solution manual - chemistry-4th ed. (mcmurry)

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Assignment 1. Questions from chapters 1 and 2 of McMurry and Fay Question numbers are from the fourth edition. Chapter 1. Chemistry: Matter and Measurement 1.1 (a) Cd (b) Sb (c) Am 1.2 (a) silver (b) rhodium (c) rhenium (d) cesium (e) argon (f) arsenic 1.3 (a) Ti, metal (b) Te, semimetal (c) Se, nonmetal

(d) Sc, metal (e) At, semimetal (f) Ar, nonmetal 1.4 The three Acoinage metals@ are copper (Cu), silver (Ag), and gold (Au). 1.5 (a) The decimal point must be shifted ten places to the right so the exponent is S10. The

result is 3.72 x 10S10 m. (b) The decimal point must be shifted eleven places to the left so the exponent is 11. The result is 1.5 x 1011 m.

1.6 (a) microgram (b) decimeter (c) picosecond

(d) kiloampere (e) millimole

1.7 o oo 5 5C = x F _ 32) = x (98.6 _ 32) = C( 37.0

9 9

oK = C + 273.15 = 37.0 + 273.15 = 310.2 K 1.8 (a) K = oC + 273.15 = S78 + 273.15 = 195.15 K = 195 K

(b) o oo o9 9F = ( x C) + 32 = ( x 158) + 32 = F = F316.4 316

5 5

(c) oC = K S 273.15 = 375 S 273.15 = 101.85oC = 102oC

o oo o9 9F = ( x C) + 32 = ( x 101.85) + 32 = F = F215.33 215

5 5

1.9 33

m 27.43 gd = = = 2.212 g/cm

V 12.40 cm

1.10 1 mL

volume = 9.37 g x = 6.32 mL1.483 g

1.11 The actual mass of the bottle and the acetone = 38.0015 g + 0.7791 g = 38.7806 g. The

measured values are 38.7798 g, 38.7795 g, and 38.7801 g. These values are both close to each other and close to the actual mass. Therefore the results are both precise and accurate.

Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________

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1.12 (a) 76.600 kg has 5 significant figures because zeros at the end of a number and after the decimal point are always significant. (b) 4.502 00 x 103 g has 6 significant figures because zeros in the middle of a number are significant and zeros at the end of a number and after the decimal point are always significant. (c) 3000 nm has 1, 2, 3, or 4 significant figures because zeros at the end of a number and before the decimal point may or may not be significant. (d) 0.003 00 mL has 3 significant figures because zeros at the beginning of a number are not significant and zeros at the end of a number and after the decimal point are always significant. (e) 18 students has an infinite number of significant figures since this is an exact number. (f) 3 x 10S5 g has 1 significant figure. (g) 47.60 mL has 4 significant figures because a zero at the end of a number and after the decimal point is always significant. (h) 2070 mi has 3 or 4 significant figures because a zero in the middle of a number is significant and a zero at the end of a number and before the decimal point may or may not be significant.

1.13 (a) Since the digit to be dropped (the second 4) is less than 5, round down. The result is

3.774 L. (b) Since the digit to be dropped (0) is less than 5, round down. The result is 255 K. (c) Since the digit to be dropped is equal to 5 with nothing following, round down. The result is 55.26 kg.

1.14 (a)

24.567 g

+ 0.044 78 g

24.611 78 g

This result should be expressed with 3 decimal places. Since the

digit to be dropped (7) is greater than 5, round up. The result is 24.612 g (5 significant figures).

(b) 4.6742 g / 0.003 71 L = 1259.89 g/L 0.003 71 has only 3 significant figures so the result of the division should have only 3 significant figures. Since the digit to be dropped (first 9) is greater than 5, round up. The result is 1260 g/L (3 significant figures), or 1.26 x 103 g/L.

(c)

0.378 mL

+ 42.3 mL

_ 1.5833 mL

41.0947 mL

This result should be expressed with 1 decimal place. Since the

digit to be dropped (9) is greater than 5, round up. The result is 41.1 mL (3 significant figures).

1.15 The level of the liquid in the thermometer is just past halfway between the 32oC and 33oC


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marks on the thermometer. The temperature is 32.6oC (3 significant figures).

1.16 (a) Calculation: oo o9 9F = ( x C) + 32 = ( x 1064) + 32 = F1947

5 5

Ballpark estimate: oF . 2 x oC if oC is large. The melting point of gold . 2000oF.

(b) r = d/2 = 3 x 10S6 m = 3 x 10S4 cm; h = 2 x 10S6 m = 2 x 10S4 cm Calculation: volume = πr2h = (3.1416)(3 x 10S4 cm)2(2 x 10S4 cm) = 6 x 10S11 cm3

Ballpark estimate: volume = πr2h . 3r2h . 3(3 x 10S4 cm)2(2 x 10S4 cm) . 5 x 10S11 cm3 1.17 1 carat = 200 mg = 200 x 10S3 g = 0.200 g

Mass of Hope Diamond in grams = 0.200 g

44.4 carats x = 8.88 g1 carat

1 ounce = 28.35 g

Mass of Hope Diamond in ounces = 1 ounce

8.88 g x = 0.313 ounces28.35 g

1.18 An LD50 value is the amount of a substance per kilogram of body weight that is a lethal

dose for 50% of the test animals.

1.19 mass of salt = 453.6 g 1 kg 4 g

155 lb x x x 1 lb 1000 g 1 kg

= 281.2 g or 300 g

Understanding Key Concepts

1.20

1.21


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1.22 red B gas; blue B 42; green B sodium 1.23 The element is americium (Am) with atomic number = 95. It is in the actinide series. 1.24 (a) Darts are clustered together (good precision) but are away from the bullseye (poor

accuracy). (b) Darts are clustered together (good precision) and hit the bullseye (good accuracy). (c) Darts are scattered (poor precision) and are away from the bullseye (poor accuracy).

1.25 (a) 34.2 mL (3 significant figures) (b) 2.68 cm (3 significant figures) 1.26

The 5 mL graduated cylinder is marked every 0.2 mL and can be read to ∀ 0.02 mL. The 50 mL graduated cylinder is marked every 2 mL and can only be read to ∀ 0.2 mL. The 5 mL graduated cylinder will give more accurate measurements.

1.27 A liquid that is less dense than another will float on top of it. The most dense liquid is

mercury, and it is at the bottom of the cylinder. Because water is less dense than mercury but more dense than vegetable oil, it is the middle liquid in the cylinder. Vegetable oil is the least dense of the three liquids and is the top liquid in the cylinder.

Additional Problems Elements and the Periodic Table 1.28 114 elements are presently known. About 90 elements occur naturally. 1.29 The rows are called periods, and the columns are called groups.


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1.30 There are 18 groups in the periodic table. They are labeled as follows:

1A, 2A, 3B, 4B, 5B, 6B, 7B, 8B (3 groups), 1B, 2B, 3A, 4A, 5A, 6A, 7A, 8A 1.31 Elements within a group have similar chemical properties.

1.32

1.33

1.34 A semimetal is an element with properties that fall between those of metals and nonmetals.

1.35 (a) The alkali metals are shiny, soft, low-melting metals that react rapidly with water to

form products that are alkaline. (b) The noble gases are gases of very low reactivity. (c) The halogens are nonmetallic and corrosive. They are found in nature only in


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combination with other elements. 1.36 Li, Na, K, Rb, and Cs 1.37 Be, Mg, Ca, Sr, and Ba 1.38 F, Cl, Br, and I 1.39 He, Ne, Ar, Kr, Xe, and Rn 1.40 (a) gadolinium, Gd (b) germanium, Ge (c) technetium, Tc (d) arsenic, As 1.41 (a) cadmium, Cd (b) iridium, Ir (c) beryllium, Be (d) tungsten, W 1.42 (a) Te, tellurium (b) Re, rhenium (c) Be, beryllium

(d) Ar, argon (e) Pu, plutonium 1.43 (a) B, boron (b) Rh, rhodium (c) Cf, californium

(d) Os, osmium (e) Ga, gallium 1.44 (a) Tin is Sn: Ti is titanium. (b) Manganese is Mn: Mg is magnesium.

(c) Potassium is K: Po is polonium. (d) The symbol for helium is He. The second letter is lowercase.

1.45 (a) The symbol for carbon is C. (b) The symbol for sodium is Na.

(c) The symbol for nitrogen is N. (d) The symbol for chlorine is Cl. Units and Significant Figures 1.46 Mass measures the amount of matter in an object, whereas weight measures the pull of

gravity on an object by the earth or other celestial body. 1.47 There are only seven fundamental (base) SI units for scientific measurement. A derived

SI unit is some combination of two or more base SI units. Base SI unit: Mass, kg; Derived SI unit: Density, kg/m3

1.48 (a) kilogram, kg (b) meter, m (c) kelvin, K (d) cubic meter, m3 1.49 (a) kilo, k (b) micro, Φ (c) giga, G (d) pico, p (e) centi, c

1.50 A Celsius degree is larger than a Fahrenheit degree by a factor of 9

5.

1.51 A kelvin and Celsius degree are the same size.


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1.52 The volume of a cubic decimeter (dm3) and a liter (L) are the same. 1.53 The volume of a cubic centimeter (cm3) and a milliliter (mL) are the same. 1.54 Only (a) is exact because it is obtained by counting. (b) and (c) are not exact because

they result from measurements.

1.55

4.8673 g

_ 4.8 g

0.0673 g

The result should contain only 1 decimal place. Since the digit to

be dropped (6) is greater than 5, round up. The result is 0.1 g. 1.56 cL is centiliter (10-2 L) 1.57 (a) deciliter (10-1 L) (b) decimeter (10-1 m)

(c) micrometer (10-6 m) (d) nanoliter (10-9 L) 1.58 1 mg = 1 x 10-3 g and 1 pg = 1 x 10-12 g

_39

_12

1 x g 1 pg10 x = 1 x pg/mg101 mg 1 x g10

35 ng = 35 x 10-9 g _9

4_12

35 x g 1 pg10 x = 3.5 x pg / 35 ng1035 ng 1 x g10

1.59 1 µL = 10-6 L 6_6

1 L = L/L10

L10

µ µ

20 mL = 20 x 10-3 L _3

4_ 6

20 x L 1 L10 x = 2 x L/ 20 mL1020 mL L10

µ µ

1.60 (a) 5 pm = 5 x 10-12 m

5 x 10-12 m x 100 cm

1 m = 5 x 10-10 cm

5 x 10-12 m x _9

1 nm

1 x m10 = 5 x 10-3 nm

(b) 8.5 cm3 x 3

1 m

100 cm

= 8.5 x 10-6 m3


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8.5 cm3 x 3

10 mm

1 cm

= 8.5 x 103 mm3

(c) 65.2 mg x _31 x g10

1 mg = 0.0652 g

65.2 mg x _31 x g10

1 mg x

_12

1 pg

1 x g10 = 6.52 x 1010 pg

1.61 (a) A liter is just slightly larger than a quart.

(b) A mile is about twice as long as a kilometer. (c) An ounce is about 30 times larger than a gram. (d) An inch is about 2.5 times larger than a centimeter.

1.62 (a) 35.0445 g has 6 significant figures because zeros in the middle of a number are

significant. (b) 59.0001 cm has 6 significant figures because zeros in the middle of a number are significant. (c) 0.030 03 kg has 4 significant figures because zeros at the beginning of a number are not significant and zeros in the middle of a number are significant. (d) 0.004 50 m has 3 significant figures because zeros at the beginning of a number are not significant and zeros at the end of a number and after the decimal point are always significant.

(e) 67,000 m2 has 2, 3, 4, or 5 significant figures because zeros at the end of a number and before the decimal point may or may not be significant.

(f) 3.8200 x 103 L has 5 significant figures because zeros at the end of a number and after the decimal point are always significant.

1.63 (a) $130.95 is an exact number and has an infinite number of significant figures.

(b) 2000.003 has 7 significant figures because zeros in the middle of a number are significant. (c) The measured quantity, 5 ft 3 in., has 2 significant figures. The 5 ft is certain and the 3 in. is an estimate.

1.64 To convert 3,666,500 m3 to scientific notation, move the decimal point 6 places to the left

and include an exponent of 106. The result is 3.6665 x 106 m3. 1.65 Since the digit to be dropped (3) is less than 5, round down. The result to 4 significant

figures is 7926 mi or 7.926 x 103 mi. Since the digit to be dropped (2) is less than 5, round down. The result to 2 significant figures is 7900 mi or 7.9 x 103 mi.

1.66 (a) To convert 453.32 mg to scientific notation, move the decimal point 2 places to the

left and include an exponent of 102. The result is 4.5332 x 102 mg.


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(b) To convert 0.000 042 1 mL to scientific notation, move the decimal point 5 places to the right and include an exponent of 10S5. The result is 4.21 x 10S5 mL. (c) To convert 667,000 g to scientific notation, move the decimal point 5 places to the left and include an exponent of 105. The result is 6.67 x 105 g.

1.67 (a) Since the exponent is a negative 3, move the decimal point 3 places to the left to get

0.003 221 mm. (b) Since the exponent is a positive 5, move the decimal point 5 places to the right to get 894,000 m. (c) Since the exponent is a negative 12, move the decimal point 12 places to the left to get 0.000 000 000 001 350 82 m3. (d) Since the exponent is a positive 2, move the decimal point 2 places to the right to get 641.00 km.

1.68 (a) Since the digit to be dropped (0) is less than 5, round down. The result is 3.567 x 104

or 35,670 m (4 significant figures). Since the digit to be dropped (the second 6) is greater than 5, round up. The result is 35,670.1 m (6 significant figures). (b) Since the digit to be dropped is 5 with nonzero digits following, round up. The result is 69 g (2 significant figures). Since the digit to be dropped (0) is less than 5, round down. The result is 68.5 g (3 significant figures). (c) Since the digit to be dropped is 5 with nothing following, round down. The result is 4.99 x 103 cm (3 significant figures). (d) Since the digit to be dropped is 5 with nothing following, round down. The result is 2.3098 x 10S4 kg (5 significant figures).

1.69 (a) Since the digit to be dropped (1) is less than 5, round down. The result is 7.000 kg.

(b) Since the digit to be dropped is 5 with nothing following, round down. The result is 1.60 km. (c) Since the digit to be dropped (1) is less than 5, round down. The result is 13.2 g/cm3. (d) Since the digit to be dropped (1) is less than 5, round down. The result is 2,300,000. or 2.300 000 x 106.

1.70 (a) 4.884 x 2.05 = 10.012

The result should contain only 3 significant figures because 2.05 contains 3 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (1) is less than 5, round down. The result is 10.0.

(b) 94.61 / 3.7 = 25.57 The result should contain only 2 significant figures because 3.7 contains 2 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (second 5) is 5 with nonzero digits following, round up. The result is 26.

(c) 3.7 / 94.61 = 0.0391 The result should contain only 2 significant figures because 3.7 contains 2 significant


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figures (the smaller number of significant figures of the two). Since the digit to be dropped (1) is less than 5, round down. The result is 0.039.

(d)

5502.3

24

+ 0.01

5526.31

This result should be expressed with no decimal places. Since the

digit to be dropped (3) is less than 5, round down. The result is 5526.

(e)

86.3

+ 1.42

_ 0.09

87.63

This result should be expressed with only 1 decimal place. Since

the digit to be dropped (3) is less than 5, round down. The result is 87.6.

(f) 5.7 x 2.31 = 13.167 The result should contain only 2 significant figures because 5.7 contains 2 significant figures (the smaller number of significant figures of the two). Since the digit to be dropped (second 1) is less than 5, round down. The result is 13.

1.71 (a) 3.41 _ 0.23 3.18

x 0.205 = x 0.205 = 0.12457 = 0.1255.233 5.233

Complete the subtraction first. The result has 2 decimal places and 3 significant figures. The result of the multiplication and division must have 3 significant figures. Since the digit to be dropped is 5 with nonzero digits following, round up.

(b) 5.556 x 2.3 5.556 x 2.3

= = 3.08 = 3.14.223 _ 0.08 4.143

Complete the subtraction first. The result of the subtraction should have 2 decimal places and 3 significant figures (an extra digit is being carried until the calculation is completed). The result of the multiplication and division must have 2 significant figures. Since the digit to be dropped (8) is greater than 5, round up.

Unit Conversions

1.72 (a) 0.25 lb x 453.59 g

1 lb = 113.4 g = 110 g


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(b) 1454 ft x 12 in 2.54 cm 1 m

x x 1 ft 1 in 100 cm

= 443.2 m

(c) 2,941,526 mi2 x 2 2

1.6093 km 1000 m x

1 mi 1 km

= 7.6181 x 1012 m2

1.73 (a) 2.54 cm 1 m

5.4 in x x = 0.14 m1 in 100 cm

(b) 1 kg

66.31 lb x = 30.08 kg2.2046 lb

(c) _3 3

_3 33.7854 L 1 x 10 m0.5521 gal x x = 2.090 x 10 m1 gal 1 L

(d) mi 1.6093 km 1000 m 1 h 1 min m

65 x x x x = 29 h 1 mi 1 km 60 min 60 s s

(e) 3

3 31 m978.3 x = 748.0 yd m

1.0936 yd

(f) 2.380 mi2 x 2 2

1.6093 km 1000 m x

1 mi 1 km

= 6.164 x 106 m2

1.74 (a) 1 acre-ft x 22

31 5280 ftmi x = 43,560 ft640 acres 1 mi

(b) 3

833

5280 ft 1 acref- ft116 x x = 3.92 x acre- ft10mi

1 mi 43,560 ft

1.75 (a) 1/ 3 ft 12 in 2.54 cm

18.6 hands x x x = 189 cm1 hand 1 ft 1 in

(b) (6 x 2.5 x15) hands3 x 3 3 3 3

31/ 3 ft 12 in 2.54 cm 1 m x x x = 0.2 m

1 hand 1 ft 1 in 100 cm

1.76 (a) 200 mg 1000 mL

x = 2000 mg/L100 mL 1 L

(b) _3

_6

200 mg 1 x g 1 g10 x x = 2000 g/mL100 mL 1 mg 1 x g10

µ µ

(c) _3200 mg 1 x g 1000 mL10 x x = 2 g/L

100 mL 1 mg 1 L

(d) _3 _ 6

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200 mg 1 x g 1000 mL 1 ng 1 x L10 10 x x x x = 2000 ng/ L 100 mL 1 mg 1 L 1 x g 1 L10

µµ

(e) 2 g/L x 5 L = 10 g


12

1.77 14 lb

8.65 stones x = 121 lb1 stone

1.78 _ 4mi 5280 ft 12 in 2.54 cm 1 h 2.5 x s cm1055 x x x x x = 0.61

h 1 mi 1 ft 1 in 3600 s 1 shake shake

1.79 1 kg

160 lb x = 72.6 kg2.2046 lb

3

20 g 1 mg72.6 kg x x

1 kg 1 x g10

µµ

= 1.452 mg = 1.5 mg

Temperature

1.80 o o9F = ( x C) + 32

5

o oo 9F = ( x 39. C) + 32 = F9 103.8

5 (goat)

oo o9F = ( x 22. C) + 32 = F72.0 2

5 (Australian spiny anteater)

1.81 For Hg: mp is o9 x (_ 38.87) + 32 = F_ 37.97

5

For Br2: mp is o9 x (_ 7.2) + 32 = F19.0

5

For Cs: mp is o9 x (28.40) + 32 = F83.12

5

For Ga: mp is o9 x (29.78) + 32 = F85.60

5

1.82 oo o5 5C = x F _ 32) = x (6192 _ 32) = 342 C( 2

9 9

K = oC + 273.15 = 3422 + 273.15 = 3695.15 K or 3695 K

1.83 oo o9 9F = ( x C) + 32 = ( x 175) + 32 = F347

5 5

1.84 Ethanol boiling point 78.5oC 173.3oF 200oE

Ethanol melting point S117.3oC S179.1oF 0oE

(a) o o

o oo o o

E E200 200 = = 1.021 E/ C [78. C _ (_117. C)] 195. C5 3 8


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(b) o o

o oo o o

E E200 200 = = 0.5675 E/ F [173. F _ (_179. F)] 352. F3 1 4

(c) oo 200E = x C + 117.3)(

195.8

H2O melting point = 0oC; oo 200E = x (0 + 117.3) = 119. E8

195.8

H2O boiling point = 100oC; oo 200E = x (100 + 117.3) = 222. E0

195.8

(d) o oo 200 200E = x F + 179.1) = x (98.6 + 179.1) = 157. E( 6

352.4 352.4

(e) oo o 352.4 352.4F = E x _ 179.1 = 130 x _ 179.1 = F50.0

200 200

Since the outside temperature is 50.0oF, I would wear a sweater or light jacket. 1.85 NH3 boiling point S33.4oC S28.1oF 100oA

NH3 melting point S77.7oC S107.9oF 0oA

(a) o o

o oo o

A A100 100 = = 2.26 A / C [_ 33.4 _ (_ 77. C)] 44. C7 3

(b) o o

o oo o

A A100 100 = = 1.25 A / F [_ 28.1 _ (_107. F)] 79. F9 8

(c) o o100A = x ( C + 77.7)

44.3

H2O melting point = 0oC; oo 100A = x (0 + 77.7) = A175

44.3

H2O boiling point = 100oC; oo 100A = x (100 + 77.7) = A401

44.3

(d) oo o100 100A = x ( F + 107.9) = x (98.6 + 107.9) = A259

79.8 79.8

Density

1.86 _3 3

31 x g 1 10 cm250 mg x = 0.25 g; V = 0.25 g x = 0.18 cm1 mg 1.40 g

3

3 3453.59 g 1 cm500 lb x = 226,795 g; V = 226,795 g x = 161,996 = 162,000 cm cm1 lb 1.40 g

1.87 For H2: 1 L

V = 1.0078 g x = 11.2 L0.0899 g

For Cl2: 1 L

V = 35.45 g x = 11.03 L3.214 g


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1.88 3 3 3

m 220.9 g g gd = = = 11.4 = 11

V (0.50 x 1.55 x 25.00) cm cm cm

1.89 d = 2.40 mm = 0.240 cm

r = d/2 = 0.120 cm and V = πr2h

32

m 0.3624 gd = = = 0.534 g/cm

V ( 3.1416 ) ( 0.120 cm ( 15.0 cm ))

1.90 3 3

m 8.763 g 8.763 g g gd = = = = 2.331 = 2.33

V (28.76 _ 25.00) mL 3.76 mL cm cm

1.91 The explosion was caused by a chemical property. Na reacts violently with H2O. General Problems 1.92 (a) selenium, Se (b) rhenium, Re (c) cobalt, Co (d) rhodium, Rh 1.93 (a) Element 117 is a halogen because it would be found directly below At in group 7A.

(b) Element 119 (c) Element 115 would be found directly below Bi and would be a metal. Element 117 might have the properties of a semimetal. (d) Element 119, at the bottom of group 1A, would likely be a soft, shiny, very reactive metal forming a +1 cation.

1.94 NaCl melting point = 1074 K

oC = K S 273.15 = 1074 S 273.15 = 800.85oC = 801oC

o oo o9 9F = ( x C) + 32 = ( x 800.85) + 32 = 1473. F = F53 1474

5 5

NaCl boiling point = 1686 K oC = K S 273.15 = 1686 S 273.15 = 1412.85oC = 1413oC

o oo o9 9F = ( x C) + 32 = ( x 1412.85) + 32 = 2575. F = F13 2575

5 5

1.95 oo o9 9F = x C + 32 = x (_ 38.9) + 32 = _ 38. F0

5 5

1.96 1 mL

V = 112.5 g x = 75.85 mL1.4832 g

1.97 lb 453.59 g 1 gal 1 L

15.28 x x x = 1.831 g / mLgal 1 lb 3.7854 L 1000 mL


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1.98 10 10453.59 g 1 mL 1 LV = 8.728 x lb x x x = 2.162 x L10 10

1 lb 1.831 g 1000 mL

1.99 2.54 cm 10 mm

0.22 in x x = 5.6 mm1 in 1 cm

1.100 (a) density = 1 lb 8 pints 1 gal 453.59 g 1 L

x x x x 1 pint 1 gal 3.7854 L 1 lb 1000 mL

= 0.95861 g/mL

(b) area in m2 = 2 2 2 221 5280 ft 12 in 2.54 cm 1 mmi1 acre x x x x x

640 acres 1 mi 1 ft 1 in 100 cm

= 4047 m2

(c) mass of wood = 3 33

3

128 12 in 2.54 cm 0.40 g 1 kgft1 cord x x x x x 1 cord 1 ft 1 in 1 1000 gcm

= 1450 kg = 1400 kg

(d) mass of oil = 42 gal 3.7854 L 1000 mL 0.85 g 1 kg

1 barrel x x x x x 1 barrel 1 gal 1 L 1 mL 1000 g

= 135.1 kg = 140 kg

(e) fat Calories = 32 servings 165 Calories 30.0 Cal from fat

0.5 gal x x x 1 gal 1 serving 100 Cal total

= 792 Cal from fat

1.101 amount of chocolate =

105 mg caffeine 1.0 ounce chocolate2.0 cups coffee x x

1 cup coffee 15 mg caffeine = 14 ounces of chocolate

14 ounces of chocolate is just under 1 pound. 1.102 (a) number of Hershey=s Kisses =

453.59 g 1 serving 9 kisses2.0 lb x x x

1 lb 41 g 1 serving = 199 kisses = 200 kisses

(b) Hershey=s Kiss volume = 41 g 1 serving 1 mL

x x 1 serving 9 kisses 1.4 g

= 3.254 mL = 3.3 mL

(c) Calories/Hershey=s Kiss = 230 Cal 1 serving

x 1 serving 9 kisses

= 25.55 Cal/kiss = 26 Cal/kiss

(d) % fat Calories = 13 g fat 9 Cal from fat 1 serving

x x x 100%1 serving 1 g fat 230 Cal total

= 51% Calories from fat

1.103 Let Y equal volume of vinegar and (422.8 cm3 S Y) equal the volume of oil.

Mass = volume x density 397.8 g = (Y x 1.006 g/cm3) + [(422.8 cm3 S Y) x 0.918 g/cm3] 397.8 g = (1.006 g/cm3)Y + 388.1 g S (0.918 g/cm3)Y


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397.8 g S 388.1 g = (1.006 g/cm3)Y S (0.918 g/cm3)Y 9.7 g = (0.088 g/cm3)Y

Y = vinegar volume = 3

9.7 g

0.088 g/cm = 110 cm3

oil volume = (422.8 cm3 S Y) = (422.8 cm3 S 110 cm3) = 313 cm3

1.104 oo 5C = x F _ 32)(

9; o oSet C = F : oo 5

C = x C _ 32)( 9

o o o

o o

o

oo

9Solve for C : C x = C _ 32

59

C x ) _ C = _ 32( 5

4C x = _ 32

55

C = (_ 32) = _ 4 C0 4

The Celsius and Fahrenheit scales Across@ at o o_ C (_ F).40 40 1.105 Cork: volume = 1.30 cm x 5.50 cm x 3.00 cm = 21.45 cm3

mass = 33

0.235 g21.45 x = 5.041 gcm

1 cm

Lead: volume = (1.15 cm)3 = 1.521 cm3

mass = 33

11.35 g1.521 x = 17.26 gcm

1 cm

total mass = 5.041 g + 17.26 g = 22.30 g total volume = 21.45 cm3 + 1.521 cm3 = 22.97 cm3

average density = 33

22.30 g = 0.971 g/cm

22.97 cm so the cork and lead will float.

1.106 Convert 8 min, 25 s to s. 60 s

8 min x 1 min

+ 25 s = 505 s

Convert 293.2 K to oF 293.2 S 273.15 = 20.05oC

oo 9F = ( x 20.05) + 32 = 68. F09

5

Final temperature = 68.09oF + o

o3. F0505 s x = 93. F3460 s

oC = o5 x (93.34 _ 32) = 34. C1

9

1.107 Ethyl alcohol density = 19.7325 g

25.00 mL = 0.7893 g/mL

total mass = metal mass + ethyl alcohol mass = 38.4704 g


17

ethyl alcohol mass = total mass S metal mass = 38.4704 g S 25.0920 g = 13.3784 g

ethyl alcohol volume = 13.3784 g x 1 mL

0.7893 g = 16.95 mL

metal volume = total volume S ethyl alcohol volume = 25.00 mL S 16.95 mL = 8.05 mL

metal density = 25.0920 g

8.05 mL = 3.12 g/mL

1.108 Average brass density = (0.670)(8.92 g/cm3) + (0.330)(7.14 g/cm3) = 8.333 g/cm3

length = 1.62 in x 2.54 cm

1 in = 4.115 cm

diameter = 0.514 in x 2.54 cm

1 in = 1.306 cm

volume = πr2h = (3.1416)[(1.306 cm)/2]2(4.115 cm) = 5.512 cm3

mass = 5.512 cm3 x 3

8.333 g

1 cm = 45.9 g

1.109 35 sv = 35 x 109 3m

s

(a) gulf stream flow = 33

93

100 cm 1 mL 60 sm35 x 10s 1 m 1 1 mincm

= 2.1 x 1018 mL/min

(b) mass of H2O = ( )18 mL 60 min 1.025 g2.1 x 24 h10

min 1 h 1 mL

= 3.1 x 1021 g = 3.1 x 1018

kg

(c) time = ( )1518

1000 mL 1 min1.0 x L10

1 L 2.1 x mL10

= 0.48 min

1.110 (a) Gallium is a metal.

(b) Indium, which is right under gallium in the periodic table, should have similar chemical properties.

(c) Ga density = 3

33

0.2133 lb 453.59 g 1 in. x x 1 in 1 lb (2.54 cm).

= 5.904 g/cm3

(d) Ga boiling point 2204oC 1000oG Ga melting point 29.78oC 0oG

o o o

o o o

G _ G G1000 0 1000 = =C _ 29. C 2174. C2204 78 22

0.4599 oG/oC

oG = 0.4599 x (oC S 29.78) oG = 0.4599 x (801 S 29.78) = 355oG


18

The melting point of sodium chloride (NaCl) on the gallium scale is 355oG.


19

2

Chapter 2. Atoms, Molecules and Ions 2.1 First, find the S:O ratio in each compound.

Substance A: S:O mass ratio = (6.00 g S) / (5.99 g O) = 1.00 Substance B: S:O mass ratio = (8.60 g S) / (12.88 g O) = 0.668 S: O mass ratio in substance A 1.00 3

= = 1.50 = S: O mass ratio in substance B 0.668 2

2.2 0.0002 in x 48

2.54 cm 1 Au atom x = 2 x Au atoms10

1 in 2.9 x cm10•

2.3 _10

19 1.5 x m 1 km 1 time101 x C atoms x x x = 37.4 times10C atom 1000 m 40,075 km

. 40 times

2.4 75

34 Se has 34 protons, 34 electrons, and (75 S 34) = 41 neutrons. 2.5 35

17Cl has (35 S 17) = 18 neutrons. 3717 Cl has (37 S 17) = 20 neutrons.

2.6 The element with 47 protons is Ag. The mass number is the sum of the protons and the

neutrons, 47 + 62 = 109. The isotope symbol is 109 47 Ag .

2.7 atomic mass = (0.6917 x 62.94 amu) + (0.3083 x 64.93 amu) = 63.55 amu

2.8 2.15 g x 22_ 24

1 amu 1 Cu x = 2.04 x Cu atoms10

1.6605 x g 63.55 amu10

2.9

H H

| |

H _ C _ N _ H

|

H phantomC

2.10 Figure (b) represents a collection of hydrogen peroxide (H2O2) molecules. 2.11 adrenaline, C9H13NO3 2.12 (a) LiBr is composed of a metal (Li) and nonmetal (Br) and is ionic.


20

(b) SiCl4 is composed of only nonmetals and is molecular. (c) BF3 is composed of only nonmetals and is molecular. (d) CaO is composed of a metal (Ca) and nonmetal (O) and is ionic.

2.13 Figure (a) most likely represents an ionic compound because there are no discrete molecules, only a regular array of two different chemical species (ions). Figure (b) most likely represents a molecular compound because discrete molecules are present.

2.14 (a) HF is an acid. In water, HF dissociates to produce H+(aq).

(b) Ca(OH)2 is a base. In water, Ca(OH)2 dissociates to produce OH!(aq). (c) LiOH is a base. In water, LiOH dissociates to produce OH!(aq). (d) HCN is an acid. In water, HCN dissociates to produce H+(aq).

2.15 (a) CsF, cesium fluoride (b) K2O, potassium oxide (c) CuO, copper(II) oxide (d) BaS, barium sulfide 2.16 (a) vanadium(III) chloride, VCl3 (b) manganese(IV) oxide, MnO2

(c) copper(II) sulfide, CuS (d) aluminum oxide, Al 2O3 2.17 red B potassium sulfide, K2S; green B strontium iodide, SrI2; blue B gallium oxide, Ga2O3 2.18 (a) NCl3, nitrogen trichloride (b) P4O6, tetraphosphorus hexoxide

(c) S2F2, disulfur difluoride (d) SeO2, selenium dioxide 2.19 (a) disulfur dichloride, S2Cl2 (b) iodine monochloride, ICl

(c) nitrogen triiodide, NI3 2.20 (a) Ca(ClO)2, calcium hypochlorite

(b) Ag2S2O3, silver(I) thiosulfate or silver thiosulfate (c) NaH2PO4, sodium dihydrogen phosphate (d) Sn(NO3)2, tin(II) nitrate (e) Pb(CH3CO2)4, lead(IV) acetate (f) (NH4)2SO4, ammonium sulfate

2.21 (a) lithium phosphate, Li3PO4 (b) magnesium hydrogen sulfate, Mg(HSO4)2

(c) manganese(II) nitrate, Mn(NO3)2 (d) chromium(III) sulfate, Cr2(SO4)3 2.22 Drawing 1 represents ionic compounds with one cation and two anions. Only (c) CaCl2 is

consistent with drawing 1. Drawing 2 represents ionic compounds with one cation and one anion. Both (a) LiBr and (b) NaNO2 are consistent with drawing 2.

2.23 (a) HIO4, periodic acid (b) HBrO2, bromous acid (c) H2CrO4, chromic acid 2.24 A normal visual image results when light from the sun or other source reflects off an

object, strikes the retina in our eye, and is converted into electrical signals that are processed by the brain. The image obtained with a scanning tunneling microscope, by contrast, is a three-dimensional, computer-generated data plot that uses tunneling current to mimic depth perception. The nature of the computer-generated image depends on the identity of the molecules or atoms on the surface, on the precision with which the probe

Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________

21

tip is made, on how the data are manipulated, and on other experimental variables.

Understanding Key Concepts 2.25 Drawing (a) represents a collection of SO2 molecules. Drawing (d) represents a mixture

of S atoms and O2 molecules. 2.26 To obey the law of mass conservation, the correct drawing must have the same number of

red and yellow spheres as in drawing (a). The correct drawing is (d).

2.27 Figures (b) and (d) illustrate the law of multiple proportions. The mass ratio is 2.

2.28. (a) alanine, C3H7NO2 (b) ethylene glycol, C2H6O2 (c) acetic acid, C2H4O2 2.29 A Na atom has 11 protons and 11 electrons [drawing (b)].

A Ca2+ ion has 20 protons and 18 electrons [drawing (c)]. A FS ion has 9 protons and 10 electrons [drawing (a)].

2.30 2.31 (a) MgSO4 (b) Li2CO3 (c) FeCl2 (d) Ca3(PO4)2 Additional Problems Atomic Theory 2.32 The law of mass conservation in terms of Dalton=s atomic theory states that chemical

reactions only rearrange the way that atoms are combined; the atoms themselves are not changed. The law of definite proportions in terms of Dalton=s atomic theory states that the chemical combination of elements to make different substances occurs when atoms join together in small, whole-number ratios.

2.33 The law of multiple proportions states that if two elements combine in different ways to

form different substances, the mass ratios are small, whole-number multiples of each other. This is very similar to Dalton=s statement that the chemical combination of elements to make different substances occurs when atoms join together in small, whole-number ratios.


22

2.34 First, find the C:H ratio in each compound. Benzene: C:H mass ratio = (4.61 g C) / (0.39 g H) = 12 Ethane: C:H mass ratio (4.00 g C) / (1.00 g H) = 4.00 Ethylene: C:H mass ratio = (4.29 g C) / (0.71 g H) = 6.0 C : H mass ratio in benzene 12 3

= = C : H mass ratio in ethane 4.00 1

C : H mass ratio in benzene 12 2 = =

C : H mass ratio in ethylene 6.0 1

C : H mass ratio in ethylene 6.0 3 = =

C : H mass ratio in ethane 4.00 2

2.35 First, find the C:O ratio in each compound.

Carbon suboxide: C:O mass ratio = (1.32 g C) / (1.18 g O) = 1.12 Carbon dioxide: C:O mass ratio = (12.00 g C) / (32.00 g O) = 0.375 C : O mass ratio in carbon suboxide 1.12 3

= = C : O mass ratio in carbon dioxide 0.375 1

2.36 (a) For benzene:

4.61 g x 23_ 24

1 amu 1 C atom x = 2.31 x C atoms10

1.6605 x g 12.011 amu10

0.39 g x 23_ 24

1 amu 1 H atom x = 2.3 x H atoms10

1.6605 x g 1.008 amu10

23

23

C 2.31 x C atoms 1 C10 = = H 2.3 x H atoms 1 H10

A possible formula for benzene is CH.

For ethane:

4.00 g x 23_ 24


1.6605 x g 12.011 amu10

1.00 g x 23_ 24


1.6605 x g 1.008 amu10

23

23


A possible formula for ethane is CH3.

For ethylene:

4.29 g x 23_ 24


1.6605 x g 12.011 amu10

0.71 g x 23_ 24


1.6605 x g 1.008 amu10

23

23



23

A possible formula for ethylene is CH2.

(b) The results in part (a) give the smallest whole-number ratio of C to H for benzene, ethane, and ethylene, and these ratios are consistent with their modern formulas.

2.37 1.32 g x 22_ 24


1.6605 x g 12.011 amu10

1.18 g x 22_ 24

1 amu 1 O atom x = 4.44 x O atoms10

1.6605 x g 15.9994 amu10

22

22

C 6.62 x C atoms 1.5 C10 = = ;O 4.44 x O atoms 1 O10

therefore the formula for carbon suboxide is C1.5O, or C3O2.

2.38 (a) _ 24 23g(1.67 x )(6.02 x H atoms) = 1.01 g10 10

H atom

This result is numerically equal to the atomic mass of H in grams.

(b) _ 24 23g(26.558 x )(6.02 x O atoms) = 16.0 g10 10

O atom

This result is numerically equal to the atomic mass of O in grams. 2.39 The mass of 6.02 x 1023 atoms is its atomic mass expressed in grams.

(a) If the atomic mass of an element is X, then 6.02 x 1023 atoms of this element weighs X grams. (b) If the mass of 6.02 x 1023 atoms of element Y is 83.80 g, then the atomic mass of Y is 83.80. Y is Kr.

2.40 Assume a 1.00 g sample of the binary compound of zinc and sulfur.

0.671 x 1.00 g = 0.671 g Zn; 0.329 x 1.00 g = 0.329 g S

0.671 g x 21_ 24

1 amu 1 Zn atom x = 6.18 x Zn atoms10

1.6605 x g 65.39 amu10

0.329 g x 21_ 24

1 amu 1 S atom x = 6.18 x S atoms10

1.6605 x g 32.066 amu10

21

21

Zn 6.18 x Zn atoms 1 Zn10 = = ;S 6.18 x S atoms 1 S10

therefore the formula is ZnS.

2.41 Assume a 1.000 g sample of one of the binary compounds.

0.3104 x 1.000 g = 0.3104 g Ti; 0.6896 x 1.000 g = 0.6896 g Cl

0.3104 g x 21_ 24

1 amu 1 Ti atom x = 3.90 x Ti atoms10

1.6605 x g 47.88 amu10

0.6896 g x 22_ 24

1 amu 1 Cl atom x = 1.17 x Cl atoms10

1.6605 x g 35.453 amu10


24

22

21

Cl 1.17 x 310 = = Ti 3.90 x 110

Assume a 1.000 g sample of the other binary compound. 0.2524 x 1.000 g = 0.2524 g Ti; 0.7476 x 1.000 g = 0.7476 g Cl

0.2524 g x 21_ 24

1 amu 1 Ti atom x = 3.17 x Ti atoms10

1.6605 x g 47.88 amu10

0.7476 g x 22_ 24

1 amu 1 Cl atom x = 1.27 x Cl atoms10

1.6605 x g 35.453 amu10

22

21

Cl 1.27 x 410 = = Ti 3.17 x 110

Elements and Atoms 2.42 The atomic number is equal to the number of protons.

The mass number is equal to the sum of the number of protons and the number of neutrons. 2.43 The atomic number is equal to the number of protons.

The atomic mass is the weighted average mass (in amu) of the various isotopes for a particular element.

2.44 Atoms of the same element that have different numbers of neutrons are called isotopes. 2.45 The mass number is equal to the sum of the number of protons and the number of

neutrons for a particular isotope. For 14

6 C , mass number = 6 protons + 8 neutrons = 14.

For 147 N , mass number = 7 protons + 7 neutrons = 14.

2.46 The subscript giving the atomic number of an atom is often left off of an isotope symbol

because one can readily look up the atomic number in the periodic table. 2.47 Te has isotopes with more neutrons than the isotopes of I. 2.48 (a) carbon, C (b) argon, Ar (c) vanadium, V 2.49 137

55 Cs 2.50 (a) 220

86 Rn (b) 210 84 Po (c) 197

79 Au 2.51 (a) 140

58 Ce (b) 6027 Co

2.52 (a) 15

7 N , 7 protons, 7 electrons, (15 S 7) = 8 neutrons


25

(b) 6027Co, 27 protons, 27 electrons, (60 S 27) = 33 neutrons

(c) 131

53 I , 53 protons, 53 electrons, (131 S 53) = 78 neutrons

(d) 142 58 Ce, 58 protons, 58 electrons, (142 S 58) = 84 neutrons

2.53 (a) 27Al , 13 protons and (27 S 13) = 14 neutrons

(b) 32S , 16 protons and (32 S 16) = 16 neutrons

(c) 64Zn , 30 protons and (64 S 30) = 34 neutrons

(d) 207Pb , 82 protons and (207 S 82) = 125 neutrons 2.54 (a) 24

12 Mg , magnesium (b) 5828Ni, nickel

(c) 10446 Pd, palladium (d) 183

74 W, tungsten 2.55 (a) 202

80 Hg, mercury (b) 19578 Pt , platinum

(c) 18476 Os, osmium (d) 209

83 Bi, bismuth

2.56 (0.199 x 10.0129 amu) + (0.801 x 11.009 31 amu) = 10.8 amu for B 2.57 (0.5184 x 106.9051 amu) + (0.4816 x 108.9048 amu) = 107.9 amu for Ag 2.58 24.305 amu = (0.7899 x 23.985 amu) + (0.1000 x 24.986 amu) + (0.1101 x Z)

Solve for Z. Z = 25.982 amu for 26Mg. 2.59 The total abundance of all three isotopes must be 100.00%. The natural abundance of

29Si is 4.67%. The natural abundance of 28Si and 30Si together must be 100.00% S 4.67% = 95.33%. Let Y be the natural abundance of 28Si and [95.33 S Y] the natural abundance of 30Si. 28.0855 amu = (0.0467 x 28.9765 amu) + (Y x 27.9769 amu)

+ ([0.9533 S Y] x 29.9738 amu)

Solve for Y. _1.842

Y = = 0.922_1.997

28Si natural abundance = 92.2% 30Si natural abundance = 95.33 S 92.2 = 3.1% Compounds and Mixtures, Molecules and Ions 2.60 (a) muddy water, heterogeneous mixture

(b) concrete, heterogeneous mixture (c) house paint, homogeneous mixture (d) a soft drink, homogeneous mixture (heterogeneous mixture if it contains CO2 bubbles)

2.61 (a) 18 karat gold, (b) window glass, and (d) liquefied air are homogeneous mixtures.

(c) Tomato juice is a heterogeneous mixture because the liquid contains solid pulp.


26

2.62 An atom is the smallest particle that retains the chemical properties of an element. A molecule is matter that results when two or more atoms are joined by covalent bonds. H and O are atoms, H2O is a water molecule.

2.63 A molecule is the unit of matter that results when two or more atoms are joined by

covalent bonds. An ion results when an atom gains or loses electrons. CH4 is a methane molecule. Na+ is the sodium cation.

2.64 A covalent bond results when two atoms share several (usually two) of their electrons.

An ionic bond results from a complete transfer of one or more electrons from one atom to another. The CBH bonds in methane (CH4) are covalent bonds. The bond in NaCl (Na+ClS) is an ionic bond.

2.65 Covalent bonds typically form between nonmetals. (a) BBBr, (c) BrBCl, and (d) OBBr

are covalent bonds. Ionic bonds typically form between a metal and a nonmetal. (b) NaBBr is an ionic bond.

2.66 Element symbols are composed of one or two letters. If the element symbol is two letters,

the first letter is uppercase and the second is lowercase. CO stands for carbon and oxygen in carbon monoxide.

2.67 (a) The formula of ammonia is NH3.

(b) The ionic solid potassium chloride has the formula KCl. (c) ClS is an anion. (d) CH4 is a neutral molecule.

2.68 (a) Be2+, 4 protons and 2 electrons (b) Rb+, 37 protons and 36 electrons

(c) Se2S, 34 protons and 36 electrons (d) Au3+, 79 protons and 76 electrons 2.69 (a) A +2 cation that has 36 electrons must have 38 protons. X = Sr.

(b) A S1 anion that has 36 electrons must have 35 protons. X = Br. 2.70 C3H8O 2.71 C3H6O3

2.72

H H H H

| | | |

H _ C _ C _ C _ C _ H

| | | |

H H H H


27

2.73 Acids and Bases 2.74 (a) HI, acid (b) CsOH, base (c) H3PO4, acid

(d) Ba(OH)2, base (e) H2CO3, acid 2.75 (a) HI, one H+ ion (b) H3PO4, three H+ ions (c) H2CO3, two H+ ions 2.76 HI(aq) H+(aq) + IS(aq); the anion is IS

H3PO4(aq) H+(aq) + H2PO4S(aq); the predominant anion is H2PO4

S H2CO3(aq) H+(aq) + HCO3

S(aq); the predominant anion is HCO3S

2.77 (b) CsOH(aq) Cs+(aq) + OHS(aq); the cation is Cs+

(d) Ba(OH)2(aq) Ba2+(aq) + 2 OHS(aq); the cation is Ba2+ Naming Compounds 2.78 (a) KCl (b) SnBr2 (c) CaO (d) BaCl2 (e) AlH3 2.79 (a) Ca(CH3CO2)2 (b) Fe(CN)2 (c) Na2Cr2O7 (d) Cr2(SO4)3 (e) Hg(ClO4)2 2.80 (a) barium ion (b) cesium ion (c) vanadium(III)

ion (d) hydrogen carbonate ion (e) ammonium ion (f) nickel(II) ion (g) nitrite ion (h) chlorite ion (i) manganese(II) ion (j) perchlorate ion

2.81 (a) carbon tetrachloride (b) chlorine dioxide

(c) dinitrogen monoxide (d) dinitrogen trioxide 2.82 (a) SO3

2S (b) PO43S (c) Zr4+ (d) CrO4

2S (e) CH3CO2S (f) S2O3

2S 2.83 (a) Zn2+ (b) Fe3+ (c) Ti4+ (d) Sn2+ (e) Hg2

2+ (f) Mn4+ (g) K+ (h) Cu2+ 2.84 (a) zinc(II) cyanide (b) iron(III) nitrite (c) titanium(IV) sulfate

(d) tin(II) phosphate (e) mercury(I) sulfide (f) manganese(IV) oxide (g) potassium periodate (h) copper(II) acetate


28

2.85 (a) magnesium sulfite (b) cobalt(II) nitrite (c) manganese(II) hydrogen carbonate (d) zinc(II) chromate

(g) aluminum sulfate (h) lithium chlorate 2.86 (a) Na+ and SO4

2S; therefore the formula is Na2SO4 (b) Ba2+ and PO4

3S; therefore the formula is Ba3(PO4)2 (c) Ga3+ and SO4

2S; therefore the formula is Ga2(SO4)3 2.87 (a) Na2O2 (b) AlBr3 (c) Cr2(SO4)3 General Problems 2.88 atomic mass = (0.205 x 69.924 amu) + (0.274 x 71.922 amu)

+ (0.078 x 72.923 amu) + (0.365 x 73.921 amu) + (0.078 x 75.921 amu) = 72.6 amu

2.89 _ 24

_ 2312.011 amu 1.6605 x g10mass of 1 C atom = x = 2.00 x g / C atom101 C atoms 1 amu

number of C atoms = _5

17_ 23

1 x g10 = 5 x C atoms102.00 x g / C atom10

time = 17

175 x C atoms10 = 2.5 x s102 C atoms / s

2.90 (a) sodium bromate (b) phosphoric acid

(c) phosphorous acid (d) vanadium(V) oxide 2.91 (a) Ca(HSO4)2 (b) SnO (c) Ru(NO3)3 (d) (NH4)2CO3 (e) HI (f) Be3(PO4)2

2.92 For NH3, 3 x 1.0079 amu H

(2.34 g N) = 0.505 g H14.0067 amu N

For N2H4, 4 x 1.0079 amu H

(2.34 g N) = 0.337 g H2 x 14.0067 amu N

2.93 3.670 g N

g N = (1.575 g H) = 10.96 g N0.5275 g H

From Problem 2.92:

for NH3, g N 2.34 g N

= = 4.63g H 0.505 g H

for N2H4, g N 2.34 g N

= = 6.94g H 0.337 g H


29

for compound X, g N 10.96 g N

= = 6.96g H 1.575 g H

; X is N2H4

2.94 TeO42S, tellurate; TeO3

2S, tellurite. TeO4

2S and TeO32S are analogous to SO4

2S and SO32S.

2.95 H2TeO4, telluric acid; H2TeO3, tellurous acid 2.96 (a) IS (b) Au3+ (c) Kr

2.97 12.0000 amu X

= 15.9994 amu 16.0000 amu

; X = 12.0005 amu for 12C prior to 1961.

2.98 39.9626 amu X

= ;15.9994 amu 16.0000 amu

X = 39.9641 amu for 40Ca prior to 1961.

2.99 (a) AsO4

3S, arsenate (b) SeO32S, selenite

(c) SeO42S, selenate (d) HAsO4

2S, hydrogen arsenate 2.100 (a) calcium-40, 40Ca

(b) Not enough information, several different isotopes can have 63 neutrons. (c) The neutral atom contains 26 electrons. The ion is iron-56, 56Fe3+. (d) Se2S

2.101 Deuterium is 2H and deuterium fluoride is 2HF.

2H has 1 proton, 1 neutron, and 1 electron. F has 9 protons, 10 neutrons, and 9 electrons. 2HF has 10 protons, 11 neutrons, and 10 electrons. Chemically, 2HF is like HF and is a weak acid.

2.102 1H35Cl has 18 protons, 18 neutrons, and 18 electrons.

1H37Cl has 18 protons, 20 neutrons, and 18 electrons. 2H35Cl has 18 protons, 19 neutrons, and 18 electrons. 2H37Cl has 18 protons, 21 neutrons, and 18 electrons. 3H35Cl has 18 protons, 20 neutrons, and 18 electrons. 3H37Cl has 18 protons, 22 neutrons, and 18 electrons.

2.103 (a) 40Ar has 18 protons, 22 neutrons, and 18 electrons

(b) 40Ca2+ has 20 protons, 20 neutrons, and 18 electrons (c) 39K+ has 19 protons, 20 neutrons, and 18 electrons (d) 35ClS has 17 protons, 18 neutrons, and 18 electrons

2.104 (a) Mg2+ and ClS, MgCl2, magnesium chloride

(b) Ca2+ and O2S, CaO, calcium oxide (c) Li+ and N3S, Li3N, lithium nitride (d) Al3+ and O2S, Al2O3, aluminum oxide


30

2.105

2.106

2.107 Mass of H2SO4 solution = 1.3028 g/mL x 40.00 mL = 52.112 g

Total mass of Zn and H2SO4 solution before reaction = 9.520 g + 52.112 g = 61.632 g Total mass of solution after the reaction = 61.338 g Because of the conservation of mass, the difference between the two masses is the mass of H2 produced. H2 mass = 61.632 g S 61.338 g = 0.294 g

H2 volume = 0.294 g H2 x 2

1 L

0.0899 g H = 3.27 L H2

2.108 Molecular mass = (8 x 12.011 amu) + (9 x 1.0079 amu) + (1 x 14.0067 amu)

+ (2 x 15.9994 amu) = 151.165 amu

2.109 mass % C = 8 x 12.011

151.165 x 100 = 63.565%

mass % H = 9 x 1.0079

151.165 x 100 = 6.0008%

mass % N = 14.0067

151.165 x 100 = 9.2658%

mass % O = 2 x 15.9994

151.165 x 100 = 21.168%


31

2.110 (a) Aspirin is likely a molecular compound because it is composed of only nonmetal

elements. (b) Assume a 100.0 g sample of aspirin. It would contain: 60.00 g C, 4.48 g H, and 35.52 g O.

_ 24

1 amu 1 C atom60.00 g x x =

1.6605 x g 12.011 amu10 3.008 x 1024 C atoms

_ 24

1 amu 1 H atom4.48 g x x =

1.6605 x g 1.008 amu10 2.68 x 1024 H atoms

_ 24

1 amu 1 O atom35.52 g x x =

1.6605 x g 15.999 amu10 1.337 x 1024 O atoms

The atom ratio in aspirin is: 2424 242.68 x 3.008 x 1.337 x 1010 10 C OH , divide each subscript by 1 x 1024

C3.008 H2.68 O1.337 , divide each subscript by the smallest, 1.337 C3.008 / 1.337 H2.68 / 1.337 O1.337 / 1.337 C2.25H2 O, multiply each subscript by 4 C(2.25 x 4) H(2 x 4) O(1 x 4) C9H8O4

2.111 (a) Because X reacts by losing electrons, it is likely to be a metal.

(b) Because Y reacts by gaining electrons, it is likely to be a nonmetal. (c) X2Y3 (d) X is likely to be in group 3A and Y is likely to be in group 6A.

2.112 65.39 amu = (0.4863 x 63.929 amu) + (0.2790 x Z) + (0.0410 x 66.927 amu) + (0.1875 x 67.925 amu) + (0.0062 x 69.925 amu)

Solve for Z. 65.39 amu = 47.00 amu + (0.2790 x Z) 65.39 amu S 47.00 amu = 18.39 amu = 0.2790 x Z 18.39 amu/0.2790 = Z Z = 65.91 amu for 66Zn

33

3

Formulas, Equations, and Moles 3.1 2 KClO3 → 2 KCl + 3 O2 3.2 (a) C6H12O6 → 2 C2H6O + 2 CO2

(b) 4 Fe + 3 O2 → 2 Fe2O3 (c) 4 NH3 + Cl2 → N2H4 + 2 NH4Cl

3.3 3 A2 + 2 B → 2 BA3 3.4 (a) Fe2O3: 2(55.85) + 3(16.00) = 159.7 amu

(b) H2SO4: 2(1.01) + 1(32.07) + 4(16.00) = 98.1 amu (c) C6H8O7: 6(12.01) + 8(1.01) + 7(16.00) = 192.1 amu (d) C16H18N2O4S: 16(12.01) + 18(1.01) + 2(14.01) + 4(16.00) + 1(32.07) = 334.4 amu

3.5 Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

0.500 mol CO mol 1.50 = OFe mol 1

CO mol 3 x OFe

3232

3.6 C5H11NO2S: 5(12.01) + 11(1.01) + 1(14.01) + 2(16.00) + 1(32.07) = 149.24 amu 3.7 C9H8O4, 180.2 amu; 500 mg = 500 x 10-3 g = 0.500 g

0.500 g x aspirin mol 10 x 2.77 = g 180.2

mol 1 3_

2.77 x 10-3 mol x moleculesaspirin 10 x 1.67 = mol 1

molecules 10 x 6.02 2123

3.8 salicylic acid, C7H6O3, 138.1 amu; acetic anhydride, C4H6O3, 102.1 amu

aspirin, C9H8O4, 180.2 amu; acetic acid, C2H4O2, 60.1 amu

4.50 g C7H6O3 x OHC mol 1OHC g 102.1

x OHC mol 1OHC mol 1

x OHC g 138.1

OHC mol 1

364

364

367

364

367

367 = 3.33 g

C4H6O3



x OHC g 138.1

OHC mol 1

489

489

367

489

367

367 = 5.87 g

C9H8O4



x OHC g 138.1

OHC mol 1

242

242

367

242

367

367 = 1.96 g

34

C2H4O2 3.9 C2H4, 28.1 amu; C2H6O, 46.1 amu

4.6 g OHC mol 1

OHC g 46.1 x

HC mol 1

OHC mol 1 x

HC g 28.1HC mol 1

x HC62

62

42

62

42

4242 = 7.5 g C2H6O

(theoretical yield)

% 63 = % 100 x g 7.5

g 4.7 = % 100 x

yield lTheoretica

yield Actual = yieldPercent

3.10 CH4, 16.04 amu; CH2Cl2, 84.93 amu; 1.85 kg = 1850 g

1850 g CH4 x ClCH mol 1ClCH g 84.93

x CH mol 1

ClCH mol 1 x

CH g 16.04CH mol 1

22

22

4

22

4

4 = 9800 g CH2Cl2

(theoretical yield) Actual yield = (9800 g)(0.431) = 4220 g CH2Cl2 3.11 Li2O, 29.9 amu: 65 kg = 65,000 g; H2O, 18.0 amu: 80.0 kg = 80,000 g

65,000 g Li2O x OLi g 29.9

OLi mol 1

2

2 = 2.17 x 103 mol Li2O

80,000 g H2O x OH g 18.0

OH mol 1

2

2 = 4.44 x 103 mol H2O

The reaction stoichiometry between Li2O and H2O is one to one. There are twice as many moles of H2O as there are moles of Li2O. Therefore, Li2O is the limiting reactant. (4.44 x 103 mol - 2.17 x 103 mol) = 2.27 x 103 mol H2O remaining

2.27 x 103 mol H2O x OH mol 1

OH g 18.0

2

2 = 40,860 g H2O = 40.9 kg = 41 kg H2O

3.12 LiOH, 23.9 amu; CO2, 44.0 amu

CO g 921 = CO mol 1CO g 44.0

x LiOH mol 1CO mol 1

x LiOH g 23.9

LiOH mol 1 x LiOH g 500.0 2

2

22

3.13 (a) A + B2 → AB2

There is a 1:1 stoichiometry between the two reactants. A is the limiting reactant because there are fewer reactant A's than there are reactant B2's. (b) 1.0 mol of AB2 can be made from 1.0 mol of A and 1.0 mol of B2.

3.14 (a) 125 mL = 0.125 L; (0.20 mol/L)(0.125 L) = 0.025 mol NaHCO3

(b) 650.0 mL = 0.6500 L; (2.50 mol/L)(0.6500 L) = 1.62 mol H2SO4 3.15 (a) NaOH, 40.0 amu; 500.0 mL = 0.5000 L

NaOH g 25.0 = NaOH mol 1

NaOH g 40.0 x L 0.500 x

L

NaOH mol 1.25

(b) C6H12O6, 180.2 amu

OHC g 67.6 = OHC mol 1OHC g 180.2

x L 1.50 x L

OHC mol 0.250 6126

6126

61266126

Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________

35

3.16 C6H12O6, 180.2 amu;

25.0 g C6H12O6 x OHC g 180.2

OHC mol 1

6126

6126 = 0.1387 mol C6H12O6

mol 0.20

L 1 x mol 0.1387 = 0.69 L; 0.69 L = 690 mL

3.17 C27H46O, 386.7 amu; 750 mL = 0.750 L

OHC g 1 = OHC mol 1

OHC g 386.7 x L 0.750 x

L

OHC mol 0.005 4627

4627

46274627

3.18 Mi x Vi = Mf x Vf; Mf = mL 400.0

mL 75.0 x M 3.50 =

V

V x M

f

ii = 0.656 M

3.19 Mi x Vi = Mf x Vf; mL 6.94 = M 18.0

mL 250.0 x M 0.500 =

M

V x M = Vi

ffi

Dilute 6.94 mL of 18.0 M H2SO4 with enough water to make 250.0 mL of solution. The resulting solution will be 0.500 M H2SO4.

3.20 50.0 mL = 0.0500 L; (0.100 mol/L)(0.0500 L) = 5.00 x 10-3 mol NaOH

5.00 x 10-3 mol NaOH x NaOH mol 2

SOH mol 1 42 = 2.50 x 10-3 mol H2SO4

volume = mol 0.250

L 1 x mol 10 x 2.50 3_ = 0.0100 L; 0.0100 L = 10.0 mL H2SO4

3.21 HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)

25.0 mL = 0.0250 L and 68.5 mL = 0.0685 L

HNO mol 10 x 3.75 = KOH mol 1HNO mol 1

x L 0.0250 x L

KOH mol 0.150 3

3_3

M 10 x 5.47 = L 0.0685

mol 10 x 3.75 =molarity HNO 2_

3_

3

3.22 From the reaction stoichiometry, moles NaOH = moles CH3CO2H

(0.200 mol/L)(0.0947 L) = 0.018 94 mol NaOH = 0.018 94 mol CH3CO2H

molarity = L 0.0250

mol 94 0.018 = 0.758 M

3.23 For dimethylhydrazine, C2H8N2, divide each subscript by 2 to obtain the empirical

formula. The empirical formula is CH4N. C2H8N2, 60.1 amu or 60.1 g/mol

% 39.9 = % 100 x g 60.1

g 12.0 x 2 = C %

% 13.4 = % 100 x g 60.1

g 1.01 x 8 = H %


36

% 46.6 = % 100 x g 60.1

g 14.0 x 2 = N %

3.24 Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains

14.25 g C, 56.93 g O, and 28.83 g Mg.

14.25 g C x C g 12.0

C mol 1 = 1.19 mol C

56.93 g O x O g 16.0

O mol 1 = 3.56 mol O

28.83 g Mg x Mg g 24.3

Mg mol 1 = 1.19 mol Mg

Mg1.19C1.19O3.56; divide each subscript by the smallest, 1.19. Mg1.19 / 1.19C1.19 / 1.19O3.56 / 1.19 The empirical formula is MgCO3.

3.25 1.161 g H2O x OH mol 1

H mol 2 x

OH g 18.0

OH mol 1

22

2 = 0.129 mol H

2.818 g CO2 x CO mol 1

C mol 1 x

CO g 44.0CO mol 1

22

2 = 0.0640 mol C

0.129 mol H x H mol 1

H g 1.01 = 0.130 g H

0.0640 mol C x C mol 1

C g 12.0 = 0.768 g C

1.00 g total - (0.130 g H + 0.768 g C) = 0.102 g O

0.102 g O x O g 16.0

O mol 1 = 0.006 38 mol O

C0.0640H0.129O0.006 38; divide each subscript by the smallest, 0.006 38. C0.0640 / 0.006 38H0.129 / 0.006 38O0.006 38 / 0.006 38 C10.03H20.22O1 The empirical formula is C10H20O.

3.26 The empirical formula is CH2O, 30 amu: molecular mass = 150 amu.

5 = amu 30

amu 150 =

mass formula empirical

massmolecular ; therefore

molecular formula = 5 x empirical formula = C(5 x 1)H(5 x 2)O(5 x 1) = C5H10O5 3.27 (a) Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample

contains 21.86 g H and 78.14 g B.

H mol 21.6 = H g 1.01

H mol 1 x H g 21.86

B mol 7.24 = B g 10.8

B mol 1 x B g 78.14

B7.24 H21.6; divide each subscript by the smaller, 7.24.


37

B7.24 / 7.24 H21.6 / 7.24 The empirical formula is BH3, 13.8 amu. 27.7 amu / 13.8 amu = 2; molecular formula = B(2 x 1)H(2 x 3) = B2H6. (b) Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains 6.71 g H, 40.00 g C, and 53.28 g O.

H mol 6.64 = H g 1.01

H mol 1 x H g 6.71

C mol 3.33 = C g 12.0

C mol 1 x C g 40.00

O mol 3.33 = O g 16.0

O mol 1 x O g 53.28

C3.33 H6.64 O3.33; divide each subscript by the smallest, 3.33. C3.33 / 3.33 H6.64 / 3.33 O3.33 / 3.33 The empirical formula is CH2O, 30.0 amu. 90.08 amu / 30.0 amu = 3; molecular formula = C(3 x 1)H(3 x 2)O(3 x 1) = C3H6O3

3.28 Main sources of error in calculating Avogadro's number by spreading oil on a pond are:

(i) the assumption that the oil molecules are tiny cubes (ii) the assumption that the oil layer is one molecule thick

(iii) the assumption of a molecular mass of 200 for the oil 3.29 area of oil = 2.0 x 107 cm2

volume of oil = 4.9 cm3 = area x 4 l = (2.0 x 107 cm2) x 4 l

l = )(4)cm 10 x (2.0

cm 4.927

3

= 6.125 x 10-8 cm

area of oil = 2.0 x 107 cm2 = l2 x N = (6.125 x 10-8 cm)2 x N

N = )cm 10 x (6.125

cm 10 x 2.028_

27

= 5.33 x 1021 oil molecules

moles of oil = (4.9 cm3) x (0.95 g/cm3) x oil g 200

oil mol 1= 0.0233 mol oil

Avogadro's number = mol 0.0233

molecules 10 x 5.33 21

= 2.3 x 1023 molecules/mole

Understanding Key Concepts 3.30 The concentration of a solution is cut in half when the volume is doubled. This is best

represented by box (b). 3.31 (c) 2 A + B2 → A2B2 3.32 The molecular formula for cytosine is C4H5N3O.

mol CO2 = = CCO 1

x cyt

C 4cyt x mol 0.001 2 0.004 mol CO2


38

mol H2O = = H 2

OH 1 x

cyt

H 5cyt x mol 0.001 2 0.0025 mol H2O

3.33 reactants, box (d), and products, box (c) 3.34 C17H18F3NO 17(12.01) + 18(1.01) + 3(19.00) + 1(14.01) + 1(16.00) = 309.36 amu 3.35 Because the two volumes are equal (let the volume = y L), the concentrations are

proportional to the number of solute ions.

OH- concentration = 1.00 M x Ly

OH 8 x

H 12

Ly _

+ = 0.67 M

3.36 (a) A2 + 3 B2 → 2 AB3; B2 is the limiting reactant because it is completely consumed.

(b) For 1.0 mol of A2, 3.0 mol of B2 are required. Because only 1.0 mol of B2 is available, B2 is the limiting reactant.

1 mol B2 x B mol 3

AB mol 2

2

3 = 2/3 mol AB3

O2 3.37 CxHy → 3 CO2 + 4 H2O; x is equal to the coefficient for CO2 and y is equal to 2

times the coefficient for H2O. The empirical formula for the hydrocarbon is C3H8. Additional Problems Balancing Equations 3.38 Equation (b) is balanced, (a) is not balanced . 3.39 (a) and (c) are not balanced, (b) is balanced.

(a) 2 Al + Fe2O3 → Al2O3 + 2 Fe (balanced) (c) 4 Au + 8 NaCN + O2 + 2 H2O → 4 NaAu(CN)2 + 4 NaOH (balanced)

3.40 (a) Mg + 2 HNO3 → H2 + Mg(NO3)2

(b) CaC2 + 2 H2O → Ca(OH)2 + C2H2 (c) 2 S + 3 O2 → 2 SO3 (d) UO2 + 4 HF → UF4 + 2 H2O

3.41 (a) 2 NH4NO3 → 2 N2 + O2 + 4 H2O

(b) C2H6O + O2 → C2H4O2 + H2O (c) C2H8N2 + 2 N2O4 → 3 N2 + 2 CO2 + 4 H2O

Molecular Masses and Moles 3.42 Hg2Cl2: 2(200.59) + 2(35.45) = 472.1 amu


39

C4H8O2: 4(12.01) + 8(1.01) + 2(16.00) = 88.1 amu CF2Cl2: 1(12.01) + 2(19.00) + 2(35.45) = 120.9 amu

3.43 (a) (1 x 30.97 amu) + (Y x 35.45 amu) = 137.3 amu; Solve for Y; Y = 3.

The formula is PCl3. (b) (10 x 12.01 amu) + (14 x 1.008 amu) + (Z x 14.01 amu) = 162.2 amu. Solve for Z; Z = 2. The formula is C10H14N2.

3.44 One mole equals the atomic mass or molecular mass in grams.

(a) Ti, 47.88 g (b) Br2, 159.81 g (c) Hg, 200.59 g (d) H2O, 18.02 g

3.45 (a) Cr mol 0.0192 = Cr g 52.0

Cr mol 1Cr x g 1.00

(b) Cl mol 0.0141 = Cl g 70.9Cl mol 1

x Cl g 1.00 22

22

(c) Au mol 08 0.005 = Au g 197.0

Au mol 1Au x g 1.00

(d) NH mol 0.0588 = NH g 17.0NH mol 1

x NH g 1.00 33

33

3.46 There are 2 ions per each formula unit of NaCl. (2.5 mol)(2 mol ions/mol) = 5.0 mol ions 3.47 There are 2 K+ ions per each formula unit of K2SO4.

K mol 2.90 = SOK mol 1K mol 2

x SOK mol 1.45 +

42

+

42

3.48 There are 3 ions (one Mg2+ and 2 Cl-) per each formula unit of MgCl2.

MgCl2, 95.2 amu

27.5 g MgCl2 x MgCl mol 1

ions mol 3 x

MgCl g 95.2

MgCl mol 1

22

2 = 0.867 mol ions

3.49 There are 3 F- anions per each formula unit of AlF3.

AlF3, 84.0 amu

35.6 g AlF3 x AlF mol 1

anions mol 3 x

AlF g 84.0AlF mol 1

33

3 = 1.27 mol F-

3.50 Molar mass = mol / g 119 = mol 0.0275

g 3.28; molecular mass = 119 amu.

3.51 Molar mass = mol 0.5731

g 221.6 = 386.7 g/mol; molecular mass = 386.7 amu.

3.52 FeSO4 , 151.9 amu; 300 mg = 0.300 g


40

FeSO mol 10 x 1.97 = FeSO g 151.9

FeSO mol 1 x FeSO g 0.300 4

3_

4

44

atoms Fe(II) 10 x 1.19 = FeSO mol 1

atoms Fe(II) 10 x 6.02 x FeSO mol 10 x 1.97 21

4

23

43_

3.53 0.0001 g C x C mol 1

atoms C 10 x 6.02 x

C g 12.0

C mol 1 23

= 5 x 1018 C atoms

3.54 C8H10N4O2, 194.2 amu; 125 mg = 0.125 g

0.125 g caffeine x = caffeine g 194.2

caffeine mol 16.44 x 10-4 mol caffeine

0.125 g caffeine x mol 1

molecules 10 x 6.022 x

caffeine g 194.2

caffeine mol 1 23

= 3.88 x 1020 caffeine

molecules

3.55 eggs of mol / g 10 x 2.7 = eggs mol 1

eggs 10 x 6.02 x

egg

g 45 25

23

3.56 (a) 1.0 g Li x Li g 6.94

Li mol 1 = 0.14 mol Li

(b) 1.0 g Au x Au g 197.0

Au mol 1 = 0.0051 mol Au

(c) penicillin G: C16H17N2O4SK, 372.5 amu

1.0 g x G penicillin g 372.5

G penicillin mol 1 = 2.7 x 10-3 mol penicillin G

3.57 (a) Na g 0.034 = Na mol 1

Na g 23.0 x Na mol 0.0015

(b) Pb g 0.31 = Pb mol 1

Pb g 207.2 x Pb mol 0.0015

(c) C16H13ClN2O, 284.7 amu

diazepam g 0.43 = diazepam mol 1

diazepam g 284.7 x diazepam mol 0.0015

Stoichiometry Calculations

3.58 TiO2, 79.88 amu; Ti kg 47.88

TiO kg 79.88 x Ti kg 100.0 2 = 166.8 kg TiO2

3.59 Fe2O3, 159.7 amu; % 69.94 = % 100 x OFe g 159.7

Fe g) 2(55.85 = Fe %

32


41

mass Fe = (0.6994)(105 kg) = 73.4 kg 3.60 (a) 2 Fe2O3 + 3 C → 4 Fe + 3 CO2

(b) Fe2O3, 159.7 amu; 525 g Fe2O3 x OFe mol 2

C mol 3 x

OFe g 159.7OFe mol 1

3232

32 = 4.93 mol C

(c) 4.93 mol C x C mol 1

C g 12.01 = 59.2 g C

3.61 (a) Fe2O3 + 3 CO → 2 Fe + 3 CO2

(b) Fe2O3, 159.7 amu; CO, 28.01 amu

CO g 1.59 = CO mol 1

CO g 28.01 x

OFe mol 1

CO mol 3 x


x OFe g 3.023232

3232

(c) CO g 141 = CO mol 1

CO g 28.01 x

OFe mol 1

CO mol 3 x OFe mol 1.68

3232

3.62 (a) 2 Mg + O2 → 2 MgO

(b) Mg, 24.30 amu; O2, 32.00 amu; MgO, 40.30 amu

25.0 g Mg x O mol 1O g 32.00

x Mg mol 2O mol 1

x Mg g 24.30

Mg mol 1

2

22 = 16.5 g O2

25.0 g Mg x MgO mol 1

MgO g 40.30 x

Mg mol 2

MgO mol 2 x

Mg g 24.30

Mg mol 1 = 41.5 g MgO

(c) 25.0 g O2 x Mg mol 1

Mg g 24.30 x

O mol 1

Mg mol 2 x

O g 32.00O mol 1

22

2 = 38.0 g Mg

25.0 g O2 x MgO mol 1

MgO g 40.30 x

O mol 1

MgO mol 2 x

O g 32.00O mol 1

22

2 = 63.0 g MgO

3.63 C2H4 + H2O → C2H6O; C2H4, 28.05 amu; H2O, 18.02 amu; C2H6O, 46.07 amu

(a) HC g 3.73 = HC mol 1HC g 28.05

x OH mol 1HC mol 1

x OH mol 0.133 4242

42

2

422

OHC g 6.13 = OHC mol 1

OHC g 46.07 x

OH mol 1

OHC mol 1 x OH mol 0.133 62

62

62

2

622

(b) OH g 6.69 = OH mol 1

OH g 18.02 x

HC mol 1

OH mol 1 x HC mol 0.371 2

2

2

42

242


OHC g 46.07 x

HC mol 1

OHC mol 1 x HC mol 0.371 62

62

62

42

6242

3.64 (a) 2 HgO → 2 Hg + O2 (b) HgO, 216.6 amu; Hg, 200.6 amu; O2, 32.0 amu

45.5 g HgO x Hg mol 1

Hg g 200.6 x

HgO mol 2

Hg mol 2 x

HgO g 216.6

HgO mol 1 = 42.1 g Hg


42

45.5 g HgO x O mol 1O g 32.00

x HgO mol 2O mol 1

x HgO g 216.6

HgO mol 1

2

22 = 3.36 g O2

(c) 33.3 g O2 x HgO mol 1

HgO g 216.6 x

O mol 1

HgO mol 2 x

O g 32.00O mol 1

22

2 = 451 g HgO

3.65 5.60 kg = 5600 g; TiCl4, 189.7 amu; TiO2, 79.88 amu

TiO kg 2.36 = TiO g 2358 = TiO mol 1TiO g 79.88

x TiCl mol 1TiO mol 1

x TiCl g 189.7

TiCl mol 1 x TiCl g 5600 22

2

2

4

2

4

44

3.66 Ag mol 0.0185 = Ag g 107.9

Ag mol 1 x Ag g 2.00

Cl mol 0.0185 = Cl g 35.45

Cl mol 1 x Cl g 0.657

Ag0.0185Cl0.0185 Divide both subscripts by 0.0185. The empirical formula is AgCl.

