CH2

o

AP-8

ANSWERS TO ODD-NUMBERED PROBLEMS

H

H

H

(c) H __{~f) '\)

H

+ CH?

H

10.59 This is an elimination reaction: H

(

H H 2N: + C I

......--+. H---
---C-NH-

The general reaction is a condensation to form an amide:

o

0

II

+

R-C - OH

o

R'-NH 2

II

• R-C-NH-R'

+

H 20

H

II

I

10.99 -+-C-N-C-+-

I

I

H H

•

10.1018 different tripeptides: Lys-Lys-Lys, Lys - Lys-Ala, Lys-Ala - Lys, Ala -Lys - Lys, Lys-Ala-Ala, Ala -Lys -Ala, Ala-Ala-Lys, Ala - Ala - Ala.

Chapter 11 11.13 0.493 atm, 375 torr, 5.00 X 104 Pa, 0.500 bar. 11.15 13.1 m. 11.17 7 .3 atm. 11.21 (a) Diagram d. (b) Diagram b. 11.2345.9 mL. 11.25 587 mmHg. 11.27 31.8 L. 11.29 1 volume of NO. 11.35 6.6 atm. 11.37 1590°C. 11.39 1.8 atm. 11.41 0.70 L. 11.4363.31 L. 11.456.1 X 10- 3 20 atm. 11.4735.1 g/mo!. 11.492.1 X 1022 N 2, 5.6 X 1021 O 2,2.7 X 10 Ar. 11.51 2.98 gIL. 11.53 SF4. 11.55 3.70 X 102 L. 11.57 88.9%. 11.59 M(s) + 3HCI(aq) • 1.5H2 (g) + MCI 3 (aq), M 20 3 and M 2 (S04h 11.61 Assuming the impurities do not produce CO 2, assuming 100% yield: 94.7% 3 11.63 C2 H sOH(l) + 30 2 (g) • 2C0 2 (g) + 3H20(l), 1.44 X 10 L. 11.67 (a) 0 .89 atm. (b) 1.4 L. 11.69349 mmHg. 11.7119.8 g Zn. 11.73 P I2 = 217 mmHg, PH, = 650 mmHg. 11.75 (a) Box 2. (b) Box 2.11.83 2 472 mls; 0 2 441 m/s; 0 3 360 mls. 11.85 RMS = 2.8 mls; Average speed = 2.7 mls. The root-mean-square value is always greater than the average value because squaring favors the larger values compared to just taking the average value. 11.8743.8 g/mol, possibly CO 2 (44.01 g/mol). 11.93 P vdW = 18.0 atm vs. Pideal = 18.5 atm. 11.95 (a) Neither the amount of gas in the tire nor its volume change appreciably. The pressure is proportional to the temperature. T herefore, as the temperature rises , the pressure increases. (b) As the paper bag is hit, its volume decreases so that its pressure increases. The popping sound occurs when the bag is broken. (c) As the balloon rises, the pressure outside decreases steadily, and the balloon expands. (d) The pressure inside the bulb is greater than 1 atm. 11.97 1.7 X 102 L. CO 2 0.49 atm, H 20 0.41 atm, N2 0.25 atm, O 2 • N 2(g) + 2H20(g). (b) 0.274 g. 0 .041 atm. 11.99 (a) NH4N0 2(s) 11.101 (a) P ii = 4.00 atm, P iii = 2.67 atm. (b) 2.67 atm, PA = 1.33 atm, PB = 1.33 atm. 11.103 (a) 9.54 atm. (b) If the tetracarbonylnickel gas tarts to decompose significantly above 43 °C, then for every mole of tetracarbony lni ckel gas decomposed, four moles of carbon monoxide would be produced. 11.105 1.3 X 1022 molecules. CO 2, H 20, N2 and O 2 , 11.107 (a) 2KCl0 3(s) • 2KCl(s) + 302(g ). (b) 6.21 L. 11.109 0.0700 M. 11.111 He = 0.16 atm, Ne = 2 .0 atm. 11.113 When the water enters the flask from the dropper, some hydrogen chloride dissolves, creating a partial vacuum. Pressure from the atmosphere forces more water up the vertical tube. 11.1157.11.117 (a) 61.2 mls. (b) 4.58 X 10- 4 s. (c) 328 mls. The nns speed of Bi = 366 mis, (average of all Bi atoms), so these were some of the slower atoms in the sample. 11.119 _0. 9 kg of oxygen, 1.58 X 104 L. 11.121 The fru it ripens more rapidly because the quantity (partial pressure) of ethylene gas inside the bag increases. 11.123 As the pen is used the amount of ink decreases,

AP-9

increasing the volume inside the pen. As the volume increases, the pressure inside the pen decreases. The hole is needed to equalize the pressure as the volume inside the pen increases. 11.1250.0821 L·atm/mol·K. 11.127 C 6H 6. 11.129 The gases inside the mine were a mixture of carbon dioxide, carbon monoxide, methane, and other harmful compounds . The low atmospheric pressure caused the gases to flow out of the mine (the gases in the mine were at a higher pressure), and the man suffocated. 11.131 (a) 4.90 L. (b) 6.0 atm. (c) 1 atm. 11.133 (a) 4 X 10- 22 atm. (b) 20 6 X 10 L. 11.13591 %.11.1371.7 X 10 12 11.139 N0 2. 11.1411.4 g. 11.143 CO 2: 8.7 X 106 Pa, N2: 3.2 X lOs Pa, S02: 1.4 X 103 Pa. 11.145 (a) (i) He atoms in both flasks have the same rms speed. (a)(ii) The atoms in the larger volume strike the walls of their container with the same average force as those in the smaller volume, but pressure in the larger volume will be less. (b) (i) Uoz is greater than £II by a factor of ~T2 I~ (b)(ii) Both the frequency of collision and the average force of collision of the atoms at Tz will be greater than those at T I . (c )(i) The rms speed of He is greater than that ofNe because He is less massive. (c)(ii) The average kinetic energies of the two gases are equal. (c) (iii) The nns speed ofthe whole collection of atoms is 1.47 X 103 mis, but the speeds of the individual atoms are random and constantly changing. 11.1473.7 nm. 0.31 nm. Water molecules are packed very closely together in the liquid, but much farther apart in steam . 11.1499.98 atm. 11.1511.41 atm. 11.153 Xmethane = 0.789, Xethane = 0.211. 11.155 (a) 8Vatom ' (b) The excluded volume is 4 times larger than the sum of volumes of 1 mole of atoms. 11.157 NO. 11.159 Reaction (b).

Chapter 12 12.9 Butane would be a liquid in w inter, and on the coldest days even propane would become a liquid. Only methane would remain gaseous. 12.11 (a) Dispersion. (b) Dispersion and dipole-dipole. (c) Dispersion and dipole-dipole. (d) Dispersion and ionic. (e) Dispersion. 12.13 (e) CH 3COOH. 12.15 I-butanol has greater intermolecular forces because it can form hydrogen bonds. 12.17 (a) Xe. (b) CS 2. (c) C12 . (d) LiP. (e) NH 3 . 12.19 (a) Dispersion and dipole-dipole, including hydrogen bonding. (b) Dispersion only. (c) Dispersion only. (d) Covalent bonds. 12.21 The compound with - N0 2 and -OH groups on adjacent carbons can form hydrogen bonds with itself (intramolecular hydrogen bonds). Such bonds do not contribute to intennolecular attraction and do not help raise the melting point of the compound. The other compound , with the - N0 2 and -OH groups on opposite sides of the ring, can form only intermolecular hydrogen bonds; therefore, it will take a higher temperature to escape into the gas phase. 12.33 Ethylene glycol has two -OH groups, allowing it to exert strong intermolecular forces through hydrogen bonding. Its viscosity should fall between ethanol (l -OH group) and glycerol (3 -OH groups). 12.41 simple cubic: one sphere, body centered cubic: two spheres, face centered cubic: four spheres. 12.43 6.20 X 1023 mol- 1 12.45458 pm. 12.47 XY3' 12.490.220 nm. 12.51 ZnO. 12.55 Molecular solid. 12.57 Molecular: Ses, HBr, CO2 , P40 6 , SiH4; covalent: Si, C. 12.59 Diamond: each carbon atom is covalently bonded to four other carbon atoms. Because these bonds are strong and uniform, diamond is a very hard substance. Graphite: the carbon atoms in each layer are linked by strong bonds, but the layers are bound by weak dispersion forces. As a result, graphite may be cleaved easily between layers and is not hard. 12.81 2.72 X 103 kJ. 12.8347.03 kllmo!. 12.85 Two phase changes occur in this process. First, the liquid is turned to solid (freezing), then the soli d • ice is turned to gas (sublimation). 12.87 When steam condenses to liquid water at 100°C, it releases a large amount of heat equal to the enthalpy of vaporization. Thus steam at 100°C exposes one to more heat than an equal amount of water at 100c C. 12.89 331 mmHg. 12.91 75.9 kJ/mo!. 12.95 Initially, the ice melts because of the increase in pressure. As the wire sinks into the ice, the water above the wire refreezes. Eventually the wire actually moves completely through the ice block w ithout cutting it in half. 12.97 (a) Ice would melt. (If heating continues, the liquid water would eventually boil and become a vapor.) (b) Liquid water would vaporize. (c) Water vapor would solidify without becoming a liquid. 12.99 (d). 12.101 Boron is covalent. 12.103 CCI 4· 12.105 0.26 cm; 7 .7 X 106 base pairs/ molecule. 12.107760 mmHg.

AP-l0

ANSWERS TO ODD-NUMBERED PROBLEMS

12.109 It has reached the critical point; the point of critical temperature (TJ and critical pressure (Pc). 12.111 Crystalline Si0 2· 12.113 (a) and (b). 12.1158.3 X 10- 3 atm. 12.117 (a) K 2S. (b) Br2' 12.119 S02 because it is polar and has greater intermolecular forces. 12.121 62.4 kJ/moi. 12.123 Smaller ions can approach polar water molecules more closely, resulting in larger ion-dipole interactions. The greater the ion-dipole interaction, the larger is the heat of hydration. 12.125 (a) 30.7 kJ/mo!. (b) 192.5 kJ/mo!. 12.127 (a) Decreases . (b) No change. (c) No change. 12.129 CaC0 3(s) - _ . CaO(s) + CO 2(g), initial state: one solid phase, final state: two solid phase components and one gas Rhase component. 12.131 The • molecules are all polar. The F atoms can form H-bonds with water and other -OH and -NH groups in the membrane, so water solubility plus easy attachment to the membrane would allow these molecules to pass the blood-brain ban·ier. 12.133 The time required to cook food depends on the boiling point of the water in which it is cooked. The boiling point of water increases when the pressure inside the cooker increases. 12.135 (a) Extra heat produced when steam condenses at 100°e. (b) Avoids extraction of ingredients by boiling in water. 12.137 (a) -2.3 K. (b) -10 atm. (c) -5 K. (d) No. 12.139 (a) Pumping allows Ar atoms to escape, thus removing heat from the liquid phase. Eventually the liquid freezes. (b) The solid-liquid line of cyclohexane is positive. Therefore, its melting point increases with pressure. (c) These droplets are super-cooled liquids. (d) When the dry ice is added to water, it sublimes. The cold CO 2 gas generated causes nearby water vapor to condense, hence the appearance offog. 12.141 (a) Two triple points: Diamond/graphite/liquid and graphite/liquid/vapor. (b) Diamond. (c) Apply high pressure at high temperature. 12.143 Ethanol mixes well with water. The mixture has a lower surface tension and readily flows out of the ear channel. 12.145 When water freezes it releases heat, helping keep the fruit warm enough not to freeze. Al so, a layer of ice is a thermal insulator. 12.147 The fuel source for the Bunsen burner is most likely methane gas. When methane bums in air, carbon dioxide and water are produced: CH4 (g) + 20ig) • COig) + 2H20 (g) . The water vapor produced during the combustion condenses to liquid water when it comes in contact with the outside of the cold beaker. 12.1496.020 X 1023 12.151 127 mmHg. 12.153 55 °e.

Chapter 13 13.9 "Like dissol ves like." Naphthalene and benzene are nonpolar, whereas CsF is ionic. 13.11 O 2 < Br2 < LiCI < CH 30H. 13.15 (a) 8.47%. (b) 17.7 %. (c) 11 %.13.17 (a) 0.0610 m. (b) 2.04 m. 13.19 (a) 1.74 m. (b) 0.87 m. (c) 6.99 m. 13.21 3.0 X 102 g. 13.23 18.3 M , 27.4 m. 13.25 X(N2) = 0.677, X(02) = 0.323. Due to the greater solubility of oxygen, it has a larger mole fraction in solution than it does in the air. 13.3345.9 g. 13.35 According to Henry's law, the solubility of a gas in a liquid increases as the pressure increases (c = kP). The soft drink tastes flat at the bottom of the mine because the carbon dioxide pressure is greater and the dis sol ved gas is not rel eased from the solution. As the miner goes up in the elevator, the atmospheric carbon dioxide pressure decreases and dissolved gas is released from his stomach. 13.37 1.0 X 10- 5 mollL. 13.392.6 X 10- 4 mollL. Total dissolved oxygen = 1.3 X 10- 3 moles, enough to metabolize about 2.2 X 10 - 4 moles of glucose or about 40 milligrams. You could not survive on meals serving only 40 mg of food ; so the "bioavailability" of oxygen is greatly enhanced by hemoglobin, whose molecules do not depend on Henry's Law to 3 carry oxygen. 13.57 l.35 X 10 g. 13.59 P elhanol = 30 mmHg, P propanol = 26.3 mmHg. 13.61 187 g. 13.630.59 m. 13.65 -5.4°e. 13.67 (a) CaCI2. (b) urea. (c) CaCI 2. 13.690.15 m C6 H I2 0 6 > 0.15 In CH 3COOH > 0.10 m Na3P04 > 0.20 In MgCI2 > 0.35 m NaC!. 13.71 NaCI: boiling point = 102.8°C, freezing point = - IO.O°C; Urea: boiling point = 102.0°C, freezing point = -7.14°e. 13.73 Both NaCI and CaCI 2 are strong electrolytes. Urea and sucrose are nonelectrolytes. The NaCI or CaCI 2 will yield more particles per mole of the solid dissolved, resulting in greater freezing point depression. Also, sucrose and urea would make a mess when the ice melts. 13.752.47.13.77 9.16 atm: 13.81282.8 gl mol , 2 4 C I9 H 3S O. 13.834.3 X 10 glmol, C24H20P4' 13.85 l.75 X 10 g/mo!. 13.87 342 glmol. 13.89 15.7%. 13.93 /:::"P = 2.05 X 10- 5 mmHg; /:::"Tr = 8.9 X 10- 5 DC; /:::"Tb = 2.5 X 10 - 5 DC; 1T = 0.889 mmHg. With the