3.67 5.0 g Al x Al g 27.0

Al mol 1 = 0.19 mol Al; 4.45 g O x

O g 16.0


Al 0.19O0.28; divide both subscripts by the smaller, 0.19. Al 0.19 / 0.19O0.28 / 0.19 Al 1O1.5; multiply both subscripts by 2 to obtain integers. The empirical formula is Al2O3.

Limiting Reactants and Reaction Yield

3.68 3.44 mol N2 x N mol 1H mol 3

2

2 = 10.3 mol H2 required.

Because there is only 1.39 mol H2, H2 is the limiting reactant.

1.39 mol H2 x NH mol 1NH g 17.03

x H mol 3

NH mol 2

3

3

2

3 = 15.8 g NH3

1.39 mol H2 x N mol 1N g 28.01

x H mol 3N mol 1

2

2

2

2 = 13.0 g N2 reacted

3.44 mol N2 x N mol 1N g 28.01

2

2 = 96.3 g N2 initially

(96.3 g - 13.0 g) = 83.3 g N2 left over 3.69 H2, 2.016 amu; Cl2, 70.91 amu; HCl 36.46 amu

H mol 1.77 = H g 2.016

H mol 1 x H g 3.56 2

2

22

Cl mol 0.126 = Cl g 70.91

Cl mol 1 x Cl g 8.94 2

2

22

Because the reaction stoichiometry between H2 and Cl2 is one to one, Cl2 is the limiting reactant.


43

HCl g 9.19 = HCl mol 1

HCl g 36.46 x

Cl mol 1

HCl mol 2 x Cl mol 0.126

22

3.70 C2H4, 28.05 amu; Cl2, 70.91 amu; C2H4Cl2, 98.96 amu

15.4 g C2H4 x HC g 28.05

HC mol 1

42

42 = 0.549 mol C2H4

3.74 g Cl2 x Cl g 70.91

Cl mol 1

2

2 = 0.0527 mol Cl2

Because the reaction stoichiometry between C2H4 and Cl2 is one to one, Cl2 is the limiting reactant.

0.0527 mol Cl2 x ClHC mol 1ClHC g 98.96

x Cl mol 1

ClHC mol 1

242

242

2

242 = 5.22 g C2H4Cl2

3.71 (a) NaCl, 58.44 amu; AgNO3, 169.9 amu; AgCl, 143.3 amu; NaNO3, 85.00 amu

NaCl + AgNO3 → AgCl + NaNO3

1.3 g NaCl x NaCl g 58.44

NaCl mol 1 = 0.022 mol NaCl

3.5 g AgNO3 x AgNO g 169.9

AgNO mol 1

3

3 = 0.021 mol AgNO3

Because the reaction stoichiometry between NaCl and AgNO3 is one to one, AgNO3 is the limiting reactant.

0.021 mol AgNO3 x AgCl mol 1

AgCl g 143.3 x

AgNO mol 1

AgCl mol 1

3

= 3.0 g AgCl

0.021 mol AgNO3 x NaNO mol 1NaNO g 85.00

x AgNO mol 1NaNO mol 1

3

3

3

3 = 1.8 g NaNO3

0.021 mol AgNO3 x NaCl mol 1

NaCl g 58.44 x

AgNO mol 1

NaCl mol 1

3

= 1.2 g NaCl reacted

(1.3 g - 1.2 g) = 0.1 g NaCl left over

(b) BaCl2, 208.2 amu; H2SO4, 98.08 amu; BaSO4, 233.4 amu; HCl, 36.46 amu BaCl2 + H2SO4 → BaSO4 + 2 HCl

2.65 g BaCl2 x BaCl g 208.2

BaCl mol 1

2

2 = 0.0127 mol BaCl2

6.78 g H2SO4 x SOH g 98.08

SOH mol 1

42

42 = 0.0691 mol H2SO4

Because the reaction stoichiometry between BaCl2 and H2SO4 is one to one, BaCl2 is the limiting reactant.

0.0127 mol BaCl2 x BaSO mol 1BaSO g 233.4

x BaCl mol 1BaSO mol 1

4

4

2

4 = 2.96 g BaSO4

0.0127 mol BaCl2 x HCl mol 1

HCl g 36.46 x

BaCl mol 1

HCl mol 2

2

= 0.926 g HCl


44

0.0127 mol BaCl2 x SOH mol 1SOH g 98.1

x BaCl mol 1

SOH mol 1

42

42

2

42 = 1.25 g H2SO4 reacted

(6.78 g - 1.25 g) = 5.53 g H2SO4 left over 3.72 CaCO3, 100.1 amu; HCl, 36.46 amu

CaCO3 + 2 HCl → CaCl2 + H2O + CO2

CaCO mol 0.0235 = CaCO g 100.1

CaCO mol 1 x CaCO g 2.35 3

3

33

HCl mol 0.0645 = HCl g 36.46

HCl mol 1 x HCl g 2.35

The reaction stoichiometry is 1 mole of CaCO3 for every 2 moles of HCl. For 0.0235 mol CaCO3, we only need 2(0.0235 mol) = 0.0470 mol HCl. We have 0.0645 mol HCl; therefore CaCO3 is the limiting reactant.

CO L 0.526 = CO mol 1

L 22.4 x

CaCO mol 1CO mol 1

x CaCO mol 0.0235 223

23

3.73 2 NaN3 → 3 N2 + 2 Na; NaN3, 65.01 amu; N2, 28.01 amu

38.5 g NaN3 x N mol 1.00

L 47.0 x

NaN mol 2N mol 3

x NaN g 65.01

NaN mol 1

23

2

3

3 = 41.8 L

3.74 CH3CO2H + C5H12O → C7H14O2 + H2O

CH3CO2H, 60.05 amu; C5H12O, 88.15 amu; C7H14O2, 130.19 amu

HCOCH mol 0.0596 = HCOCH g 60.05

HCOCH mol 1 x HCOCH g 3.58 23

23

2323

OHC mol 0.0539 = OHC g 88.15

OHC mol 1 x OHC g 4.75 125

125

125125

Because the reaction stoichiometry between CH3CO2H and C5H12O is one to one, isopentyl alcohol (C5H12O) is the limiting reactant.


OHC g 130.19 x

OHC mol 1OHC mol 1

x OHC mol 0.0539 21472147

2147

125

2147125

7.02 g C7H14O2 is the theoretical yield. Actual yield = (7.02 g)(0.45) = 3.2 g. 3.75 K2PtCl4 + 2 NH3 → 2 KCl + Pt(NH3)2Cl2

K2PtCl4, 415.1 amu; NH3, 17.03 amu; Pt(NH3)2Cl2, 300.0 amu

55.8 g K2PtCl4 x PtClK g 415.1

PtClK mol 1

42

42 = 0.134 mol K2PtCl4

35.6 g NH3 x NH g 17.03

NH mol 1

3

3 = 2.09 mol NH3

Only 2(0.134) = 0.268 mol NH3 are needed to react with 0.134 mol K2PtCl4. Therefore, the NH3 is in excess and K2PtCl4 is the limiting reactant.


45

0.134 mol K2PtCl4 x Cl)NHPt( mol 1Cl)NHPt( g 300.0

x PtClK mol 1

Cl)NHPt( mol 1

223

223

42

223 = 40.2 g Pt(NH3)2Cl2

40.2 g Pt(NH3)2Cl2 is the theoretical yield. Actual yield = (40.2 g)(0.95) = 38 g Pt(NH3)2Cl2.

3.76 CH3CO2H + C5H12O → C7H14O2 + H2O

CH3CO2H, 60.05 amu; C5H12O, 88.15 amu; C7H14O2, 130.19 amu

1.87 g CH3CO2H x HCOCH g 60.05

HCOCH mol 1

23

23 = 0.0311 mol CH3CO2H

2.31 g C5H12O x OHC g 88.15

OHC mol 1

125

125 = 0.0262 mol C5H12O

Because the reaction stoichiometry between CH3CO2H and C5H12O is one to one, isopentyl alcohol (C5H12O) is the limiting reactant.

0.0262 mol C5H12O x OHC mol 1

OHC g 130.19 x

OHC mol 1OHC mol 1

2147

2147

125

2147 = 3.41 g C7H14O2

3.41 g C7H14O2 is the theoretical yield.

% Yield = 100% x g 3.41

g 2.96 = 100% x

yield lTheoretica

yield Actual = 86.8%

3.77 K2PtCl4 + 2 NH3 → 2 KCl + Pt(NH3)2Cl2

K2PtCl4, 415.1 amu; NH3, 17.03 amu; Pt(NH3)2Cl2, 300.0 amu

3.42 g K2PtCl4 x PtClK g 415.1

PtClK mol 1

42

42 = 0.008 24 mol K2PtCl4

1.61 g NH3 x NH g 17.03

NH mol 1

3

3 = 0.0945 mol NH3

Only 2 x (0.008 24) = 0.0165 mol of NH3 are needed to react with 0.008 24 mol K2PtCl4. Therefore, the NH3 is in excess and K2PtCl4 is the limiting reactant.

0.008 24 mol K2PtCl4 x Cl)NHPt( mol 1Cl)NHPt( g 300.0

x PtClK mol 1

Cl)NHPt( mol 1

223

223

42

223 = 2.47 g Pt(NH3)2Cl2

2.47 g Pt(NH3)2Cl2 is the theoretical yield. 2.08 g Pt(NH3)2Cl2 is the actual yield.

% Yield = yield lTheoretica

yield Actual x 100% =

g 2.47

g 2.08 x 100% = 84.2%

Molarity, Solution Stoichiometry, Dilution, and Tit ration

3.78 (a) 35.0 mL = 0.0350 L; HNO mol 0.0420 = L 0.0350 x L

HNO mol 1.2003

3

(b) 175 mL = 0.175 L; OHC mol 0.12 = L 0.175 x L

OHC mol 0.676126

6126

3.79 (a) C2H6O, 46.07 amu; 250.0 mL = 0.2500 L


46

L 0.2500 x L

OHC mol 0.600 62 = 0.150 mol C2H6O

(0.150 mol)(46.07 g/mol) = 6.91 g C2H6O

(b) H3BO3, 61.83 amu; 167 mL = 0.167 L

L 0.167 x L

BOH mol 0.200 33 = 0.0334 mol H3BO3

(0.0334 mol)(61.83 g/mol) = 2.07 g H3BO3 3.80 BaCl2, 208.2 amu

BaCl mol 0.0720 = BaCl g 208.2

BaCl mol 1 x BaCl g 15.0 2

2

22

mL 160 = L 0.16 L; 0.16 = mol 0.45

L 1.0 x mol 0.0720

3.81 0.0171 mol KOH x KOH mol 0.350

L 1.00 = 0.0489 L; 0.0489 L = 48.9 mL

3.82 NaCl, 58.4 amu; 400 mg = 0.400 g; 100 mL = 0.100 L


NaCl mol 1 = 0.006 85 mol NaCl

molarity = L 0.100

mol 85 0.006 = 0.0685 M

3.83 C6H12O6, 180.2 amu; 90 mg = 0.090 g; 100 mL = 0.100 L

OHC mol 50 0.000 = OHC g 180.2

OHC mol 1 x OHC g 0.090 6126

6126

61266126

molarity = L 0.100

mol 50 0.000 = 0.0050 M = 5.0 x 10-3 M

3.84 NaCl, 58.4 amu; KCl, 74.6 amu; CaCl2, 111.0 amu; 500 mL = 0.500 L


NaCl mol 1 = 0.0736 mol NaCl

0.150 g KCl x KCl g 74.6

KCl mol 1 = 0.002 01 mol KCl

0.165 g CaCl2 x CaCl g 111.0

CaCl mol 1

2

2 = 0.001 49 mol CaCl2

0.0736 mol + 0.002 01 mol + 2(0.001 49 mol) = 0.0786 mol Cl-


47

Na+ molarity = L 0.500

mol 0.0736 = 0.147 M

Ca2+ molarity = L 0.500

mol 49 0.001 = 0.002 98 M

K+ molarity = L 0.500

mol 01 0.002 = 0.004 02 M

Cl- molarity = L 0.500

mol 0.0786 = 0.157 M

3.85 3.045 g Cu x Cu g 63.546

Cu mol 1 = 0.047 92 mol Cu; 50.0 mL = 0.0500 L

Cu(NO3)2 molarity = L 0.0500

mol 92 0.047 = 0.958 M

3.86 Mf x Vf = Mi x Vi; HCl M 1.71 = mL 250.0

mL 35.7 x M 12.0 =

V

V x M = Mf

iif

3.87 Mf x Vf = Mi x Vi; mL 426 = M 0.0150

mL 70.00 x M 0.0913 =

M

V x M = Vf

iif

3.88 2 HBr(aq) + K2CO3(aq) → 2 KBr(aq) + CO2(g) + H2O(l)

K2CO3, 138.2 amu; 450 mL = 0.450 L

L 0.450 x L

HBr mol 0.500 = 0.225 mol HBr

0.225 mol HBr x COK mol 1COK g 138.2

x HBr mol 2COK mol 1

32

3232 = 15.5 g K2CO3

3.89 2 C4H10S + NaOCl → C8H18S2 + NaCl + H2O C4H10S, 90.19 amu; 5.00 mL = 0.005 00 L

L

NaOCl mol 0.0985 x 0.005 00 L = 4.925 x 10-4 mol NaOCl

4.925 x 10-4 mol NaOCl x SHC mol 1

SHC g 90.19 x

NaOCl mol 1

SHC mol 2

104

104104 = 0.0888 g C4H10S

3.90 H2C2O4, 90.04 amu

KMnO mol 0.0143 = OCH mol 5

KMnO mol 2 x

OCH g 90.04OCH mol 1

x OCH g 3.225 4422

4

422

422422

L 0.0572 = mol 0.250

L 1 x mol 0.0143 = 57.2 mL

3.91 H2C2O4, 90.04 amu; 400.0 mL = 0.4000 L; 25.0 mL = 0.0250 L


48

OCH mol 0.133 = OCH g 90.04

OCH mol 1 x OCH g 12.0 422

422

422422

molarity = OCH M 0.333 = L 0.4000

mol 0.133422

H2C2O4(aq) + 2 KOH(aq) → K2C2O4(aq) + 2 H2O(l)

L 0.0250 x L

OHC mol 0.333 422 = 0.008 32 mol H2C2O4

0.008 32 mol H2C2O4 x OCH mol 1

KOH mol 2

422

= 0.0166 mol KOH

mol 0.100

L 1 x mol 0.0166 = 0.166 L; 0.166 L = 166 mL

Formulas and Elemental Analysis 3.92 CH4N2O, 60.1 amu

% 20.0 = % 100 x g 60.1

C g 12.0 = C %

% 6.72 = % 100 x g 60.1

H g 1.01 x 4 = H %

% 46.6 = % 100 x g 60.1

N g 14.0 x 2 = N %

% 26.6 = % 100 x g 60.1

O g 16.0 = O %

3.93 (a) Cu2(OH)2CO3, 221.1 amu

% Cu = 100% x g 221.1

Cu g 63.5 x 2 = 57.4%

% O = 100% x g 221.1

O g 16.0 x 5 = 36.2%

% C = 100% x g 221.1

C g 12.0 = 5.43%

% H = 100% x g 221.1

H g 1.01 x 2 = 0.91%

(b) C8H9NO2, 151.2 amu

% C = 100% x g 151.2

C g 12.0 x 8 = 63.5%

% H = 100% x g 151.2

H g 1.01 x 9 = 6.01%


49

% N = 100% x g 151.2

N g 14.0 = 9.26%

% O = 100% x g 151.2

O g 16.0 x 2 = 21.2%

(c) Fe4[Fe(CN)6]3, 859.2 amu

% Fe = 100% x g 859.2

Fe g 55.85 x 7 = 45.50%

% C = 100% x g 859.2

C g 12.01 x 18 = 25.16%

% N = 100% x g 859.2

N g 14.01 x 18 = 29.35%

3.94 Assume a 100.0 g sample. From the percent composition data, a 100.0 g sample contains

24.25 g F and 75.75 g Sn.

F mol 1.276 = F g 19.00

F mol 1 x F g 24.25

Sn mol 0.6382 = Sn g 118.7

Sn mol 1Sn x g 75.75

Sn0.6382F1.276; divide each subscript by the smaller, 0.6382. Sn0.6382 / 0.6382F1.276 / 0.6382 The empirical formula is SnF2.

3.95 (a) Assume a 100.0 g sample of ibuprofen. From the percent composition data, a 100.0 g

sample contains 75.69 g C, 15.51 g O, and 8.80 g H.

75.69 g C x C g 12.01

C mol 1 = 6.302 mol C

15.51 g O x O g 16.00

O mol 1 = 0.9694 mol O

8.80 g H x H g 1.01

H mol 1 = 8.71 mol H

C6.302H8.71O0.9694, divide each subscript by the smallest, 0.9694. C6.302 / 0.9694H8.71 / 0.9694O0.9694 / 0.9694

C6.5H9O; multiply each subscript by 2 to obtain integers. The empirical formula is C13H18O2.

(b) Assume a 100.0 g sample of tetraethyllead. From the percent composition data, a 100.0 g sample contains 29.71 g C, 6.23 g H, and 64.06 g Pb.

29.71 g C x C g 12.01

C mol 1 = 2.474 mol C

6.23 g H x H g 1.01



50

64.06 g Pb x Pb g 207.2

Pb mol 1 = 0.3092 mol Pb

Pb0.3092C2.474H6.17; divide each subscript by the smallest, 0.3092. Pb0.3092 / 0.3092C2.474 / 0.3092H6.17 / 0.3092 The empirical formula is PbC8H20.

(c) Assume a 100.0 g sample of zircon. From the percent composition data, a 100.0 g sample contains 34.91 g O, 15.32 g Si, and 49.76 g Zr.

34.91 g O x O g 16.00

O mol 1 = 2.182 mol O

15.32 g Si x Si g 28.09

Si mol 1 = 0.5454 mol Si

49.76 g Zr x Zrg 91.22

Zrmol 1 = 0.5455 mol Zr

Zr0.5455Si0.5454O2.182; divide each subscript by the smallest, 0.5454. Zr0.5455 / 0.5454Si0.5454 / 0.5454O2.182 / 0.5454 The empirical formula is ZrSiO4.

3.96 Mass of toluene sample = 45.62 mg = 0.045 62 g; mass of CO2 = 152.5 mg = 0.1525 g;

mass of H2O = 35.67 mg = 0.035 67 g

C mol 465 0.003 = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

x CO g 0.152522

22

mass C = 0.003 465 mol C x = C mol 1

C g 12.0110.041 62 g C

H mol 959 0.003 = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1 x OH g 67 0.035

22

22

mass H = 0.003 959 mol H x = H mol 1

H g 1.0080.003 991 g H

The (mass C + mass H) = 0.041 62 g + 0.003 991 g = 0.045 61 g. The calculated mass of (C + H) essentially equals the mass of the toluene sample, this means that toluene contains only C and H and no other elements. C0.003 465H0.003 959; divide each subscript by the smaller, 0.003 465. C0.003 465 / 0.003 465H0.003 959 / 0.003 465 CH1.14; multiply each subscript by 7 to obtain integers. The empirical formula is C7H8.

3.97 5.024 mg = 0.005 024 g; 13.90 mg = 0.013 90 g; 6.048 mg = 0.006 048 g

0.013 90 g CO2 x CO mol 1

C mol 1 x

CO g 44.01CO mol 1

22

2 = 3.158 x 10-4 mol C

0.006 048 g H2O x OH mol 1

H mol 2 x

OH g 18.02

OH mol 1

22

2 = 6.713 x 10-4 mol H


51

3.158 x 10-4 mol C x C mol 1

C g 12.01 = 0.003 793 g C

6.713 x 10-4 mol H x H mol 1

H g 1.008 = 0.000 676 7 g H

mass N = 0.005 024 g - (0.003 793 g + 0.000 676 7 g) = 0.000 554 g N

0.000 554 g N x N g 14.01

N mol 1 = 3.95 x 10-5 mol N

Scale each mol quantity to eliminate exponents. C3.158H6.713N0.395; divide each subscript by the smallest, 0.395. C3.158 / 0.395H6.713 / 0.395N0.395 / 0.395 The empirical formula is C8H17N.

3.98 Let X equal the molecular mass of cytochrome c.

X

amu 55.847 = 0.0043 ; amu 13,000 =

0.0043

amu 55.847 = X

3.99 Let X equal the molecular mass of nitrogenase.

0.000 872 = X

amu 95.94 x 2; X =

872 0.000

amu 95.94 x 2 = 220,000 amu

3.100 Let X equal the molecular mass of disilane.

X

amu 28.09 x 2 = 0.9028 ; amu 62.23 =

0.9028

amu 28.09 x 2 = X

62.23 amu - 2(Si atomic mass) = 62.23 amu - 2(28.09 amu) = 6.05 amu 6.05 amu is the total mass of H atoms.

atoms H 6 = amu 1.01

atom H 1 x amu 6.05 ; Disilane is Si2H6.

3.101 Let X equal the molecular mass of MS2.

0.4006 = X

amu 32.07 x 2; X =

0.4006

amu 32.07 x 2 = 160.1 amu

Atomic mass of M = 160.1 amu - 2(S atomic mass) = 160.1 amu - 2(32.07 amu) = 95.96 amu

M is Mo. General Problems 3.102 (a) C6H12O6, 180.2 amu

39.99% = % 100 x g 180.2

C g 12.01 x 6 = C %


52

6.713% = % 100 x g 180.2

H g 1.008 x 12 = H %

% 53.27 = % 100 x g 180.2

O g 16.00 x 6 = O %

(b) H2SO4, 98.08 amu

% 2.055 = % 100 x g 98.08

H g 1.008 x 2 = H %

% 32.70 = % 100 x g 98.08

S g 32.07 = S %

% 65.25 = % 100 x g 98.08

O g 16.00 x 4 = O %

(c) KMnO4, 158.0 amu

% 24.75 = % 100 x g 158.0

K g 39.10 =K %

% 34.77 = % 100 x g 158.0

Mn g 54.94 =Mn %

% 40.51 = % 100 x g 158.0

O g 16.00 x 4 = O %

(d) C7H5NO3S, 183.2 amu

% 45.89 = % 100 x g 183.2

C g 12.01 x 7 = C %

% 2.751 = % 100 x g 183.2

H g 1.008 x 5 = H %

% 7.647 = % 100 x g 183.2

N g 14.01 = N %

% 26.20 = % 100 x g 183.2

O g 16.00 x 3 = O %

% 17.51 = % 100 x g 183.2

S g 32.07 = S %

3.103 (a) Assume a 100.0 g sample of aspirin. From the percent composition data, a 100.0 g

sample contains 60.00 g C, 35.52 g O, and 4.48 g H.

60.00 g C x C g 12.01

C mol 1 = 4.996 mol C

35.52 g O x O g 16.00

O mol 1 = 2.220 mol O


53

4.48 g H x H g 1.01


C4.996H4.44O2.220; divide each subscript by the smallest, 2.220. C4.996 / 2.220H4.44 / 2.220O2.220 / 2.220 C2.25H2O1; multiply each subscript by 4 to obtain integers. The empirical formula is C9H8O4.

(b) Assume a 100.0 g sample of ilmenite. From the percent composition data, a 100.0 g sample contains 31.63 g O, 31.56 g Ti, and 36.81 g Fe.

31.63 g O x O g 16.00

O mol 1 = 1.977 mol O

31.56 g Ti x Ti g 47.88

Ti mol 1 = 0.6591 mol Ti

36.81 g Fe x Fe g 55.85

Fe mol 1 = 0.6591 mol Fe

Fe0.6591Ti0.6591O1.977; divide each subscript by the smallest, 0.6591. Fe0.6591 / 0.6591Ti0.6591 / 0.6591O1.977 / 0.6591 The empirical formula is FeTiO3.

(c) Assume a 100.0 g sample of sodium thiosulfate. From the percent composition data, a 100.0 g sample contains 30.36 g O, 29.08 g Na, and 40.56 g S.

30.36 g O x O g 16.00

O mol 1 = 1.897 mol O

29.08 g Na x Na g 22.99

Na mol 1 = 1.265 mol Na

40.56 g S x S g 32.07

S mol 1 = 1.265 mol S

Na1.265S1.265O1.897; divide each subscript by the smallest, 1.265. Na1.265 / 1.265S1.265 / 1.265O1.897 / 1.265 NaSO1.5; multiply each subscript by 2 to obtain integers. The empirical formula is Na2S2O3.

3.104 (a) SiCl4 + 2 H2O → SiO2 + 4 HCl

(b) P4O10 + 6 H2O → 4 H3PO4 (c) CaCN2 + 3 H2O → CaCO3 + 2 NH3 (d) 3 NO2 + H2O → 2 HNO3 + NO

3.105 NaH, 24.00 amu; B2H6, 27.67 amu; NaBH4, 37.83 amu

2 NaH + B2H6 → 2 NaBH4

8.55 g NaH x NaH g 24.00

NaH mol 1 = 0.356 mol NaH


54

6.75 g B2H6 x HB g 27.67

HB mol 1

62

62 = 0.244 mol B2H6

For 0.244 mol B2H6, 2 x (0.244) = 0.488 mol NaH are needed. Because only 0.356 mol of NaH is available, NaH is the limiting reactant.

0.356 mol NaH x NaBH mol 1NaBH g 37.83

x NaH mol 2

NaBH mol 2

4

44 = 13.5 g NaBH4 produced

0.356 mol NaH x HB mol 1HB g 27.67

x NaH mol 2

HB mol 1

62

6262 = 4.93 g B2H6 reacted

B2H6 left over = 6.75 g - 4.93 g = 1.82 g B2H6 3.106 Assume a 100.0 g sample of ferrocene. From the percent composition data, a 100.0 g

sample contains 5.42 g H, 64.56 g C, and 30.02 g Fe.

5.42 g H x H g 1.01


64.56 g C x C g 12.01

C mol 1 = 5.376 mol C

30.02 g Fe x Fe g 55.85


C5.376H5.37Fe0.5375; divide each subscript by the smallest, 0.5375. C5.376 / 0.5375H5.37 / 0.5375Fe0.5375 / 0.5375 The empirical formula is C10H10Fe.

3.107 Mass of 1 HCl molecule = (36.5 molecule

amu)(1.6605 x 10-24

amu

g) = 6.06 x 10-23 g/molecule

Avogadro's number =

g/molecule 10 x 6.06

g/mol 36.523_

= 6.02 x 1023 molecules/mol

3.108 Na2SO4, 142.04 amu; Na3PO4, 163.94 amu; Li2SO4, 109.95 amu; 100.00 mL = 0.10000 L

0.550 g Na2SO4 x SONa g 142.04

SONa mol 1

42

42 = 0.003 872 mol Na2SO4

1.188 g Na3PO4 x PONa g 163.94

PONa mol 1

43

43 = 0.007 247 mol Na3PO4

0.223 g Li2SO4 x SOLi g 109.95

SOLi mol 1

42

42 = 0.002 028 mol Li2SO4

Na+ molarity = L 00 0.100

mol) 247 0.007 x (3 + mol) 872 0.003 x (2 = 0.295 M

Li + molarity = L 00 0.100

mol 028 0.002 x 2 = 0.0406 M

SO42- molarity =

L 00 0.100

mol) 028 0.002 x (1 + mol) 872 0.003 x (1 = 0.0590 M


55

PO43- molarity =

L 00 0.100

mol 247 0.007 x 1 = 0.0725 M

3.109 23.46 mg = 0.023 46 g; 20.42 mg = 0.02042 g; 33.27 mg = 0.033 27 g

C mol 10 x 7.560 = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

x CO g 27 0.033 4_

22

22

H mol 10 x 2.266 = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1 x OH g 42 0.020 3_

22

22

C g 080 0.009 = C mol 1

C g 12.01 x C mol 10 x 7.560 4_

H g 284 0.002 = H mol 1

H g 1.008 x H mol 10 x 2.266 3_

mass O = 0.023 46 g - (0.009 080 g + 0.002 284 g) = 0.012 10 g O

O mol 10 x 7.563 = O g 16.00

O mol 1 x O g 10 0.012 4_

Scale each mol quantity to eliminate exponents. C0.7560H2.266O0.7563; divide each subscript by the smallest, 0.7560. C0.7560 / 0.7560H2.266 / 0.7560O0.7563 / 0.7560 The empirical formula is CH3O, 31.0 amu. 62.0 amu / 31.0 amu = 2; molecular formula = C(2 x 1)H(2 x 3)O(2 x 1) = C2H6O2

3.110 High resolution mass spectrometry is capable of measuring the mass of molecules with a

particular isotopic composition. 3.111 (a) CO(NH2)2(aq) + 6 HOCl(aq) → 2 NCl3(aq) + CO2(aq) + 5 H2O(l)

(b) 2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s) → P4(g) + 6 CaSiO3(l) + 10 CO(g) 3.112 The combustion reaction is: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

C8H18, 114.23 amu; CO2, 44.01 amu

pounds CO2 = xHC g 114.23

HC mol 1 x

mL 1HC g 0.703

x L 1

mL 1000 x

gal 1

L 3.7854 x gal 1.00

188

188188

= g 453.59

lb 1 x

CO mol 1CO g 44.01

x HC mol 2

CO mol 16

2

2

188

2 18.1 pounds CO2

3.113 The reaction is: CaCO3 + 2 HCl → CaCl2 + CO2 + H2O

CaCO3, 100.09 amu; CO2, 44.01 amu

mol CaCO3 = = CaCO g 100.09

CaCO mol 1 x CaCO g 6.35

3

33 0.0634 mol CaCO3

mol HCl = = L 1

HCl mol 0.31 x

mL 1000

L 1 x HCl mL 500.0 0.155 mol HCl

Determine the limiting reactant.

mol HCl needed = 0.0634 mol CaCO3 x = CaCO mol 1

HCl mol 2

3

0.127 mol HCl needed


56

Because we have excess HCl, CaCO3 is the limiting reactant.

mass CO2 = = CO mol 1CO g 44.01

x CaCO mol 1CO mol 1

x CaCO mol 0.06342

2

3

23 2.79 g CO2

3.114 AgCl, 143.32 amu; CO2, 44.01 amu; H2O, 18.02 amu

mol Cl in 1.00 g of X = = AgCl mol 1

Cl mol 1 x

AgCl g 143.32

AgCl mol 1 x AgCl g 1.95 0.0136 mol Cl

mass Cl = 0.0136 mol Cl x = Cl mol 1

Cl g 35.453 0.482 g Cl

mol C in 1.00 g of X = = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

x CO g 0.90022

22 0.0204 mol C

mass C = 0.0204 mol C x = C mol 1

C g 12.011 0.245 g C

mol H in 1.00 g of X = = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1 x OH g 0.735

22

22 0.0816 mol H

mass H = 0.0816 mol H x = H mol 1

H g 1.008 0.0823 g H

mass N = 1.00 g - mass Cl - mass C - mass H = 1.00 - 0.482 g - 0.245 g - 0.0823 g = 0.19 g N

mol N in 1.00 g of X = 0.19 g N x = N g 14.01

N mol 1 0.014 mol N

Determine empirical formula. C0.0204H0.0816N0.014Cl0.0136, divide each subscript by the smallest, 0.0136. C0.0204 / 0.0136H0.0816 / 0.0136N0.014 / 0.0136Cl0.0136 / 0.0136

C1.5H6NCl, multiply each subscript by 2 to get integers. The empirical formula is C3H12N2Cl2.

3.115 CaCO3, 100.09 amu

% Ca = = 100% x g 100.09

Ca g 40.08 40.04%

% C = = 100% x g 100.09

C g 12.01 12.00%

% O = = 100% x g 100.09

O g 16.00 x 3 47.96%

Because the mass %’s for the pulverized rock are different from the mass %’s for pure CaCO3 calculated here, the pulverized rock cannot be pure CaCO3.

3.116 Let SA stand for salicylic acid.

mol C in 1.00 g of SA = = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

x CO g 2.2322

22 0.0507 mol C


57


C g 12.011 0.609 g C

mol H in 1.00 g of SA = = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1 x OH g 0.39

22

22 0.043 mol H


H g 1.008 0.043 g H

mass O = 1.00 g - mass C - mass H = 1.00 - 0.609 g - 0.043 g = 0.35 g O

mol O in 1.00 g of = 0.35 g N x = O g 16.00

O mol 1 0.022 mol O

Determine empirical formula. C0.0507H0.043O0.022, divide each subscript by the smallest, 0.022. C0.0507 / 0.022H0.043 / 0.022O0.022 / 0.022 C2.3H2O, multiply each subscript by 3 to get integers. The empirical formula is C7H6O3. The empirical formula mass = 138.12 g/mol

Because salicylic acid has only one acidic hydrogen, there is a 1 to 1 mol ratio between salicylic acid and NaOH in the acid-base titration.

mol SA in 1.00 g SA = 72.4 mL x = NaOH mol 1

SA mol 1 x

L 1

NaOH mol 0.100 x

mL 1000

L 1

0.00724 mol SA

SA molar mass = mol 0.00724

g 1.00 = 138 g/mol

Because the empirical formula mass and the molar mass are the same, the empirical formula is the molecular formula for salicylic acid.

3.117 (a) mol C = 4.83 g CO2 x = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

22

2 0.110 mol C


C g 12.0111.32 g C

mol H = 1.48 g H2O x = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1

22

2 0.164 mol H


H g 1.0080.165 g H

109.8 mL = 0.1098 L mol NaOH = (0.1098 L)(1.00 mol/L) = 0.110 mol NaOH H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)

mol H2SO4 = 0.110 mol NaOH x = NaOH mol 2

SOH mol 1 42 0.0550 mol H2SO4

mol S = 0.0550 mol H2SO4 x = SOH mol 1

S mol 1

42

0.0550 mol S


58

mass S = 0.0550 mol S x = S mol 1

S g 32.061.76 g S

mass O = 5.00 g - mass C - mass H - mass S = 5.00 g -1.32 g - 0.165 g - 1.76 g = 1.75 g O

mol O = 1.75 g O x = O g 16.00

O mol 10.109 mol O

C0.110H0.164O0.109S0.0550 Divide all subscripts by the smallest. C0.110 / 0.0550H0.164 / 0.0550O0.109 / 0.0550S0.0550 / 0.0550 The empirical formula is C2H3O2S. The empirical formula mass = 91.1 g/mol (b) 54.9 mL = 0.0549 L mol NaOH = (0.0549 L)(1.00 mol/L) = 0.0549 mol NaOH Because X has two acidic hydrogens, two mol of NaOH are required to titrate 1 mol of X.

mol X = 0.0549 mol NaOH x = NaOH mol 2

X mol 10.0274 mol X

X molar mass = mol 0.0274

g 5.00 = 182 g/mol

Because the molar mass is twice the empirical formula mass, the molecular formula is twice the empirical formula. The molecular formula is C(2 x 2)H(2 x 3)O(2 x 2)S(2 x 1) = C4H6O4S2

3.118 Let X equal the mass of benzoic acid and Y the mass of gallic acid in the 1.00 g mixture.

Therefore, X + Y = 1.00 g. Because both acids contain only one acidic hydrogen, there is a 1 to 1 mol ratio between each acid and NaOH in the acid-base titration. In the titration, mol benzoic acid + mol gallic acid = mol NaOH

Therefore, = GA g 170

GA mol 1 x Y +

BA g 122

BA mol 1 x X mol NaOH

mol NaOH = 14.7 mL x = L 1

NaOH mol 0.500 x

mL 1000

L 1 0.00735 mol NaOH

We have two unknowns, X and Y, and two equations. X + Y = 1.00 g

= GA g 170

GA mol 1 x Y +

BA g 122

BA mol 1 x X 0.00735 mol NaOH

Rearrange to get X = 1.00 g - Y and then substitute it into the equation above to solve for Y.

= GA g 170

GA mol 1 x Y +

BA g 122

BA mol 1 x Y) _ g (1.00 0.00735 mol NaOH

mol 0.00735 = g 170

mol Y +

g 122

mol Y _

122

mol 1

mol 10 x 8.47 _ = 122

mol 1 _ mol 0.00735 =

g 170

mol Y +

g 122

mol Y _ 4 _

mol 10 x 8.47 _ = g) g)(122 (170

g) mol)(122 (Y + g) mol)(170 Y (_ 4 _


59

; mol 10 x 8.47 _ = g 20740

mol Y 48 _ 4 _ 10 x 8.47 = g 20740

Y 48 4 _

g 0.366 = 48

)10 x g)(8.47 (20740 = Y

4 _

X = 1.00 g - 0.366 g = 0.634 g In the 1.00 g mixture there is 0.63 g of benzoic acid and 0.37 g of gallic acid.

3.119 C2H6O, 46.07 amu; H2O, 18.02 amu

Let X = mass of H2O in the 10.00 g sample. Let Y = mass of ethanol (C2H6O) in the 10.00 g sample. X + Y = 10.00 g and Y = 10.00 g - X mass of collected H2O = 11.27 g

mass of collected H2O = X +

OH mol 1

OH g 18.02 x

OHC mol 1

OH mol 3 x

OHC g 46.07

OHC mol 1 x Y

2

2

62

2

62

62

Substitute for Y.

11.27 g = X +

OH mol 1

OH g 18.02 x

OHC mol 1

OH mol 3 x

OHC g 46.07

OHC mol 1 x X) _ g (10.00

2

2

62

2

62

62

11.27 g = X + (10.00 g - X)(1.173) 11.27 g = X + 11.73 g - 1.173 X 0.173 X = 11.73 g - 11.27 g = 0.46 g

X = = 0.173

g 0.46 2.7 g H2O

Y = 10.00 g - X = 10.00 g - 2.7 g = 7.3 g C2H6O 3.120 FeO, 71.85 amu; Fe2O3, 159.7 amu

Let X equal the mass of FeO and Y the mass of Fe2O3 in the 10.0 g mixture. Therefore, X + Y = 10.0 g.

mol Fe = 7.43 g x Fe g 55.85


mol FeO + 2 x mol Fe2O3 = 0.133 mol Fe

= OFe g 159.7

OFe mol 1 x Y x 2 +

FeO g 71.85

FeO mol 1 x X

32

32

0.133 mol Fe


= OFe g 159.7

OFe mol 1 x Y x 2 +

FeO g 71.85

FeO mol 1 x Y) _ g (10.0

32

32

0.133 mol Fe

= g 159.7

mol Y 2 +

g 71.85

mol Y _

71.85

mol 10.0 0.133 mol

71.85

mol 10.0 _ mol 0.133 =

g 159.7

mol Y 2 +

g 71.85

mol Y _ = - 0.0062 mol

= g) g)(159.7 (71.85

g) mol)(71.85 Y (2 + g) mol)(159.7 Y (_ - 0.0062 mol


60

; mol 0.0062 _ = g 11474

mol Y 16.0 _ 0.0062 =

g 11474

Y 16.0

Y = (0.0062)(11474 g)/16.0 = 4.44 g = 4.4 g Fe2O3

X = 10.0 g - Y = 10.0 g - 4.4 g = 5.6 g FeO

3.121 AgCl, 143.32 amu; Find the mass of Cl in 1.68 g of AgCl.

mol Cl in 1.68 g of AgCl = = AgCl mol 1

Cl mol 1 x

AgCl g 143.32

AgCl mol 1 x AgCl g 1.68 0.0117 mol Cl


Cl g 35.453 0.415 g Cl

All of the Cl in AgCl came from XCl3. Find the mass of X in 0.634 g of XCl3. Mass of X = 0.634 g - 0.415 g = 0.219 g X

0.0117 mol Cl x Cl mol 3

X mol 1 = 0.00390 mol X

molar mass of X = mol 0.00390

g 0.219 = 56.2 g/mol; X = Fe

3.122 C6H12O6 + 6 O2 → 6 CO2 + 6 H2O; C6H12O6, 180.16 amu; CO2, 44.01 amu

66.3 g C6H12O6 x CO mol 1CO g 44.01

x OHC mol 1

CO mol 6 x

OHC g 180.16OHC mol 1

2

2

6126

2

6126

6126 = 97.2 g CO2

66.3 g C6H12O6 x CO mol 1CO L 25.4

x OHC mol 1

CO mol 6 x

OHC g 180.16OHC mol 1

2

2

6126

2

6126

6126 = 56.1 L CO2

3.123 H2C2O4, 90.04 amu; 22.35 mL = 0.02235 L

0.5170 g H2C2O2 x OCH mol 5

KMnO mol 2 x

OCH g 90.04OCH mol 1

422

4

422

422 = 0.002 297 mol KMnO4

KMnO4 molarity = L 35 0.022KMnO mol 297 0.002 4 = 0.1028 M

3.124 Mass of Cu = 2.196 g; mass of S = 2.748 g - 2.196 g = 0.552 g S

(a) %Cu = g 2.748

g 2.196 x 100% = 79.91%

%S = g 2.748

g 0.552 x 100% = 20.1%

(b) 2.196 g Cu x Cu g 63.55

Cu mol 1 = 0.034 55 mol Cu

0.552 g S x S g 32.07

S mol 1 = 0.0172 mol S


61

Cu0.03455S0.0172; divide each subscript by the smaller, 0.0172. Cu0.03455 / 0.0172S0.0172 / 0.0172 The empirical formula is Cu2S.

(c) Cu2S, 159.16 amu

ions Cu mol 1

ions Cu 10 x 6.022 x

SCu mol 1

ions Cu mol 2 x

SCu g 159.16

SCu mol 1 x

cm 1

SCu g 5.6+

+23

2

+

2

2

3

2

= 4.2 x 1022 Cu+ ions/cm3 3.125 Mass of added Cl = mass of XCl5 - mass of XCl3 = 13.233 g - 8.729 g = 4.504 g

mass of Cl in XCl5 = 5 Cl’s x lsC 2

g 4.504′

= 11.26 g Cl

mass of X in XCl5 = 13.233 g - 11.26 g = 1.973 g X

11.26 g Cl x Cl g 35.45

Cl mol 1 = 0.3176 mol Cl

0.3176 mol Cl x Cl mol 5

X mol 1 = 0.063 52 mol X

molar mass of X = X mol 52 0.063

X g 1.973 = 31.1 g/mol; atomic mass =31.1 amu, X = P

3.126 PCl3, 137.33 amu; PCl5, 208.24 amu

Let Y = mass of PCl3 in the mixture, and (10.00 - Y) = mass of PCl5 in the mixture.

fraction Cl in PCl3 = g/mol 137.33

g/mol) (3)(35.453 = 0.774 48

fraction Cl in PCl5 = g/mol 208.24

g/mol) (5)(35.453 = 0.851 25

(mass of Cl in PCl3) + (mass of Cl in PCl5) = mass of Cl in the mixture 0.774 48Y + 0.851 25(10.00 g - Y) = (0.8104)(10.00 g) Y = 5.32 g PCl3 and 10.00 - Y = 4.68 g PCl5

3.127 100.00 mL = 0.100 00 L; 71.02 mL = 0.071 02 L

mol H2SO4 = L

SOH mol 0.1083 42 x 0.100 00 L = 0.010 83 mol H2SO4

mol NaOH = L

NaOH mol 0.1241 x 0.071 02 L = 0.008 814 mol NaOH

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

mol H2SO4 reacted with NaOH = 0.008 814 mol NaOH x NaOH mol 2

SOH mol 1 42 = 0.004 407 mol H2SO4

mol H2SO4 reacted with MCO3 = 0.010 83 mol - 0.004 407 mol = 0.006 423 mol H2SO4

mol H2SO4 reacted with MCO3 = mol CO32- in MCO3 = mol CO2 produced = 0.006 423 mol CO2

(a) CO32-, 60.01 amu; 0.006 423 mol CO3

2- x CO mol 1CO g 60.01

_23

_23 = 0.3854 g CO3

2-

mass of M = 1.268 g - 0.3854 g = 0.8826 g M


62

molar mass of M = mol 423 0.006

g 0.8826 = 137.4 g/mol; M is Ba

(b) 0.006 423 mol CO2 x g 1.799

L 1 x

CO mol 1CO g 44.01

2

2 = 0.1571 L CO2

3.128 NH4NO3, 80.04 amu; (NH4)2HPO4, 132.06 amu

Assume you have a 100.0 g sample of the mixture. Let X = grams of NH4NO3 and (100.0 - X) = grams of (NH4)2HPO4. Both compounds contain 2 nitrogen atoms per formula unit. Because the mass % N in the sample is 30.43%, the 100.0 g sample contains 30.43 g N.

mol NH4NO3 = g 80.04NONH mol 1

x (X) 34

mol (NH4)2HPO4 = g 132.06HPO)NH( mol 1

x X) _ (100.0 424

mass N =

g 132.06HPO)NH( mol 1

x X) _ (100.0 + g 80.04NONH mol 1

x (X) 42434 x

N mol 1

N g 14.0067 x

cmpds ammonium mol 1

N mol 2= 30.43 g

Solve for X.

7)(2)(14.006132.06

X _ 100.0 +

80.04

X

= 30.43

132.06

X _ 100.0 +

80.04

X= 1.08627

2.06)(80.04)(13

X)(80.04) _ (100.0 + )(132.06)(X = 1.08627

(132.06)(X) + (100.0 - X)(80.04) = (1.08627)(80.04)(132.06) 132.06X + 8004 - 80.04X = 11481.96 132.06X - 80.04X = 11481.96 - 8004 52.02X = 3477.96

X = 52.02

3477.96 = 66.86 g NH4NO3

(100.0 - X) = (100.0 - 66.86) = 33.14 g (NH4)2HPO4

2.018 = g 33.14

g 66.86 =

mass

mass

HPO)NH(

NONH

424

34

The mass ratio of NH4NO3 to (NH4)2HPO4 in the mixture is 2 to 1. 3.129 Na2CO3 → Na2O + CO2; Na2CO3, 106 amu; Na2O, 62 amu

CaCO3 → CaO + CO2; CaCO3, 100 amu; CaO, 56 amu In a 0.35 kg sample of glass there would be:

0.12 x 0.35 kg = 0.042 kg = 42 g of Na2O 0.13 x 0.35 kg = 0.045 kg = 45 g of CaO


63

350 g - 42 g - 45 g = 263 g of SiO2

mass Na2CO3 = 42 g Na2O x CONa mol 1CONa g 106

x ONa mol 1

CONa mol 1 x

ONa g 62

ONa mol 1

32

32

2

32

2

2 = 72 g

Na2CO3

mass CaCO3 = 45 g CaO x CaCO mol 1CaCO g 100

x CaO mol 1

CaCO mol 1 x

CaO g 56

CaO mol 1

3

33 = 80 g CaCO3

To make 0.35 kg of glass, start with 72 g Na2CO3, 80 g CaCO3, and 263 g SiO2.

3.130 (a) 56.0 mL = 0.0560 L

mol X2 = (0.0560 L X2)

L 22.41

mol 1= 0.00250 mol X2

mass X2 = 1.12 g MX2 - 0.720 g MX = 0.40 g X2

molar mass X2 = mol 0.00250

g 0.40 = 160 g/mol

atomic mass of X = 160/2 = 80 amu; X is Br.

(b) mol MX = 0.00250 mol X2 x X mol 1

MX mol 2

2

= 0.00500 mol MX

mass of X in MX = 0.00500 mol MX x X mol 1

X g 80 x

MX mol 1

X mol 1 = 0.40 g X

mass of M in MX = 0.720 g MX - 0.40 g X = 0.32 g M

molar mass M = mol 0.00500

g 0.32 = 64 g/mol

atomic mass of X = 64 amu; M is Cu.

65

4

Reactions in Aqueous Solution 4.1 (a) precipitation (b) redox (c) acid-base neutralization 4.2 FeBr3 contains 3 Br- ions. The molar concentration of Br- ions = 3 x 0.225 M = 0.675 M 4.3 A2Y is the strongest electrolyte because it is completely dissociated into ions.

A2X is the weakest electrolyte because it is the least dissociated of the three substances. 4.4 (a) Ionic equation: 2 Ag+(aq) + 2 NO3

-(aq) + 2 Na+(aq) + CrO42-(aq) → Ag2CrO4(s) + 2 Na+(aq) + 2 NO3

-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: 2 Ag+(aq) + CrO4

2-(aq) → Ag2CrO4(s) (b) Ionic equation: 2 H+(aq) + SO4

2-(aq) + MgCO3(s) → H2O(l) + CO2(g) + Mg2+(aq) + SO42-(aq)

Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: 2 H+(aq) + MgCO3(s) → H2O(l) + CO2(g) + Mg2+(aq) (c) Ionic equation: Hg2

2+(aq) + 2 NO3-(aq) + 2 NH4

+(aq) + 2 Cl-(aq) → Hg2Cl2(s) + 2 NH4+(aq) + 2 NO3

-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: Hg2

2+(aq) + 2 Cl-(aq) → Hg2Cl2(s) 4.5 (a) CdCO3, insoluble (b) MgO, insoluble (c) Na2S, soluble

(d) PbSO4, insoluble (e) (NH4)3PO4, soluble (f) HgCl2, soluble 4.6 (a) Ionic equation:

Ni2+(aq) + 2 Cl-(aq) + 2 NH4+(aq) + S2-(aq) → NiS(s) + 2 NH4

+(aq) + 2 Cl-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: Ni2+(aq) + S2-(aq) → NiS(s) (b) Ionic equation: 2 Na+(aq) + CrO4

2-(aq) + Pb2+(aq) + 2 NO3-(aq) → PbCrO4(s) + 2 Na+(aq) + 2 NO3

-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: Pb2+(aq) + CrO4

2-(aq) → PbCrO4(s) (c) Ionic equation: 2 Ag+(aq) + 2 ClO4

-(aq) + Ca2+(aq) + 2 Br-(aq) → 2 AgBr(s) + Ca2+(aq) + 2 ClO4-(aq)

Delete spectator ions from the ionic equation, and reduce coefficients to get the net ionic equation. Net ionic equation: Ag+(aq) + Br-(aq) → AgBr(s) (d) Ionic equation: Zn2+(aq) + 2 Cl-(aq) + 2 K+(aq) + CO3

2-(aq) → ZnCO3(s) + 2 K+(aq) + 2 Cl-(aq) Delete spectator ions from the ionic equation to get the net ionic equation.

66

Net ionic equation: Zn2+(aq) + CO32-(aq) → ZnCO3(s)

4.7 3 CaCl2(aq) + 2 Na3PO4(aq) → Ca3(PO4)2(s) + 6 NaCl(aq) Ionic equation: 3 Ca2+(aq) + 6 Cl-(aq) + 6 Na+(aq) + 2 PO4

3-(aq) → Ca3(PO4)2(s) + 6 Na+(aq) + 6 Cl-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: 3 Ca2+(aq) + 2 PO4

3-(aq) → Ca3(PO4)2(s) 4.8 A precipitate results from the reaction. The precipitate contains cations and anions in a

3:2 ratio. The precipitate is either Mg3(PO4)2 or Zn3(PO4)2. 4.9 (a) Ionic equation:

2 Cs+(aq) + 2 OH-(aq) + 2 H+(aq) + SO42-(aq) → 2 Cs+(aq) + SO4

2-(aq) + 2 H2O(l) Delete spectator ions from the ionic equation, and reduce coefficients to get the net ionic equation. Net ionic equation: H+(aq) + OH-(aq) → H2O(l)

(b) Ionic equation: Ca2+(aq) + 2 OH-(aq) + 2 CH3CO2H(aq) → Ca2+(aq) + 2 CH3CO2

-(aq) + 2 H2O(l) Delete spectator ions from the ionic equation, and reduce coefficients to get the net ionic equation. Net ionic equation: CH3CO2H(aq) + OH-(aq) → CH3CO2

-(aq) + H2O(l) 4.10 HY is the strongest acid because it is completely dissociated.

HX is the weakest acid because it is the least dissociated. 4.11 (a) SnCl4: Cl -1, Sn +4 (b) CrO3: O -2, Cr +6

(c) VOCl3: O -2, Cl -1, V +5 (d) V2O3: O -2, V +3 (e) HNO3: O -2, H +1, N +5 (f) FeSO4: O -2, S +6, Fe +2

4.12 2 Cu2+(aq) + 4 I-(aq) → 2 CuI(s) + I2(aq)

oxidation numbers: Cu2+ +2; I- -1; CuI: Cu +1, I -1; I2: 0 oxidizing agent (oxidation number decreases), Cu2+ reducing agent (oxidation number increases) , I-

4.13 (a) SnO2(s) + 2 C(s) → Sn(s) + 2 CO(g)

C is oxidized (its oxidation number increases from 0 to +2). C is the reducing agent. The Sn in SnO2 is reduced (its oxidation number decreases from +4 to 0). SnO2 is the oxidizing agent. (b) Sn2+(aq) + 2 Fe3+(aq) → Sn4+(aq) + 2 Fe2+(aq) Sn2+ is oxidized (its oxidation number increases from +2 to +4). Sn2+ is the reducing agent. Fe3+ is reduced (its oxidation number decreases from +3 to +2). Fe3+ is the oxidizing agent. (c) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) The N in NH3 is oxidized (its oxidation number increases from -3 to +2). NH3 is the reducing agent. Each O in O2 is reduced (its oxidation number decreases from 0 to -2). O2 is the oxidizing agent.

Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________

67

4.14 (a) Pt is below H in the activity series; therefore NO REACTION.

(b) Mg is below Ca in the activity series; therefore NO REACTION. 4.15 Because B will reduce A+, B is above A in the activity series. Because B will not reduce

C+, C is above B in the activity series. Therefore C must be above A in the activity series and C will reduce A+.

4.16

8 H+(aq) + Cr2O72-(aq) + I-(aq) → 2 Cr3+(aq) + IO3

-(aq) + 4 H2O(l)

4.17

2 MnO4-(aq) + Br-(aq) → 2 MnO2(s) + BrO3

-(aq) 2 H+(aq) + 2 MnO4

-(aq) + Br-(aq) → 2 MnO2(s) + BrO3-(aq) + H2O(l)

2 H+(aq) + 2 OH-(aq) + 2 MnO4-(aq) + Br-(aq) →

2 MnO2(s) + BrO3-(aq) + H2O(l) + 2 OH-(aq)

2 H2O(l) + 2 MnO4-(aq) + Br-(aq) → 2 MnO2(s) + BrO3

-(aq) + H2O(l) + 2 OH-(aq) H2O(l) + 2 MnO4

-(aq) + Br-(aq) → 2 MnO2(s) + BrO3-(aq) + 2 OH-(aq)

4.18 (a) MnO4

-(aq) → MnO2(s) (reduction) IO3

-(aq) → IO4-(aq) (oxidation)

(b) NO3

-(aq) → NO2(g) (reduction) SO2(aq) → SO4

2-(aq) (oxidation) 4.19 NO3

-(aq) + Cu(s) → NO(g) + Cu2+(aq) [Cu(s) → Cu2+(aq) + 2 e-] x 3 (oxidation half reaction)

NO3

-(aq) → NO(g) NO3

-(aq) → NO(g) + 2 H2O(l) 4 H+(aq) + NO3

-(aq) → NO(g) + 2 H2O(l) [3 e- + 4 H+(aq) + NO3

-(aq) → NO(g) + 2 H2O(l)] x 2 (reduction half reaction)


68

Combine the two half reactions. 2 NO3

-(aq) + 8 H+(aq) + 3 Cu(s) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) 4.20 Fe(OH)2(s) + O2(g) → Fe(OH)3(s)

[Fe(OH)2(s) + OH-(aq) → Fe(OH)3(s) + e-] x 4 (oxidation half reaction)

O2(g) → 2 H2O(l) 4 H+(aq) + O2(g) → 2 H2O(l) 4 e- + 4 H+(aq) + O2(g) → 2 H2O(l) 4 e- + 4 H+(aq) + 4 OH-(aq) + O2(g) → 2 H2O(l) + 4 OH-(aq) 4 e- + 4 H2O(l) + O2(g) → 2 H2O(l) + 4 OH-(aq) 4 e- + 2 H2O(l) + O2(g) → 4 OH-(aq) (reduction half reaction)

Combine the two half reactions. 4 Fe(OH)2(s) + 4 OH-(aq) + 2 H2O(l) + O2(g) → 4 Fe(OH)3(s) + 4 OH-(aq) 4 Fe(OH)2(s) + 2 H2O(l) + O2(g) → 4 Fe(OH)3(s)

4.21 31.50 mL = 0.031 50 L; 10.00 mL = 0.010 00 L

0.031 50 L x BrO mol 1Fe mol 6

x L 1

BrO mol 0.105_3

+2_3 = 1.98 x 10-2 mol Fe2+

molarity = L 00 0.010

Fe mol 10 x 1.98 +22_

= 1.98 M Fe2+ solution

4.22 The Na2S2O3, or hypo, is used to solubilize the remaining unreduced AgBr on the film so

that it is no longer sensitive to light. The reaction is AgBr(s) + 2 S2O3

2-(aq) → Ag(S2O3)23-(aq) + Br-(aq)

4.23 To convert this negative image into the final printed photograph, the entire photographic

procedure is repeated a second time. Light is passed through the negative image onto special photographic paper that is coated with the same kind of gelatin–AgBr emulsion used on the original film. Developing the photographic paper with hydroquinone and fixing the image with sodium thiosulfate reverses the negative image, and a final, positive image is produced.

Understanding Key Concepts 4.24 (a) 2 Na+(aq) + CO3

2-(aq) does not form a precipitate. This is represented by box (1). (b) Ba2+(aq) + CrO4

2-(aq) → BaCrO4(s). This is represented by box (2). (c) 2 Ag+(aq) + SO4

2-(aq) → Ag2SO4(s). This is represented by box (3). 4.25 In the precipitate there are two cations (blue) for each anion (green). Looking at the ions

in the list, the anion must have a -2 charge and the cation a +1 charge for charge neutrality of the precipitate. The cation must be Ag+ because all Na+ salts are soluble. Ag2CrO4 and Ag2CO3 are insoluble and consistent with the observed result.


69

4.26 One OH- will react with each available H+ on the acid forming H2O. The acid is

identified by how many of the 12 OH- react with three molecules of each acid. (a) Three HF's react with three OH-, leaving nine OH- unreacted (box 2). (b) Three H2SO3's react with six OH-, leaving six OH- unreacted (box 3). (c) Three H3PO4's react with nine OH-, leaving three OH- unreacted (box 1).

4.27 The concentration in the buret is three times that in the flask. The NaOCl concentration

is 0.040 M. Because the I- concentration in the buret is three times the OCl- concentration in the flask and the reaction requires 2 I- ions per OCl- ion, 2/3 or 67% of the I- solution from the buret must be added to the flask to react with all of the OCl-.

4.28 (a) Ionic equation:

K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → AgCl(s) + K+(aq) + NO3

-(aq) (b) Ionic equation: HF(aq) + K+(aq) + OH-(aq) → K+(aq) + F-(aq) + H2O(l) (c) Ionic equation: Ba2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + SO4

2-(aq) → BaSO4(s) + 2 Na+(aq) + 2 Cl-(aq)

Reaction (c) would have the highest initial conductivity because of the 3 net ions for each BaCl2 (a strong electrolyte). Reaction (b) would have have the lowest (almost zero) initial conductivity because HF is a very weak acid/electrolyte. Reaction (a) would have an intermediate initial conductivity between that for reactions (b) and (c). Figure (1) is for reaction (a); figure (2) is for reaction (b); and figure (3) is for reaction (c).

4.29 (a) Sr+ + At → Sr + At+ No reaction.

(b) Si + At+ → Si+ + At Reaction would occur. (c) Sr + Si+ → Sr+ + Si Reaction would occur.

Additional Problems Aqueous Reactions and Net Ionic Equations 4.30 (a) precipitation (b) redox (c) acid-base neutralization 4.31 (a) redox (b) precipitation (c) acid-base neutralization 4.32 (a) Ionic equation:

Hg2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 I-(aq) → 2 Na+(aq) + 2 NO3

-(aq) + HgI2(s) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: Hg2+(aq) + 2 I-(aq) → HgI2(s)

Heat (b) 2 HgO(s) → 2 Hg(l) + O2(g) (c) Ionic equation:


70

H3PO4(aq) + 3 K+(aq) + 3 OH-(aq) → 3 K+(aq) + PO43-(aq) + 3 H2O(l)

Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: H3PO4(aq) + 3 OH-(aq) → PO4

3-(aq) + 3 H2O(l) 4.33 (a) S8(s) + 8 O2(g) → 8 SO2(g)

(b) Ionic equation: Ni2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + S2-(aq) → NiS(s) + 2 Na+(aq) + 2 Cl-(aq) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: Ni2+(aq) + S2-(aq) → NiS(s) (c) Ionic equation: 2 CH3CO2H(aq) + Ba2+(aq) + 2 OH-(aq) → 2 CH3CO2

-(aq) + Ba2+(aq) + 2 H2O(l) Delete spectator ions from the ionic equation to get the net ionic equation.

Net ionic equation: CH3CO2H(aq) + OH-(aq) → CH3CO2-(aq) + H2O(l)

4.34 Ba(OH)2 is soluble in aqueous solution, dissociates into Ba2+(aq) and 2 OH-(aq), and

conducts electricity. In aqueous solution H2SO4 dissociates into H+(aq) and HSO4-(aq).

H2SO4 solutions conduct electricity. When equal molar solutions of Ba(OH)2 and H2SO4

are mixed, the insoluble BaSO4 is formed along with two H2O. In water BaSO4 does not produce any appreciable amount of ions and the mixture does not conduct electricity.

4.35 H2O is polar and a good H+ acceptor. It allows the polar HCl to dissociate into ions in

aqueous solution: HCl + H2O → H3O+ + Cl-.

CHCl3 is not very polar and not a H+ acceptor and so does not allow the polar HCl to dissociate into ions.

4.36 (a) HBr, strong electrolyte (b) HF, weak electrolyte

(c) NaClO4, strong electrolyte (d) (NH4)2CO3, strong electrolyte (e) NH3, weak electrolyte (f) C2H5OH, nonelectrolyte

4.37 It is possible for a molecular compound to be a strong electrolyte. For example, HCl is a

molecular compound when pure but dissociates completely to give H+ and Cl- ions when it dissolves in water.

4.38 (a) K2CO3 contains 3 ions (2 K+ and 1 CO3

2-). The molar concentration of ions = 3 x 0.750 M = 2.25 M. (b) AlCl3 contains 4 ions (1 Al3+ and 3 Cl-). The molar concentration of ions = 4 x 0.355 M = 1.42 M.

4.39 (a) CH3OH is a nonelectrolyte. The ion concentration from CH3OH is zero.

(b) HClO4 is a strong acid. HClO4(aq) → H+(aq) + ClO4

-(aq) In solution, there are 2 moles of ions per mole of HClO4. The molar concentration of ions = 2 × 0.225 M = 0.450 M.

Precipitation Reactions and Solubility Rules


71

4.40 (a) Ag2O, insoluble (b) Ba(NO3)2, soluble (c) SnCO3, insoluble (d) Fe2O3, insoluble

4.41 (a) ZnS, insoluble (b) Au2(CO3)3, insoluble

(c) PbCl2, insoluble (soluble in hot water) (d) MnO2, insoluble 4.42 (a) No precipitate will form. (b) FeCl2(aq) + 2 KOH(aq) → Fe(OH)2(s) + 2 KCl(aq)

(c) No precipitate will form. (d) No precipitate will form. 4.43 (a) MnCl2(aq) + Na2S(aq) → MnS(s) + 2 NaCl(aq)

(b) No precipitate will form. (c) 3 Hg(NO3)2(aq) + 2 Na3PO4(aq) → Hg3(PO4)2(s) + 6 NaNO3(aq) (d) Ba(NO3)2(aq) + 2 KOH(aq) → Ba(OH)2(s) + 2 KNO3(aq)

4.44 (a) Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq)

(b) 3 MgCl2(aq) + 2 K3PO4(aq) → Mg3(PO4)2(s) + 6 KCl(aq) (c) ZnSO4(aq) + Na2CrO4(aq) → ZnCrO4(s) + Na2SO4(aq)

4.45 (a) AlCl3(aq) + 3 NaOH(aq) → Al(OH)3(s) + 3 NaCl(aq)

(b) Fe(NO3)2(aq) + Na2S(aq) → FeS(s) + 2 NaNO3(aq) (c) CoSO4(aq) + K2CO3(aq) → CoCO3(s) + K2SO4(aq)

4.46 Add HCl(aq); it will selectively precipitate AgCl(s). 4.47 Add Na2SO4(aq); it will selectively precipitate BaSO4(s). 4.48 Ag+ is eliminated because it would have precipitated as AgCl(s); Ba2+ is eliminated

because it would have precipitated as BaSO4(s). The solution might contain Cs+ and/or NH4

+. Neither of these will precipitate with OH–, SO42–, or Cl–.

4.49 Cl- is eliminated because it would have precipitated as AgCl(s). OH- is eliminated

because it would have precipitated as either AgOH(s) or Cu(OH)2(s). SO42- is eliminated

because it would have precipitated as BaSO4(s). The solution might contain NO3-

because all nitrates are soluble. Acids, Bases, and Neutralization Reactions 4.50 Add the solution to an active metal, such as magnesium. Bubbles of H2 gas indicate the

presence of an acid. 4.51 We use a double arrow to show the dissociation of a weak acid or weak base in aqueous

solution to indicate the equilibrium between reactants and products. 4.52 (a) 2 H+(aq) + 2 ClO4

-(aq) + Ca2+(aq) + 2 OH-(aq) → Ca2+(aq) + 2 ClO4-(aq) + 2 H2O(l)


72

(b) CH3CO2H(aq) + Na+(aq) + OH-(aq) → CH3CO2-(aq) + Na+(aq) + H2O(l)

4.53 (a) 2 HF(aq) + Ca2+(aq) + 2 OH-(aq) → Ca2+(aq) + 2 F-(aq) + 2 H2O(l)

(b) Mg(OH)2(s) + 2 H+(aq) + 2 NO3-(aq) → Mg2+(aq) + 2 NO3

-(aq) + 2 H2O(l) 4.54 (a) LiOH(aq) + HI(aq) → LiI(aq) + H2O(l)

Ionic equation: Li+(aq) + OH-(aq) + H+(aq) + I-(aq) → Li+(aq) + I-(aq) + H2O(l) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: H+(aq) + OH-(aq) → H2O(l)

(b) 2 HBr(aq) + Ca(OH)2(aq) → CaBr2(aq) + 2 H2O(l) Ionic equation: 2 H+(aq) + 2 Br-(aq) + Ca2+(aq) + 2 OH-(aq) → Ca2+(aq) + 2 Br-(aq) + 2 H2O(l) Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: H+(aq) + OH-(aq) → H2O(l)

4.55 (a) 2 Fe(OH)3(s) + 3 H2SO4(aq) → Fe2(SO4)3(aq) + 6 H2O(l)

Ionic equation and net ionic equation are the same. 2 Fe(OH)3(s) + 3 H+(aq) + 3 HSO4

-(aq) → 2 Fe3+(aq) + 3 SO42-(aq) + 6 H2O(l)

(b) HClO3(aq) + NaOH(aq) → NaClO3(aq) + H2O(l) Ionic equation H+(aq) + ClO3

-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + ClO3-(aq) + H2O(l)

Delete spectator ions from the ionic equation to get the net ionic equation. Net ionic equation: H+(aq) + OH-(aq) → H2O(l)

Redox Reactions and Oxidation Numbers 4.56 The best reducing agents are at the bottom left of the periodic table. The best oxidizing

agents are at the top right of the periodic table (excluding the noble gases). 4.57 The most easily reduced elements in the periodic table are in the top-right corner,

excluding group 8A. The most easily oxidized elements in the periodic table are in the bottom-left corner.

4.58 (a) An oxidizing agent gains electrons.

(b) A reducing agent loses electrons. (c) A substance undergoing oxidation loses electrons. (d) A substance undergoing reduction gains electrons.

4.59 (a) In a redox reaction, the oxidation number decreases for an oxidizing agent.

(b) In a redox reaction, the oxidation number increases for a reducing agent. (c) In a redox reaction, the oxidation number increases for a substance undergoing oxidation. (d) In a redox reaction, the oxidation number decreases for a substance undergoing reduction.


73

4.60 (a) NO2 O -2, N +4 (b) SO3 O -2, S +6 (c) COCl2 O -2, Cl -1, C +4 (d) CH2Cl2 Cl -1, H +1, C 0 (e) KClO3 O -2, K +1, Cl +5 (f) HNO3 O -2, H +1, N +5

4.61 (a) VOCl3 O -2, Cl -1, V +5

(b) CuSO4 O -2, S +6, Cu +2 (c) CH2O O -2, H +1, C 0 (d) Mn2O7 O -2, Mn +7 (e) OsO4 O -2, Os +8 (f) H2PtCl6 Cl -1, H +1, Pt +4

4.62 (a) ClO3

- O -2, Cl +5 (b) SO32- O -2, S +4

(c) C2O42- O -2, C +3 (d) NO2

- O -2, N +3 (e) BrO- O -2, Br +1 (f) AsO4

3- O -2, As +5 4.63 (a) Cr(OH)4

- O -2, H +1, Cr +3 (b) S2O3

2- O -2, S +2 (c) NO3

- O -2, N +5 (d) MnO4

2- O -2, Mn +6 (e) HPO4

2- O -2, H +1, P +5 (f) V2O7

4- O -2, V +5 4.64 (a) Ca(s) + Sn2+(aq) → Ca2+(aq) + Sn(s)

Ca(s) is oxidized (oxidation number increases from 0 to +2). Sn2+(aq) is reduced (oxidation number decreases from +2 to 0). (b) ICl(s) + H2O(l) → HCl(aq) + HOI(aq) No oxidation numbers change. The reaction is not a redox reaction.

4.65 (a) Si(s) + 2 Cl2(g) → SiCl4(l)

Si(s) is oxidized (oxidation number increases from 0 to +4). Cl2(g) is reduced (oxidation number decreases from 0 to -1). (b) Cl2(g) + 2 NaBr(aq) → Br2(aq) + 2 NaCl(aq) Br-(aq) is oxidized (oxidation number increases from -1 to 0). Cl2(g) is reduced (oxidation number decreases from 0 to -1).

4.66 (a) Zn is below Na+; therefore no reaction.

(b) Pt is below H+; therefore no reaction. (c) Au is below Ag+; therefore no reaction. (d) Ag is above Au3+; the reaction is Au3+(aq) + 3 Ag(s) → 3 Ag+(aq) + Au(s).

4.67 Sr is more metallic than Sb because it is in the same period and to the left of Sb on the

periodic table. Sr is the better reducing agent. 2 Sb3+(aq) + 3 Sr(s) → 2 Sb(s) + 3 Sr2+(aq) will occur, the reverse will not.