exception of osmotic pressure, these changes in colligative properties are essentially negligible. 13.95 Levels before: 7.77 X 10- 6 mol glu/mL blood, 3.89 X 10 - 2 mol glu total, 7.01 g total. Levels after: 1.33 X 10- 5 mol glu/mL blood, 6.66 X 10- 2 mol glu total, 12.0 g total. 13.97 Water migrates through the semi-permeable cell walls of the cucumber into the concentrated salt sol ution. 13.993.5. 13.101 X = water, Y = NaCl, Z = urea. 13.103 The pill is in a hypotonic solution. Consequently, by osmosis, water moves across the semi-permeable membrane into the pill. The increase in pressure pushes the elastic membrane to the right, causing the drug to exit through the small holes at a constant rate. 13.105 34 atm. Reverse osmosis involves no phase changes and is usually cheaper than di stillation or freezing. 13.1070.295 mol/L, - 0.55°e. 13.109 (a) 2.14 4 3 X 10 glmol. (b) 4.5 X 10 glmol.13.111 (a) 99 mL. (c) 9.9.13.113 (a) At reduced pressure, the solution is supersaturated with CO2, The excess CO 2 bubbles out of solution. (b) As the escaping CO 2 cools it expands, condensing water vapor in the air to form fog. 13.115 (a) Runoff of the salt solution into the soil increases the salinity of the soil. If the soil becomes hypeltonic relative to the tree cells, osmosis would reverse, and the tree would lose water to the soil and eventually die of dehydration. (b) Assuming the collecting duct acts as a semipellneable membrane, water would flow from the urine into the hypertonic fluid , thus retuming water to the body. 13.117 2 M beaker = 67 mL, 1 M beaker = 33 mL. 13.119 12.3 M. 13.121 14.2%.13.123 (a) Solubility decreases with increasing lattice energy. (b) Ionic compounds are more soluble in a polar solvent. (c) Solubility increases with enthalpy of hydration of the cation and anion. 13.125 1.43 glmL, 37.0 m. 13.127 NH3 can fOlln hydrogen bonds with water; NCl 3 cannot. 13.1293%. 13.131 Egg yolk contains lecithins which solubilize oil in water (See Figure 13.18 of the text). The nonpolar oil becomes soluble in water because the nonpolar tails of lecithin dissolve in the oil, and the polar heads of the lecithin molecules dissolve in polar water (like dissolves like). 13.133 Yes. 13.135 As the water freezes , dissolved minerals in the water precipitate from solution. The minerals refract light and create an opaque appearance. 13.137 Assume both naphthalene and X are non-volatile, 179.6 glmol. 13.139 l.9 m. 13.141 P~ = ? 2 1.9 X 10- mmHg, PB = 4.0 X 10 mmHg. 13.143 -0.738°C. 0

Chapter 14 14.5 (a) Rate = _ /:::"[H2l

= _ /:::"[I2l = ~ /:::,,[HIl

M (b) Rate = _~ /:::"[Bcl

5

M

.

2 M

= _ I:::,,[ BID 'l l = _~ /:::"[Wl = ~ /:::"[Br2l

M

/:::"t

6

M

3

M

14.7 (a) 0.066 Mis. (b) 0.033 Mis. 14.158.1 X 10- 6 Mis. 14.17 The · reaction is first order in A and first order overall; k = 0.213 S-I. 14.19 (a) 2. (b) O. (c) 2.5. (d) 3.14.21 First order, k = 1.19 X 10- 4 S- I . 14.27 30 min. 14.29 (a) 0.034 M. (b) 17 sand 23 s. 14.31 4.52 X 10- 10 Mis, 5.38 X 106 s. 14.33 (a) 4:3:6. (b) The relative rates would be unaffected, each absolute rate would decrease by 50%. (c) I: I: 1. 14.41 103 kllmo!. 14.43 k = 3.0 X 103 S - 1 14.45 51.8 kllmo!. 14.47 131 kllmo!. For maximum freshness, fish should be frozen immediately after capture and kept frozen until cooked. 14.49 Diagram (a). 14.59 (a) Second-order. (b) The first step is the slower (rate-determining) step. 14.61 Mechanism I can be discarded; Mechanisms II and III are possible. 14.71 Higher temperatures may disrupt the intricate three dimensional structure of the enzyme, thereby reducing or totally destroying its catalytic activity. 14.73 Temperature, energy of activation, concentration of reactants, and a catalyst. 14.75 Temperature must be specified. 14.77 k = 0.035 S- l 14.79 280

250 ~

c

o

~ ~

Cl..

220 190 160

14?

130

o

---------------------

50

100

150 I(S)

200

250

300 , t v,

= 270 s.

ANSWERS TO ODD-NUMBERED PROBLEMS 14.81 Since the methanol contains no oxygen-I 8, the oxygen atom must come from the phosphate group and not the water. The mechanism must involve a bond-breaking process like:

rate

II

I

14.83 Most transition metals have several stable oxidation states. This allows the metal atoms to act as either a source or a receptor of electrons in a broad range of reactions. 14.85 (a) Rate = k[CH 3COCH 3l [H +l. (b) 3.8 X 10- 3 M - IS-I. (c) k = k l k2/L I .14.87 (1) Fe3+ oxidizes 1- : 2Fe3+ + 21 • 2Fe2+ + 12; (2) Fe2+ reduces S l O~- : 2Fe l + + S2 0~- - ... 2Fe3+ + 2S0~- ; overall reaction: + S20~• . I2 + 2S0~(Fe3+ undergoes a redox cycle: Fe3+ • Fe2+ • Fe3+). The uncatalyzed reaction is slow because both 1- and S 2 0 ~- are negatively charged which makes their mutual approach unfavorable. 14.89 (a) rate = k[Al o = k,

2r

[A]

,

[Alo,

slope = k

Time

[Al o t =-"'(b) 112 2k. (c) t = 2tl/2' 14.91 There are three gases present and we can measure only the total pressure of the gases. To measure the partial pressure of azomethane at a particular time, we must withdraw a sample of the mixture, analyze and determine the mole fractions. Then, Pa zomethane = X azomethane

X

[NOF . At very low

k2

rate = - ' - - -- -=I

O- H

= -kt +

l

1

kl[NOF[H zl

CH-O+P - O-H

[Al

k 2 [H l

=

k

[H2l, kl [H 2 l < < I and

o 3

=

k [NOF[H?l I -

AP- 11

P total .

14.93

,•

14.99 :2 2.50,--------

- - - - ----:;.------, Initial Rate vs. Cone.

:::;;:: ~

8o

-

2.00

o o - 1.50 x

" 1.00 '" e::: ~

..,& -

0.50 JL-- - - r - - -- - - - , - - - - , -- ----1 0.50 1.00 1.50 2.00 2.50 [Dinitrogen Pentoxide] (M)

rate = k[N 1 0 5 ], k = 1.0 X 10- 5 S - I. 14.101 The red bromine vapor absorbs photons of blue light and dissociates to form bromine atoms: Brl • 2Bf". The bromine atoms collide with methane molecules • HBr + ·CH 3. The and abstract hydrogen atoms: Br· + CH4 methyl radical then reacts with Brl, giving the observed product and regenerating a bromine atom to start the process over again: ·CH3 + Br2 • CH3Br + Bf" , Bf" + CH4 • HBr + ·CH 3 and so on. 14.103 Lowering the temperature would slow all chemical reactions, which would be especially important for those that might damage the brain. 14.105 (a) Rate = k[X][yf (b) k = 0.019 M - lS - I. 14.107 Second order, k = 2.4 X 107 M - I S- 1 14.109 During the first five minutes or so the engine is relatively cold, so the exhaust gases will not fully react with the components of the catalytic converter. Remember, for almost all reactions, the rate of reaction increases with temperature. 14.111 (a) Ea has a large value. (b) Ea = O. Orientation factor is not important. 14.113 5.7 X 105 yr. 14.115 (a) Catalyst: Mn2+ ; intermediate: Mn3+ . First step is rate-determining. (b) Without the catalyst, the reaction would be a termolecular one involving 3 cations ! (Tl + and two Ce4 +). The reaction would be slow. (c) The catalyst is a homogeneous catalyst because it has the same phase (aqueou s) as the reactants. 14.1170.45 atm. 14.119 (a) 6.[Bl /M = kl [Al - k2[Bl. (b) [Bl = (k llk2)[Al. 14.121 (a) k = 0 .0247 yr- I. (b) 9.8 X 10- 4 (c) 186 yr. 14.123 (a) Three steps. (b) Two intermediates. (c) The third step is rate determining . (d) The reaction overall is exothermic. 14.125 1.8 X 103 K. 14.127 (a) 2.5 X 10- 5 Mis. (b) 2.5 X 10- 5 Mi s. (c) 8.3 X 10- 6 M . 1 14.129 11 = 0, ti l? = C I = C[Alo; [Al o 11 =

1

t

1

112

= C --:- = C· [Al g ,

1

=C-[Al a 14.131 (a) 1.13 X 10- 3 Mlmin. (b) 6.83 X 10- 4 Mlmin; 8.8 X 10- 3 M. 14.133 Second order, 0.42IM·min. 14.135 kl is 60% larger than k l • 11 =

2,

t ill

Chapter 15 2XG

•

•

Gl + X l 2EG

Reaction progress

14.95 (a) A catalyst works by changing the reaction mechanism, thus lowering the activation energy. (b) A catalyst changes the reaction mechanism. (c) A catalyst does not change the enthalpy of reaction . (d) A catalyst increases the forward rate of reaction. (e) A catalyst increases the reverse rate of reaction. 14.97 At very high [Hl l , kl [H 2l > > 1 and

15.11 1.08 X 10 7 15.21 (1 ) Diagram (a), (2) Diagram (d). Volume cancels in the Kc expression. 15.232.40 X 1033 15.253.5 X 10- 7 15.27 (a) 8.2 X 10- 2. (b) 0.29.15.29 Kp = 0.105, Kc = 2.05 X 10- 3. 15.317.09 X 10- 3 15.33 Kp = 9.6 X 10- 3, Kc = 3.9 X 10 - 4 .15.35 4.0 X 10- 6 15.375.6 X 1013 15.39 The equilibrium pressure is less than the initial pressure. 15.41 0.173 mol H 2. 15.43 [H2l = [Brl l = 1.80 X 10- 4 M. [HBrl = 0.267 M. 15.45 P eoCl , = 0.408 atm, P eo =

AP-12

ANSWERS TO ODD-NUMBERED PROBLEMS

PC!, = 0.352 atm. 15.47 Pco = 1.96 atm, P co , = 2.54 atm. 15.49 The forward reaction will not occur. 15.55 (a) The equilibrium would shift to the right. (b) The equilibrium would be unaffected. (c) The equilibrium would be unaffected. 15.57 (a) No effect. (b) No effect. (c) Shift to the left. (d) No effect. (e) Shift to the left. 15.59 (a) Shift to the right. (b) Shift to the left. (c) Shift to the right. (d)Shift to the left. (e) A catalyst has no effect on equilibrium position. 15.61 No change. 15.63 (a) Shift to right. (b) No effect. (c) No effect. (d) Shift to left. (e) Shift to the right [because NaOH(aq) reacts with CO(g) to produce NaHC0 3(aq)]. (f) Shift to the left [because HCl(aq) reacts ;.'lith CaC0 3 (s) to produce COz(g) and other products]. (g) Shift to the right (the decomposition of CaC0 3 is endothermic). 15.65 (a) 203(g) • • 30 2(g), !:ili0 = -284.4 kJ. (b) Equilibrium would shift to the left. 15.67 (a) P 0 = 0.24 atm; PC!, = 0.12 atm. (b) 0.017.15.69 (a) No effect. (b) More CO 2 and H 20. 15.71 (a) 8 X 10- 44 . (b) A mixture of H z and Oz can be kept at room temperature because of a very large activation energy. 15.73 Kp = 1.7, 5 PA = 0.69 atm, P B = 0.81 atm. 15.75 1.5 X 10 15.77 PH, = 0.28 atm, PC!, = 0.051 atm, P HCI = 1.67 atm. 15.795.0 X 10 1 atm. i5.81 0 .0384. 15.83 328 atm. 15.85 6.3 X 10- 4 15.87 P N, = 0.860 atm. PH, = 0 .366 atm. P NH 3 = 4.40 X 10- 3 atm. 15.89 (a) 1.16. (b) 53.7%.15.91 (it) 0.49 atm. (b) 0.23 (23 %). (c) 0.037 (3.7 %) . (d) Greater than 0 .037 mol. 15.93 [H2 ] = 0.070 M. [12] = 0.1 82 M. [HI] = 0.825 M. 15.95 (c) N2 0 i colorless) - _ . 2N02(brown) is consistent with the observations. The reaction is endothermic so heating darkens the color. Above 150°C, the O 2 breaks • 2NO(g) + 0 2(g) . An up into colorless NO and O2 : 2N0 2(g) increase in pressure shifts the equilibrium to the left, restoring the color by producing N0 2 . 15.97 (a) 4.2 X 10- 4 . (b) 0.83. (c) 1.1. (d) 2.3 X 103 and 2.1 X lO- z. 15.99 (a) Color deepens. (b) Increases. (c) Decreases. (d) Increases. (e) Unchanged. 15.101 Potassium is more volatile than sodium. Therefore, its removal shifts the equilibrium from left to right. 15.1033.6 X 10- 2 15.105 (a) Shifts to the right. (b) Shifts to the right. (c) No change. (d) No change. (e) No change. (f) Shifts to the left. 15.107 (a) 1.85 X 10- 16 (b) 1.02 X 10- 14 , 1.01 X 10- 7 M. 15.109 P ,0, = 0.09 atm, P NO, = 0.100 atm. 15.111 (a) 1.03 atm. (b) 0.39 atm. (c) 1.67 atm. (d) 0.620- (62.0%).15.113 (a) Kp = 2 .6 X 10- 6, Kc = 1.1 X 10- 7. (b) 2.2 g, 22 mg/m3, yes. 15.115 There is a temporary dynamic equilibrium between the melting ice cubes and the freezing of water between the ice cubes. 15.117 [NH 3] = 0.042 M , [N 2] = 0 .086 M, [Hz] = 0 .26 M . ? ~4 x-? 15.119 (a) Kp = ? P. (b) If P increases, the fraction z (and I-xI - x therefore, x ) must decrease. Equilibrium shifts to the left to produce less NO z and more N 2 0 4, as predicted. 15.121 P sop , = 3.58 atm, P so , = PC!, = 2.71 atm. 15.1234.0.

Chapter 16 16.3 (a) both. (b) base. (c) acid. (d) base. (e) acid. (f) base. (g) neither. (h) base. (i) acid. CD acid. 16.5 (a) N0 2 . (b) HS0 4 . (c) HS - . (d) CN- . (e) HCOO - .16.7 (a) CH2CICOO- . (b) 10 4, (c) H 2 P0 4 . (d) HPO ~- . (e) PO ~- . (f) HSO 4' (g) SO~- . (h) 10 3 , (i) SO ~- . (j) NH 3. (k) HS - . (1) S2- . (m) ClO- .16.17 7. 1 X 10- 12 M. 16.19 (a) 3.00. (b) 13.89. 16.21 (a) 3.8 X 10- 3 M. (b) 6.2 X 10- 12 M. (c) 1.1 X 10- 7 M . (d) 1.0 X 10- 15 M. 16.23 pH < 7: [H +] > 1.0 X 10- 7 M, acidic; pH > 7: [H +] < 1.0 X 10- 7 M, basic; pH = 7: [H +] = 1.0 X 10- 7 M , neutral. 16.252.5 X 10- 5 M. 16.27 0.0022 g. 16.33 (a) - 0.0086. (b) 1.46. (c) 5.82.16.35 (a) 6.17 X 10- 5 M. (b) 2.82 X 10-4 M. (c) 0.105 M. 16.37 (a) pOH = -0.093, pH = 14.09. (b) pOH = 0.36, pH = 13.64. (c) pOH = 1.07, pH = 12.93.16.39 (a) 1.1 X 10- 3 M. (b) 5.5 X 10- 4 M. 16.47 (a) Strong. (b) Weak. (c) Weak. (d) Weak. (e) Strong. 16.492.59.16.515.17.16.534.80 X 10- 9.16.55 1.8 X 10- 3 M. 16.576.80. 16.61 6.97 X 10- 7. 16.63 11.98.16.65 (a) A has the smallest Kb value. (b) B- is the strongest base. 16.692.0 X 10- 5; 1.4 X 10- 11 ; 5.6 X 10- 10 ; 2.4 X 10- 8 16.71 (1) c. (2) band d. 16.73 1.40 (0.040 M HCI), 1.31 (0.040 M H 2S0 4), 16.75 1.0 X 10- 4 M , 1.0 X 10- 4 M, 4.8 X 10- 11 M. 16.77 1.00. 16.81 (a) H 2 S04 > H2Se04' (b) H 3P0 4 > H 3As04. 16.83 The conjugate bases are C 6 H sO - from phenol and CH 30 - from methanol. The C 6H 50 - is stabili zed by resonance:

o

o

0-

•

•

•

o

•

•

•

The CH 30 - ion has no such resonance stabilization. A more stable conjugate base means an increase in the strength of the acid. 16.89 (a) Neutral. (b) Basic. (c) Acidic. (d) Acidic. 16.91 HZ < HY < HX. 16.93 pH > 7 .16.95 4.82.16.975.39.16.101 (a) Al20 3 < BaO < K 20. (b) Cr03 < CrZ0 3 < CrO. 16.103 Al(OH)3(s) + OW(aq) • Al(OH)4 (aq), Lewis acid-base reaction. 16.107 AICl 3 (AI3+) is a Lewis acid, Clis a Lewis base. 16.109 CO 2 , S02, and BCI 3 (other answers are possible). 16.111 (a) Acid = AlBr3, base = Br- . (b) Acid = Cr, base = CO. (c) Acid 2 = Cu +, base = CN- . 16.113 (b) represents a strong acid. (c) represents a weak acid. (d) represents a very weak acid. 16.115 In theory, the products will be CH3COO - (aq) and HCI (aq) but this reaction will not occur to any measurable extent. 16.117 pH = 1.70, % ionization = 2.26%. 16.119 (c). 16.121 (a) For the forward reaction NH ~ and NH3 are the conjugate acid and base pair, respectively. For the reverse reaction NH3 and NH2 are the conjugate acid and base pair, respectively. (b) H + corresponds to NH!; OH- corresponds to NH 2 · For the neutral solution, [NH:] = [NH z ].

16.123 K = a

K = [W] = a

[W][A-]

[HA]

K

, [HA] "" 0.1 M, and [A ] "" 0.1 M. Therefore,

w and [OW] =

[OW]

_

K

w . 16.125 1.7 X 1010.

Ka

16.127 (a) H- (basel ) + H 20 (acid2 ) • OH- (basez) + H z (acid I) ' (b) H- is the reducing agent and H 20 is the oxidizing agent. 16.129 2.8 X 10- 2 16.131 PH 3 is a weaker base than NH 3. 16.133 (a) HN02 . (b) HF. (c) BF3. (d) NH3. (e) H 2 S0 3. (f) HC0 3 and COj-. The reactions for (f) are: HC0 3 (aq) + H +(aq) • COz(g) + H 20(l), CO~-(aq) + 2H+(aq) • COz(g) + H zO(I). 16.135 (a) trigo nal pyramidal. (b) H 40 2+ does not exist because the positively charged H30+ has no affinity to accept the positive H + ion. If H 40 2+ existed, it would have a tetrahedral geometry. 16.137 The equations are: Clz(g) + H 20(l) • • HCI(aq) + HCIO(aq), HCl(aq) + AgN0 3(aq). • AgCl(s) + HN0 3(aq). In the presence of OH- ions, the first equation is shifted to the • H 20. Therefore, the concentration right: H + (from HCl) + OHof HCIO increases. (The 'bleaching action' is due to CIO - ions.) 16.13911.80.16.141 (a) 0.181 (18.1 %). (b) 4.63 X 10- 3.16.1434.26. 16.1457.2 X 10- 3 g. 16.147 1.000. 16.149 (a) The pH of the solution ofHA would be lower. (b) The electrical conductance of the HA solution would be greater. (c) The rate of hydrogen evolution from the HA solution would be greater. 16.151 1.4 X 10- 4 16.153 2.7 X 10- 3 g. 16.155 (a)

+ H 2 0 (l) • NH 3(aq) + OW(aq), and N 3-(aq) + 3HzO(l) - _ . NH 3(aq) + 30H - (aq). (b) N 3- . 16.157 In inhaling the smelling NH2 (aq)

salt, some of the powder dissolves in the basic solution. The ammonium • NH3(aq) ions react with the base as follows: NH; (aq) + OH-(aq) + H 2 0. It is the pungent odor of ammonia that prevents a person from fainting. 16.159 (c). 16.16121 mL. 16.163 Mg. 16.165 Both NaF and SnF2 provide F - ions in solution. The F - ions replace OH- ions during the remineralization process 5Ca2+ + +F• CaS(P04)3F (fluorapatite). Because F - is a weaker base than OH- , fluorapatite is more resistant to attacks by acids compared to hydroxyapatite.

3Pol-

Chapter 17 17.5 (a) 2.57. (b) 4.44.17.9 (c) and (d). 17.11 8.89. 17.130.024.17.15 0.58.17.179.25 and 9.18.17.19 Na2A1NaHA. 17.21 (1) a, b, and d; (2) a (highest concentration). 17.27202 g!mol. 17.290.25 M. 17.31 (a) 1.10 X 102 g/mol. (b) 1.6 X 10- 6.17.335.82.17.35 (a) 2.87. (b) 4.56. (c) 5.34. (d) 8.78. (e) 12.10. 17.37 (a) Cresol red or /phenolphthalein. (b) Most of the indicators in Table 17.3 are suitable for a strong acid-strong base titration . Exceptions are thymol blue and. to a lesser extent, bromophenol blue and methyl orange. (c) Bromophenol blue, methyl orange, methyl

ANSWERS TO ODD-NUMBERED PROBLEMS

red, or chlorophenol blue. 17.39 Red. 17.41 (1) Diagram (c), (2) Diagram (b), (3) Diagram (d), (4) Diagram (a). 17.49 (a) 9.1 X 10 - 9 M. (b) 7.4 X 10 - 8 M. 17.511.8 X lO - /l. 17.533.3 X lO-93 17.559.52.17.57 Yes. 17.63 (a) 1.3 X lO- z M. (b) 2.2 X lO - 4 M. (c) 3.3 X lO-3 M. 17.65 (a) 1.0 X lO- s M. (b) 1.1 X 10- 10 M. 17.67 (b), (c), (d), and (e) 17.69 (a) l.6 X lO - z M. (b) 1.6 X lO - 6 M. 17.71 Fe(OH)z will precipitate. 17.73 [Cd2+ ] = 1.1 X lO - 18 M , [Cd(CN)~- ] = 4.2 X lO-3 M, [CW] = 0.48 M. 17.75 3.5 X lO- s M. 17.77 (a) The equations are as follows: Culz(s), • Cuz +(aq) + 2I - (aq), Cu2+ (aq) + 4NH 3(aq) • • [Cu(NH 3)4]2+(aq). The ammonia combines with the Cu2+ ions formed in the first step to form the complex ion [Cu(NH 3 )4f+, effectively removing the Cu z+ ions, causing the first equilibrium to shift to the right (resulting in more Culz dissolving). (b) Similar to part (a): AgBr(s) • • Ag+(aq) + Br- (aq), Ag+(aq) + 2CN - (aq) • • [Ag(CNhnaq). (c) Similar to parts (a) and (b): HgClz(s). • Hgz+(aq) + 2CI - (aq), Hg2+ (aq) + 4Cl -(aq)' • [HgCI 4 ]z- (aq). 17.81 At pH values between 2.68 and 8.11, Fe(OH)3 will precipitate but Zn(OH)z will not. 17.830.011 M. 17.85 Chloride ion will precipitate Ag + but not Cu2+ . So, dissolve some solid in H 2 0 and add HCI. If a precipitate forms, the salt was AgN03' A flame test will also work: Cu2+ gives a green flame test. 17.872.51 to 4.41. 17.89 1.3 M. 17.91 [Na+] = 0.0835 M, [HCOO-] = 0.0500 M , [OW] = 0.0335 M, [H+] = 3.0 X lO-13 M, [HCOOH] = 8.8 X 10- 11 M. 17.93 Most likely the increase in solubility is due to complex ion formation: Cd(OH)z(s) + 20H - (aq)' • Cd(OH)~- (aq). This is a Lewis acid-base reaction. 17.95 (d). 17.97 [Ag+] = 2.0 X lO-9 M , [Cn = 0.080 M, [Znz+] = 0.070 M, [N0 3 ] = 0.060 M .17.990.035 gIL. 17.101 2.4 X lO - l3. 17.103 1.0 X lO- s M. Ba(N03) is too soluble to be used for this purpose. 17.105 (a) AgBr precipitates first. (b) [Ag+] = 1.8 X lO- 7 M. (c) 0.0018 %.17.1073.0 X lO- 8 17.109 (a) H + + OH• HzO: K = l.0 X 10 14. (b) H + + NH3 • NH~ :

K =

1

K

=

•

1

5.6 X 10-10

= l.8 X 10 9

Broken into two equations: CH 3COOH

• CH 3COO - + H+: Ka

Broken into two equations: • CH 3COO- + H +: K. NH3 + H + --.+. NH~: 11 K~ CH 3COOH

s

K= K a _ 1.8x10- =3.2 x I04 K' 5.6 X 10- 10

•

17.111 (a) 500 mL of 0.40 M CH 3COOH mixed with 500 mL of 0.40 M CH 3COONa. (b) 500 mL of 0.80 M CH 3COOH mixed with 500 mL of 0.40 M NaOH. (c) 500 mL of 0.80 M CH 3COONa mixed with 500 mL of 0.40 M HCl. 17.113 (a) Increase. (b) No change. (c) No change. (d) pK. very large. 17.115 (a) Add sulfate. Na2S04 is soluble, BaS04 is not. (b) Add sulfide. K 2S is soluble, PbS is not. (c) Add iodide. ZnI2 is soluble, HgI2 is not. 17.117 The amphoteric oxides cannot be used to prepare buffer solutions because they are insoluble in water. 17.119 The ionized polyphenols have a dark color. In the presence of citric acid from lemon juice, the anions are converted to the lighter-colored acids. 17.121 Yes. 17.123 (c) . 17.125 Precipitation would be minimized by decreasing pH. 17.127 At pH = 1.0, the predominant species is +NH 3 -CH2-COOH. At pH = 7.0, the predominant species is +NH 3 -CH z-COO- . At pH = 12.0, the predominant species is NH2 - CHz-COO- . 17.129 (a) pH = 4.74. The pH of a buffer does not change upon dilution. (b) Before dilution, pH = 2.52; after dilution, pH = 3.02.17.1314.75.17.133 (a) The strongest acid group (with the lowest pK.) ionizes first, followed by the successively weaker acids. They are, in order: COOH: pK. = 1.82,

AP-' 3

NH +: pK. = 6.00, NHj : pKa = 9.17. (b) The dipolar ion is the product of the second ionization . (c) pi = 7.59. (d) The pair shown in the secon d ionization, since the pKa for that pair is closest to the required pH of 7.4.

Chapter 18 18.9 The probability that all the molecules will end up in the same fla sk (either the flask on the left or the fla sk on the right) is (a) 0.125. (b) 1.95 X 10- 3 (c) 1.24 X lO -6o. The probability that all the molecules will end up in one particular flask is (a) 0.063. (b) 9.8 X 10- 4 (c) 6.2 X lO-6 1. 18.11 (a) Negative. (b) Positive. (c) Positive. (d) Negative. 18.15 (c) < (d) < (e) < (a) < (b). 18.17 (a) 47.5 JI K . mol. (b) - 12.51 JI K . mol. (c) -242 .8 JI K· mol. 18.23 (a) -1139 kJI K . mol. (b) - 140.0 kJI K . mo!. (c) - 2935.0 kJ/K . mol. 18.25 (a) Spontaneous at all temperatures. (b) Spontaneous below 111 K. 18.27 Fusion: 1.00 X 102 JI K . mol, vaporization: 93 .6 JI K . mol. 18.29 -226.6 kJ/mol. 18.31 75 .9 kJ of Gibbs free energy released. 18.350.35. 18.3779 kJ/mo!. 18.39 (a) 39 kJ/mol, 1 X lO-7. (b) 48 kJ/mol. 18.41 (a) 1.6 X lO-23 atm. (b) 0.535 atm. 18.4323.6 rrunHg. 18.4793. 18.49 When Humpty broke into pieces, he became more disordered (spontaneously). Humpty couldn't be put together again because all the King's horses and all the King's men could not reverse the spontaneous process. (Too great a negative entropy change would have been required.) 18.51 E and H. 18.53 42°C. 18.55 (a) t:.H is positive, t:.S is positive, t:.G is negative. (b) t:.H is positive, t:.S is positive, t:.G is zero. (c) t:.H is positive, t:.S is positive, t:.G is positive. 18.57 t:.S is positive. 18.59 (a) Trouton's rule is a statement about t:.S~.P (See Equation 18.9). In most substances, the molecules are in constant and random motion in both the liquid and gas phases, so t:.S ~ap = 90 J/mol·K. (b) But in ethanol and water, there is less randomness of the molecules due to the network of H -bonds, so t:.S~ap is greater. 18.61 • 2CO z(g) + N z(g) . (b) CO is the reducing (a) 2CO(g) + 2NO(g) agent. NO is the oxidizing agent. (c) 3 X lO 120. (d) 1.2 X lO1 4, reaction proceeds to the right. (e) No. 18.632.6 X 10- 9 18.65 703°C. 18.6738 kJ . 18.69174 kJ/mol. 18.71 (a) Positive. (b) Negative. (c) Positive. (d) Positive. 18.73625 K. We assume that t:.Ho and t:.So do not depend on temperature. 18.75 No. A negative t:.Go tells us that a reaction has the potential to happen, but gives no indication of the rate. 18.77 (a) - lO6.4 kJ/mol, 4 X 10 18 (b) - 53.2 kJ/mol, 2 X lO9. 18.79 Talking involves various biological processes (to provide the necessary energy) that lead to an increase in the entropy of the universe. Since the overall process (talking) is spontaneous, the entropy of the uni verse must increase. 18.81 (a) 86.7 kJ/mol. (b) 4 X 10- 31. (c) 3 X lO-7 (d) Lightning supplies the energy necessary to drive thi s reaction, converting the two most abundant gases in the atmosphere into NO(g). The NO gas dissolves in the rain, which carries it into the soil where it is converted into nitrate and nitrite by bacterial action. This "fixed" nitrogen is a necessary nutrient for plants. 18.83 673.2 K.18.85 (a) 7.6 X lO I4 (b) K = 4.1 X 10- 12 18.87 (a) Disproportionation redox. (b) 8.2 X lOIS. (c) Less effective. 18.89 1.8 X 10 70. 18.91 t:.Ssys = 91.1 J/K, t:.Ssurr = -9 1.1 J/K, t:.Suni v = O. Conclusion: the system is at equilibrium. 18.93 t:.G = 8.5 kJ/mol. 18.95 (a) CH 3COOH, t:.Go = 27 kJ/mol, CH2CICOOH, t:.G o = 16 kJ/mo!. (b) Entropy dominates. (c) Breaking and making of specific 0- H bonds. Other contributions include solvent separation and ion solvation. (d) The CH3COO - ion, which is smaller than the CHzClCOO - ion, can participate in hydration to a greate~ extent, leading to more ordered solutions. 18.97 Xeo, = 0.55 Xeo = 0.45 . 18.99249 J/K. 18.101 3 X - 13 s. 18.103 t:.Ssys = - 327 JI K . mol, t:.Ssurr = 1918 JI K . mol, t:.SlIni v 10 = 1591 JI K . mol. 18.105 q and ware not state functions. 18.107 t:.G, t:.S, and t:.H are all negati ve. 18.109 (a) S = 5.76 J/K (mol. (b) The fact that the actual residual entropy is 4.2 J/K(mol means that the orientation is not totally random. 18.111 t:,.H0 = 33.89 kJ/mol, t:.so = 96.4 J/mol'K, t:.Go = 5.2 kJ/mo!.