4.68 (a) “Any element higher in the activity series will react with the ion of any element lower


74

in the activity series.” A + B+ → A+ + B; therefore A is higher than B. C+ + D → no reaction; therefore C is higher than D. B + D+ → B+ + D; therefore B is higher than D. B + C+ → B+ + C; therefore B is higher than C. The net result is A > B > C > D. (b) (1) C is below A+; therefore no reaction. (2) D is below A+; therefore no reaction.

4.69 (a) “Any element higher in the activity series will react with the ion of any element lower in the activity series.” 2 A + B2+ → 2 A+ + B; therefore A is higher than B. B + D2+ → B2+ + D; therefore B is higher than D. A+ + C → no reaction; therefore A is higher than C. 2 C + B2+ → 2 C+ + B; therefore C is higher than B. The net result is A > C > B > D. (b) (1) D is below A+; therefore no reaction. (2) C is above D2+; therefore the reaction will occur.

Balancing Redox Reactions 4.70 (a) N oxidation number decreases from +5 to +2; reduction.

(b) Zn oxidation number increases from 0 to +2; oxidation. (c) Ti oxidation number increases from +3 to +4; oxidation. (d) Sn oxidation number decreases from +4 to +2; reduction.

4.71 (a) O oxidation number decreases from 0 to -2; reduction.

(b) O oxidation number increases from -1 to 0; oxidation. (c) Mn oxidation number decreases from +7 to +6; reduction. (d) C oxidation number increases from -2 to 0; oxidation.

4.72 (a) NO3

-(aq) → NO(g) NO3

-(aq) → NO(g) + 2 H2O(l) 4 H+(aq) + NO3

-(aq) → NO(g) + 2 H2O(l) 3 e- + 4 H+(aq) + NO3

-(aq) → NO(g) + 2 H2O(l)

(b) Zn(s) → Zn2+(aq) + 2 e-

(c) Ti3+(aq) → TiO2(s) Ti3+(aq) + 2 H2O(l) → TiO2(s) Ti3+(aq) + 2 H2O(l) → TiO2(s) + 4 H+(aq) Ti3+(aq) + 2 H2O(l) → TiO2(s) + 4 H+(aq) + e-

(d) Sn4+(aq) + 2 e- → Sn2+(aq)

4.73 (a) O2(g) → OH-(aq)


75

O2(g) → OH-(aq) + H2O(l) 3 H+(aq) + O2(g) → OH-(aq) + H2O(l) 3 H+(aq) + 3 OH-(aq) + O2(g) → 4 OH-(aq) + H2O(l) 3 H2O(l) + O2(g) → 4 OH-(aq) + H2O(l) 4 e- + 2 H2O(l) + O2(g) → 4 OH-(aq)

(b) H2O2(aq) → O2(g)

H2O2(aq) → O2(g) + 2 H+(aq)

2 OH-(aq) + H2O2(aq) → O2(g) + 2 H+(aq) + 2 OH-(aq) 2 OH-(aq) + H2O2(aq) → O2(g) + 2 H2O(l) + 2 e-

(c) MnO4

-(aq) → MnO42-(aq)

MnO4-(aq) + e- → MnO4

2-(aq)

(d) CH3OH(aq) → CH2O(aq) CH3OH(aq) → CH2O(aq) + 2 H+(aq) CH3OH(aq) + 2 OH-(aq) → CH2O(aq) + 2 H+(aq) + 2 OH-(aq) CH3OH(aq) + 2 OH-(aq) → CH2O(aq) + 2 H2O(l) CH3OH(aq) + 2 OH-(aq) → CH2O(aq) + 2 H2O(l) + 2 e-

4.74 (a) Te(s) + NO3

-(aq) → TeO2(s) + NO(g) oxidation: Te(s) → TeO2(s) reduction: NO3

-(aq) → NO(g) (b) H2O2(aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) oxidation: Fe2+(aq) → Fe3+(aq) reduction: H2O2(aq) → H2O(l)

4.75 (a) Mn(s) + NO3

-(aq) → Mn2+(aq) + NO2(g) oxidation: Mn(s) → Mn2+(aq) reduction: NO3

-(aq) → NO2(g) (b) Mn3+(aq) → MnO2(s) + Mn2+(aq) oxidation: Mn3+(aq) → MnO2(s) reduction: Mn3+(aq) → Mn2+(aq)

4.76 (a) Cr2O7

2-(aq) → Cr3+(aq) Cr2O7

2-(aq) → 2 Cr3+(aq) Cr2O7

2-(aq) → 2 Cr3+(aq) + 7 H2O(l) 14 H+(aq) + Cr2O7


2-(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l)

(b) CrO42-(aq) → Cr(OH)4

-(aq) 4 H+(aq) + CrO4

2-(aq) → Cr(OH)4-(aq)

4 H+(aq) + 4 OH-(aq) + CrO42-(aq) → Cr(OH)4

-(aq) + 4 OH-(aq)


76

4 H2O(l) + CrO42-(aq) → Cr(OH)4

-(aq) + 4 OH-(aq) 4 H2O(l) + CrO4

2-(aq) + 3 e- → Cr(OH)4-(aq) + 4 OH-(aq)

(c) Bi3+(aq) → BiO3

-(aq) Bi3+(aq) + 3 H2O(l) → BiO3

-(aq) Bi3+(aq) + 3 H2O(l) → BiO3

-(aq) + 6 H+(aq) Bi3+(aq) + 3 H2O(l) + 6 OH-(aq) → BiO3

-(aq) + 6 H+(aq) + 6 OH-(aq) Bi3+(aq) + 3 H2O(l) + 6 OH-(aq) → BiO3

-(aq) + 6 H2O(l) Bi3+(aq) + 6 OH-(aq) → BiO3

-(aq) + 3 H2O(l) Bi3+(aq) + 6 OH-(aq) → BiO3

-(aq) + 3 H2O(l) + 2 e-

(d) ClO-(aq) → Cl-(aq) ClO-(aq) → Cl-(aq) + H2O(l) 2 H+(aq) + ClO-(aq) → Cl-(aq) + H2O(l) 2 H+(aq) + 2 OH-(aq) + ClO-(aq) → Cl-(aq) + H2O(l) + 2 OH-(aq) 2 H2O(l) + ClO-(aq) → Cl-(aq) + H2O(l) + 2 OH-(aq) H2O(l) + ClO-(aq) → Cl-(aq) + 2 OH-(aq) H2O(l) + ClO-(aq) + 2 e- → Cl-(aq) + 2 OH-(aq)

4.77 (a) VO2+(aq) → V3+(aq)

VO2+(aq) → V3+(aq) + H2O(l) 2 H+(aq) + VO2+(aq) → V3+(aq) + H2O(l) 2 H+(aq) + VO2+(aq) + e- → V3+(aq) + H2O(l)

(b) Ni(OH)2(s) → Ni2O3(s)

2 Ni(OH)2(s) → Ni2O3(s) + H2O(l) 2 Ni(OH)2(s) → Ni2O3(s) + H2O(l) + 2 H+(aq) 2 Ni(OH)2(s) + 2 OH-(aq) → Ni2O3(s) + H2O(l) + 2 H+(aq) + 2 OH-(aq) 2 Ni(OH)2(s) + 2 OH-(aq) → Ni2O3(s) + 3 H2O(l) + 2 e-

(c) NO3

-(aq) → NO2(g) NO3

-(aq) → NO2(g) + H2O(l) 2 H+(aq) + NO3

-(aq) → NO2(g) + H2O(l) 2 H+(aq) + NO3

-(aq) + e- → NO2(g) + H2O(l)

(d) Br2(aq) → BrO3-(aq)

Br2(aq) → 2 BrO3-(aq)

Br2(aq) + 6 H2O(l) → 2 BrO3-(aq)

Br2(aq) + 6 H2O(l) → 2 BrO3-(aq) + 12 H+(aq)

Br2(aq) + 6 H2O(l) + 12 OH-(aq) → 2 BrO3-(aq) + 12 H+(aq) + 12 OH-(aq)

Br2(aq) + 6 H2O(l) + 12 OH-(aq) → 2 BrO3-(aq) + 12 H2O(l)

Br2(aq) + 12 OH-(aq) → 2 BrO3-(aq) + 6 H2O(l) + 10 e-

4.78 (a) MnO4

-(aq) → MnO2(s)


77

MnO4-(aq) → MnO2(s) + 2 H2O(l)

4 H+(aq) + MnO4-(aq) → MnO2(s) + 2 H2O(l)

[4 H+(aq) + MnO4-(aq) +3 e- →MnO2(s) + 2 H2O(l)] x 2 (reduction half reaction)

IO3

-(aq) → IO4-(aq)

H2O(l) + IO3-(aq) → IO4

-(aq) H2O(l) + IO3

-(aq) → IO4-(aq) + 2 H+(aq)

[H2O(l) + IO3-(aq) → IO4

-(aq) + 2 H+(aq) + 2 e-] x 3 (oxidation half reaction) Combine the two half reactions. 8 H+(aq) + 3 H2O(l) + 2 MnO4

-(aq) + 3 IO3-(aq) →

6 H+(aq) + 4 H2O(l) + 2 MnO2(s) + 3 IO4-(aq)

2 H+(aq) + 2 MnO4-(aq) + 3 IO3

-(aq) → 2 MnO2(s) + 3 IO4-(aq) + H2O(l)

2 H+(aq) + 2 OH-(aq) + 2 MnO4

-(aq) + 3 IO3-(aq) →

2 MnO2(s) + 3 IO4-(aq) + H2O(l) + 2 OH-(aq)

2 H2O(l) + 2 MnO4-(aq) + 3 IO3

-(aq) → 2 MnO2(s) + 3 IO4

-(aq) + H2O(l) + 2 OH-(aq) H2O(l) + 2 MnO4

-(aq) + 3 IO3-(aq) → 2 MnO2(s) + 3 IO4

-(aq) + 2 OH-(aq)

(b) Cu(OH)2(s) → Cu(s) Cu(OH)2(s) → Cu(s) + 2 H2O(l) 2 H+(aq) + Cu(OH)2(s) → Cu(s) + 2 H2O(l) [2 H+(aq) + Cu(OH)2(s) + 2 e- → Cu(s) + 2 H2O(l)] x 2 (reduction half reaction)

N2H4(aq) → N2(g) N2H4(aq) → N2(g) + 4 H+(aq) N2H4(aq) → N2(g) + 4 H+(aq) + 4 e- (oxidation half reaction)

Combine the two half reactions. 4 H+(aq) + 2 Cu(OH)2(s) + N2H4(aq) → 2 Cu(s) + 4 H2O(l) + N2(g) + 4 H+(aq) 2 Cu(OH)2(s) + N2H4(aq) → 2 Cu(s) + 4 H2O(l) + N2(g)

(c) Fe(OH)2(s) → Fe(OH)3(s)

Fe(OH)2(s) + H2O(l) → Fe(OH)3(s) Fe(OH)2(s) + H2O(l) → Fe(OH)3(s) + H+(aq) [Fe(OH)2(s) + H2O(l) → Fe(OH)3(s) + H+(aq) + e-] x 3 (oxidation half reaction)

CrO4

2-(aq) → Cr(OH)4-(aq)

4 H+(aq) + CrO42-(aq) → Cr(OH)4

-(aq) 4 H+(aq) + CrO4

2-(aq) + 3 e- → Cr(OH)4-(aq) (reduction half reaction)

Combine the two half reactions. 3 Fe(OH)2(s) + 3 H2O(l) + 4 H+(aq) + CrO4

2-(aq) → 3 Fe(OH)3(s) + 3 H+(aq) + Cr(OH)4

-(aq)


78

3 Fe(OH)2(s) + 3 H2O(l) + H+(aq) + CrO42-(aq) → 3 Fe(OH)3(s) + Cr(OH)4

-(aq) 3 Fe(OH)2(s) + 3 H2O(l) + H+(aq) + OH-(aq) + CrO4

2-(aq) → 3 Fe(OH)3(s) + Cr(OH)4

-(aq) + OH-(aq) 3 Fe(OH)2(s) + 4 H2O(l) + CrO4

2-(aq) → 3 Fe(OH)3(s) + Cr(OH)4-(aq) + OH-(aq)

(d) ClO4

-(aq) → ClO2-(aq)

ClO4-(aq) → ClO2

-(aq) + 2 H2O(l) 4 H+(aq) + ClO4

-(aq) → ClO2-(aq) + 2 H2O(l)

4 H+(aq) + ClO4-(aq) + 4 e- → ClO2

-(aq) + 2 H2O(l) (reduction half reaction)

H2O2(aq) → O2(g) H2O2(aq) → O2(g) + 2 H+(aq) [H2O2(aq) → O2(g) + 2 H+(aq) + 2 e-] x 2 (oxidation half reaction)

Combine the two half reactions. 4 H+(aq) + ClO4

-(aq) + 2 H2O2(aq) → ClO2-(aq) + 2 H2O(l) + 2 O2(g) + 4 H+(aq)

ClO4-(aq) + 2 H2O2(aq) → ClO2

-(aq) + 2 H2O(l) + 2 O2(g) 4.79 (a) S2O3

2-(aq) → S4O62-(aq)

2 S2O32-(aq) → S4O6

2-(aq) 2 S2O3

2-(aq) → S4O62-(aq) + 2 e- (oxidation half reaction)

I2(aq) → I-(aq) I2(aq) → 2 I-(aq) I2(aq) + 2 e- → 2 I-(aq) (reduction half reaction)

Combine the two half reactions. 2 S2O3

2-(aq) + I2(aq) → S4O62-(aq) + 2 I-(aq)

(b) Mn2+(aq) → MnO2(s)

Mn2+(aq) + 2 H2O(l) → MnO2(s) Mn2+(aq) + 2 H2O(l) → MnO2(s) + 4 H+(aq) Mn2+(aq) + 2 H2O(l) → MnO2(s) + 4 H+(aq) + 2 e- (oxidation half reaction)

H2O2(aq) → 2 H2O(l) 2 H+(aq) + H2O2(aq) → 2 H2O(l) 2 H+(aq) + H2O2(aq) + 2 e- → 2 H2O(l) (reduction half reaction)

Combine the two half reactions. Mn2+(aq) + 2 H2O(l) + 2 H+(aq) + H2O2(aq) → MnO2(s) + 4 H+(aq) + 2 H2O(l) Mn2+(aq) + H2O2(aq) → MnO2(s) + 2 H+(aq) Mn2+(aq) + H2O2(aq) + 2 OH-(aq) → MnO2(s) + 2 H+(aq) + 2 OH-(aq) Mn2+(aq) + H2O2(aq) + 2 OH-(aq) → MnO2(s) + 2 H2O(l)


79

(c) Zn(s) → Zn(OH)4

2-(aq) 4 H2O(l) + Zn(s) → Zn(OH)4

2-(aq) 4 H2O(l) + Zn(s) → Zn(OH)4

2-(aq) + 4 H+(aq) [4 H2O(l) + Zn(s) → Zn(OH)4

2-(aq) + 4 H+(aq) + 2 e-] x 4 (oxidation half reaction)

NO3-(aq) → NH3(aq)

NO3-(aq) → NH3(aq) + 3 H2O(l)

9 H+(aq) + NO3-(aq) → NH3(aq) + 3 H2O(l)

9 H+(aq) + NO3-(aq) + 8 e- → NH3(aq) + 3 H2O(l) (reduction half reaction)

Combine the two half reactions. 16 H2O(l) + 4 Zn(s) + 9 H+(aq) + NO3

-(aq) → 4 Zn(OH)4

2-(aq) + 16 H+(aq) + NH3(aq) + 3 H2O(l) 13 H2O(l) + 4 Zn(s) + NO3

-(aq) → 4 Zn(OH)42-(aq) + 7 H+(aq) + NH3(aq)

13 H2O(l) + 4 Zn(s) + NO3-(aq) + 7 OH-(aq) →

4 Zn(OH)42-(aq) + 7 H+(aq) + 7 OH-(aq) + NH3(aq)

13 H2O(l) + 4 Zn(s) + NO3-(aq) + 7 OH-(aq) →

4 Zn(OH)42-(aq) + 7 H2O(l) + NH3(aq)

6 H2O(l) + 4 Zn(s) + NO3-(aq) + 7 OH-(aq) → 4 Zn(OH)4

2-(aq) + NH3(aq)

(d) Bi(OH)3(s) → Bi(s) Bi(OH)3(s) → Bi(s) + 3 H2O(l) 3 H+(aq) + Bi(OH)3(s) → Bi(s) + 3 H2O(l) [3 H+(aq) + Bi(OH)3(s) + 3 e- → Bi(s) + 3 H2O(l)] x 2 (reduction half reaction)

Sn(OH)3

-(aq) → Sn(OH)62-(aq)

Sn(OH)3-(aq) + 3 H2O(l) → Sn(OH)6

2-(aq) Sn(OH)3

-(aq) + 3 H2O(l) → Sn(OH)62-(aq) + 3 H+(aq)

[Sn(OH)3-(aq) + 3 H2O(l) → Sn(OH)6


Combine the two half reactions. 6 H+(aq) + 2 Bi(OH)3(s) + 3 Sn(OH)3

-(aq) + 9 H2O(l) → 2 Bi(s) + 6 H2O(l) + 3 Sn(OH)6

2-(aq) + 9 H+(aq) 2 Bi(OH)3(s) + 3 Sn(OH)3

-(aq) + 3 H2O(l) →2 Bi(s) + 3 Sn(OH)62-(aq) + 3 H+(aq)

2 Bi(OH)3(s) + 3 Sn(OH)3-(aq) + 3 H2O(l) + 3 OH-(aq) →

2 Bi(s) + 3 Sn(OH)62-(aq) + 3 H+(aq) + 3 OH-(aq)

2 Bi(OH)3(s) + 3 Sn(OH)3-(aq) + 3 H2O(l) + 3 OH-(aq) →

2 Bi(s) + 3 Sn(OH)62-(aq) + 3 H2O(l)

2 Bi(OH)3(s) + 3 Sn(OH)3-(aq) + 3 OH-(aq) → 2 Bi(s) + 3 Sn(OH)6

2-(aq) 4.80 (a) Zn(s) → Zn2+(aq)

Zn(s) → Zn2+(aq) + 2 e- (oxidation half reaction)


80

VO2+(aq) → V3+(aq) VO2+(aq) → V3+(aq) + H2O(l) 2 H+(aq) + VO2+(aq) → V3+(aq) + H2O(l) [2 H+(aq) + VO2+(aq) + e- → V3+(aq) + H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. Zn(s) + 2 VO2+(aq) + 4 H+(aq) → Zn2+(aq) + 2 V3+(aq) + 2 H2O(l)

(b) Ag(s) → Ag+(aq)

Ag(s) → Ag+(aq) + e- (oxidation half reaction)

NO3-(aq) → NO2(g)

NO3-(aq) → NO2(g) + H2O(l)

2 H+(aq) + NO3-(aq) → NO2(g) + H2O(l)

2 H+(aq) + NO3-(aq) + e- → NO2(g) + H2O(l) (reduction half reaction)

Combine the two half reactions. 2 H+(aq) + Ag(s) + NO3

-(aq) → Ag+(aq) + NO2(g) + H2O(l)

(c) Mg(s) → Mg2+(aq) [Mg(s) → Mg2+(aq) + 2 e- ] x 3 (oxidation half reaction)

VO4

3-(aq) → V2+(aq) VO4

3-(aq) → V2+(aq) + 4 H2O(l) 8 H+(aq) + VO4

3-(aq) → V2+(aq) + 4 H2O(l) [8 H+(aq) + VO4

3-(aq) + 3 e- → V2+(aq) + 4 H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 3 Mg(s) + 16 H+(aq) + 2 VO4

3-(aq) → 3 Mg2+(aq) + 2 V2+(aq) + 8 H2O(l)

(d) I-(aq) → I3-(aq)

3 I-(aq) → I3-(aq)

[3 I-(aq) → I3-(aq) + 2 e-] x 8 (oxidation half reaction)

IO3

-(aq) → I3-(aq)

3 IO3-(aq) → I3

-(aq) 3 IO3

-(aq) → I3-(aq) + 9 H2O(l)

18 H+(aq) + 3 IO3-(aq) → I3

-(aq) + 9 H2O(l) 18 H+(aq) + 3 IO3

-(aq) + 16 e- → I3-(aq) + 9 H2O(l) (reduction half reaction)

Combine the two half reactions. 18 H+(aq) + 3 IO3

-(aq) + 24 I-(aq) → 9 I3-(aq) + 9 H2O(l)

Divide each coefficient by 3. 6 H+(aq) + IO3

-(aq) + 8 I-(aq) → 3 I3-(aq) + 3 H2O(l)


81

4.81 (a) MnO4

-(aq) → Mn2+(aq) MnO4

-(aq) → Mn2+(aq) + 4 H2O(l) 8 H+(aq) + MnO4

-(aq) → Mn2+(aq) + 4 H2O(l) [8 H+(aq) + MnO4

-(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)] x 4 (reduction half reaction)

C2H5OH(aq) → CH3CO2H(aq) C2H5OH(aq) + H2O(l) → CH3CO2H(aq) C2H5OH(aq) + H2O(l) → CH3CO2H(aq) + 4 H+(aq) [C2H5OH(aq) + H2O(l) → CH3CO2H(aq) + 4 H+(aq) + 4 e-] x 5

(oxidation half reaction) Combine the two half reactions. 32 H+(aq) + 4 MnO4

-(aq) + 5 C2H5OH(aq) + 5 H2O(l) → 4 Mn2+(aq) + 16 H2O(l) + 5 CH3CO2H(aq) + 20 H+(aq)

12 H+(aq) + 4 MnO4-(aq) + 5 C2H5OH(aq) →

4 Mn2+(aq) + 11 H2O(l) + 5 CH3CO2H(aq)

(b) Cr2O72-(aq) → Cr3+(aq)

Cr2O72-(aq) → 2 Cr3+(aq)

Cr2O72-(aq) → 2 Cr3+(aq) + 7 H2O(l)

14 H+(aq) + Cr2O7


2-(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) (reduction half reaction)

H2O2(aq) → O2(g) H2O2(aq) → O2(g) + 2 H+(aq) [H2O2(aq) → O2(g) + 2 H+(aq) + 2 e-] x 3 (oxidation half reaction)

Combine the two half reactions. 14 H+(aq) + Cr2O7

2-(aq) + 3 H2O2(aq) → 2 Cr3+(aq) + 7 H2O(l) + 3 O2(g) + 6 H+(aq)

8 H+(aq) + Cr2O72-(aq) + 3 H2O2(aq) → 2 Cr3+(aq) + 7 H2O(l) + 3 O2(g)

(c) Sn2+(aq) → Sn4+(aq)

[Sn2+(aq) → Sn4+(aq) + 2 e-] x 4 (oxidation half reaction)

IO4-(aq) → I-(aq)

IO4-(aq) → I-(aq) + 4 H2O(l)

8 H+(aq) + IO4-(aq) → I-(aq) + 4 H2O(l)

8 H+(aq) + IO4-(aq) + 8 e- → I-(aq) + 4 H2O(l) (reduction half reaction)

Combine the two half reactions. 4 Sn2+(aq) + 8 H+(aq) + IO4

-(aq) → 4 Sn4+(aq) + I-(aq) + 4 H2O(l)


82

(d) PbO2(s) + Cl-(aq) → PbCl2(s) PbO2(s) + 2 Cl-(aq) → PbCl2(s) PbO2(s) + 2 Cl-(aq) → PbCl2(s) + 2 H2O(l) PbO2(s) + 4 H+(aq) + 2 Cl-(aq) → PbCl2(s) + 2 H2O(l) [PbO2(s) + 4 H+(aq) + 2 Cl-(aq) + 2 e- → PbCl2(s) + 2 H2O(l)] x 2

(reduction half reaction) H2O(l) → O2(g) 2 H2O(l) → O2(g) 2 H2O(l) → O2(g) + 4 H+(aq) 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e- (oxidation half reaction)

Combine the two half reactions. 2 PbO2(s) + 8 H+(aq) + 4 Cl-(aq) + 2 H2O(l) →

2 PbCl2(s) + 4 H2O(l) + O2(g) + 4 H+(aq) 2 PbO2(s) + 4 H+(aq) + 4 Cl-(aq) → 2 PbCl2(s) + 2 H2O(l) + O2(g)

Redox Titrations 4.82 I2(aq) + 2 S2O3

2-(aq) → S4O62-(aq) + 2 I-(aq); 35.20 mL = 0.032 50 L

0.035 20 L x I g 0.670 = I mol 1I g 253.8

x OS mol 2I mol 1

x L

OS mol 0.1502

2

2

_232

2_2

32

4.83 2.486 g I2 x I mol 1OS mol 2

x I g 253.8

I mol 1

2

_232

2

2 = 1.959 x 10-2 mol S2O32-

mol 0.250

L 1 x mol 10 x 1.959 2_ = 0.0784 L; 0.0784 L = 78.4 mL

4.84 3 H3AsO3(aq) + BrO3

-(aq) → Br-(aq) + 3 H3AsO4(aq) 22.35 mL = 0.022 35 L and 50.00 mL = 0.050 00 L

0.022 35 L x AsOH mol 10 x 6.70 = BrO mol 1

AsOH mol 3 x

LBrO mol 0.100

333_

_3

33_3

molarity = As(III) M 0.134 = L 00 0.050

mol 10 x 6.70 3_

4.85 As2O3, 197.84 amu; 28.55 mL = 0.028 55 L

1.550 g As2O3 x AsOH mol 3

BrO mol 1 x

OAs mol 1AsOH mol 2

x OAs g 197.84

OAs mol 1

33

_3

32

33

32

32

= 5.223 x 10-3 mol BrO3-; KBrO3 molarity =

L 55 0.028

mol 10 x 5.223 3_

= 0.1829 M

4.86 2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq); 13.28 mL = 0.013 28 L


83

0.013 28 L x Fe g 0.1506 = Fe mol 1

Fe g 55.847 x

Sn mol 1Fe mol 2

x L

Sn mol 0.1015 +3+3

+3

+2

+3+2

mass % Fe = 80.32% = 100% x g 0.1875

g 0.1506

4.87 Fe2O3, 159.69 amu; 23.84 mL = 0.023 84 L

1.4855 g Fe2O3 x Fe mol 2Sn mol 1

x OFe mol 1

Fe mol 2 x


+3

+2

32

+3

32

32 = 0.009 302 mol

Sn2+

Sn2+ molarity = L 84 0.023

mol 302 0.009 = 0.3902 M

4.88 C2H5OH(aq) + 2 Cr2O7

2-(aq) + 16 H+(aq) → 2 CO2(g) + 4 Cr3+(aq) + 11 H2O(l) C2H5OH, 46.07 amu; 8.76 mL = 0.008 76 L

OHHC g 07 0.010 =

OHHC mol 1

OHHC g 46.07 x

OCr mol 2

OHHC mol 1 x

LOCr mol 88 0.049

x L 76 0.008

52

52

52

_272

52_2

72

mass % C2H5OH = 0.101% = 100% x g 10.002

g 07 0.010

4.89 21.08 mL = 0.021 08 L

0.021 08 L x

Ca mol 1Ca g 40.08

x OCH mol 1

Ca mol 1 x

MnO mol 2OCH mol 5

x overLMnO mol 10 x 9.88+2

+2

422

+2

_4

422_4

4_

= 0.002 09 g = 2.09 mg General Problems 4.90 (a) [Fe(CN)6]

3-(aq) → Fe(CN)6]4-(aq)

([Fe(CN)6]3-(aq) + e- → [Fe(CN)6]

4-(aq)) x 4 (reduction half reaction)

N2H4(aq) → N2(g) N2H4(aq) → N2(g) + 4 H+(aq) N2H4(aq) → N2(g) + 4 H+(aq) + 4 e- N2H4(aq) + 4 OH-(aq) → N2(g) + 4 H+(aq) + 4 OH-(aq) + 4 e- N2H4(aq) + 4 OH-(aq) → N2(g) + 4 H2O(l) + 4 e- (oxidation half reaction)

Combine the two half reactions. 4 [Fe(CN)6]

3-(aq) + N2H4(aq) + 4 OH-(aq) → 4 [Fe(CN)6]

4-(aq) + N2(g) + 4 H2O(l)


84

(b) Cl2(g) → Cl-(aq) Cl2(g) → 2 Cl-(aq) Cl2(g) + 2 e- → 2 Cl-(aq) (reduction half reaction)

SeO3

2-(aq) → SeO42-(aq)

SeO32-(aq) + H2O(l) → SeO4

2-(aq) SeO3

2-(aq) + H2O(l) → SeO42-(aq) + 2 H+(aq)

SeO32-(aq) + H2O(l) → SeO4

2-(aq) + 2 H+(aq) + 2 e- SeO3

2-(aq) + H2O(l) + 2 OH-(aq) → SeO42-(aq) + 2 H+(aq) + 2 OH-(aq) + 2 e-

SeO32-(aq) + H2O(l) + 2 OH-(aq) → SeO4

2-(aq) + 2 H2O(l) + 2 e- SeO3

2-(aq) + 2 OH-(aq) → SeO42-(aq) + H2O(l) + 2 e- (oxidation half reaction)

Combine the two half reactions. SeO3

2-(aq) + Cl2(g) + 2 OH-(aq) → SeO42-(aq) + 2 Cl-(aq) + H2O(l)

(c) CoCl2(aq) → Co(OH)3(s) + Cl-(aq)

CoCl2(aq) → Co(OH)3(s) + 2 Cl-(aq) CoCl2(aq) + 3 H2O(l) → Co(OH)3(s) + 2 Cl-(aq) CoCl2(aq) + 3 H2O(l) → Co(OH)3(s) + 2 Cl-(aq) + 3 H+(aq) [CoCl2(aq) + 3 H2O(l) → Co(OH)3(s) + 2 Cl-(aq) + 3 H+(aq) + e-] x 2

(oxidation half reaction) HO2

-(aq) → H2O(l) HO2

-(aq) → 2 H2O(l) 3 H+(aq) + HO2

-(aq) → 2 H2O(l) 3 H+(aq) + HO2

-(aq) + 2 e- → 2 H2O(l) (reduction half reaction) Combine the two half reactions. 2 CoCl2(aq) + 6 H2O(l) + 3 H+(aq) + HO2

-(aq) → 2 Co(OH)3(s) + 4 Cl-(aq) + 6 H+(aq) + 2 H2O(l)

2 CoCl2(aq) + 4 H2O(l) + HO2-(aq) → 2 Co(OH)3(s) + 4 Cl-(aq) + 3 H+(aq)

2 CoCl2(aq) + 4 H2O(l) + HO2-(aq) + 3 OH-(aq) →

2 Co(OH)3(s) + 4 Cl-(aq) + 3 H+(aq) + 3 OH-(aq) 2 CoCl2(aq) + 4 H2O(l) + HO2

-(aq) + 3 OH-(aq) → 2 Co(OH)3(s) + 4 Cl-(aq) + 3 H2O(l)

2 CoCl2(aq) + H2O(l) + HO2-(aq) + 3 OH-(aq) → 2 Co(OH)3(s) + 4 Cl-(aq)

4.91 57.91 mL = 0.057 91 L

0.057 91 L x Fe g 0.3292 = Fe mol 1Fe g 55.85

x Ce mol 1Fe mol 1

x overLCe mol 0.1018 +2+2

+2

+4

+2+4

mass % Fe = g 1.2284

g 0.3292 x 100% = 26.80%

4.92 (a) C2H6 H +1, C -3

(b) Na2B4O7 O -2, Na +1, B +3 (c) Mg2SiO4 O -2, Mg +2, Si +4


85

4.93 (a) PbO2(s) → Pb2+(aq)

PbO2(s) → Pb2+(aq) + 2 H2O(l) 4 H+(aq) + PbO2(s) → Pb2+(aq) + 2 H2O(l) [4 H+(aq) + PbO2(s) + 2 e- → Pb2+(aq) + 2 H2O(l)] x 5 (reduction half reaction)

Mn2+(aq) → MnO4

-(aq) 4 H2O(l) + Mn2+(aq) → MnO4

-(aq) 4 H2O(l) + Mn2+(aq) → MnO4

-(aq) + 8 H+(aq) [4 H2O(l) + Mn2+(aq) → MnO4

-(aq) + 8 H+(aq) + 5 e-] x 2 (oxidation half reaction)

Combine the two half reactions. 20 H+(aq) + 5 PbO2(s) + 8 H2O(l) + 2 Mn2+(aq) →

5 Pb2+(aq) + 10 H2O(l) + 2 MnO4-(aq) + 16 H+(aq)

4 H+(aq) + 5 PbO2(s) + 2 Mn2+(aq) → 5 Pb2+(aq) + 2 H2O(l) + 2 MnO4-(aq)

(b) As2O3(s) → H3AsO4(aq)

As2O3(s) → 2 H3AsO4(aq) 5 H2O(l) + As2O3(s) → 2 H3AsO4(aq) 5 H2O(l) + As2O3(s) → 2 H3AsO4(aq) + 4 H+(aq) 5 H2O(l) + As2O3(s) → 2 H3AsO4(aq) + 4 H+(aq) + 4 e- (oxidation half

reaction)

NO3-(aq) → HNO2(aq)

NO3-(aq) → HNO2(aq) + H2O(l)

3 H+(aq) + NO3-(aq) → HNO2(aq) + H2O(l)

[3 H+(aq) + NO3-(aq) + 2 e- → HNO2(aq) + H2O(l)] x 2

(reduction half reaction) Combine the two half reactions. 5 H2O(l) + As2O3(s) + 6 H+(aq) + 2 NO3

-(aq) → 2 H3AsO4(aq) + 4 H+(aq) + 2 HNO2(aq) + 2 H2O(l)

3 H2O(l) + As2O3(s) + 2 H+(aq) + 2 NO3-(aq) → 2 H3AsO4(aq) + 2 HNO2(aq)

(c) Br2(aq) → Br-(aq)

Br2(aq) → 2 Br-(aq) Br2(aq) + 2 e- → 2 Br-(aq) (reduction half reaction)

SO2(g) → HSO4

-(aq) 2 H2O(l) + SO2(g) → HSO4

-(aq) 2 H2O(l) + SO2(g) → HSO4

-(aq) + 3 H+(aq) 2 H2O(l) + SO2(g) → HSO4

-(aq) + 3 H+(aq) + 2 e- (oxidation half reaction)


86

Combine the two half reactions. 2 H2O(l) + Br2(aq) + SO2(g) → 2 Br-(aq) + HSO4

-(aq) + 3 H+(aq)

(d) I-(aq) → I2(s) 2 I-(aq) → I2(s) 2 I-(aq) → I2(s) + 2 e- (oxidation half reaction)

NO2

-(aq) → NO(g) NO2

-(aq) → NO(g) + H2O(l) 2 H+(aq) + NO2

-(aq) → NO(g) + H2O(l) [2 H+(aq) + NO2

-(aq) + e- → NO(g) + H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 4 H+(aq) + 2 NO2

-(aq) + 2 I-(aq) → 2 NO(g) + I2(s) + 2 H2O(l) 4.94 (a) “Any element higher in the activity series will react with the ion of any element lower

in the activity series.” C + B+ → C+ + B; therefore C is higher than B. A+ + D → no reaction; therefore A is higher than D. C+ + A → no reaction; therefore C is higher than A. D + B+ → D+ + B; therefore D is higher than B. The net result is C > A > D > B. (b) (1) The reaction, A+ + C → A + C+, will occur because C is above A in the activity

series. (2) The reaction, A+ + B → A + B+, will not occur because B is below A in the

activity series. 4.95 (a) Ksp = [Ag+]2[CrO4

2-]

(b) Ag2CrO4(s) _ 2 Ag+(aq) + CrO42-(aq)

2x x In a saturated solution 2x = [Ag+] and x = [CrO4

2-]. Ksp = [Ag+]2[CrO4

2-] = 1.1 x 10-12 = (2x)2(x) = 4x3; Solve for x; x = 6.5 x 10-5 M [Ag+] = 2x = 2(6.5 x 10-5 M) = 1.3 x 10-4 M; [CrO4

2-] = x = 6.5 x 10-5 M

4.96 MgF2(s) _ Mg2+(aq) + 2 F-(aq) x 2x

[Mg2+] = x = 2.6 x 10-4 M and [F-] = 2x = 2(2.6 x 10-4 M) = 5.2 x 10-4 M in a saturated solution. Ksp = [Mg2+][F-]2 = (2.6 x 10-4 M)(5.2 x 10-4 M)2 = 7.0 x 10-11

4.97 65.20 mL = 0.065 20 L


87

1.926 g succinic acid x acid succinic g 118.1

acid succinic mol 1 = 0.016 31 mol succinic acid

0.5000 L 1

NaOH mol x 0.065 20 L = 0.032 60 mol NaOH

acid succinic mol 31 0.016

NaOH mol 60 0.032 = 2; therefore succinic acid has two acidic hydrogens.

4.98 (a) Add HCl to precipitate Hg2Cl2. Hg2

2+(aq) + 2Cl-(aq) → Hg2Cl2(s) (b) Add H2SO4 to precipitate PbSO4. Pb2+(aq) + SO4

2-(aq) → PbSO4(s) (c) Add Na2CO3 to precipitate CaCO3. Ca2+(aq) + CO3

2-(aq) → CaCO3(s) (d) Add Na2SO4 to precipitate BaSO4. Ba2+(aq) + SO4

2-(aq) → BaSO4(s) 4.99 (a) Add AgNO3 to precipitate AgCl. Ag+(aq) + Cl-(aq) → AgCl(s)

(b) Add NiCl2 to precipitate NiS. Ni2+(aq) + S2-(aq) → NiS(s) (c) Add CaCl2 to precipitate CaCO3. Ca2+(aq) + CO3

2-(aq) → CaCO3(s) (d) Add MgCl2 to precipitate Mg(OH)2. Mg2+(aq) + 2 OH-(aq) → Mg(OH)2(s)

4.100 All four reactions are redox reactions.

(a) Mn(OH)2(s) → Mn(OH)3(s) Mn(OH)2(s) + OH-(aq) → Mn(OH)3(s) [Mn(OH)2(s) + OH-(aq) → Mn(OH)3(s) + e-] x 2 (oxidation half reaction)

H2O2(aq) → 2 H2O(l) 2 H+(aq) + H2O2(aq) → 2 H2O(l) 2 e- + 2 H+(aq) + H2O2(aq) → 2 H2O(l) 2 e- + 2 OH-(aq) + 2 H+(aq) + H2O2(aq) → 2 H2O(l) + 2 OH-(aq) 2 e- + 2 H2O(l) + H2O2(aq) → 2 H2O(l) + 2 OH-(aq) 2 e- + H2O2(aq) → 2 OH-(aq) (reduction half reaction)

Combine the two half reactions. 2 Mn(OH)2(s) + 2 OH-(aq) + H2O2(aq) → 2 Mn(OH)3(s) + 2 OH-(aq) 2 Mn(OH)2(s) + H2O2(aq) → 2 Mn(OH)3(s)

(b) [MnO4

2-(aq) → MnO4-(aq) + e- ] x 2 (oxidation half reaction)

MnO4

2-(aq) → MnO2(s) MnO4

2-(aq) → MnO2(s) + 2 H2O(l) 4 H+(aq) + MnO4

2-(aq) → MnO2(s) + 2 H2O(l) 2 e- + 4 H+(aq) + MnO4

2-(aq) → MnO2(s) + 2 H2O(l) (reduction half reaction)

Combine the two half reactions. 4 H+(aq) + 3 MnO4

2-(aq) → MnO2(s) + 2 MnO4-(aq) + 2 H2O(l)


88

(c) I-(aq) → I3-(aq)

3 I-(aq) → I3-(aq)

[3 I-(aq) → I3-(aq) + 2 e- ] x 8 (oxidation half reaction)

IO3

-(aq) → I3-(aq)

3 IO3-(aq) → I3

-(aq) 3 IO3

-(aq) → I3-(aq) + 9 H2O(l)

18 H+(aq) + 3 IO3-(aq) → I3

-(aq) + 9 H2O(l) 16 e- + 18 H+(aq) + 3 IO3

-(aq) → I3-(aq) + 9 H2O(l) (reduction half reaction)

Combine the two half reactions. 24 I-(aq) + 3 IO3

-(aq) + 18 H+(aq) → 9 I3-(aq) + 9 H2O(l)

Divide all coefficients by 3. 8 I-(aq) + IO3

-(aq) + 6 H+(aq) → 3 I3-(aq) + 3 H2O(l)

(d) P(s) → HPO3

2-(aq) 3 H2O(l) + P(s) → HPO3

2-(aq) 3 H2O(l) + P(s) → HPO3

2-(aq) + 5 H+(aq) [3 H2O(l) + P(s) → HPO3

2-(aq) + 5 H+(aq) + 3 e- ] x 2 (oxidation half reaction)

PO43-(aq) → HPO3

2-(aq) PO4

3-(aq) → HPO32-(aq) + H2O(l)

3 H+(aq) + PO43-(aq) → HPO3

2-(aq) + H2O(l) [2 e- + 3 H+(aq) + PO4

3-(aq) → HPO32-(aq) + H2O(l)] x 3

(reduction half reaction) Combine the two half reactions and add OH-. 6 H2O(l) + 2 P(s) + 9 H+(aq) + 3 PO4

3-(aq) → 5 HPO3

2-(aq) + 10 H+(aq) + 3 H2O(l) 3 H2O(l) + 2 P(s) + 3 PO4

3-(aq) → 5 HPO32-(aq) + H+(aq)

3 H2O(l) + 2 P(s) + 3 PO43-(aq) + OH-(aq) →

5 HPO32-(aq) + H+(aq) + OH-(aq)

3 H2O(l) + 2 P(s) + 3 PO4

3-(aq) + OH-(aq) → 5 HPO32-(aq) + H2O(l)

2 H2O(l) + 2 P(s) + 3 PO43-(aq) + OH-(aq) → 5 HPO3

2-(aq) 4.101 100.0 mL = 0.1000 L; 47.14 mL = 0.047 14 L

mol HCl and HBr = mol H+ = 0.1235 L 1

NaOH mol x 0.047 14 L = 5.8218 x 10-3 mol

mass of AgCl and AgBr = 0.9974 g; mol Ag = mol H+ = 5.8218 x 10-3 mol

mass of Ag = 5.8218 x 10-3 mol Ag x Ag mol 1

Ag g 107.87 = 0.6280 g Ag

mass of Cl and Br = 0.9974 g - 0.6280 g = 0.3694 g of Cl and Br Let Y = moles Cl and Z = moles Br in 0.3694 g of Cl and Br. Let (Y + Z) = moles Ag in 0.6280 g Ag.


89

For Ag: 0.6280 g = (Y + Z) x 107.87 g For Cl and Br: 0.3694 g = (Y x 35.453 g) + (Z x 79.904 g) Solve the simultaneous equations for Y and Z.

Rearrange the Ag equation:

Z_

g 107.87

g 0.6280 = Y

Substitute for Y in the Cl and Br equation above and solve for Z.

0.3694 g =

g 35.453 x Z_

g 107.87

g 0.6280 + (Z x 79.904 g)

Z = 44.451

0.1630 = 3.667 x 10-3

Y =

Z_

g 107.87

g 0.6280 =

10 x 3.667 _

g 107.87

g 0.6280 3_ = 2.155 x 10-3

HCl molarity = L 0.1000

mol 10 x 2.155 3_

= 0.021 55 M

HBr molarity = L 0.1000

mol 10 x 3.667 3_

= 0.036 67 M

4.102 (a) S4O6

2-(aq) → H2S(aq) S4O6

2-(aq) → 4 H2S(aq) S4O6

2-(aq) → 4 H2S(aq) + 6 H2O(l) 20 H+(aq) + S4O6

2-(aq) → 4 H2S(aq) + 6 H2O(l) 18 e- + 20 H+(aq) + S4O6

2-(aq) → 4 H2S(aq) + 6 H2O(l) (reduction half reaction)

Al(s) → Al3+(aq) [Al(s) → Al3+(aq) + 3 e-] x 6 (oxidation half reaction)

Combine the two half reactions. 20 H+(aq) + S4O6

2-(aq) + 6 Al(s) → 4 H2S(aq) + 6 Al3+(aq) + 6 H2O(l)

(b) S2O32-(aq) → S4O6

2-(aq) 2 S2O3

2-(aq) → S4O62-(aq)

[2 S2O32-(aq) → S4O6

2-(aq) + 2 e-] x 3 (oxidation half reaction)

Cr2O72-(aq) → Cr3+(aq)

Cr2O72-(aq) → 2 Cr3+(aq)

Cr2O72-(aq) → 2 Cr3+(aq) + 7 H2O(l)

14 H+(aq) + Cr2O72-(aq) → 2 Cr3+(aq) + 7 H2O(l)

6 e- + 14 H+(aq) + Cr2O72-(aq) → 2 Cr3+(aq) + 7 H2O(l) (reduction half reaction)

Combine the two half reactions.


90

14 H+(aq) + 6 S2O32-(aq) + Cr2O7

2-(aq) → 3 S4O62-(aq) + 2 Cr3+(aq) + 7 H2O(l)

(c) ClO3

-(aq) → Cl-(aq) ClO3

-(aq) → Cl-(aq) + 3 H2O(l) 6 H+(aq) + ClO3

-(aq) → Cl-(aq) + 3 H2O(l) [6 e- + 6 H+(aq) + ClO3

-(aq) → Cl-(aq) + 3 H2O(l)] x 14 (reduction half reaction)

As2S3(s) → H2AsO4-(aq) + HSO4

-(aq) As2S3(s) → 2 H2AsO4

-(aq) + 3 HSO4-(aq)

20 H2O(l) + As2S3(s) → 2 H2AsO4-(aq) + 3 HSO4

-(aq) 20 H2O(l) + As2S3(s) → 2 H2AsO4

-(aq) + 3 HSO4-(aq) + 33 H+(aq)

[20 H2O(l) + As2S3(s) → 2 H2AsO4

-(aq) + 3 HSO4-(aq) + 33 H+(aq) + 28 e-] x 3

(oxidation half reaction)

Combine the two half reactions. 84 H+(aq) + 60 H2O(l) + 14 ClO3

-(aq) + 3 As2S3(s) → 14 Cl-(aq) + 6 H2AsO4

-(aq) + 9 HSO4-(aq) + 42 H2O(l) + 99 H+(aq)

18 H2O(l) + 14 ClO3-(aq) + 3 As2S3(s) →

14 Cl-(aq) + 6 H2AsO4-(aq) + 9 HSO4

-(aq) + 15 H+(aq)

(d) IO3-(aq) → I-(aq)

IO3-(aq) → I-(aq) + 3 H2O(l)

6 H+(aq) + IO3-(aq) → I-(aq) + 3 H2O(l)

[6 e- + 6 H+(aq) + IO3-(aq) → I-(aq) + 3 H2O(l)] x 7 (reduction half reaction)

Re(s) → ReO4

-(aq) 4 H2O(l) + Re(s) → ReO4

-(aq) 4 H2O(l) + Re(s) → ReO4

-(aq) + 8 H+(aq) [4 H2O(l) + Re(s) → ReO4

-(aq) + 8 H+(aq) + 7 e-] x 6 (oxidation half reaction)

Combine the two half reactions. 42 H+(aq) + 24 H2O(l) + 7 IO3

-(aq) + 6 Re(s) → 7 I-(aq) + 6 ReO4

-(aq) + 21 H2O(l) + 48 H+(aq) 3 H2O(l) + 7 IO3

-(aq) + 6 Re(s) → 7 I-(aq) + 6 ReO4-(aq) + 6 H+(aq)

(e) HSO4-(aq) + Pb3O4(s) → PbSO4(s)

3 HSO4-(aq) + Pb3O4(s) → 3 PbSO4(s)

3 HSO4-(aq) + Pb3O4(s) → 3 PbSO4(s) + 4 H2O(l)

5 H+(aq) + 3 HSO4-(aq) + Pb3O4(s) → 3 PbSO4(s) + 4 H2O(l)

[2 e- + 5 H+(aq) + 3 HSO4-(aq) + Pb3O4(s) → 3 PbSO4(s) + 4 H2O(l)] x 10

(reduction half reaction)

As4(s) → H2AsO4-(aq)

As4(s) → 4 H2AsO4-(aq)


91

16 H2O(l) + As4(s) → 4 H2AsO4-(aq)

16 H2O(l) + As4(s) → 4 H2AsO4-(aq) + 24 H+(aq)

16 H2O(l) + As4(s) → 4 H2AsO4-(aq) + 24 H+(aq) + 20 e- (oxidation half reaction)

Combine the two half reactions. 26 H+(aq) + 30 HSO4

-(aq) + As4(s) + 10 Pb3O4(s) → 4 H2AsO4

-(aq) + 30 PbSO4(s) + 24 H2O(l)

(f) HNO2(aq) → NO3-(aq)

H2O(l) + HNO2(aq) → NO3-(aq)

H2O(l) + HNO2(aq) → NO3-(aq) + 3 H+(aq)

H2O(l) + HNO2(aq) → NO3-(aq) + 3 H+(aq) + 2 e- (oxidation half reaction)

HNO2(aq) → NO(g) HNO2(aq) → NO(g) + H2O(l) H+(aq) + HNO2(aq) → NO(g) + H2O(l) [1 e- + H+(aq) + HNO2(aq) → NO(g) + H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 3 HNO2(aq) → NO3

-(aq) + 2 NO(g) + H2O(l) + H+(aq) 4.103 (a) C4H4O6

2-(aq) → CO32-(aq)

C4H4O62-(aq) → 4 CO3

2-(aq) C4H4O6

2-(aq) + 6 H2O(l) → 4 CO32-(aq)

C4H4O6

2-(aq) + 6 H2O(l) → 4 CO32-(aq) + 16 H+(aq)

[C4H4O62-(aq) + 6 H2O(l) → 4 CO3


ClO3

-(aq) → Cl-(aq) ClO3

-(aq) → Cl-(aq) + 3 H2O(l) ClO3

-(aq) + 6 H+(aq) → Cl-(aq) + 3 H2O(l) [6 e- + ClO3

-(aq) + 6 H+(aq) → Cl-(aq) + 3 H2O(l)] x 5 (reduction half reaction)

Combine the two half reactions. 3 C4H4O6

2-(aq) + 18 H2O(l) + 5 ClO3-(aq) + 30 H+(aq) →

12 CO32-(aq) + 48 H+(aq) + 5 Cl-(aq) + 15 H2O(l)

3 C4H4O62-(aq) + 3 H2O(l) + 5 ClO3

-(aq) → 12 CO32-(aq) + 18 H+(aq) + 5 Cl-(aq)

3 C4H4O62-(aq) + 3 H2O(l) + 5 ClO3

-(aq) + 18 OH-(aq) → 12 CO3

2-(aq) + 18 H+(aq) + 18 OH-(aq) + 5 Cl-(aq) 3 C4H4O6

2-(aq) + 3 H2O(l) + 5 ClO3-(aq) + 18 OH-(aq) →

12 CO32-(aq) + 18 H2O(aq) + 5 Cl-(aq)

3 C4H4O62-(aq) + 5 ClO3

-(aq) + 18 OH-(aq) → 12 CO32-(aq) + 15 H2O(l) + 5 Cl-(aq)


92

(b) Al(s) → Al(OH)4-(aq)

Al(s) + 4 OH-(aq) → Al(OH)4-(aq)

[Al(s) + 4 OH-(aq) → Al(OH)4-(aq) + 3 e-] x 11 (oxidation half reaction)

BiONO3(s) → Bi(s) + NH3(aq) BiONO3(s) → Bi(s) + NH3(aq) + 4 H2O(l) BiONO3(s) + 11 H+(aq) → Bi(s) + NH3(aq) + 4 H2O(l) [BiONO3(s) + 11 H+(aq) + 11 e- → Bi(s) + NH3(aq) + 4 H2O(l)] x 3

(reduction half reaction)

Combine the two half reactions. 11 Al(s) + 44 OH-(aq) + 3 BiONO3(s) + 33 H+(aq) →

11 Al(OH)4-(aq) + 3 Bi(s) + 3 NH3(aq) + 12 H2O(l)

11 Al(s) + 11 OH-(aq) + 3 BiONO3(s) + 33 H2O(l) → 11 Al(OH)4

-(aq) + 3 Bi(s) + 3 NH3(aq) + 12 H2O(l) 11 Al(s) + 11 OH-(aq) + 3 BiONO3(s) + 21 H2O(l) →

11 Al(OH)4-(aq) + 3 Bi(s) + 3 NH3(aq)

(c) H2O2(aq) → O2(g) H2O2(aq) → O2(g) + 2 H+(aq) [H2O2(aq) → O2(g) + 2 H+(aq) + 2 e-] x 4 (oxidation half reaction)

Cl2O7(aq) → ClO2

-(aq) Cl2O7(aq) → 2 ClO2

-(aq) Cl2O7(aq) → 2 ClO2

-(aq) + 3 H2O(l) Cl2O7(aq) + 6 H+(aq) → 2 ClO2

-(aq) + 3 H2O(l) Cl2O7(aq) + 6 H+(aq) + 8 e- → 2 ClO2

-(aq) + 3 H2O(l) (reduction half reaction) Combine the two half reactions. 4 H2O2(aq) + Cl2O7(aq) + 6 H+(aq) → 4 O2(g) + 8 H+(aq) + 2 ClO2

-(aq) + 3 H2O(l)

4 H2O2(aq) + Cl2O7(aq) → 4 O2(g) + 2 H+(aq) + 2 ClO2-(aq) + 3 H2O(l)

4 H2O2(aq) + Cl2O7(aq) + 2 OH-(aq) → 4 O2(g) + 2 H+(aq) + 2 OH-(aq) + 2 ClO2

-(aq) + 3 H2O(l) 4 H2O2(aq) + Cl2O7(aq) + 2 OH-(aq) → 4 O2(g) + 2 ClO2

-(aq) + 5 H2O(l)

(d) Tl2O3(s) → TlOH(s) Tl2O3(s) → 2 TlOH(s) Tl2O3(s) → 2 TlOH(s) + H2O(l) Tl2O3(s) + 4 H+(aq) → 2 TlOH(s) + H2O(l) Tl2O3(s) + 4 H+(aq) + 4 e- → 2 TlOH(s) + H2O(l) (reduction half reaction) NH2OH(aq) → N2(g) 2 NH2OH(aq) → N2(g) 2 NH2OH(aq) → N2(g) + 2 H2O(l) 2 NH2OH(aq) → N2(g) + 2 H2O(l) + 2 H+(aq)


93

[2 NH2OH(aq) → N2(g) + 2 H2O(l) + 2 H+(aq) + 2 e-] x 2 (oxidation half reaction)

Combine the two half reactions. Tl2O3(s) + 4 H+(aq) + 4 NH2OH(aq) → 2 TlOH(s) + 2 N2(g) + 5 H2O(l) + 4 H+(aq) Tl2O3(s) + 4 NH2OH(aq) → 2 TlOH(s) + 2 N2(g) + 5 H2O(l)

(e) Cu(NH3)4

2+(aq) → Cu(s) + 4 NH3(aq) Cu(NH3)4

2+(aq) + 2 e- → Cu(s) + 4 NH3(aq) (reduction half reaction)

S2O42-(aq) → SO3

2-(aq) S2O4

2-(aq) → 2 SO32-(aq)

S2O42-(aq) + 2 H2O(l) → 2 SO3

2-(aq) S2O4

2-(aq) + 2 H2O(l) → 2 SO32-(aq) + 4 H+(aq)

S2O42-(aq) + 2 H2O(l) → 2 SO3

2-(aq) + 4 H+(aq) + 2 e- (oxidation half reaction)

Combine the two half reactions. Cu(NH3)4

2+(aq) + S2O42-(aq) + 2 H2O(l) → Cu(s) + 4 NH3(aq) + 2 SO3

2-(aq) + 4 H+(aq) Cu(NH3)4

2+(aq) + S2O42-(aq) + 2 H2O(l) + 4 OH-(aq) →

Cu(s) + 4 NH3(aq) + 2 SO32-(aq) + 4 H+(aq) + 4 OH-(aq)

Cu(NH3)42+(aq) + S2O4

2-(aq) + 2 H2O(l) + 4 OH-(aq) → Cu(s) + 4 NH3(aq) + 2 SO3

2-(aq) + 4 H2O(l) Cu(NH3)4

2+(aq) + S2O42-(aq) + 4 OH-(aq) →

Cu(s) + 4 NH3(aq) + 2 SO32-(aq) + 2 H2O(l)

(f) Mn(OH)2(s) → MnO2(s) Mn(OH)2(s) → MnO2(s) + 2 H+(aq) [Mn(OH)2(s) → MnO2(s) + 2 H+(aq) + 2 e-] x 3 (oxidation half reaction)

MnO4

-(aq) → MnO2(s) MnO4

-(aq) → MnO2(s) + 2 H2O(l) MnO4

-(aq) + 4 H+(aq) → MnO2(s) + 2 H2O(l) [MnO4

-(aq) + 4 H+(aq) + 3 e- → MnO2(s) + 2 H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 3 Mn(OH)2(s) + 2 MnO4

-(aq) + 8 H+(aq) → 5 MnO2(s) + 6 H+(aq) + 4 H2O(l) 3 Mn(OH)2(s) + 2 MnO4

-(aq) + 2 H+(aq) → 5 MnO2(s) + 4 H2O(l) 3 Mn(OH)2(s) + 2 MnO4

-(aq) + 2 H+(aq) + 2 OH-(aq) → 5 MnO2(s) + 4 H2O(l) + 2 OH-(aq)

3 Mn(OH)2(s) + 2 MnO4-(aq) + 2 H2O(l) → 5 MnO2(s) + 4 H2O(l) + 2 OH-(aq)

3 Mn(OH)2(s) + 2 MnO4-(aq) → 5 MnO2(s) + 2 H2O(l) + 2 OH-(aq)

4.104 CuO, 79.55 amu; Cu2O, 143.09 amu

Let X equal the mass of CuO and Y the mass of Cu2O in the 10.50 g mixture. Therefore,


94

X + Y = 10.50 g.

mol Cu = 8.66 g x Cu g 63.546

Cu mol 1 = 0.1363 mol Cu

mol CuO + 2 x mol Cu2O = 0.1363 mol Cu

= OCu g 143.09

OCu mol 1 x Y x 2 +

CuO g 79.55

CuO mol 1 x X

2

2

0.1363 mol Cu


= OCu g 143.09

OCu mol 1 x Y x 2 +

CuO g 79.55

CuO mol 1 x Y) _ g (10.50

2

2

0.1363 mol Cu

= g 143.09

mol Y 2 +

g 79.55

mol Y _

79.55

mol 10.50 0.1363 mol

79.55

mol 10.50 _ mol 0.1363 =

g 143.09

mol Y 2 +

g 79.55

mol Y _ = 0.0043 mol

= g) g)(143.09 (79.55

g) mol)(79.55 Y (2 + g) 9mol)(143.0 Y (_ 0.0043 mol

; mol 0.0043 = g 11383

mol Y 16.01 0.0043 =

g 11383

Y 16.01

Y = (0.0043)(11383 g)/16.01 = 3.06 g Cu2O

X = 10.50 g - Y = 10.50 g - 3.06 g = 7.44 g CuO 4.105 (a) PbI2, 461.01 amu

Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) 75.0 mL = 0.0750 L and 100.0 mL = 0.1000 L mol Pb(NO3)2 = (0.0750 L)(0.100 mol/L) = 7.50 x 10-3 mol Pb(NO3)2 mol KI = (0.1000 L)(0.190 mol/L) = 1.90 x 10-2 mol KI

mols KI needed = 7.50 x 10-3 mol Pb(NO3)2 x = )NOPb( mol 1

KI mol 2

23

1.50 x 10-2 mol KI

There is an excess of KI, so Pb(NO3)2 is the limiting reactant.

mass PbI2 = 7.50 x 10-3 mol Pb(NO3)2 x = PbI mol 1

PbI g 461.01 x

)NOPb( mol 1PbI mol 1

2

2

23

2 3.46 g PbI2

(b) Because Pb(NO3)2 is the limiting reactant, Pb2+ is totally consumed and [Pb2+] = 0. mol K+ = mol KI = 1.90 x 10-2 mol

mol NO3- = 7.50 x 10-3 mol Pb(NO3)2 x =

)NOPb( mol 1NO mol 2

23

_3 0.0150 mol NO3

-

mol I- = (initial mol KI) - (mol KI needed) = 0.0190 mol - 0.0150 mol = 0.0040 mol I-

total volume = 0.0750 L + 0.1000 L = 0.1750 L


95

[K+] = = L 0.1750

mol 0.01900.109 M

[NO3-] = =

L 0.1750

mol 0.01500.0857 M

[I -] = = L 0.1750

mol 0.00400.023 M

Multi-Concept Problems 4.106 NaOH, 40.00 amu; Ba(OH)2, 171.34 amu

Let X equal the mass of NaOH and Y the mass of Ba(OH)2 in the 10.0 g mixture. Therefore, X + Y = 10.0 g.

mol HCl = 108.9 mL x = L 1

HCl mol 1.50 x

mL 1000

L 1 0.163 mol HCl

mol NaOH + 2 x mol Ba(OH)2 = 0.163 mol HCl

= )Ba(OH g 171.34

)Ba(OH mol 1 x Y x 2 +

NaOH g 40.00

NaOH mol 1 x X

2

2

0.163 mol HCl


= )Ba(OH g 171.34

)Ba(OH mol 1 x Y x 2 +

NaOH g 40.00

NaOH mol 1 x Y) _ g (10.0

2

2

0.163 mol HCl

= g 171.34

mol Y 2 +

g 40.00

mol Y _

40.00

mol 10.00 0.163 mol

40.00

mol 10.00 _ mol 0.163 =

g 171.34

mol Y 2 +

g 40.00

mol Y _ = -0.087 mol

= g) g)(171.34 (40.00

g) mol)(40.00 Y (2 + g) 4mol)(171.3 Y (_ -0.087 mol

; mol 0.087 _ = g 6853.6

mol Y 91.34 _ 0.087 =

g 6853.6

Y 91.34

Y = (0.087)(6853.6 g)/91.34 = 6.5 g Ba(OH)2 X = 10.0 g - Y = 10.0 g - 6.5 g = 3.5 g NaOH

4.107 100.0 mL = 0.1000 L and 50.0 mL = 0.0500 L

mol Na2SO4 = (0.1000 L)(0.100 mol/L) = 0.0100 mol Na2SO4 mol SO4

2- = mol Na2SO4 = 0.0100 mol SO42-

mol Na+ = 0.0100 mol Na2SO4 x = SONa mol 1

Na mol 2

42

+

0.0200 mol Na+

mol ZnCl2 = (0.0500 L)(0.300 mol/L) = 0.0150 mol ZnCl2 mol Zn2+ = mol ZnCl2 = 0.0150 mol Zn2+

mol Cl- = 0.0150 mol ZnCl2 x = ZnCl mol 1

Cl mol 2

2

_

0.0300 mol Cl-

mol Ba(CN)2 = (0.1000 L)(0.200 mol/L) = 0.0200 mol Ba(CN)2


96

mol Ba2+ = mol Ba(CN)2 = 0.0200 mol Ba2+

mol CN- = 0.0200 mol Ba(CN)2 x = )Ba(CN mol 1

CN mol 2

2

_

0.0400 mol CN-

The following two reactions will take place to form precipitates. Zn2+(aq) + 2 CN-(aq) → Zn(CN)2(s) Ba2+(aq) + SO4

2-(aq) → BaSO4(s)

For Zn2+, mol CN- needed = 0.0150 mol Zn2+ x = Zn mol 1CN mol 2

+2

_

0.0300 mol CN- needed

CN- is in excess, so Zn2+ is the limiting reactant and is totally consumed. mol CN- remaining after reaction= 0.0400 mol - 0.0300 mol = 0.0100 mol CN-

For Ba2+, mol SO4

2- needed = mol Ba2+ = 0.0200 mol SO42- needed

Ba2+ is in excess, so SO42- is the limiting reactant and is totally consumed.

mol Ba2+ remaining after reaction = 0.0200 mol - 0.0100 mol = 0.0100 mol Ba2+ total volume = 0.1000 L + 0.0500 L + 0.1000 L = 0.2500 L [Zn2+] = 0 [SO4

2-] = 0

[Na+] = = L 0.2500

mol 0.02000.0800 M

[Cl -] = = L 0.2500

mol 0.0300 0.120 M

[CN-] = = L 0.2500

mol 0.01000.0400 M

[Ba2+] = = L 0.2500

mol 0.01000.0400 M

4.108 KNO3, 101.10 amu; BaCl2, 208.24 amu; NaCl, 58.44 amu; BaSO4, 233.40 amu;

AgCl, 143.32 amu (a) The two precipitates are BaSO4(s) and AgCl(s). (b) H2SO4 only reacts with BaCl2. H2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2 HCl(aq) Calculate the number of moles of BaCl2 in 100.0 g of the mixture.

mol BaCl2 = 67.3 g BaSO4 x = BaSO mol 1BaCl mol 1

x BaSO g 233.40

BaSO mol 1

4

2

4

4 0.288 mol BaCl2

Calculate mass and moles of BaCl2 in 250.0 g sample.

mass BaCl2 = 0.288 mol BaCl2 x = g 100.0

g 250.0 x

BaCl mol 1BaCl g 208.24

2

2 150. g BaCl2

mol BaCl2 = 150. g BaCl2 x = BaCl g 208.24

BaCl mol 1

2

2 0.720 mol BaCl2

AgNO3 reacts with both NaCl and BaCl2 in the remaining 150.0 g of the mixture. 3 AgNO3(aq) + NaCl(aq) + BaCl2(aq) → 3 AgCl(s) + NaNO3(aq) + Ba(NO3)2(aq) Calculate the moles of AgCl that would have been produced from the 250.0 g mixture.


97

mol AgCl = 197.6 g AgCl x = g 150.0

g 250.0 x

AgCl g 143.32

AgCl mol 1 2.30 mol AgCl

mol AgCl = 2 x (mol BaCl2) + mol NaCl Calculate the moles and mass of NaCl in the 250.0 g mixture. 2.30 mol AgCl = 2 x 0.720 mol BaCl2 + mol NaCl mol NaCl = 2.30 mol - 2(0.720 mol) = 0.86 mol NaCl

mass NaCl = 0.86 mol NaCl x = NaCl mol 1

NaCl g 58.44 50. g NaCl

Calculate the mass of KNO3 in the 250.0 g mixture. total mass = mass BaCl2 + mass NaCl + mass KNO3 250.0 g = 150. g BaCl2 + 50. g NaCl + mass KNO3 mass KNO3 = 250.0 g - 150. g BaCl2 - 50. g NaCl = 50. g KNO3

4.109 100.0 mL = 0.1000 L; 50.0 mL = 0.0500 L; 250.0 mL = 0.2500 L

After step (2): BaCl2(aq) + 2 AgNO3(aq) → AgCl(s) + Ba(NO3)2(aq) mol BaCl2 = (0.1000 L)(0.100 mol/L) = 0.0100 mol BaCl2 mol Ba2+ = mol BaCl2 = 0.0100 mol Ba2+

mol Cl- = 0.0100 mol BaCl2 x = BaCl mol 1

Cl mol 2

2

_

0.0200 mol Cl-

mol AgNO3 = (0.0500 L)(0.100 mol/L) = 0.00500 mol AgNO3 mol Ag+ = mol AgNO3 = 0.00500 mol Ag+ mol NO3

- = mol AgNO3 = 0.00500 mol NO3-

0.00500 mol Ag+ requires only 0.00500 mol Cl-, so Ag+ is the limiting reactant and totally consumed. mol Cl- remaining after reaction = 0.0200 mol - 0.00500 mol = 0.0150 mol Cl-

After step (3): Ba2+(aq) + H2SO4(aq) → BaSO4(s) + 2 H+(aq) mol H2SO4 = (0.0500 L)(0.100 mol/L) = 0.00500 mol H2SO4 mol SO4

2- = mol H2SO4 = 0.00500 mol SO42-

mol H+ = 0.00500 mol H2SO4 x = SOH mol 1H mol 2

42

+

0.0100 mol H+

0.0100 mol Ba2+ requires 0.0100 mol SO42-, so SO4

2- is the limiting reactant and is totally consumed. mol Ba2+ remaining after reaction = 0.0100 mol - 0.00500 mol = 0.00500 mol Ba2+

After step (4): NH3(aq) + H+(aq) → NH4

+(aq) mol NH3 = (0.2500 L)(0.100 mol/L) = 0.0250 mol NH3 0.0250 mol NH3 requires 0.0250 mol H+, so H+ is the limiting reactant and is totally consumed. mol NH3 remaining after reaction = 0.0250 mol - 0.0100 mol = 0.0150 mol NH3 mol NH4

+ = mol H+ before reaction = 0.0100 mol NH4+


98

total volume = 0.1000 L + 0.0500 L + 0.0500 L + 0.2500 L = 0.4500 L

[Ba2+] = = L 0.4500

mol 0.005000.0111 M

[Cl-] = = L 0.4500

mol 0.01500.0333 M

[NO3-] = =

L 0.4500

mol 0.00500 0.0111 M

[NH3] = = L 0.4500

mol 0.01500.0333 M

[NH4+] = =

L 0.4500

mol 0.01000.0222 M

4.110 (a) Cr2+(aq) + Cr2O7

2-(aq) → Cr3+(aq) [Cr2+(aq) → Cr3+(aq) + e-] x 6 (oxidation half reaction)

Cr2O7

2-(aq) → Cr3+(aq) Cr2O7

2-(aq) → 2 Cr3+(aq) Cr2O7


2-(aq) → 2 Cr3+(aq) + 7 H2O(l) 6 e- + 14 H+(aq) + Cr2O7

2-(aq) → 2 Cr3+(aq) + 7 H2O(l) (reduction half reaction)

Combine the two half reactions. 14 H+(aq) + Cr2O7

2-(aq) + 6 Cr2+(aq) → 8 Cr3+(aq) + 7 H2O(l)

(b) total volume = 100.0 ml + 20.0 mL = 120.0 mL = 0.1200 L Initial moles:

0.120 L 1

)NOCr( mol 23 x 0.1000 L = 0.0120 mol Cr(NO3)2

0.500 L 1

HNO mol 3 x 0.1000 L = 0.0500 mol HNO3

0.250 L 1

OCrK mol 722 x 0.0200 L = 0.005 00 mol K2Cr2O7

Check for the limiting reactant. 0.0120 mol of Cr2+ requires (0.0120)/6 = 0.00200 mol Cr2O7

2- and (14/6)(0.0120) = 0.0280 mol H+. Both are in excess of the required amounts, so Cr2+ is the limiting reactant.

14 H+(aq) + Cr2O7

2-(aq) + 6 Cr2+(aq) → 8 Cr3+(aq) + 7 H2O(l) Initial moles 0.0500 0.00500 0.0120 0 Change -14x -x -6x +8x Because Cr2+ is the limiting reactant, 6x = 0.0120 and x = 0.00200 Final moles 0.0220 0.00300 0 0.00160


99

mol K+ = 0.00500 mol K2Cr2O7 x OCrK mol 1

K mol 2

722

+

= 0.0100 mol K+

mol NO3- = 0.0120 mol Cr(NO3)2 x

)NOCr( mol 1NO mol 2

23

_3

+ 0.0500 mol HNO3 x HNO mol 1NO mol 1

3

_3 = 0.0740 mol NO3

-

mol H+ = 0.0220 mol; mol Cr2O72- = 0.00300 mol; mol Cr3+ = 0.01600 mol

Check for charge neutrality. Total moles of +charge = 0.0100 + 0.0220 + 3 x (0.01600) = 0.0800 mol +charge Total moles of -charge = 0.0740 + 2 x (0.00300) = 0.0800 mol -charge The charges balance and there is electrical neutrality in the solution after the reaction.