Chapter 19 19.1 (a) 2H+ + HzO z + 2Fe2+ • 2Fe3+ + 2H20. (b ) 6H+ + 2HN0 3 + 3Cu • 3Cu2+ + 2NO + 4H 2 0. (c) H 20 + 2Mn0 4 + 3CN•

AP-14

ANSWERS TO ODD-NUMBERED PROBLEMS

2Mn02 + 3CNO- + 20H - . (d) 60H - + 3Br2 • Br0 3 + 3H2 0 + 5Br- . (e) 2SzO~- + 12 • S4 0 ~- + 21- . 19.11 3Ag+(l.0 M) + Al(s) - _ . 3Ag(s) + AlH ( l.O M), E~ell = 2.46 V. 19.13 CI2 (g) and Mn0 4 (aq) . 19.15 (a) Spontaneous. (b) Not spontaneous. (c) Not spontaneous. (d) Spontaneous. 19.17 (a) Li. (b) H 2. (c) Fe2+ . (d) Br- . 19.213 X 1054 19.23 (a) 2 X 10 18 (b) 3 X 108. (c) 3 X 1062 . 19.25 - 81 kJ, 2 X 10 14 19.29 1.09 V. 19.31 E~ell = 0.76 V; E eel! = 0.78 V. 19.33 6.9 X 10- 38 19.39 l.09 V. 19.43 12.2 g Mg. 19.45 It is less expensive to prepare 1 ton of sodi um by electrolysis. 19.470.012 F. 19.495.33 g Cu, 13.4 g Br2' 19.517.70 X 103 C. 19.53 l. 84 kg/h. 19.5563.3 g/mol. 19.5727.0 g/mol. 19.63 (a) (i) H 2(g) • 2H +(aq) + 2e - ; Ni2+(aq) + 2e• Ni(s); (ii) Hz(g) + N i2+ (aq) • 2H +(aq) + Ni(s); (iii) Reaction will proceed to the left. (b) (i) 2Cqaq) • Clz(g) + 2e-; 5e- + 8H +(aq) + Mn0 4 (aq) • Mn2+(aq) + 4H2 0(l); (ii) l6H +(aq) + 2Mn0 4 (a q) + lOCqaq) • 2Mn2+(aq) + 8H 20 (l) + 5CI2(g); (iii) Reaction will proceed to the right. (c) (i) Cr(s) • CrH(aq) + 3e - ; Zn2+ (a q) + 2e • Zn(s); (ii) 2Cr(s) + 3Zn2+(aq) • 2Cr3+(aq) + 3Zn(s), (iii) 2CrH (aq) + 3Zn(s) • 2Cr(s) + 3Zn2+(aq). Reaction will proceed to the left. 19.650.00944 g S02' 19.67 (a) 2Mn0 4 + 6H+ + 5H20 2 • 2Mn2+ + 8H20 + 502, (b) 0.0602 M. 19.690.232 mg Ca/mL blood. 19.715 X 10- 13 . 19.73 (a) 3.14 V. (b) 3.13 V. 19.75 E eeu = 0.035 V. 19.77 Mercury(I) is Hg;+. 19.79 [Mg2+] = 0.0500 M, and Mg(s) + -= l.44 g. [Ag ] = 7 X 10-» M. 19.81 (a) Hydrogen gas, 0.206 L. (b) 6.09 X 1023 e- /mol e- . 19.83 (a) -1356.8 kJ/mol. (b) 1.17 V. 19.85 +3 .19.876.8 kJ/mol , 0.064.19.891.4 A. 19.91 +4.19.93 l.60 X 10- 19 Cle - . 19.95 Cells of higher voltage require very reactive oxidizing and red ucing agents, which are difficult to handle. Batteries made up of several cells in series are easier to use. 19.972 X 1020 19.99 (a) E~ed for X is negative ( - 0.25 V), E~ed for Y is positive (+0.34 V). (b) E~ll = 0.59 V. 19.101 (a) Gold does not tarnish in air because the reduction potential for oxygen is not sufficiently positive to result in the oxidation of gold. (b) Yes, E~ell = 0.19 V. (c) 2Au(s) + 3F2(g) • 2AuF3(aq) . 19.103 [Fe2+] = 0.0920 M, [FeH ] = 0.0680 M. 19.105 The two half reactions are: H 20 2 (aq) + 2H +(aq) + 2e • 2H20(l), 1.77 V; H 20 2(aq) • Oz(g) + 2H +(aq) + 2e - , -0.68 V. Overall: 2H 20z(aq) • 2H 20 (l) + 0 2(g), EO= 1.09 V (spontaneous) . 19.107 (a) Unchanged. (b) Unchanged. (c) Squared. (d) Doubled. (e) Doubled. 19.109 As [H+] increases, F 2(g) does become a stronger oxidizing agent. 19.111 4.4 X 102 atm. 19.113 (a) Half-reactions: 1120 2(g) + 2e • 2 0 - (aq) , Zn(s) • Zn2+(aq) + 2e - . Overall: Zn(s) + 1I20z(g) • ZnO (s), E~ell = 1.65 V. (b) 1.63 V. (c) 4.87 X 103 kJ/kg Zn. (d) 64 L of air. 19.115 -3.05 V. 19.117 1 X 10- 14 19.119 (a) 3600 C. (b) 105 A·h. (c) E~ell = 2.01 V, t1Go = -388 kJ/mol. 19.121 $217 .19.123 -0.037 V. 19.125 2 X 10 37

Chapter 20 14 20.5 (a) ~~ Na. (b) : H. (c) 6 n. (d) ~~ Fe. (e) __~f3. 20.13 2.72 X 10 glcm 3 . 20.15 (a) Nl. (b) Se. (c) Cd. 20.174.85 X 10 L kg. 20.19 (a) 6.30 X 10- 12 J, 9.00 X 10- 13 J/nucleon. (b) 4.78 X 10- 11 J, 1.37 X 10- 12 J/nucleon. 20.23 (a) 2~~ Th " ) 2~~ Ra ~) 2~~ Ac ~) 2~g Th. (b) 2~~U

" ) 23 1Th 90

~

) 231 Pa 91

" ) 2~~ Ac.

Np ") 233 Pa ~) 233 U " ) 229 Th (c) 237 93 91 92 90' 20.254.89 X 10 19 atoms. 20.273.09 X 103 yr. 20.29 A = 0 mol, B = 0.25 mol, C = 0, D = 0.75 mol. 20.31 5.5 dpm. 20.33 Mass ratio UlPb = 43.3:1. 20.37 (a) 14N(ex,p)17 0. (b) 9Be (ex,n) 12 c. (c) 238U(d,2n)238Np. 20.39 (a) 4oCa(d,p)4ICa. (b) 32S(n,pi2p' (c) 239Pu (ex,n)242 Cm. ) 1§6 Hg ) I j~ Au + :p. 20.53 Thefact that 20.41 1§~ Hg + 6n the radioi sotope appears only in the 12 shows that the 10 3 is formed only from the 10 4' 20.55 Add iron-59 to the person's diet, and allow a few days for the iron-59 isotope to be incorporated into the person's body. Isolate red blood cells from a blood sample and monitor radioactivity from the hemoglobin molecules present in the red blood cells. 20.61 3.96 X 10 15 20.63 65 .3 yr. 20.65 70.5 dpm . 2067 (a) 235 U+0 In -7 14°6 Ba + 3 In + 36 93Kr ' • 92 5 0

+ 0In -7 144CS + 90 55 37 Rb + 2 0In . U + In -7 87Br + 146 La + 3 In (c) 235 92 0 35 57 0 . U (b) 235 92

U + 0In -7 160 72 Zn + 4 0In . (d) 235 92 62 Sm + 30 20.69 (a) ~ H -7 _~ f3 + ~ He (b) 2~~ Pu -7 i ex + 2~~U 0 (.) + I~I Xe (c) 135311-7 -11-' )4 (d) 251 98 Cf -7 42ex + 247 96 Cm .

20.71 Because both Ca and Sr belong to Group 2A, radioactive strontium that has been ingested into the human body becomes concentrated in bones (replacing Ca) and can damage blood cell production. 20.73 Normally the human body concentrates iodine in the thyroid gland. The purpose of the large doses of KI is to displace radioactive iodine from the thyroid and allow its excretion from the body. 20.75 (a) 2~~ Bi + i ex

) 2~~ At + 2 6n.

Bi(ex 2n) 211 At ' (b) 209 83' 85 20.77 2.77 X 103 yr. 20.790.069%.20.81 (a) 5.59 X 10- 15 J and 2.84 X 10- 13 J. (b) 0.024 mol. (c) 4.26 X 106 kJ. 20.83 2.8 X 10 14 .20.856.1 X 1023 atoms/mol. 20.87 (a) 1.73 X 10- 12 J. (b) The a particle will move away faster because it is smaller. 20.89 U-238, t1/2 = 4.5 X 109 yr and Th-232, t ll2 = 1.4 X 1010 yr. They are still present because of their long half lives. 20.91 8.3 X 10- 4 nm. 20.93 3H. 20.95 A small scale chain reaction (fission of 235 U ). 20.97 2.1 X 102 g/mol. 20.99 (a) r = roA 113 (ro is a proportionality constant). (b) 1.7 X 10- 42 m 3 20.101 0.49 rem.

Chapter 21 21.5 X = 3.30 X 10- 4, ppm = 330.21.7 In the stratosphere, the air temperature ri ses with altitude. This warming effect is the result of exothermic reactions triggered by UV radiation from the sun. 21.11 260 nm. 21.21 3.2 X 10 12 kg 0 3, 4.0 X 1037 molecules 0 3, 21.23 CC4 + HF • HCl + CFCl3 (Freon-II), CFCl 3 + HF • HCl + CF2Cl 2 (Freon-12). 21.25479 kJ/mol, this is sufficient to break the C-Cl bond, but not enough to break the C- F bond. + .. :Cl-O-N-O: o.

••

..

..

II

..

• ••

••

4

21.27 : 0 : , .. (NH 2hCO(s) + H 20 (I) . The reaction should be run at high pressure. 24.43 The oxidation state of N in nitric acid is + 5, the highest oxidation state for N. N can be easily reduced to ox idation state + 3. 24.45 (a) NH 4 N0 3(s) • N 20(g) + 2H 20(I). (b) 2KN0 3(s) • 2KN0 2(s) + 02(g) . (c) • PbO(s) + 2N0 2 (g) + OzCg) . 24.47 KN0 3 (s) + Pb(N0 3M s) C(s) • KN0 2(s) + CO(g), 48.0 g KN0 2. 24.49 (a) D.G~ = 31 86.7 kJ/mol. (b) K = Kp = Kc = 4 X 10- . 24.51 125 g/mol, P4 . 24.53 • 2N 2 0 S + 4HP0 3, 60.4 g. 24.55 Sp 3 24.63 4HN0 3 + P4 0 lO 34 - 198.3 kJ/mol , K = Kp = Kc = 6 X 10 24.65 (a) To exclude light. (b) 0.371 L. 24.67 F = - 1, 0= 0.24.69 (a) HCOOH(l) . • CO (g) + H 20(I). (b) 4H 3POil) . • P40 IO (S) + 6H20 (l) . (c) 2HN0 3(l) +.==' N 20 S(g) + H 20 (I). (d) 2HCl0 3 (l) • • CI20 s(l) + H 20 (l) . 24.71 To form OF6 there would have to be six bonds (twelve electrons) around the oxygen atom. This would violate the octet rule. 24.73 35 g C12 . 24.75 9H 2SOiaq) + 8Nal(aq) • 412(s) + H 2S(g) + 8NaHS0 4 (aq) + 4H 20(l). 24.79 (a) H-f.:----H- f.: (b) W----H- --:f.f 24.81 (a) Linear. (b) Tetrahedral. (c) Trigonal bipyramidal. (d) See-saw. 24.83 25 .3 L C12. 24.852.81 L. 24.87 120 5(s) + 5CO(g) • 5C0 2(g ) + 12(s), iodine is • H 3POiaq) + reduced and carbon is oxidized. 24.89 (a) 2H 3P0 3(aq) PH 3(g) + 0 2(g) . (b) Li4 C(s) + 4HCI(aq) • 4LiCl(aq) + CH 4 (g). (c) 2HI(g) + 2HN0 2(aq) , 12 (s) + 2NO (g) + 2H 20 (I). (d) H 2S(g) + 2CI 2 (g) • 2HCI(g) + SCI 2(1) . 24.91 (a) SiC14 . (b) F-. (c) F. (d) CO 2, 24.93 O. 24.95 PC!; , tetrahedral , Sp3 hybrids; PC1 6 , octahedra!, sp 3d2 hybrids. 24.97 K 298 = 9.61 X 10- 22 , Km = 1.2 X 10- 15 24.99 The glass is etched (dissolved) uniformly by the reaction 6HF(aq) + Si0 2(s) ----+. H 2SiF 6(aq) + 2H20 (l) . 24.101 1.18.24.1030. 833 gIL. The molar mass derived from the observed density is 74.41, which suggests that the molecules are associated to some extent in the gas phase. This makes sense due to strong hydrogen bonding in HF.