K+ molarity = L 0.1200

K mol 0.0100 +

= 0.0833 M

NO3- molarity =

L 0.1200NO mol 0.0740 _

3 = 0.617 M

H+ molarity = L 0.1200

H mol 0.0220 +

= 0.183 M

Cr2O72- molarity =

L 0.1200OCr mol 0.00300 _2

72 = 0.0250 M

Cr3+ molarity = L 0.1200Cr mol 0.0160 +3

= 0.133 M

4.111 (a) (1) I-(aq) → I3

-(aq) 3 I-(aq) → I3

-(aq) 3 I-(aq) → I3

-(aq) + 2 e- (oxidation half reaction)

HNO2(aq) → NO(g) HNO2(aq) → NO(g) + H2O(l) H+(aq) + HNO2(aq) → NO(g) + H2O(l) [e- + H+(aq) + HNO2(aq) → NO(g) + H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 3 I-(aq) + 2 H+(aq) + 2 HNO2(aq) → I3

-(aq) + 2 NO(g) + 2 H2O(l)

(2) S2O32-(aq) → S4O6

2-(aq) 2 S2O3

2-(aq) → S4O62-(aq)

2 S2O32-(aq) → S4O6

2-(aq) + 2 e- (oxidation half reaction)

I3-(aq) → I-(aq)


100

I3-(aq) → 3 I-(aq)

2 e- + I3-(aq) → 3 I-(aq) (reduction half reaction)

Combine the two half reactions. 2 S2O3

2-(aq) + I3-(aq) → S4O6

2-(aq) + 3 I-(aq)

(b) 18.77 mL = 0.018 77 L; NO2-, 46.01 amu

0.1500 L 1

OS mol _232 x 0.018 77 L = 0.002 815 5 mol S2O3

2-

mass NO2- = 0.002 815 5 mol S2O3

2- x xOS mol 2I mol 1

_232

_3 x

I mol 1NO mol 2

_3

_2

NO mol 1NO g 46.01

_2

_2 = 0.1295 g NO2

-

mass % NO2- =

g 2.935

g 0.1295 x 100% = 4.412%

4.112 (a) (1) Cu(s) → Cu2+(aq)

[Cu(s) → Cu2+(aq) + 2 e-] x 3 (oxidation half reaction)

NO3-(aq) → NO(g)

NO3-(aq) → NO(g) + 2 H2O(l)

4 H+(aq) + NO3-(aq) → NO(g) + 2 H2O(l)

[3 e- + 4 H+(aq) + NO3-(aq) → NO(g) + 2 H2O(l)] x 2

(reduction half reaction Combine the two half reactions. 3 Cu(s) + 8 H+(aq) + 2 NO3

-(aq) → 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) (2) Cu2+(aq) + SCN-(aq) → CuSCN(s)

[e- + Cu2+(aq) + SCN-(aq) → CuSCN(s)] x 2 (reduction half reaction)

HSO3-(aq) → HSO4

-(aq) H2O(l) + HSO3

-(aq) → HSO4-(aq)

H2O(l) + HSO3-(aq) → HSO4

-(aq) + 2 H+(aq) H2O(l) + HSO3

-(aq) → HSO4-(aq) + 2 H+(aq) + 2 e-

(oxidation half reaction) Combine the two half reactions. 2 Cu2+(aq) + 2 SCN-(aq) + H2O(l) + HSO3

-(aq) → 2 CuSCN(s) + HSO4

-(aq) + 2 H+(aq)

(3) Cu+(aq) → Cu2+(aq) [Cu+(aq) → Cu2+(aq) + e-] x 10 (oxidation half reaction)

IO3

-(aq) → I2(aq) 2 IO3

-(aq) → I2(aq)


101

2 IO3-(aq) → I2(aq) + 6 H2O(l)

12 H+(aq) + 2 IO3-(aq) → I2(aq) + 6 H2O(l)

10 e- + 12 H+(aq) + 2 IO3-(aq) → I2(aq) + 6 H2O(l)

(reduction half reaction) Combine the two half reactions. 10 Cu+(aq) + 12 H+(aq) + 2 IO3

-(aq) → 10 Cu2+(aq) + I2(aq) + 6 H2O(l)

(4) I2(aq) → I-(aq) I2(aq) → 2 I-(aq) 2 e- + I2(aq) → 2 I-(aq) (reduction half reaction)

S2O3

2-(aq) → S4O62-(aq)

2 S2O32-(aq) → S4O6

2-(aq) 2 S2O3

2-(aq) → S4O62-(aq) + 2 e- (oxidation half reaction)

Combine the two half reactions. I2(aq) + 2 S2O3

2-(aq) → 2 I-(aq) + S4O62-(aq)

(5) 2 ZnNH4PO4 → Zn2P2O7 + H2O + 2 NH3

(b) 10.82 mL = 0.01082 L

mol S2O32- = (0.1220 mol/L)(0.01082 L) = 0.00132 mol S2O3

2-

mol I2 = 0.00132 mol S2O32- x

OS mol 2I mol 1

_232

2 = 6.60 x 10-4 mol I2

mol Cu+ = 6.60 x 10-4 mol I2 x I mol 1Cu mol 10

2

+

= 6.60 x 10-3 mol Cu+ (Cu)

g Cu = (6.60 x 10-3 mol)(63.546 g/mol) = 0.419 g Cu

mass % Cu in brass = 100% x brass g 0.544

Cu g 0.419= 77.1% Cu

(c) Zn2P2O7, 304.72 amu

mass % Zn in Zn2P2O7 = 100% x g 304.72

g 65.39 x 2= 42.92%

mass of Zn in Zn2P2O7 = (0.4292)(0.246 g) = 0.106 g Zn

mass % Zn in brass = 100% x brass g 0.544

Zng 0.106= 19.5% Zn

4.113 (a) BaSO4, 233.38 amu

mol S = 7.19 g BaSO4 x = BaSO mol 1

S mol 1 x

BaSO g 233.38BaSO mol 1

44

4 0.0308 mol S

theoretical mol S = = 0.913

S mol 0.03080.0337 mol S

(b) Assume n = 1:


102

mol Cl in MCl5 = 0.0337 mol S x = S mol 1

Cl mol 5 0.168 mol Cl


Cl g 35.453 5.97 g Cl

This is impossible because the initial mass of MCl5 was only 4.61 g.

Assume n = 2:




Cl g 35.453 2.99 g Cl

mass M = 4.61g - 2.99 g = 1.62 g M

mol M = 0.0337 mol S x = S mol 2

M mol 10.0168 mol

M molar mass = = mol 0.0168

g 1.6296.4 g/mol; M atomic mass = 96.4 amu

96.4 is reasonable and suggests that M is Mo

Assume n = 3:




Cl g 35.453 1.99 g Cl

mass M = 4.61g - 1.99 g = 2.62 g M


M mol 10.0112 mol

M molar mass = = mol 0.0112

g 2.62 234 g/mol; M atomic mass = 234 amu

234 is between Pa and U, which is highly unlikely for a lubricant.

Assume n = 4:




Cl g 35.453 1.49 g Cl

mass M = 4.61g - 1.49 g = 3.12 g M


M mol 1 0.00842 mol

M molar mass = = mol 42 0.008

g 3.12 371 g/mol; M atomic mass = 371 amu

No known elements have a mass as great as 371 amu. (c) M is most likely Mo and the metal sulfide is MoS2. (d) (1) 2 MoCl5(s) + 5 Na2S(s) → 2 MoS2(s) + S(l) + 10 NaCl(s)


103

(2) 2 MoS2(s) + 7 O2(g) → 2 MoO3(s) + 4 SO2(g) (3) SO2(g) + 2 Fe3+(aq) + 2 H2O(l) → 2 Fe2+(aq) + SO4

2-(aq) + 4 H+(aq) (4) SO4

2-(aq) + Ba2+(aq) → BaSO4(s)

103

Assignment 2. Questions from chapter 5 of McMurry and Fay Question numbers are from the fourth edition. Chapter 5. Periodicity and Atomic Structure

5.1 Gamma ray 8

_11

c 3.00 x m/s10 = = 3.56 x m10

νλ

= 8.43 x 1018 sS1 = 8.43 x 1018 Hz

Radar wave 8

_ 2

c 3.00 x m/s10 = = 10.3 x m10

νλ

= 2.91 x 109 sS1 = 2.91 x 109 Hz

5.2 v = 102.5 MHz = 102.5 x 106 Hz = 102.5 x 106 sS1

8

6 _1

c 3.00 x m / s10 = = = 2.93 m102.5 x 10 s

λν

v = 9.55 x 1017 Hz = 9.55 x 1017 sS1 8

_1017 _1

c 3.00 x m / s10 = = = 3.14 x m109.55 x 10 s

λν

5.3 The wave with the shorter wavelength (b) has the higher frequency. The wave with the

larger amplitude (b) represents the more intense beam of light. The wave with the shorter wavelength (b) represents blue light. The wave with the longer wavelength (a) represents red light.

5.4 Balmer series: m = 2; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 22

1 1 1 = R _

72

λ

; 1

= λ

2.519 x 10S3 nmS1; λ = 397.0 nm

5.5 Paschen series: m = 3; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

3 4

λ

; 1

λ = 5.333 x 10S4 nmS1; λ = 1875 nm

5.6 Paschen series: m = 3; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

3

λ ∞

; 1

λ = 1.219 x 10S3 nmS1; λ = 820.4 nm

5.7 λ = 91.2 nm = 91.2 x 10S9 m

8

_9

c 3.00 x m/s10 = = 91.2 x m10

νλ

= 3.29 x 1015 sS1

E = hv = (6.626 x 10S34 J≅s)(3.29 x 1015 sS1) = 2.18 x 10S18 J/photon E = (2.18 x 10S18 J/photon)(6.022 x 1023 photons/mol) = 1.31 x 106 J/mol = 1310 kJ/mol

5.8 IR, λ = 1.55 x 10S6 m

8_34 23

_ 6

c 3.00 x m / s10E = h = (6.626 x J s) (6.022 x / mol)10 101.55 x m10

• λ

Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________

104

E = 7.72 x 104 J/mol = 77.2 kJ/mol

UV, λ = 250 nm = 250 x 10S9 m 8

_34 23_9

c 3.00 x m / s10E = h = (6.626 x J s) (6.022 x / mol)10 10250 x m10

• λ

E = 4.79 x 105 J/mol = 479 kJ/mol

X ray, λ = 5.49 nm = 5.49 x 10S9 m 8

_34 23_9


• λ

E = 2.18 x 107 J/mol = 2.18 x 104 kJ/mol

5.9 _34 _12h 6.626 x kg 10 sm = =

mv (1150 kg)(24.6 m/s)λ = 2.34 x 10S38 m

5.10 (∆x)(∆mv) h

4

≥π

; uncertainty in velocity = (45 m/s)(0.02) = 0.9 m/s

_34 _12_34h 6.626 x kg 10 sm x = = 5 x m10

4 ( mv) 4 (0.120 kg)(0.9 m/s)∆ ≥

π ∆ π

5.11 n l ml Orbital No. of Orbitals

5 0 0 5s 1 1 S1, 0, +1 5p 3 2 S2, S1, 0, +1, +2 5d 5 3 S3, S2, S1, 0, +1, +2, +3 5f 7 4 S4, S3, S2, S1, 0, +1, +2, +3, +4 5g 9

There are 25 possible orbitals in the fifth shell. 5.12 (a) 2p (b) 4f (c) 3d 5.13 (a) 3s orbital: n = 3, l = 0, ml = 0

(b) 2p orbital: n = 2, l = 1, ml = S1, 0, +1 (c) 4d orbital: n = 4, l = 2, ml = S2, S1, 0, +1, +2

5.14 The g orbitals have four nodal planes. 5.15 The figure represents a d orbital, n = 4 and l = 2. 5.16 m = 1, n = 4; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

1

λ ∞

; 1 1

= R 1 λ

= 1.097 x 10S2 nmS1; λ = 91.2 nm


105

E = 8

_34 23_9

3.00 x m/s10(6.626 x J s) (6.022 x / mol)10 1091.2 x m10

•

E = 1.31 x 106 J/mol = 1.31 x 103 kJ/mol 5.17 (a) Ti, 1s2 2s2 2p6 3s2 3p6 4s2 3d2 or [Ar] 4s2 3d2

[Ar] __ __ __ 4s 3d

(b) Zn, 1s2 2s2 2p6 3s2 3p6 4s2 3d10 or [Ar] 4s2 3d10 [Ar]

4s 3d

(c) Sn, 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 or [Kr] 5s2 4d10 5p2 [Kr]

5s 4d 5p

(d) Pb, [Xe] 6s2 4f 14 5d10 6p2 5.18 For Na+, 1s2 2s2 2p6; for ClS, 1s2 2s2 2p6 3s2 3p6 5.19 The ground-state electron configuration contains 28 electrons. The atom is Ni. 5.20 Cr, Cu, Nb, Mo, Ru, Rh, Pd, Ag, La, Ce, Gd, Pt, Au, Ac, Th, Pa, U, Np, Cm 5.21 (a) Ba; atoms get larger as you go down a group.

(b) W; atoms get smaller as you go across a period. (c) Sn; atoms get larger as you go down a group. (d) Ce; atoms get smaller as you go across a period.

5.22 The aurora borealis begins on the surface of the sun with a massive solar flare. These

flares eject a solar "gas" of energetic protons and electrons that reach earth after about 2 days and are then attracted toward the north and south magnetic poles. The energetic electrons are deflected by the earth's magnetic field into a series of sheetlike beams. The electrons then collide with O2 and N2 molecules in the upper atmosphere, exciting them, ionizing them, and breaking them apart into O and N atoms. The energetically excited atoms, ions, and molecules generated by collisions with electrons emit energy of characteristic wavelengths when they decay to their ground states. The O2

+ ions emit a red light around 630 nm; N2

+ ions emit violet and blue light at 391.4 nm and 470.0 nm; and O atoms emit a greenish-yellow light at 557.7 nm and a deep red light at 630.0 nm.



106

5.23

5.24 5.25 The wave with the larger amplitude (a) has the greater intensity. The wave with the

shorter wavelength (a) has the higher energy radiation. The wave with the shorter wavelength (a) represents yellow light. The wave with the longer wavelength (b) represents infrared radiation.

5.26 [Ar] 4s2 3d10 4p1 is Ga. 5.27 There are 34 total electrons in the atom, so there are also 34 protons in the nucleus. The

atom is selenium (Se) Se, [Ar]

4s 3d 4p

5.28 Ca and Br are in the same period, with Br to the far right of Ca. Ca is larger than Br. Sr is directly below Ca in the same group, and is larger than Ca. The result is Sr (215 pm) > Ca (197 pm) > Br (114 pm)

5.29 (a) 3py n = 3, l = 1 (b) 2z4d n = 4, l = 2

Additional Problems Electromagnetic Radiation


107

5.30 Violet has the higher frequency and energy. Red has the higher wavelength. 5.31 Ultraviolet light has the higher frequency and the higher energy. Infrared light has the

higher wavelength.

5.32 8

15 _1

c 3.00 x m/s10 = = 5.5 x 10 s

λν

= 5.5 x 10S8 m

5.33 8

10 _1 10_3

c 3.00 x m/s10 = = = 6.93 x = 6.93 x Hz10 s 104.33 x m10

λν

5.34 (a) v = 99.5 MHz = 99.5 x 106 sS1

E = hv = (6.626 x 10S34 J≅s)(99.5 x 106 sS1)(6.022 x 1023 /mol) E = 3.97 Η 10S2 J/mol = 3.97 Η 10S5 kJ/mol v = 1150 kHz = 1150 x 103 sS1 E = hv = (6.626 x 10S34 J≅s)(1150 x 103 sS1)(6.022 x 1023 /mol) E = 4.589 x 10S4 J/mol = 4.589 x 10S7 kJ/mol The FM radio wave (99.5 MHz) has the higher energy. (b) λ = 3.44 x 10S9 m

8_34 23

_9


• λ

E = 3.48 x 107 J/mol = 3.48 x 104 kJ/mol λ = 6.71 Η 10S2 m

8_34 23

_ 2


• λ

E = 1.78 J/mol = 1.78 x 10S3 kJ/mol The X ray (λ = 3.44 x 10S9 m) has the higher energy.

5.35 v = 400 MHz = 400 x 106 sS1

E = (6.626 x 10S34 J≅s)(400 x 106 sS1)(6.02 x 1023/mol) = 0.160 J/mol = 1.60 x 10S4 kJ/mol

5.36 (a) E = 90.5 kJ/mol x 23

1000 J 1 mol x

1 kJ 6.02 x 10 = 1.50 x 10S19 J

_19

_34

E 1.50 x J10 = = h 6.626 x J s10

ν•

= 2.27 x 1014 sS1

8

14 _1

c 3.00 x m/s10 = = 2.27 x 10 s

λν

= 1.32 x 10S6 m = 1320 x 10S9 m = 1320 nm, near IR

(b) E = 8.05 x 10S4 kJ/mol x 23

1000 J 1 mol x

1 kJ 6.02 x 10 = 1.34 x 10S24 J


108

_ 24

_34

E 1.34 x J10 = = h 6.626 x J s10

ν•

= 2.02 x 109 sS1

8

9 _1

c 3.00 x m/s10 = = 2.02 x 10 s

λν

= 0.149 m, radio wave

(c) E = 1.83 x 103 kJ/mol x 23

1000 J 1 mol x

1 kJ 6.02 x 10 = 3.04 x 10S18 J

_18

_34

E 3.04 x J10 = = h 6.626 x J s10

ν•

= 4.59 x 1015 sS1

8

15 _1

c 3.00 x m/s10 = = 4.59 x 10 s

λν

= 6.54 x 10S8 m = 65.4 x 10S9 m = 65.4 nm, UV

5.37 (a) _34 19 _1 231 kJE = h = (6.626 x J s)(5.97 x ) (6.022 x / mol)10 10 s 10

1000 J ν •

E = 2.38 x 107 kJ/mol

(b) _34 6 _1 231 kJE = h = (6.626 x Jcdot s)(1.26 x ) (6.022 x / mol)10 10 s 10

1000 J ν

E = 5.03 x 10S7 kJ/mol

(c) 8

_34 232

3.00 x m / s 1 kJ10E = h = (6.626 x J s) (6.022 x / mol)10 102.57 x m 1000 J10

ν •

E = 4.66 x 10S7 kJ/mol

5.38 _34 _12

_31 8

h 6.626 x kg 10 sm = = mv (9.11 x kg)(0.99 x 3.00 x m/s)10 10

λ = 2.45 x 10S12 m, γ ray

5.39 _34 _12

_ 27 8

h 6.626 x kg 10 sm = = mv (1.673 x kg)(0.25 x 3.00 x m/s)10 10

λ = 5.28 x 10S15 m, γ ray

5.40 156 km/h = 156 x 103 m/3600 s = 43.3 m/s; 145 g = 0.145 kg

_34 _12_34h 6.626 x kg 10 sm = = = 1.06 x m10

mv (0.145 kg)(43.3 m/s)λ

5.41 1.55 mg = 1.55 x 10S3 g = 1.55 x 10S6 kg

_34 _12

_6

h 6.626 x kg 10 sm = = mv (1.55 x kg)(1.38 m/s)10

λ = 3.10 x 10S28 m

5.42 145 g = 0.145 kg; 0.500 nm = 0.500 x 10S9 m


109

_34 _12_ 24

_9

h 6.626 x kg 10 smv = = = 9.14 x m /s10m (0.145 kg)(0.500 x m)10λ

5.43 750 nm = 750 x 10S9 m

_34 _12

_31 _9

h 6.626 x kg 10 smv = = m (9.11 x kg) (750 x m)10 10λ

= 970 m/s

Atomic Spectra 5.44 For n = 3; λ = 656.3 nm = 656.3 x 10S9 m

8_34 23

_9

c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10656.3 x m 1000 J10

E = 182.3 kJ / mol

λ

For n = 4; λ = 486.1 nm = 486.1 x 10S9 m

8_34 23

_9


E = 246.1 kJ / mol

λ

For n = 5; λ = 434.0 nm = 434.0 x 10S9 m

8_34 23

_9


E = 275.6 kJ / mol

λ

5.45 m = 2, n = 4; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

2

λ ∞

; 1 1

= R 4 λ

= 2.74 x 10S3 nmS1; λ = 364.6 nm

5.46 From problem 5.45, for n = 4, λ = 364.6 nm = 364.6 x 10S9 m

8_34 23

_9


E = 328.1 kJ / mol

λ

5.47 Brackett series: m = 4, n = 5; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 22

1 1 1 = R _

54

λ

= 2.468 x 10S4 nmS1; λ = 4051 nm

8_34 23

_9

c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 104051 x m 1000 J10

E = 29.55 kJ / mol, IR

λ

Brackett series: m = 4, n = 6; R = 1.097 x 10S2 nmS1


110

2 2

1 1 1 = R _

m n

λ

; 22

1 1 1 = R _

64

λ

= 3.809 x 10S4 nmS1; λ = 2625 nm

8_34 23

_9


E = 45.60 kJ / mol, IR

λ

5.48 λ = 330 nm = 330 x 10S9 m

8_34 23

_9


E = 363 kJ / mol

λ

5.49 795 nm = 795 x 10S9 m

8_34 23

_9


E = 151 kJ / mol

λ

Orbitals and Quantum Numbers 5.50 n is the principal quantum number. The size and energy level of an orbital depends on n.

l is the angular-momentum quantum number. l defines the three-dimensional shape of an orbital. ml is the magnetic quantum number. ml defines the spatial orientation of an orbital. ms is the spin quantum number. ms indicates the spin of the electron and can have either of two values, +2 or S2.

5.51 The Heisenberg uncertainty principle states that one can never know both the position and

the velocity of an electron beyond a certain level of precision. This means we cannot think of electrons circling the nucleus in specific orbital paths, but we can think of electrons as being found in certain three-dimensional regions of space around the nucleus, called orbitals.

5.52 The probability of finding the electron drops off rapidly as distance from the nucleus

increases, although it never drops to zero, even at large distances. As a result, there is no definite boundary or size for an orbital. However, we usually imagine the boundary surface of an orbital enclosing the volume where an electron spends 95% of its time.

5.53 A 4s orbital has three nodal surfaces.


111

4s orbital 5.54 Part of the electron-nucleus attraction is canceled by the electron-electron repulsion, an

effect we describe by saying that the electrons are shielded from the nucleus by the other electrons. The net nuclear charge actually felt by an electron is called the effective nuclear charge, Zeff, and is often substantially lower than the actual nuclear charge, Zactual.

Zeff = Zactual S electron shielding 5.55 Electron shielding gives rise to energy differences among 3s, 3p, and 3d orbitals in

multielectron atoms because of the differences in orbital shape. For example, the 3s orbital is spherical and has a large probability density near the nucleus, while the 3p orbital is dumbbell shaped with a node at the nucleus. An electron in a 3s orbital can penetrate closer to the nucleus than an electron in a 3p orbital can and feels less of a shielding effect from other electrons. Generally, for any given value of the principal quantum number n, a lower value of l corresponds to a higher value of Zeff and to a lower energy for the orbital.

5.56 (a) 4s n = 4; l = 0; ml = 0; ms = ∀2

(b) 3p n = 3; l = 1; ml = S1, 0, +1; ms = ∀2 (c) 5f n = 5; l = 3; ml = S3, S2, S1, 0, +1, +2, +3; ms = ∀2 (d) 5d n = 5; l = 2; ml = S2, S1, 0, +1, +2; ms = ∀2

5.57 (a) 3s (b) 2p (c) 4f (d) 4d 5.58 (a) is not allowed because for l = 0, ml = 0 only.

(b) is allowed. (c) is not allowed because for n = 4, l = 0, 1, 2, or 3 only.

5.59 Co 1s2 2s2 2p6 3s2 3p6 4s2 3d7

(a) is not allowed because for l = 0, ml = 0 only. (b) is not allowed because n = 4 and l = 2 is for a 4d orbital. (c) is allowed because n = 3 and l = 1 is for a 3p orbital.

5.60 For n = 5, the maximum number of electrons will occur when the 5g orbital is filled:

[Rn] 7s2 5f 14 6d10 7p6 8s2 5g18 = 138 electrons 5.61 n = 4, l = 0 is a 4s orbital. The electron configuration is 1s2 2s2 2p6 3s2 3p6 4s2. The

number of electrons is 20. 5.62 0.68 g = 0.68 x 10S3 kg

h

( x)( mv) 4

∆ ∆ ≥π

; _34 _12

_31_3

h 6.626 x kg 10 sm x = = 8 x m104 ( mv) 4 (0.68 x kg)(0.1 m / s)10

∆ ≥π ∆ π


112

5.63 4.0026 amu x _ 271.660 540 x kg10

1 amu = 6.6465 x 10S27 kg;

h( x)( mv)

4 ∆ ∆ ≥

π

_34 _12

_ 27 3

h 6.626 x kg 10 sm x = 4 ( mv) 4 (6.6465 x kg)(0.01 x 1.36 x m / s)10 10

∆ ≥π ∆ π

= 5.833 x 10S10 m

Electron Configurations 5.64 The number of elements in successive periods of the periodic table increases by the

progression 2, 8, 18, 32 because the principal quantum number n increases by 1 from one period to the next. As the principal quantum number increases, the number of orbitals in a shell increases. The progression of elements parallels the number of electrons in a particular shell.

5.65 The n and l quantum numbers determine the energy level of an orbital in a multielectron

atom. 5.66 (a) 5d (b) 4s (c) 6s 5.67 (a) 2p < 3p < 5s < 4d (b) 2s < 4s < 3d < 4p (c) 3d < 4p < 5p < 6s 5.68 (a) 3d after 4s (b) 4p after 3d (c) 6d after 5f (d) 6s after 5p 5.69 (a) 3s before 3p (b) 3d before 4p (c) 6s before 4f (d) 4f before 5d 5.70 (a) Ti, Z = 22 1s2 2s2 2p6 3s2 3p6 4s2 3d2

(b) Ru, Z = 44 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d6 (c) Sn, Z = 50 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 (d) Sr, Z = 38 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 (e) Se, Z = 34 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

5.71 (a) Z = 55, Cs [Kr] 5s2 4d10 5p6 6s1 (b) Z = 40, Zr [Kr] 5s2 4d2

(c) Z = 80, Hg [Xe] 6s2 4f 14 5d10 (d) Z = 62, Sm [Xe] 6s2 4f 6 5.72 (a) Rb, Z = 37 [Kr]

5s (b) W, Z = 74 [Xe]

6s 4f 5d

(c) Ge, Z = 32 [Ar] 4s 3d 4p

(d) Zr, Z = 40 [Kr] 5s 4d

5.73 (a) Z = 25, Mn [Ar] 4s 3d

(b) Z = 56, Ba [Xe]


113

6s

(c) Z = 28, Ni [Ar] 4s 3d

(d) Z = 47, Ag [Kr] 5s 4d 5.74 4s > 4d > 4f 5.75 K < Ca < Se < Kr 5.76 Z = 116 [Rn] 7s2 5f 14 6d10 7p4 5.77 Z = 119 [Rn] 7s2 5f 14 6d10 7p6 8s1 5.78 (a) O 1s2 2s2 2p4 2 unpaired eS

2p (b) Si 1s2 2s2 2p6 3s2 3p2 2 unpaired eS

3p (c) K [Ar] 4s1 1 unpaired eS

(d) As [Ar] 4s2 3d10 4p3 3 unpaired eS

4p 5.79 (a) Z = 31, Ga (b) Z = 46, Pd 5.80 Order of orbital filling: 1s2s2p3s3p4s3d4p5s4d5p6s4f5d6p7s5f6d7p8s5g

Z = 121 5.81 A g orbital would begin filling at atomic number = 121 (see 5.80). There are nine g

orbitals that can each hold two electrons. The first element to have a filled g orbital would be atomic number = 138.

Atomic Radii and Periodic Properties 5.82 Atomic radii increase down a group because the electron shells are farther away from the

nucleus. 5.83 Across a period, the effective nuclear charge increases, causing a decrease in atomic radii. 5.84 F < O < S 5.85 (a) K, lower in group 1A (b) Ta, lower in group 5B

(c) V, farther to the left in same period


114

(d) Ba, four periods lower and only one group to the right 5.86 Mg has a higher ionization energy than Na because Mg has a higher Zeff and a smaller

size. 5.87 F has a higher electron affinity than C because of a higher effective nuclear charge and

room in the valence shell for the additional electron. In addition, FS achieves a noble gas electron configuration.

General Problems 5.88 Balmer series: m = 2; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 22

1 1 1 = R _

62

λ

= 2.438 x 10S3 nmS1

λ = 410.2 nm = 410.2 x 10S9 m 8

_34 23_9


E = 291.6 kJ / mol

λ

5.89 Pfund series: m = 5; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

n = 6, 2 2

1 1 1 = R _

5 6

λ

= 1.341 x 10S4 nmS1; λ = 7458 nm = 7458 x 10S9 m

8_34 23

_9


E = 16.04 kJ / mol

λ

n = 7, 2 2

1 1 1 = R _

5 7

λ

= 2.149 x 10S4 nmS1; λ = 4653 nm = 4653 x 10S9 m

8_34 23

_9


E = 25.71 kJ / mol

λ

These lines in the Pfund series are in the infrared region of the electromagnetic spectrum. 5.90 Pfund series: m = 5, n = 4; R = 1.097 x 10S2 nmS1

2 2

1 1 1 1 = R _ = R

255

λ ∞

= 4.388 x 10S4 nmS1; λ = 2279 nm

5.91 (a) 1923

kJ 1000 J 1 molE = 142 = 2.36 x J10

mol 1 kJ 6.02 x 10•


115

cE = h

λ,

34 8

19

hc (6.626 x J s)(3.00 x m / s)10 10 = = E 2.36 x J10

•

•

•λ

λ = 8.42 x 10S7 m (infrared)

(b) 2 2323

kJ 1000 J 1 molE = 4.55 x = 7.56 x J10 10

mol 1 kJ 6.02 x 10• •

cE = h

λ,

34 8

23

hc (6.626 x J s)(3.00 x m / s)10 10 = = E 7.56 x J10

•

•

•λ

λ = 2.63 x 10S3 m (microwave)

(c) 4 1723

kJ 1000 J 1 molE = 4.81 x = 7.99 x J10 10

mol 1 kJ 6.02 x 10•

cE = h

λ,

34 8

17

hc (6.626 x J s)(3.00 x m / s)10 10 = = E 7.99 x J10

•

•

•λ

λ = 2.49 x 10S9 m (X ray)

5.92 (a) E = hv = (6.626 x 10S34 J≅s)(3.79 x 1011 sS1)1 kJ

1000 J

(6.022 x 1023/mol) = 0.151 kJ/mol

(b) E = hv = (6.626 x 10S34 J≅s)(5.45 x 104 sS1)1 kJ

1000 J

(6.022 x 1023/mol) = 2.17 x 10S8

kJ/mol

(c) E = hv = (6.626 x 10S34 J≅s)8

_5

3.00 x m/s 1 kJ104.11 x m 1000 J10

(6.022 x 1023/mol) = 2.91 kJ/mol

5.93 v = 9,192,631,770 sS1 = 9.19263 x 109 sS1

E = hv = (6.626 x 10S34 J≅s)(9.19263 x 109 sS1)1 kJ

1000 J

(6.022 x 1023/mol) = 3.668 x 10S3

kJ/mol 5.94 (a) Ra [Rn] 7s2 [Rn] 7s

(b) Sc [Ar] 4s2 3d1 [Ar] 4s 3d

(c) Lr [Rn] 7s2 5f14 6d1 [Rn]

7s 5f 6d (d) B [He] 2s2 2p1 [He]

2s 2p (e) Te [Kr] 5s2 4d10 5p4 [Kr]

5s 4d 5p


116

5.95 (a) row 1 n = 1, l = 0 1s 2 elements

l = 1 1p 6 elements l = 2 1d 10 elements

row 2 n = 2, l = 0 2s 2 elements

l = 1 2p 6 elements l = 2 2d 10 elements l = 3 2f 14 elements

There would be 50 elements in the first two rows. (b) There would be 18 elements in the first row [see (a) above]. The fifth element in the second row would have atomic number = 23. (c) Z = 12

1s 1p 1d

5.96 206.5 kJ = 206.5 x 103 J; E = 3

23

206.5 x J 1 mol10 x 1 mol 6.022 x 10

= 3.429 x 10S19 J

cE = h

λ,

_34 8

_19

hc (6.626 x J s)(3.00 x m / s)10 10 = = E 3.429 x J10

•λ = 5.797 x 10S7 m = 580. nm

5.97 780 nm is at the red end of the visible region of the electromagnetic spectrum.

780 nm = 780 x 10S9 m 8

_34 23_9

c 3.00 x m / s 1 kJ10E = h = (6.626 x J s) (6.022 x / mol) = 153 kJ / mol10 10780 x m 1000 J10

• λ

5.98 (a) Sr, Z = 38 [Kr]

5s (b) Cd, Z = 48 [Kr]

5s 4d (c) Z = 22, Ti [Ar]

4s 3d (d) Z = 34, Se [Ar]

4s 3d 4p 5.99 La ([Xe] 6s2 5d1) is directly below Y ([Kr] 5s2 4d1) in the periodic table. Both have

similar valence electron configurations, but for La the valence electrons are one shell farther out leading to its larger radius. Although Hf ([Xe] 6s2 4f 14 5d2) is directly below Zr ([Kr] 5s2 4d2) in the periodic table, Zr and Hf have almost identical atomic radii because the 4f electrons in Hf are not effective in shielding the valence electrons. The valence electrons in Hf are drawn in closer to the nucleus by the higher Zeff.

5.100 For K, Zeff = 2(418.8 kJ/mol)( )4

1312 kJ/mol = 2.26; For Kr, Zeff =

2(1350.7 kJ/mol)( )41312 kJ/mol

= 4.06


117

5.101 75 W = 75 J/s; 550 nm = 550 x 10S9 m; (0.05)(75 J/s) = 3.75 J/s

8_34 19

_9

c 3.00 x m / s10E = h = (6.626 x J s) = 3.61 x J / photon10 10550 x m10

• • λ

number of photons = 1919

3.75 J / s = 1.0 x photons / s10

3.61 x J / photon10•

5.102 q = (350 g)(4.184 J/g≅oC)(95oC S 20oC) = 109,830 J

λ = 15.0 cm = 15.0 x 10S2 m

E = (6.626 x 10S34 J≅s)8

_ 2

3.00 x m/s1015.0 x m10

= 1.33 x 10S24 J/photon

number of photons = _ 24

109,830 J

1.33 x J/photon10 = 8.3 x 1028 photons

5.103 1923

kJ 1000 J 1 molE = 310 = 5.15 x J10

mol 1 kJ 6.022 x 10•

cE = h

λ,

_34 8

_19

hc (6.626 x J s)(3.00 x m / s)10 10 = = E 5.15 x J10

•λ = 3.86 x 10S7 m = 386 nm

5.104 48.2 nm = 48.2 x 10S9 m

8 23_34 3

_9

3.00 x m / s 1 kJ 6.022 x 10 10E(photon) = 6.626 x J s x x x = 2.48 x kJ/mol10 1048.2 x m 1000 J mol10

•

23

2_31 6K

3K

1 kJ 6.022 x 10 = E(electron) = ‰ (9.109 x kg) (2.371 x m / s)10E 101000 J mol

= 1.54 x kJ/mol10E

E(photon) = Ei + EK; Ei = E(photon) B EK = (2.48 x 103) S (1.54 x 103) = 940 kJ/mol 5.105 Charge on electron = 1.602 x 10S19 C; 1 V ≅C = 1 J = 1 kg m2/s2

(a) EK = (30,000 V)(1.602 x 10S19 C) = 4.806 x 10S15 J

EK = 2mv2; v = _15 22

K

_31

2 2 x 4.806 x kg /10 sE m = m 9.109 x kg10

= 1.03 x 108 m/s

_34 2

_31 8

h 6.626 x kg /s10 m = = mv (9.109 x kg)(1.03 x m/s)10 10

λ = 7.06 x 10S12 m

(b) 8

_34 15_10

c 3.00 x m / s10E = h = (6.626 x J s) = 1.29 x J / photon10 101.54 x m10

• • λ


118

5.106 Substitute the equation for the orbit radius, r, into the equation for the energy level, E, to

get 2 22

22oo

_ _Ze eZE = = 2 a nan2

Z

Let E1 be the energy of an electron in a lower orbit and E2 the energy of an electron in a higher orbit. The difference between the two energy levels is

∆E = E2 S E1 = 2 22 2

2 22 1o o

_ _e eZ Z _ 2 2 a an n

= 2 22 2

2 22 1o o

_ e eZ Z + 2 2 a an n

= 2 22 2

2 21 2o o

e eZ Z _ 2 2 a an n

∆E = 22

2 21 2o

1 1eZ _ 2a n n

Because Z, e, and ao are constants, this equation shows that ∆E is proportional to

2 21 2

1 1 _

n n

where n1 and n2 are integers with n2 > n1. This is similar to the Balmer-

Rydberg equation where 1/λ or v for the emission spectra of atoms is proportional to

2 2

1 1 _

m n

where m and n are integers with n > m.

5.107 (a) 0 0 0 0 0 0 0 0 0 ___

1s 2s 2p 3s 3p 4s Two partially filled orbitals. (b) The element in the 3rd column and 4th row under these new rules would have an atomic number of 30 and be in the s-block.

5.108 (a) 3d, n = 3, l = 2

(b) 2p, n = 2, l = 1, ml = S1, 0, +1 3p, n = 3, l = 1, ml = S1, 0, +1 3d, n = 3, l = 2, ml = S2, S1, 0, +1, +2 (c) N, 1s2 2s2 2p3 so the 3s, 3p, and 3d orbitals are empty. (d) C, 1s2 2s2 2p2 so the 1s and 2s orbitals are filled. (e) Be, 1s2 2s2 so the 2s orbital contains the outermost electrons. (f) 2p and 3p ( ) and 3d ( ).

5.109 λ = 1.03 x 10S7 m = 103 x 10S9 m = 103 nm

2 2

1 1 1 = R _

m n

λ

, R = 1.097 x 10S2 nmS1

_ 2 _12 2

1 1 1 = (1.097 x ) _ 10 nm

103 nm 1 n

, solve for n.

_ 2 _1 2

(1/103 nm) 1 _ 1 = _

(1.097 x )10 nm n


119

2

1 = 0.115

n; 2 1

= n0.115

; 1

n = =0.115

2.95

The electron jumps to the third shell.

5.110 (a) E = hv; v = _19

_34

E 7.21 x J10 = =h 6.626 x J s10 •

1.09 x 1015 sS1

(b) E(photon) = Ei + EK; from (a), Ei = 7.21 x 10S19 J

E(photon) = 8

_34 19_ 7

c 3.00 x m / s10h = (6.626 x J s) = 7.95 x J10 102.50 x m10

• • λ

EK = E(photon) S Ei = (7.95 x 10S19 J) S (7.21 x 10S19 J) = 7.4 x 10S20 J Calculate the electron velocity from the kinetic energy, EK. EK = 7.4 x 10S20 J = 7.4 x 10S20 kgΑm2/s2 = 2mv2 = 2(9.109 x 10S31 kg)v2

v = _ 20 22

_31

2 x (7.4 x kg / )10 sm9.109 x kg10

• = 4.0 x 105 m/s

deBroglie wavelength = _34 2

_31 5

h 6.626 x kg /s10 m = m v (9.109 x kg)(4.0 x m/s)10 10

• = 1.8 x 10S9 m = 1.8

nm

Multi-Concept Problem

5.111 (a) E = hv; v = _16

_34

E 4.70 x J10 = =h 6.626 x J s10 •

7.09 x 1017 sS1

(b) λ = 8

17 _1

c 3.00 x m/s10 = =7.09 x 10 sν

4.23 x 10S10 m = 0.423 x 10S9 m = 0.423 nm

(c) λ = h

mv; v =

_34 2

_31 _10

h 6.626 x kg /s10 m = =m (9.11 x kg)(4.23 x m)10 10

•λ

1.72 x 106 m/s

(d) KE = 22 _31 6(9.11 x kg)(1.72 x m/s)mv 10 10 = =

2 2 1.35 x 10S18 kg Α m2/s2 = 1.35 x 10S18 J

5.112 (a) 5f subshell: n = 5, l = 3, ml = S3, S2, S1, 0, +1, +2, +3

3d subshell: n = 3, l = 2, ml = S2, S1, 0, +1, +2 (b) In the H atom the subshells in a particular energy level are all degenerate, i.e., all have the same energy. Therefore, you only need to consider the principal quantum number, n, to calculate the wavelength emitted for an electron that drops from the 5f to the 3d subshell. m = 3, n = 5; R = 1.097 x 10S2 nmS1

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

3 5

λ

; 1

= λ

7.801 x 10S4 nmS1; λ = 1282 nm

(c) m = 3, n = 4; R = 1.097 x 10S2 nmS1


120

2 2

1 1 1 = R _

m n

λ

; 2 2

1 1 1 = R _

3

λ ∞

; 2

1 1 = R

3

λ

= 1.219 x 10S3 nmS1; λ = 820.3 nm

E = 8

_34 23_9

3.00 x m/s10(6.626 x J s) (6.022 x / mol)10 10820.3 x m10

•

= 1.46 x 105 J/mol = 146

kJ/mol 5.113 (a) [Kr] 5s2 4d10 5p6 (b) [Kr] 5s2 4d10 5p5 6s1

(c) Both Xe* and Cs have a single electron in the 6s orbital with similar effective nuclear charges. Therefore the 6s electrons in both cases are held with similar strengths and require almost the same energy to remove.

5.114 (a) Cl2, 70.91 amu

M + Cl2 MCl2

mol Cl2 = 0.8092 g Cl2 x 2

2

1 mol Cl70.91 g Cl

= 0.01141 mol Cl2

mol M = 0.01141 mol Cl2 x 2

1 mol M

1 mol Cl = 0.01141 mol M

molar mass of M = 1.000 g

0.01141 mol = 87.64 g/mol

atomic mass of M = 87.64 amu; M = Sr

(b) q = 9.46 kJ

0.01141 mol = 829 kJ/mol

5.115 (a) H3MO3(aq) H3MO4(aq)

H3MO3(aq) + H2O(l) H3MO4(aq) H3MO3(aq) + H2O(l) H3MO4(aq) + 2 H+(aq) [H3MO3(aq) + H2O(l) H3MO4(aq) + 2 H+(aq) + 2 eS] x 5 (oxidation half reaction)

MnO4

S(aq) Mn2+(aq) MnO4

S(aq) Mn2+(aq) + 4 H2O(l) MnO4

S(aq) + 8 H+(aq) Mn2+(aq) + 4 H2O(l) [MnO4

S(aq) + 8 H+(aq) + 5 eS Mn2+(aq) + 4 H2O(l)] x 2 (reduction half reaction)

Combine the two half reactions. 5 H3MO3(aq) + 5 H2O(l) + 2 MnO4

S(aq) + 16 H+(aq) 5 H3MO4(aq) + 10 H+(aq) + 2 Mn2+(aq) + 8 H2O(l)

5 H3MO3(aq) + 2 MnO4S(aq) + 6 H+(aq) 5 H3MO4(aq) + 2 Mn2+(aq) + 3 H2O(l)

(b) 10.7 mL = 0.0107 L mol MnO4

S = (0.0107 L)(0.100 mol/L) = 1.07 x 10S3 mol MnO4S


121

mol H3MO3 = 1.07 x 10S3 mol MnO4S x 3 3

_4

5 mol MOH =2 mol MnO

2.67 x 10S3 mol H3MO3

mol M2O3 = 2.67 x 10S3 mol H3MO3 x 2 3

3 3

1 mol OM =2 mol MOH

1.34 x 10S3 mol M2O3

(c) mol M in M2O3 = 1.34 x 10S3 mol M2O3 x 2 3

2 mol M =

1 mol OM 2.68 x 10S3 mol M

M molar mass = _3

0.200 g =

2.68 x mol10 74.6 g/mol; M atomic mass = 74.6 amu

M is As.

(d) E = hv = (6.626 x 10S34 JΑs )(9.07 x 1014 sS1) = 6.01 x 10S19 J/photon E = (6.01 x 10S19 J/photon)(6.022 x 1023 photons/mol)(1 kJ/1000 J) = 362 kJ/mol

123

6

Ionic Bonds and Some Main-Group Chemistry

6.1 (a) Ra2+ [Rn] (b) La3+ [Xe] (c) Ti4+ [Ar] (d) N3- [Ne]

Each ion has the ground-state electron configuration of the noble gas closest to it in the periodic table.

6.2 The neutral atom contains 30 e- and is Zn. The ion is Zn2+. 6.3 (a) O2-; decrease in effective nuclear charge and an increase in electron-electron

repulsions lead to the larger anion. (b) S; atoms get larger as you go down a group. (c) Fe; in Fe3+ electrons are removed from a larger valence shell and there is an increase in effective nuclear charge leading to the smaller cation. (d) H-; decrease in effective nuclear charge and an increase in electron-electron repulsions lead to the larger anion.

6.4 K+ is smaller than neutral K because the ion has one less electron. K+ and Cl- are

isoelectronic, but K+ is smaller than Cl- because of its higher effective nuclear charge. K is larger than Cl- because K has one additional electron and that electron begins the next shell (period). K+, r = 133 pm; Cl-, r = 184 pm; K, r = 227 pm

6.5 (a) Br (b) S (c) Se (d) Ne 6.6 (a) Be 1s2 2s2 N 1s2 2s2 2p3

Be would have the larger third ionization energy because this electron would come from the 1s orbital. (b) Ga [Ar] 4s2 3d10 4p1 Ge [Ar] 4s2 3d10 4p2 Ga would have the larger fourth ionization energy because this electron would come from the 3d orbitals.

6.7 (b) Cl has the highest Ei1 and smallest Ei4. 6.8 Ca (red) would have the largest third ionization energy of the three because the electron

being removed is from a filled valence shell. For Al (green) and Kr (blue), the electron being removed is from a partially filled valence shell. The third ionization energy for Kr would be larger than that for Al because the electron being removed from Kr is coming out of a set of filled 4p orbitals while the electron being removed from Al is coming out of a half-filled 3s orbital. In addition, Zeff is larger for Kr than for Al. The ease of losing its third electron is Al < Kr < Ca.

Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________

124

6.9 Cr [Ar] 4s1 3d5 Mn [Ar] 4s2 3d5 Fe [Ar] 4s2 3d6 Cr can accept an electron into a 4s orbital. The 4s orbital is lower in energy than a 3d orbital. Both Mn and Fe accept the added electron into a 3d orbital that contains an electron, but Mn has a lower value of Zeff. Therefore, Mn has a less negative Eea than either Cr or Fe.

6.10 The least favorable Eea is for Kr (red) because it is a noble gas with filled set of 4p

orbitals. The most favorable Eea is for Ge (blue) because the 4p orbitals would become half filled. In addition, Zeff is larger for Ge than it is for K (green).

6.11 (a) KCl has the higher lattice energy because of the smaller K+.

(b) CaF2 has the higher lattice energy because of the smaller Ca2+. (c) CaO has the higher lattice energy because of the higher charge on both the cation and anion.

6.12 K(s) → K(g) +89.2 kJ/mol

K(g) → K+(g) + e- +418.8 kJ/mol ½ [F2(g) → 2 F(g)] +79 kJ/mol F(g) + e- → F-(g) -328 kJ/mol K+(g) + F-(g) → KF(s) -821 kJ/mol

Sum = -562 kJ/mol for K(s) + ½ F2(g) → KF(s) 6.13 The anions are larger than the cations. Cl- is larger than O2- because it is below it in the

periodic table. Therefore, (a) is NaCl and (b) is MgO. Because of the higher ion charge and shorter cation – anion distance, MgO has the larger lattice energy.

6.14 (a) Li2O, O -2 (b) K2O2, O -1 (c) CsO2, O -½ 6.15 (a) 2 Cs(s) + 2 H2O(l) → 2 Cs+(aq) + 2 OH-(aq) + H2(g)

(b) Na(s) + N2(g) → N. R. (c) Rb(s) + O2(g) → RbO2(s) (d) 2 K(s) + 2 NH3(g) → 2 KNH2(s) + H2(g) (e) 2 Rb(s) + H2(g) → 2 RbH(s)

6.16 (a) Be(s) + Br2(l) → BeBr2(s)

(b) Sr(s) + 2 H2O(l) → Sr(OH)2(aq) + H2(g) (c) 2 Mg(s) + O2(g) → 2 MgO(s)

6.17 BeCl2(s) + 2 K(s) → Be(s) + 2 KCl(s) 6.18 Mg(s) + S(s) → MgS(s); In MgS, the oxidation number of S is -2. 6.19 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g)

H+ gains electrons and is the oxidizing agent. Al loses electrons and is the reducing agent.


125

6.20 2 Al(s) + 3 S(s) → Al2S3(s) 6.21 (a) Br2(l) + Cl2(g) → 2 BrCl(g)

(b) 2 Al(s) + 3 F2(g) → 2 AlF3(s) (c) H2(g) + I2(s) → 2 HI(g)

6.22 Br2(l) + 2 NaI(s) → 2 NaBr(s) + I2(s)

Br2 gains electrons and is the oxidizing agent. I- (from NaI) loses electrons and is the reducing agent.

6.23 (a) XeF2 F -1, Xe +2 (b) XeF4 F -1, Xe +4

(c) XeOF4 F -1, O -2, Xe +6 6.24 (a) Rb would lose one electron and adopt the Kr noble-gas configuration.

(b) Ba would lose two electrons and adopt the Xe noble-gas configuration. (c) Ga would lose three electrons and adopt an Ar-like noble-gas configuration (note that Ga3+ has ten 3d electrons in addition to the two 3s and six 3p electrons). (d) F would gain one electron and adopt the Ne noble-gas configuration.

6.25 Group 6A elements will gain 2 electrons. 6.26 Only about 10% of current world salt production comes from evaporation of seawater.

Most salt is obtained by mining the vast deposits of halite, or rock salt, formed by evaporation of ancient inland seas. These salt beds can be up to hundreds of meters thick and may occur anywhere from a few meters to thousands of meters below the earth's surface.


6.27 6.28 (a) shows an extended array, which represents an ionic compound.

(b) shows discrete units, which represent a covalent compound.


126

6.29

6.30

(a) Al3+ (b) Cr3+ (c) Sn2+ (d) Ag+ 6.31 The first sphere gets larger on going from reactant to product. This is consistent with it

being a nonmetal gaining an electron and becoming an anion. The second sphere gets smaller on going from reactant to product. This is consistent with it being a metal losing an electron and becoming a cation.

6.32 (a) I2 (b) Na (c) NaCl (d) Cl2 6.33 (c) has the largest lattice energy because the charges are closest together.

(a) has the smallest lattice energy because the charges are farthest apart. 6.34 Green CBr4: C, +4; Br, -1

Blue SrF2: Sr, +2; F, -1 Red PbS: Pb, +2; S, -2 or PbS2: Pb, +4; S, -2


127

6.35 Additional Problems Ionization Energy and Electron Affinity 6.36 (a) La3+, [Xe] (b) Ag+, [Kr] 4d10 (c) Sn2+, [Kr] 5s2 4d10 6.37 (a) Se2-, [Kr] (b) N3-, [Ne] 6.38 Cr2+ [Ar] 3d4 ↑ ↑ ↑ ↑

3d Fe2+ [Ar] 3d6 ↑↓ ↑ ↑ ↑ ↑

3d 6.39 Z = 30, Zn 6.40 Ionization energies have a positive sign because energy is required to remove an electron

from an atom of any element. 6.41 Electron affinities have a negative sign because energy is released when an electron is

added. 6.42 The largest Ei1 are found in Group 8A because of the largest values of Zeff.

The smallest Ei1 are found in Group 1A because of the smallest values of Zeff. 6.43 Fr would have the smallest ionization energy, and He would have the largest. 6.44 (a) K [Ar] 4s1 Ca [Ar] 4s2

Ca has the smaller second ionization energy because it is easier to remove the second 4s valence electron in Ca than it is to remove the second electron in K from the filled 3p orbitals. (b) Ca [Ar] 4s2 Ga [Ar] 4s2 3d10 4p1 Ca has the larger third ionization energy because it is more difficult to remove the third electron in Ca from the filled 3p orbitals than it is to remove the third electron (second 4s valence electron) from Ga.


128

6.45 Sn has a smaller fourth ionization energy than Sb because of a smaller Zeff. Br has a larger sixth ionization energy than Se because of a larger Zeff.

6.46 (a) 1s2 2s2 2p6 3s2 3p3 is P (b) 1s2 2s2 2p6 3s2 3p6 is Ar (c) 1s2 2s2 2p6 3s2 3p6 4s2 is Ca

Ar has the highest Ei2. Ar has a higher Zeff than P. The 4s electrons in Ca are easier to remove than any 3p electrons. Ar has the lowest Ei7. It is difficult to remove 3p electrons from Ca, and it is difficult to remove 2p electrons from P.

6.47 The atom in the third row with the lowest Ei4 is the 4A element, Si. 1s2 2s2 2p6 3s2 3p2 6.48 Using Figure 6.3 as a reference:

Lowest Ei1 Highest Ei1 (a) K Li (b) B Cl (c) Ca Cl

6.49 (a) Group 2A (b) Group 6A 6.50 The relationship between the electron affinity of a univalent cation and the ionization

energy of the neutral atom is that they have the same magnitude but opposite sign. 6.51 The relationship between the ionization energy of a univalent anion and the electron

affinity of the neutral atom is that they have the same magnitude but opposite sign. 6.52 Na+ has a more negative electron affinity than either Na or Cl because of its positive charge. 6.53 Br would have a more negative electron affinity than Br- because Br- has no room in its

valence shell for an additional electron. 6.54 Energy is usually released when an electron is added to a neutral atom but absorbed when

an electron is removed from a neutral atom because of the positive Zeff. 6.55 Ei1 increases steadily across the periodic table from Group 1A to Group 8A because

electrons are being removed from the same shell and Zeff is increasing. The electron affinity increases irregularly from 1A to 7A and then falls dramatically for Group 8A because the additional electron goes into the next higher shell.

6.56 (a) F; nonmetals have more negative electron affinities than metals.

(b) Na; Ne (noble gas) has a positive electron affinity. (c) Br; nonmetals have more negative electron affinities than metals.

6.57 Zn, Cd, and Hg all have filled s and d subshells. An additional electron would have to go

into the higher energy p subshell. This is unfavorable and results in near-zero electron affinities.


129

Lattice Energy and Ionic Bonds 6.58 MgCl2 > LiCl > KCl > KBr 6.59 AlBr3 > CaO > MgBr2 > LiBr 6.60 Li → Li+ + e- +520 kJ/mol

Br + e- → Br- -325 kJ/mol +195 kJ/mol

6.61 The total energy = (376 kJ/mol) + (-349 kJ/mol) = +27 kJ/mol, which is unfavorable

because it is positive. 6.62 Li(s) → Li(g) +159.4 kJ/mol

Li(s) → Li(g) + e- +520 kJ/mol ½ [Br2(l) → Br2(g)] +15.4 kJ/mol ½ [Br2(g) → 2 Br(g)] +112 kJ/mol Br(g) + e- → Br-(g) -325 kJ/mol Li +(g) + Br-(g) → LiBr(s) -807 kJ/mol

Sum = -325 kJ/mol for Li(s) + ½ Br2(l) → LiBr(s) 6.63 (a) Li(s) → Li(g) +159.4 kJ/mol

Li(g) → Li+(g) + e- +520 kJ/mol ½[F2(g) → 2 F(g)] +79 kJ/mol F(g) + e- → F-(g) -328 kJ/mol Li+(g) + F-(g) → LiF(s) -1036 kJ/mol

Sum = -606 kJ/mol for Li(s) + ½ F2(g) → LiF(s)

(b) Ca(s) → Ca(g) +178.2 kJ/mol Ca(g) → Ca+(g) + e- +589.8 kJ/mol Ca+(g) → Ca2+(g) + e- +1145 kJ/mol F2(g) → 2 F(g) +158 kJ/mol 2[F(g) + e- → F-(g)] 2(-328) kJ/mol Ca2+(g) + 2 F- → CaF2(s) -2630 kJ/mol

Sum = -1215 kJ/mol for Ca(s) + F2(g) → CaF2(s) 6.64 Na(s) → Na(g) +107.3 kJ/mol

Na(g) → Na+(g) + e- +495.8 kJ/mol ½ [H2(g) → 2 H(g)] ½(+435.9) kJ/mol H(g) + e- → H-(g) -72.8 kJ/mol Na+(g) + H-(g) → NaH(s) -U


130

Sum = -60 kJ/mol for Na(s) + ½ H2(g) → NaH(s)

- U = - 60 - 107.3 - 495.8 - 435.9/2 + 72.8 = -808 kJ/mol; U = 808 kJ/mol 6.65 Ca(s) → Ca(g) +178.2 kJ/mol

Ca(g) → Ca+(g) + e- +589.8 kJ/mol Ca+(g) → Ca2+(g) + e- +1145 kJ/mol H2(g) → 2 H(g) +435.9 kJ/mol 2[H(g) + e- → H-(g)] 2(-72.8) kJ/mol Ca2+(g) + 2 H-(g) → CaH2(s) -U

Sum = -186.2 kJ/mol for Ca(s) + H2(g) → CaH2(s)

- U = -186.2 - 178.2 - 589.8 - 1145 - 435.9 + 2(72.8) = -2390 kJ/mol; U = 2390 kJ/mol 6.66 Cs(s) → Cs(g) +76.1 kJ/mol

Cs(g) → Cs+(g) + e- +375.7 kJ/mol ½ [F2(g) → 2 F(g)] +79 kJ/mol F(g) + e- → F-(g) -328 kJ/mol Cs+(g) + F-(g) → CsF(s) -740 kJ/mol

Sum = -537 kJ/mol for Cs(s) + ½ F2(g) → CsF(s) 6.67 Cs(s) → Cs(g) +76.1 kJ/mol

Cs(g) → Cs+(g) + e- +375.7 kJ/mol Cs+(g) → Cs2+(g) + e- +2422 kJ/mol F2(g) → 2 F(g) +158 kJ/mol 2[F(g) + e- → F-(g)] 2(-328) kJ/mol Cs2+(g) + 2 F-(g) → CsF2(s) -2347 kJ/mol

Sum = +29 kJ/mol for Cs(s) + F2(g) → CsF2(s)

The overall reaction absorbs 29 kJ/mol. In the reaction of cesium with fluorine, CsF will form because the overall energy for the formation of CsF is negative, whereas it is positive for CsF2.

6.68 Ca(s) → Ca(g) +178.2 kJ/mol

Ca(g) → Ca+(g) + e- +589.8 kJ/mol ½[Cl2(g) → 2 Cl(g)] +121.5 kJ/mol Cl(g) + e- → Cl-(g) -348.6 kJ/mol Ca+(g) + Cl-(g) → CaCl(s) -717 kJ/mol

Sum = -176 kJ/mol for Ca(s) + ½ Cl2(g) → CaCl(s) 6.69 Ca(s) → Ca(g) + e- +178.2 kJ/mol


131

Ca(g) → Ca+(g) + e- +589.8 kJ/mol Ca+(g) → Ca2+(g) +1145 kJ/mol Cl2(g) → 2 Cl(g) +243 kJ/mol 2[Cl(g) + e- → Cl-(g)] 2(-348.6) kJ/mol Ca2+(g) + 2 Cl-(g) → CaCl2(s) -2258 kJ/mol

Sum = -799 kJ/mol for Ca(s) + Cl2(g) → CaCl2(s)

In the reaction of calcium with chlorine, CaCl2 will form because the overall energy for the formation of CaCl2 is much more negative than for the formation of CaCl.

6.70 6.71

Main-Group Chemistry 6.72 Solids: I2; Liquids: Br2; Gases: F2, Cl2, He, Ne, Ar, Kr, Xe 6.73 (a) Li is used in automotive grease. Li2CO3 is a manic depressive drug.


132

(b) K salts are used in plant fertilizers. (c) SrCO3 is used in color TV picture tubes. Sr salts are used for red fireworks. (d) Liquid He (bp = 4.2 K) is used for low temperature studies and for cooling superconducting magnets.

6.74 (a) At is in Group 7A. The trend going down the group is gas → liquid → solid. At,

being at the bottom of the group, should be a solid. (b) At would likely be dark, like I2, maybe with a metallic sheen. (c) At is likely to react with Na just like the other halogens, yielding NaAt.

6.75 Predicted for Fr: melting point ≈ 23 oC boiling point ≈ 650 oC

density ≈ 2 g/cm3 atomic radius ≈ 275 pm

6.76 (a) (g)Cl + Na(l) 2 NaCl 2 2

CaClin iselectrolys

C580

_2

°

(b) (g)O 3 + Al(l) 4 OAl 2 2

AlFNain iselectrolys

C980

32 _63

°

(c) Ar is obtained from the distillation of liquid air. (d) 2 Br-(aq) + Cl2(g) → Br2(l) + 2 Cl-(aq)

6.77 Group 1A metals react by losing an electron. Down Group 1A, the valence electron is

more easily removed. This trend parallels chemical reactivity. Group 7A nonmetals react by gaining an electron. The electron affinity generally increases up the group. This trend parallels chemical reactivity.

6.78 Main-group elements tend to undergo reactions that leave them with eight valence

electrons. That is, main-group elements react so that they attain a noble gas electron configuration with filled s and p sublevels in their valence electron shell. The octet rule works for valence shell electrons because taking electrons away from a filled octet is difficult because they are tightly held by a high Zeff; adding more electrons to a filled octet is difficult because, with s and p sublevels full, there is no low-energy orbital available.

6.79 Main-group nonmetals in the third period and below occasionally break the octet rule. 6.80 (a) 2 K(s) + H2(g) → 2 KH(s)

(b) 2 K(s) + 2 H2O(l) → 2 K+(aq) + 2 OH-(aq) + H2(g) (c) 2 K(s) + 2 NH3(g) → 2 KNH2(s) + H2(g) (d) 2 K(s) + Br2(l) → 2 KBr(s)


133

(e) K(s) + N2(g) → N. R. (f) K(s) + O2(g) → KO2(s)

6.81 (a) Ca(s) + H2(g) → CaH2(s)

(b) Ca(s) + 2 H2O(l) → Ca2+(aq) + 2 OH-(aq) + H2(g) (c) Ca(s) + He(g) → N. R. (d) Ca(s) + Br2(l) → CaBr2(s) (e) 2 Ca(s) + O2(g) → 2 CaO(s)

6.82 (a) Cl2(g) + H2(g) → 2 HCl(g)

(b) Cl2(g) + Ar(g) → N. R. (c) Cl2(g) + Br2(l) → 2 BrCl(g) (d) Cl2(g) + N2(g) → N. R.

6.83 (a) 2 Cl-(aq) + F2(g) → 2 F-(aq) + Cl2(g)

F2 gains electrons and is the oxidizing agent. Cl- loses electrons and is the reducing agent. (b) 2 Br-(aq) + I2(s) → N. R. (c) 2 I-(aq) + Br2(aq) → 2 Br-(aq) + I2(aq) Br2 gains electrons and is the oxidizing agent. I- loses electrons and is the reducing agent.

6.84 AlCl3 + 3 Na → Al + 3 NaCl

Al 3+ (from AlCl3) gains electrons and is reduced. Na loses electrons and is oxidized. 6.85 2 Mg(s) + O2(g) → 2 MgO(s)

MgO(s) + H2O(l) → Mg(OH)2(aq) 6.86 CaIO3, 215.0 amu; 1.00 kg = 1000 g

% I = g 215.0

g 126.9 x 100% = 59.02%; (0.5902)(1000 g) = 590 g I2

6.87 2 Li(s) + 2 H2O(l) → 2 LiOH(aq) + H2(g); 455 mL = 0.455 L

mass of H2 = (0.0893 g/L)(0.455 L) = 0.0406 g H2

Li g 0.280 = Li mol 1

Li g 6.94 x

H mol 1

Li mol 2 x

H g 2.016H mol 1

x H g 0.040622

22

6.88 Ca(s) + H2(g) → CaH2(s); H2, 2.016 amu; CaH2, 42.09 amu

5.65 g Ca x Ca g 40.08

Ca mol 1 = 0.141 mol Ca

3.15 L H2 x 0.0893 H g 2.016

H mol 1 x

L 1H g

2

22 = 0.140 mol H2

Because the reaction stoichiometry between Ca and H2 is one to one, H2 is the limiting reactant.


134

0.140 mol H2 x CaH mol 1CaH g 42.09

x H mol 1

CaH mol 1

2

2

2

2 x 0.943 = 5.56 g CaH2

6.89 6 Li(s) + N2(g) → 2 Li3N(s); N2, 28.01 amu; Li3N, 34.83 amu

2.87 g Li x N g 1.25

L 1 x

N mol 1N g 28.01

x Li mol 6N mol 1

x Li g 6.941

Li mol 1

22

22 = 1.54 L N2

6.90 (a) Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)

H+ gains electrons and is the oxidizing agent. Mg loses electrons and is the reducing agent. (b) Kr(g) + F2(g) → KrF2(s) F2 gains electrons and is the oxidizing agent. Kr loses electrons and is the reducing agent. (c) I2(s) + 3 Cl2(g) → 2 ICl3(l) Cl2 gains electrons and is the oxidizing agent. I2 loses electrons and is the reducing agent.

6.91 (a) 2 XeF2(s) + 2 H2O(l) → 2 Xe(g) + 4 HF(aq) + O2(g)

Xe in XeF2 gains electrons and is the oxidizing agent. O in H2O loses electrons and is the reducing agent. (b) NaH(s) + H2O(l) → Na+(aq) + OH-(aq) + H2(g) H in H2O gains electrons and is the oxidizing agent. H in NaH loses electrons and is the reducing agent. (c) 2 TiCl4(l) + H2(g) → 2 TiCl3(s) + 2 HCl(g) Ti in TiCl4 gains electrons and is the oxidizing agent. H2 loses electrons and is the reducing agent.

General Problems 6.92 Cu2+ has fewer electrons and a larger effective nuclear charge; therefore it has the smaller

ionic radius. 6.93 S2- > Ca2+ > Sc3+ > Ti4+, Zeff increases on going from S2- to Ti4+. 6.94 Mg(s) → Mg(g) +147.7 kJ/mol

Mg(g) → Mg+(g) + e- +737.7 kJ/mol ½[F2(g) → 2 F(g)] +79 kJ/mol F(g) + e- → F-(g) -328 kJ/mol Mg+(g) + F-(g) → MgF(s) -930 kJ/mol

Sum = -294 kJ/mol for Mg(s) + ½ F2(g) → MgF(s)

Mg(s) → Mg(g) +147.7 kJ/mol Mg(g) → Mg+(g) + e- +737.7 kJ/mol


135

Mg+(g) → Mg2+(g) + e- +1450.7 kJ/mol F2(g) → 2 F(g) +158 kJ/mol 2[F(g) + e- → F-(g)] 2(-328) kJ/mol Mg2+(g) + 2 F-(g) → MgF2(s) -2952 kJ/mol

Sum = -1114 kJ/mol for Mg(s) + F2(g) → MgF2(s) 6.95 In the reaction of magnesium with fluorine, MgF2 will form because the overall energy

for the formation of MgF2 is much more negative than for the formation of MgF. 6.96 (a) Na is used in table salt (NaCl), glass, rubber, and pharmaceutical agents.

(b) Mg is used as a structural material when alloyed with Al. (c) F2 is used in the manufacture of Teflon, (C2F4)n, and in toothpaste as SnF2.

6.97 (a) (g)F + (g)H HF(l) 2 22

iselectrolys

C100

_°

(b) (s)OAl + Ca(l) 3 Al(l) 2 + CaO(l) 3 32

eraturehigh temp

_

(c) (g)Cl + Na(l) 2 NaCl(l) 2 2

CaClin iselectrolys

C580

_2

°

6.98 (a) 2 Li(s) + H2(g) → 2 LiH(s)

(b) 2 Li(s) + 2 H2O(l) → 2 Li+(aq) + 2 OH-(aq) + H2(g) (c) 2 Li(s) + 2 NH3(g) → 2 LiNH2(s) + H2(g) (d) 2 Li(s) + Br2(l) → 2 LiBr(s) (e) 6 Li(s) + N2(g) → 2 Li3N(s) (f) 4 Li(s) + O2(g) → 2 Li2O(s)

6.99 (a) F2(g) + H2(g) → 2 HF(g) (b) F2(g) + 2 Na(s) → 2 NaF(s)

(c) F2(g) + Br2(l) → 2 BrF(g) (d) F2(g) + 2 NaBr(s) → 2 NaF(g) + Br2(l) 6.100 When moving diagonally down and right on the periodic table, the increase in atomic

radius caused by going to a larger shell is offset by a decrease caused by a higher Zeff. Thus, there is little net change.

6.101 Na(s) → Na(g) +107.3 kJ/mol

Na(g) + e- → Na-(g) -52.9 kJ/mol ½[Cl2(g) → 2 Cl(g)] +122 kJ/mol Cl(g) → Cl+(g) + e- +1251 kJ/mol Na-(g) + Cl+(g) → ClNa(s) -787 kJ/mol

Sum = +640 kJ/mol for Na(s) + ½ Cl2(g) → Cl+Na-(s) The formation of Cl+Na- from its elements is not favored because the net energy change is


136

positive whereas for the formation of Na+Cl- it is negative.

6.102 6.103 94.2 mL = 0.0942 L

0.0942 L Cl2 x Cl mol 1Cl mol 2

x L 22.4Cl mol 1

2

_2 = 8.41 x 10-3 mol Cl-

Possible formulas for the metal halide are MCl, MCl2, MCl3, etc. For MCl, mol M = mol Cl- = 8.41 x 10-3 mol M

molar mass of M = mol 10 x 8.41

g 0.7193_

= 85.5 g/mol

For MCl2, mol M = 8.41 x 10-3 mol Cl- x Cl mol 2

M mol 1_1111 = 4.20 x 10-3 mol M


g 0.7193_

= 171 g/mol

For MCl3, mol M = 8.41 x 10-3 mol Cl- x Cl mol 3

M mol 1_

= 2.80 x 10-3 mol M


g 0.7193_

= 257 g/mol

The best match for a metal is with 85.5 g/mol, which is Rb. 6.104 Mg(s) → Mg(g) +147.7 kJ/mol

Mg(g) → Mg+(g) + e- +738 kJ/mol Mg+(g) → Mg2+(g) + e- +1451 kJ/mol ½[O2(g) → 2 O(g)] +249.2 kJ/mol O(g) + e- → O-(g) -141.0 kJ/mol O-(g) + e- → O2-(g) Eea2 Mg2+(g) + O2-(g) → MgO(s) -3791 kJ/mol Mg(s) + ½O2(g) → MgO(s) -601.7 kJ/mol

147.7 + 738 + 1451 + 249.2 - 141.0 + Eea2 - 3791 = -601.7

Eea2 = -147.7 - 738 - 1451 - 249.2 + 141.0 + 3791 - 601.7 = +744 kJ/mol Because Eea2 is positive, O2- is not stable in the gas phase. It is stable in MgO because of the large lattice energy that results from the +2 and -2 charge of the ions and their small size.


137

6.105 (a) (i) Ra because it is farthest down (7th period) in the periodic table.

(ii) In because it is farthest down (5th period) in the periodic table. (b) (i) Tl and Po are farthest down (6th period) but Tl is larger because it is to the left of Po and thus has the smaller ionization energy. (ii) Cs and Bi are farthest down (6th period) but Cs is larger because it is to the left of Bi and thus has the smaller ionization energy.

6.106 (a) The more negative the Eea, the greater the tendency of the atom to accept an electron,

and the more stable the anion that results. Be, N, O, and F are all second row elements. F has the most negative Eea of the group because the anion that forms, F-, has a complete octet of electrons and its nucleus has the highest effective nuclear charge. (b) Se2- and Rb+ are below O2- and F- in the periodic table and are the larger of the four. Se2- and Rb+ are isoelectronic, but Rb+ has the higher effective nuclear charge so it is smaller. Therefore Se2- is the largest of the four ions.

6.107 Ca(s) → Ca(g) +178 kJ/mol

Ca(g) → Ca+(g) +590 kJ/mol Ca+(g) → Ca2+(g) +1145 kJ/mol 2 C(s) → 2 C(g) 2(+717 kJ/mol) 2 C(g) → C2(g) -614 kJ/mol C2(g) → C2

-(g) -315 kJ/mol C2

-(g) → C22-(g) +410 kJ/mol

Ca2+(g) + C22-(g) → CaC2(s) -U

Ca(s) + 2 C(s) → CaC2(s) -60 kJ/mol -U = -60 -178 - 590 - 1145 - 2(717) + 614 + 315 - 410 = -2888 kJ/mol U = 2888 kJ/mol

6.108 Cr(s) → Cr(g) +397 kJ/mol Cr(g) → Cr+(g) +652 kJ/mol Cr+(g) → Cr2+(g) +1588 kJ/mol Cr2+(g) → Cr3+(g) +2882 kJ/mol ½(I2(s) → I2(g)) +62/2 kJ/mol ½ (I2(g) → 2 I(g)) +151/2 kJ/mol I(g) + e- → I-(g) -295 kJ/mol Cl2(g) → 2 Cl(g) +243 kJ/mol 2(Cl(g) + e- → Cl-(g)) 2(-349) kJ/mol

Cr3+(g) + 2 Cl-(g) + I-(g) → CrCl2I(s) -U Cr(s) + Cl2(g) + ½ I2(g) → CrCl2I(s) - 420 kJ/mol

-U = - 420 - 397 - 652 - 1588 - 2882 - 62/2 - 151/2 + 295 - 243 + 2(349) = -5295.5 kJ/mol U = 5295 kJ/mol Multi-Concept Problems 6.109 (a) E = (703 kJ/mol)(1000 J/1 kJ)/(6.022 x 1023 photons/mol) = 1.17 x 10-18 J/photon


138

E = λhc

= J 10 x 1.17

m/s) 10 x s)(3.00 J 10 x (6.626 =

E

hc =

18_

834_ •λ 1.70 x 10-7 m = 170 x 10-9 m = 170 nm

(b) Bi [Xe] 6s2 4f 14 5d10 6p3 Bi+ [Xe] 6s2 4f 14 5d10 6p2 (c) n = 6, l = 1 (d) Element 115 would be directly below Bi in the periodic table. The valence electron is farther from the nucleus and less strongly held than in Bi. The ionization energy for element 115 would be less than that for Bi.

6.110 (a) Fe [Ar] 4s2 3d6

Fe2+ [Ar] 3d6 Fe3+ [Ar] 3d5 (b) A 3d electron is removed on going from Fe2+ to Fe3+. For the 3d electron, n = 3 and l =

2.