Chapter 25 25.3 The monomer must have a triple bond. 25.5 There are two possible polymers, but if they are long enough, the difference would be negligible:

°II

H?N-CH-C-O

-

I

°II

H

°II

H

N-CH-C-O-N-CH-C-O

I

I

°

H II N-CH-C-OH

I

CH?

CH-OH

CH 2

CH-OH

SH

CH 3

SH

CH 3

I -

I

I

I

II

°II

HN-CH-C-O 2

I

°

°

H II H II N-CH-C-O-N-CH-C-O

°

H II N -CH-C-OH

I

CH-OH

I CH? I -

I CH-OH I

CH 2

CH 3

SH

CH 3

SH

I

I

II

25.9 (1) Sc(s) + 2C2H sOH(l) • Sc(OC 2H s)(alc) + 2H+(alc) ("alc" indicates a solution in alcohol); (2) Sc(OC2Hs)(alc) + 2H 20(I) • Sc(OHMs) + 2C 2H sOH(alc); (3) Sc(OHM s) • ScO(s) + 2H 20(g). 25.11 Bakelite is best described as a thermosetting composite polymer. 25.15 No. These polymers are too flexible, and liquid crystals require long, relati vely rigid molecules. 25.19 Alternating condensation copolymer of the polyester class. 25.21 Metal amalgams expand with age; composite fillings tend to shrink. 25.25 sp2 25.27 Dispersion forces. 25.31 (a) 4+ 5: n-type. (b) 4+ 3: p-type. 25.35 Bi2Sr2Cu06' 25.37 Two are +2 ([Ar]3£), one is + 3 ( [Ar]3d 8) . The +3 oxidation state is unusual

° ° II II HO-C-fCH:m--C-OH

for copper. 25.39 H2N-fCH2)8 NH2 25.41 In a plastic (organic) polymer: covalent, disulfide (covalent), H-bonds and dispersion forces; in ceramics, mostly ionic and network covalent. 25.43 Fluoroapatite is less soluble than hydroxyapatite, particularly in acidic solutions. Dental fillings must also be insoluble. 25.45 The molecule is long, flat, and rigid, so it should form a liquid crystal.

Answers To

PRE-PROFESSIONAL PRACTICE EXAM PROBLEMS

Chapter 1

Chapter 8

Chapter 15

Chapter 22

I.b 2.d 3.e 4.a

I.e 2.a 3. a 4. b

I.b 2.a 3.b 4.d

I.e 2. a 3. b 4. b

Chapter 2

Chapter 9

Chapter 16

Chapter 23

I.e 2.d 3. a 4. b

I.e 2. b 3. b 4. d

I.d 2. b 3. a 4. b

1. e 2.d 3. b 4.a

Chapter 3

Chapter 10

Chapter 17

Chapter 24

l.d 2.b 3.b 4.e

I.e 2.d 3.a 4.e

I.e 2. e 3. a 4.d

La 2. b 3.e 4. b

Chapter 4

Chapter 11

Chapter 18

Chapter 25

La 2. b 3.a 4.e

Lb 2.a 3. b 4.d

Ld 2. b 3.a 4.e

Lb 2.d 3. b 4.d

Chapter 5

Chapter 12

Chapter 19

Ld 2.a 3.e 4. b

I.e 2. b 3.d 4.a

Lb 2.e 3.d 4.d

Chapter 6

Chapter 13

Chapter 20

I.b 2.a 3.e 4.e

Lb 2.d 3.e 4.e

La 2.e 3.d 4.a

Chapter 7

Chapter 14

Chapter 21

I.b 2.a 3. b 4.d

La 2. b 3.e 4. b

Lb 2.d 3.e 4.a

AP-17

Index A absolute entropies, 735, 736 absolute temperature scale, 424 absolute zero, 10, 424 absorbed, 531 acceptor impurities, 890 accuracy, 18-19 acid-base indicators, 698-700 acid-base reactions, 121-122, 125-126 acid-base neutralization, 124 Br¢nsted acids and bases, 122-124 strong acids and bases, 122 acid-base titrations, 144-146, 690-691 acid-base indicators, 698-700 strong acid-strong base titrations, 691-692 strong acid-weak base titrations, 696-698 weak acid-strong base titrations, 693-696 acid ionization constant, 648 acid rain, 838, 843-845 acids in molecular compounds, 50, 51 oxoacids, 59 strong, 112, 113, 122 acids and bases, 634-635 acid-base properties of oxides and hydroxides, 667- 668 acid-base properties of salt solutions, 662-667 acid-base properties of water, 637-639 Br¢nsted, 636-637 conjugate acid-base pairs, 654-657 diprotic and polyprotic acids, 657-659 Lewis acids and bases, 668-670 molecular structure and acid strength, 660-662 pH scale, 639- 644 strong, 644-647 weak acids and acid ionization constants, 647-652 weak bases and base ionization constants, 652-654 actinide series, 225 action potential, 776 activated complex, 564 activation energy, 563 active metals, 130, 258 active site, 575 active transport, 643 activity series, 130, 131 actual yield, 96 addition polymerization, 936

addition polymers, 395-396, 397, 936-941 addition reactions, 386-388 adenosine diphosphate (ADP), 389 adenosine triphosphate (ATP), 389 adhesion, 468 adsorbed, 531 aerosol, 845 Agriculture, U.S. Department of, 159 airbag,433 alcohols, 367, 370, 372 aldehydes, 367, 370, 373 aliphatic compounds, 365 alkali metals, 45, 891 - 893 alkaline batteries, 777 alkaline earth metals, 45, 893-895 alkanes, 51, 52, 365, 368 alkyl group, 366, 367, 368 alkynes, 943 allotropes, 48, 910 alloys, 883 alpha (a) particles, 38 alpha (a) rays, 38 altitude sickness, 448 aluminum, 895-897 alums, 897 amalgam defined, 883 dental, 759, 950 amide group, 367, 370 amide ions, 912 amide linkages, 464 amides, 367, 370, 373 amines, 367, 370, 373 amino acid residue, 464 amino acids, 370, 397-399 amino group, 367, 370 ammonia, 912 amorphous solids, 484 amount-volume relationship, 425-426 amphoteric, 266, 637 amplitude, 194 analogues, 363 angular momentum quantum number, 213 , 214 anions, 55 anisotropic, 947 Annan, Kofi, 829 anode, 36, 763, 766 antacids, 642- 643 antibonding molecular orbital, 343 aqueous solutions, 506 acid-base reactions, 121-126

aqueous defined, 77 aqueous reactions and chemical analysis, 142-147 concentration of solutions, 136-142 general properties of, 112-116 oxidation-reduction reactions, 126- 136 precipitation reactions, 117-121 Aristotle, 34 Armstrong, Lance, 73 aromatic compounds, 365 Arrhenius acid, 123 Arrhenius base, 123 Arrhenius equation, 564-565 Arrhenius, Svante, 122 arsenic, 905 artificial joints, 951- 952 ascorbic acid, 33, 635, 671 Aston, F. w., 43 atactic, 941 atmosphere Earth 's, 830-833 phenomena in the outer layers, 833- 834 atmospheres (atm), 418 atmospheric pressure, 417 atomic bomb, 812 atomic ions, 55-56 atomic line spectra, 201-202 atomic mass, 46 atomic mass unit (amu), 46 atomic number (Z), 40, 239 atomic orbitals, 212, 216, 219 d orbitals and other higher-energy orbitals. 217-218 energies of orbitals, 218, 219-220 forbitals,218 hybridization of, 327-334, 351 P orbitals, 217 s orbitals, 216-217 atomic radius, 246-247, 255 atomic theory, 34-36 atomic weight, 46 atoms, 4 atomic mass scale and average atomic mass.. 46-47 atomic number, mass number, and isOlope5_ 40-43 atomic theory, 34-36 defined, 36 hydrogen atom, 200- 208, 212 structure of, 36-40 attracti on, 160

1-2

INDEX

attractive forces, 245, 457 Aufbau principle, 221 aurora australis, 833 aurora borealis, 833 autoionization of water, 638 average reaction rates, 544-546 Avogadro, Amedeo, 425 Avogadro's law, 425-426, 440 Avogadro's number, 82 Axel, Richard, 313 axial, 316

•

B balanced equations, 77-78 ball-and-stick models (molecular art), 5 Balmer, Johann, 202 band theory, 888 band theory of conductivity, 888 conductors, 888-889 semiconductors, 889-890 barometer, 418, 419 base Br0nsted acids and bases, 122-124 defined, 112 strong acids and bases, 122 base ionization constants, 652 basic oxygen process, 885-886 batteries defined, 777 dry cells and alkaline, 777 fuel cells, 778-779 lead storage, 777-778 lithium-ion, 778 Becquerel, Antoine, 38 belt of stability, 800 bends, the, 438 beta-particle emission, 800-801 beta (f3) particles, 39 beta (f3) rays, 39 bidentate ligands, 861 bimolecular, 569 binary hydrides, 907-908 binary molecular compounds, 49, 51 biological catalysts, 575-577 biological concentration cells, 776 biological polymers, 397-401 biomedical materials, 949-952 bioterrorism agents, 3, 23 bipolar disorder, 267 birth defects, 363 blackbody radiation, 197 blast furnace, 884, 885 blood, maintaining pH of, 689-690 blood alcohol concentration (BAC), 111 blood doping, 591 blood plasma, 689, 690 body-centered cubic cell, 474 Bohr, Niels, 202 . Bohr's theory of the hydrogen atom, 200-201, 208,212 atomic line spectra, 201-202 line spectrum of hydrogen, 202-207 boiling point, 485 boiling-point elevation, 519-520 Boltzmann constant, 729

bomb calorimetry, 175 bond angle, 316, 319 bond dipoles, 32 1, 322 bond enthalpy, 286, 300-303, 324 bond length, 285 bond order, 343-344 bonding molecular orbital, 342 bonding theories, 351-353 bone, radioactive, ?65 Born-Haber cycle, 282-283 boron neutron capture therapy (BNCT), 797 Boyle, Robert, 421 Boyle's law, 421-423, 440 Bragg equation, 476 Breathalyzer test, Ill, 148 breeder reactors, 814 bromine, 927 Br0nsted acids and bases, 123, 636-637 Br0nsted base, 123 Br0nsted, Johannes, 123 Buck, Linda, 313 buckyballs, 953 buffer, 683 buffer solutions, 683 with a specific pH, 688-689 calculating the pH of a buffer, 684-688 buret, 8, 9 bums, 489 burst lung, 415

c Cade, John, 267 Cade, Mary, 115 Cade, Robert, 114-115 caffeine, 653 calcium, 894-895 calcium ascorbate, 671 calories caloric content of food, 159 calorie (cal) defined, 162 Calorie (Cal) defined, 163 calorimeter, l7 5 calorimetry bomb (constant-volume), l75-177 constant-pressure, l72-l74 defined, 171 specific heat and heat capacity, 171- 172 cancer drugs, 73 nuclear chemistry in treating, 797 smoking and, 819-820 capillary action, 468 carbides, 910 carbocation, 387 carbohydrates, 80 carbon, 364-365, 910-911 carbon-14, 563 , 806 carbon dating, 806 carbon dioxide, 838-842, 848, 911 carbon disulfide, 921 carbon monoxide, 348-349, 848, 911 carbon nanotubes, 953 carbonyl group, 367, 370 carborundum, 910 carboxy group, 367, 370

carboxylic acids, 367, 370, 372, 662 Carothers, Wallace, 397 cast iron, 885 catalysis, 573-577 , 616-617 catalyst, 573 Category A agents, 3, 23, 298 catenation, 365,910 cathode, 36, 763, 766 cathode ray tube, 36-37 cathodic protection, 785 cations, 55 , 56, 714 cell potential, 764 Celsius scale, 10-11 , 424 cementite, 886 Centers for Disease Control and Prevention (CDC), 3, 23 central science, 4 ceramic matrix composites, 946 ceramics, 945-946 Chadwick, James, 40 chain reactions, 396 chalco gens, 45 Charles, Jacques, 423 Charles's and Guy-Lussac 's law, 423-425 Charles's law, 424, 440, 441 Chauvin, Yves, 935 chelating agents, 861 chelation therapy, 857, 873 chemical analysis, 873 chemical bonding, basic concepts, 276-277 bond enthalpy, 300-303 Born-Haber cycle, 282-284 comparison of ionic and covalent compounds, 286 covalent bonding, 284-286 drawing Lewis structures, 291-292 e1ectronegativity and polarity, 286-290 exceptions to the octet rule, 296-300 ionic bonding, 279-284 lattice energy, 280-281 Lewis dot symbols, 278-279 Lewis structures, 284-285 Lewis structures and formal charge, 292-295 multiple bonds, 285-286 resonance, 295-296 chemical bonds, 47 chemical change, 15 chemical energy, 160 chemical equations, 76 balancing, 77-81 calculations with balanced, 89-91 interpreting and writing, 76-77 chemical equilibrium, 113, 590-591 concept of equilibrium, 592-594 equilibrium constant, 594-598 equilibrium expressions, 599-605 equilibrium expressions in problem solving, 606-611 factors that affect, 611-6l7 free energy and, 742-746 chemical formulas, 48 chemical kinetics, 542-543 catalysis, 573-577 dependence of reactant concentration on time, 555-562