(c) E(J/photon) = kJ 1

J 1000 x

photons 10 x 6.022

photons mol 1 x kJ/mol 2952

23= 4.90 x 10-18 J/photon

E = λch

J 10 x 4.90

m/s) 10 x s)(3.00 J 10 x (6.626 =

E

ch =

18_

834_ •λ = 4.06 x 10-8 m = 40.6 x 10-9 m = 40.6

nm (d) Ru is directly below Fe in the periodic table and the two metals have similar electron configurations. The electron removed from Ru to go from Ru2+ to Ru3+ is a 4d electron. The electron with the higher principal quantum number, n = 4, is farther from the nucleus, less tightly held, and requires less energy to remove.

6.111 (a) 58.4 nm = 58.4 x 10-9 m

E(photon) = 6.626 x 10-34 J⋅s x mol

10 x 6.022 x

J 1000

kJ 1 x

m 10 x 58.4

m/s 10 x 3.00 23

9-

8

= 2049

kJ/mol

EK = E(electron) = ½(9.109 x 10-31 kg)(2.450 x 106 m/s)2

mol10 x 6.022

J 1000

kJ 1 23

EK = 1646 kJ/mol E(photon) = Ei + EK; Ei = E(photon) - EK = 2049 - 1646 = 403 kJ/mol

(b) 142 nm = 142 x 10-9 m

kJ/mol 843 = mol

10 x 6.022 x

J 1000

kJ 1 x

m 10 x 142

s / m 10 x 3.00 x s J 10 x 6.626 = E(photon)

23

9_

834_ •


139

kJ/mol 422 = E

mol10 x 6.022

J 1000

kJ 1s) / m 10 x (1.240 kg) 10 x (9.109 ‰ = )E(electron = E

K

236 231_

K

E(photon) = Ei + EK; Ei = E(photon) – EK = 843 – 422 = 421 kJ/mol 6.112 AgCl, 143.32 amu

(a) mass Cl in AgCl = 1.126 g AgCl x AgCl g 143.32

Cl g 35.453 = 0.279 g Cl

%Cl in alkaline earth chloride = g 0.436

Cl g 0.279 x 100% = 64.0% Cl

(b) Because M is an alkaline earth metal, M is a 2+ cation. For MCl2, mass of M = 0.436 g - 0.279 g = 0.157 g M

mol M = 0.279 g Cl x Cl mol 2

M mol 1 x

Cl g 35.453

Cl mol 1 = 0.003 93 mol M

molar mass for M = mol 93 0.003

g 0.157 = 39.9 g/mol; M = Ca

(c) Ca(s) + Cl2(g) → CaCl2(s) CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)2(aq)

(d) 1.005 g Ca x Ca g 40.078

Ca mol 1 = 0.0251 mol Ca

1.91 x 1022 Cl2 molecules x molecules Cl 10 x 6.022

Cl mol 1

223

2 = 0.0317 mol Cl2

Because the stoichiometry between Ca and Cl2 is one to one, the Cl2 is in excess.

Mass Cl2 unreacted = (0.0317 - 0.0251) mol Cl2 x Cl mol 1Cl g 70.91

2

2 = 0.47 g Cl2 unreacted

6.113 (a) (i) M2O3(s) + 3 C(s) + 3 Cl2(g) → 2 MCl3(l) + 3 CO(g)

(ii) 2 MCl3(l) + 3 H2(g) → 2 M(s) + 6 HCl(g) (b) HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) 144.2 mL = 0.1442 L mol NaOH = (0.511 mol/L)(0.1442 L) = 0.07369 mol NaOH

mol HCl = 0.07369 mol NaOH x NaOH mol 1

HCl mol 1 = 0.07369 mol HCl

mol M = 0.07369 mol HCl x HCl mol 6

M mol 2 = 0.02456 mol M

mol M2O3 = 0.02456 mol M x M mol 2

MCl mol 2 3 x MCl mol 2

OM mol 1

3

32 = 0.01228 mol M2O3


140

molar mass M2O3 = mol 0.01228

g 0.855 = 69.6 g/mol; molecular mass M2O3 = 69.6 amu

atomic mass of M = 2

amu) 16.0 x (3 _amu 69.6 = 10.8 amu; M = B

(c) mass of M = 0.02456 mol M x M mol 1

M g 10.81 = 0.265 g M

6.114 (a) Sr(s) → Sr(g) +164.44 kJ/mol

Sr(g) → Sr+(g) + e- +549.5 kJ/mol Sr+(g) → Sr2+(g) + e- +1064.2 kJ/mol Cl2(g) → 2 Cl(g) +243 kJ/mol 2[Cl(g) + e- → Cl-(g)] 2(-348.6) kJ/mol Sr2+(g) + 2 Cl-(g) → SrCl2(s) -2156 kJ/mol

Sum = -832 kJ/mol for Sr(s) + Cl2(g) → SrCl2(s) (b) Sr, 87.62 amu; Cl2, 70.91 amu; SrCl2, 158.53 amu

20.0 g Sr x Sr g 87.62

Sr mol 1 = 0.228 mol Sr and 25.0 g Cl2 x

Cl g 70.91Cl mol 1

2

2 = 0.353 mol Cl2

Because there is a 1:1 stoichiometry between the reactants, the one with the smaller mole amount is the limiting reactant. Sr is the limiting reactant.

0.228 mol Sr x SrCl mol 1

SrCl g 158.53 x

Sr mol 1SrCl mol 1

2

22 = 36.1 g SrCl2

(c) 0.228 mol SrCl2 x SrCl mol 1

kJ 832 _

2

= -190 kJ

190 kJ is released during the reaction of 20.0 g of Sr with 25.0 g Cl2. 6.115 (a) The alkali metal is Li because it is the only alkali metal to form the nitride, M3N.

(b) 4 Li(s) + O2(g) → 2 Li2O(s) 6 Li(s) + N2(g) → 2 Li3N(s) Li 2O(s) + H2O(l) → 2 LiOH(aq) Li 3N(s) + 3 H2O(l) → NH3(aq) + 3 LiOH(aq) (c) 96.8 mL = 0.0968 L mol HCl = (0.0968 L)(0.100 mol/L) = 0.009 68 mol HCl Note, that the HCl neutralized only 20% (0.20) of the total sample. Let Y = mol Li2O and let Z = mol Li3N mol HCl = 2Y + 4Z = 0.009 68 mol

mol Li = 2Y + 3Z = (0.20)(0.265 g Li) x = Li g 6.941

Li mol 10.007 64 mol Li

2Y + 4Z = 0.009 68 2Y + 3Z = 0.007 64 2Y = 0.007 64 - 3Z (substitute 2Y into the first equation and solve for Z) 0.007 64 - 3Z + 4Z = 0.009 68 Z = 0.009 68 - 0.007 64 = 0.002 04 mol Li3N


141

Y = (0.007 64 - 3Z)/2 = [0.007 64 - 3(0.002 04)]/2 = 0.000 76 mol Li2O

= mol 04 0.002 + mol 76 0.000

mol 76 0.000 =

NLi mol + OLi mol

OLi mol = X

32

2OLi 2

0.271

= X NLi3 1.000 - 0.271 = 0.729

141

7

Covalent Bonds and Molecular Structure

7.1 (a) SiCl4 chlorine EN = 3.0 silicon EN = 1.8

∆EN = 1.2 The Si–Cl bond is polar covalent.

(b) CsBr bromine EN = 2.8 cesium EN = 0.7

∆EN = 2.1 The Cs+Br- bond is ionic.

(c) FeBr3 bromine EN = 2.8 iron EN = 1.8

∆EN = 1.0 The Fe–Br bond is polar covalent.

(d) CH4 carbon EN = 2.5 hydrogen EN = 2.1

∆EN = 0.4 The C–H bond is polar covalent. 7.2 (a) CCl4 chlorine EN = 3.0 carbon EN = 2.5

∆EN = 0.5

(b) BaCl2 chlorine EN = 3.0 barium EN = 0.9

∆EN = 2.1

(c) TiCl3 chlorine EN = 3.0 titanium EN = 1.5

∆EN = 1.5

(d) Cl2O oxygen EN = 3.5 chlorine EN = 3.0

∆EN = 0.5

Increasing ionic character: CCl4 ~ ClO2 < TiCl3 < BaCl2 7.3 H is positively polarized (blue). O is negatively polarized (red). This is consistent with

the electronegativity values for O (3.5) and H (2.1). The more negatively polarized atom should be the one with the larger electronegativity.

Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________

142

7.4 (a) (b)

7.5

7.6 (a) (b) (c)

(d) (e) (f)

7.7

7.8 Molecular formula: C4H5N3O;

7.9

7.10 (a) (b)

(c) (d)


143

7.11 (a) (b) (c)

(d)

7.12

7.13 (a)

(b)

(c)

(d)

7.14


144

7.15 For nitrogen: Isolated nitrogen valence electrons 5

Bound nitrogen bonding electrons 8 Bound nitrogen nonbonding electrons 0 Formal charge = 5 - ½(8) - 0 = +1

For singly bound Isolated oxygen valence electrons 6 oxygen: Bound oxygen bonding electrons 2

Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1

For doubly bound Isolated oxygen valence electrons 6 oxygen: Bound oxygen bonding electrons 4

Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0

7.16 (a) For nitrogen: Isolated nitrogen valence electrons 5

Bound nitrogen bonding electrons 4 Bound nitrogen nonbonding electrons 4 Formal charge = 5 - ½(4) - 4 = -1

For carbon: Isolated carbon valence electrons 4

Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½ (8) - 0 = 0

For oxygen: Isolated oxygen valence electrons 6

Bound oxygen bonding electrons 4 Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0

(b) For left oxygen: Isolated oxygen valence electrons 6

Bound oxygen bonding electrons 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1

For central Isolated oxygen valence electrons 6 oxygen: Bound oxygen bonding electrons 6

Bound oxygen nonbonding electrons 2 Formal charge = 6 - ½(6) - 2 = +1


145

For right Isolated oxygen valence electrons 6 oxygen: Bound oxygen bonding electrons 4


7.17 Number of Number of

Bonded Atoms Lone Pairs Shape (a) O3 2 1 bent (b) H3O

+ 3 1 trigonal pyramidal (c) XeF2 2 3 linear (d) PF6

- 6 0 octahedral (e) XeOF4 5 1 square pyramidal (f) AlH4

- 4 0 tetrahedral (g) BF4

- 4 0 tetrahedral (h) SiCl4 4 0 tetrahedral (i) ICl4

- 4 2 square planar (j) AlCl 3 3 0 trigonal planar

7.18 7.19 (a) tetrahedral (b) seesaw

7.20 Each C is sp3 hybridized. The C–C bond is formed by the overlap of one singly occupied sp3 hybrid orbital from each C. The C–H bonds are formed by the overlap of one singly occupied sp3 orbital on C with a singly occupied H 1s orbital.

7.21


146

The carbon in formaldehyde is sp2 hybridized.

7.22

In HCN the carbon is sp hybridized. 7.23 The central I in I3

- has two single bonds and three lone pairs of electrons. The hybridization of the central I is sp3d. A sketch of the ion showing the orbitals involved in bonding is shown below.

7.24 Single Bonds Lone Pairs Hybridization of S

SF2 2 2 sp3 SF4 4 1 sp3d SF6 6 0 sp3d2


147

7.25 (a) sp (b) sp3d 7.26 For He2

+ σ*1s ↑ σ1s ↑↓

He2+ Bond order = 2/1 =

2

1 _ 2 =

2

electrons gantibondin

ofnumber _

electrons bonding

ofnumber

He2+ should be stable with a bond order of 1/2.

7.27 For B2

σ*2p π*2p σ2p π2p ↑ ↑ σ*2s ↑↓ σ2s ↑↓

B2 Bond order = 1 = 2

2 _ 4 =

2


ofnumber _

electrons bonding

ofnumber

B2 is paramagnetic because it has two unpaired electrons in the π2p molecular orbitals.

For C2 σ*2p π*2p σ2p π2p ↑↓ ↑↓ σ*2s ↑↓ σ2s ↑↓

C2 Bond order = 2 = 2

2_ 6; C2 is diamagnetic because all electrons are paired.

7.28 7.29 Handed biomolecules have specific shapes that only match complementary-shaped

receptor sites in living systems. The mirror-image forms of the molecules can’t fit into the receptor sites and thus don’t elicit the same biological response.

7.30 The mirror image of molecule (a) has the same shape as (a) and is identical to it in all

respects so there is no handedness associated with it. The mirror image of molecule (b) is different than (b) so there is a handedness to this molecule.


148

Understanding Key Concepts 7.31 As the electrostatic potential maps are drawn, the Li and Cl are at the tops of each map.

The red area is for a negatively polarized region (associated with Cl). The blue area is for a positively polarized region (associated with Li). Map (a) is for CH3Cl and Map (b) is for CH3Li.

7.32 (a) square pyramidal (b) trigonal pyramidal

(c) square planar (d) trigonal planar 7.33 (a) trigonal bipyramidal (b) tetrahedral

(c) square pyramidal (4 ligands in the horizontal plane, including one hidden) 7.34 Molecular model (c) does not have a tetrahedral central atom. It is square planar. 7.35 (a) sp2 (b) sp3d2 (c) sp3 7.36 (a) C8H9NO2

(b) & (c)

7.37 (a) C13H10N2O4 (b) and (c)

All carbons that have only single bonds are sp3 hybridized and have a tetrahedral geometry. All carbons that have double bonds are sp2 hybridized and have a trigonal planar geometry. The two nitrogens are sp2 hybridized and have a trigonal planar geometry.


149

Additional Problems Electronegativity and Polar Covalent Bonds 7.38 Electronegativity increases from left to right across a period and decreases down a group. 7.39 Z = 119 would be below francium and have a very low electronegativity. 7.40 K < Li < Mg < Pb < C < Br 7.41 Cl > C > Cu > Ca > Cs 7.42 (a) HF fluorine EN = 4.0

hydrogen EN = 2.1 ∆EN = 1.9 HF is polar covalent.

(b) HI iodine EN = 2.5

hydrogen EN = 2.1 ∆EN = 0.4 HI is polar covalent.

(c) PdCl2 chlorine EN = 3.0

palladium EN = 2.2 ∆EN = 0.8 PdCl2 is polar covalent.

(d) BBr3 bromine EN = 2.8

boron EN = 2.0 ∆EN = 0.8 BBr3 is polar covalent.

(e) NaOH Na+ – OH- is ionic OH- oxygen EN = 3.5

hydrogen EN = 2.1 ∆EN = 1.4 OH- is polar covalent.

(f) CH3Li lithium EN = 1.0

carbon EN = 2.5 ∆EN = 1.5 CH3Li is polar covalent.

7.43 The electronegativity for each element is shown in parentheses.

(a) C (2.5), H (2.1), Cl (3.0): The C–Cl bond is more polar than the C–H bond because of the larger electronegativity difference between the bonded atoms. (b) Si (1.8), Li (1.0), Cl (3.0): The Si–Cl bond is more polar than the Si–Li bond because of the larger electronegativity difference between the bonded atoms. (c) N (3.0), Cl (3.0), Mg (1.2): The N–Mg bond is more polar than the N–Cl bond because of the larger electronegativity difference between the bonded atoms.


150

7.44 (a) H

+ _

C

_ δδ

Cl

_ _

C

+ δδ (b)

Li

+ _

Si

_ δδ

Cl

_ _

Si

+ δδ

(c) N – Cl Mg

+ _

N

_ δδ

7.45 (a) H

+ _

F

_ δδ (b)

H

+ _

I

_ δδ (c)

Pd

+ _

Cl

_ δδ

(d) B

+ _

Br

_ δδ (e)

H

+ _

O

_ δδ

Electron-Dot Structures and Resonance 7.46 The octet rule states that main-group elements tend to react so that they attain a noble gas

electron configuration with filled s and p sublevels (8 electrons) in their valence electron shells. The transition metals are characterized by partially filled d orbitals that can be used to expand their valence shell beyond the normal octet of electrons.

7.47 (a) AlCl3 Al has only 6 electrons around it. (b) PCl5 P has 10 electrons around it.

7.48 (a) (b) (c)

(d) (e) (f)

7.49 (a) (b) (c)

(d) (e)


151

(f)

7.50 (a)

(b)

(c)

7.51 (a)

(b)

(c)

(d)

7.52

7.53 ; CS2 has two double bonds. 7.54 (a) yes (b) yes (c) yes (d) yes 7.55 (a) yes (b) no (c) yes 7.56 (a) The anion has 32 valence electrons. Each Cl has seven valence electrons (28 total).

The minus one charge on the anion accounts for one valence electron. This leaves three valence electrons for X. X is Al. (b) The cation has eight valence electrons. Each H has one valence electron (4 total).


152

X is left with four valence electrons. Since this is a cation, one valence electron was removed from X. X has five valence electrons. X is P.

7.57 (a) This fourth-row element has six valence electrons. It is Se.

(b) This fourth-row element has eight valence electrons. It is Kr.

7.58 (a) (b)

7.59 (a) (b) Formal Charges

7.60 For carbon: Isolated carbon valence electrons 4

Bound carbon bonding electrons 6 Bound carbon nonbonding electrons 2 Formal charge = 4 - ½(6) - 2 = -1


Bound oxygen bonding electrons 6 Bound oxygen nonbonding electrons 2 Formal charge = 6 - ½(6) - 2 = +1

7.61 (a) For hydrogen: Isolated hydrogen valence electrons 1

Bound hydrogen bonding electrons 2 Bound hydrogen nonbonding electrons 0 Formal charge = 1 - ½(2) - 0 = 0

For nitrogen: Isolated nitrogen valence electrons 5

Bound nitrogen bonding electrons 6 Bound nitrogen nonbonding electrons 2 Formal charge = 5 - ½(6) - 2 = 0


Bound oxygen bonding electrons 4


153


(b) For hydrogen: Isolated hydrogen valence electrons 1



Bound nitrogen bonding electrons 4 Bound nitrogen nonbonding electrons 4 Formal charge = 5 - ½(4) - 4 = -1


Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0

(c) For chlorine: Isolated chlorine valence electrons 7

Bound chlorine bonding electrons 2 Bound chlorine nonbonding electrons 6 Formal charge = 7 - ½(2) - 6 = 0

For oxygen: Isolated oxygen valence electrons 6 Bound oxygen bonding electrons 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1

For phosphorus: Isolated phosphorus valence electrons 5

Bound phosphorus bonding electrons 8 Bound phosphorus nonbonding electrons 0 Formal charge = 5 - ½(8) - 0 = +1

7.62 For both oxygens: Isolated oxygen valence electrons 6


154


For chlorine: Isolated chlorine valence electrons 7

Bound chlorine bonding electrons 4 Bound chlorine nonbonding electrons 4 Formal charge = 7 - ½(4) - 4 = +1

For left oxygen: Isolated oxygen valence electrons 6


For right oxygen: Isolated oxygen valence electrons 6


For chlorine: Isolated chlorine valence electrons 7

Bound chlorine bonding electrons 6 Bound chlorine nonbonding electrons 4 Formal charge = 7 - ½(6) - 4 = 0

7.63 For sulfur: Isolated sulfur valence electrons 6

Bound sulfur bonding electrons 8 Bound sulfur nonbonding electrons 2 Formal charge = 6 - ½(8) - 2 = 0

For doubly Isolated oxygen valence electrons 6 bound oxygen: Bound oxygen bonding electrons 4


For oxygen Isolated oxygen valence electrons 6 bound to Bound oxygen bonding electrons 4 hydrogen: Bound oxygen nonbonding electrons 4

Formal charge = 6 - ½(4) - 4 = 0


155

For hydrogen: Isolated hydrogen valence electrons 1 Bound hydrogen bonding electrons 2 Bound hydrogen nonbonding electrons 0 Formal charge = 1 - ½(2) - 0 = 0

For sulfur: Isolated sulfur valence electrons 6

Bound sulfur bonding electrons 6 Bound sulfur nonbonding electrons 2 Formal charge = 6 - ½(6) - 2 = +1

For oxygen not Isolated oxygen valence electrons 6 bound to Bound oxygen bonding electrons 2 hydrogen: Bound oxygen nonbonding electrons 6

Formal charge = 6 - ½(2) - 6 = - 1

For oxygen Isolated oxygen valence electrons 6 bound Bound oxygen bonding electrons 4 to hydrogen: Bound oxygen nonbonding electrons 4

Formal charge = 6 - ½(4) - 4 = 0

For hydrogen: Isolated hydrogen valence electrons 1 Bound hydrogen bonding electrons 2 Bound hydrogen nonbonding electrons 0 Formal charge = 1 - ½(2) - 0 = 0

7.64 (a) For hydrogen: Isolated hydrogen valence electrons 1


For nitrogen: Isolated nitrogen valence electrons 5 (central) Bound nitrogen bonding electrons 8

Bound nitrogen nonbonding electrons 0 Formal charge = 5 - ½(8) - 0 = +1

For nitrogen: Isolated nitrogen valence electrons 5 (terminal) Bound nitrogen bonding electrons 4

Bound nitrogen nonbonding electrons 4 Formal charge = 5 - ½(4) - 4 = -1


156

For carbon: Isolated carbon valence electrons 4 Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0

(b) For hydrogen: Isolated hydrogen valence electrons 1


For nitrogen: Isolated nitrogen valence electrons 5 (central) Bound nitrogen bonding electrons 6

Bound nitrogen nonbonding electrons 2 Formal charge = 5 - ½(6) - 2 = 0

For nitrogen: Isolated nitrogen valence electrons 5 (terminal) Bound nitrogen bonding electrons 4

Bound nitrogen nonbonding electrons 4 Formal charge = 5 - ½(4) - 4 = -1


Bound carbon bonding electrons 6 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(6) - 0 = +1

Structure (a) is more important because of the octet of electrons around carbon.

7.65 For oxygen: Isolated oxygen valence electrons 6


For left carbon: Isolated carbon valence electrons 4



157

For right carbon: Isolated carbon valence electrons 4 Bound carbon bonding electrons 6 Bound carbon nonbonding electrons 2 Formal charge = 4 - ½(6) - 2 = -1



For left carbon: Isolated carbon valence electrons 4


For right carbon: Isolated carbon valence electrons 4


The second structure is more important because of the -1 formal charge on the more electronegative oxygen.

The VSEPR Model 7.66 From data in Table 7.4:

(a) trigonal planar (b) trigonal bipyramidal (c) linear (d) octahedral 7.67 From data in Table 7.4:

(a) T shaped (b) bent (c) square planar 7.68 From data in Table 7.4:

(a) tetrahedral, 4 (b) octahedral, 6 (c) bent, 3 or 4 (d) linear, 2 or 5 (e) square pyramidal, 6 (f) trigonal pyramidal, 4

7.69 From data in Table 7.4:

(a) seesaw, 5 (b) square planar, 6 (c) trigonal bipyramidal, 5 (d) T shaped, 5 (e) trigonal planar, 3 (f) linear, 2 or 5


158


Bonded Atoms Lone Pairs Shape (a) H2Se 2 2 bent (b) TiCl4 4 0 tetrahedral (c) O3 2 1 bent (d) GaH3 3 0 trigonal planar


Bonded Atoms Lone Pairs Shape (a) XeO4 4 0 tetrahedral (b) SO2Cl2 4 0 tetrahedral (c) OsO4 4 0 tetrahedral (d) SeO2 2 1 bent


Bonded Atoms Lone Pairs Shape (a) SbF5 5 0 trigonal bipyramidal (b) IF4

+ 4 1 see saw (c) SeO3

2- 3 1 trigonal pyramidal (d) CrO4

2- 4 0 tetrahedral 7.73 Number of Number of

Bonded Atoms Lone Pairs Shape (a) NO3

- 3 0 trigonal planar (b) NO2

+ 2 0 linear (c) NO2

- 2 1 bent 7.74 Number of Number of

Bonded Atoms Lone Pairs Shape (a) PO4

3- 4 0 tetrahedral (b) MnO4

- 4 0 tetrahedral (c) SO4

2- 4 0 tetrahedral (d) SO3

2- 3 1 trigonal pyramidal (e) ClO4

- 4 0 tetrahedral (f) SCN- 2 0 linear (C is the central atom)


Bonded Atoms Lone Pairs Shape (a) XeF3

+ 3 2 T shaped (b) SF3

+ 3 1 trigonal pyramidal (c) ClF2

+ 2 2 bent


159

(d) CH3+ 3 0 trigonal planar

7.76 (a) In SF2 the sulfur is bound to two fluorines and contains two lone pairs of electrons.

SF2 is bent and the F–S–F bond angle is approximately 109°. (b) In N2H2 each nitrogen is bound to the other nitrogen and one hydrogen. Each nitrogen has one lone pair of electrons. The H–N–N bond angle is approximately 120°. (c) In KrF4 the krypton is bound to four fluorines and contains two lone pairs of electrons. KrF4 is square planar, and the F–Kr–F bond angle is 90°. (d) In NOCl the nitrogen is bound to one oxygen and one chlorine and contains one lone pair of electrons. NOCl is bent, and the Cl–N–O bond angle is approximately 120°.

7.77 (a) In PCl6

- the phosphorus is bound to six chlorines. There are no lone pairs of electrons on the phosphorus. PCl6

- is octahedral, and the Cl–P–Cl bond angle is 90o. (b) In ICl2

- the iodine is bound to two chlorines and contains three lone pairs of electrons. ICl2

- is linear, and the Cl–I–Cl bond angle is 180o. (c) In SO4

2- the sulfur is bound to four oxygens. There are no lone pairs of electrons on the sulfur. SO4

2- is tetrahedral, and the O–S–O bond angle is 109.5o. (d) In BO3

3- the boron is bound to three oxygens. There are no lone pairs of electrons on the boron. BO3

3- is trigonal planar, and the O–B–O bond angle is 120o.

7.78 H – Ca – H ~ 120o Cb – Cc – N 180o H – Ca – Cb ~ 120o Ca – Cb – H ~ 120o Ca – Cb – Cc ~ 120o H – Cb – Cc ~ 120o

7.79 7.80 All six carbons in cyclohexane are bonded to two other carbons and two hydrogens (i.e.

four charge clouds). The geometry about each carbon is tetrahedral with a C–C–C bond angle of approximately 109°. Because the geometry about each carbon is tetrahedral, the cyclohexane ring cannot be flat.

7.81 All six carbon atoms are sp2 hybridized and the bond angles are ~120o. The geometry

about each carbon is trigonal planar. Hybrid Orbitals and Molecular Orbital Theory 7.82 In a π bond, the shared electrons occupy a region above and below a line connecting the

two nuclei. A σ bond has its shared electrons located along the axis between the two


160

nuclei. 7.83 Electrons in a bonding molecular orbital spend most of their time in the region between

the two nuclei, helping to bond the atoms together. Electrons in an antibonding molecular orbital cannot occupy the central region between the nuclei and cannot contribute to bonding.

7.84 See Table 7.5.

(a) sp (b) sp3d (c) sp3d2 (d) sp3 7.85 See Table 7.5.

(a) tetrahedral (b) octahedral (c) linear 7.86 See Table 7.5.

(a) sp3 (b) sp3d2 (c) sp2 or sp3 (d) sp or sp3d (e) sp3d2 7.87 See Tables 7.4 and 7.5.

(a) seesaw, 5 charge clouds, sp3d (b) square planar, 6 charge clouds, sp3d2 (c) trigonal bipyramidal, 5 charge clouds, sp3d (d) T shaped, 5 charge clouds, sp3d (e) trigonal planar, 3 charge clouds, sp2

7.88 (a) sp2 (b) sp3 (c) sp3d2 (d) sp2 7.89 (a) sp3 (b) sp2 (c) sp2 (d) sp3

7.90 The C is sp2 hybridized and the N atoms are sp3 hybridized.

7.91 (a) (b) H–C–H, ~109o; O–C–O, ~120o; H–N–H, ~107o (c) N, sp3; left C, sp3; right C, sp2

7.92 σ*2p

π*2p ↑ ↑ ↑ ↑↓ ↑ π2p ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ σ2p ↑↓ ↑↓ ↑↓


161

σ*2s ↑↓ ↑↓ ↑↓ σ2s ↑↓ ↑↓ ↑↓

O2+ O2 O2

-

Bond order = 2


ofnumber _

electrons bonding

ofnumber

2.5 = 2

3 _ 8 =order bond O+

2 ; 2 = 2

4 _ 8 =order bond O2

1.5 = 2

5 _ 8 =order bond O_

2

All are stable with bond orders between 1.5 and 2.5. All have unpaired electrons. 7.93 σ*2p

π*2p ↑ σ2p ↑ ↑↓ ↑↓ π2p ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ σ*2s ↑↓ ↑↓ ↑↓ σ2s ↑↓ ↑↓ ↑↓

N2+ N2 N2

-

Bond order = 2


ofnumber _

electrons bonding

ofnumber

2.5 = 2

2 _ 7 =order bond N+

2 ; 3 = 2

2 _ 8 =order bond N2

2.5 = 2

3 _ 8 =order bond N_

2

All are stable with bond orders of either 3 or 2.5. N2+ and N2

- contain unpaired electrons.

7.94 p orbitals in allyl cation

allyl cation showing only the σ bonds (each C is sp2 hybridized)


162

delocalized MO model for π bonding in the allyl cation

7.95 p orbitals in NO2-

NO2- showing only the σ bonds (N is sp2 hybridized)

delocalized MO model for π bonding in NO2-

General Problems

7.96 7.97 In ascorbic acid (Problem 7.96) all carbons that have only single bonds are sp3

hybridized. The three carbons that have double bonds are sp2 hybridized. 7.98 Every carbon is sp2 hybridized. There are 18 σ bonds and 5 π bonds.

7.99


163

For For oxygen: Isolated oxygen valence electrons 6 (top) Bound oxygen bonding electrons 2


For oxygen: Isolated oxygen valence electrons 6 (middle) Bound oxygen bonding electrons 4


For oxygen: Isolated oxygen valence electrons 6 (left) Bound oxygen bonding electrons 4


For oxygen: Isolated oxygen valence electrons 6 (right) Bound oxygen bonding electrons 2









Bound oxygen nonbonding electrons 6


164

Formal charge = 6 - ½(2) - 6 = -1








Bound oxygen nonbonding electrons 2 Formal charge = 6 - ½(6) - 2 = +1





For sulfur: Isolated sulfur valence electrons 6 Bound sulfur bonding electrons 8 Bound sulfur nonbonding electrons 0 Formal charge = 6 - ½(8) - 0 = +2


165

7.100

For For hydrogen: Isolated hydrogen valence electrons 1


For carbon: Isolated carbon valence electrons 4 (left) Bound carbon bonding electrons 8

Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0


Bound nitrogen bonding electrons 6 Bound nitrogen nonbonding electrons 2 Formal charge = 5 - ½(6) - 2 = 0

For carbon: Isolated carbon valence electrons 4 (right) Bound carbon bonding electrons 8




For For hydrogen: Isolated hydrogen valence electrons 1


For carbon: Isolated carbon valence electrons 4 (left) Bound carbon bonding electrons 8



166


Bound nitrogen bonding electrons 8 Bound nitrogen nonbonding electrons 0 Formal charge = 5 - ½(8) - 0 = +1

For carbon: Isolated carbon valence electrons 4 (right) Bound carbon bonding electrons 8




7.101 They are geometric isomers not resonance forms. In resonance forms the atoms have the same geometrical arrangement.

7.102 (a) For boron: Isolated boron valence electrons 3

Bound boron bonding electrons 8 Bound boron nonbonding electrons 0 Formal charge = 3 – ½(8) – 0 = -1


Bound oxygen bonding electrons 6 Bound oxygen nonbonding electrons 2 Formal charge = 6 – ½ (6) – 2 = +1

(b) In BF3 the B has three bonding pairs of electrons and no lone pairs. The B is sp2 hybridized and BF3 is trigonal planar.

is bent about the oxygen because of two bonding pairs and two lone pairs of electrons. The O is sp3 hybridized. In the product, B is sp3 hybridized (with four bonding pairs of electrons), and the geometry about it is tetrahedral. The O is also sp3 hybridized (with three bonding pairs and one lone pair of electrons), and the geometry about it is trigonal pyramidal.

7.103 Both the B and N are sp2 hybridized. All bond angles are ~120o. The overall geometry of


167

the molecule is planar. 7.104 The triply bonded carbon atoms are sp hybridized. The theoretical bond angle for C–C≡C

is 180o. Benzyne is so reactive because the C–C≡C bond angle is closer to 120o and is very strained.

7.105 (a) (b) (c)

7.106 7.107 Li2 σ*2s

σ2s ↑↓ σ*1s ↑↓ σ1s ↑↓

Li 2 Bond order = 1 = 2

2 _ 4 =

2


ofnumber _

electrons bonding

ofnumber

The bond order for Li2 is 1, and the molecule is likely to be stable. 7.108 C2

2-

σ*2p π*2p σ2p ↑↓ π2p ↑↓ ↑↓ σ*2s ↑↓ σ2s ↑↓

Bond order = 2


ofnumber _

electrons bonding

ofnumber

3 = 2

2 _ 8 =order bond C _2

2 ; there is a triple bond between the two carbons.

7.109 (a) (b) (c)

Structure (a) is different from structures (b) and (c) because both chlorines are on the same carbon. Structures (b) and (c) are different because in (b) both chlorines are on the same side of the molecule (“cis”) and in (c) they are on opposite sides of the molecule


168

(“trans”). There is no rotation around the carbon-carbon double bond. 7.110 CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)

Energy change = D (Reactant bonds) - D (Product bonds) Energy change = [4 DC-H + DCl-Cl] - [3 DC-H + DC-Cl + DH-Cl] Energy change = [(4 mol)(410 kJ/mol) + (1 mol)(243 kJ/mol] - [(3 mol)(410 kJ/mol) + (1 mol)(330 kJ/mol) + (1 mol)(432 kJ/mol)] = -109 kJ

7.111

7.112 (a) (b) (c)

(d) (e) (f)

(g) (h)

Structures (a) – (d) make more important contributions to the resonance hybrid because of only -1 and 0 formal charges on the oxygens.

7.113 (a) (1) (2) (3) (b) Structure (1) makes the greatest contribution to the resonance hybrid because of the -1 formal charge on the oxygen. Structure (3) makes the least contribution to the resonance hybrid because of the +1 formal charge on the oxygen. (c) and (d) OCN- is linear because the C has 2 charge clouds. It is sp hybridized in all three resonance structures. It forms two π bonds.


169

7.114

21 σ bonds 5 π bonds Each C with a double bond is sp2 hybridized. The –CH3 carbon is sp3 hybridized.

7.115 7.116 (a) σ*3p

π*3p ↑ ↑ ↑↓ ↑↓ π3p ↑↓ ↑↓ ↑↓ ↑↓ σ3p ↑↓ ↑↓ σ*3s ↑↓ ↑↓ σ3s ↑↓ ↑↓

S2 S22-

(b) S2 would be paramagnetic with two unpaired electrons in the π*3p MOs.

(c) Bond order = 2


ofnumber _

electrons bonding

ofnumber

2 = 2

4 _ 8 =order bond S2

(d) 1 = 2

6 _ 8 =order bond S _2

2

The two added electrons go into the antibonding π*3p MOs, the bond order drops from 2 to 1, and the bond length in S2

2- should be longer than the bond length in S2.


170

7.117 (a) CO σ*2p π*2p σ2p ↑↓ π2p ↑↓ ↑↓ σ*2s ↑↓ σ2s ↑↓

(b) All electrons are paired, CO is diamagnetic.

(c)

Bond order = 2


ofnumber _

electrons bonding

ofnumber

3 = 2

2 _ 8 =order bond CO

The bond order here matches that predicted by the electron-dot structure ( ).

(d)

7.118 (a) The left S has 5 electron clouds (4 bonding, 1 lone pair). The S is sp3d hybridized and the geometry about this S is seesaw. The right S has 4 electron clouds (2 bonding, 2 lone pairs). The S is sp3 hybridized and the geometry about this S is bent.

(b) The left C has 4 electron clouds (4 bonding, 0 lone pairs). This C is sp3 hybridized and its geometry is tetrahedral. The right C has 3 electron clouds (3 bonding, 0 lone pairs). This C is sp2 hybridized and its geometry is trigonal planar. The central two C’s have 2 electron clouds (2 bonding, 0 lone pairs). These C’s are sp hybridized and the geometry about both is linear.


171

7.119 Multi-Concept Problems

7.120 (a) (b) The oxygen in OH has a half-filled 2p orbital that can accept the additional electron. For a 2p orbital, n = 2 and l = 1. (c) The electron affinity for OH is slightly more negative than for an O atom because when OH gains an additional electron, it achieves an octet configuration.

7.121 (a) (4 orbitals)(3 electrons) = 12 outer-shell electrons

(b) 3 electrons

(c) 1s3 2s3 2p6;

(d) (e) σ*2p

π*2p ↑↓ ↑ π2p ↑↓↑ ↑↓↑ σ2p ↑↓↑ σ*2s ↑↓↑ σ2s ↑↓↑

Bond order = 3


ofnumber _

electrons bonding

ofnumber

2 = 3

6 _ 12 =order bond X2


172

7.122 (a) (b) Each Cr atom has 6 pairs of electrons around it. The likely geometry about each Cr atom is tetrahedral because each Cr has 4 charge clouds.

7.123 (a) XOCl2 + 2 H2O → 2 HCl + H2XO3

(b) 96.1 mL = 0.0961 L mol NaOH = (0.1225 mol/L)(0.0961 L) = 0.01177 mol NaOH

mol H+ = 0.01177 mol NaOH x NaOH mol 1

H mol 1 +

= 0.01177 mol H+

Of the total H+ concentration, half comes from HCl and half comes from H2XO3.

mol H2XO3 = 2

H mol 0.01177 +

x H mol 2XOH mol 1

+

32 = 2.943 x 10-3 mol H2XO3

mol XOCl2 = 2.943 x 10-3 mol H2XO3 x XOH mol 1

XOCl mol 1

32

2 = 2.943 x 10-3 mol XOCl2

molar mass XOCl2 = XOCl mol 10 x 2.943

XOCl g 0.350

23_

2 = 118.9 g/mol

molecular mass of XOCl2 = 118.9 amu atomic mass of X = 118.9 amu - 16.0 amu - 2(35.45 amu) = 32.0 amu: X = S

(c)

(d) trigonal pyramidal

7.124 (a)

(b) All three molecules are planar. The first two structures are polar because they both have an unsymmetrical distribution of atoms about the center of the molecule (the middle of the double bond), and bond polarities do not cancel. Structure 3 is nonpolar because the H’s and Cl’s, respectively, are symmetrically distributed about the center of the molecule, both being opposite each other. In this arrangement, bond polarities cancel. (c) 200 nm = 200 x 10-9 m

mol) / 10 x (6.022 m 10 x 200

s) / m 10 x (3.00 s) J 10 x (6.626 =

ch = E 23

9_

834_ •λ

E = 5.99 x 105 J/mol = 599 kJ/mol


173

(d) The π bond must be broken before rotation can occur.

173

8

Thermochemistry: Chemical Energy

8.1 Convert lb to kg. kg 1043 = g 1000

kg 1 x

lb 1

g 453.59 x lb 2300

Convert mi/h to m/s. m/s 24.6 = s 3600

h 1 x

km1

m 1000 x

mi 0.62137

km 1 x

h

mi 55

1 kg⋅ m2/s2 = 1 J; E = ½mv2 = ½(1043 kg)(24.6 m/s)2 = 3.2 x 105 J

E = 3.2 x 105 J x kJ 10 x 3.2 = J 1000

kJ 1 2

8.2 (a) and (b) are state functions; (c) is not. 8.3 ∆V = (4.3 L - 8.6 L) = - 4.3 L

w = -P∆V = -(44 atm)( - 4.3 L) = +189.2 L⋅ atm

w = (189.2 L⋅ atm)(101 atm L

J

•) = +1.9 x 104 J

The positive sign for the work indicates that the surroundings does work on the system. Energy flows into the system.

8.4 w = -P∆V = - (2.5 atm)(3 L - 2 L) = - 2.5 L⋅atm

w = (-2.5 L⋅atm)

• atm L

J 101 = -252.5 J = -250 J = -0.25 kJ

The negative sign indicates that the expanding system loses work energy and does work on the surroundings.

8.5 (a) w = -P∆V is positive and P∆V is negative for this reaction because the system

volume is decreased at constant pressure. (b) P∆V is small compared to ∆E. ∆H = ∆E + P∆V; ∆H is negative. Its value is slightly more negative than ∆E.

8.6 ∆Ho = - 484 H mol 2

kJ

2

P∆V = (1.00 atm)(-5.6 L) = -5.6 L⋅ atm

P∆V = (-5.6 L⋅ atm)(101 atm L

J

•) = -565.6 J = -570 J = -0.57 kJ

w = -P∆V = 570 J = 0.57 kJ

∆H = H mol 0.50

kJ 121 _

2

∆E = ∆H - P∆V = -121 kJ - (-0.57 kJ) = -120.43 kJ = -120 kJ 8.7 ∆V = 448 L and assume P = 1.00 atm

Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________

174

w = -P∆V = -(1.00 atm)(448 L) = - 448 L⋅ atm

w = -(448 L⋅ atm)(101 atm L

J

• ) = - 4.52 x 104 J

w = - 4.52 x 104 J x J 1000

kJ 1 = - 45.2 kJ

8.8 (a) C3H8, 44.10 amu; ∆Ho = -2219 kJ/mol C3H8

15.5 g x HC mol 1

kJ 2219- x

HC g 44.10HC mol 1

8383

83 = -780. kJ

780. kJ of heat is evolved.

(b) Ba(OH)2 ⋅ 8 H2O, 315.5 amu; ∆Ho = +80.3 kJ/mol Ba(OH)2 ⋅ 8 H2O

4.88 g x OH 8 )Ba(OH mol 1

kJ 80.3 x

OH 8 )Ba(OH g 315.5

OH 8 )Ba(OH mol 1

2222

22

•••

= +1.24 kJ

1.24 kJ of heat is absorbed. 8.9 CH3NO2, 61.04 amu

q = = NOCH mol 4

kJ 2441.6 x

NOCH g 61.04NOCH mol 1

x NOCH g 100.02323

2323 1.000 x 103 kJ

8.10 q = (specific heat) x m x ∆T = (4.18 C g

Jo•

)(350 g)(3oC - 25oC) = -3.2 x 104 J

q = -3.2 x 104 J x J 1000

kJ 1 = -32 kJ

8.11 q = (specific heat) x m x ∆T; specific heat = C) g)(10 (75

J 96 =

T x m

qo∆

= 0.13 J/(g ⋅ oC)

8.12 25.0 mL = 0.0250 L and 50.0 mL = 0.0500 L

mol H2SO4 = (1.00 mol/L)(0.0250 L) = 0.0250 mol H2SO4 mol NaOH = (1.00 mol/L)(0.0500 L) = 0.0500 mol NaOH NaOH and H2SO4 are present in a 2:1 mol ratio. This matches the stoichiometric ratio in the balanced equation. q = (specific heat) x m x ∆T m = (25.0 mL + 50.0 mL)(1.00 g/mL) = 75.0 g

J 2790 = C) 25.0 _ C g)(33.9 )(75.0C g

J (4.18 = q oo

o•

mol H2SO4 = 0.0250 L x 1.00 L

mol H2SO4 = 0.0250 mol H2SO4

Heat evolved per mole of H2SO4 SOH mol / J 10 x 1.1 = SOH mol 0.0250

J 10 x 2.79 = 42

5

42

3

Because the reaction evolves heat, the sign for ∆H is negative.


175

∆H = -1.1 x 105 J x J 1000

kJ 1 = -1.1 x 102 kJ

8.13 CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ∆Ho

1 = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ∆Ho

2 = -104 kJ Sum CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)

∆Ho = ∆Ho1 + ∆Ho

2 = -202 kJ 8.14 (a) A + 2 B → D; ∆Ho = -100 kJ + (-50 kJ) = -150 kJ

(b) The red arrow corresponds to step 1: A + B → C The green arrow corresponds to step 2: C + B → D The blue arrow corresponds to the overall reaction. (c) The top energy level represents A + 2 B. The middle energy level represents C + B. The bottom energy level represents D.

8.15 8.16 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

∆Horxn = [4 ∆Ho

f (NO) + 6 ∆Hof (H2O)] - [4 ∆Ho

f (NH3)] ∆Ho

rxn = [(4 mol)(90.2 kJ/mol) + (6 mol)(- 241.8 kJ/mol)] - [(4 mol)(- 46.1 kJ/mol)] ∆Ho

rxn = -905.6 kJ 8.17 6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g)

∆Horxn = ∆Ho

f(C6H12O6) - [6 ∆Hof(CO2) + 6 ∆Ho

f(H2O(l))] ∆Ho

rxn = [(1 mol)(-1260 kJ/mol)] - [(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] ∆Ho

rxn = +2815.8 kJ = +2816 kJ 8.18 H2C=CH2(g) + H2O(g) → C2H5OH(g)

∆Horxn = D (Reactant bonds) - D (Product bonds)

∆Horxn = (DC=C + 4 DC-H + 2 DO-H) - (DC-C + DC-O + 5 DC-H + DO-H)

∆Horxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol) + (2 mol)(460 kJ/mol)]

- [(1 mol)(350 kJ/mol) + (1 mol)( 350 kJ/mol) + (5 mol)(410 kJ/mol) + (1 mol)(460 kJ/mol)] ∆Ho

rxn = -39 kJ 8.19 2 NH3(g) + Cl2(g) → N2H4(g) + 2 HCl(g)


∆Horxn = (6 DN-H + DCl-Cl) - (DN-N + 4 DN-H + 2 DH-Cl)

∆Horxn = [(6 mol)(390 kJ/mol) + (1 mol)(243 kJ/mol)]


176

- [(1 mol)(240 kJ/mol) + (4 mol)(390 kJ/mol) + (2 mol)(432 kJ/mol)] = -81 kJ

8.20 C4H10(l) + 2

13 O2(g) → 4 CO2(g) + 5 H2O(g)

∆Horxn = [4 ∆Ho

f (CO2) + 5 ∆Hof (H2O)] - ∆Ho

f (C4H10) ∆Ho

rxn = [(4 mol)(-393.5 kJ/mol) + (5 mol)(-241.8 kJ/mol)] - [(1 mol)(-147.5 kJ/mol)] ∆Ho

rxn = -2635.5 kJ ∆Ho

C = -2635.5 kJ/mol

C4H10, 58.12 amu; ∆HoC = g / kJ 45.35 _ =

g 58.12

mol 1

mol

kJ 2635.5_

∆HoC = mL / kJ 26.3 _ =

mL

g 0.579

g

kJ 45.35 _

8.21 ∆So is negative because the reaction decreases the number of moles of gaseous molecules. 8.22 The reaction proceeds from a solid and a gas (reactants) to all gas (product). This is more

disordered and ∆So is positive. 8.23 (a) Because ∆Go is negative, the reaction is spontaneous.

(b) Because ∆Go is positive, the reaction is nonspontaneous. 8.24 ∆Go = ∆Ho - T∆So = (-92.2 kJ) - (298 K)(-0.199 kJ/K) = -32.9 kJ

Because ∆Go is negative, the reaction is spontaneous. Set ∆Go = 0 and solve for T. ∆Go = 0 = ∆Ho - T∆So

T = kJ/K 0.199_

kJ 92.2_ =

S

Ho

o

∆∆

= 463 K = 190oC

8.25 (a) 2 A2 + B2 → 2 A2B

(b) Because the reaction is exothermic, ∆H is negative. There are more reactant molecules than product molecules. The randomness of the system decreases on going from reactant to product, therefore ∆S is negative. (c) Because ∆G = ∆H - T∆S, a reaction with both ∆H and ∆S negative is favored at low temperatures where the negative ∆H term is larger than the positive - T∆S, and ∆G is negative.

Understanding Key Concepts 8.26. (a) w = -P∆V, ∆V > 0; therefore w < 0 and the system is doing work on the

surroundings. (b) Since the temperature has increased there has been an enthalpy change. The system evolved heat, the reaction is exothermic, and ∆H < 0.


177

8.27

8.28

8.29 8.30 VP + E = H ∆∆∆

VP = E _ H ∆∆∆

∆V = kJ 10 x 101

atm L 1 x

atm 1

kJ)] 34.8(_ _ kJ 35.0[_ =

P

E _ H3_

•∆∆= -2 L

∆V = -2 L = Vfinal - Vinitial = Vfinal - 5 L; Vfinal = -2 L - (-5 L) = 3 L The volume decreases from 5 L to 3 L.

8.31 ∆Ho = +55 kJ

∆So is positive because the chemical system becomes more disordered in going from


178

reactant to products. ∆Go = ∆Ho - T∆So; For the reaction to be spontaneous, ∆Go must be negative. Because ∆Ho and ∆So are both positive, the reaction is spontaneous at some higher temperatures but nonspontaneous at some lower temperatures.

8.32 The change is the spontaneous conversion of a liquid to a gas. ∆G is negative because the

change is spontaneous. The conversion of a liquid to a gas is endothermic, therefore ∆H is positive. ∆S is positive because the gas is more disordered than the liquid.

8.33 (a) 2 A3 → 3 A2

(b) Because the reaction is spontaneous, ∆G is negative. ∆S is positive because the number of molecules increases in going from reactant to products. ∆H could be either positive or negative and the reaction would still be spontaneous. ∆H is probably positive because there is more bond breaking than bond making

Additional Problems Heat, Work, and Energy 8.34 Heat is the energy transferred from one object to another as the result of a temperature

difference between them. Temperature is a measure of the kinetic energy of molecular motion. Energy is the capacity to do work or supply heat. Work is defined as the distance moved times the force that opposes the motion (w = d x F). Kinetic energy is the energy of motion. Potential energy is stored energy.

8.35 Internal energy is the sum of kinetic and potential energies for each particle in the system.

8.36 Car: EK = ½(1400 kg)

s 3600

m 10 x 115 3 2

= 7.1 x 105 J

Truck: EK = ½(12,000 kg)

s 3600

m 10 x 38 3 2

= 6.7 x 105 J

The car has more kinetic energy. 8.37 Heat = q = 7.1 x 105 J (from Problem 8.34)

q = (specific heat) x m x ∆T

m = waterof g 10 x 5.7 =

C) 20 _ C (50 C g

J 4.18

J 10 x 7.1 =

T x heat) (specific

q 3

ooo

5

•∆

8.38 w = -P∆V = -(3.6 atm)(3.4 L - 3.2 L) = -0.72 L ⋅ atm


179

w = (-0.72 L ⋅ atm)

• atm L 1

J 101 = -72.7 J = -70 J; The energy change is negative.

8.39 Vinitial = 50.0 mL + 50 mL = 100.0 mL = 0.1000 L

Vfinal = 50.0 mL = 0.0500 L ∆V = Vfinal - Vinitial = (0.0500 L - 0.1000 L) = -0.0500 L w = -P∆V = -(1.5 atm)(-0.0500 L) = +0.075 L⋅ atm

w = (+0.075 L ⋅ atm)

• atm L 1

J 101 = +7.6 J

The positive sign for the work indicates that the surroundings does work on the system. Energy flows into the system.

Energy and Enthalpy 8.40 ∆E = qv is the heat change associated with a reaction at constant volume. Since ∆V = 0,

no PV work is done. ∆H = qp is the heat change associated with a reaction at constant pressure. Since ∆V ≠ 0, PV work can also be done.

8.41 ∆H is negative for an exothermic reaction. ∆H is positive for an endothermic reaction. 8.42 ∆H = ∆E + P∆V; ∆H and ∆E are nearly equal when there are no gases involved in a

chemical reaction, or, if gases are involved, ∆V = 0 (that is, there are the same number of reactant and product gas molecules).

8.43 Heat is lost on going from H2O(g) → H2O(l) → H2O(s).

H2O(g) has the highest enthalpy content. H2O(s) has the lowest enthalpy content. 8.44 P∆V = -7.6 J (from Problem 8.39)

∆H = ∆E + P∆V ∆E = ∆H - P∆V = -0.31 kJ - (- 7.6 x 10-3 kJ) = -0.30 kJ

8.45 ∆H = -244 kJ and w = -P∆V = 35 kJ; therefore P∆V = -35 kJ

∆E = ∆H - P∆V = -244 kJ - (-35 kJ) = -209 kJ For the system: ∆H = -244 kJ and ∆E = -209 kJ ∆H and ∆E for the surroundings are just the opposite of what they are for the system. For the surroundings: ∆H = 244 kJ and ∆E = 209 kJ

8.46 ∆H = -1255.5 kJ/mol C2H2; C2H2, 26.04 amu

w = -P∆V = -(1.00 atm)(-2.80 L) = 2.80 L⋅ atm

w = (2.80 L⋅ atm)

• atm L 1

J 101 = 283 J = 0.283 kJ

6.50 g x HC g 26.04

HC mol 1

22

22 = 0.250 mol C2H2


180

q = (-1255.5 kJ/mol)(0.250 mol) = -314 kJ ∆E = ∆H - P∆V = -314 kJ - (-0.283 kJ) = -314 kJ

8.47 C2H4, 28.05 amu; HCl, 36.46 amu

w = -P∆V = -(1.00 atm)(-71.5 L) = 71.5 L⋅ atm

w = (71.5 L⋅ atm)

• atm L 1

J 101 = 7222 J = 7.22 kJ

89.5 g C2H4 x HC g 28.05

HC mol 1

42

42 = 3.19 mol C2H4

125 g HCl x HCl g 36.46


Because the reaction stoichiometry between C2H4 and HCl is one to one, C2H4 is the limiting reactant. ∆Ho = -72.3 kJ/mol C2H4 q = (-72.3 kJ/mol)(3.19 mol) = -231 kJ ∆E = ∆H - P∆V = -231 kJ - (-7.22 kJ) = -224 kJ

8.48 C4H10O, 74.12 amu; mass of C4H10O = g 71.38 = mL) g/mL)(100 (0.7138

mol C4H10O = mol 0.9626 = g 74.12

mol 1 x g 71.38

q = n x ∆Hvap = 0.9626 mol x 26.5 kJ/mol = 25.5 kJ 8.49 Assume 100 mL of H2O = 100 g; H2O, 18.02 amu

100 g x OH mol 1

kJ 40.7 x

OH g 18.02

OH mol 1

22

2 = 226 kJ

The heat to vaporize 100 mL of H2O is much greater than that to vaporize 100 mL of diethyl ether.

8.50 Al, 26.98 amu

mol Al = 5.00 g x mol 0.1853 = g 26.98

mol 1

q = n x ∆Ho = 0.1853 mol Al x kJ 131_ = Al mol 2

kJ 1408.4_; 131 kJ is released.

8.51 Na, 22.99 amu; ∆Ho = -368.4 kJ/2 mol Na = -184.2 kJ/mol Na

1.00 g Na x Na mol 1

kJ 184.2_ x

Na g 22.99

Na mol 1 = -8.01 kJ

8.01 kJ of heat is evolved. The reaction is exothermic. 8.52 Fe2O3, 159.7 amu

mol Fe2O3 = 2.50 g x mol 65 0.015 = g 159.7

mol 1


181

q = n x ∆Ho = 0.015 65 mol Fe2O3 x kJ 0.388_ = OFe mol 1

kJ 24.8_

32

; 0.388 kJ is evolved.

Because ∆H is negative, the reaction is exothermic. 8.53 CaO, 56.08 amu

mol CaO = 233.0 g x mol 4.155 = g 56.08

mol 1

q = n x ∆Ho = 4.155 mol CaO x kJ 1931 = CaO mol 1

kJ 464.8; 1931 kJ is absorbed.

Because ∆H is positive, the reaction is endothermic. Calorimetry and Heat Capacity 8.54 Heat capacity is the amount of heat required to raise the temperature of a substance a

given amount. Specific heat is the amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1oC.

8.55 A measurement carried out in a bomb calorimeter is done at constant volume and

therefore ∆E is obtained. 8.56 Na, 22.99 amu

specific heat = 28.2 C) g( / J 1.23 = g 22.99

mol 1 x

C mol

J oo

••

8.57 q = (specific heat) x m x ∆T

specific heat = C) g)(5.20 (33.0

J 89.7 =

T x m

qo∆

= 0.523 J/(g ⋅ oC)

Cm = [0.523 J/(g ⋅ oC)](47.88 g/mol) = 25.0 J/(mol ⋅ oC) 8.58 Mass of solution = 50.0 g + 1.045 g = 51.0 g


kJ 1.56 = J 10 x 1.56 = C) 25.0 _ C g)(32.3 (51.0C g

J 4.18 = q 3oo

o

•

CaO, 56.08 amu; mol CaO = 1.045 g x mol 63 0.018 = g 56.08

mol 1

Heat evolved per mole of CaO CaO mol / kJ 83.7 = mol 63 0.018

kJ 1.56 =

Because the reaction evolves heat, the sign for ∆H is negative. ∆H = -83.7 kJ 8.59 C6H6, 78.11 amu; 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)


182

∆E = qV = q _ OH2= - C)48g)(7. (250.0

C g

J 4.18 o

o

•= -7817 J = -7.82 kJ

0.187 g C6H6 x HC g 78.11

HC mol 1

66

66 = 0.002 39 mol C6H6

∆E(per mole) = (-7.82 kJ)/(0.002 39 mol) = -3.27 x 103 kJ/mol ∆E (per gram C6H6) = (-3.27 x 103 kJ/mol)/(78.11 g/mol) = - 41.9 kJ/g

8.60 NaOH, 40.00 amu; HCl, 36.46 amu

8.00 g NaOH x NaOH g 40.00

NaOH mol 1 = 0.200 mol NaOH

8.00 g HCl x HCl g 36.46


Because the reaction stoichiometry between NaOH and HCl is one to one, the NaOH is the limiting reactant.

qP = -qsoln = -(specific heat) x m x ∆T = - C)025. _ C5g)(33. (316C g

J 4.18 oo

o

• = -11.2 kJ

∆H = qp/n = (-11.2 kJ)/(0.200 mol) = -56 kJ/mol When 10.00 g of HCl in 248.0 g of water is added the same temperature increase is observed because the mass of NaOH is the same and it is still the limiting reactant. The mass of the solution is also the same.

8.61 NH4NO3, 80.04 amu; assume 125 mL = 125 g H2O

50.0 g NH4NO3 x NONH g 80.04

NONH mol 1

34

34 = 0.625 mol NH4NO3

qp = ∆H x n = (+25.7 kJ/mol)(0.625 mol) = 16.1 kJ = 16,100 J qsoln = -qp = -16,100 J qsoln = (specific heat) x m x ∆T

∆T =

g) 125 + g (50C g

J 4.18

J 16,100_ =

m x heat) (specific

q

o

soln

•

= -22.0oC

∆T = -22.0oC = Tfinal - Tinitial = Tfinal - 25.0oC Tfinal = -22.0oC + 25.0oC = 3.0oC

Hess's Law and Heats of Formation 8.62 The standard state of an element is its most stable form at 1 atm and the specified

temperature, usually 25oC. 8.63 A compound’s standard heat of formation is the amount of heat associated with the

formation of 1 mole of a compound from its elements (in their standard states). 8.64 Hess’s Law – the overall enthalpy change for a reaction is equal to the sum of the

enthalpy changes for the individual steps in the reaction.


183

Hess’s Law works because of the law of conservation of energy. 8.65 Elements always have ∆Ho

f = 0 because the standard state of elements is the reference point from which all enthalpy changes are measured.

8.66 S(s) + O2(g) → SO2(g) ∆Ho

1 = -296.8 kJ SO2 + ½ O2(g) → SO3(g) ∆Ho

2 = -98.9 kJ Sum S(s) + 3/2 O2(g) → SO3(g) ∆Ho

3 = ∆Ho1 + ∆Ho

2 ∆Ho

f = ∆Ho3 = -296.8 kJ + (-98.9 kJ) = -395.7 kJ/mol

8.67 ∆Ho

rxn = [12 ∆Hof(CO2) + 6 ∆Ho

f(H2O)] - [2 ∆Hof(C6H6)]

-6534 kJ = [(12 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] - [(2 mol)(∆Hof(C6H6))]

Solve for ∆Hof(C6H6).

-6534 kJ = -6436.8 kJ - [(2 mol)(∆Hof(C6H6))]

97.2 kJ = (2 mol)(∆Hof(C6H6))

∆Hof(C6H6) = +48.6 kJ/mol

8.68 SO3(g) + H2O(l) → H2SO4(aq) ∆Ho

1 = -227.8 kJ H2(g) + ½ O2(g) → H2O(l) ∆Ho

2 = ∆Hof = -285.8 kJ

S(s) + 3/2 O2(g) → SO3(g) ∆Ho3 = ∆Ho

f = -395.7 Sum S(s) + H2(g) + 2 O2(g) → H2SO4(aq) ∆Ho

f (H2SO4) = ? ∆Ho

f (H2SO4) = ∆Ho1 + ∆Ho

2 + ∆Ho3 = -909.3 kJ

8.69 ∆Ho

rxn = [∆Hof(CH3CO2H) + ∆Ho

f(H2O)] - ∆Hof(CH3CH2OH)

∆Horxn = [(1 mol)(-484.5 kJ/mol) + (1mol)(-285.8 kJ/mol)] - [(1 mol)(-277.7 kJ/mol)]

∆Horxn = - 492.6 kJ

8.70 C8H8(l) + 10 O2(g) → 8 CO2(g) + 4 H2O(l)

∆Horxn = ∆Ho

c = - 4395.2 kJ ∆Ho

rxn = [8 ∆Hof(CO2) + 4 ∆Ho

f(H2O)] - ∆Hof(C8H8)

- 4395.2 kJ = [(8 mol)(-393.5 kJ/mol) + (4 mol)(-285.8 kJ/mol)]- [(1 mol)(∆Hof(C8H8))]

Solve for ∆Hof(C8H8)

- 4395.2 kJ = - 4291.2 kJ - (1 mol)(∆Hof(C8H8))

-104.0 kJ = -(1 mol)(∆Hof(C8H8))

∆Hof(C8H8) = mol / kJ 104.0 + =

mol 1_

kJ 104.0_

8.71 C5H12O(l) + 15/2 O2(g) → 5 CO2(g) + 6 H2O(l)

∆Horxn = [5 ∆Ho

f(CO2) + 6 ∆Hof(H2O)] - ∆Ho

f(C5H12O) ∆Ho

rxn = [(5 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] - [(1 mol)(-313.6 kJ/mol)] ∆Ho

rxn = -3369 kJ 8.72 ∆Ho

rxn = ∆Hof(MTBE) - [∆Ho

f(2-Methylpropene) + ∆Hof(CH3OH)]

-57.8 kJ = -313.6 kJ - [(1 mol)(∆Hof(2-Methylpropene)) + (-238.7 kJ)]


184

Solve for ∆Hof(2-Methylpropene).

-17.1 kJ = (1 mol)(∆Hof(2-Methylpropene))

∆Hof(2-Methylpropene) = -17.1 kJ/mol

8.73 C51H88O6(l) + 70 O2(g) → 51 CO2(g) + 44 H2O(l)

∆Horxn = [51 ∆Ho

f(CO2) + 44 ∆Hof(H2O)] - ∆Ho

f(C51H88O6) ∆Ho

rxn = [(51 mol)(-393.5 kJ/mol) + (44 mol)(-285.8 kJ/mol)] - [(1 mol)(-1310 kJ/mol)] ∆Ho

rxn = -3.133 x 104 kJ/mol C51H88O6 C51H88O6, 797.25 amu

q = -3.133 x 104 mL / kJ 37_ = mL

g 0.94 x

g 797.25

mol 1 x

mol

kJ; 37 kJ released per mL

Bond Dissociation Energies 8.74 H2C=CH2(g) + H2(g) → CH3CH3(g)


∆Horxn = (DC=C + 4 DC-H+ DH-H) - (6 DC-H + DC-C)

∆Horxn = [(1 mol)(611 kJ/mol) + (4 mol)(410 kJ/mol) + (1 mol)(436 kJ/mol)]

- [(6 mol)(410 kJ/mol) + (1 mol)(350 kJ/mol)] = -123 kJ 8.75 CH3CH=CH2 + H2O → CH3CH(OH)CH3


∆Horxn = (DC=C + DC-C + 6 DC-H + 2 DO-H) - (2 DC-C + 7 DC-H + DC-O + DO-H)

∆Horxn = [(1 mol)(611 kJ/mol) + (1 mol)(350 kJ/mol) + (6 mol)(410 kJ/mol)

+ (2 mol)(460 kJ/mol)] - [(2 mol)(350 kJ/mol) + (7 mol)(410 kJ/mol) + (1 mol)(350 kJ/mol) + (1 mol)(460 kJ/mol)] = -39 kJ

8.76 C4H10 + 13/2 O2 → 4 CO2 + 5 H2O


∆Horxn = (3 DC-C + 10 DC-H + 13/2 DO=O) - (8 DC=O + 10 DO-H)

∆Horxn = [(3 mol)(350 kJ/mol) + (10 mol)(410 kJ/mol) + (13/2 mol)(498 kJ/mol)]

- [(8 mol)(804 kJ/mol) + (10 mol)(460 kJ/mol)] = -2645 kJ 8.77 CH3CO2H + CH3CH2OH → CH3CO2CH2CH3 + H2O


∆Horxn = (DC=O + 2 DC-O + 8 DC-H + 2 DO-H) - (DC=O + 2 DC-O + 8 DC-H + 2 DO-H) = 0 kJ

Free Energy and Entropy 8.78 Entropy is a measure of molecular disorder. 8.79 ∆G = ∆H - T∆S

∆H is usually more important because it is usually much larger than T∆S. 8.80 A reaction can be spontaneous yet endothermic if ∆S is positive (more disorder) and the

T∆S term is larger than ∆H.


185

8.81 A reaction can be nonspontaneous yet exothermic if ∆S is negative (more order) and the

temperature is high enough so that the T∆S term is more negative than ∆H. 8.82 (a) positive (more disorder) (b) negative (more order) 8.83 (a) positive (more disorder) (b) negative (more order)

(c) positive (more disorder) 8.84 (a) zero (equilibrium) (b) zero (equilibrium)

(c) negative (spontaneous) 8.85 Because the mixing of gas molecules is spontaneous, ∆G is negative. The mixture of gas

molecules is more disordered so ∆S is positive. For the diffusion of gases, ∆H is approximately zero.

8.86 ∆S is positive. The reaction increases the total number of molecules. 8.87 ∆S < 0. The reaction decreases the number of gas molecules. 8.88 ∆G = ∆H - T∆S

(a) ∆G = - 48 kJ - (400 K)(135 x 10-3 kJ/K) = -102 kJ ∆G < 0, spontaneous; ∆H < 0, exothermic. (b) ∆G = - 48 kJ - (400 K)(-135 x 10-3 kJ/K) = +6 kJ ∆G > 0, nonspontaneous; ∆H < 0, exothermic. (c) ∆G = +48 kJ - (400 K)(135 x 10-3 kJ/K) = -6 kJ ∆G < 0, spontaneous; ∆H > 0, endothermic. (d) ∆G = +48 kJ - (400 K)(-135 x 10-3 kJ/K) = +102 kJ ∆G > 0, nonspontaneous; ∆H > 0, endothermic.

8.89 ∆G = ∆H - T∆S

(a) ∆G = -128 kJ - (500 K)(35 x 10-3 kJ/K) = -146 kJ ∆G < 0, spontaneous; ∆H < 0, exothermic (b) ∆G = +67 kJ - (250 K)(-140 x 10-3 kJ/K) = +102 kJ ∆G > 0, nonspontaneous; ∆H > 0, endothermic (c) ∆G = +75 kJ - (800 K)(95 x 10-3 kJ/K) = -1 kJ ∆G < 0, spontaneous; ∆H > 0, endothermic

8.90 ∆G = ∆H - T∆S; Set ∆G = 0 and solve for T (the crossover temperature).

T = S

H

∆∆

= K / kJ 0.058_

kJ 33_ = 570 K

8.91 Because ∆H > 0 and ∆S < 0, the reaction is nonspontaneous at all temperatures. There is

no crossover temperature.


186

8.92 (a) ∆H < 0 and ∆S > 0; reaction is spontaneous at all temperatures.

(b) ∆H < 0 and ∆S < 0; reaction has a crossover temperature. (c) ∆H > 0 and ∆S > 0; reaction has a crossover temperature. (d) ∆H > 0 and ∆S < 0; reaction is nonspontaneous at all temperatures.

8.93 (a) ∆H < 0 and ∆S < 0. The reaction is favored by enthalpy but not by entropy.

∆Go = ∆Ho - T∆So = -217.5 kJ/mol - (298 K)[-233.9 x 10-3 kJ/(K ⋅ mol)] = -147.8 kJ (b) The reaction has a crossover temperature. Set ∆G = 0 and solve for T (the crossover temperature). ∆Go = 0 = ∆Ho - T∆So

T = S

Ho

o

∆∆

= mol) kJ/(K 10 x 233.9

kJ/mol 217.53_ •

= 929.9 K

8.94 T = -114.1oC = 273.15 + (-114.1) = 159.0 K

∆Gfus = ∆Hfus - T∆Sfus; ∆G = 0 at the melting point temperature. Set ∆G = 0 and solve for ∆Sfus. ∆G = 0 = ∆Hfus - T∆Sfus

∆Sfus = K 159.0

kJ/mol 5.02 =

THfus∆

= 0.0316 kJ/(K⋅mol) = 31.6 J/(K⋅mol)

8.95 T = 61.2oC = 273.15 + (61.2) = 334.4 K

∆Gvap = ∆Hvap - T∆Svap; ∆G = 0 at the boiling point temperature. Set ∆G = 0 and solve for ∆Svap. ∆G = 0 = ∆Hvap - T∆Svap

∆Svap = K 334.4

kJ/mol 29.2 =

THvap∆

= 0.0873 kJ/(K⋅mol) = 87.3 J/(K⋅mol)

General Problems 8.96 Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)

mol Mg = 1.50 g x Mg mol 0.0617 = g 24.3

mol 1

mol HCl = 0.200 L x 6.00 HCl mol 1.20 = L

mol

There is an excess of HCl. Mg is the limiting reactant.

C) 25.0 _ C (42.9C

J 776 + ) 25.0 _ C g)(42.9 (200

C g

J 4.18 = q oo

ooCo

o

•=

2.89 x 104 J

q = 2.89 x 104 J x kJ 28.9 = J 1000

kJ 1

Heat evolved per mole of Mg mol / kJ 468 = mol 0.0617

kJ 28.9 =


187

Because the reaction evolves heat, the sign for ∆H is negative. ∆H = - 468 kJ 8.97 (a) C(s) + CO2(g) → 2 CO(g)

∆Horxn = [2 ∆Ho

f(CO)] - ∆Hof(CO2)

∆Horxn = [(2 mol)(-110.5 kJ/mol)] - [(1 mol)(-393.5 kJ/mol)] = +172.5 kJ

(b) 2 H2O2(aq) → 2 H2O(l) + O2(g) ∆Ho

rxn = [2 ∆Hof(H2O)] - [2 ∆Ho

f(H2O2)] ∆Ho

rxn = [(2 mol)(-285.8 kJ/mol)] - [(2 mol)(-191.2 kJ/mol)] = -189.2 kJ (c) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ∆Ho

rxn = [3 ∆Hof(CO2)] - [∆Ho

f(Fe2O3) + 3 ∆Hof(CO)]

∆Horxn = [(3 mol)(-393.5 kJ/mol)]

- [(1 mol)(-824.2 kJ/mol) + (3 mol)(-110.5 kJ/mol)] = -24.8 kJ 8.98 2 NO(g) + O2(g) → 2 NO2 (g) ∆Ho

1 = 2(-57.0 kJ) 2 NO2 (g) → N2O4(g) ∆Ho

2 = -57.2 kJ Sum 2 NO(g) + O2(g) → N2O4(g)

∆Ho = ∆Ho1 + ∆Ho

2 = -171.2 kJ 8.99 ∆G = ∆H - T∆S; at equilibrium ∆G = 0. Set ∆G = 0 and solve for T.

∆G = 0 = ∆H - T∆S

T = S

H

∆∆

= mol) kJ/(K 10 x 93.2

kJ/mol 30.913_ •

= 332 K = 59oC

8.100 ∆Gfus = ∆Hfus - T∆Sfus; at the melting point ∆G = 0. Set ∆G = 0 and solve for T (the

melting point). ∆G = 0 = ∆Hfus - T∆Sfus

T = S

H

fus

fus

∆∆

= K 279 = kJ/K 0.0357

kJ 9.95

8.101 HgS(s) + O2(g) → Hg(l) + SO2(g)

(a) ∆Horxn = ∆Ho

f(SO2) - ∆Hof(HgS)

∆Horxn = [(1 mol)(-296.8 kJ/mol)] - [(1 mol)(-58.2 kJ/mol)] = -238.6 kJ

(b) and (c) Because ∆H < 0 and ∆S > 0, the reaction is spontaneous at all temperatures. 8.102 ∆Ho

rxn = D (Reactant bonds) - D (Product bonds) (a) 2 CH4(g) → C2H6(g) + H2(g) ∆Ho

rxn = (8 DC-H) - (DC-C + 6 DC-H + DH-H) ∆Ho

rxn = [(8 mol)(410 kJ/mol)] - [(1 mol)(350 kJ/mol) + (6 mol)(410 kJ/mol) + (1 mol)(436 kJ/mol)] = +34 kJ (b) C2H6(g) + F2(g) → C2H5F(g) + HF(g) ∆Ho

rxn = (6 DC-H + DC-C + DF-F) - (5 DC-H + DC-C + DC-F + DH-F) ∆Ho

rxn = [(6 mol)(410 kJ/mol) + (1 mol)(350 kJ/mol) + (1 mol)(159 kJ/mol)]


188

- [(5 mol)(410 kJ/mol) + (1 mol)(350 kJ/mol) + (1 mol)(450 kJ/mol) + (1 mol)(570 kJ/mol)] = - 451 kJ (c) N2(g) + 3 H2(g) → 2 NH3(g) The bond dissociation energy for N2 is 945 kJ/mol. ∆Ho

rxn = (D N2 + 3 DH-H) - (6 DN-H)

∆Horxn = [(1 mol)(945 kJ/mol) + (3 mol)(436 kJ/mol)] - [(6 mol)(390 kJ/mol)] = -87 kJ

8.103 (a) ∆Ho

rxn = ∆Hof(CH3OH) - ∆Ho

f(CO) ∆Ho

rxn = [(1 mol)(-238.7 kJ/mol)] - [(1 mol)(-110.5 kJ/mol)] = -128.2 kJ (b) ∆Go = ∆Ho - T∆So = -128.2 kJ - (298 K)(-332 x 10-3 kJ/K) = -29.3 kJ (c) Step 1 is spontaneous since ∆Go < 0. (d) ∆Ho, because it is larger than T∆So. (e) Set ∆G = 0 and solve for T. ∆G = 0 = ∆H - T∆S

T = S

H

∆∆

= kJ/K 10 x 332

kJ 128.23_

= 386 K; The reaction is spontaneous below 386 K.

(f) ∆Horxn = ∆Ho

f(CH4) - ∆Hof(CH3OH)

∆Horxn = [(1 mol)(-74.8 kJ/mol)] - [(1 mol)(-238.7 kJ/mol)] = +163.9 kJ

(g) ∆Go = ∆Ho - T∆So = +163.9 kJ - (298 K)(162 x 10-3 kJ/K) = +115.6 kJ (h) Step 2 is nonspontaneous since ∆Go > 0. (i) ∆Ho, because it is larger than T∆So. (j) Set ∆G = 0 and solve for T. ∆G = 0 = ∆H - T∆S

T = S

H

∆∆

= kJ/K 10 x 162

kJ 163.93_

= 1012 K; The reaction is spontaneous above 1012 K.

(k) ∆Gooverall = ∆Go

1 + ∆Go2 = -29.3 kJ + 115.6 kJ = +86.3 kJ

∆Hooverall = ∆Ho

1 + ∆Ho2 = -128.2 kJ + 163.9 kJ = +35.7 kJ

∆Sooverall = ∆So

1 + ∆So2 = -332 J/K + 162 J/K = -170 J/K

(l) The overall reaction is nonspontaneous since ∆Gooverall > 0.

(m) The two reactions should be run separately. Run step 1 below 386 K and run step 2 above 1012 K.

8.104 (a) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

(b) C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g) ∆Ho

rxn = ∆Hoc = -5456.6 kJ

∆Horxn = [8 ∆Ho

f(CO2) + 9 ∆Hof(H2O)] - ∆Ho

f(C8H18) -5456.6 kJ = [(8 mol)(-393.5 kJ/mol) +(9 mol)(-241.8 kJ/mol)] - [(1 mol)(∆Ho

f(C8H18))] Solve for ∆Ho

f(C8H18). -5456.6 kJ = -5324 kJ - [(1 mol)(∆Ho

f(C8H18))] -132.4 kJ = -(1 mol)(∆Ho

f(C8H18)) ∆Ho

f(C8H18) = +132.4 kJ/mol 8.105 Assume 1.00 kg of H2O; 1 kg⋅m2/s2 = 1 J

Ep = (1.00 kg)(9.81 m/s2)(739 m) = 7250 kg⋅m2/s2 = 7250 J


189

q = specific heat x m x ∆T

∆T = C) J/(g g)(4.18 (1000

J 7250 =

heat specific x m

qo•

= 1.73oC (temperature rise)

8.106 (a) ∆Stotal = ∆Ssystem + ∆Ssurr and ∆Ssurr = -∆H/T

∆Stotal = ∆Ssystem + (-∆H/T) = ∆Ssystem - ∆H/T ∆Ssystem = ∆Stotal + ∆H/T ∆G = ∆H - T∆S (substitute ∆Ssystem for ∆S in this equation) ∆G = ∆H - T(∆Stotal + ∆H/T) = -T∆Stotal ∆G = -T∆Stotal For a spontaneous reaction, if ∆Stotal > 0 then ∆G < 0. (b) ∆Go = ∆Ho - T∆So

∆Ho = ∆Go + T∆So

∆Ssurr = -K 298

mol))] J/(K 210 K)(_ (298 + J/mol 10 x [2879 _ =

T

]ST + G[ _ =

TH 3ooo •∆∆∆

∆Ssurr = -9451 J/(K⋅ mol)

8.107 3/2 NO2(g) + 1/2 H2O(l) → HNO3(aq) + 1/2 NO(g) ∆Ho1 =

2

kJ 138.4_

3/2 NO(g) + 3/4 O2(g) → 3/2 NO2(g) ∆Ho2 =

4

kJ) 114.0(3)(_

1/2 N2(g) + 3/2 H2(g) → NH3(g) ∆Ho3 = - 46.1 kJ

NH3(g) + 5/4 O2(g) → NO(g) + 6/4 H2O(l) ∆Ho4 =

4

kJ 1169.6_

H2O(l) → 1/2 O2(g) + H2(g) ∆Ho5 = +285.8 kJ

Sum 1/2 H2(g) + 1/2 N2(g) + 3/2 O2(g) → HNO3(aq) ∆Ho = -207.4 kJ 8.108 2 CH4(g) + 4 O2(g) → 2 CO2(g) + 4 H2O(l) ∆Ho

1 = 2(-890.3 kJ) C2H6(g) → C2H4(g) + H2(g) ∆Ho

2 = +137.0 kJ

2 CO2(g) + 3 H2O(l) → C2H6(g) + 7/2 O2(g) ∆Ho3 =

2

kJ 3119.4

H2O(l) → H2(g) + 1/2 O2(g) ∆Ho4 = +285.8 kJ

Sum 2 CH4(g) → C2H4(g) + 2 H2(g) ∆Ho = +201.9 kJ 8.109 qMo = (110.0 g)(specific heat Mo)(28.0oC - 100.0oC)

= q OH2(150.0 g)[4.184 J/(g ⋅ oC)](28.0oC - 24.6oC)

qMo = -q OH2

(110.0 g)(specific heat Mo)(28.0oC - 100.0oC) = - (150.0 g)[4.184 J/(g ⋅ oC)](28.0oC - 24.6oC)

specific heat Mo = = C)0100. _ C0g)(28. (110.0

C)624. _ C0C)](28. J/(g g)[4.184 (150.0 _oo

ooo•0.27 J/(g ⋅ oC)

8.110 qice tea = -qice


190

qice tea = = C) 80.0 _ C g)(10.0 )(400.0C g

J (4.18 oo

o• -1.17 x 105 J

H2O, 18.02 amu

qice = 1.17 x 105 J =

OH g 18.02

OH mol 1 x m

kJ 1

J 1000kJ/mol) (6.01

2

2ice

C)00. _ C0)(10.m(C) (g

J 4.18 + oo

iceo

•

Solve for the mass of ice, mice. 1.17 x 105 J = (3.34 x 102 J/g)(mice) + (41.8 J/g)(mice) = (3.76 x 102 J/g)(mice)

mice = = J/g 10 x 3.76

J 10 x 1.172

5

311 g of ice

8.111 There is a large excess of NaOH.

5.00 mL = 0.005 00 L mol citric acid = (0.005 00 L)(0.64 mol/L) = 0.0032 mol citric acid

= q OH2 (51.6 g)[4.0 J/(g ⋅ oC)](27.9oC - 26.0oC) = 392 J

q _ = q OHrxn 2= -392 J

∆H = = J 1000

kJ 1 x

mol 0.0032

J 392 _ -123 kJ/mol = -120 kJ/mol citric acid

8.112 CsOH(aq) + HCl(aq) → CsCl(aq) + H2O(l)

mol CsOH = 0.100 L x = L 1.00

CsOH mol 0.2000.0200 mol CsOH

mol HCl = 0.050 L x = L 1.00

HCl mol 0.400 0.0200 mol HCl

The reactants were mixed in equal mole amounts. Total volume = 150 mL and has a mass of 150 g

qsolution = C)5022. _ C28g)(24. (150C) (g

J 4.2 oo

o

•= 1121 J

qreaction = -qsolution = -1121 J

∆H = = J 1000

kJ 1 x

CsOH mol 0.0200

J 1121_ =

CsOH mol

qreaction -56 kJ/mol CsOH

8.113 NaNO3, 84.99 amu; KF, 58.10 amu

For NaNO3(s) → NaNO3(aq), q = 20.5 kJ/mol x = NaNO g 84.99

NaNO mol 1

3

3 0.241 kJ/g

For KF(s) → KF(aq), q = -17.7 kJ/mol x = KF g 58.10

KF mol 1 -0.305 kJ/g

= qsoln (110.0 g)[4.18 J/(g ⋅ oC)](2.22oC) = 1021 J = 1.02 kJ

q _ = q solnrxn = -1.02 kJ

Let X = mass of NaNO3 and Y = mass of KF


191

X + Y = 10.0 g, so Y = 10.0 g - X = qrxn -1.02 kJ = X(0.241 kJ/g) + Y(- 0.305 kJ/g) (substitute for Y and solve for X)

-1.02 kJ = X(0.241 kJ/g) + (10.0 g - X)(- 0.305 kJ/g) -1.02 kJ = (0.241 kJ)X - 3.05 kJ + (0.305 kJ)X 2.03 kJ = (0.546 kJ)X

X = = kJ 0.546

kJ 2.033.72 g NaNO3

Y = 10.0 g - X = 10.0 g - 3.72 g = 6.28 g KF = 6.3 g KF 8.114 ∆H

4 CO(g) + 2 O2(g) → 4 CO2(g) 2(-566.0 kJ) 2 NO2(g) → 2 NO(g) + O2(g) +114.0 kJ 2 NO(g) → O2(g) + N2(g) 2(- 90.2 kJ) 4 CO(g) + 2 NO2(g) → 4 CO2(g) + N2(g) - 1198.4 kJ

Multi-Concept Problems 8.115 (a) Each S has 2 bonding pairs and 2 lone pairs of electrons. Each S is sp3 hybridized

and the geometry around each S is bent. (b) ∆H = D(reactant bonds) - D(product bonds) ∆H = (8 DS-S) - (4 DS=S) = +99 kJ ∆H = [(8 mol)(225 kJ/mol) - [(4 mol)(DS=S)] = +99 kJ - (4 mol)(DS=S) = 99 kJ - 1800 kJ = -1701 kJ DS=S = (1701 kJ)/(4 mol) = 425 kJ/mol (c) σ*3p

π*3p ↑ ↑ π3p ↑↓ ↑↓ σ3p ↑↓ σ*3s ↑↓ σ3s ↑↓

S2 S2 should be paramagnetic with two unpaired electrons in the π*3p MOs.

8.116 (a)

(b) C(g) + ½ O2(g) + Cl2(g) → COCl2(g) ∆Ho

f = ∆Hof(C(g)) + (½ DO=O + DCl-Cl) - (DC=O + 2 DC-Cl)

∆Hof = (716.7 kJ) + [(½ mol)(498 kJ/mol) + (1 mol)(243 kJ/mol)]

- [(1 mol)(732 kJ/mol) + (2 mol)(330 kJ/mol)] ∆Ho

rxn = - 183 kJ per mol COCl2 From Appendix B, ∆Ho

f(COCl2) = -219.1 kJ/mol The calculation of ∆Ho

f from bond energies is only an estimate because the bond energies


192

are average values derived from many different compounds. 8.117 (a) (1) 2 CH3CO2H(l) + Na2CO3(s) → 2 CH3CO2Na(aq) + CO2(g) + H2O(l)

(2) CH3CO2H(l) + NaHCO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)

(b) CH3CO2H, 60.05 amu; Na2CO3, 105.99 amu; NaHCO3, 84.01 amu

1 gal x mL 1

HCOCH g 1.049 x

L 1

mL 1000 x

gal 1

L 3.7854 23 = 3971 g CH3CO2H

3971 g CH3CO2H x HCOCH g 60.05

HCOCH mol 1

23

23 = 66.13 mol CH3CO2H

For reaction (1)

g 1000

kg 1 x

CONa mol 1CONa g 105.99

x HCOCH mol 2

CONa mol 1 x HCOCH mol 66.13

32

32

23

3223 = 3.505 kg

Na2CO3 For reaction (2)

g 1000

kg 1 x

NaHCO mol 1NaHCO g 84.01

x HCOCH mol 1

NaHCO mol 1 x HCOCH mol 66.13

3

3

23

323 = 5.556 kg

NaHCO3

(c) 2 CH3CO2H(l) + Na2CO3(s) → 2 CH3CO2Na(aq) + CO2(g) + H2O(l) ∆Ho

rxn = [2 ∆Hof(CH3CO2Na) + ∆Ho

f(CO2) + ∆Hof(H2O)]

- [2 ∆Hof(CH3CO2H) + ∆Ho

f(Na2CO3)] ∆Ho

rxn = [(2 mol)(-726.1 kJ/mol) + (1 mol)(-393.5 kJ/mol) + (1 mol)(-285.8 kJ/mol)] - [(2 mol)(- 484.5 kJ/mol) + (1 mol)(-1130.7 kJ/mol)]

∆Horxn = -31.8 kJ for 2 mol CH3CO2H

Heat = -HCOCH mol 2

kJ 31.8

23

x 66.13 mol CH3CO2H = -1050 kJ (liberated)

CH3CO2H(l) + NaHCO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l) ∆Ho

rxn = [∆Hof(CH3CO2Na) + ∆Ho

f(CO2) + ∆Hof(H2O)]

- [∆Hof(CH3CO2H) + ∆Ho

f(NaHCO3)] ∆Ho

rxn = [(1 mol)(-726.1 kJ/mol) + (1 mol)(-393.5 kJ/mol) + (1 mol)(-285.8 kJ/mol)] - [(1 mol)(- 484.5 kJ/mol) + (1 mol)(-950.8 kJ/mol)]

∆Horxn = +29.9 kJ for 1 mol CH3CO2H

q = HCOCH mol 1

kJ 29.9

23

x 66.13 mol CH3CO2H = +1980 kJ (absorbed)

8.118 (a) 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)

(b) ∆Horxn = [2 ∆Ho

f(KOH)] - [2 ∆Hof(H2O)]

∆Horxn = [(2 mol)(- 482.4 kJ/mol)] - [(2 mol)(-285.8 kJ/mol)] = -393.2 kJ

(c) The reaction produces 393.2 kJ/ 2 mol K = 196.6 kJ/ mol K. Assume that the mass of the water does not change and that the specific heat = 4.18 J/(g⋅oC) for the solution that is produced.


193

q = 7.55 g K x kJ 1

J 1000 x

K mol 1

kJ 196.6 x

K g 39.10

K mol 1 = 3.80 x 104 J


∆T = g) C)](400.0 J/(g [4.18

J 10 x 3.80 =

m x heat) (specific

qo

4

• = 22.7oC

∆T = Tfinal - Tinitial Tfinal = ∆T + Tinitial = 22.7oC + 25.0oC = 47.7oC

(d) 7.55 g K x K mol 2

KOH mol 2 x

K g 39.10

K mol 1 = 0.193 mol KOH

Assume that the mass of the solution does not change during the reaction and that the solution has a density of 1.00 g/mL.

solution volume = 400.0 g x mL 1000

L 1 x

g 1

mL 1.00 = 0.400 L

molarity = L 0.400

KOH mol 0.193 = 0.483 M

2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)

0.193 mol KOH x SOH mol 0.554

mL 1000 x

KOH mol 2SOH mol 1

42

42 = 174 mL of 0.554 M H2SO4

8.119 (a) Each N is sp3 hybridized and the geometry about each N is trigonal pyramidal.

(b) 2/4 NH3(g) + 3/4 N2O(g) → N2(g) + 3/4 H2O(l) ∆Ho1 =

4

kJ 1011.2_

1/4 H2O(l) + 1/4 N2H4(l) → 1/8 O2(g) + 1/2 NH3(g) ∆Ho2 =

8

kJ 286+

3/4 N2H4(l) + 3/4 H2O(l) → 3/4 N2O(g) + 9/4 H2(g) ∆Ho3 =

4

kJ) (3)(+314

9/4 H2(g) + 9/8 O2(g) → 9/4 H2O(l) ∆Ho4 =

4

kJ) 285.8(9)(_

Sum N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ∆Ho = -622 kJ

(c) N2H4, 32.045 amu

mol N2H4 = 100.0 g N2H4 x HN g 32.045

HN mol 1

42

42 = 3.12 mol N2H4

q = (3.12 mol N2H4)(622 kJ/mol) = 1940 kJ


194

8.120 Assume 100.0 g of Y.

mol F = 61.7 g F x = F g 19.00

F mol 1 3.25 mol F

mol Cl = 38.3 g Cl x = Cl g 35.45


Cl1.08F3.25, divide each subscript by the smaller of the two, 1.08. Cl1.08 / 1.08F3.25 / 1.08 ClF3 (a) Y is ClF3 and X is ClF

(b) There are five electron clouds around the Cl (3 bonding and 2 lone pairs). The geometry is T-shaped.

(c) ∆H

Cl2O(g) + 3 OF2(g) → 2 O2(g) + 2 ClF3(g) -533.4 kJ 2 ClF(g) + O2(g) → Cl2O(g) + OF2(g) +205.6 kJ O2(g) + 2 F2(g) → 2 OF2(g) 2(24.7 kJ) 2 ClF(g) + 2 F2(g) → 2 ClF3(g) -278.4 kJ

Divide reaction coefficients and ∆H by 2.

ClF(g) + F2(g) → ClF3(g) ∆H = -278.4 kJ/2 = -139.2 kJ/mol ClF3

(d) ClF, 54.45 amu

q = 25.0 g ClF x = 0.875 x ClF mol 1

kJ 139.2_ x

ClF g 54.45

ClF mol 1 -55.9 kJ

55.9 kJ is released in this reaction.

195

9

Gases: Their Properties and Behavior 9.1 1.00 atm = 14.7 psi

1.00 mm Hg x atm 1

psi 14.7 x

Hg mm 760

atm 1 = 1.93 x 10-2 psi

9.2 1.00 atmosphere pressure can support a column of Hg 0.760 m high. Because the density

of H2O is 1.00 g/mL and that of Hg is 13.6 g/mL, 1.00 atmosphere pressure can support a column of H2O 13.6 times higher than that of Hg. The column of H2O supported by 1.00 atmosphere will be (0.760 m)(13.6) = 10.3 m.

9.3 The pressure in the flask is less than 0.975 atm because the liquid level is higher on the

side connected to the flask. The 24.7 cm of Hg is the difference between the two pressures.

Pressure difference = 24.7 cm Hg x 1

76 0

atm. cm Hg

= 0.325 atm

Pressure in flask = 0.975 atm - 0.325 atm = 0.650 atm 9.4 The pressure in the flask is greater than 750 mm Hg because the liquid level is lower on

the side connected to the flask.

Pressure difference = 25 cm Hg x Hg cm 1

Hg mm 10= 250 mm Hg

Pressure in flask = 750 mm Hg + 250 mm Hg = 1000 mm Hg 9.5 (a) Assume an initial volume of 1.00 L.

First consider the volume change resulting from a change in the number of moles with the pressure and temperature constant.

n

V = n

V

f

f

i

i ; Vf = mol 0.3

mol) L)(0.225 (1.00 =

n

n V

i

fi = 0.75 L

Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant.

T

V = T

V

f

f

i

i ; Vf = K 300

K) L)(400 (0.75 =

T

T V

i

fi = 1.0 L

There is no net change in the volume as a result of the decrease in the number of moles of gas and a temperature increase.

Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________

196

(b) Assume an initial volume of 1.00 L. First consider the volume change resulting from a change in the number of moles with the pressure and temperature constant.

n

V = n

V

f

f

i

i ; Vf = mol 0.3

mol) L)(0.225 (1.00 =

n

n V

i

fi = 0.75 L

Now consider the volume change from 0.75 L as a result of a change in temperature with the number of moles and the pressure constant.

T

V = T

V

f

f

i

i ; Vf = K 300

K) L)(200 (0.75 =

T

T V

i

fi = 0.5 L

The volume would be cut in half as a result of the decrease in the number of moles of gas and a temperature decrease.

9.6 n = K) (273.15

mol K atm L

06 0.082

L) 10 x atm)(1.000 (1.000 =

RT

PV 5

••

= 4.461 x 103 mol CH4

CH4, 16.04 amu; mass CH4 = (4.461 x 103 mol)

mol 1

g 16.04 = 7.155 x 104 g CH4

9.7 C3H8, 44.10 amu; V = 350 mL = 0.350 L; T = 20oC = 293 K

n = 3.2 g x HC g 44.10

HC mol 1

83

83 = 0.073 mol C3H8

P = V

nRT =

L 0.350

K) (293mol K atm L

06 0.082mol) (0.073

••

= 5.0 atm

9.8 P = 1.51 x 104 kPa x kPa 101.325

atm 1 = 149 atm; T = 25.0oC = 298 K

n = K) (298

mol K atm L

06 0.082

L) atm)(43.8 (149 =

RT

PV

••

= 267 mol He

9.9 The volume and number of moles of gas remain constant.

T

P = T

P = V

nR

f

f

i

i ; Tf = atm 2.15

K) atm)(273 (2.37 =

P

T P

i

if = 301 K = 28oC

9.10 (a) The temperature has increased by about 10% (from 300 K to 325 K) while the


197

amount and the pressure are unchanged. Thus, the volume should increase by about 10%.

(b) The temperature has increased by a factor of 1.5 (from 300 K to 450 K) and the pressure has increased by a factor of 3 (from 0.9 atm to 2.7 atm) while the amount is unchanged. Thus, the volume should decrease by half (1.5/3 = 0.5).

(c) Both the amount and the pressure have increased by a factor of 3 (from 0.075 mol to 0.22 mol and from 0.9 atm to 2.7 atm) while the temperature is unchanged. Thus, the volume is unchanged.

9.11 CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)

CaCO3, 100.1 amu; CO2, 44.01 amu

mole CO2 = 33.7 g CaCO3 x CaCO mol 1CO mol 1

x CaCO g 100.1

CaCO mol 1

3

2

3

3 = 0.337 mol CO2

mass CO2 = 0.337 mol CO2 x CO mol 1CO g 44.01

2

2 = 14.8 g CO2

V = atm 1.00

K) (273mol K atm L

06 0.082mol) (0.337 =

P

nRT

••

= 7.55 L

9.12 C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)


198

npropane = K) (298

mol K atm L

06 0.082

L) atm)(15.0 (4.5 =

RT

PV

••

= 2.76 mol C3H8

2.76 mol C3H8 x HC mol 1

CO mol 3

83

2 = 8.28 mol CO2

V = atm 1.00

K) (273mol K atm L

06 0.082mol) (8.28 =

P

nRT

••

= 186 L = 190 L

9.13 n = K) (273

mol K atm L

06 0.082

L) atm)(1.00 (1.00 =

RT

PV

••

= 0.0446 mol

molar mass = mol 0.0446

g 1.52 = 34.1 g/mol; molecular mass = 34.1 amu

Na2S(aq) + 2 HCl(aq) → H2S(g) + 2 NaCl(aq) The foul-smelling gas is H2S, hydrogen sulfide.

9.14 12.45 g H2 x H g 2.016

H mol 1

2

2 = 6.176 mol H2

60.67 g N2 x N g 28.01

N mol 1

2

2 = 2.166 mol N2

2.38 g NH3 x NH g 17.03

NH mol 1

3

3 = 0.140 mol NH3

ntotal = n + n + n NHNH 322 = 6.176 mol + 2.166 mol + 0.140 mol = 8.482 mol

mol 8.482

mol 6.176 = XH2

= 0.7281; mol 8.482

mol 2.166 = XN2

= 0.2554; mol 8.482

mol 0.140 = XNH3

= 0.0165

9.15 mol 8.482 = ntotal (from Problem 9.14). T = 90oC = 363 K

L 10.00

K) (363mol K atm L

06 0.082mol) (8.482 =

V

RTn = Ptotal

total

••

= 25.27 atm

P X = P totalHH 22• = (0.7281)(25.27 atm) = 18.4 atm

P X = P totalNN 22• = (0.2554)(25.27 atm) = 6.45 atm

P X = P totalNHNH 33• = (0.0165)(25.27 atm) = 0.417 atm

9.16 X = P OHOH 22

⋅ PTotal = (0.0287)(0.977 atm) = 0.0280 atm

9.17 The number of moles of each gas is proportional to the number of each of the different

gas molecules in the container. ntotal = nred + nyellow + ngreen = 6 + 2 + 4 = 12


199

Xred = 12

6 =

n

n

total

red = 0.500; Xyellow = 12

2 =

n

n

total

yellow = 0.167; Xgreen = 12

4 =

n

n

total

green =

0.333 Pred = Xred ⋅ Ptotal = (0.500)(600 mm Hg) = 300 mm Hg Pyellow = Xyellow ⋅ Ptotal = (0.167)(600 mm Hg) = 100 mm Hg Pgreen = Xgreen ⋅ Ptotal = (0.333)(600 mm Hg) = 200 mm Hg

9.18 M

RT3 =u , M = molar mass, R = 8.314 J/(K ⋅ mol), 1 J = 1 kg ⋅ m2/s2

at 37oC = 310 K, m/s 525 = mol / kg 10 x 28.01

K 310 x mol)K s/(m kg 8.314 x 3 =u

3_

22

at -25oC = 248 K, m/s 470 = mol / kg 10 x 28.01

K 248 x mol)K s/(m kg 8.314 x 3 =u

3_

22

9.19 M

RT3 =u , M = molar mass, R = 8.314 J/(K ⋅ mol), 1 J = 1 kg ⋅ m2/s2

O2, 32.00 amu, 32.00 x 10-3 kg/mol

u = 580 mi/h x = s 60

min 1 x

min 60

hr 1 x

km 1

m 1000 x

mi 1

km 1.6093 259 m/s

M

RT3 =u ; u2 =

M

T R 3

T = = mol) K s/m kg (3)(8.314

kg/mol) 10 x (32.00 )m/s (259 =

R 3

M u22

3_22

••• 86.1 K

T = 86.1 - 273.15 = -187.0oC

9.20 (a) 32.0

83.8 =

M

M = Kr rateO rate

O

Kr2

2

; Kr rateO rate 2 = 1.62

O2 diffuses 1.62 times faster than Kr.

(b) 26.0

28.0 =

M

M = N rateHC rate

HC

N

2

22

22

2 ; N rateHC rate

2

22 = 1.04

C2H2 diffuses 1.04 times faster than N2.

9.21 1.05 = 20

22 =

Ne M

Ne M =

Ne rate

Ne rate20

22

22

20

; 1.02 = 21

22 =

Ne M

Ne M =

Ne rate

Ne rate21

22

22

21

Thus, the relative rates of diffusion are Ne(1.00) > Ne(1.02) > Ne(1.05) 222120 .

9.22 P = L) (0.600

K) (300mol K atm L

06 0.082mol) (0.500 =

V

nRT

••

= 20.5 atm


200

V

n a _

bn _ V

T Rn = P

2

2

P = )L (0.600

)mol (0.500mol

atm L 1.35

_ L/mol)] 7mol)(0.038 (0.500 _ L) [(0.600

K) (300mol K atm L

06 0.082mol) (0.500

2

2

2

2

•

••

= 20.3

atm 9.23 The amount of ozone is assumed to be constant.

Therefore nR = T

V P = T

V P

f

ff

i

ii

Because V ∝ h, then T

h P = T

h P

f

ff

i

ii where h is the thickness of the O3 layer.

hf = m) 10 x (20K 230

K 273

atm 1

atm 10 x 1.6 = h x

T

T x P

P 39_

ii

f

f

i

= 3.8 x 10-5 m

(Actually, V = 4πr2h, where r = the radius of the earth. When you go out ~30 km to get to the ozone layer, the change in r2 is less than 1%. Therefore you can neglect the change in r2 and assume that V is proportional to h.)

9.24 For ether, the MAC = Hg mm 760

Hg mm 15 x 100% = 2.0%

9.25 (a) Let X = partial pressure of chloroform.

MAC = Hg mm 760

X x 100% = 0.77%

Solve for X. X = 760 mm Hg x 100%

0.77% = 5.9 mm Hg

(b) CHCl3, 119.4 amu

PV = nRT; n = ( )

K) (273mol K atm L

06 0.082

L) (10.0Hg mm 5.9 =

T R

V P

••

= 0.00347 mol CHCl3

mass CHCl3 = 0.00347 mol CHCl3 x CHCl mol 1CHCl g 119.4

3

3 = 0.41 g CHCl3


9.26 (a) The volume of a gas is proportional to the kelvin temperature at constant pressure. As the temperature increases from 300 K to 450 K, the volume will increase by a factor of 1.5.


201

(b) The volume of a gas is inversely proportional to pressure at constant temperature. As the pressure increases from 1 atm to 2 atm, the volume will decrease by a factor of 2.

(c) PV = nRT; The amount of gas (n) is constant.

Therefore nR = T

V P = T

V P

f

ff

i

ii .

Assume Vi = 1 L and solve for Vf.

L 1 = V = atm) K)(2 (300

K) L)(200 atm)(1 (3 =

P T

T V Pf

fi

fii

There is no change in volume. 9.27 If the sample remains a gas at 150 K, then drawing (c) represents the gas at this

temperature. The gas molecules still fill the container. 9.28 The two gases should mix randomly and homogeneously and this is best represented by

drawing (c). 9.29 The two gases will be equally distributed among the three flasks.

9.30 The gas pressure in the bulb in mm Hg is equal to the difference in the height of the Hg in

the two arms of the manometer.


202

9.31 A When stopcock A is opened, the pressure in the flask will equal the external pressure, and the level of mercury will be the same in both arms of the manometer.

9.32 (a) Because there are more yellow gas molecules than there are blue, the yellow gas

molecules have the higher average speed. (b) Each rate is proportional to the number of effused gas molecules of each type. Myellow = 25 amu

M

M = rate

rate

blue

yellow

yellow

blue ; M

amu 25 =

6

5

blue

; M

amu 25 =

6

5

blue

2

; Mblue =

65

amu 252 = 36

amu

9.33

(a) T

P = T

P

1

1

2

2 ; = K 298

Hg) mm K)(760 (248 =

T

)P)(T( = P1

122 632 mm Hg

The column of Hg will rise to ~130 mm Hg inside the tube (drawing 1). The pressure inside the tube (632 mm Hg) plus the pressure of 130 mm Hg equals the external pressure, ~760 mm Hg. (b) The column of Hg will rise to ~760 mm (drawing 2), which is equal to the external pressure, ~760 mm Hg.


203

(c) The pressure inside the tube is equal to the external pressure and the Hg level inside the tube will be the same as in the dish (drawing 3).

Additional Problems Gases and Gas Pressure 9.34 Temperature is a measure of the average kinetic energy of gas particles. 9.35 Gases are much more compressible than solids or liquids because there is a large amount

of empty space between individual gas molecules.

9.36 P = 480 mm Hg x Hg mm 760

atm 1.00 = 0.632 atm

P = 480 mm Hg x Hg mm 760

Pa 101,325 = 6.40 x 104 Pa

9.37 P = 352 torr x Pa 1000

kPa 1 x

torr760

Pa 101,325 = 46.9 kPa

P = 0.255 atm x atm 1.00

Hg mm 760 = 194 mm Hg

P = 0.0382 mm Hg x Hg mm 760

Pa 101,325 = 5.09 Pa

9.38 Pflask > 754.3 mm Hg; Pflask = 754.3 mm Hg + 176 mm Hg = 930 mm Hg 9.39 Pflask < 1.021 atm (see Figure 9.4)

Pdifference = 28.3 cm Hg x Hg cm 76.0

atm 1.00 = 0.372 atm

Pflask = 1.021 atm - Pdifference = 1.021 atm - 0.372 atm = 0.649 atm 9.40 Pflask > 752.3 mm Hg (see Figure 9.4)

If the pressure in the flask can support a column of ethyl alcohol (d = 0.7893 g/mL) 55.1 cm high, then it can only support a column of Hg that is much shorter because of the higher density of Hg.

55.1 cm x g/mL 13.546

g/mL 0.7893 = 3.21 cm Hg = 32.1 mm Hg

Pflask = 752.3 mm Hg + 32.1 mm Hg = 784.4 mm Hg

Pflask = 784.4 mm Hg x Hg mm 760

Pa 101,325 = 1.046 x 105 Pa

9.41 Compute the height of a column of CHCl3 that 1.00 atm can support.

760 mm Hg x g/mL 1.4832

g/mL 13.546 = 6941 mm CHCl3; therefore 1.00 atm = 6941 mm CHCl3

The pressure in the flask is less than atmospheric pressure.


204

Patm - Pflask = 0.849 atm - 0.788 atm = 0.061 atm

0.061 atm x atm 1.00

CHCl mm 6941 3 = 423 mm CHCl3

The chloroform will be 423 mm higher in the manometer arm connected to the flask. 9.42 % Volume

N2 78.08 O2 20.95 Ar 0.93 CO2 0.037 The % volume for a particular gas is proportional to the number of molecules of that gas in a mixture of gases. Average molecular mass of air = (0.7808)(mol. mass N2) + (0.2095)(mol. mass O2)

+ (0.0093)(at. mass Ar) + (0.000 37)(mol. mass CO2) = (0.7808)(28.01 amu) + (0.2095)(32.00 amu)

+ (0.0093)(39.95 amu) + (0.000 37)(44.01 amu) = 28.96 amu 9.43 The % volume for a particular gas is proportional to the number of molecules of that gas

in a mixture of gases. Average molecular mass of a diving-gas = (0.020)(mol. mass O2) + (0.980)(at. mass He) = (0.020)(32.00 amu) + (0.980)(4.00 amu) = 4.56 amu

The Gas Laws

9.44 (a) T

P = T

P = V

nR

f

f

i

i ; T

T P

i

fi = Pf

Let Pi = 1 atm, Ti = 100 K, Tf = 300 K

Pf = K) (100

K) atm)(300 (1 =

T

T P

i

fi = 3 atm

The pressure would triple.

(b) n

P = n

P = V

RT

f

f

i

i ; n

n P

i

fi = Pf

Let Pi = 1 atm, ni = 3 mol, nf = 1 mol

Pf = mol) (3

mol) atm)(1 (1 =

n

n P

i

fi = 3

1 atm

The pressure would be 3

1 the initial pressure.

(c) nRT = PiV i = PfVf; V

V P

f

ii = Pf

Let Pi = 1 atm, Vi = 1 L, Vf = 1 - 0.45 L = 0.55 L


205

Pf = L) (0.55

L) atm)(1 (1 =

V

V P

f

ii = 1.8 atm

The pressure would increase by 1.8 times.

(d) nR = T

V P = T

V P

f

ff

i

ii ; V T

T V P

fi

fii = Pf

Let Pi = 1 atm, Vi = 1 L, Ti = 200 K, Vf = 3 L, Ti = 100 K

Pf = L) K)(3 (200

K) L)(100 atm)(1 (1 =

V T

T V P

fi

fii = 0.17 atm

The pressure would be 0.17 times the initial pressure.

9.45 (a) T

V = T

V = P

nR

f

f

i

i ; V = T

TVf

i

fi

Let Vi = 1 L, Ti = 400 K, Tf = 200 K

K) (400

K) L)(200 (1 =

T

TV = Vi

fif = 0.5 L

The volume would be halved.

(b) n

V = n

V = P

RT

f

f

i

i ; V = n

n Vf

i

fi

Let Vi = 1 L, ni = 4 mol, nf = 5 mol

mol) (4

mol) L)(5 (1 =

n

n V = Vi

fif = 1.25 L

The volume would increase by 1/4.

(c) nRT = Pi Vi = Pf Vf; V = P

V Pf

f

ii

Let Vi = 1 L, Pi = 4 atm, Pf = 1 atm

atm) (1

L) atm)(1 (4 =

P

V P = Vf

iif = 4 L

The volume would increase by a factor of 4.

(d) T

V P = T

V P = nRf

ff

i

ii ; V = T P

T V Pf

if

fii

Let Vi = 1 L, Ti = 200 K, Tf = 400 K, Pi = 1 atm, Pf = 2atm

K) atm)(200 (2

K) L)(400 atm)(1 (1 =

T P

T V P = Vif

fiif = 1 L

There is no volume change. 9.46 They all contain the same number of gas molecules. 9.47 For air, T = 50oC = 323 K.


206

n = K) (323

mol K atm L

06 0.082

L) (2.50Hg mm 760

atm 1.00 x Hg mm 750

= RT

PV

••

= 0.0931 mol air

For CO2, T = -10oC = 263 K

n = K) (263

mol K atm L

06 0.082

L) (2.16Hg mm 760


= RT

PV

••

= 0.101 mol CO2

Because the number of moles of CO2 is larger than the number of moles of air, the CO2 sample contains more molecules.

9.48 n and T are constant; therefore nRT = VP = VP ffii

Vf = atm) (1.02

L) atm)(49.0 (150 =

P

VP

f

ii = 7210 L

n and P are constant; therefore T

V = T

V = P

nR

f

f

i

i

Vf = K) (293

K) L)(308 (49.0 =

T

T V

i

fi = 51.5 L

9.49 Ti = 20oC = 293 K; nR = T

V P = T

V P

f

ff

i

ii

Vf = atm) K)(1.00 (293

K) L)(273 atm)(8.0 (140 =

P T

T V P

fi

fii = 1.0 x 103 L

9.50 15.0 g CO2 x CO g 44.0CO mol 1

2

2 = 0.341 mol CO2

P = L) (0.30

K) (300mol K atm L

06 0.082mol) (0.341 =

V

nRT

••

= 27.98 atm

27.98 atm x atm 1

Hg mm 760 = 2.1 x 104 mm Hg

9.51 20.0 g N2 x N g 28.0N mol 1

2

2 = 0.714 mol N2

T =

••

mol K atm L

06 0.082mol) (0.714

L) atm)(0.40 (6.0 =

nR

PV = 41 K


207

9.52 L 1cm 1000

x atoms 10 x 6.02

H mol 1 x

cm

atom H 1 3

233 = 1.7 x 10-21 mol H/L

P = L) (1

K) (100mol K atm L

06 0.082mol) 10 x (1.7 =

V

nRT21_

••

= 1.4 x 10-20 atm

P = 1.4 x 10-20 atm x atm 1.0

Hg mm 760 = 1 x 10-17 mm Hg

9.53 CH4, 16.04 amu; 5.54 kg = 5.54 x 103 g; T = 20oC = 293 K

P = L) (43.8

K) (293mol K atm L

06 0.082g 16.04

mol 1 x g 10 x 5.54

= V

nRT3

••

= 189.6 atm

P = 189.6 atm x Pa 1000

Pak 1 x

atm 1

Pa 101,325 = 1.92 x 104 kPa

9.54 n = K)(293

mol K atm L

06 0.082

L) (43.8Pa 101,325

atm 1 x

Pak 1Pa 1000

x kPa 17,180

= RT

PV

••

= 308.9 mol

mass Ar = 308.9 mol x mol 1

g 39.948 = 12340 g = 1.23 x 104 g

9.55 P = 13,800 kPa x Pa 101,325

atm 1 x

kPa 1

Pa 1000 = 136.2 atm

n and T are constant; therefore nRT = PiV i = PfVf

Vf = atm) (1.25

L) atm)(2.30 (136.2 =

P

V P

f

ii = 250.6 L

250.6 L x L 1.5

balloon 1 = 167 balloons

Gas Stoichiometry 9.56 For steam, T = 123.0oC = 396 K

n = K) (396

mol K atm L

06 0.082

L) atm)(15.0 (0.93 =

RT

PV

••

= 0.43 mol steam

For ice, H2O, 18.02 amu; n = 10.5 g x g 18.02

mol 1 = 0.583 mol ice

Because the number of moles of ice is larger than the number of moles of steam, the ice contains more H2O molecules.

9.57 T = 85.0oC = 358 K


208

nAr = K) (358

mol K atm L

06 0.082

L) (3.14Hg mm 760

atm 1.00 x Hg mm 1111

= RT

PV

••

= 0.156 mol Ar

Cl2, 70.91 amu; Cl g 70.91

Cl mol 1 x Cl g 11.07 = n

2

22Cl2 = 0.156 mol Cl2

There are equal numbers of Ar atoms and Cl2 molecules in their respective samples. 9.58 The containers are identical. Both containers contain the same number of gas molecules.

Weigh the containers. Because the molecular mass for O2 is greater than the molecular mass for H2, the heavier container contains O2.

9.59 Assuming that you can see through the flask, Cl2 gas is greenish and He is colorless.

9.60 room volume = 4.0 m x 5.0 m x 2.5 m x m 10

L 133_ = 5.0 x 104 L

ntotal = K) (273

mol K atm L

06 0.082

L) 10 x atm)(5.0 (1.0 =

RT

PV 4

••

= 2.23 x 103 mol

nO2 = (0.2095)ntotal = (0.2095)(2.23 x 103 mol) = 467 mol O2

mass O2 = 467 mol x mol 1

g 32.0 = 1.5 x 104 g O2

9.61 0.25 g O2 x O g 32.0O mol 1

2

2 = 7.8 x 10-3 mol O2

V = atm 1.0

K) (310mol K atm L

06 0.082mol) 10 x (7.8 =

P

nRT3_

••

= 0.198 L = 0.200 L = 200 mL O2

9.62 (a) CH4, 16.04 amu; d = L 22.4

g 16.04 = 0.716 g/L

(b) CO2, 44.01 amu; d = L 22.4

g 44.01 = 1.96 g/L

(c) O2, 32.00 amu; d = L 22.4

g 32.00 = 1.43 g/L

(d) UF6, 352.0 amu; d = L 22.4

g 352.0 = 15.7 g/L

9.63 Average molar mass = (0.270)(molar mass F2) + (0.730)(molar mass He)

= (0.270)(38.00 g/mol) + (0.730)(4.003 g/mol) = 13.18 g/mol Assume 1.00 mole of the gas mixture. T = 27.5oC = 300.6 K


209

V =

••

Hg mm 760atm 1.00

x Hg mm 714

K) (300.6mol K atm L

06 0.082mol) (1.00 =

P

nRT = 26.3 L

d = L 26.3

g 13.18 = 0.501 g/L

9.64 n = K) (295.5

mol K atm L

06 0.082

L) (1.500Hg mm 760


= RT

PV

••

= 0.0290 mol



9.65 (a) Assume 1.000 L gas sample

n = K) (273

mol K atm L

06 0.082

L) atm)(1.000 (1.00 =

RT

PV

••

= 0.0446 mol



(b) Assume 1.000 L gas sample

n = K) (298

mol K atm L

06 0.082

L) (1.000Hg mm 760


= RT

PV

••

= 0.0405 mol



9.66 2 HgO(s) → 2 Hg(l) + O2(g); HgO, 216.59 amu

10.57 g HgO x HgO mol 2O mol 1

x HgO g 216.59

HgO mol 1 2 = 0.024 40 mol O2

V = atm 1.000


06 0.082mol) 40 (0.024 =

P

nRT

••

= 0.5469 L

9.67 2 HgO(s) → 2 Hg(l) + O2(g); HgO, 216.59 amu

mass HgO = 0.0155 mol O2 x HgO mol 1

HgO g 216.59 x

O mol 1

HgO mol 2

2

= 6.71 g HgO

9.68 Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)


210

(a) 25.5 g Zn x Znmol 1H mol 1

x Zng 65.39

Znmol 1 2 = 0.390 mol H2

V =

••

Hg mm 760atm 1.00

x Hg mm 742

K) (288mol K atm L

06 0.082mol) (0.390 =

P

nRT = 9.44 L

(b) n = K) (303.15

mol K atm L

06 0.082

L) (5.00Hg mm 760


= RT

PV

••

= 0.092 56 mol H2

0.092 56 mol H2 x Znmol 1

Zng 65.39 x

H mol 1

Znmol 1

2

= 6.05 g Zn

9.69 2 NH4NO3(s) → 2 N2(g) + 4 H2O(g) + O2(g); NH4NO3, 80.04 amu

Total moles of gas = 450 g NH4NO3 x NONH mol 2

gas mol 7 x

NONH g 80.04

NONH mol 1

3434

34 = 19.68

mol T = 450oC = 723 K

V = atm) (1.00

K) (723mol K atm L

06 0.082mol) (19.68 =

P

nRT

••

= 1.17 x 103 L

9.70 (a) V24h = (4.50 L/min)(60 min/h)(24 h/day) = 6480 L

VCO2 = (0.034)V24h = (0.034)(6480 L) = 220 L

n = K) (298

mol K atm L

06 0.082

L) (220Hg mm 760


= RT

PV

••

= 8.70 mol CO2

8.70 mol CO2 x CO mol 1CO g 44.01

2

2 = 383 g = 380 g CO2

(b) 2 Na2O2(s) + 2 CO2(g) → 2 Na2CO3(s) + O2(g); Na2O2, 77.98 amu 3.65 kg = 3650 g

3650 g Na2O2 x CO mol 8.70

day 1 x

ONa mol 2CO mol 2

x ONa g 77.98

ONa mol 1

222

2

22

22 = 5.4 days

9.71 2 TiCl4(g) + H2(g) → 2 TiCl3(s) + 2 HCl(g); TiCl4, 189.69 amu

(a) T = 435oC = 708 K


211

K) (708mol K atm L

06 0.082

L) (155Hg mm 760


= RT

PV = nH2

••

= 2.79 mol H2

2.79 mol H2 x TiCl mol 1

TiCl g 189.69 x

H mol 1TiCl mol 2

4

4

2

4 = 1058 g = 1060 g TiCl4

(b) nHCl = 2.79 mol H2 x H mol 1

HCl mol 2

2

= 5.58 mol HCl

V = atm) (1.00

K) (273mol K atm L

06 0.082mol) (5.58 =

P

RTnHCl

••

= 125 L HCl

Dalton's Law and Mole Fraction 9.72 Because of Avogadro's Law (V ∝ n), the % volumes are also % moles.

% mole N2 78.08 O2 20.95 Ar 0.93 CO2 0.037 In decimal form, % mole = mole fraction.

P X = P totalNN 22

• = (0.7808)(1.000 atm) = 0.7808 atm

P X = P totalOO 22• = (0.2095)(1.000 atm) = 0.2095 atm

P X = P totalArAr • = (0.0093)(1.000 atm) = 0.0093 atm

P X = P totalCOCO 22• = (0.000 37)(1.000 atm) = 0.000 37 atm

Pressures of the rest are negligible.

9.73 0.94; = mol 100

mol 94 = XCH4

P X = P totalCHCH 44• = (0.94)(1.48 atm) = 1.4 atm

0.040; = mol 100

mol 4 = X HC 62

P X = P totalHCHC 6262• = (0.040)(1.48 atm) = 0.059 atm

0.015; = mol 100

mol 1.5 = X HC 83


0.0050; = mol 100

mol 0.5 = X HC 104


9.74 Assume a 100.0 g sample. g CO2 = 1.00 g and g O2 = 99.0 g

mol CO2 = 1.00 g CO2 x CO g 44.01

CO mol 1

2

2 = 0.0227 mol CO2


212

mol O2 = 99.0 g O2 x O g 32.00

O mol 1

2

2 = 3.094 mol O2

ntotal = 3.094 mol + 0.0227 mol = 3.117 mol

mol 3.117

mol 3.094 = XO2

= 0.993; mol 3.117

mol 0.0227 = XCO2

= 0.007 28

P X = P totalOO 22• = (0.993)(0.977 atm) = 0.970 atm

P X = P totalCOCO 22• = (0.007 28)(0.977 atm) = 0.007 11 atm

9.75 From Problem 9.76: XHCI = 0.026, 0.88 = X 0.094, = X NeH2

PHCI = XHCl ⋅ Ptotal = (0.026)(13,800 kPa) = 3.6 x 102 kPa P X = P totalHH 22

• = (0.094)(13,800 kPa) = 1.3 x 103 kPa

PNe = XNe ⋅ Ptotal = (0.88)(13,800) kPa) = 1.2 x 104 kPa 9.76 Assume a 100.0 g sample.

g HCl = (0.0500)(100.0 g) = 5.00 g; 5.00 g HCl x HCl g 36.5


g H2 = (0.0100)(100.0 g) = 1.00 g; 1.00 g H2 x H g 2.016

H mol 1

2

2 = 0.496 mol H2

g Ne = (0.94)(100.0 g) = 94 g; 94 g Ne x Ne g 20.18

Ne mol 1 = 4.66 mol Ne

ntotal = 0.137 + 0.496 + 4.66 = 5.3 mol

mol 5.3

mol 0.137 = XHCl = 0.026;

mol 5.3

mol 0.496 = XH2

= 0.094; mol 5.3

mol 4.66 = XNe = 0.88

9.77 Assume a 1.000 L gas sample.

n = K) (273.15

mol K atm L

06 0.082

L) atm)(1.000 (1.000 =

RT

PV

••

= 0.044 61 mol

average molar mass = mol 61 0.044

g 1.413 = 31.67 g/mol

31.67 = M x)_ (1 + M x NAr 2••

31.67 = (x)(39.948) + (1 - x)(28.013) Solve for x: x = 0.3064, 1 - x = 0.6936 The mixture contains 30.64% Ar and 69.36% N2. Assume 100 moles of gas.

mol 100

mol 30.64 = XAr = 0.3064;

mol 100

mol 69.36 = XN2

= 0.6936


213

9.78 Ptotal = P + P OHH 22; P _ P = P OHtotalH 22

= 747 mm Hg - 23.8 mm Hg = 723 mm Hg

n = K) (298

mol K atm L

06 0.082

L) (3.557Hg mm 760


= RT

PV

••

= 0.1384 mol H2

0.1384 mol H2 x Mg mol 1

Mg g 24.3 x

H mol 1

Mg mol 1

2

= 3.36 g Mg

9.79 Ptotal = P + P OHCl 22

= 755 mm Hg

P _ P = P OHtotalCl 22= 755 mm Hg - 28.7 mm Hg = 726.3 mm Hg

(a) Hg mm 755

Hg mm 726.3 =

P

P = Xtotal

ClCl

2

2 = 0.962

(b) NaCl, 58.44 amu

2

2

1

21 1

726 3 760 0 5970 0232

0 082 27 273Cl

Cl (g)

p V ( . mmHg/ mmHg atm )( . L)n . molCl

RT . LatmK mol ( )K

−

− −= = =+

0.0232 mol Cl2 x 2

2 mol NaCl 58.44 g NaCl x

1 mol 1 mol NaClCl = 2.71 g NaCl

Kinetic-Molecular Theory and Graham's Law 9.80 The kinetic-molecular theory is based on the following assumptions:

1. A gas consists of tiny particles, either atoms or molecules, moving about at random. 2. The volume of the particles themselves is negligible compared with the total volume of the gas; most of the volume of a gas is empty space. 3. The gas particles act independently; there are no attractive or repulsive forces between particles. 4. Collisions of the gas particles, either with other particles or with the walls of the container, are elastic; that is, the total kinetic energy of the gas particles is constant at constant T. 5. The average kinetic energy of the gas particles is proportional to the Kelvin temperature of the sample.

9.81 Diffusion – The mixing of different gases by random molecular motion and with frequent

collisions. Effusion – The process in which gas molecules escape through a tiny hole in a membrane without collisions.

9.82 Heat is the energy transferred from one object to another as the result of a temperature

difference between them.


214

Temperature is a measure of the kinetic energy of molecular motion. 9.83 The atomic mass of He is much less than the molecular mass of the major components of

air (N2 and O2). The rate of effusion of He through the balloon skin is much faster.

9.84 mol / kg 10 x 28.0

K 220 x mol)K s/(m kg 8.314 x 3 =

M

RT 3 =u

3_

22

= 443 m/s

9.85 For Br2: mol / kg 10 x 159.8

K 293 x mol)K s/(m kg 8.314 x 3 =

M

RT3 =u

3_

22

= 214 m/s

For Xe: u = kg/mol 10 x 131.3

T x mol)K s/(m kg 8.314 x 3 = m/s 214

3_

22

Square both sides of the equation and solve for T.

45796 m2/s2 = kg/mol 10 x 131.3

T x mol)K s/(m kg 8.314 x 33_

22

T = 241 K = -32oC

9.86 For H2, kg/mol 10 x 2.02

K 150 x mol)K s/(m kg 8.314 x 3 =

M

RT 3 =u

3_

22

= 1360 m/s

For He, mol / kg 10 x 4.00

K 648 x mol)K s/(m kg 8.314 x 3 =u

3_

22

= 2010 m/s

He at 375oC has the higher average speed. 9.87 UF6, 352.02 amu; T = 25oC = 298 K

kg/mol 10 x 352.02

K 298 x mol)K s/(m kg 8.314 x 3 =

M

RT 3 =u

3_

22

= 145 m/s

Ferrari

145 s 3600

hr 1 x

km 1

m 1000 x

mi 1

km 1.6093 x

hr

mi = 64.8 m/s

The UF6 molecule has the higher average speed.

9.88 M

M = rate

rate

H

X

X

H

2

2 ; 2.02M =

1

2.92 X ; M = 2.02 2.92 X

MX = ) 2.02 2.92( 2 = 17.2 g/mol; molecular mass = 17.2 amu


215

9.89 M

M = Zrate

Xe rate

Xe

Z ; 131.29

M = 1.86

1 Z ; M = 1.86

131.29Z ; M =

)(1.86

131.29Z2

MZ = 37.9 g/mol; molecular mass = 37.9 amu; The gas could be F2. 9.90 HCl, 36.5 amu; F2, 38.0 amu; Ar, 39.9 amu

36.5

39.9 =

M

M = Ar rate

HCl rate

HCl

Ar = 1.05 38.0

39.9 =

M

M = Ar rateF rate

F

Ar2

2

= 1.02

The relative rates of diffusion are HCl(1.05) > F2(1.02) > Ar(1.00). 9.91 Because CO and N2 have the same mass, they will have the same diffusion rates.

9.92 u = mol / kg 10 x 4.00

T x mol)K s/(m kg 8.314 x 3 = m/s 45

3_

22


mol / kg 10 x 4.00

T x mol)K s/(m kg 8.314 x 3 = s/m 2025

3_

2222

T = 0.325 K = -272.83oC (near absolute zero)

9.93 230 km/h x s 3600

h 1 x

km 1

m 1000 = 63.9 m/s

u = 63.9 m/s = mol / kg 10 x 32.0

T x mol)K s/(m kg 8.314 x 33_

22


4083 m2/s2 = mol / kg 10 x 32.0

T x mol)K s/(m kg 8.314 x 33_

22

T = 5.24 K = -268oC General Problems

9.94 70.0

74.0 =

Cl MCl M

= Cl rateCl rate

235

237

237

235

= 1.03

72.0

74.0 =

Cl Cl MCl M

= Cl rate

Cl Cl rate3735

237

237

3735

= 1.01

The relative rates of diffusion are (1.00)Cl > Cl(1.01) Cl > (1.03)Cl 2373735

235 .

9.95 Average molecular mass of air = 28.96 amu; CO2, 44.01 amu

P = 760 mm Hg x g/mol 28.96

g/mol 44.01 = 1155 mm Hg


216

9.96 atm) (75

K) (1050mol K atm L

06 0.082mol) (1.00 =

P

nRT = V

••

= 1.1 L

9.97 1 atm = 1033.228 g/cm2

column height = (1033.228 g/cm2)(1 cm3/0.89g) = 1200 cm = 12 m

9.98 n = K) (293

mol K atm L

06 0.082

L) atm)(7.35 (2.15 =

RT

PV

••

= 0.657 mol Ar

0.657 mol Ar x Ar mol 1

Ar g 39.948 = 26.2 g Ar

mtotal = 478.1 g + 26.2 g = 504.3 g 9.99 This is initially a Boyle's Law problem, because only P and V are changing while n and T

remain fixed. The initial volume for each gas is the volume of their individual bulbs. The final volume for each gas is the total volume of the three bulbs. nRT = PiV i = PfVf; Vf = 1.50 + 1.00 + 2.00 = 4.50 L

For CO2: L) (4.50

L) atm)(1.50 (2.13 =

V

V P = Pf

iif = 0.710 atm

For H2: L) (4.50

L) atm)(1.00 (0.861 =

V

V P = Pf

iif = 0.191 atm

For Ar: L) (4.50

L) atm)(2.00 (1.15 =

V

V P = Pf

iif = 0.511 atm

From Dalton's Law, Ptotal = P + P + P ArHCO 22

Ptotal = 0.710 atm + 0.191 atm + 0.511 atm = 1.412 atm 9.100 (a) Bulb A contains CO2(g) and N2(g); Bulb B contains CO2(g), N2(g), and H2O(s).

(b) Initial moles of gas = n = K) (298

mol K atm L

06 0.082

L) (1.000Hg mm 760


= RT

PV

••

Initial moles of gas = 0.030 35 mol

mol gas in Bulb A = n = K) (298

mol K atm L

06 0.082

L) (1.000Hg mm 760


= RT

PV

••

= 0.011 78 mol


217

mol gas in Bulb B = n = K) (203

mol K atm L

06 0.082

L) (1.000Hg mm 760


= RT

PV

••

= 0.017 29 mol

n OH2 = ninitial - nA - nB = 0.030 35 - 0.011 78 - 0.017 29 = 0.001 28 mol = 0.0013 mol H2O

(c) Bulb A contains N2(g). Bulb B contains N2(g) and H2O(s). Bulb C contains N2(g) and CO2(s).

(d) nA = K) (298

mol K atm L

06 0.082

L) (1.000Hg mm 760

atm 1.00 x Hg mm 33.5

= RT

PV

••

= 0.001 803 mol

nB = K) (203

mol K atm L

06 0.082

L) (1.000Hg mm 760

atm 1.00 x Hg mm 33.5

= RT

PV

••

= 0.002 646 mol

nC = K) (83

mol K atm L

06 0.082

L) (1.000Hg mm 760

atm 1.00 x Hg mm 33.5

= RT

PV

••

= 0.006 472 mol

nN2 = nA + nB + nC = 0.001 803 + 0.002 646 + 0.006 472 = 0.010 92 mol N2

(e) nCO2

= ninitial - n OH2 - nN2

= 0.030 35 - 0.0013 - 0.010 92 = 0.0181 mol CO2

9.101 C3H5N3O9, 227.1 amu

(a) moles C3H5N3O9 = 1.00 g x g 227.1

mol 1 = 0.004 40 mol

nair = K) (293

mol K atm L

06 0.082

L) atm)(0.500 (1.00 =

RT

PV

••

= 0.0208 mol air

(b) moles gas from C3H5N3O9 = 0.004 40 mol x nitro mol 4

gas mol 29

moles gas from C3H5N3O9 = 0.0319 mol gas from C3H5N3O9 ntotal = 0.0319 mol + 0.0208 mol = 0.0527 mol


218

(c) P = L) (0.500

K) (698mol K atm L

06 0.082mol) (0.0527 =

V

nRT

••

= 6.04 atm

9.102 NH3, 17.03 amu; mol NH3 = 45.0 g x g 17.03

mol 1 = 2.64 mol

V

nRT = P or

V

an _ nb) _ (V

nRT = P

2

2

(a) At T = 0oC = 273 K

P = L) (1.000

K) (273mol K atm L

06 0.082mol) (2.64

••

= 59.1 atm

P = )L (1.000

)mol (2.64mol

atm L 4.17

_ L/mol)] 1mol)(0.037 (2.64 _ L) [(1.000

K) (273mol K atm L

06 0.082mol) (2.64

2

2

2

2

•

••

P = 65.6 atm - 29.1 atm = 36.5 atm

(b) At T = 50oC = 323 K

P = L) (1.000

K) (323mol K atm L

06 0.082mol) (2.64

••

= 70.0 atm

P = )L (1.000

)mol (2.64mol

atm L 4.17

_ L/mol)] 1mol)(0.037 (2.64 _ L) [(1.000

K) (323mol K atm L

06 0.082mol) (2.64

2

2

2

2

•

••

P = 77.6 atm - 29.1 atm = 48.5 atm

(c) At T = 100oC = 373 K

P = L) (1.000

K) (373mol K atm L

06 0.082mol) (2.64

••

= 80.8 atm

P = )L (1.000

)mol (2.64mol

atm L 4.17

_ L/mol)] 1mol)(0.037 (2.64 _ L) [(1.000

K) (373mol K atm L

06 0.082mol) (2.64

2

2

2

2

•

••

P = 89.6 atm - 29.1 atm = 60.5 atm At the three temperatures, the van der Waals equation predicts a much lower pressure than does the ideal gas law. This is likely due to the fact that NH3 can hydrogen bond leading to strong intermolecular forces.


219

9.103 (a) ntotal = K) (293

mol K atm L

06 0.082

L) (0.500Hg mm 760


= RT

PV

••

= 0.007 06 mol

(b) nB = K) (293

mol K atm L

06 0.082

L) (0.250Hg mm 760

atm 1 x Hg mm 344

= RT

PV

••

= 0.004 71 moles

(c) d = L 0.250

g 0.218 = 0.872 g/L

(d) molar mass = mol 71 0.004

g 0.218 = 46.3 g/mol, NO2; mol. mass = 46.3 amu

(e) Hg2CO3(s) + 6 HNO3(aq) → 2 Hg(NO3)2(aq) + 3 H2O(l) + CO2(g) + 2 NO2(g) 9.104 CO2, 44.01 amu

mol CO2 = 500.0 g CO2 x CO g 44.01

CO mol 1

2

2 = 11.36 mol CO2

PV = nRT

V

nRT = P =

L) (0.800

K) (700mol K atm L

06 0.082mol) (11.36

••

= 816 atm

9.105 (a) Let x = mol CnH2n + 2 in reaction mixture.

Combustion of CnH2n + 2 → nCO2 + (n +1)H2O needs 2

1 +n 3 =

2

1 +n +n

mol O2

Balanced equation is: CnH2n + 2(g) +

2

1 +n 3O2(g) → nCO2(g) + (n + 1)H2O(g)

In going from reactants to products, the increase in the number of moles is

[n + (n + 1)] -

2

1 +n 3 + 1 =

2

1 _n per mol of CnH2n + 2 reacted.

Before reaction: total mol = K) (298.15

mol K atm L

06 0.082

L) 0atm)(0.400 (2.000 =

RT

PV

••

= 0.032 70 mol

After reaction: total mol = K) (398.15

mol K atm L

06 0.082

L) 0atm)(0.400 (2.983 =

RT

PV

••

= 0.036 52 mol

Difference = 0.032 70 mol - 0.036 52 mol = 0.003 82 mol


220

Increase in number of mol =

2

1 _n x = 0.003 82 mol; x =

1 _n

82) 2(0.003

Also x = g/mol 2)] +n 1.008(2 +n [12.01

g 0.148 =

massmolar HC g 2 +n 2n

So 1 _n

82) 2(0.003 =

2.016 +n 14.026

0.148; 0.148 n - 0.148 = 0.107 n + 0.0154

0.041 n = 0.163; n = 0.041

0.163 = 4.0

CnH2n + 2 is C4H10 (butane); molar mass = (4)(12.01) + (10)(1.008) = 58.12 g/mol

(b) 0.148 g C4H10 x HC g 58.12

HC mol 1

104

104 = 0.002 55 mol C4H10

mol O2 initially = total mol - mol C4H10 = 0.032 70 mol - 0.002 55 mol = 0.030 15 mol O2

atm) (2.000mol 70 0.032

mol 55 0.002 = P

n

n = P initialtotal

HCHC

104

104

= 0.156 atm

atm) (2.000mol 70 0.032

mol 15 0.030 = P

n

n = P initialtotal

OO

2

2

= 1.844 atm

(c) C4H10(g) + 2

13 O2 → 4 CO2(g) + 5 H2O(g)

0.002 55 mol C4H10 x HC mol 1

CO mol 4

104

2 = 0.0102 mol CO2

0.002 55 mol C4H10 x HC mol 1

OH mol 5

104

2 = 0.012 75 mol H2O

mol O2 unreacted = total mol after reaction - mol CO2 - mol H2O = 0.03652 mol - 0.0102 mol - 0.01275 = 0.01357 mol O2

atm) (2.983mol 52 0.036

mol 0.0102 = P

n

n = P finaltotal

COCO

2

2

= 0.833 atm

atm) (2.983mol 52 0.036

mol 75 0.012 = P

n

n = P finaltotal

OHOH

2

2

= 1.041 atm

atm) (2.983mol 52 0.036

mol 57 0.013 = P

n

n = P finaltotal

OO

2

2

= 1.108 atm

9.106 (a) average molecular mass for natural gas

= (0.915)(16.04 amu) + (0.085)(30.07 amu) = 17.2 amu

total moles of gas = 15.50 g x gas g 17.2

gas mol 1 = 0.901 mol gas


221

(b) P = L) (15.00

K) (293mol K atm L

06 0.082mol) (0.901

••

= 1.44 atm

(c) P X = P totalCHCH 44

• = (1.44 atm)(0.915) = 1.32 atm

P X = P totalHCHC 6262• = (1.44 atm)(0.085) = 0.12 atm

(d) ∆Hcombustion(CH4) = -802.3 kJ/mol and ∆Hcombustion(C2H6) = -1427.7 kJ/mol Heat liberated = (0.915)(0.901 mol)(-802.3 kJ/mol)

+ (0.085)(0.901)(-1427.7 kJ/mol) = -771 kJ 9.107 PV = nRT

= K) (373.1

mol K atm L

06 0.082

L) atm)(10.0 (3.00 =

RT

PV = n (initial) total

••

0.980 mol

= K) (373.1

mol K atm L

06 0.082

L) atm)(10.0 (2.40 =

RT

PV = n (final) total

••

0.784 mol

CS2(g) + 3 O2(g) → CO2(g) + 2 SO2(g)

before reaction (mol) y 0.980 - y 0 0 change (mol) -x -3x +x +2x after reaction (mol) y - x = 0 0.980 - y - 3x x 2x

= n (final) total (y - x) + (0.980 - y - 3x) + x + 2x = 0.784 mol

0.980 mol - 4x + 3x = 0.784 mol x = 0.980 mol - 0.784 mol = 0.196 mol mol CO2 = x = 0.196 mol

L) (10.0


06 0.082mol) (0.196 =

V

nRT = PCO2

••

= 0.600 atm

mol SO2 = 2x = 2(0.196 mol) = 0.392 mol

L) (10.0


06 0.082mol) (0.392 =

V

nRT = PSO2

••

= 1.20 atm

mol O2 = 0.980 mol - y - 3x = 0.980 mol - x - 3x = 0.980 - 4(0.196 mol) = 0.196 mol = P = P COO 22

0.600 atm

9.108 (a) T = 0oC = 273 K; PV = nRT

nQ = K) (273

mol K atm L

06 0.082

L) 0atm)(0.050 (0.229 =

RT

PV

••

= 5.11 x 10-4 mol Q


222

Q molar mass = = Q mol 10 x 5.11

Q g 0.1004_

196 g/mol

Xe molar mass = 131.3 g/mol On molar mass = 196 g/mol - 131.3 g/mol = 65 g/mol So, n = 4 and XeO4 is the likely formula for Q. (b) XeO4(g) → Xe(g) + 2 O2(g) After decomposition, P + P = P OXeTotal 2

.

Because of the stoichiometry of the decomposition reaction, the partial pressure of O2 is twice the partial pressure of Xe. Let x = PXe and 2x = PO2

. P + P = P OXeTotal 2 = x + 2x = 3x = 0.941 atm

x = = 3

atm 0.941 0.314 atm

PXe = x = 0.314 atm; PO2 = P _ P XeTotal = 0.941 atm - 0.314 atm = 0.627 atm

9.109 Ca(ClO3)2, 206.98 amu; Ca(ClO)2, 142.98 amu

(a) Ca(ClO3)2(s) → CaCl2(s) + 3 O2(g) Ca(ClO)2(s) → CaCl2(s) + O2(g) (b) T = 700oC = 700 + 273 = 973 K PV = nRT

= K) (973

mol K atm L

06 0.082

L) atm)(10.0 (1.00 =

RT

PV = nO2

••

0.125 mol O2

Let X = mol Ca(ClO3)2 and let Y = mol Ca(ClO)2 X(206.98 g/mol) + Y(142.98 g/mol) = 10.0 g 3X + Y = 0.125 mol, so Y = 0.125 mol - 3X (substitute for Y and solve for X) X(206.98 g/mol) + (0.125 mol - 3X)(142.98 g/mol) = 10.0 g X(206.98 g/mol) + 17.9 g - X(428.94 g/mol) = 10.0 g X(206.98 g/mol) - X(428.94 g/mol) = 10.0 g - 17.9 g = -7.9 g X(-221.96 g/mol) = -7.9 g X = (-7.9 g)/(-221.96 g/mol) = 0.0356 mol Ca(ClO3)2 Y = 0.125 mol - 3X; Y = 0.125 mol - 3(0.0356 mol) = 0.0182 mol Ca(ClO)2

mass Ca(ClO3)2 = 0.0356 mol Ca(ClO3)2 x = )ClOCa( mol 1

)ClOCa( g 206.98

23

23 7.4 g Ca(ClO3)2

mass Ca(ClO)2 = 10.0 g - 7.4 g = 2.6 g Ca(ClO)2 9.110 PCl3, 137.3 amu; O2, 32.00 amu; POCl3, 153.3 amu

2 PCl3(g) + O2(g) → 2 POCl3(g)

mol PCl3 = 25.0 g x = PCl g 137.3

PCl mol 1

3

3 0.182 mol PCl3

mol O2 = 3.00 g x = O g 32.00

O mol 1

2

2 0.0937 mol O2


223

Check for limiting reactant.

mol O2 needed = 0.182 mol PCl3 x = PCl mol 2O mol 1

3

2 0.0910 mol O2 needed

There is a slight excess of O2. PCl3 is the limiting reactant.

mol POCl3 = 0.182 mol PCl3 x = PCl mol 2

POCl mol 2

3

3 0.182 mol POCl3

mol O2 left over = 0.0937 mol - 0.0910 mol = 0.0027 mol O2 left over T = 200.0oC = 200.0 + 273.15 = 473.1 K; PV = nRT

V

nRT = P = =

L) (5.00


06 0.082mol) 0.0027 + mol (0.182

••

1.43 atm

9.111 (a) T = 225oC = 225 + 273 = 498 K

PV = nRT

= P oNOCl =

V

nRT=

L) (400.0

K) (498mol K atm L

06 0.082mol) (2.00

••

0.204 atm

2 NOCl(g) → 2 NO(g) + Cl2(g)

initial (atm) 0.204 0 0 change (atm) -2x +2x +x equil (atm) 0.204 - 2x 2x x

Ptotal (after rxn) = (0.204 atm - 2x) + 2x + x = 0.246 atm x = 0.246 atm - 0.204 atm = 0.042 atm

= PNO 2x = 2(0.042) = 0.084 atm

= PCl2 x = 0.042 atm

= PNOCl 0.204 - 2x = 0.204 - 2(0.042) = 0.120 atm

(b) % NOCl decomposed = = % 100 x atm 0.204

atm 0.084 = % 100 x

P

x2o

NOCl

41%

9.112 O2, 32.00 amu; O3, 48.00 amu

3 O2(g) → 2 O3(g) initial (atm) 32.00 0 change (atm) -3x +2x after rxn (atm) 32.00 - 3x 2x

P + P = P OOTotal 32

= 30.64 atm = 32.00 atm - 3x + 2x = 32.00 atm - x

x = 32.00 atm - 30.64 atm = 1.36 atm = PO2

32.00 - 3x = 32.00 - 3(1.36 atm) = 27.92 atm

= PO32x = 2(1.36 atm) = 2.72 atm


224

T = 25oC = 25 + 273 = 298 K; PV = nRT

nO2=

K) (298mol K atm L

06 0.082

L) atm)(10.00 (27.92 =

RT

PV

••

= 11.42 mol O2

nO3=

K) (298mol K atm L

06 0.082

L) atm)(10.00 (2.72 =

RT

PV

••

= 1.11 mol O3

mass O2 = 11.42 mol O2 x = O mol 1O g 32.00

2

2 365.4 g O2

mass O3 = 1.11 mol O3 x = O mol 1O g 48.00

3

3 53.3 g O3

total mass = 365.4 g + 53.3 g = 418.7 g

mass % O3 = = 100% x g 418.7

g 53.3 =

mass totalO mass 3 12.7 %

9.113 CaCO3, 100.09 amu; CaO, 56.08 amu

mol CaO (or CO2) = 25.0 g CaCO3 x = CaCO mol 1

COor CaO mol 1 x

CaCO g 100.09CaCO mol 1

3

2

3

3 0.250

mol

mass CaO = 0.250 mol CaO x = CaO mol 1

CaO g 56.0814.02 g CaO

(a) 500.0 mL = 0.5000 L PV = nRT; = nCO2

0.250 mol

= PCO2 =

V

nRT=

L) (0.5000

K) (1500mol K atm L

06 0.082mol) (0.250

••

61.5 atm

(b) = VCaO (14.02 g)(3.34 g/mL) = 4.20 mL V = 500.0 mL - 4.20 mL = 495.8 mL = 0.4958 L

V

n a _

bn _ V

T Rn = P

2

2

P = )L (0.4958

)mol (0.250mol

atm L 3.59

_ L/mol)] 7mol)(0.042 (0.250 _ L) [(0.4958

K) (1500mol K atm L

06 0.082mol) (0.250

2

2

2

2

•

••

= 62.5 atm

Multi-Concept Problems 9.114 CO2, 44.01 amu

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ∆Ho = -802 kJ (a) 1.00 atm of CH4 only requires 2.00 atm O2, therefore O2 is in excess. T = 300oC = 300 + 273 = 573 K; PV = nRT


225

nCH4=

K) (573mol K atm L

06 0.082

L) atm)(4.00 (1.00 =

RT

PV

••

= 0.0851 mol CH4

nO2=

K) (573mol K atm L

06 0.082

L) atm)(4.00 (4.00 =

RT

PV

••

= 0.340 mol O2

mass CO2 = 0.0851 mol CH4 x = CO mol 1CO g 44.01

x CH mol 1CO mol 1

2

2

4

2 3.75 g CO2

(b) qrxn = 0.0851 mol CH4 x = CH mol 1

kJ 802 _

4

-68.3 kJ

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) initial (mol) 0.0851 0.340 0 0 change (mol) -0.0851 -2(0.0851) +0.0851 +2(0.0851) after rxn (mol) 0 0.340 - 2(0.0851) 0.0851 0.170

total moles of gas = 0.340 ml - 2(0.0851) mol + 0.0851 mol + 0.170 mol = 0.425 mol gas

qrxn = -68.3 kJ x = kJ 1

J 1000 - 68,300 J

qvessel = -qrxn = 68,300 J = (0.425 mol)(21 J/(mol ⋅ oC))( t f - 300oC) + (14.500

kg) C)300 _ tC))( J/(g (0.449kg 1

g 1000 of

o•

Solve for tf .