INDEX

dependence of reaction rate on reactant concentration, 551-555 dependence of reaction rate on temperature, 562- 568 reaction mechanisms, 568-573 reaction rates, 544-550 chemical properties, 15 chemical reactions, 4, 7 comparison of nuclear reactions and, 798 energy changes in, 160-161 chemical reduction , 884 chemistry defined, 4 study of, 4-6 chemotherapy, 73, 873 chiral, 383 chiral switching, 385 chloralkali process, 923 chlorine, 926-927 chlorofluorocarbons (CFCs), 829, 835, 837-838, 841 , 842 cholesteric, 947 cidofovir, 23 cis isomers, 337, 382, 865-866 cisplatin, 73 Clausius-Clapeyron equation, 470 closed system, 163, 164 closest packing, 475, 478-479 coal, 910 cohesion, 468 coinage metals, 265 colligative properties, 517 collision theory, 562-564 colloids, 530- 532 color, 868-869 combination reactions, 98, 133, 134 combustion analysis of compounds, 87-88 determination of empirical formula, 87 determination of molecular formula, 88 combustion reactions, 98, 134 common ion effect, 682-683, 705-7l0 complex ion, 710 complex ion formation, 710-712 composite materials, 946 compounds, 6-7. See also inorganic compounds ; organic compounds insoluble, 118 ionic, 55 , 57-61 , 116, 117-119 molecular, 47, 49- 51 , 116,417 percent composition of, 75-76 soluble, 118 compressibility, gas, 440 concentration cells, 775-776 concentration of solutions concentration defined, 136 dilution, 137, 140 molarity, 136-137, 138-139 preparing a solution from a soli d, 138-139 solution stoichiometry, 141-142 condensation, 470 condensation polymers, 396-397, 941-944 condensation reactions, 396 condensed structural formulas , 376 condensed structures, 376 conducting polymers, 943-944 conduction band, 954

conductivity, 113-114 conductors, 888-889 conjugate acid, 636 conjugate acid-base pairs, 654-657 conjugate base, 636 conjugate pair, 636 constant-pressure calorimetry, 172- 174 constant-volume (bomb) calorimetry, 175-177 constant-volume calorimetry, 175-177 constituent elements, 7 constitutional isomerism, 382 constructive interference, 195, 196 conversion factors, 20 coordinate covalent bonds, 297, 858 coordination chemistry, 856-857 applications of coordination compounds, 873-874 coordination compounds, 858-864 coordination compounds, defined, 858 crystal field theory, 867-872 naming coordination compounds, 862-864 reactions of coordination compounds, 872-873 structure of coordination compounds, 864-867 coordination numbers, 473, 861 copolymers, 396, 939 copper, 898 core electrons, 244 con'osion, 784-785 coulomb (C), 289 Coulomb's law, 245, 386, 800 covalent bond, 284 types of covalent bonds, 285-286 covalent bonding, 284 covalent compounds, 286, 925 covalent crystals, 482, 483 covalent hydrides, 908 covalent radius, 246 cowpox, 3 crenation, 524-525 critical mass, 812 critical pressure, 486, 487 critical temperature, 486, 487 cross-links, 937 Crutzen, Paul, 829 crystal field splitting, 867, 868 crystal field theory, 867 color, 868-869 crystal field splitting in octahedral complexes, 867-868 magnetic properties, 869-871 tetrahedral and square-planar complexes, 871 crystal structure, 472 closest packing, 475, 478-479 packing spheres, 473-475 structures of crystals, 476-477 types of crystals, 480-484 unit cells, 472-473 crystalline solid, 472, 507 cubic cells, 474-475 cubic close-packed (ccp) structure, 478 Cunningham, Orville, 438 curie (Ci), 818, 847 Curie, Marie, 38 cyanides, 911

D d-block elements, ions of, 253-254 d orbitals, 330- 334 Dalton, John, 34-36 Dalton's law of partial pressures, 434-435, 440,442 dative bond, 297 Davisson, Clinton Joseph, 210 de Broglie hypothesis, 208- 210 de Broglie wavelength, 209 Debye (D), 289 decomposition reactions, 98, 133- 134 decompression sickness (DCS), 438 definite proportions, law of, 35 degenerate, 221 delocalized bonding, 351-353 delocalized bonds, 352 delocalized electrons, 944 demineralization, 681 Democritus, 34, 35 density, 12- 13, 15 dental implants, 950 dental pain, electrochemistry of, 759 deoxyribonucleic acid (DNA), 400, 401 deposition, 488 derived units, 12-13 destructive interference, 195, 196,208 detergents, 874 dextrorotatory isomer, 384, 866 dialysis, 505 diamagnetic, 342 diatomic molecules, 48 diffraction , electron, 210 diffusion, 444 dilution, 137, 140 dimensional analysis, 20 diodes, 955 dipole-dipole interactions, 462-463 dipole moment, 289-290 dipoles, bond, 321, 322 diprotic acids, 123,657-659 directionality, chemical bonds and, 325 dispersion forces, 465-467 displacement reactions, 130 disproportionation reaction, 134, 917 dissociation, 112, 523, 529 distillation, 887 distribution, 729 diving, 415, 438, 448 donor atoms, 860 donor impurities, 890 doping, 889, 955 double bonds, 285, 336, 338 double-sli experiment, nature of light and, 195-196 Douglas, Dwayne, 114-115 drugs chiral switching in, 385-386 organic chemistry and, 363 single-isomer versions, 385- 386 drunk driving, III dry cells, 777 dynamic chemical equilibrium, 113 dynamic equilibrium, 470

1-3

1-4

INDEX

E Earth alkaline earth metals, 45, 893-895 atmosphere, 830-833 elements on, 45 effective collision, 562 effective nuclear charge (Zeff)' 244 effusion, 444, 445 • Einstein, Albert, 193, 198, 199,802,8 10 elastomers, 937 electrocardiogram (EeG), 776 electrochemistry, 758-759 balancing redox reactions, 760- 762 batteries, 777-779 corrosion, 784-785 electrolysis, 780-784 galvanic cells, 763-764 spontaneity of redox reactions under conditions other than standard state, 773-777 spontaneity of redox reactions under standard-state conditions, 770-773 standard reduction potentials, 764-770 electrodes, 763 electrolysis of an aqueous sodium chloride solution, 781-782 defined, 780 in metals purification, 887-888 of molten sodium chloride, 780 quantitative applications of, 782-783 of water, 780-781 electrolyte solutions, 523-524 electrolytes in aqueous solutions, 112-114 compounds as, 116 defined, 112 electrolytic cells, 780 electrolytic reduction, 884 electromagnetic spectrum, 194 electromagnetic wave, 195 electromotive force (emf), 764 electron affinity (EA), 250-251 electron capture, 801 electron configuration Aufbau principle, 221 defined, 219 energies of atomic orbitals in many-electron systems, 219-220 general rules for writing, 222 Hund's rule, 221-222 Pauli exclusion principle, 220-221 and the periodic table, 223-226 electron density, 212 electron-domain geometry, 316-319 defined, 316 electron domains, 314-315 electron spin quantum number, 214-215 electronegativity, 287-290 electronic structure of atoms, 193 electrons, 39, 40, 41 diffraction of, 210 discovery of, 36-38 free, 203

odd number of, 297-298 in orbitals, 324 electrophile addition, 387 electrophiles, 386 electrospinning, 951 electrostatic energy, 160 elemental iron, 33 elementary reactions, 568-569 elements. See also periodic table defined, 5, 6 elimination reactions, 394 emission spectra, 200, 20 I , 202 emission spectrum of hydrogen, 203, 204-205 empirical formula mass, 75, 87 empirical formulas, 51-52, 53, 54-55 determination of, 87 from percent composition, 85-86 emulsifier/emulsifying agent, 532 enantiomers, 383-3 84, 385-386, 866 end point, 698 endothermic process, 161 , 169 entropy changes, 734 ionic bonds, 279 in solution-formation, 507 endpoints, 144 energy defined, 160 energy changes in chemical reactions, 160--161 forms of, 160 ionization energy, 247-249 nuclear binding energy, 801-803 quantization of, 197-198 in solution formation , 508-509 units of, 161-163 English units, 8 enthalpy bond, 286, 300-303,324 defined, 168 and enthalpy changes, 168-169 reactions under constant-volume or constant-pressure conditions, 167-168 thermochemical equations, 169-17 1 enthalpy of reactions, 169 entropy, 508, 724-726 defined, 726 entropy changes in a system, 730--733 entropy changes in the surroundings, 734-735 living systems in, 725 microstates, 727-729 probability, 726-727 standard, 729-730 third law of thermodynamics, 735-736 environmental chemistry, 828-829 acid rain, 843-845 depletion of ozone in the stratosphere, 835-838 Earth's atmosphere, 830-833 greenhouse effect, 838-843 indoor pollution, 846-848 phenomena in the outer layers of the atmosphere, 833-834 photochemical smog, 845-846 volcanoes, 838 Environmental Protection Agency (EPA), 847 enzymes, biological catalysts, 575-577

equatorial, 316 equilibrium. See also chemical equilibrium concepto~ 592-594 defined, 592 and tooth decay, 681 equilibrium constants calculating, 595-597 defined, 594-595 magnitude of the, 598 units in, 622 equilibrium expressions, 594, 599 calculating equilibrium concentrations, 607-611 containing gases, 603-604 heterogeneous equilibria, 599-600 manipulating, 600-602 predicting the direction of a reaction, 606- 607 in problem solving, 606-6ll equilibrium process, 733 equilibrium vapor pressure, 470 equivalence points, 144 erythrocytes, 689-690 esters, 367, 370, 372 ethyl group, 370 ethylene, 749 evaporation, 469-470 exact numbers, 15 excess reactants, 92 excited state, 203 exothermic process, 161 entropy changes, 734 ionic bonds, 279 in solution-formation, 507 expanded octets, 298-300 extensive properties, 15

F f

orbitals, 218 face-centered cubic cell, 474 factor-label method, 20 Fahrenheit, Daniel Gabriel, 11 Fahrenheit temperature scale, 11 families, in periodic table, 44 Faraday constant, 771 Faraday, Michael, 782 fat soluble vitamins, 509-5ll . . ferromagnetic metals, 883 ferrous sulfate, 33 first law of thermodynamics, 164-165,725 first-order reactions, 555-556 fission, 811 flotation, 883 fluoresce, 38 fluoride, 533, 716 fluorine, 921-922 fluoroapatite, 716 Food and Drug Administration (FDA), U.S., 23,63,159,267,363,402 food labels, 159 . formal charge, 292-295 formaldehyde, 848 formation constants (Kf), 710- 711 formic acid, 49 formula mass, 74 formula weight, 74

INDEX

fossil fuels, 843-844 fractional precipitation, 712-713 Frasch process, 919 free electrons, 203 free energy and chemical equilibrium, 742-746 defined, 737, 771 Gibbs free energy, 736-742 free radicals, 297, 819 freezing, 486 freezing point, 486-487 freezing-point depression, 521-522 Freon, 835 frequency, 194 fuel cells, 778-779 Fuller, R. Buckminster, 953 fullerenes, 953 functional groups, 51, 53 fusion, 486

G galvanic cells, 763-764 galvanization, 785 gamma (y) rays, 194 Gamow Bag, 448 Gamow, Igor, 448 gangue, 883 gas, 6, 7 gas constant (R), 427 gas embolism, 415 gas laws, 421-426 application to the, 440-445 defined, 421 gases characteristics of, 417 deviation from ideal behavior, 445-447 diving and the property of, 415 equilibrium expressions containing, 603-604 gas laws, 421-426, 440-445 gas mixtures, 434 439 gas pressure, 417-420 ideal gas equation, 427-430 kinetic molecular theory of gases, 439-445 properties of, 416-421 reactions with gaseous reactants and products, 430-433 Gatorade, 114-115 geiger counter, 818 Geiger, Hans, 39 General Conference on Weights and Measures, 8 geodesic dome, 953 geometric isomers, 865 geometrical isomers, 382 Gerlach, Walther, 215 Germer, Lester Halbert, 210 Gibbs free energy, 737 glass, 484, 485 glassware, volumetric, 9 Goodyear, Charles, 937 graduated cylinder, 8, 9 Graham's law, 444 graphene, 953

graphite, 952-953 Graves, Ray, 115 gravimetric analysis, 142-143 gravitational constant, 418 Greek prefixes, 49 greenhouse effect, 838-843 ground state, 203, 223, 224 Group lA elements, 258, 264-265 Group 1B elements, 264-265 Group 2A elements, 259 Group 3A elements, 260 Group 4A elements, 261 Group 5A elements, 261-262 Group 6A elements, 262, 263 Group 7 A elements, 262, 263 Group 8A elements, 264 group number, 291 groups, in periodic table, 44 Grubbs, Robert H., 935 Guldberg, Cato, 595, 596 Guy-Lussac, Joseph, 423-425

H Haber process, 611, 612 half-cell potentials, 764 half-cell reactions, 766-768 half-cells, 763 half-life, 558-559, 806-807 half-reaction method, 131-132, 760 half-reactions, 127,760-761 halides, 262 Hall process, 895-896 halogens, 45, 906, 921-922 compounds of, 925-926 properties of, 922, 923-925 uses of, 926-927 heart medication, explosives and, 277 heat specific heat and heat capacity, 171-172 thermal energy and, 161 and work, 165-167 heat capacity, 171- 172 heavy water, 814,908 Heisenberg uncertainty principle, 211 Heisenberg, Werner, 211 hemochromatosis, 63 hemodialysis, 527 hemoglobin, 617, 848 hemolysis, 525 Henderson-Hasselbalch equation, 685, 688,694 Henry's law, 516 Henry's law constant, 516 Hertz (Hz), 194 Hess's law, 177-178 heteroatoms, 377 heterogeneous catalysis, 574-575 heterogeneous equilibria, 599-600 heterogeneous mixtures, 7 heteronuclear diatomic molecules, 48 hexagonal close-packed (hcp) structure, 478 high-density polyethylene (HDPE), 936-937 high-spin complex, 870 high-temperature superconductor, 956 homogeneous catalysis, 575

homogeneous mixtures, 7 homonuclear diatomic molecules, 4 Hund's rule, 221-222, 869-870 hybridization defined,327 in molecules containing multiple bon - . 335-341 hybridization of atomic orbitals, 327. 3 - 1 sand p orbitals, 328-330 s, p, and d orbitals, 330-334 hydrated ions, 120 hydrates, 61 hydration, 117, 118,391,508 hydrazine, 912-913 hydride ions, 257 hydrocarbons, 51 hydrocyanic acid, 911 hydrofluoric acid, 922 hydrogen, 906-907 binary hydrides, 907-908 hydrogen economy, 909-910 hydrogenation, 909 isotopes of, 908-909 in the periodic table, 257 hydrogen atom Bohr's theory of the, 200-208 quantum mechanical description of the. _: : hydrogen bomb, 816 hydrogen bonding, 463-464 hydrogen displacement, 133 hydrogen fluoride, 922 hydrogen peroxide, 917-918 hydrogen sulfide, 919-920 hydrohalic acids, 660 hydrolysis, 389 hydronium ion, 123 hydrophilic, 530, 531-532 hydrophobic, 530, 531- 532 hydroxide ion, 123 hydroxy group, 367, 370 hyperbaric oxygen therapy, 438 hypertonic, 525 hypothermia heat capacity and, 174 phase changes and, 489 hypothesis,S hypotonic, 525 hypoxic tents, 591, 623

I ideal behavior (gas), 445 factors that cause deviation from , 4-+ van der Waals equation, 445 447 ideal gas, 427 ideal gas equation, 427 ideal solution, 519 implosion, 816 in phase, 195 incomplete octets, 297 indicators, 144 indoor pollution, 846-848 inert complex, 872 inert gases, 264 inexact numbers, 15 initial rate,S 51

.---

1-6

INDEX

inorganic compounds, 51, 61-62 defined, 364 insoluble compoun ds, 118 instantaneous dipole, 466 instantaneous rates, 546- 548 insulators, 889 integrated rate law, 556 intensive properties, 15 interatomic, 5 interference, 195 • interference pattern, 195, 196 intermediates, 568 intermolecular bonding, 286 intermolecular forces defined, 462 dipole-dipole interactions, 462-463 dispersion forces, 465-467 hydrogen bonding, 463-464 ion-dipole interactions, 467 solutions and, 507-508 internal energy (U), 165 International System of Units, 9 International Union of Pure and Applied Chemistry, 242 internuclear axis, 335 interstitial hydrides, 908 intramolecular bonding, 286 intravenous fluid s, 524-526 iodine, 927 ion-dipole interactions, 467 ion pairs, 523 ion-product constant (Kw), 638 ionic bonding, 279 ioni c bonds, 286, 288 ionic compounds defined, 55, 11 6 formulas of, 57- 58 halides as, 925 lattice energy of, 280-281 naming, 58-61 solubility products of, 702 in water, solubility guidelines, 117- 119 ionic crystals, 480-482, 483 ionic equations, 120 net, 120 ionic hydrides, 907-908 ionic radius, 254-256 ioni zable hydrogen atoms, 51 ionization, electrolytes and, 112 ionization, percent, 529, 649-650 ionization constants acid, 647- 652 base, 652-654 ionization energy (fE) , 247-249 ionosphere, 833 ion-product constant (K,.), 638 • Ions atomic, 55-5 6 common ion effect, 682-683, 705- 710 complex ion formation, 710- 712 defined, 43, 55 electron configuration of, 252-254 formulas of ionic compounds, 57-58 ionic and covalent compounds, 286 ionic bonding, 279 ionic compounds, 55, 57-61 , 116, 117-119

Kekule structures, 376 Kelvin, Lord (William Thomason), 424 Kel vin temperature scale, 10- 11 , 424 ketones, 367, 370, 373 kilojoule (kJ), 162 kinetic energy, 160 kinetic isotope effect, 909 kinetic lability, 872 kinetic molecular theory, 439 kinetic molecular theory of gases, 439-445

Lewis, Gilbert, 278, 284 Lewis structures, 284-285 drawing, 291 - 292 and formal charge, 292-295 Lewis theory of bonding, 284, 324, 35 1 Libby, Willard F., 563 ligand exchange, 872 Ligands, 860-861 light, nature of, 194, 197 double-slit experiment, 195-196 electromagnetic spectrum, 195 propel1ies of waves, 194-195 light water reactors, 812-8 13 liming, 845 limiting reactant, 92 determining the, 93-95 reaction yield, 96-97 Lind, James, 635 line spectra, 201 liquid crystals, 947-949 liquids, 6, 7 gases as, 417 liquid-vapor phase transition, 485-486 properties of, 468-472 solid-liquid phase transition, 486-488 surface tension, 468 vapor pressure, 469-472 viscosity, 468-469 liter, 12 lithium, 267 lithium-ion batteries, 778 Ii ving systems entropy in, 725 thermodynamics in, 746-748 localized bonds, 352, 378 London's dispersion forces , 466 lone pairs, 284, 285 low-density polyethylene (LDPE), 936-937 low-spin complex, 870

L

M

labile complexes, 872 lanthanide (rare earth) series, 224-225 lasers, 207-208 laser pointers, 199 in medicine, 193 LASIK surgery, 193 lattice, 57 lattice energy, 279, 280-281 lattice points, 472 lattice structure, 472 laughing gas, 913 law of conservation of energy, 160 law of conservation of mass, 36 law of definite proportions, 35 law of mass action, 595 law of multiple proportions, 35 law of octaves, 238 law, scientific, 5 Le Chiltelier's principle, 611-612 lead poisoning, 857 lead storage batteries, 777- 778 levorotatory isomer, 384, 866 Lewis acids and bases, 668-670 Lewis dot symbols, 278-279

macroscopic level, 5 magic numbers, 800 magma, 838 magnesium, 894 magnetic confinement, 816 magnetic quantum number, 213- 214 main group elements, 241 manometer, 419 Mars Climate Orbiter, 14 Marsden, Ernest, 39 mass, 9-10 atomic, 46 law of conservation of mass, 36 molecular and formula, 74 percent composition by mass, 75 of reactants and products, 90-91 mass defect, 802 mass-energy equivalence relationship, Einstein 's, 802 mass number (A), 40-41 mass spectrometer, 43 matter classification of, 6-8 defined, 4

naming ionic compounds, 58-61 polyatomic, 56-57 solubility and separation of, 712- 715 iron, 63, 884-885 iron deficiency, 33 iron deficiency anemia (IDA), 33 isoelectronic, 253 isoelectronic series, 255-256 isolated system, 163, l75 isomerism, 382- 386 isomerization reactions, 394 isopropy I group, 370 isotactic, 941 isotonic, 523 isotopes, 41-42 chemical analysis, 817 hydrogen, 908- 909 in medicine, 817-818 parent and daughter, 805 isotropic, 947

J Jenner, Edward, 3 joule (J), 161-162 Joule, James, 161

K

•

INDEX

properties of, 14--15 states of, 6, 7 Maxwell, James Clerk, 195,442 measured numbers, 16- 18 measurement, uncertainty in, 15- 19 medicine lasers in, 193 nuclear, 817-818 Meissner effect, 956 melting, 486 melting point, 486 membrane potential, 776 Mendeleev, Dimitri, 238-239 meniscus, 468 Menkes disease, 881 mercury, 418-420, 422 mesosphere, 833 metabolic acidosis, 543 metabolism, stoichiometry of, 80 metal matrix composites, 946 metallic character, 252 metallic crystals, 483 metallic radius, 246 metalloids, 44, 244, 252, 905, 906 metallurgy, 873 , 880-881 alkali metals, 891-893 alkaline earth metals, 893- 895 aluminum, 895-897 band theory of conductivity, 888- 890 defined,883 of iron, 884-885 metallurgical processes, 883-888 metals in human biology, 881 occurrence of metals, 882 periodic trends in metallic properties, 890-891 preparation of the ore, 883 production of metals, 883-884 purification of metals, 887-888 steelmaking, 885-887 metals, 44, 45, 56 in chemical equations, 243- 244 oxidation of metals in aqueous solutions, 130 oxides of, 667-668 qualitative analysi s of metal ions in solution, 714-715 metathesis method, 935 meterstick, 8 methanol, 578 methanol poisoning, 543 methyl group, 367, 370 metric system, 8, 12 Meyer, Lothar, 238 microstates, 728- 729 millicurie (mCi), 818 Millikan, R. A. , 37 milliliter (mL), 12 millimeters mercury (mmHg), 418 minerals, 882 miscible, 508 mixtures, 7-8 gas, 434 439 moderators, 813 modern materials, 934-935 biomedical materials, 949-952 ceramics and composite materials, 945-946 .

liquid crystals, 947- 949 nanotechnology, 952-953 polymers, 936-945 semiconductors, 954--956 superconductors, 956- 957 molal boiling-point elevation constant, 520 molal freezing-point depression constant, 521 molality, 512 molar concentration, 136 molar heat of fu sion, 488 molar heat of sublimation, 488 molar heat of vaporization, 485-486 molar mass, 84 molar solubility, 701 molarity (M), 136, 138- 139 mole and molar masses determining molar mass, 84 empirical formula from percent composition, 85-86 interconverting mass, moles, and numbers of particles, 84--85 the mole, 82- 84 mole defined, 82 moles of reactants and products, 89- 90 mole fractions, 435-436, 513 , 518 molecular art, 5 molecular compounds, 47 , 49-51 , 116,417 molecular crystals, 483 molecular equations, 119 molecular formula, determination of, 88 molecular geometry, 314 defined, 316 deviation from ideal bond angles, 319 electron-domain geometry and, 316- 319 geometry of molecules with more than one central atom, 319- 320 and polarity, 321-323 VSEPR model, 314--315 molecular level, 5 molecular mass, 74 molecular orbital theory, 342, 351 - 353 bond order, 343- 344 bonding and antibonding molecular orbitals, 342- 343 defined,342 molecular orbital diagrams, 346-347 , 349 molecular orbitals, 342-345 molecular parity, 321 molecular speed, 442-444 molecular weight, 74 molecularity, 568-569 molecules, 4, 5, 47-48 with delocalized bonding, 351-353 empirical formulas, 51, 53, 54--55 hybridization and multiple bonds, 335-341 molecular formulas , 48 naming molecular compounds, 49- 51 , 52, 53 , 374 representation of organic molecules, 375- 381 Molina, Mario, 829 molybdenum, 883 momentum (p), 211 monatomic cations, 55 monatomic ions, 55 , 56 monatomic molecules, 47 Mond process, 887

monodentate ligands, 861 monomers, 395, 936 monoprotic acids, 123 Montreal Protocol, 829, 837 Moseley, Henry, 239 mUltiple bonds, 285-286 multiple proportions, law of, 35 musk,354

N n-type semiconductors, 890, 955 nanofibers, 951 nanotechnology, 952-953 nanotubes, 953 NASA,14 natural rubber, 936-937 Nelmes, Sarah, 3 nematic, 947 Nernst equation, 773-775 net ionic equations, 120 neutralization reactions, 124 neutron-to-proton ratio (nip), 800 neutrons, 40, 41 Newlands, John, 238 Newton 's laws of motion, 212 newtons (N) , 14,418 nitric acid, 913-9 14 nitric oxide, 304, 845-846, 913 nitrogen, 911- 914 nitrogen dioxide, 913 nitrogen fixation, 831, 913 nitroglycerin, 277 nitrous oxide, 913 Nobel, Alfred, 277 Nobel Prize, 277, 313, 563 , 829, 935 noble gas core, 223 noble gases, 45, 241, 244 noble metals, 130 nodes, 208 nomenclature, chemical, 49 nonconductors, 954 nonelectrolytes, 112 nonmetals, 44 carbon, 910- 911 in chemical equations, 244 general properties of, 906 halogens, 921-927 • hydrogen, 906-910 nitrogen, 911- 914 oxides of, 667-668 oxygen,9l6- 918 phosphorus, 914--916 sulfur, 918- 921 nonpolar,,288, 321 nonpolar covalent bonds, 288 nonspontaneous processes, 726 nonvolatile, 517 normal boiling point, 485 northem lights, 833 nuclear binding energy, 801 - 803 nuclear chain reaction, 812 nuclear chemistry, 796-797 biological effects of radiation, 818-8 19 natural radioactivity, 804-808 nuclear fi ssion, 811 - 815

1-7

1-8

INDEX

nuclear chemistry-continued nuclear fusion , 815-817 nuclear stability, 799-804 nuclear transmutation, 808-811 nuclei and nuclear reactions, 798-799 uses of isotopes, 817-818 nuclear fission, 811-815 nuclear fusion, 815-817 nuclear medicine, 817- 818 nuclear reactors, 812-815 • nuclear transmutation, 798, 808-811 nucleic acids, 397, 400, 401 nucleons, 41 nucleophile addition, 388 nucleophiles, 386, 387 nucleotides, 400, 401 nucleus, 40 utrition Facts labels, 159 nylon, 396-397

o octet rule, 284, 285 exceptions to, 296-300 oil-drop experiment, Millikan, 38 old quantum theory, 212 olefin metathesis, 935 olfactory receptors, 313 open system, 163, 164 optical isomers, 383, 865, 866 orbitals electrons in, 324 hybridization of atomic, 327-334, 351 molecular orbital theory, 342-351 ore, 882 organic chemistry carbon, 364-365 classes of organic compounds, 365-375 and drugs, 363 isomerism, 382-386 organic polymers, 395-401 organic reactions, 386-395 representation of organic molecules, 375-381 organic compounds, 50- 51 classes of, 365-375 defined, 364 naming, 368-369 osmosis, 522, 527 osmotic pressure, 522- 523 out of phase, 195 overvoltage, 782 oxidation, 127 oxidation number, 127 oxidation-reduction reactions, 126-127, 394 balancing simple redox equations, 130-133 oxidation numbers, 127-130 oxidation of metals in aqueous solutions, 130 oxidation state, 127 oxides acid-base properties of hydroxides and, 667-668 basic and amphoteric hydroxides, 668 of metals and nonmetals, 667-668 variation of third-period elements, 265-266 oxidizing agent, 127 oxoacids, 59, 660-661

oxoanions, 59 oxygen, 916-918 oxygen transport, 874 ozone, 918 CFCs and, 829, 835 depletion of stratospheric, 835-836 polar ozone holes, 836-838

p p orbitals, 217, 328-334

p-type semiconductors, 890, 955 packing spheres, 473-475 paramagnetic, 342 partial charges, 289 partial pressure, 434 particle accelerator, 810 particles (N), 84 pascal (Pa), 418 passivation, 785 patina, 785 Pauli exclusion principle, 220-221, 870 Pauling, Linus, 287 peptide bonds, 398, 399 per mole of the reaction, 169 percent by mass, 512 percent composition by mass, 75, 85-86 percent dissociation, 529 percent ionic character, 288 percent yield, 96 periodic table, 44 46, 135 atomic radius, 246-247, 255 classification of elements, 241-243 comparing ionic radius with atomic radius, 255 comparison of Group lA and Group IE elements, 264-265 development of, 238-240 effective nuclear charge, 244-245 electron affinity (EA), 250-251 electron configuration of ions, 252-254 electron configurations and the, 223-226 elements essential for life, 239-240 gaseous elements in, 416 general trends in chemical properties, 257 Group lA elements, 258 Group lB elements, 264-265 Group 2A elements, 259 Group 3A elements, 260 Group 4A elements, 261 Group 5A elements, 261-262 Group 6A elements, 262, 263 Group 7 A elements, 262, 263 Group 8A elements, 264 ionic radius, 254-256 ionization energy (IE), 247-249 ions of d-block elements, 253-254 ions of main group elements, 253 isoelectronic series, 255-256 main group elements, 256-266 metallic character, 252 modern, 240-244 numbers at top of, 242 periodic trends in metallic properties, 890--891 periodic trends in properties, 245, 246-252, 256-266

properties of other main group elements, 260-264 properties of the active metals, 258-259 properties of transition metals, 858-859 representing free elements in chemical equations, 243-244 salt and salt substitutes, 237 variation in properties of oxides within a period, 265-266 periodicity, 238 periods, in periodic table, 44 peroxide, 917-918 pH, 639 of acid-base titrations, 691 acid ionization constants and, 648-651 base ionization constants and, 653-654 buffer solutions and, 684-689 mean precipitation pH in the U.S., 843 pH scale, 639-644 phase boundary line, 491 phase changes, 484 485,486 dangers of, 489-491 liquid-vapor phase transition, 485-486 solid-liquid phase transition, 486-488 solid-vapor phase transition, 488 phase diagrams, 491-493 Phipps, James, 3 phosphine, 914-915 phosphorus, 914-916 photochemical smog, 845-846 photodecomposition, 830 photodissociation, 835 photoelectric effect, 198-199 photons, 198-199 photosynthesis, 830, 832 physical change, 15 physical properties, 14-15 pi bonds, 336, 340-341 pig iron, 885 pipet, 8, 9 Planck, Max, 193, 197, 198 plasma, 816 Platinol, 73 Plato, 34 pOH scale, 641 polar, 286 polar covalent bonds, 286, 288 polar molecules, 117, 321,462 polar ozone holes, 836-838 polar stratospheric clouds (PSCs), 836, 837 polar vortex, 836 polarimeter, 866 polarity, molecular geometry and, 321-323 polarized light, 384, 385, 866, 867 polarized molecules, 466 pollution acid rain, 843-845 greenhouse effect, 838-843 indoor, 846-848 photochemical smog, 845-846 polyamides, 942 polyatomic ions, 56-57 poly atomic molecules, 48, 840 polyesters, 942 polyisoprene, 936-937 polymer matrix composites, 946