68,300 J = (8.925 J/oC + 6510 J/oC)( tf - 300oC) = (6519 J/oC)( tf

- 300oC)

= C/J 6519

J 68,300o

10.5oC = (t f - 300oC)

300oC + 10.5oC = t f

tf = 310oC

(c) T = 310oC = 310 + 273 = 583 K

V

nRT = PCO2

= = L) (4.00

K) (583mol K atm L

06 0.082mol) (0.0851

••

1.02 atm

9.115 X + 3 O2 → 2 CO2 + 3 H2O

(a) X = C2H6O C2H6O + 3 O2 → 2 CO2 + 3 H2O (b) It is an empirical formula because it is the smallest whole number ratio of atoms. It is also a molecular formula because any higher multiple such as C4H12O2 does not correspond to a stable electron-dot structure.


226

(c) (d) C2H6O, 46.07 amu

mol C2H6O = 5.000 g C2H6O x OHC g 46.07

OHC mol 1

62

62 = 0.1085 mol C2H6O

∆Hcombustion = mol 0.1085

kJ 144.2_= -1328.6 kJ/mol

∆Hcombustion = [2 ∆Hof(CO2) + 3 ∆Ho

f(H2O)] - ∆Hof(C2H6O)

∆Ho

f(C2H6O) = [2 ∆Hof(CO2) + 3 ∆Ho

f(H2O)] - ∆Hcombustion = [(2mol)(-393.5 kJ/mol) + (3 mol)(-241.8 kJ/mol)] - (-1328.6 kJ) = -183.8 kJ/mol

9.116 (a) 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)

(b) 4.6 x 1010 L C8H18 x mL 1

g 0.792 x

L 1

mL 1000 = 3.64 x 1013 g C8H18

3.64 x 1013 g C8H18 x HC mol 2

CO mol 16 x

HC g 114.2HC mol 1

188

2

188

188 = 2.55 x 1012 mol CO2

2.55 x 1012 mol CO2 x g 1000

kg 1 x

CO mol 1CO g 44.0

2

2 = 1.1 x 1011 kg CO2

(c) atm) (1.00

K) (273mol K atm L

06 0.082mol) 10 x (2.55 =

P

nRT = V

12

••

= 5.7 x 1013 L of CO2

(d) 12.5 moles of O2 are needed for each mole of isooctane (from part a).

12.5 mol O2 = (0.210)(nair); nair = 0.210

mol 12.5 = 59.5 mol air

atm) (1.00

K) (273mol K atm L

06 0.082mol) (59.5 =

P

nRT = V

••

= 1.33 x 103 L

9.117 (a) Freezing point of H2O on the Rankine scale is (9/5)(273.15) = 492oR.

(b) mol R

atm L 0.0456 =

)2mol)(49 (1.00

L) 4atm)(22.41 (1.00 =

nT

PV = R

o ••

∨

(c) P = )L (0.4000

)mol (2.50mol

atm L 2.253

_ L/mol)] 78mol)(0.042 (2.50 _ L) [(0.4000

R) (525mol R

atm L 0.0456mol) (2.50

2

2

2

2o

o

•

••

P = 204.2 atm - 88.0 atm = 116 atm


227

9.118 n = K) (2223

mol K atm L

06 0.082

L) atm)(1323 (1 =

RT

PV

••

= 7.25 mol of all gases

(a) 0.004 00 mol “nitro” x _nitro_ mol 1

gases mol 7.25 = 0.0290 mol hot gases

(b) n = K) (263

mol K atm L

06 0.082

L) (0.500Hg mm 760


= RT

PV

••

= 0.0190 mol B + C + D

nA = ntotal - n(B+C+D) = 0.0290 - 0.0190 = 0.0100 mol A; A = H2O

(c) n = K) (298

mol K atm L

06 0.082

L) (0.500Hg mm 760


= RT

PV

••

= 0.007 00 mol C + D

nB = n(B+C+D) - n(C+D) = 0.0190 - 0.007 00 = 0.0120 mol B; B = CO2

(d) n = K) (298

mol K atm L

06 0.082

L) (0.500Hg mm 760


= RT

PV

••

= 0.006 00 mol D

nC = n(C+D) - nD = 0.007 00 - 0.006 00 = 0.001 00 mol C; C = O2

molar mass D = mol 00 0.006

g 0.168 = 28.0 g/mol; D = N2

(e) 0.004 C3H5N3O9(l) → 0.0100 H2O(g) + 0.012 CO2(g) + 0.001 O2(g) + 0.006 N2(g) Multiply each coefficient by 1000 to obtain integers. 4 C3H5N3O9(l) → 10 H2O(g) + 12 CO2(g) + O2(g) + 6 N2(g)

9.119 CO2, 44.01 amu; H2O, 18.02 amu

(a) mol C = 0.3744 g CO2 x = CO mol 1

C mol 1 x

CO g 44.01CO mol 1

22

2 0.008 507 mol C

mass C = 0.008 507 mol C x = C mol 1

C g 12.011 0.1022 g C

mol H = 0.1838 g H2O x = OH mol 1

H mol 2 x

OH g 18.02

OH mol 1

22

2 0.020 400 mol H


228

mass H = 0.020 400 mol H x = H mol 1

H g 1.008 0.02056 g H

mass O = 0.1500 g - 0.1022 g - 0.02056 g = 0.0272 g O

mol O = 0.0272 g O x = O g 16.00

O mol 1 0.001 70 mol O

C0.008 507H0.020 400O0.001 70 Divide each subscript by the smallest, 0.001 70. C0.008 507 / 0.001 70H0.020 400 / 0.001 70O0.001 70 / 0.001 70 The empirical formula is C5H12O. The empirical formula mass is 88 g/mol.

(b) 1 atm = 101,325 Pa; T = 54.8oC = 54.8 + 273.15 = 327.9 K PV = nRT

n = = K) (327.9

mol K atm L

06 0.082

L) (1.00kPa 101.325

atm 1.00 x kPa 100.0

= RT

PV

••

0.0367 mol methyl tert-butyl ether

methyl tert-butyl ether molar mass = = mol 0.0367

g 3.233 88.1 g/mol

The empirical formula mass and the molar mass are the same, so the molecular formula and empirical formula are the same. C5H12O is the molecular formula and 88.15 amu is the molecular mass for methyl tert-butyl ether.

(c) C5H12O(l) + 15/2 O2(g) → 5 CO2(g) + 6 H2O(l)

(d) ∆Ho

combustion = [5 ∆Hof (CO2) + 6 ∆Ho

f (H2O(l))] - ∆Hof (C5H12O) = -3368.7 kJ

-3368.7 kJ = [(5 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] - (1 mol)∆Hof (C5H12O)

(1 mol)∆Hof (C5H12O) = [(5 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)] + 3368.7 kJ

∆Hof (C5H12O) = -313.6 kJ/mol

229

10

Liquids, Solids, and Changes of State

10.1 µ = Q x r = (1.60 x 10-19 C)(92 x 10-12 m)

• m C 10 x 3.336

D 130_

= 4.41 D

% ionic character for HF = D 4.41

D 1.82 x 100% = 41%

HF has more ionic character than HCl. HCl has only 17% ionic character. 10.2 (a) SF6 has polar covalent bonds but the molecule is symmetrical (octahedral). The

individual bond polarities cancel, and the molecule has no dipole moment. (b) H2C=CH2 can be assumed to have nonpolar C–H bonds. In addition, the molecule is symmetrical. The molecule has no dipole moment.

(c) The C–Cl bonds in CHCl3 are polar covalent bonds, and the molecule is polar.

(d) The C–Cl bonds in CH2Cl2 are polar covalent bonds, and the molecule is polar.

10.3 10.4 The N atom is electron rich (red) because of its high electronegativity. The H atoms are

electron poor (blue) because they are less electronegative. 10.5 (a) Of the four substances, only HNO3 has a net dipole moment.

(b) Only HNO3 can hydrogen bond. (c) Ar has fewer electrons than Cl2 and CCl4, and has the smallest dispersion forces.

10.6 H2S dipole-dipole, dispersion

230

CH3OH hydrogen bonding, dipole-dipole, dispersion C2H6 dispersion Ar dispersion Ar < C2H6 < H2S < CH3OH

10.7 (a) CO2(s) → CO2(g), ∆S is positive.

(b) H2O(g) → H2O(l), ∆S is negative. (c) ∆S is positive (more disorder).

10.8 ∆G = ∆H - T∆S; at the boiling point (phase change), ∆G = 0.

∆H = T∆S; T = mol) kJ/(K 10 x 87.5

kJ/mol 29.2 =

S

H3_

vap

vap

•∆∆

= 334 K

10.9 The boiling point is the temperature where the vapor pressure of a liquid equals the

external pressure. P1 = 760 mm Hg; P2 = 260 mm Hg; T1 = 80.1oC ∆Hvap = 30.8 kJ/mol

∆T

1 _

T

1

RH + Pln = Pln

21

vap12

T

1 _

T

1 =

H

R)Pln _ P(ln

21vap12

∆

Solve for T2 (the boiling point for benzene at 260 mm Hg).

T

1 =

H

R)Pln _ P(ln _

T

1

2vap12

1

∆

T

1 =

J/mol 30,800mol K

J 8.3145

ln(760)] _ )260[ln( _ K 353.2

1

2

•

T

1

2

= 0.003 121 K-1; T2 = 320 K = 47oC (boiling point is lower at lower pressure)

10.10 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 400 mm Hg; T1 = 41.0 oC = 314.2 K P2 = 760 mm Hg; T2 = 331.9 K

∆Hvap =

•

K 331.91

_ K 314.2

1mol KJ

8.3145 (400)]ln _ (760)[ln = 31,442 J/mol = 31.4 kJ/mol

10.11 (a) 1/8 atom at 8 corners and 1 atom at body center = 2 atoms

Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________

231

(b) 1/8 atom at 8 corners and 1/2 atom at 6 faces = 4 atoms

10.12 For a simple cube, d = 2r; r = 2

pm 334 =

2

d = 167 pm

10.13 For a simple cube, there is one atom per unit cell.

mass of one Po atom = atoms 10 x 6.022

mol 1 x g/mol 209

23 = 3.4706 x 10-22 g/atom

unit cell edge = d = 334 pm = 334 x 10-12 m = 3.34 x 10-8 cm unit cell volume = d3 = (3.34 x 10-8 cm)3 = 3.7260 x 10-23 cm3

density = cm 10 x 3.7260

g 10 x 3.4706 =

volume

mass323_

22_

= 9.31 g/cm3

10.14 There are several possibilities. Here's one.

10.15 For CuCl:

1/8 Cl- at 8 corners and 1/2 Cl- at 6 faces = 4 Cl- (4 minuses) 4 Cu+ inside (4 pluses) For BaCl2: 1/8 Ba2+ at 8 corners and 1/2 Ba2+ at 6 faces = 4 Ba2+ (8 pluses) 8 Cl- inside (8 minuses)

10.16 (a) In the unit cell there is a rhenium atom at each corner of the cube. The number of

rhenium atoms in the unit cell = 1/8 Re at 8 corners = 1 Re atom. In the unit cell there is an oxygen atom in the center of each edge of the cube. The number of oxygen atoms in the unit cell = 1/4 O on 12 edges = 3 O atoms. (b) ReO3 (c) Each oxide has a -2 charge and there are three of them for a total charge of -6. The charge (oxidation state) of rhenium must be +6 to balance the negative charge of the oxides. (d) Each oxygen atom is surrounded by two rhenium atoms. The geometry is linear. (e) Each rhenium atom is surrounded by six oxygen atoms. The geometry is octahedral.

10.17 The minimum pressure at which liquid CO2 can exist is its triple point pressure of 5.11

atm.


232

10.18 (a) CO2(s) → CO2(g) (b) CO2(l) → CO2(g) (c) CO2(g) → CO2(l) → supercritical CO2

10.19 (a)

(b) Gallium has two triple points. The one below 1 atm is a solid, liquid, vapor triple point. The one at 104 atm is a solid(1), solid(2), liquid triple point. (c) Increasing the pressure favors the liquid phase, giving the solid/liquid boundary a negative slope. At 1 atm pressure the liquid phase is more dense than the solid phase.

10.20 The molecules in a liquid crystal can move around, as in viscous liquids, but they have a

restricted range of motion, as in solids. 10.21 Liquid crystal molecules have a rigid rodlike shape with a length four to eight times

greater than their diameter. Understanding Key Concepts 10.22 The electronegative O atoms are electron rich (red), while the rest of the molecule is

electron poor (blue). 10.23 (a) cubic closest-packed (b) simple cubic

(c) hexagonal closest-packed (d) body-centered cubic 10.24 (a) cubic closest-packed

(b) 1/8 S2- at 8 corners and 1/2 S2- at 6 faces = 4 S2-; 4 Zn2+ inside 10.25 (a) 1/8 Ca2+ at 8 corners = 1 Ca2+; 1/2 O2- at 6 faces = 3 O2-; 1 Ti4+ inside

The formula for perovskite is CaTiO3. (b) The oxidation number of Ti is +4 to maintain charge neutrality in the unit cell.

10.26 (a) normal boiling point ≈ 300 K; normal melting point ≈ 180 K

(b) (i) solid (ii) gas (iii) supercritical fluid


233

10.27 10.28 Here are two possibilities.

10.29 (a), (c), (d)

(b) There are three triple points. (e) The solid phase that is stable at the higher pressure is more dense. The more dense phase is diamond.


234

Additional Problems Dipole Moments and Intermolecular Forces 10.30 If a molecule has polar covalent bonds, the molecular shape (and location of lone pairs of

electrons) determines whether a molecule has a dipole moment or not. The molecular shape will determine whether the bond dipoles cancel or not.

10.31 Dipole-dipole forces arise between molecules that have permanent dipole moments.

London dispersion forces arise between molecules as a result of induced temporary dipoles. 10.32 (a) CHCl3 has a permanent dipole moment. Dipole-dipole intermolecular forces are

important. London dispersion forces are also present. (b) O2 has no dipole moment. London dispersion intermolecular forces are important. (c) polyethylene, CnH2n+2. London dispersion intermolecular forces are important. (d) CH3OH has a permanent dipole moment. Dipole-dipole intermolecular forces and hydrogen bonding are important. London dispersion forces are also present.

10.33 (a) Xe has no dipole-dipole forces (b) HF has the largest hydrogen bond forces

(c) Xe has the largest dispersion forces 10.34 For CH3OH and CH4, dispersion forces are small. CH3OH can hydrogen bond; CH4

cannot. This accounts for the large difference in boiling points. For 1-decanol and decane, dispersion forces are comparable and relatively large along the C–H chain. 1-decanol can hydrogen bond; decane cannot. This accounts for the 55oC higher boiling point for 1-decanol.

10.35 (a) C8H18 has the larger dispersion forces because of its longer hydrocarbon chain.

(b) HI has the larger dispersion forces because of the larger, more polarizable iodine. (c) H2Se has the larger dispersion forces because of the more polarizable and less electronegative Se.

10.36 (a) (b)

(c) (d)


235

10.37 (a) (b)

(c) (d)

10.38 SO2 is bent and the individual bond dipole moments add to give the molecule a net dipole moment. CO2 is linear and the individual bond dipole moments point in opposite directions to cancel each other out. CO2 has no net dipole moment.

10.39 In both PCl3 and PCl5 the P–Cl bond is polar covalent. PCl3 is trigonal pyramidal and the bond dipoles add to give the molecule a net dipole moment. PCl5 is trigonal bipyramidal and the bond dipoles cancel. PCl5 has no dipole moment.

10.40

10.41 Vapor Pressure and Changes of State 10.42 ∆Hvap is usually larger than ∆Hfusion because ∆Hvap is the heat required to overcome all

intermolecular forces. 10.43 Sublimation is the direct conversion of a solid to a gas. A solid can also be converted to

a gas in two steps; melting followed by vaporization. The energy to convert a solid to a gas must be the same regardless of the path. Therefore ∆Hsubl = ∆Hfusion + ∆Hvap.


236

10.44 (a) Hg(l) → Hg(g)

(b) no change of state, Hg remains a liquid (c) Hg(g) → Hg(l) → Hg(s)

10.45 (a) solid I2 melts to form liquid I2 (b) no change of state, I2 remains a liquid 10.46 As the pressure over the liquid H2O is lowered, H2O vapor is removed by the pump. As

H2O vapor is removed, more of the liquid H2O is converted to H2O vapor. This conversion is an endothermic process and the temperature decreases. The combination of both a decrease in pressure and temperature takes the system across the liquid/solid boundary in the phase diagram so the H2O that remains turns to ice.

10.47 The normal boiling point for ether is relatively low (34.6oC). As the pressure is reduced

by the pump, the relatively high vapor pressure of the ether equals the external pressure produced by the pump and the liquid boils.

10.48 H2O, 18.02 amu; 5.00 g H2O x OH g 18.02

OH mol 1

2

2 = 0.2775 mol H2O

q1 = (0.2775 mol)[36.6 x 10-3 kJ/(K ⋅ mol)](273 K - 263 K) = 0.1016 kJ q2 = (0.2775 mol)(6.01 kJ/mol) = 1.668 kJ q3 = (0.2775 mol)(75.3 x 10-3 kJ/(K ⋅ mol)](303 K - 273 K) = 0.6269 kJ qtotal = q1 + q2 + q3 = 2.40 kJ; 2.40 kJ of heat is required.

10.49 H2O, 18.02 amu; 15.3 g H2O x OH g 18.02

OH mol 1

2

2 = 0.8491 mol H2O

q1 = (0.8491 mol)[33.6 x 10-3 kJ/(K ⋅ mol)](373 K - 388 K) = -0.4279 kJ q2 = -(0.8491 mol)(40.67 kJ/mol) = -34.53 kJ q3 = (0.8491 mol)[75.3 x 10-3 kJ/(K ⋅ mol)](348 K - 373 K) = -1.598 kJ qtotal = q1 + q2 + q3 = -36.6 kJ; 36.6 kJ of heat is released.

10.50 H2O, 18.02 amu; 7.55 g H2O x OH g 18.02

OH mol 1

2

2 = 0.4190 mol H2O

q1 = (0.4190 mol)[75.3 x 10-3 kJ/(K ⋅ mol)](273.15 K - 306.65 K) = -1.057 kJ q2 = -(0.4190 mol)(6.01 kJ/mol) = -2.518 kJ q3 = (0.4190 mol)[36.6 x 10-3 kJ/(K ⋅ mol)](263.15 K - 273.15 K) = -0.1534 kJ qtotal = q1 + q2 + q3 = -3.73 kJ; 3.73 kJ of heat is released.

10.51 C2H5OH, 46.07 amu; 25.0 g C2H5OH x OHHC g 46.07

OHHC mol 1

52

52 = 0.543 mol C2H5OH

q1 = (0.543 mol)[65.7 x 10-3 kJ/(K ⋅ mol)](351.55 K - 366.15 K) = -0.521 kJ


237

q2 = -(0.543 mol)(38.56 kJ/mol) = -20.94 kJ q3 = (0.543 mol)[113 x 10-3 kJ/(K ⋅ mol)](263.15 K - 351.55 K) = -5.42 kJ qtotal = q1 + q2 + q3 = -26.9 kJ; 26.9 kJ of heat is released.

10.52

10.53 10.54 boiling point = 218oC = 491 K

∆G = ∆Hvap - T∆Svap; At the boiling point (phase change), ∆G = 0

∆Hvap = T∆Svap; ∆Svap = K 491

kJ/mol 43.3 =

THvap∆

= 0.0882 kJ/(K⋅ mol) = 88.2 J/(K⋅ mol)

10.55 K 371

kJ/mol 2.64 =

TH = S

fusfus

∆∆ = 0.007 12 kJ/(K⋅ mol) = 7.12 J/(K⋅ mol)

10.56 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

T1 = -5.1oC = 268.0 K; P1 = 100 mm Hg T2 = 46.5oC = 319.6 K; P2 = 760 mm Hg


238

∆Hvap =

•

K 319.61

_ K 268.0

1mol)] kJ/(K 10 x 145(100)][8.3ln _ (760)[ln 3_

= 28.0 kJ/mol

10.57 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 100 mm Hg; T1 = 5.4oC = 278.6 K P2 = 760 mm Hg; T2 = 56.8oC = 330.0 K

∆Hvap =

•

K 330.01

_ K 278.6

1mol)] kJ/(K 10 x .3145ln(100)][8 _ [ln(760) 3_

= 30.2 kJ/mol

10.58 ln P2 = ln P1 +

∆T

1 _

T

1

RH

21

vap

∆Hvap = 28.0 kJ/mol P1 = 100 mm Hg; T1 = -5.1oC = 268.0 K; T2 = 20.0oC = 293.2 K Solve for P2.

ln P2 = ln (100) +

• K 293.2

1 _

K 268.0

1

mol)] kJ/(K 10 x [8.3145

kJ/mol 28.03_

ln P2 = 5.6852; P2 = e5.6852 = 294.5 mm Hg = 294 mm Hg

10.59 ln P2 = ln P1 +

∆T

1 _

T

1

RH

21

vap

∆Hvap = 30.2 kJ/mol P1 = 100 mm Hg; T1 = 5.4oC = 278.6 K; T2 = 30.0oC = 303.2 K Solve for P2.

ln P2 = ln (100) +

• K 303.2

1 _

K 278.6

1

mol)] kJ/(K 10 x [8.3145

kJ/mol 30.23_

ln P2 = 5.6630; P2 = e5.6620 = 288.0 mm Hg = 288 mm Hg 10.60 T(K) Pvap(mm Hg) ln Pvap 1/T

263 80.1 4.383 0.003 802 273 133.6 4.8949 0.003 663 283 213.3 5.3627 0.003 534 293 329.6 5.7979 0.003 413


239

303 495.4 6.2054 0.003 300 313 724.4 6.5853 0.003 195

ln Pvap = T

1

RH _ vap

∆ + C; C = 18.2

slope = -3628 K = RH _ vap∆

∆Hvap = (3628 K)(R) = (3628 K)[8.3145 x 10-3 kJ/(K⋅ mol)] = 30.1 kJ/mol

10.61 T(K) Pvap(mm Hg) ln Pvap 1/T

500 39.3 3.671 0.002 000 520 68.5 4.227 0.001 923 540 114.4 4.7397 0.001 852 560 191.6 5.2554 0.001 786 580 286.4 5.6574 0.001 724 600 432.3 6.0691 0.001 667

ln Pvap = T

1

RH _ vap

∆ + C; C = 18.1

slope = -7219 K = RH _ vap∆

∆Hvap = (7219 K)(R) = (7219 K)[8.3145 x 10-3 kJ/(K⋅ mol)] = 60.0 kJ/mol 10.62 ∆Hvap = 30.1 kJ/mol 10.63 ∆Hvap = 60.0 kJ/mol

10.64 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 80.1 mm Hg; T1 = 263 K P2 = 724.4 mm Hg; T2 = 313 K


240

∆Hvap =

•

K 3131

_ K 263

1mol)] kJ/(K 10 x 3145(80.1)][8.ln _ (724.4)[ln 3_

= 30.1 kJ/mol

The calculated ∆Hvap and that obtained from the plot in Problem 10.62 are the same.

10.65 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 39.3 mm Hg; T1 = 500 K P2 = 432.3 mm Hg; T2 = 600 K

∆Hvap =

•

K 6001

_ K 500

1mol)] kJ/(K 10 x 8.3145ln(39.3)][ _ [ln(432.3) 3_

= 59.8 kJ/mol

The calculated ∆Hvap and that obtained from the plot in Problem 10.63 are consistent with each other. The value from the slope is 60.0 kJ/mol

Structures of Solids 10.66 molecular solid, CO2, I2; metallic solid, any metallic element;

covalent network solid, diamond; ionic solid, NaCl 10.67 molecular solid, covalent molecules; metallic solid, metal atoms;

covalent network solid, nonmetal atoms; ionic solid, cations and anions 10.68 The unit cell is the smallest repeating unit in a crystal. 10.69 From Table 10.10.

Hexagonal and cubic closest packing are the most efficient because 74% of the available space is used. Simple cubic packing is the least efficient because only 52% of the available space is used.

10.70 Cu is face-centered cubic. d = 362 pm; r = 8

)pm (362 =

8d

22

= 128 pm

362 pm = 362 x 10-12 m = 3.62 x 10-8 cm unit cell volume = (3.62 x 10-8 cm)3 = 4.74 x 10-23 cm3

mass of one Cu atom = 63.55 g/mol x atom 10 x 6.022

mol 123

= 1.055 x 10-22 g/atom

Cu is face-centered cubic; there are therefore four Cu atoms in the unit cell. unit cell mass = (4 atoms)(1.055 x 10-22 g/atom) = 4.22 x 10-22 g


g10 x 4.22 =

volume

mass323_

22_

= 8.90 g/cm3


241

10.71 Pb is face-centered cubic. d = 495 pm = 4.95 x 10-8 cm

r = 8

)pm (495 =

8d

22

= 175 pm

unit cell volume = (4.95 x 10-8 cm)3 = 1.2129 x 10-22 cm3

mass of one Pb atom = atoms 10 x 6.022

mol 1 x g/mol 207.2

23 = 3.4407 x 10-22 g/atom

Pb is face-centered cubic; there are therefore four Pb atoms in the unit cell.


g) 10 x 4(3.4407 =

volume

mass322_

22_

= 11.3 g/cm3

10.72 mass of one Al atom = 26.98 g/mol x atom 10 x 6.022

mol 123

= 4.480 x 10-23 g/atom

Al is face-centered cubic; there are therefore four Al atoms in the unit cell. unit cell mass = (4 atoms)(4.480 x 10-23 g/atom) = 1.792 x 10-22 g

density = volume

mass

unit cell volume = density

mass cellunit =

cmg/ 2.699

g 10 x 1.7923

22_

= 6.640 x 10-23 cm3

unit cell edge = d = 3 323_ cm 10 x 6.640 = 4.049 x 10-8 cm

d = 4.049 x 10-8 cm x cm 100

m1 = 4.049 x 10-10 m = 404.9 x 10-12 m = 404.9 pm

10.73 W is body-centered cubic. d = 317 pm

a = edge = d; b = face diagonal; c = body diagonal b2 = 2a2 c2 = a2 + b2 c2 = a2 + 2a2 = 3a2

c = a 3

unit cell body diagonal = pm) (317 3 = d 3 = 549 pm 10.74 unit cell body diagonal = 4r = 549 pm

For W, r = 4

pm 549 = 137 pm

10.75 mass of one Na atom = atoms 10 x 6.022

mol 1 x g/mol 23.0

23 = 3.82 x 10-23 g/atom

Because Na is body-centered cubic; there are two Na atoms in the unit cell. unit cell mass = 2(3.82 x 10-23 g) = 7.64 x 10-23 g

unit cell volume = cmg/ 0.971

g 10 x 7.64 =

density

mass cellunit 3

23_

= 7.87 x 10-23 cm3

unit cell edge = d = 3 323_ cm 10 x 7.87 = 4.29 x 10-8 cm = 429 pm


242

4R = d 3 ; R = 4

pm) (429 3 =

4

d 3 = 186 pm

10.76 mass of one Ti atom = 47.88 g/mol x atoms 10 x 6.022

mol 123

= 7.951 x 10-23 g/atom

r = 144.8 pm = 144.8 x 10-12 m

r = 144.8 x 10-12 m x m 1

cm 100 = 1.448 x 10-8 cm

Calculate the volume and then the density for Ti assuming it is primitive cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell. For primitive cubic:

d = 2r; volume = d3 = [2(1.448 x 10-8 cm)]3 = 2.429 x 10-23 cm3


g 10 x 7.951 =

volume

mass cellunit 323_

23_

= 3.273 g/cm3

For face-centered cubic:

d = 2 2 r; volume = d3 = [2 2 (1.448 x 10-8 cm)]3 = 6.870 x 10-23 cm3


g) 10 x 4(7.951323_

23_

= 4.630 g/cm3

For body-centered cubic: From Problems 10.73 and 10.74,

d = 3

r4; volume = d3 =

3

cm) 10 x 4(1.448 8_ 3

= 3.739 x 10-23 cm3


g) 10 x 2(7.951323_

23_

= 4.253 g/cm3

The calculated density for a face-centered cube (4.630 g/cm3) is closest to the actual density of 4.54 g/cm3. Ti crystallizes in the face-centered cubic unit cell.

10.77 mass of one Ca = atom 10 x 6.022

mol 1 x g/mol 40.08

23 = 6.656 x 10-23 g/atom

unit cell edge = d = 558.2 pm = 5.582 x 10-8 cm unit cell volume = d3 = (5.582 x 10-8 cm)3 = 1.739 x 10-22 cm3 unit cell mass = (1.739 x 10-22 cm3)(1.55 g/cm3) = 2.695 x 10-22 g

(a) number of Ca atoms in unit cell = atom Ca one of mass

mass cellunit

= g/atom 10 x 6.656

g 10 x 2.69523_

22_

= 4.05 = 4 Ca atoms

(b) Because the unit cell contains 4 Ca atoms, the unit cell is face-centered cubic. 10.78 Six Na+ ions touch each H- ion and six H- ions touch each Na+ ion.


243

10.79 For CsCl: (1/8 x 8 corners), so 1 Cl- and 1 minus per unit cell

1 Cs+ inside, so 1 plus per unit cell 10.80 Na+ H- Na+

← 488 pm → unit cell edge = d = 488 pm; Na–H bond = d/2 = 244 pm 10.81 See Problem 10.73.

body diagonal = d 3 = pm) (412.3 3 = 714.12 pm Cs–Cl bond = body diagonal/2 = (714.12 pm)/2 = 357.1 pm Cs–Cl bond length = rCs+ + rCl_

357.1 pm = rCs+ + rCl_

357.1 pm = rCs+ + 181 pm

rCs+ = 357.1 pm - 181 pm = 176 pm

Phase Diagrams 10.82 (a) gas (b) liquid (c) solid 10.83 (a) H2O(l) → H2O(s)

(b) 380oC is above the critical temperature; therefore, the water cannot be liquefied. At the higher pressure, it will behave as a supercritical fluid.

10.84


244

10.85 10.86 (a) Br2(s) (b) Br2(l) 10.87 (a) O2(l) (b) O2 - supercritical fluid 10.88 Solid O2 does not melt when pressure is applied because the solid is denser than the

liquid and the solid/liquid boundary in the phase diagram slopes to the right. 10.89 Ammonia can be liquefied at 25oC because this temperature is below Tc (132.5oC).

Methane cannot be liquefied at 25oC because this temperature is above Tc (-82.1oC). Sulfur dioxide can be liquefied at 25oC because this temperature is below Tc (157.8oC).

10.90

The starting phase is benzene as a solid, and the final phase is benzene as a gas.


245

10.91

The starting phase is a gas, and the final phase is a liquid.

10.92 solid → liquid → supercritical fluid → liquid → solid → gas 10.93 gas → solid → liquid → gas → liquid General Problems 10.94 Because chlorine is larger than fluorine, the charge separation is larger in CH3Cl

compared to CH3F resulting in CH3Cl having a slightly larger dipole moment. 10.95 Because Ar crystallizes in a face-centered cubic unit cell, there are four Ar atoms in the

unit cell.

mass of one Ar atom = 39.95 g/mol x atom 10 x 6.022

mol 123

= 6.634 x 10-23 g/atom

unit cell mass = 4 atoms x mass of one Ar atom = 4 atoms x 6.634 x 10-23 g/atom = 2.654 x 10-22 g

density = volume

mass

unit cell volume = density

mass cellunit =

cmg/ 1.623

g 10 x 2.6543

22_

= 1.635 x 10-22 cm3

unit cell edge = d = 3 322_ cm 10 x 1.635 = 5.468 x 10-8 cm

d = 5.468 x 10-8 cm x cm 100

m1 = 5.468 x 10-10 m = 546.8 x 10-12 m = 546.8 pm


246

r = 8

)pm (546.8 =

8d

22

= 193.3 pm

10.96 7.50 g x g 200.6

mol 1 = 0.037 39 mol Hg

q1 = (0.037 39 mol)[28.2 x 10-3 kJ/(K ⋅ mol)](234.2 K - 223.2 K) = 0.011 60 kJ q2 = (0.037 39 mol)(2.33 kJ/mol) = 0.087 12 kJ q3 = (0.037 39 mol)[27.9 x 10-3 kJ/(K ⋅ mol)](323.2 K - 234.2 K) = 0.092 84 kJ qtotal = q1 + q2 + q3 = 0.192 kJ; 0.192 kJ of heat is required.

10.97

10.98 ln P2 = ln P1 +

∆T

1 _

T

1

RH

21

vap

∆Hvap = 40.67 kJ/mol At 1 atm, H2O boils at 100oC; therefore set T1 = 100oC = 373 K, and P1 = 1.00 atm. Let T2 = 95oC = 368 K, and solve for P2. (P2 is the atmospheric pressure in Denver.)

ln P2 = ln(1) +

• K 368

1 _

K 373

1

mol)] kJ/(K 0l x [8.3145

kJ/mol 40.673_

ln P2 = -0.1782; P2 = e-0.1782 = 0.837 atm

10.99 10.100 ∆G = ∆H - T∆S; at the melting point (phase change), ∆G = 0.

∆H = T∆S; T = mol) kJ/(K 10 x 9.79

kJ/mol 9.037 =

S

H3_

fus

fus

•∆∆

= 923 K = 650oC

10.101 melting point = -23.2oC = 250.0 K

∆G = ∆Hfusion - T∆Sfusion At the melting point (phase change), ∆G = 0 ∆Hfusion = T∆Sfusion

∆Sfusion = K 250.0

kJ/mol 9.37 =

THfusion∆

= 0.0375 kJ/(K ⋅ mol) = 37.5 J/(K ⋅ mol)


247

10.102 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 40.0 mm Hg; T1 = -81.6oC =191.6 K P2 = 400 mm Hg; T2 = -43.9oC = 229.2 K

∆Hvap =

•

K 229.21

_ K 191.6

1mol K

kJ10 x 8.3145(40.0)]ln _ (400)[ln 3_

= 22.36 kJ/mol

Using ∆Hvap = 22.36 kJ/mol

∆T

1 _

T

1

RH + Pln = Pln

21

vap12

T

1 _

T

1 =

H

R)Pln _ P(ln

21vap12

∆

T

1 =

H

R)Pln _ P(ln _

T

1

2vap12

1

∆

P1 = 40.0 mm Hg; T1 = 191.6 K P2 = 760 mm Hg Solve for T2 (the normal boiling point).

T

1 =

kJ/mol 22.36mol K

kJ10 x 8.3145

ln(40.0)] _ )760[ln( _ K 191.6

1

2

3_

•

T

1

2

= 0.004 124 33; T2 = 242.46 K = -30.7oC

10.103 (a)

∆T

1 _

T

1

RH + Pln = Pln

21

vap12

T

1 _

T

1 =

H

R)Pln _ P(ln

21vap12

∆

T

1 =

H

R)Pln _ P(ln _

T

1

2vap12

1

∆

P1 = 100.0 mm Hg; T1 = -23oC = 250 K P2 = 760.0 mm Hg Solve for T2, the normal boiling point for CCl3F.


248

T

1 =

kJ/mol 24.77mol K

kJ10 x 8.3145

ln(100.0)] _ )760.0[ln( _ K 250

1

2

3_

•

T

1

2

= 0.003 319; T2 = 301.3 K = 28.1oC

(b) ∆Svap = K 301.3

kJ/mol 24.77 =

THvap∆

= 0.082 21 kJ/(K ⋅ mol) = 82.2 J/(K ⋅ mol)

10.104 ∆Hvap =

T

1 _

T

1

)(R)Pln _ P(ln

21

12

P1 = 100 mm Hg; T1 = -110.3oC = 162.85 K P2 = 760 mm Hg; T2 = -88.5oC = 184.65 K

∆Hvap =

•

K 184.651

_ K 162.85

1mol K

kJ10 x 8.3145(100)]ln _ (760)[ln 3_

= 23.3 kJ/mol

10.105

∆T

1 _

T

1

RH + Pln = Pln

21

vap12

T

1 _

T

1 =

H

R)Pln _ P(ln

21vap12

∆

T

1 =

H

R)Pln _ P(ln _

T

1

2vap12

1

∆

P1 = 760 mm Hg; T1 = 56.2oC = 329.4 K P2 = 105 mm Hg Solve for T2.

T

1 =

kJ/mol 29.1mol K

kJ10 x 8.3145

ln(760)] _ )105[ln( _ K 329.4

1

2

3_

•

T

1

2

= 0.003 601; T2 = 277.7 K = 4.5oC


249

10.106 Kr cannot be liquified at room temperature because room temperature is above Tc (-63oC). 10.107 (a) Kr(l) (b) supercritical Kr 10.108 For a body-centered cube

4r = 3 edge; edge = 3

r4

volume of sphere = r3

4 3π

volume of unit cell =

3

r43

= 3 3r 64 3

volume of 2 spheres =

r

3

4 2 3π = r

3

8 3π

% volume occupied =

33r 64

r 38

3

3π x 100% = 68%

10.109 From Problem 10.73, 4r = 3 d; r = 4

pm) (2873 =

4

d 3 = 124 pm

10.110 unit cell edge = d = 287 pm = 287 x 10-12 m = 2.87 x 10-8 cm

unit cell volume = d3 = (2.87 x 10-8 cm)3 = 2.364 x 10-23 cm3 unit cell mass = (2.364 x 10-23 cm3)(7.86 g/cm3) = 1.858 x 10-22 g Fe is body-centered cubic; therefore there are two Fe atoms per unit cell.

mass of one Fe atom = atoms Fe 2

g 10 x 1.858 22_

= 9.290 x 10-23 g/atom

Avogadro's number = g 10 x 9.290

atom 1 x g/mol 55.85

23_ = 6.01 x 1023 atoms/mol


250

10.111 unit cell edge = d = 408 pm = 408 x 10-12 m = 4.08 x 10-8 cm

unit cell volume = (4.08 x 10-8 cm)3 = 6.792 x 10-23 cm3 unit cell mass = (10.50 g/cm3)(6.792 x 10-23 cm3) = 7.132 x 10-22 g Ag is face-centered cubic; therefore there are four Ag atoms in the unit cell.

mass of one Ag atom = atoms Ag 4

g10 x 7.132 22_

= 1.783 x 10-22 g/atom

Avogadro's number = g 10 x 1.783

atom 1 x g/mol 107.9

22_ = 6.05 x 1023 atoms/mol

10.112 (a) unit cell edge = r2 + r2 NaCl +_ = 2(181 pm) + 2(97 pm) = 556 pm

(b) unit cell edge = d = 556 pm = 556 x 10-12 m = 5.56 x 10-8 cm unit cell volume = (5.56 x 10-8 cm)3 = 1.719 x 10-22 cm3 The unit cell contains 4 Na+ ions and 4 Cl- ions.

mass of one Na+ ion = 22.99 g/mol x ions 10 x 6.022

mol 123

= 3.818 x 10-23 g/Na+

mass of one Cl- ion = 35.45 g/mol x ions 10 x 6.022

mol 123

= 5.887 x 10-23 g/Cl-

unit cell mass = 4(3.818 x 10-23 g) + 4(5.887 x 10-23 g) = 3.882 x 10-22 g


g 10 x 3.882 =

volumecellunit

mass cellunit 322_

22_

= 2.26 g/cm3

10.113 (a) (1/2 Nb/face)(6 faces) = 3 Nb; (1/4 O/edge)(12 edges) = 3 O

(b) NbO (c) The oxidation state of Nb is +2.

10.114 Al2O3, ionic (greater lattice energy than NaCl because of higher ion charges);

F2, dispersion; H2O, dipole-dipole, H–bonding; Br2, dispersion (larger and more polarizable than F2), ICl, dipole-dipole, NaCl, ionic

rank according to normal boiling points: F2 < Br2 < ICl < H2O < NaCl < Al2O3

10.115 Ag2Te, 343. 33 amu; 529 pm = 529 x 10-12 m = 529 x 10-10 cm

unit cell volume = (529 x 10-10 cm)3 = 1.48 x 10-22 cm3 unit cell mass = (1.48 x 10-22 cm3)(7.70 g/cm3) = 1.14 x 10-21 g

mass of one Ag2Te = = mol / units formula TeAg 10 x 6.022

mol / TeAg g 343.33

223

2 5.70 x 10-22 g Ag2Te/formula

unit

Ag2Te formula units/unit cell = = TeAgg/ 10 x 5.70

cellg/unit 10 x 1.14

222_

21_

2 Ag2Te/unit cell

Ag/unit cell = = TeAg

Ag 2 x

cellunit

TeAg 2

2

2 4 Ag/unit cell


251

10.116 (a)

(b) (i) solid (ii) gas (iii) liquid (iv) liquid (v) solid Multi-Concept Problems 10.117 C2H5OH(l) → C2H5OH(g)

Calculate ∆H and ∆S for this process and assume they do not change as a function of temperature. ∆Ho = ∆Ho

f(C2H5OH(g)) - ∆Hof(C2H5OH(l))

∆Ho = [(1 mol)(-235.1 kJ/mol)] - [(1 mol)(-277.7 kJ/mol)] = 42.6 kJ ∆So = So(C2H5OH(g)) - So(C2H5OH(l)) ∆So = [(1 mol)(282.6 J/(K ⋅ mol))] - [(1 mol)(161 J/(K ⋅ mol))] = 122 J/K ∆So = 122 x 10-3 kJ/K) ∆Go = ∆Ho - T∆So and at the boiling point, ∆G = 0 0 = ∆Ho - Tbp∆So Tbp∆So = ∆Ho

Tbp = = kJ/K 10 x 122

kJ 42.6 =

S H

3_o

o

∆∆

349 K

Tbp = 349 - 273 = 76oC

ln P2 = ln P1 +

∆T

1 _

T

1

RH

21

vap

∆Hvap = 42.6 kJ/mol At 1 atm, C2H5OH boils at 349 K; therefore set T1 = 349 K, and P1 = 1.00 atm. Let T2 = 25oC = 298 K, and solve for P2. P2 is the vapor pressure of C2H5OH at 25oC.

ln P2 = ln(1.00) +

• K 298

1 _

K 349

1

mol)] kJ/(K 0l x [8.3145

kJ/mol 42.63_

ln P2 = -2.512; P2 = e-2.512 = 0.0811 atm


252

P2 = 0.0811 atm x = atm 1.00

Hg mm 760 61.6 mm Hg

10.118 (a) Let the formula of magnetite be FexOy, then FexOy + y CO → x Fe + y CO2

=y = nCO2

K) (298mol K atm L

06 0.082

L) (1.136Hg mm 760


= RT

PV

••

= 0.04590 mol CO2

0.04590 mol CO2 = mol of O in FexOy

mass of O in FexOy = 0.04590 mol O x O mol 1

O g 16.0 = 0.7345 g O

mass of Fe in FexOy = 2.660 g - 0.7345 g = 1.926 g Fe

(b) mol Fe in magnetite = 1.926 g Fe x Fe g 55.85

Fe mol 1= 0.0345 mol Fe

formula of magnetite: Fe 0.0345 O 0.0459 (divide each subscript by the smaller) Fe 0.0345 / 0.0345 O 0.0459 / 0.0345 FeO 1.33 (multiply both subscripts by 3) Fe (1 x 3) O (1.33 x 3); Fe3O4 (c) unit cell edge = d = 839 pm = 839 x 10-12 m

d = 839 x 10-12 m x m 1

cm 100 = 8.39 x 10-8 cm

unit cell volume = d3 = (8.39 x 10-8 cm)3 = 5.91 x 10-22 cm3 unit cell mass = (5.91 x 10-22 cm3)(5.20 g/cm3) = 3.07 x 10-21 g

mass of Fe in unit cell = g) 10 x (3.07g 2.660

Fe g 1.926 21_

= 2.22 x 10-21 g Fe

mass of O in unit cell = g) 10 x (3.07g 2.660

O g 0.7345 21_

= 8.47 x 10-22 g O

Fe atoms in unit cell = g/mol 55.847

atoms/mol 10 x 6.022 x g 10 x 2.22

2321_ = 24 Fe atoms

O atoms in unit cell = g/mol 16.00

atoms/mol 10 x 6.022 x g 10 x 8.47

2322_ = 32 O atoms

10.119 (a) K) (296

mol K atm L

0.08206

L) (4.00Hg mm 760


= RT

PV = nH2

••

= 0.160 mol H2

M = Group 3A metal; 2 M(s) + 6 H+(aq) → 2 M3+(aq) + 3 H2(g)


253

nM = 0.160 mol H2 x H mol 3

M mol 2

2

= 0.107 mol M

mass M = 1.07 cm3 x 2.70 g/cm3 = 2.89 g M

molar mass M = M mol 0.107

M g 2.89 = 27.0 g/mol; The Group 3A metal is Al

(b) mass of one Al atom = 26.98 g/mol x atoms 10 x 6.022

mol 123

= 4.48 x 10-23 g/atom

unit cell edge = d = 404 pm = 404 x 10-12 m

d = 404 x 10-12 m x m 1

cm 100 = 4.04 x 10-8 cm

unit cell volume = d3 = (4.04 x 10-8 cm)3 = 6.59 x 10-23 cm3 Calculate the density of Al assuming it is primitive cubic, body-centered cubic, and face-centered cubic. Compare the calculated density with the actual density to identify the unit cell. For primitive cubic:


atom) g/Al 10 x Al)(4.48 (1 =

volumecellunit

mass cellunit 323_

23_

= 0.680 g/cm3

For body-centered cubic:


atom) g/Al 10 x Al)(4.48 (2 =

volumecellunit

mass cellunit 323_

23_

= 1.36 g/cm3

For face-centered cubic:


atom) g/Al 10 x Al)(4.48 (4 =

volumecellunit

mass cellunit 323_

23_

= 2.72 g/cm3

The calculated density for a face-centered cube (2.72 g/cm3) is closest to the actual density of 2.70 g/cm3. Al crystallizes in the face-centered cubic unit cell.

(c) 8

)pm (404 =

8d =r

22

= 143 pm

10.120 (a) M = alkali metal; 500.0 mL = 0.5000 L; 802oC = 1075 K

K) (1075mol K atm L

06 0.082

L) (0.5000Hg mm 760

atm 1.00 x Hg mm 12.5

= RT

PV = nM

••

= 9.32 x 10-5 mol M

1.62 mm = 1.62 x 10-3 m; crystal volume = (1.62 x 10-3 m)3 = 4.25 x 10-9 m3 M atoms in crystal = (9.32 x 10-5 mol)(6.022 x 1023 atoms/mol) = 5.61 x 1019 M atoms Because M is body-centered cubic, only 68% (Table 10.10) of the total volume is occupied by M atoms.

volume of M atom = atoms M 10 x 5.61

m) 10 x 5(0.68)(4.219

9_

= 5.15 x 10-29 m3/M atom


254

volume of a sphere = 3

4πr3

rM = 3

4

3(volume)

π = 3

329_

4

)m 10 x 3(5.15

π = 2.31 x 10-10 m = 231 x 10-12 m = 231 pm

(b) The radius of 231 pm is closest to that of K. (c) 1.62 mm = 0.162 cm

density of solid = )cm (0.162

g/mol) mol)(39.1 10 x (9.323

5_

= 0.857 g/cm3

density of vapor = cm 500.0

g/mol) mol)(39.1 10 x (9.323

5_

= 7.29 x 10-6 g/cm3

10.121 (a)

K) (298mol K atm L

06 0.082

L) (0.500Hg mm 760


= RT

PV = nX2

••

= 0.0203 mol X2

M(s) + 1/2 X2(g) → MX(s)

mol M = 0.0203 mol X2 x X mol 2/1

M mol 1

2

= 0.0406 mol M

molar mass M = M mol 0.0406

M g 1.588= 39.1 g/mol; atomic mass = 39.1 amu ; M = K

(b) From Figure 6.1, the radius for K+ is ~140 pm. unit cell edge = 535 pm = r2 + r2 XK _+

2

pm) 2(140 _ pm 535 =

2r2 _ pm 535

= r KX

+

_ = 128 pm

From Figure 6.2, X- = F- (c) Because the cation and anion are of comparable size, the anions are not in contact with each other.

(d) unit cell contents: 1/8 F- at 8 corners and 1/2 F- at 6 faces = 4 F- 1/4 K+ at 12 edges and 1 K+ inside = 4 K+

mass of one K+ = /molK 10 x 6.022

g/mol 39.098+23

= 6.493 x 10-23 g/K+


255

mass of one F- = /molF 10 x 6.022

g/mol 18.998_23

= 3.155 x 10-23 g/F-

unit cell mass = (4 K+)(6.493 x 10-23 g/K+) + (4 F-)(3.155 x 10-23 g/F-) = 3.859 x 10-22 g unit cell volume = [(535 x 10-12 m)(100 cm/m)]3 = 1.531 x 10-22 cm3


g 10 x 3.859 =

cellunit of volume

cellunit of mass322_

22_

= 2.52 g/cm3

(e) K(s) + 1/2 F2(g) → KF(s) is a formation reaction.

∆Hof(KF) =

mol 0.0406

kJ 22.83_= -562 kJ/mol

255

11

Solutions and Their Properties 11.1 Toluene is nonpolar and is insoluble in water.

Br2 is nonpolar but because of its size is polarizable and is soluble in water. KBr is an ionic compound and is very soluble in water. toluene < Br2 < KBr (solubility in H2O)

11.2 (a) Na+ has the larger (more negative) hydration energy because the Na+ ion is smaller

than the Cs+ ion and water molecules can approach more closely and bind more tightly to the Na+ ion. (b) Ba2+ has the larger (more negative) hydration energy because of its higher charge.

11.3 NaCl, 58.44 amu; 1.00 mol NaCl = 58.44 g

1.00 L H2O = 1000 mL = 1000 g (assuming a density of 1.00 g/mL)

mass % NaCl = 100% x g 58.44 + g 1000

g 58.44 = 5.52 mass %

11.4 ppm = solution of mass total

CO of mass 2 x 106 ppm

total mass of solution = density x volume = (1.3 g/L)(1.0 L) = 1.3 g

35 ppm = g 1.3CO of mass 2 x 106 ppm

mass of CO2 = ppm 10

g) ppm)(1.3 (356

= 4.6 x 10-5 g CO2

11.5 Assume 1.00 L of sea water.

mass of 1.00 L = (1000 mL)(1.025 g/mL) = 1025 g

100% x g 1025

NaCl mass = 3.50 mass %; mass NaCl =

100

3.50 x g 1025 = 35.88 g

There are 35.88 g NaCl per 1.00 L of solution.

M = L 1.00

NaCl g 58.44NaCl mol 1

x NaCl g 35.88

= 0.614 M

11.6 C27H46O, 386.7 amu; CHCl3, 119.4 amu; 40.0 g x g 1000

kg 1 = 0.0400 kg

molality = kg 0.0400

g 386.7mol 1

x g 0.385

= CHCl kg

OHC mol

3

4627

= 0.0249 mol/kg = 0.0249 m

Chapter 11 - Solutions and Their Properties ______________________________________________________________________________

256

CHCl mol + OHC mol

OHC mol = X

34627

4627OHC 4627

X OHC 4627 =

g 119.4mol 1

x g 40.0 + g 386.7

mol 1 x g 0.385

g 386.7mol 1

x g 0.385

= 2.96 x 10-3

11.7 CH3CO2Na, 82.03 amu

kg H2O =

NaCOCH mol 0.500

OH kg 1Na)COCH mol (0.150

23

223 = 0.300 kg H2O

mass CH3CO2Na = 0.150 mol CH3CO2Na x NaCOCH mol 1

NaCOCH g 82.03

23

23 = 12.3 g CH3CO2Na

mass of solution needed = 300 g + 12.3 g = 312 g 11.8 Assume you have a solution with 1.000 kg (1000 g) of H2O. If this solution is 0.258 m,

then it must also contain 0.258 mol glucose.

mass of glucose = 0.258 mol x mol 1

g 180.2 = 46.5 g glucose

mass of solution = 1000 g + 46.5 g = 1046.5 g density = 1.0173 g/mL

volume of solution = g 1.0173

mL 1 x g 1046.5 = 1028.7 mL

volume = 1028.7 mL x mL 1000

L 1 = 1.029 L; molarity =

L 1.029

mol 0.258 = 0.251 M

11.9 Assume 1.00 L of solution.

mass of 1.00 L = (1.0042 g/mL)(1000 mL) = 1004.2 g of solution

0.500 mol CH3CO2H x HCOCH mol 1

HCOCH g 60.05

23

23 = 30.02 g CH3CO2H

1004.2 g - 30.02 g = 974.2 g = 0.9742 kg of H2O; molality = kg 0.9742

mol 0.500 = 0.513 m

11.10 Assume you have 100.0 g of seawater.

mass NaCl = (0.0350)(100.0 g) = 3.50 g NaCl mass H2O = 100.0 g - 3.50 g = 96.5 g H2O

NaCl, 58.44 amu; mol NaCl = 3.50 g x g 58.44

mol 1 = 0.0599 mol NaCl

mass H2O = 96.5 g x g 1000

kg 1 = 0.0965 kg H2O; molality =

kg 0.0965

mol 0.0599 = 0.621 m


257

11.11 M = k ⋅ P; k = atm 1.0

M 10 x 3.2 =

P

M 2_

= 3.2 x 10-2 mol/(L⋅ atm)

11.12 (a) M = k ⋅ P = [3.2 x 10-2 mol/(L ⋅ atm)](2.5 atm) = 0.080 M

(b) M = k ⋅ P = [3.2 x 10-2 mol/(L ⋅ atm)](4.0 x 10-4 atm) = 1.3 x 10-5 M 11.13 C7H6O2, 122.1 amu; C2H6O, 46.07 amu

g 122.1mol 1

x g 5.00 + g 46.07

mol 1 x g 100

g 46.07mol 1

x g 100

= OHC mol + OHC mol

OHC mol = X

26762

62solv = 0.981

Psoln = Psolv ⋅ Xsolv = (100.5 mm Hg)(0.981) = 98.6 mm Hg

11.14 Psoln = Psolv ⋅ Xsolv; P

P = Xsolv

solnsolv =

Hg mm 55.3

Hg mm 1.30) _ (55.3 = 0.976

NaBr dissociates into two ions in aqueous solution.

Xsolv = Br mol + Na mol + OH mol

OH mol_+

2

2

Xsolv = 0.976 =

Br mol x + Na mol x + g 18.02

mol 1 x g 250

g 18.02mol 1

x g 250

_+

0.976 = molx 2 + mol 13.9

mol 13.9; solve for x.

0.976(13.9 mol + 2x mol) = 13.9 mol 13.566 mol + 1.952 x mol = 13.9 mol 1.952 x mol = 13.9 mol - 13.566 mol

x mol = = 1.952

mol 13.566 _ mol 13.9 0.171 mol

x = 0.171 mol Na+ = 0.171 mol Br- = 0.171 mol NaBr

NaBr, 102.9 amu; mass NaBr = 0.171 mol x mol 1

g 102.9 = 17.6 g NaBr

11.15 At any given temperature, the vapor pressure of a solution is lower than the vapor

pressure of the pure solvent. The upper curve represents the vapor pressure of the pure solvent. The lower curve represents the vapor pressure of the solution.

11.16 C2H5OH, 46.07 amu; H2O, 18.02 amu

(a) OHHC g 46.07

OHHC mol 1 x OHHC g 25.0

52

5252 = 0.5426 mol C2H5OH


258

100.0 g H2O x OH g 18.02

OH mol 1

2

2 = 5.549 mol H2O

mol 5.549 + mol 0.5426

mol 0.5426 = X OHHC 52

= 0.08907

mol 5.549 + mol 0.5426

mol 5.549 = X OH2

= 0.9109

Psoln = P X + P X oOHOH

oOHHCOHHC 225252

Psoln = (0.08907)(61.2 mm Hg) + (0.9109)(23.8 mm Hg) = 27.1 mm Hg

(b) OHHC g 46.07

OHHC mol 1 x OHHC g 100

52

5252 = 2.171 mol C2H6O

25.0 g H2O x OH g 18.02

OH mol 1

2

2 = 1.387 mol H2O

mol 1.387 + mol 2.171

mol 2.171 = X OHHC 52

= 0.6102

mol 1.387 + mol 2.171

mol 1.387 = X OH2

= 0.3898

Psoln = P X + P X oOHOH

oOHHCOHHC 225252

Psoln = (0.6102)(61.2 mm Hg) + (0.3898)(23.8 mm Hg) = 46.6 mm Hg 11.17 (a) Because the vapor pressure of the solution (red curve) is higher than that of the first

liquid (green curve), the vapor pressure of the second liquid must be higher than that of the solution (red curve). Because the second liquid has a higher vapor pressure than the first liquid, the second liquid has a lower boiling point.

(b)

11.18 C9H8O4, 180.2 amu; CHCl3 is the solvent. For CHCl3, Kb = 3.63 mol

kg C o •

75.00 g x g 1000

kg 1 = 0.075 00 kg


259

∆Tb = Kb ⋅ m =

•kg 00 0.075

g 180.2mol 1

x g 1.50

mol

kg C 3.63o

= 0.40oC

Solution boiling point = 61.7oC + ∆Tb = 61.7oC + 0.40oC = 62.1oC 11.19 MgCl2, 95.21 amu

110 g x g 1000

kg 1 = 0.110 kg

∆Tf = Kf ⋅ m ⋅ i = (2.7)kg 0.110

g 95.21mol 1

x g 7.40

mol

kg C 1.86o

• = 3.55oC

Solution freezing point = 0.00oC - ∆Tf = 0.00oC - 3.55oC = -3.55oC 11.20 ∆Tf = Kf ⋅ m ⋅ i; For KBr, i = 2.

Solution freezing point = -2.95oC = 0.00oC - ∆Tf; ∆Tf = 2.95oC

m =

(2)mol

kg C 1.86

C952. =

i K

To

o

f

f

••∆

= 0.793 mol/kg = 0.793 m

11.21 HCl, 36.46 amu; ∆Tf = Kf ⋅ m ⋅ i

190 g x g 1000

kg 1 = 0.190 kg

Solution freezing point = - 4.65oC = 0.00oC - ∆Tf; ∆Tf = 4.65oC

i =

••

∆

kg 0.190g 36.46

mol 1 x g 9.12

molkg C 1.86

C654. =

m K

T

o

o

f

f = 1.9

11.22 The red curve represents the vapor pressure of pure chloroform.

(a) The normal boiling point for a liquid is the temperature where the vapor pressure of the liquid equals 1 atm (760 mm Hg). The approximate boiling point of pure chloroform is 62oC. (b) The approximate boiling point of the solution is 69oC. ∆Tb = 69oC - 62oC = 7oC


260

∆Tb = Kb ⋅ m; m =

molkg C 3.63

C7 = K

To

o

b

b

•∆

= 2 mol/kg = 2 m

11.23 For CaCl2 there are 3 ions (solute particles)/CaCl2

Π = MRT; For CaCl2, Π = 3MRT

Π = (3)(0.125 mol/L) K) (310mol K

atm L 06 0.082

••

= 9.54 atm

11.24 Π = MRT; M = K) (300

mol K atm L

06 0.082

atm) (3.85 =

RT

••

Π = 0.156 M

11.25 ∆Tf = Kf ⋅ m; m =

molkg C 37.7

C102. =

K

To

o

f

f

•∆

= 0.0557 mol/kg = 0.0557 m

35.00 g x g 1000

kg 1 = 0.03500 kg

mol = 0.0557 = kg 0.03500 x kg

mol0.001 95 mol naphthalene

molar mass of naphthalene = = enaphthalen mol 95 0.001

enaphthalen g 0.250 128 g/mol

11.26 Π = MRT; M = K) (298

mol K atm L

0.08206

Hg mm 760atm 1

x Hg mm 149

= RT

••

Π = 8.02 x 10-3 M

300.0 mL = 0.3000 L (8.02 x 10-3 mol/L)(0.3000 L) = 0.002 406 mol sucrose

molar mass of sucrose = = sucrose mol 406 0.002

sucrose g 0.822 342 g/mol


261

11.27 (a) and (c)

(b) The mixture will begin to boil at ~50oC. (d) After two cycles of boiling and condensing, the approximate composition of the liquid would 90% dichloromethane and 10% chloroform.

11.28 Both solvent molecules and small solute particles can pass through a semipermeable dialysis membrane. Only large colloidal particles such as proteins can’t pass through. Only solvent molecules can pass through a semipermeable membrane used for osmosis.

Understanding Key Concepts 11.29 The upper curve is pure ether.

(a) The normal boiling point for ether is the temperature where the upper curve intersects the 760 mm Hg line, ~ 37oC. (b) ∆Tb _3oC

∆Tb = Kb ⋅ m; m =

molkg C 2.02

C3 = K

To

o

b

b

•∆

_1.5 mol/kg _1.5 m

11.30 (a) < (b) < (c) 11.31 At any given temperature, the vapor pressure of a mixture of two pure liquids falls

between the individual vapor pressures of the two pure liquids themselves. Because the vapor pressure of the mixture is greater than the vapor pressure of the solvent, the second liquid is more volatile (has a higher vapor pressure) than the solvent.

11.32 Assume that only the blue (open) spheres (solvent) can pass through the semipermeable membrane. There will be a net transfer of solvent from the right compartment (pure solvent) to the left compartment (solution) to achieve equilibrium.

11.33 At point 1, the temperature should be near the boiling point of the lower boiling solvent,

CHCl3, approximately 62oC.


262

At point 3, the temperature should be about halfway between the two boiling points at approximately 70oC. At point 2, the temperature should be about halfway between the temperatures at points 1 and 3, approximately 66oC.

11.34 The vapor pressure of the NaCl solution is lower than that of pure H2O. More H2O

molecules will go into the vapor from the pure H2O than from the NaCl solution. More H2O vapor molecules will go into the NaCl solution than into pure H2O. The result is represented by (b).

11.35 (b) ~95oC Additional Problems Solutions and Energy Changes 11.36 The surface area of a solid plays an important role in determining how rapidly a solid

dissolves. The larger the surface area, the more solid-solvent interactions, and the more rapidly the solid will dissolve. Powdered NaCl has a much larger surface area than a large block of NaCl, and it will dissolve more rapidly.

11.37 (a) a gas in a liquid – carbonated soft drink

(b) a solid in a solid – metal alloys (14-karat gold) (c) a liquid in a solid – dental amalgam (Hg in Ag)

11.38 Substances tend to dissolve when the solute and solvent have the same type and

magnitude of intermolecular forces; thus the rule of thumb "like dissolves like." 11.39 Both Br2 and CCl4 are nonpolar, and intermolecular forces for both are dispersion forces.

H2O is a polar molecule with dipole-dipole forces and hydrogen bonding. Therefore, Br2 is more soluble in CCl4.

11.40 Energy is required to overcome intermolecular forces holding solute particles together in

the crystal. For an ionic solid, this is the lattice energy. Substances with higher lattice energies tend to be less soluble than substances with lower lattice energies.

11.41 SO4

2- has the larger hydration energy because of its higher charge. Both SO42- and ClO4

- are comparable in size, so size is not a factor.


263

11.42 Ethyl alcohol and water are both polar with small dispersion forces. They both can hydrogen bond, and are miscible. Pentyl alcohol is slightly polar and can hydrogen bond. It has, however, a relatively large dispersion force because of its size, which limits its water solubility.

11.43 The intermolecular forces associated with octane are dispersion forces. Both pentyl

alcohol and methyl alcohol can hydrogen bond. Pentyl alcohol has relatively large dispersion forces because of its size. Methyl alcohol does not. Pentyl alcohol is soluble in octane; methyl alcohol is not.

11.44 CaCl2, 110.98 amu

For a 1.00 m solution: heat released = 81,300 J mass of solution = 1000 g H2O + 110.98 g CaCl2 = 1110.98 g

∆T = solution) of heat)(mass (specific

q =

g) 8g)](1110.9 J/(K [4.18

J 81,300

• = 17.5 K =

17.5oC Final temperature = 25.0oC + 17.5oC = 42.5oC

11.45 NH4ClO4, 117.48 amu

For a 1.00 m solution: heat absorbed = 33,500 J mass of solution = 1000 g H2O + 117.48 g NH4ClO4 = 1117.48 g

∆T = solution) of heat)(mass (specific

q =

g) 8g)](1117.4 J/(K [4.18

J 33,500_

• = -7.2 K =

-7.2oC Final temperature = 25.0oC - 7.2oC = 17.8oC

Units of Concentration

11.46 molarity = solution of liters

solute of moles; molality =

solvent of kg

solute of moles

11.47 A saturated solution contains enough solute so that there is an equilibrium between

dissolved solute and undissolved solid. A supersaturated solution contains a greater-than-equilibrium amount of solute.

11.48 (a) Dissolve 0.150 mol of glucose in water; dilute to 1.00 L.

(b) Dissolve 1.135 mol of KBr in 1.00 kg of H2O. (c) Mix together 0.15 mol of CH3OH with 0.85 mol of H2O.

11.49 (a) Dissolve 15.5 mg urea in 100 mL water

(b) Choose a K+ salt, say KCl, and dissolve 0.0075 mol (0.559 g) in water; dilute to100 mL.


264

11.50 C7H6O2, 122.12 amu, 165 mL = 0.165 L mol C7H6O2 = (0.0268 mol/L)(0.165 L) = 0.004 42 mol

mass C7H6O2 = 0.004 42 mol x mol 1

g 122.12 = 0.540 g

Dissolve 4.42 x 10-3 mol (0.540 g) of C7H6O2 in enough CHCl3 to make 165 mL of solution.

11.51 C7H6O2, 122.12 amu

0.0268 mol C7H6O2 x OHC mol 1

OHC g 122.12

267

267 = 3.27 g C7H6O2

Dissolve 3.27 g of C7H6O2 in 1.000 kg of CHCl3, and take 165 mL of the solution. 11.52 (a) KCl, 74.6 amu

A 0.500 M KCl solution contains 37.3 g of KCl per 1.00 L of solution. A 0.500 mass % KCl solution contains 5.00 g of KCl per 995 g of water. The 0.500 M KCl solution is more concentrated (that is, it contains more solute per amount of solvent). (b) Both solutions contain the same amount of solute. The 1.75 M solution contains less solvent than the 1.75 m solution. The 1.75 M solution is more concentrated.

11.53 (a) KI, 166.00 amu; KBr, 119.00 amu; assume 1.000 L = 1000 mL = 1000 g solution

10 ppm = g 1000

KI mass x 106 ; mass KI = 0.010 g

10,000 ppb = g 1000

KBr mass x 109 ; mass KBr = 0.010 g

Both solutions contain the same mass of solute in the same amount of solvent. Because the molar mass of KBr is less than that of KI, the number of moles of KBr is larger than the number of moles of KI. The KBr solution has a higher molarity than the KI solution. (b) Because the mass % of the two solutions is the same, they both contain the same mass of solute and solution. Because the molar mass of KCl is less than that of citric acid, the number of moles of KCl is larger than the number of moles of citric acid. The KCl solution has a higher molarity than the citric acid solution.

11.54 (a) C6H8O7, 192.12 amu

0.655 mol C6H8O7 x OHC mol 1

OHC g 192.12

786

786 = 126 g C6H8O7

mass % C6H8O7 = 100% x g 1000 + g 126

g 126 = 11.2 mass %

(b) 0.135 mg = 0.135 x 10-3 g (5.00 mL H2O)(1.00 g/mL) = 5.00 g H2O

mass % KBr = 100% x g 5.00 + g) 10 x (0.135

g 10 x 0.1353_

3_

= 0.002 70 mass % KBr


265

(c) mass % aspirin = 100% x g 145 + g 5.50

g 5.50 = 3.65 mass % aspirin

11.55 (a) molality = kg 1.00

mol 0.655 = 0.655 m

(b) KBr, 119.00 amu; 5.00 g = 0.005 00 kg

molality = kg 00 0.005

g 119.00mol 1

x g 10 x 0.135 3_

= 2.27 x 10-4 mol/kg = 2.27 x 10-4 m

(c) C9H8O4, 180.16 amu; 145 g = 0.145 kg

molality = kg 0.145

g 180.16mol 1

x g 5.50

= 0.211 mol/kg = 0.211 m

11.56 X P = P OtotalO 33

•

atm 10 x 1.3

atm 10 x 1.6 =

P

P = X 2_

9_

total

OO

3

3 = 1.2 x 10-7

Assume one mole of air (29 g/mol) mol O3 = nair ⋅ XO3

= (1 mol)(1.2 x 10-7) = 1.2 x 10-7 mol O3

O3, 48.00 amu; mass O3 = 1.2 x 10-7 mol x mol 1

g 48.0 = 5.8 x 10-6 g O3

ppm O3 = g 29

g 10 x 5.8 6_

x 106 = 0.20 ppm

11.57 Assume 1 mL of blood weighs 1 g. 1 dL = 0.1 L = 100 mL = 100 g

ppb = g 100

g 10 x 10 6_

x 109 = 100 ppb

11.58 (a) H2SO4, 98.08 amu; molality = kg 1.30

g 98.08mol 1

x g 25.0

= 0.196 mol/kg = 0.196 m

(b) C10H14N2, 162.23 amu; CH2Cl2, 84.93 amu

2.25 g C10H14N2 x NHC g 162.23

NHC mol 1

21410

21410 = 0.0139 mol C10H14N2

80.0 g CH2Cl2 x ClCH g 84.93

ClCH mol 1

22

22 = 0.942 mol CH2Cl2


266

mol 0.0139 + mol 0.942

mol 0.0139 = X NHC 21410

= 0.0145

mol 0.0139 + mol 0.942

mol 0.942 = X ClCH 22

= 0.985

11.59 NaOCl, 74.44 amu

A 5.0 mass % aqueous solution of NaOCl contains 5.0 g NaOCl and 95 g H2O.

molality = kg 0.095

g 74.44mol 1

x g 5.0

= 0.71 mol/kg = 0.71 m

5.0 g NaOCl x NaOCl g 74.44

NaOCl mol 1 = 0.0672 mol NaOCl

95 g H2O x OH g 18.02

OH mol 1

2

2 = 5.27 mol H2O

mol 0.0672 + mol 5.27

mol 0.0672 = XNaOCl = 0.013

11.60 16.0 mass % = OH g 84.0 + SOH g 16.0

SOH g 16.0

242

42

H2SO4, 98.08 amu; density = 1.1094 g/mL


mL 1 x g 100.0 = 90.14 mL = 0.090 14 L


g 98.08mol 1

x g 16.0

= 1.81 M

11.61 C2H6O2, 62.07 amu

A 40.0 mass % aqueous solution of C2H6O2 contains 40.0 g C2H6O2 and 60.0 g H2O. density = 1.0514 g/mL


mL 1 x g 100.0 = 95.1 mL = 0.0951 L

molarity = L 0.0951

g 62.07mol 1

xg 40.0

= 6.78 M

11.62 molality = kg 0.0600

g 62.07mol 1

x g 40.0

= 10.7 mol/kg = 10.7 m


267

11.63 molality = kg 0.0840

g 98.08mol 1

x g 16.0

= 1.94 mol/kg = 1.94 m

11.64 C19H21NO3, 311.34 amu; 1.5 mg = 1.5 x 10-3 g

1.3 x 10-3 mol/kg solvent of kg

g 311.34mol 1

x g 10 x 1.5

=

3_

; solve for kg of solvent.

kg of solvent = = mol/kg 10 x 1.3

g 311.34mol 1

x g 10 x 1.5

3_

3_

0.0037 kg

Because the solution is very dilute, kg of solvent ≈ kg of solution.

g of solution = (0.0037 kg)

kg 1

g 1000 = 3.7 g

11.65 C12H22O11, 342.30 amu

32.5 g C12H22O11 x OHC g 342.30

OHC mol 1

112212

112212 = 0.0949 mol C12H22O11

0.850 m = 0.850 mol/kg OH of kg

mol 0.0949 =

2

; kg of H2O = mol/kg 0.850

mol 0.0949 = 0.112 kg

mass of H2O = 0.112 kg x kg 1

g 1000 = 112 g H2O

11.66 C6H12O6, 180.16 amu; H2O, 18.02 amu; Assume 1.00 L of solution.

mass of solution = (1000 mL)(1.0624 g/mL) = 1062.4 g

mass of solute = 0.944 mol x mol 1

g 180.16 = 170.1 g C6H12O6

mass of H2O = 1062.4 g - 170.1 g = 892.3 g H2O

mol C6H12O6 = 0.944 mol; mol H2O = 892.3 g x g 18.02

mol 1 = 49.5 mol

(a) mol 49.5 + mol 0.944

mol 0.944 =

OH mol + OHC molOHC mol

= X26126

6126OHC 6126

= 0.0187

(b) mass % = g 1062.4

g 170.1 = 100% x

solution of mass totalOHC mass 6126 x 100% = 16.0%

(c) molality = kg 0.8923

mol 0.944 =

OH kgOHC mol

2

6126 = 1.06 mol/kg = 1.06 m

11.67 C12H22O11, 342.30 amu; Assume 1.00 L of solution.

mass of solution = (1000 mL)(1.0432 g/mL) = 1043.2 g


268

mass of solute = 0.335 mol C12H22O11 x OHC mol 1

OHC g 342.30

112212

112212 = 114.7 g C12H22O11

mass of H2O = 1043.2 g - 114.7 g = 928.5 g H2O

mol C12H22O11 = 0.335 mol; 928.5 g H2O x OH g 18.02

OH mol 1

2

2 = 51.53 mol H2O

mol 0.335 + mol 51.53

mol 0.335 = X OHC 112212

= 0.006 46

mass % C12H22O11 = 100% x g 1043.2

g 114.7 = 11.0 mass % C12H22O11


mol 0.335 = 0.361 mol/kg = 0.361 m

Solubility and Henry's Law

11.68 M = k ⋅ P = (0.091 atm L

mol

•)(0.75 atm) = 0.068 M

11.69 M = k ⋅ P; k = atm 1.00

M 0.195 =

P

M = 0.195 mol/(L⋅ atm)

P = 25.5 mm Hg x Hg mm 760

atm 1.00 = 0.0336 atm

M = k ⋅ P = (0.195 atm L

mol

•)(0.0336 atm) = 6.55 x 10-3 M

11.70 M = k ⋅ P

Calculate k: k = atm 1.00

mol/L 10 x 2.21 =

P

M 3_

= 2.21 x 10-3 atm L

mol

•

Convert 4 mg/L to mol/L: 4 mg = 4 x 10-3 g

O2 molarity = L 1.00

g 32.00mol 1

x g 10 x 4 3_

= 1.25 x 10-4 M

atm Lmol

10 x 2.21

Lmol

10 x 1.25 =

k

M = P

3_

4_

O2

•

= 0.06 atm

11.71 k = 1.93 x 10-3 mol/(L ⋅ atm)

M = k ⋅ P = [1.93 x 10-3 mol/(L ⋅ atm)]

Hg mm 760

atm 1.00 x Hg mm 68 = 1.73 x 10-4 mol/L


269

1.73 x 10-4 mol/L x g 10 x 1

mg 1 x

O mol 1O g 32.00

3_2

2 = 5.5 mg/L

11.72 [Xe] = 10 mmol/L = 0.010 M at STP

M = k ⋅ P; k = atm 1.0

M 0.010 =

P

M = 0.010 mol/(L⋅ atm)

11.73 Assuming H2O as the solvent, NH3 does not obey Henry's law because NH3 can both

hydrogen bond and react with H2O. Colligative Properties 11.74 The difference in entropy between the solvent in a solution and a pure solvent is

responsible for colligative properties. 11.75 Osmotic pressure is the amount of pressure that needs to be applied to cause

osmosis to stop. 11.76 NaCl is a nonvolatile solute. Methyl alcohol is a volatile solute. When NaCl is added to

water, the vapor pressure of the solution is decreased, which means that the boiling point of the solution will increase. When methyl alcohol is added to water, the vapor pressure of the solution is increased which means that the boiling point of the solution will decrease.