INDEX

polymers addition, 395-396, 397, 936-941 biological,397-401 condensation, 396-397, 941-944 defined,395,936,941 electrically conducting, 943-944 organic, 395-401 polypeptides, 399 polyprotic acids, 59, 123- 124,657-659 polysaccharides, 397, 400 polystyrene, 938 polyvinyl chloride (PVC), 938 positron, 798 potassium hydroxide, 893 potassium nitrate, 893 potential energy, 160 pounds, 14 precipitate, 117 precipitation, fractional, 712-713 precipitation reactions in aqueous solutions, 117-121 defined,117 solubility and, 704-705 precision, 18-19 prefixes, Greek, 49 pressure calculation of, 418 Dalton's law of partial pressures, 434 435 defined,418 equilibrium and, 614-615 measurement of, 418-420 pressure-volume relationship, 421-423 solubility and, 515-517 using partial pressure to solve problems, 436-438 primary ami des, 373 primary amines, 373 primary pollutants, 845 primary structure, protein, 465 principal quantum number, 213, 214 probability, 726-727 probability density, 216- 2l7 problem solving, using units and, 20-22 products, in chemical reactions, 77 proteins, 397, 399-400 protein structure, 464 465 proton acceptor, 123 proton donor, 123 protons, 40, 41 Proust, Joseph, 35 pure covalent bonds, 286, 288 pyrometallurgy, 883-884

qualitative analysis, 714-715 qualitative properties, 14 quantitative properties, 8, 14 quantum mechanics, 210-211 beginning of, 212 quantum mechanical description of the hydrogen atom, 212 Schrodinger equation, 212 uncertainty principle, 211 quantum numbers, 212, 213-214 electron spin quantum number, 214-215

quantum theory, 193, 197 developments in, 212 photons and the photoelectric effect, 198-200 quantization of energy, 197- 198 quantum defined, 197 quaternary structure, protein, 465 quicklime, 844-845

R racemic mixture, 384, 866 racemization, 391 radial probability distribution, 216-2l7 radiation, 36 biological effects of, 818-819 ultraviolet (UV), 830 radiation absorbed dose (rad), 818 radicals, 297, 298, 819 radio waves, 194, 195- 196 radioactive bone, 265 radioactive decay dating based on, 806-808 defined, 798 kinetics of, 805-806 radioactive decay series, 804-805 radioactivity, 38-39,798 radiocarbon dating, 563, 806-808 radon, 820, 846-847 randomness, 508 Raoult's law, 517 rate constant, 547 rate-determining step, 569-571 rate law, 551 rate of reaction, 548 reactants. See also limiting reactant defined, 77 gaseous, 430-433 mass of reactants and products, 90-91 moles of reactants and products, 89-90 reversible, 592 reaction mechanism, 568 reaction order, 551 reaction quotient, 595 reaction rates, 544 Arrhenius equation, 564-·565 average, 544-546 catalysis, 573-577 collision theory, 562-564 elementary reactions, 568-569 experimental determination of the rate law, 551-555 first-order reactions, 555-559 instantaneous rate, 546-548 rate-determining step, 569- 572 rate law, 551 reaction mechanisms, 568- 573 second-order reactions, 560-561 stoichiometry and, 548-550 reactions organic, 386-395 types of, 98 red phosphorus, 914 redox reactions, 126 balancing, 760-762 balancing simple redox equations, 130-133

1-9

spontaneity under conditions other than standard state, 773-777 spontaneity under standard-state conditions. 770-773 types of, 133-134 reduced iron, 33 reducing agent, 127 reduction, 127 Reinitzer, Frederick, 947 relative biological effectiveness (REE), 81 8 remineralization, 681 representative elements, 241 repulsion, 160 repulsive force, 245 resonance,295-296,378-381 resonance stabilized, 378 resonance structures, 295, 378 reversible process, 592 revised metric system (SI) units, 8, 9 rhombic sulfur, 919 ribonucleic acid (RNA), 400, 401 roentgen equivalent for man (rem), 81 Roman numerals, cations with, 56 Rontgen, Wilhelm, 38 root-mean-square (rms) speed, 442 Rosenberg, Barnett, 73 Rowland, F. Sherwood, 829 rubber, 936-938 Rutherford, Ernest, 39, 40, 193 Rydberg equation, 202 Rydberg, Johan, 202

5 s orbitals, 216-217, 328-334 salt in acid-base reactions, 124 and salt substitutes, 237 salt bridge, 763-764 salt hydrolysis, 662 salt solutions acid-base properties of, 662-667 acidic, 664-665 basic, 662-663 cations and anions, 666 neutral, 665-666 saturated solutions, 506 scanning tunneling microscope (STM . 9'::: Schrock, Richard R., 935 Schrodinger equation, 212 Schrodinger, Erwin, 212 scientific measurement, 8- 13 scientific method, 3, 5-6 scurvy, 635 second law of thermodynamics, 725. , '}-- ~: second-order reactions, 560 secondary pollutants, 845 secondary structure, protein, 465 semiconductors, 889-890, 954-9 -6 semipermeable membrane, 522 sequestrants, 874 shell, 213 shielding, 244 shielding constant, 245 SI base units, 9, 10, 12 sickle cell disease, 461

1-10

INDEX

sigma bonds, 335, 338 sigma molecular orbitals, 343 significant figures, 15- 16 sil ver fluoride, 922 simple cubic cell, 473-474 single bonds, 285 sintering, 945- 946 skeletal structures, 376-378 slag, 885 smallpox, 3, 23 • smectic, 947 smell, olfactory receptors, 313 smog, photochemical , 845-846 smoking, radioactivity in tobacco, 819-820 SNI reactions, 390-391 soda ash, 893 sodium carbonate, 893 sodium chloride electrolysis of an aqueous sodium chloride solution, 78 1-782 electrolysis of molten, 780 sodium hydroxide, 893 sodium nitrate, 893 soft tiss ue materials, 950-951 sol-gel process, 946 solar flares , 833 solids, 6, 7. See also crystal structure amorphous, 484 gases as, 417 preparing a soluti on from, 138-139 solid-liquid phase transition, 486-488 solid-vapor phase transition, 488 solubility defined, 117,506,701 factors that affect, 515- 517 , 705-7 12 guidelines for ionic compounds, 117-119 separation of ions in, 712- 715 vitamin, 509- 511 solubility equilibria, 700-705 solubility product constant, 701 soluble compounds, 118 solutes, 112, 506 solution stoichiometry, 141-142 solutions, 112 colligative properties, 51 7-529 colloids, 530- 532 concentration units, 511-514 energy and entropy in solution formation, 508-509 factors that affect solubility, 515- 517 intermolecular forces, 507- 508 types of, 506-507 solvation, 507 solvents, 112,506 solving problems, using units and, 20-22 space-filling models (molecular art), 5 space shuttles, mystery glow of, 834 specific heat, 171-172 spectator ions, 120 spectrochemical series, 869 speed of light (c), 194 splitting, of energy levels, 219 spontaneous processes, 726 stability constant, 7l 0 standard atmospheric pressure, 419

standard enthalpies offormation, 179-1 8 1 standard enthalpy of reaction, 179 standard entropy, 729-730, 736 standard free-energy of formation , 739 standard free-energy of reaction, 738 standard hydrogen electrode (SHE), 765 standard reduction potential, 765 standard solution, 144 standard states, 622 standard temperature and pressure (STP), 428 standing/stationary waves, 208, 209 state functi ons, 163 state of a system, 163 steam burns, 489 steelmaking, 885-887 stereoisomerism, 382-384 stereoisomers, 382, 865 Stern, Otto, 215 Stock system, 56 stoichiometric amount, 90 stoichiometric coefficients, 77 stoichiometry and reaction rates, 548-550 solution, 141-142 stoichiometry, ratios of combination, 72- 73 calculations with balanced chemical equations, 89-91 chemical equations, 76- 81 combustion anal ysis, 86-89 limiting reactants, 92-97 mole and molar masses, 82-86 molecular and formula masses, 74 percent composition of compounds, 75-76 reaction types, 98 stomach, pH balance, 642-643 stratosphere defined,833 depletion of ozone in, 835-838 strong acid-strong base titrations, 691-692 strong acid-weak base titrations, 696- 698 strong acids, 644-645 strong bases, 645-647 strong conjugate acid, 655 strong conjugate base, 655 strong electrolytes in aqueous solutions, 112-116, 120 defined, 112 strong-field ligands, 869 structural formulas, 48 structural isomers, 321, 382 subatom ic particles, 36, 798 subcritical mass, 812 sublimation, 488 subshell,213 substances defined,6 equilibrium and addition/removal of, 612-614 substituents, 368, 374 substitution reactions, 388-390 substrate, 575 sulfur, 918-921 sul fur dioxide, 843- 845, 920-921 sulfur hexafluoride, 9? I sulfur trioxide, 921 su lfuric acid, 921

superconducting transition temperature, 956 superconductors, 956- 957 supercooling, 488 supercritical fluids, 486 supersaturated solutions, 506, 507 surface tension, 468 su tToundings, 160 sutures, 951 sy mbols, element, 4 1 symbols, Lewis dot, 278 syndiotactic, 941 systems, 160

T tacticity, 941 Tefl on, 939 temperature, 10- 11 , 15 absolute, 10 equilibrium and, 61 5-6 16, 6 18- 619 solubility and, 515 temperature-volume relationship, 423-425 tempering, 886 termolecular, 569 tertiary structure, protein, 465 testicular cancer, 73 thalidomide, 363, 390, 402 theoretical yield, 96 theory, scien ti fi c, 5 thermal energy, 160 thermal pollution, 515 thermite reaction, 897 thermochemical equations, 169- 171 thermochemistry, 158-159, 182- 183 calorimetry, 171- 177 defined, 161 energy and energy changes, 160-163 enthalpy, 167-1 7 1 Hess's law, 177-178 standard enthalpies of formation, 179-181 thermodynamics, 163- 167 thermodynamics defined, 163 first law of thermodynamics, 164-165,725 in living systems, 746-748 second law of thermodynamics, 725, 733-735 states and state functions, 163- 164 third law of thermodynamics, 735- 736 work and heat, 165-167 thermonuclear bomb, 816 thermonuclear reactions, 815 thermoplastic, 936 thermosetting, 936 thermosphere, 833 third law of therm odynamics, 735-736 Thomson, George Paget, 210 Thomson, J. 1., 37, 39 three-dimensional (3D) movies, 385 threshold frequency, 198 titrations, 144. See also acid-base titrations tobacco, radioactivity in, 819-820 tooth decay, 681 totT,418 trace elements, 240 tracers, 817

INDEX

trans isomers, 337, 382, 865-866 transition elements, 45 transition metals, 45, 56, 223, 241, 858-859 transition rate, 564 transition state, 564 transmutation elements, 809 transmutation, nuclear, 798, 808- 811 triple bonds, 285, 338 triple point, 491 , 492 triprotic acids, 123 troposphere, 832- 833 Tyndall effect, 530, 531

u ultraviolet (UV) radiation, 830 uncertain digits, 15 uncertainty in measurement, 15-19 uncertainty principle, 211 unimolecular, 569 unit cells, 472-473 units, 8 unsaturated solutions, 506 uranium decay series, 804-805

v vaccination, 3, 23 vaccinia immune globulin (VIG), 23 valence band, 888, 954 valence bond theory, 324-326, 351 valence electrons, 243 valence-shell electron-pair repulsion (VSEPR) model, 314- 315, 351 van der Waals equation, 445-447 van der Waals forces, 362 van der Waals, J, D., 445 vanadium oxide, 921 van't Hoff factor, 523 vapor, solid-vapor phase transition, 488

vapor pressure, 437, 469-472 vapor-pressure lowering, 517-519 vaporization, 470 visible light, 194 vision, 394-395 vitamin C, 635, 671 vitamin solubility, 509-511 volatile, 469, 519 volcanoes, 838 voltaic cells, 763 volume amount-volume relationship, 425-426 constant-volume calorimetry, l75-177 equilibrium and, 614-615, 620-621 pressure-volume relationship, 421-423 reactions under constant-volume! constant-pressure conditions, 167-168 temperature-volume relationship, 423-425 volumetric flask, 8, 9 von Laue, Max Theodor Felix, 476 VSEPR model, 314-315, 351 vulcanization, 937

wavelength, 194 waves, properties of, 194- 195 weak acid-strong base titrations, 693--696 weak acids, 647 weak bases, 652 weak conjugate acid, 655 weak conjugate base, 654 weak electrolytes in aqueous solutions, 11 2- 116 defined, 112 weak-field ligands, 869 weight, 9 Werner, Alfred, 858 Werner's coordination theory. 858 white phosphorus, 914 · Wilson disease, 881 Wohler, Friedrich, 364 work, and heat, 165-167 World Anti-Doping Agency (WADA '. '::91. ' ':':: World Health Organization (WHO). "'. "':

x X-ray diffraction, 210, 476, 477 X-rays, 38, 194,239

Waage, Peter, 595, 596 water acid-base properties of, 637-639 electrolysis of, 780-781 molecule, 48 physical states of, 6, 7 water gas, 907 water-soluble, 112 water-insoluble compounds, 11 8 water-soluble compounds, 11 8 wave function, 212 wave mechanics, 212 wave properties of matter, 208 de Broglie hypothesis, 208-210 diffraction of electrons, 210

• •

y yield actual, 96 percent, 96 theoretical, 96

z zeroth-order reaction, 561 zone refining, 888

,

Avogadro's number Electron charge (e) Electron mass Paraday constant Gas constant (R)

Planck's constant (h) Proton mass Neutron mass Speed of light in a vacuum

tera (T) giga (G) mega (M) kilo (k) deci (d)

10 12 9 10 106 3 10 10- 1

6.0221367 X 1023 1.6022 X 10- 19 C 9.109387 X 10- 28 g 96,485.3 Clmo1 e0.08206 L . atrnlK . mol 8.314 J/K· mol 62.36 L . torr/K . mol 1.987 cal/K . mol 6.6256 X 10- 34 J . s 1.672623 X 10- 24 g 1.674928 X 10- 24 g 8 2.99792458 X 10 rnIs

centi (c) milli (m) micro (IL) nano (n) pico (p)

10- 2 10- 3 10- 6 10- 9 10- 12

lIb = 453.6 g 1 in = 2.54 cm (exactly) 1 mi = 1.609 km 1 km = 0.6215 mi 1pm = 1 X 1O- 12 m= 1 X lO- lO cm 1 atm = 760 rnmHg = 760 torr = 101,325 N/m2 = 101 ,325 Pa 1 cal = 4.184 J (exactly) 1 L . atm = 101.325 J IJ=1CX1V ?OC = (OP _ 32 0 P) X 5°C . 9°P ?OP = 9°P X (OC) 5°C

+ 32°P

?K = (OC + 273.1 5°C)

( 1\~

•

,