11.77 When 100 mL of 9 M H2SO4 at 0oC is added to 100 mL of liquid water at 0oC, the

temperature rises because ∆Hsoln for H2SO4 is exothermic. When 100 mL of 9 M H2SO4 at 0oC is added to 100 g of solid ice at 0oC, some of the ice will melt (an endothermic process) and the temperature will fall because the H2SO4 (solute) lowers the freezing point of the ice/water mixture.

11.78 11.79 Molality is a temperature independent concentration unit. For freezing point depression

and boiling point elevation, molality is used so that the solute concentration is independent of temperature changes. Molarity is temperature dependent. Molarity can be used for osmotic pressure because osmotic pressure is measured at a fixed


270

temperature. 11.80 (a) CH4N2O, 60.06 amu; H2O, 18.02 amu

10.0 g CH4N2O x ONCH g 60.06

ONCH mol 1

24

24 = 0.167 mol CH4N2O

150.0 g H2O x OH g 18.02

OH mol 1

2

2 = 8.32 mol H2O

mol 0.167 + mol 8.32

mol 8.32 = X OH2

= 0.980

Psoln = X P OHo

OH 22• = (71.93 mm Hg)(0.980) = 70.5 mm Hg

(b) LiCl, 42.39 amu; 10.0 g LiCl x LiCl g 42.39

LiCl mol 1 = 0.236 mol LiCl

LiCl dissociates into Li+(aq) and Cl-(aq) in H2O. mol Li+ = mol Cl- = mol LiCl = 0.236 mol

150.0 g H2O x OH g 18.02

OH mol 1

2

2 = 8.32 mol H2O

mol 0.236 + mol 0.236 + mol 8.32

mol 8.32 = X OH2

= 0.946

Psoln = X P OHo

OH 22• =(71.93 mm Hg)(0.946) = 68.0 mm Hg

11.81 C6H12O6, 180.16 amu; CH3OH, 32.04 amu

16.0 g C6H12O6 x OHC g 180.16

OHC mol 1

6126

6126 = 0.0888 mol C6H12O6

80.0 g CH3OH x OHCH g 32.04

OHCH mol 1

3

3 = 2.50 mol CH3OH

mol 0.0888 + mol 2.50

mol 2.50 = X OHCH3

= 0.966

Psoln = X P OHCHo

OHCH 33• = (140 mm Hg)(0.966) = 135 mm Hg

11.82 For H2O, Kb = 0.51mol

kg C o •; 150.0 g = 0.1500 kg

(a) ∆Tb = Kb ⋅ m =

•kg 0.1500

mol 0.167

mol

kg C 0.51o

= 0.57oC

Solution boiling point = 100.00oC + ∆Tb = 100.00oC + 0.57oC = 100.57oC

(b) ∆Tb = Kb ⋅ m =

•kg 0.1500

mol) 2(0.236

mol

kg C 0.51o

= 1.6oC

Solution boiling point = 100.00oC + ∆Tb = 100.00oC + 1.6oC = 101.6oC

11.83 For H2O, Kf = 1.86 mol

kg C o •; 150.0 g = 0.1500 kg


271

(a) ∆Tf = Kf ⋅ m =

•kg 0.1500

mol 0.167

mol

kg C 1.86o

= 2.07oC

Solution freezing point = 0.00oC - ∆Tf = 0.00oC - 2.07oC = -2.07oC

(b) ∆Tf = Kf ⋅ m =

•kg 0.1500

mol) 2(0.236

mol

kg C 1.86o

= 5.85oC

Solution freezing point = 0.00oC - ∆Tf = 0.00oC - 5.85oC = -5.85oC 11.84 ∆Tf = Kf ⋅ m ⋅ i

Solution freezing point = - 4.3oC = 0.00oC - ∆Tf; ∆Tf = 4.3oC

i =

mol/kg) (1.0mol

kg C 1.86

C34. =

m K

To

o

f

f

••∆

= 2.3

11.85 ∆Tb = Kb ⋅ m ⋅ i = = 85)mol/kg)(1. (0.75mol

kg C 0.51o

•0.71oC

Solution boiling point = 100.00oC + ∆Tb = 100.00oC + 0.71oC = 100.71oC 11.86 Acetone, C3H6O, 58.08 amu, Po

OHC 63 = 285 mm Hg

Ethyl acetate, C4H8O2, 88.11 amu, PoOHC 284

= 118 mm Hg

25.0 g C3H6O x OHC g 58.08

OHC mol 1

63

63 = 0.430 mol C3H6O

25.0 g C4H8O2 x OHC g 88.11

OHC mol 1

284

284 = 0.284 mol C4H8O2

mol 0.284 + mol 0.430

mol 0.430 = X OHC 63

= 0.602; mol 0.284 + mol 0.430

mol 0.284 = X OHC 284

= 0.398

Psoln = X P + X P OHCo

OHCOHCo

OHC 2842846363••

Psoln = (285 mm Hg)(0.602) + (118 mm Hg)(0.398) = 219 mm Hg 11.87 CHCl3, 119.38 amu, Po

CHCl3 = 205 mm Hg; CH2Cl2, 84.93 amu, PoClCH 22

= 415 mm Hg

15.0 g CHCl3 x CHCl g 119.38

CHCl mol 1

3

3 = 0.126 mol CHCl3

37.5 g CH2Cl2 x ClCH g 84.93

ClCH mol 1

22

22 = 0.442 mol CH2Cl2

mol 0.442 + mol 0.126

mol 0.126 = XCHCl3 = 0.222;

mol 0.442 + mol 0.126

mol 0.442 = X ClCH 22

= 0.778

Psoln = X P + X P ClCHo

ClCHCHCloCHCl 222233

••

Psoln = (205 mm Hg)(0.222) + (415 mm Hg)(0.778) = 368 mm Hg


272

11.88 In the liquid, Xacetone = 0.602 and Xethyl acetate = 0.398 In the vapor, PTotal = 219 mm Hg Pacetone = Po

acetone⋅ Xacetone = (285 mm Hg)(0.602) = 172 mm Hg

Pethyl acetate = Poacetate ethyl ⋅ Xethyl acetate = (118 mm Hg)(0.398) = 47 mm Hg

Hg mm 219

Hg mm 172 =

P

P = Xtotal

acetoneacetone = 0.785;

Hg mm 219

Hg mm 47 =

P

P = Xtotal

acetate ethylacetate ethyl = 0.215

11.89 In the liquid, XCHCl3 = 0.222 and X ClCH 22

= 0.778

In the vapor, Ptotal = 368 mm Hg X P = P CHCl

oCHClCHCl 333

• = (205 mm Hg)(0.222) = 45.5 mm Hg

X P = P ClCHo

ClCHClCH 222222• = (415 mm Hg)(0.778) = 323 mm Hg

Hg mm 368

Hg mm 45.5 =

P

P = Xtotal

CHClCHCl

3

3 = 0.124;

Hg mm 368

Hg mm 323 =

P

P = Xtotal

ClCHClCH

22

22 = 0.876

11.90 C9H8O4, 180.16 amu; 215 g = 0.215 kg

∆Tb = Kb ⋅ m = 0.47oC; Kb =

∆

kg 0.215g 180.16

mol 1 x g 5.00

C470. =

mT o

b = 3.6 mol

kg C o •

11.91 C6H8O6, 176.13 amu; 50.0 g = 0.0500 kg

∆Tf = Kf ⋅ m = 1.33oC; Kf =

∆

kg 0.0500g 176.13

mol 1 x g 3.00

C 1.33 =

mT o

f = 3.90 mol

kg C o •

11.92 ∆Tb = Kb ⋅ m = 1.76oC; m =

molkg C 3.07

C61.7 =

K

To

o

b

b

•∆

= 0.573 m

11.93 C6H12O6, 180.16 amu

For ethyl alcohol, Kb = 1.22 mol

kg C o •; 285 g = 0.285 kg

∆Tb = Kb ⋅ m =

•kg 0.285

g 180.16mol 1

x g 26.0

mol

kg C 1.22o

= 0.618oC

Solution boiling point = normal boiling point + ∆Tb = 79.1oC Normal boiling point = 79.1oC - ∆Tb = 79.1oC - 0.618oC = 78.5oC


273

11.94 Π = MRT (a) NaCl 58.44 amu; 350.0 mL = 0.3500 L There are 2 moles of ions/mole of NaCl

Π = (2) K) (323mol K

atm L 06 0.082

L 0.3500g 58.44

mol1 x g 5.00

••

= 13.0 atm

(b) CH3CO2Na, 82.03 amu; 55.0 mL = 0.0550 L There are 2 moles of ions/mole of CH3CO2Na

Π = (2) K)(283mol K

atm L 06 0.082

L 0.0550g 82.03

mol 1 x g 6.33

••

= 65.2 atm

11.95 Π = MRT =

L 60 0.006g 5990

mol 1 x g 10 x 11.5 3_

K)(298mol K

atm L 06 0.082

••

= 0.007 11 atm

Π = 0.007 11 atm x atm 1

Hg mm 760 = 5.41 mm Hg

height of H2O column = 5.41 mm Hg x Hg mm 1.00

OH mm 13.534 2 = 73.2 mm

height of H2O column = 73.2 mm x mm 1000

m 1 = 0.0732 m

11.96 Π = MRT; M = K) (300

mol K atm L

06 0.082

atm 4.85 =

RT

••

Π = 0.197 M

11.97 Π = MRT; M = K) (310

mol K atm L

06 0.082

atm 7.7 =

RT

••

Π = 0.30 M

Uses of Colligative Properties 11.98 Osmotic pressure is most often used for the determination of molecular mass because, of

the four colligative properties, osmotic pressure gives the largest colligative property change per mole of solute.

11.99 C6H12O6 does not dissociate in aqueous solution. LiCl and NaCl both dissociate into two

solute particles per formula unit in aqueous solution. CaCl2 dissociates into three solute particles per formula unit in aqueous solution. Assume that you have 1.00 g of each


274

substance. Calculate the number of moles of solute particles in 1.00 g of each substance.

C6H12O6, 180.2 amu; moles solute particles = 1.00 g x g 180.2

mol 1= 0.005 55 moles

LiCl, 42.4 amu; moles solute particles = 2

g 42.4

mol 1 x g 1.00 = 0.0472 moles

NaCl, 58.4 amu; moles solute particles = 2

g 58.4

mol 1 x g 1.00 = 0.0342 moles

CaCl2, 111.0 amu; moles solute particles = 3

g 111.0

mol 1 x g 1.00 = 0.0270 moles

LiCl produces more solute particles/gram than any of the other three substances. LiCl would be the most efficient per unit mass.

11.100 Π = 407.2 mm Hg x Hg mm 760

atm 1 = 0.5358 atm

Π = MRT; M = K) (298.15

mol K atm L

06 0.082

atm 0.5358 =

RT

••

Π = 0.021 90 M

200.0 mL x = mL 1000

L 1 0.2000 L

mol cellobiose = (0.2000 L)(0.021 90 mol/L) = 4.380 x 10-3 mol

molar mass of cellobiose = = cellobiose mol 10 x 4.380

cellobiose g 1.5003_

342.5 g/mol

molecular mass = 342.5 amu

11.101 height of Hg column = 32.9 cm H2O x OH cm 13.534

Hg cm 1.00

2

= 2.43 cm Hg

Π = 2.43 cm Hg x Hg cm 76.0

atm 1.00 = 0.0320 atm

Π = MRT; M = M 31 0.001 = K) (298

mol K atm L

06 0.082

atm 0.0320 =

RT

••

Π

20.0 mL x = mL 1000

L 1 0.0200 L

15.0 mg x = mg 1000

g 1.00 0.0150 g

mol met-enkephalin = (0.0200 L)(0.001 31 mol/L) = 2.62 x 10-5 mol


275

molar mass of met-enkephalin = = enkephalinmet- mol 10 x 2.62

enkephalinmet- g 0.01505_

573 g/mol

molecular mass = 573 amu 11.102 HCl is a strong electrolyte in H2O and completely dissociates into two solute particles

per each HCl. HF is a weak electrolyte in H2O. Only a few percent of the HF molecules dissociates into ions.

11.103 Na2SO4, 142.0 amu; m = kg 1.00

g 142.0mol 1

x g 71

= 0.50 mol/kg = 0.50 m

∆Tb = Kb ⋅ m =

•mol

kg C 0.51o

(0.50 m) = 0.26oC

The experimental ∆T is approximately 3 times that predicted by the equation above because Na2SO4 dissociates into three solute particles (2 Na+ and SO4

2-) in aqueous solution.

11.104 First, determine the empirical formula:

Assume 100.0 g of β-carotene.

10.51% H 10.51 g H x H g 1.008

H mol 1 = 10.43 mol H

89.49% C 89.49 g C x C g 12.01


C7.45H10.43; Divide each subscript by the smaller, 7.45. C7.45 / 7.45H10.43 / 7.45 CH1.4 Multiply each subscript by 5 to obtain integers. Empirical formula is C5H7, 67.1 amu. Second, calculate the molecular mass:

∆Tf = Kf ⋅ m; m = mol/kg 0.0310 =

molkg C 37.7

C171. =

K

To

o

f

f

•∆

= 0.0310 m

1.50 g x = g 1000

kg11.50 x 10-3 kg

mol β-carotene = (1.50 x 10-3 kg)(0.0310 mol/kg) = 4.65 x 10-5 mol

molar mass of β-carotene = = carotene- mol 10 x 4.65

carotene- g 0.02505_ β

β538 g/mol

molecular mass = 538 amu

Finally, determine the molecular formula: Divide the molecular mass by the empirical formula mass.


276

8 = amu 67.1

amu 538; molecular formula is C(8 x 5)H(8 x 7), or C40H56

11.105 First, determine the empirical formula:

Assume a 100.0 g sample of lysine.

49.29% C 49.29 g C x C g 12.011


9.65% H 9.65 g H x H g 1.008


19.16% N 19.16 g N x N g 14.007

N mol 1 = 1.37 mol N

21.89% O 21.89 g O x O g 15.999


C4.10H9.57N1.37O1.37; Divide each subscript by the smallest, 1.37. C4.10 / 1.37H9.57 / 1.37N1.37 / 1.37O1.37 / 1.37 Empirical formula is C3H7NO, 73.09 amu Second, calculate the molecular mass:

∆Tf = Kf ⋅ m = 1.37 oC; m =

molkg C 8.00

C 1.37 =

K

To

o

f

f

•∆

= 0.171 mol/kg = 0.171 m

1.200 g x = g 1000

kg1 1.200 x 10-3 kg

30.0 mg x = mg 1000

g 1.00 0.0300 g

mol lysine = (1.200 x 10-3 kg)(0.171 mol/kg) = 2.05 x 10-4 mol

molar mass of lysine = = lysine mol 10 x 2.05

lysine g 0.03004_

146 g/mol

molecular mass = 146 amu Finally, determine the molecular formula: Divide the molecular mass by the empirical formula mass.

amu 73.09

amu 146 = 2; molecular formula is C(2 x 3)H(2 x 7)N(2 x 1)O(2 x 1), or C6H14N2O2

General Problems

11.106 Kf for snow (H2O) is 1.86 mol

kg C o •. Reasonable amounts of salt are capable of

lowering the freezing point (∆Tf) of the snow below an air temperature of -2oC. Reasonable amounts of salt, however, are not capable of causing a ∆Tf of more than 30oC which would be required if it is to melt snow when the air temperature is -30oC.


277

11.107 KBr, 119.00 amu; for KBr, i = 2

125 g x = g 1000

kg1 0.125 kg

∆Tb = 103.2 oC - 100.0 oC = 3.2 oC

∆Tb = Kb ⋅ m ⋅ i; m =

(2)mol

kg C 0.51

C 3.2 =

2 K

To

o

b

b

••∆

= 3.137 mol/kg = 3.137 m

mol KBr = (0.125 kg)(3.137 mol/kg) = 0.392 mol KBr

mass of KBr = 0.392 mol KBr x = KBr mol 1

KBr g 119.00 47 g KBr

11.108 C2H6O2, 62.07 amu; ∆Tf = 22.0oC

∆Tf = Kf ⋅ m; m =

molkg C 1.86

C022. =

K

To

o

f

f

•∆

= 11.8 mol/kg = 11.8 m

mol C2H6O2 = (3.55 kg)(11.8 mol/kg) = 41.9 mol C2H6O2

mass C2H6O2 = 41.9 mol C2H6O2 x = OHC mol 1OHC g 62.07

262

262 2.60 x 103 g C2H6O2

11.109 The vapor pressure of toluene is lower than the vapor pressure of benzene at the same

temperature. When 1 mL of toluene is added to 100 mL of benzene, the vapor pressure of the solution decreases, which means that the boiling point of the solution will increase. When 1 mL of benzene is added to 100 mL of toluene, the vapor pressure of the solution increases, which means that the boiling point of the solution will decrease.

11.110 When solid CaCl2 is added to liquid water, the temperature rises because ∆Hsoln for

CaCl2 is exothermic. When solid CaCl2 is added to ice at 0oC, some of the ice will melt (an endothermic process) and the temperature will fall because the CaCl2 lowers the freezing point of an ice/water mixture.

11.111 AgCl, 143.32 amu; there are 2 ions/AgCl

Π = 2MRT

Π = 2

L 0.001g 143.32

mol 1 x g 10 x 0.007 3_

K) (278mol K

atm L 06 0.082

••

= 0.002 atm

11.112 C10H8, 128.17 amu; ∆Tf = 0.35oC


278

∆Tf = Kf ⋅ m; m = mol/kg 0.0684 =

molkg C 5.12

C350. =

K

To

o

f

f

•∆

= 0.0684 m

150.0 g x = g 1000

kg1 0.1500 kg

mol C10H8 = (0.1500 kg)(0.0684 mol/kg) = 0.0103 mol C10H8

mass C10H8 = 0.0103 mol C10H8 x = HC mol 1

HC g 128.17

810

810 1.3 g C10H8

11.113 Br2, 159.81 amu; CCl4, 153.82 amu

kPa 101.325

Hg mm 760 x kPa 30.5 = Po

Br2 = 228.8 mm Hg

kPa 101.325

Hg mm 760 x kPa 16.5 = Po

CCl4 = 123.8 mm Hg

1.50 g Br2 x Br g 159.81

Br mol 1

2

2 = 9.39 x 10-3 mol Br2

145.0 g CCl4 x CCl g 153.82

CCl mol 1

4

4 = 0.943 mol CCl4

mol) 10 x (9.39 + mol) (0.943

mol 10 x 9.39 = X 3_

3_

Br2 = 0.009 86

mol) 10 x (9.39 + mol) (0.943

mol 0.943 = X 3_CCl4 = 0.990

Psoln = X P + X P CCloCClBr

oBr 4422

••

Psoln = (228.8 mm Hg)(0.009 86) + (123.8 mm Hg)(0.990) = 125 mm Hg 11.114 NaCl, 58.44 amu; there are 2 ions/NaCl

A 3.5 mass % aqueous solution of NaCl contains 3.5 g NaCl and 96.5 g H2O.


g 58.44mol 1

x g 3.5

= 0.62 mol/kg = 0.62 m

∆Tf = Kf ⋅ 2 ⋅ m = mol/kg) (2)(0.62mol

kg C 1.86o

• = 2.3oC

Solution freezing point = 0.0oC - ∆Tf = 0.0oC - 2.3oC = -2.3oC

∆Tb = Kb ⋅ 2 ⋅ m = mol/kg) (2)(0.62mol

kg C 0.51o

• = 0.63oC

Solution boiling point = 100.00oC + ∆Tb = 100.00oC + 0.63oC = 100.63oC 11.115 (a) Assume a total mass of solution of 1000.0 g.


279

ppm = solution of mass total

ion solute of mass x 106

For each ion: mass of solute ion = 10

g) .0(ppm)(10006

Ion Mass Moles Cl- 19.0 g 0.536 mol Na+ 10.5 g 0.457 mol SO4

2- 2.65 g 0.0276 mol Mg2+ 1.35 g 0.0555 mol Ca2+ 0.400 g 0.009 98 mol K+ 0.380 g 0.009 72 mol HCO3

- 0.140 g 0.002 29 mol Br- 0.065 g 0.000 81 mol Total 34.5 g 1.099 mol

Mass of H2O = 1000.0 g - 34.5 g = 965.5 g H2O = 0.9655 kg H2O


mol 1.099 = 1.138 mol/kg = 1.138 m

(b) Assume M = m for a dilute solution.

Π = MRT = (1.138 mol/L)

••

mol K

atm L 06 0.082 (300 K) = 28.0 atm

11.116 (a) 90 mass % isopropyl alcohol = 100% x OH of mass + g 10.5

g 10.5

2

Solve for the mass of H2O.

mass of H2O = g 10.5 _ 90

100 x g 10.5

= 1.2 g

mass of solution = 10.5 g + 1.2 g = 11.7 g 11.7 g of rubbing alcohol contains 10.5 g of isopropyl alcohol. (b) C3H8O, 60.10 amu mass C3H8O = (0.90)(50.0 g) = 45 g

45 g C3H8O x OHC g 60.10

OHC mol 1

83

83 = 0.75 mol C3H8O

11.117 C6H12O6, 180.16 amu; 50.0 mL = 0.0500 L; 17.5 mg = 17.5 x 10-3 g

Π = MRT

T =

••

Π

mol K atm L

06 0.082L 0.0500

g 180.16mol 1

x g10 x 17.5

Hg mm 760atm 1

x Hg mm 37.8

= MR

3_

= 312 K


280

11.118 First, determine the empirical formula.

3.47 mg = 3.47 x 10-3 g sample 10.10 mg = 10.10 x 10-3 g CO2 2.76 mg = 2.76 x 10-3 g H2O

mass C = 10.10 x 10-3 g CO2 x CO g 44.01

C g 12.01

2

= 2.76 x 10-3 g C

mass H = 2.76 x 10-3 g H2O x OH g 18.02

H g 1.008 x 2

2

= 3.09 x 10-4 g H

mass O = 3.47 x 10-3 g - 2.76 x 10-3 g C - 3.09 x 10-4 g H = 4.01 x 10-4 g O

2.76 x 10-3 g C x C g 12.01

C mol 1 = 2.30 x 10-4 mol C

3.09 x 10-4 g H x H g 1.008

H mol 1 = 3.07 x 10-4 mol H

4.01 x 10-4 g O x O g 16.00

O mol 1 = 2.51 x 10-5 mol O = 0.251 x 10-4 mol O

To simplify the empirical formula, divide each mol quantity by 10-4. C2.30H3.07O0.251; Divide all subscripts by the smallest, 0.251. C2.30 / 0.251H3.07 / 0.251O0.251 / 0.251 C9.16H12.23O, empirical formula is C9H12O (136 amu)

Second, determine the molecular mass.

7.55 mg = 7.55 x 10-3 g estradiol; 0.500 g x g 1000

kg 1 = 5.00 x 10-4 kg camphor

∆Tf = Kf ⋅ m; m =

molkg C 37.7

C102. =

K

To

o

f

f

•∆

= 0.0557 mol/kg = 0.0557 m

m = solvent kg

estradiol mol

mol estradiol = m x (kg solvent) = (0.0557 mol/kg)(5.00 x 10-4 kg) = 2.79 x 10-5 mol

molar mass = estradiol mol 10 x 2.79

estradiol g 10 x 7.555_

3_

= 271 g/mol; molecular mass = 271 amu

Finally, determine the molecular formula: Divide the molecular mass by the empirical formula mass.

amu 136

amu 271 = 2; molecular formula is C(2 x 9)H(2 x 12)O(2 x 1), or C18H24O2

11.119 CCl3CO2H(aq) _ H+(aq) + CCl3CO2-(aq)

1.00 - x x x


281

∆Tf = Kf ⋅ m; m =

molkg C 1.86

C532. =

K

To

o

f

f

•∆

= 1.36 m

1.36 = 1.00 - x + x + x = 1 + x; x = 0.36 36% of the acid molecules are dissociated.

11.120 (a) H2SO4, 98.08 amu; 2.238 mol H2SO4 x SOH mol 1SOH g 98.08

42

42 = 219.50 g H2SO4

mass of 2.238 m solution = 219.50 g H2SO4 + 1000 g H2O = 1219.50 g

volume of 2.238 m solution = 1219.50 g x g 1.1243

mL 1.0000 = 1084.68 mL = 1.0847 L

molarity of 2.238 m solution = L 1.0847

mol 2.238 = 2.063 M

The molarity of the H2SO4 solution is less than the molarity of the BaCl2 solution. Because equal volumes of the two solutions are mixed, H2SO4 is the limiting reactant and the number of moles of H2SO4 determines the number of moles of BaSO4 produced as the white precipitate.

(0.05000 L) x (2.063 mol H2SO4/L) x BaSO mol 1

BaSO g 233.39 x

SOH mol 1BaSO mol 1

4

4

42

4 = 24.07 g BaSO4

(b) More precipitate will form because of the excess BaCl2 in the solution. 11.121 KCl, 74.55amu; KNO3, 101.10 amu; Ba(NO3)2, 261.34 amu

Π = MRT; M = K) (298

mol K atm L

06 0.082

Hg mm 760atm 1.00

x Hg mm 744.7

= RT

••

Π = 0.040 07 M

0.040 07 M = L 0.500

ions mol; mol ions = (0.040 07 mol/L)(0.500 L) = 0.020 035 mol ions

mass Cl = 1.000 g x 0.2092 = 0.2092 g Cl

mass KCl = KCl mol 1

KCl g 74.55 x

Cl mol 1

KCl mol 1 x

Cl g 35.453

Cl mol 1 x Cl g 0.2092 = 0.440 g

KCl

mol ions from KCl = 0.440 g KCl x KCl mol 1

ions mol 2 x

KCl g 74.55

KCl mol 1 = 0.0118 mol ions

mol ions from KNO3 and Ba(NO3)2 = 0.020 035 - 0.0118 = 0.008 235 mol ions Let x = mass KNO3 and y = mass Ba(NO3)2 x + y = 1.000 g - 0.440 g = 0.560 g

3261.34

y + 2

101.10

x

= 0.008 235 mol ions

x = 0.560 - y 0.0198x + 0.0115y = 0.008 235


282

0.0198(0.560 - y) + 0.0115y = 0.008 235 0.011 09 - 0.0198y + 0.0115y = 0.008 235

0.002 855 = 0.0083y; y = 0.0083

855 0.002 = 0.3440; x = 0.560 - 0.3440 = 0.216

mass % KCl = 100% x g 1.000

g 0.440 = 44.0%

mass % KNO3 = 100% x g 1.000

g 0.216 = 21.6%

mass % Ba(NO3)2 = 100% x g 1.000

g 0.344 = 34.4%

11.122 Let x = X OH2

and y = X OHCH3 and assume ntotal = 1.00 mol

(14.5 mm Hg)x + (82.5 mm Hg)y = 39.4 mm Hg (26.8 mm Hg)x + (140.3 mm Hg)y = 68.2 mm Hg

x = 26.8

y140.3 _ 68.2

26.8

y)140.3 _ 14.5(68.2 + 82.5y = 39.4

26.8

y)2034.35 _ (988.9 + 82.5y = 39.4

36.90 - 75.91y + 82.5y = 39.4; 6.59 = 2.5; y = 6.59

2.5 = 0.3794

x = 26.8

94)]140.3(0.37 _ [68.2 = 0.5586

XLiCl = 1 - X OH2 - X OHCH3

= 1 - 0.5586 - 0.3794 = 0.0620

The mole fraction equals the number of moles of each component because ntotal = 1.00 mol.

mass LiCl = 0.0620 mol LiCl x LiCl mol 1

LiCl g 42.39 = 2.6 g LiCl

mass H2O = 0.5588 mol H2O x OH mol 1

OH g 18.02

2

2 = 10.1 g H2O

mass CH3OH = 0.3794 mol CH3OH x OHCH mol 1

OHCH g 32.04

3

3 = 12.2 g CH3OH

total mass = 2.6 g + 10.1 g + 12.2 g = 24.9 g

mass % LiCl = 100% x g 24.9

g 2.6 = 10%

mass % H2O = 100% x g 24.9

g 10.1 = 41%


283

mass % CH3OH = 100% x g 24.9

g 12.2 = 49%

11.123 KI, 166.00 amu

∆Tf = Kf ⋅ m ⋅ i; m =

(2)mol

kg C 1.86

C951. =

i K

To

o

f

f

••∆

= 0.524 mol/kg = 0.524 m

Π = i ⋅ MRT; M = = K) (298

mol K atm L

06 0.082(2)

atm 25.0 =

T R i

•••

Π 0.511 M = 0.511

mol/L 1.000 L of solution contains 0.511 mol KI and is 0.524 m.

mass KI = 0.511 mol KI x = KI mol 1

KI g 166.00 84.83 g KI

Calculate the mass of solvent in this solution.

0.524 m = 0.524 mol/kg = solvent of mass

mol 0.511

mass of solvent = = mol/kg 0.524

mol 0.511 0.9752 kg = 975.2 g

mass of solution = mass KI + mass of solvent = 84.83 g + 975.2 g = 1060 g

density = = mL 1000

g 1060 1.06 g/mL

11.124 Solution freezing point = -1.03oC = 0.00oC - ∆Tf; ∆Tf = 1.03oC

∆Tf = Kf ⋅ m; m =

molkg C 1.86

C031. =

K

To

o

f

f

•∆

= 0.554 mol/kg = 0.554 m

Π = MRT; M = K) (298

mol K atm L

06 0.082

atm) (12.16 =

RT

••

Π = 0.497

L

mol

Assume 1.000 L = 1000 mL of solution. mass of solution = (1000 mL)(1.063 g/mL) = 1063 g

mass of H2O in 1000 mL of solution = xsolute of mol 0.554

OH g 1000 2 0.497 mol = 897 g H2O

mass of solute = total mass - mass of H2O = 1063 g - 897 g = 166 g solute

molar mass = = mol 0.497

g 166 334 g/mol

11.125 C6H6, 78.11 amu


284

299 mm Hg = PoHC 66

⋅ X HC 66 + Po

X ⋅ XX

299 mm Hg = (395 mm Hg) ⋅ X HC 66 + (96 mm Hg) ⋅ XX

X HC 66 + XX = 1; XX = 1 - X HC 66

299 mm Hg = (395 mm Hg) ⋅ X HC 66 + (96 mm Hg)(1 - X HC 66

)

299 mm Hg = (395 mm Hg) ⋅ X HC 66 + 96 mm Hg - (96 mm Hg) ⋅ X HC 66

299 mm Hg - 96 mm Hg = (395 mm Hg) ⋅ X HC 66 - (96 mm Hg) ⋅ X HC 66

203 mm Hg = (299 mm Hg) ⋅ X HC 66

X HC 66 = 203 mm Hg/299 mm Hg = 0.679

Assume the mixture contains 1.00 mol (78.11 g) of C6H6. Then a 50/50 mixture will also contain 78.11 g of X.

X HC 66 =

X mol + HC mol 1HC mol 1

66

66 = 0.679

1 mol C6H6 = (0.679)(1 mol C6H6 + mol X)

0.679HC mol 1 66 = 1 mol C6H6 + mol X

0.679HC mol 1 66 - 1 mol C6H6 = mol X

mol X = 0.473 mol molar mass X = 78.11 g/0.473 mol = 165 g/mol

11.126 (a) NaCl, 58.44 amu; CaCl2, 110.98 amu; H2O, 18.02 amu

mol NaCl = 100.0 g NaCl x = NaCl g 58.44

NaCl mol 1 1.711 mol NaCl

mol CaCl2 = 100.0 g CaCl2 x = CaCl g 110.98

CaCl mol 1

2

2 0.9011 mol CaCl2

mass of solution = (1000 mL)(1.15 g/mL) = 1150 g mass of H2O in solution = mass of solution - mass NaCl - mass CaCl2 = 1150 g - 100.0 g - 100.0 g = 950 g

= 950 g x = g 1000

kg 1 0.950 kg

∆Tb = Kb ⋅ ) i m + i m ( CaClNaCl 2••

∆Tb = = kg 0.950

3) CaCl mol (0.9011 + 2) NaCl mol (1.711

mol

kg C 0.51 2o

••

• 3.3oC

solution boiling point = 100.0oC + ∆Tb = 100.0oC + 3.3oC = 103.3oC

(b) mol H2O = 950 g H2O x = OH g 18.02

OH mol 1

2

2 52.7 mol H2O

PSolution = Po X OH2•

PSolution = Po


285

•••

3) CaCl mol (0.9011 + 2) NaCl mol (1.711 + O)H mol (52.7

OH mol 52.7

22

2

PSolution = (23.8 mm Hg)(0.896) = 21.3 mm Hg 11.127 HIO3, 175.91 amu

mass of 1.00 L solution = (1.00 x 103 mL)(1.07 g/mL) = 1.07 x 103 g 1.00 L of solution contains 1 mol (175.91 g) HIO3. mass of H2O = 1070 g - 175.91 g = 894 g = 0.894 kg m = 1.00 mol/0.894 kg = 1.12 m ∆Tf = Kf ⋅ m ⋅ i

i = =

mol/kg) (1.12mol

kg C 1.86

C782. =

m K

To

o

f

f

••∆

1.33; The acid is 33% dissociated.

11.128 (a) KI, 166.00 amu

Assume you have 1.000 L of 1.24 M solution. mass of solution = (1000 mL)(1.15 g/mL) = 1150 g

mass of KI in solution = 1.24 mol KI x = KI mol 1

KI g 166.00 206 g KI

mass of H2O in solution = mass of solution - mass KI = 1150 g - 206 g = 944 g

= 944 g x = g 1000

kg 1 0.944 kg

molality = = OH kg 0.944

KI mol 1.24

2

1.31 m

(b) For KI, i = 2 assuming complete dissociation.

∆Tf = Kf ⋅ m ⋅ i = )(2)m (1.31mol

kg C 1.86o

• = 4.87oC

Solution freezing point = 0.00oC - ∆Tf = 0.00oC - 4.87oC = - 4.87oC

(c) i = =

mol/kg) (1.31mol

kg C 1.86

C464. =

m K

To

o

f

f

••∆

1.83

Because the calculated i is only 1.83 and not 2, the percent dissociation for KI is 83%. 11.129 (a) For NaCl, i = 2 and for MgCl2, i = 3; T = 25oC = 25 + 273 = 298 K

Π = i ⋅ MRT = [(2)(0.470 mol/L) + (3)(0.068 mol/L)] K) (298mol K

atm L 06 0.082

••

= 28.0 atm

(b) Calculate the molarity for an osmotic pressure = 100.0 atm.


286

Π = MRT; M = K) (298

mol K atm L

06 0.082

atm) (100.0 =

RT

••

Π = 4.09 mol/L

Mconc x Vconc = Mdil x Vdil

Vconc = = M

V x M

conc

dildil = mol/L 4.09

L) 00mol/L)](1. (3)(0.068 + mol/L) [(2)(0.4700.28 L

A volume of 1.00 L of seawater can reduced to 0.28 L by an osmotic pressure of 100.0 atm. The volume of fresh water that can be obtained is (1.00 L - 0.28 L) = 0.72 L.

11.130 NaCl, 58.44 amu; C12H22O11, 342.3 amu

Let X = mass NaCl and Y = mass C12H22O11, then X + Y = 100.0 g.

500.0 g x g 1000

kg 1 = 0.5000 kg

Solution freezing point = -2.25oC = 0.00oC - ∆Tf; = ∆Tf = 0.00oC + 2.25oC = 2.25oC ∆Tf = Kf ⋅ ) m + i m ( OHCNaCl 112212

•

∆Tb =

•

•kg 0.5000

)OHC (mol + 2) NaCl (mol

mol

kg C 1.86 112212o

= 2.25oC

mol NaCl = X g NaCl x NaCl g 58.44

NaCl mol 1 = X/58.44 mol

mol C12H22O11 = Y g C12H22O11 x OHC g 342.3

OHC mol 1

112212

112212 = Y/342.3 mol

∆Tb =

•

•kg 0.5000

mol) 342.3)((Y/ + mol) 2 58.44)((X/

mol

kg C 1.86o

= 2.25oC

X = 100 - Y

•

•kg 0.5000

mol] 342.3)[(Y/ + mol] 2 58.44]/Y) _ [(100

mol

kg C 1.86o

= 2.25oC

kg 0.5000

mol 342.3)](Y/ + 58.44)Y/(2 _ 58.44)/[(200 =

•mol

kg C 1.86

C52.2o

o

= 1.21 mol/kg

kg 0.5000

mol Y)](0.0313 _ [(3.42)= 1.21 mol/kg

[(3.42) - (0.0313Y)] = (0.5000 kg)(1.21) = 0.605 - 0.0313 Y = 0.605 - 3.42 = - 2.81 Y = (- 2.81)/(- 0.0313) = 89.9 g of C12H22O11 X = 100.0 g - Y = 100.0 g - 89.9 g = 10.1 g of NaCl


287

Multi-Concept Problems 11.131 (a) 382.6 mL = 0.3826 L; 20.0oC = 293.2 K

PV = nRT

nH2 =


06 0.082

L) (0.3826Hg mm 760

atm 1.0 x Hg mm 755

= T R

V P

••

= 0.0158 mol H2

(b) M + x HCl → x/2 H2 + MClx

moles HCl reacted = 0.0158 mol H2 x H mol 2x/

HCl molx

2

= 0.0316 mol HCl

moles Cl reacted = moles HCl reacted = 0.0316 mol Cl

mass Cl = 0.0316 mol Cl x Cl mol 1

Cl g 35.453= 1.120 g Cl

mass MClx = mass M + mass Cl = 1.385 g + 1.120 g = 2.505 g MClx

(c) ∆Tf = Kf ⋅ m; m = mol/kg 1.90 =

molkg C 1.86

C533. =

K

To

o

f

f

•∆

= 1.90 m

(d) 25.0 g = 0.0250 kg

1.90 m = kg

mol 1.90 =

kg 0.0250

ions molx

mol ions = (1.90 mol/kg)(0.0250 kg) = 0.0475 mol ions (e) mol M = mol ions - mol Cl = 0.0475 mol - 0.0316 mol = 0.0159 mol M

mol 0.0159

mol 0.0316 =

M

Cl= 2, the formula is MCl2.



(f) atomic mass of M = 157.5 amu - 2(35.453 amu) = 86.6 amu; M = Sr 11.132 (a) 20.00 mL = 0.02000 L

mol NaOH = (0.02000 L)(2.00 mol/L) = 0.0400 mol NaOH

mol CO2 = 0.0400 mol NaOH x NaOH mol 2CO mol 1 2 = 0.0200 mol CO2

mol C = 0.0200 mol CO2 x CO mol 1

C mol 1

2

= 0.0200 mol C

mass C = 0.0200 mol C x C mol 1

C g 12.011 = 0.240 g C

mass H = mass of compound - mass of C = 0.270 g - 0.240 g = 0.030 g H


288

mol H = 0.030 g H x H g 1.008

H mol 1 = 0.030 mol H

The mole ratio of C and H in the molecule is C0.0200 H0.030. C0.0200 H0.030, divide both subscripts by the smaller of the two, 0.0200. C0.0200 / 0.0200 H0.030 / 0.0200 C1H1.5, multiply both subscripts by 2. C(2 x 1) H(2 x 1.5) C2H3 (27.05 amu) is the empirical formula.

(b) ∆Tf = Kf ⋅ m; m =

molkg C 37.7

C)9177. _ C8(179. =

K

To

oo

f

f

•∆

= 0.050 mol/kg = 0.050 m

50.0 g x = g 1000

kg 1 0.0500 kg

mol solute = (0.050 mol/kg)(0.0500 kg) = 0.0025 mol


g 0.270 = 108 g/mol; molecular mass = 108 amu

(c) To find the molecular formula, first divide the molecular mass by the mass of the empirical formula unit.

27

108 = 4

Multiply the subscripts in the empirical formula by the result of this division, 4. C(4 x 2) H(4 x 3) C8H12 is the molecular formula of the compound.

11.133 CO2, 44.01 amu; H2O, 18.02 amu

mol C = 106.43 mg CO2 x mg 1000

g 1 x

CO g 44.01CO mol 1

2

2 x CO mol 1

C mol 1

2

= 0.002 418 mol

C

mass C = 0.002 418 mol C x C mol 1

C g 12.011 = 0.029 04 g C

mol H = 32.100 mg H2O x mg 1000

g 1 x

OH g 18.02

OH mol 1

2

2 x OH mol 1

H mol 2

2

= 0.003 563 mol

H

mass H = 0.003 563 mol H x H mol 1

H g 1.008 = 0.003 592 g H

mass O =

mg 1000

g 1 x mg 36.72 - 0.029 04 g C - 0.003 592 g H = 0.004 088 g O

mol O = 0.004 088 g O x O g 16.00

O mol 1 = 0.000 255 5 mol O

C0.002 418H0.003 563O0.000 255 5 Divide all subscripts by the smallest, 0.000 255 5. C0.002 418 / 0.000 255 5H0.003 563 / 0.000 255 5O0.000 255 5 / 0.000 255 5


289

C9.5H14O Multiply all subscripts by 2. C(9.5 x 2)H(14 x 2)O(1 x 2) Empirical formula is C19H28O2, 288 amu

T = 25oC = 25 + 273 = 298 K

Π = MRT; M = K) (298

mol K atm L

06 0.082

Hg mm 760atm 1

x Hg mm 21.5

= RT

••

Π = 0.001 16 mol/L

15.0 mL = 0.0150 L mol solute = (0.001 16 mol/L)(0.0150 L) = 1.74 x 10-5 mol

molar mass = = mol 10 x 1.74

mg 1000g 1

x mg 5.00

5_

287 g/mol

The molar mass and the empirical formula mass are essentially identical, so the molecular formula and the empirical formula are the same. The molecular formula is C19H28O2.

11.134 AgCl, 143.32 amu

Solution freezing point = - 4.42oC = 0.00oC - ∆Tf; ∆Tf = 0.00oC + 4.42oC = 4.42oC ∆Tf = Kf ⋅ m

total ion m =

molkg C 1.86

C424. =

K

To

o

f

f

•∆

= 2.376 mol/kg = 2.376 m

150.0 g x = g 1000

kg 1 0.1500 kg

total mol of ions = (2.376 mol/kg)(0.1500 kg) = 0.3564 mol of ions An excess of AgNO3 reacts with all Cl- to produce 27.575 g AgCl.

total mol Cl- = 27.575 g AgCl x = AgCl mol 1Cl mol 1

x AgCl g 143.32

AgCl mol 1 _

0.1924 mol Cl-

Let P = mol XCl and Q = mol YCl2. 0.3564 mol ions = 2 x mol XCl + 3 x mol YCl2 = (2 x P) + (3 x Q) 0.1924 mol Cl- = mol XCl + 2 x mol YCl2 = P + (2 x Q) P = 0.1924 - (2 x Q) 0.3564 = 2 x [0.1924 - (2 x Q)] + (3 x Q) = 0.3848 - (4 x Q) + (3 x Q) Q = 0.3848 - 0.3564 = 0.0284 mol YCl2 P = 0.1924 - (2 x Q) = 0.1924 - (2 x 0.0284) = 0.1356 mol XCl

mass Cl in XCl = 0.1356 mol XCl x = Cl mol 1

Cl g 35.453 x

XCl mol 1

Cl mol 1 4.81 g Cl


290

mass Cl in YCl2 = 0.0284 mol YCl2 x = Cl mol 1

Cl g 35.453 x

YCl mol 1

Cl mol 2

2

2.01 g Cl

total mass of XCl and YCl2 = 8.900 g mass of X + Y = total mass - mass Cl = 8.900 g - 4.81 g - 2.01 g = 2.08 g X is an alkali metal and there are 0.1356 mol of X in XCl. If X = Li, then mass of X = (0.1356 mol)(6.941 g/mol) = 0.941 g If X = Na, then mass of X = (0.1356 mol)(22.99 g/mol) = 3.12 g but this is not possible because 3.12 g is greater than the total mass of X + Y. Therefore, X is Li. mass of Y = 2.08 - mass of X = 2.08 g - 0.941 g = 1.14 g Y is an alkaline earth metal and there are 0.0284 mol of Y in YCl2. molar mass of Y = 1.14 g/0.0284 mol = 40.1 g/mol. Therefore, Y is Ca.

mass LiCl = 0.1356 mol LiCl x = LiCl mol 1

LiCl g 42.395.75 g LiCl

mass CaCl2 = 0.0284 mol CaCl2 x = CaCl mol 1

CaCl g 110.98

2

2 3.15 g CaCl2

291

12

Chemical Kinetics 12.1 3 I-(aq) + H3AsO4(aq) + 2 H+(aq) → I3

-(aq) + H3AsO3(aq) + H2O(l)

(a) - t

]I[ _

∆∆

= 4.8 x 10-4 M/s

M/s) 10 x (4.83

1 =

t

]I[ _3

1 =

t

] I[ 4___

3

∆∆

∆∆

= 1.6 x 10-4 M/s

(b) -

∆∆

∆∆

t

] I[2 = t

] H[ _3

+

= (2)(1.6 x 10-4 M/s) = 3.2 x 10-4 M/s

12.2 2 N2O5(g) → 4 NO2(g) + O2(g)

time [N2O5] [O2] 200 s 0.0142 M 0.0029 M 300 s 0.0120 M 0.0040 M

Rate of decomposition of N2O5 = s 200 _ s 300

M 0.0142 _ M 0.0120 _ =

t

]ON[ _ 52

∆∆

= 2.2 x 10-5

M/s

Rate of formation of O2 = s 200 _ s 300

M 0.0029 _ M 0.0040 =

t

]O[ 2

∆∆

= 1.1 x 10-5 M/s

12.3 Rate = k[BrO3

-][Br -][H+]2 1st order in BrO3

-, 1st order in Br-, 2nd order in H+, 4th order overall Rate = k[H2][I 2], 1st order in H2, 1st order in I2, 2nd order overall Rate = k[CH3CHO]3/2, 3/2 order in CH3CHO, 3/2 order overall

12.4 H2O2(aq) + 3 I-(aq) + 2 H+(aq) → I3

-(aq) + 2 H2O(l)

Rate = t

]I[ _3

∆∆

= k[H2O2]m[I -]n

(a) M/s 10 x 1.15

M/s 10 x 2.30 =

Rate

Rate4_

4_

1

3 = 2 M 0.100

M 0.200 =

]OH[

]OH[

122

322 = 2

Because both ratios are the same, m = 1.

M/s 10 x 1.15

M/s 10 x 2.30 =

Rate

Rate4_

4_

1

2 = 2 M 0.100

M 0.200 =

]I[

]I[

1_

2_

= 2

Because both ratios are the same, n = 1. The rate law is: Rate = k[H2O2][I

-]

(b) k = ]I][OH[

Rate_

22

Chapter 12 - Chemical Kinetics _____________________________________________________________________________

292

Using data from Experiment 1: k = M) M)(0.100 (0.100

M/s 10 x 1.15 4_

= 1.15 x 10-2 /(M ⋅ s)

(c) Rate = k[H2O2][I-] = [1.15 x 10-2/(M ⋅ s)](0.300 M)(0.400 M) = 1.38 x 10-3 M/s

12.5 Rate Law Units of k

Rate = k[(CH3)3CBr] 1/s Rate = k[Br2] 1/s Rate = k[BrO3

-][Br -][H+]2 1/(M3 ⋅ s) Rate = k[H2][I 2] 1/(M ⋅ s) Rate = [CH3CHO]3/2 1/(M1/2 ⋅ s)

12.6 (a) The reactions in vessels (a) and (b) have the same rate, the same number of B

molecules, but different numbers of A molecules. Therefore, the rate does not depend on A and its reaction order is zero. The same conclusion can be drawn from the reactions in vessels (c) and (d). The rate for the reaction in vessel (c) is four times the rate for the reaction in vessel (a). Vessel (c) has twice as many B molecules than does vessel (a). Because the rate quadruples when the concentration of B doubles, the reaction order for B is two. (b) rate = k[B]2

12.7 (a) ln kt _ = ]Br)NH[Co(

]Br)NH[Co(

o+2

53

t+2

53

k = 6.3 x 10-6/s; t = 10.0 h x h 1

s 3600 = 36,000 s

ln[Co(NH3)5Br2+] t = kt _ + ln[Co(NH3)5Br2+]o

ln[Co(NH3)5Br2+] t = s) /s)(36,00010 x (6.3 _ 6_ + ln(0.100) ln[Co(NH3)5Br2+] t = -2.5294; After 10.0 h, [Co(NH3)5Br2+] = e-2.5294 = 0.080 M

(b) [Co(NH3)5Br2+]o = 0.100 M If 75% of the Co(NH3)5Br2+ reacts then 25% remains. [Co(NH3)5Br2+] t = (0.25)(0.100 M) = 0.025 M

ln kt _ = ]Br)NH[Co(

]Br)NH[Co(

o+2

53

t+2

53 ; t = k _

]Br)NH[Co(]Br)NH[Co(

ln o

+253

t+2

53

t = /s)10 x (6.3 _

0.1000.025

ln

6_

= 2.2 x 105 s; t = 2.2 x 105 s x s 3600

h 1 = 61 h


293

12.8

Slope = -0.03989/min = -6.6 x 10-4/s and k = -slope A plot of ln[cyclopropane] versus time is linear, indicating that the data fit the equation for a first-order reaction. k = 6.6 x 10-4/s (0.040/min)

12.9 (a) k = 1.8 x 10-5/s

t1/2 = /s10 x 1.8

0.693 =

k

0.6935_

= 38,500 s; t1/2 = 38,500 s x s 3600

h 1 = 11 h

t1/2 t1/2 t1/2 t1/2 (b) 0.30 M → 0.15 M → 0.075 M → 0.0375 M → 0.019 M (c) Because 25% of the initial concentration corresponds to 1/4 or (1/2)2 of the initial concentration, the time required is two half-lives: t = 2t1/2 = 2(11 h) = 22 h

12.10 After one half-life, there would be four A molecules remaining. After two half-lives,

there would be two A molecules remaining. This is represented by the drawing at t = 10 min. 10 min is equal to two half-lives, therefore, t1/2 = 5 min for this reaction. After 15 min (three half-lives) only one A molecule would remain.

12.11


294

(a) A plot of 1/[HI] versus time is linear. The reaction is second-order. (b) k = slope = 0.0308/(M ⋅ min)

(c) min 260 = M 0.500

1 _

M 0.100

1

min) (M / 0.0308

1 =

][HI

1 _

][HI

1

k

1 =t

ot

•

(d) It requires one half-life (t1/2) for the [HI] to drop from 0.400 M to 0.200 M.

t1/2 = M) 0min)](0.40/(Mcdot [0.0308

1 =

]k[HI

1

o

= 81.2 min

12.12 (a) NO2(g) + F2(g) → NO2F(g) + F(g)

F(g) + NO2(g) → NO2F(g) Overall reaction 2 NO2(g) + F2(g) → 2 NO2F(g) Because F(g) is produced in the first reaction and consumed in the second, it is a reaction intermediate. (b) In each reaction there are two reactants, so each elementary reaction is bimolecular.

12.13 (a) Rate = k[O3][O] (b) Rate = k[Br]2[Ar] (c) Rate = k[Co(CN)5(H2O)2-] 12.14 Co(CN)5(H2O)2-(aq) → Co(CN)5

2-(aq) + H2O(l) (slow) Co(CN)5

2-(aq) + I-(aq) → Co(CN)5I3-(aq) (fast)

Overall reaction Co(CN)5(H2O)2-(aq) + I-(aq) → Co(CN)5I3-(aq) + H2O(l)

The predicted rate law for the overall reaction is the rate law for the first (slow) elementary reaction: Rate = k[Co(CN)5(H2O)2-] The predicted rate law is in accord with the observed rate law.

12.15 (a) Ea = 100 kJ/mol - 20 kJ/mol = 80 kJ/mol

(b) The reaction is endothermic because the energy of the products is higher than the energy of the reactants.

(c)

12.16 (a)

T

1 _

T

1

RE _

= k

kln 12

a

1

2

k1 = 3.7 x 10-5/s, T1 = 25oC = 298 K k2 = 1.7 x 10-3/s, T2 = 55oC = 328 K


295

Ea =

T

1 _

T

1

R ]kln _ k[ln _

12

12

Ea =

K 2981

_ K 328

1mol)]kJ/(Kcdot 10 x )][8.31410 x ln(3.7 _ )10 x [ln(1.7

_3_5_3_

= 104 kJ/mol

(b) k1 = 3.7 x 10-5/s, T1 = 25oC = 298 K solve for k2, T2 = 35oC = 308 K

ln k2 = kln + T

1 _

T

1

RE _

112

a

ln k2 = )10 x (3.7ln + K 298

1 _

K 308

1

mol) kJ/(K 10 x 8.314

kJ/mol 104_ 5_3_

•

ln k2 = -8.84; k2 = e-8.84 = 1.4 x 10-4/s 12.17 Assume that concentration is proportional to the number of each molecule in a box.

(a) From boxes (1) and (2), the concentration of A doubles, B and C2 remain the same and the rate does not change. This means the reaction is zeroth-order in A. From boxes (1) and (3), the concentration of C2 doubles, A and B remain the same and the rate doubles. This means the reaction is first-order in C2. From boxes (1) and (4), the concentration of B triples, A and C2 remain the same and the rate triples. This means the reaction is first-order in B. (b) Rate = k [B][C2] (c) B + C2 → BC2 (slow)

A + BC2 → AC + BC A + BC → AC + B 2 A + C2 → 2 AC (overall)

(d) B doesn’t appear in the overall reaction because it is consumed in the first step and regenerated in the third step. B is therefore a catalyst. BC2 and BC are intermediates because they are formed in one step and then consumed in a subsequent step in the reaction.

12.18 Nitroglycerin contains three nitro groups per molecule. Because the bonds in nitro groups

are relatively weak (about 200 kJ/mol) and because the explosion products (CO2, N2, H2O, and O2) are extremely stable, a great deal of energy is released (very exothermic) during an explosion.

12.19 Secondary explosives are generally less sensitive to heat and shock than primary

explosives. This would indicate that secondary explosives should have a higher activation energy than primary explosives.

12.20 C5H8N4O12(s) → 4 CO2(g) + 4 H2O(g) + 2 N2(g) + C(s)


296

∆Horxn = [4 ∆Ho

f (CO2) + 4 ∆Hof (H2O)] - ∆Ho

f (C5H8N4O12) ∆Ho

rxn = [(4 mol)(-393.5 kJ/mol) + (4 mol)(-241.8 kJ/mol)] - [(1 mol)(537 kJ/mol)] ∆Ho

rxn = -3078 kJ 12.21 C5H8N4O12(s) → 4 CO2(g) + 4 H2O(g) + 2 N2(g) + C(s)

C5H8N4O12, 316.14 amu; 1.54 kg = 1.54 x 103 g; 800oC = 1073 K From the reaction, 1 mole of PETN produces 10 moles of gas.

mol gas = 1.54 x 103 g PETN x PETN mol 1

gas mol 10 x

PETN g 316.14

PETN mol 1 = 48.7 mol

PV = nRT

V = atm 0.975

K) (1073mol K atm L

06 0.082mol) (48.7 =

P

nRT

••

= 4.40 x 103 L

Understanding Key Concepts 12.22 (a) Because Rate = k[A][B], the rate is proportional to the product of the number of A

molecules and the number of B molecules. The relative rates of the reaction in vessels (a) – (d) are 2 : 1 : 4 : 2. (b) Because the same reaction takes place in each vessel, the k's are all the same.

12.23 (a) Because Rate = k[A], the rate is proportional to the number of A molecules in each

reaction vessel. The relative rates of the reaction are 2 : 4 : 3. (b) For a first-order reaction, half-lives are independent of concentration. The half-lives are the same. (c) Concentrations will double, rates will double, and half-lives will be unaffected.

12.24 (a) For the first-order reaction, half of the A molecules are converted to B molecules

each minute.

(b) Because half of the A molecules are converted to B molecules in 1 min, the half-life is 1 minute.

12.25 (a) Two molecules of A are converted to two molecules of B every minute. This means

the rate is constant throughout the course of the reaction. The reaction is zeroth-order.


297

(b) (c) Rate = k

k = = s 60

min 1 x

min

molecules 10 x 6.022mol 1

L 1.0molecules 10 x 6.0

(2)23

21

3.3 x 10-4 M/s

12.26 (a) Because the half-life is inversely proportional to the concentration of A molecules,

the reaction is second-order in A. (b) Rate = k[A]2 (c) The second box represents the passing of one half-life, and the third box represents the passing of a second half-life for a second-order reaction. A relative value of k can be calculated.

k = (1)(16)

1 =

[A]t

1

2/1

= 0.0625

t1/2 in going from box 3 to box 4 is: t1/2 = )(0.0625)(4

1 =

k[A]

1 = 4 min

(For fourth box, t = 7 min)

12.27 (a) bimolecular (b) unimolecular (c) termolecular 12.28 (a) BC + D → B + CD

(b) 1. B–C + D (reactants), A (catalyst); 2. B---C---A (transition state), D (reactant); 3. A–C (intermediate), B (product), D (reactant); 4. A---C---D (transition state), B (product); 5. A (catalyst), C–D + B (products) (c) The first step is rate determining because the first maximum in the potential energy curve is greater than the second (relative) maximum; Rate = k[A][BC] (d) Endothermic

12.29


298

Additional Problems Reaction Rates

12.30 M/s or s L

mol

•

12.31 molecules/(cm3 ⋅ s)

12.32 (a) Rate = min 0.0 _min 5.0

M 0.098 _ M 0.080 _ =

t

ane][cycloprop _

∆∆

= 3.6 x 10-3 M/min

Rate = 3.6 x 10-3 s 60

min 1 x

min

M = 6.0 x 10-5 M/s

(b) Rate = min 5.0l _min 20.0

M 0.054 _ M 0.044 _ =

t

ane][cycloprop_

∆∆

= 2.0 x 10-3 M/min

Rate = 2.0 x 10-3 s 60

min 1 x

min

M = 3.3 x 10-5 M/s

12.33 (a) Rate = s 50 _ s 100

M) 10 x (6.58 _ M) 10 x (5.59 _ =

Deltat

]NO[ _

3_3_2∆

= 2.0 x 10-5 M/s

(b) Rate = s 100 _ s 150

M) 10 x (5.59 _ M) 10 x (4.85 _ =

Deltat

]NO[ _

3_3_2∆

= 1.5 x 10-5 M/s

12.34

(a) The instantaneous rate of decomposition of N2O5 at t = 200 s is determined from the slope of the curve at t = 200 s.

Rate = s 100 _ s 300

M) 10 x (1.69 _ M) 10 x (1.20 _ = slope _ =

Deltat

]ON[ _2_2_

52∆ = 2.4 x 10-5 M/s

(b) The initial rate of decomposition of N2O5 is determined from the slope of the curve at t = 0 s. This is equivalent to the slope of the curve from 0 s to 100 s because in this time interval the curve is almost linear.

Initial rate = - s 0 _ s 100

M) 10 x (2.00 _ M) 10 x (1.69 _ = slope _ =

t

]ON[ 2_2_52

∆∆

= 3.1 x 10-5 M/s


299

12.35 (a) The instantaneous rate of decomposition of NO2 at t = 100 s is determined from the slope of the curve at t = 100 s.

Rate = s 20 _ s 190

M) 10 x (7.00 _ M) 10 x (4.00 _ = slope _ =

Deltat

]NO[ _

3_3_2∆

= 1.8 x 10-5 M/s

(b) The initial rate of decomposition of NO2 is determined from the slope of the curve at t = 0 s. This is equivalent to the slope of the curve from 0 s to 50 s because in this time interval the curve is almost linear.

Initial rate = -s 0 _ s 50

M) 10 x (8.00 _ M) 10 x (6.58 _ = slope _ =

t

]NO[ 3_3_2

∆∆

= 2.8 x 10-5 M/s

12.36 (a) t

]N[ 3 _ = t

]H[ _ 22

∆∆

∆∆

; The rate of consumption of H2 is 3 times faster.

(b) t

]N[ 2 _ = t

]NH[ 23

∆∆

∆∆

; The rate of formation of NH3 is 2 times faster.

12.37 (a) t

]NH[

4

5 _ =

t

]O[ _ 32

∆∆

∆∆

; The rate of consumption of O2 is 1.25 times faster.

(b) t

]NH[ _ =

t

[NO] 3

∆∆

∆∆

; The rate of formation of NO is the same.

t

]NH[

4

6 _ =

t

O]H[ 32

∆∆

∆∆

; The rate of formation of H2O is 1.5 times faster.

12.38 N2(g) + 3 H2(g) → 2 NH3(g); - t

]NH[

2

1 =

t

]H[ 3

1 _ =

t

]N[ 322

∆∆

∆∆

∆∆

12.39 (a) t

] OS[ 3 _ = t

]I[ __2

82_

∆∆

∆∆

= 3(1.5 x 10-3 M/s) = 4.5 x 10-3 M/s

(b) t

] OS[ 2 _ = t

] SO[ _282

_24

∆∆

∆∆

= 2(1.5 x 10-3 M/s) = 3.0 x 10-3 M/s

Rate Laws 12.40 Rate = k[NO]2[Br2]; 2nd order in NO; 1st order in Br2; 3rd order overall 12.41 Rate = k[CHCl3][Cl 2]

1/2; 1st order in CHCl3; 1/2 order in Cl2; 3/2 order overall

12.42 Rate = k[H2][ICl]; units for k are s mol

L

• or 1/(M ⋅ s)


300

12.43 Rate = k[NO]2[H2], units for k are 1/(M2 ⋅ s) 12.44 (a) Rate = k[CH3Br][OH-]

(b) Because the reaction is first-order in OH-, if the [OH-] is decreased by a factor of 5, the rate will also decrease by a factor of 5. (c) Because the reaction is first-order in each reactant, if both reactant concentrations are doubled, the rate will increase by a factor of 2 x 2 = 4.

12.45 (a) Rate = k[Br-][BrO3

-][H+]2 (b) The overall reaction order is 1 + 1 + 2 = 4. (c) Because the reaction is second-order in H+, if the [H+] is tripled, the rate will increase by a factor of 32 = 9. (d) Because the reaction is first-order in both Br- and BrO3

-, if both reactant concentrations are halved, the rate will decrease by a factor of 4 (1/2 x 1/2 = 1/4).

12.46 (a) Rate = k[CH3COCH3]

m

m =

10 x 6.010 x 9.0

ln

10 x 5.210 x 7.8

ln

=

]COCHCH[]COCHCH[

ln

RateRateln

3_

3_

5_

5_

133

233

1

2

= 1; Rate = k[CH3COCH3]

(b) From Experiment 1: k = M 10 x 6.0

M/s 10 x 5.2 =

]COCHCH[

Rate3_

5_

33

= 8.7 x 10-3/s

(c) Rate = k[CH3COCH3] = (8.7 x 10-3/s)(1.8 x 10-3M) = 1.6 x 10-5 M/s 12.47 (a) Rate = k[CH3NNCH3]

m

m =

10 x 2.410 x 8.0

ln

10 x 6.010 x 2.0

ln

=

]NNCHCH[]NNCHCH[

ln

RateRateln

2_

3_

6_

6_

133

233

1

2

= 1; Rate = k[CH3NNCH3]

(b) From Experiment 1: k = M 10 x 2.4

M/s 10 x 6.0 =

]NNCHCH[

Rate2_

6_

33

= 2.5 x 10-4/s

(c) Rate = k[CH3NNCH3] = (2.5 x 10-4/s)(0.020 M) = 5.0 x 10-6 M/s 12.48 (a) Rate = k[NH4

+]m[NO2-]n


301

m =

0.240.12

ln

10 x 7.210 x 3.6

ln

=

]NH[]NH[

ln

RateRateln

6_

6_

1+4

2+4

1

2

= 1; n =

0.100.15

ln

10 x 3.610 x 5.4

ln

=

]NO[]NO[

ln

RateRateln

6_

6_

2_2

3_2

2

3

= 1

Rate = k[NH4+][NO2

-]

(b) From Experiment 1: k = M) M)(0.10 (0.24

M/s 10 x 7.2 =

]NO][NH[

Rate 6_

_2

+4

= 3.0 x 10-4/(M ⋅ s)

(c) Rate = k[NH4+][NO2

-] = [3.0 x 10-4/(M ⋅ s)](0.39 M)(0.052 M) = 6.1 x 10-6 M/s 12.49 (a) Rate = k[NO]m[Cl2]

n

m =

0.130.26

ln

10 x 1.010 x 4.0

ln

=

][NO][NO

ln

RateRateln

2_

2_

1

2

1

2

= 2; n =

0.200.10

ln

10 x 1.010 x 5.0

ln

=

]Cl[]Cl[

ln

RateRateln

2_

3_

12

32

1

3

= 1

Rate = k[NO]2[Cl2]

(b) From Experiment 1: k = M) (0.20)M (0.13

M/s 10 x 1.0 =

]Cl[][NO

Rate2

2_

22

= 3.0/(M2 ⋅ s)

(c) Rate = k[NO]2[Cl2] = [3.0/(M2 ⋅ s)](0.12 M)2(0.12 M) = 5.2 x 10—3 M/s Integrated Rate Law; Half-Life

12.50 ln kt _ = ]HC[

]HC[

063

t63 , k = 6.7 x 10-4/s

(a) t = 30 min x min 1

s 60 = 1800 s

ln[C3H6] t = kt _ + ln[C3H6]0 = s) /s)(180010 x (6.7 _ 4_ + ln(0.0500) = - 4.202 [C3H6] t = e- 4.202 = 0.015 M

(b) t = k _

]HC[]HC[ln 063

t63

= /s)10 x (6.7 _

0.05000.0100

ln

4_

= 2402 s; t = 2402 s x s 60

min 1 = 40 min

(c) [C3H6]0 = 0.0500 M; If 25% of the C3H6 reacts then 75% remains. [C3H6] t = (0.75)(0.0500 M) = 0.0375 M.

t = k _

]HC[]HC[ln 063

t63

= /s)10 x (6.7 _

0.05000.0375

ln

4_

= 429 s; t = 429 s x s 60

min 1 = 7.2 min


302

12.51 ln kt _ = ]NCCH[

]NCCH[

03

t3 , k = 5.11 x 10-5/s

(a) t = 2.00 hr x min 1

s 60 x

hr 1

min 60 = 7200 s

ln[CH3NC]t = kt _ + ln[CH3NC]0 = s) /s)(720010 x (5.11 _ 5_ + ln(0.0340) = -3.749 [CH3NC]t = e-3.749 = 0.0235 M

(b) t = k _

]NCCH[]NCCH[

ln 03

t3

= /s)10 x 5.11 _

0.03400.0300

ln

5_

= 2449 s; t = 2449 s x s 60

min 1 = 40.8 min

(c) [CH3NC]0 = 0.0340 M; If 20% of the CH3NC reacts then 80% remains. [CH3NC]t = (0.80)(0.0340 M) = 0.0272 M.

t = k _

]NCCH[]NCCH[

ln 03

t3

= /s)10 x (5.11 _

0.03400.0272

ln

5_

= 4367 s; t = 4367 s x s 60

min 1 = 72.8 min

12.52 t1/2 = /s10 x 6.7

0.693 =

k

0.6934_

= 1034 s = 17 min

t = /s10 x 6.7 _

(0.0500).0500)(0.0625)(0

ln =

k _

]HC[]HC[ln

4_o63

t63

= 4140 s

t = 4140 s x s 60

min 1 = 69 min

t1/2 t1/2 t1/2 t1/2 This is also 4 half-lives. 100 → 50 → 25 → 12.5 → 6.25

12.53 t1/2 = /s10 x 5.11

0.693 =

k

0.6935_

= 13,562 s

t1/2 = 13,562 s x min 60

hr 1 x

s 60

min 1 = 3.77 hr

t = /s10 x 5.11 _

(0.0340)0340)(0.125)(0.

ln =

k _

]NCCH[]NCCH[

ln

5_o3

t3

= 40,694 s

t = 40,694 s x min 60

hr 1 x

s 60

min 1 = 11.3 hr

t1/2 t1/2 t1/2 This is also 3 half-lives. 100 → 50 → 25 → 12.5


303

12.54 t1/2 = 8.0 h

t1/2 t1/2 0.60 M → 0.30 M → 0.15 M requires 2 half-lives so it will take 16.0 h.

12.55 t1/2 = 3.33 h

t1/2 t1/2 t1/2 t1/2 0.800 M → 0.400 M → 0.200 M → 0.100 M → 0.0500 M requires 4 half-lives so it will take 13.3 h.

12.56 kt = ]HC[

1 _

]HC[

1

064t64

, k = 4.0 x 10-2/(M ⋅ s)

(a) t = 1.00 h x min 1

s 60 x

hr 1

min 60 = 3600 s

]HC[

1 +kt =

]HC[

1

064t64

= M 0.0200

1 + s) s))(3600 /(M10 x (4.0 2_ •

]HC[

1

t64

= 194/M and [C4H6] = 5.2 x 10-3 M

(b) t =

]HC[

1 _

]HC[

1

k

1

064t64

t =

• M) (0.0200

1 _

M) (0.0020

1

s) /(M10 x 4.0

12_

= 11,250 s

t = 11,250 s x min 60

hr 1 x

s 60

min 1 = 3.1 h

12.57 kt = ][HI

1 _

][HI

1

0t

, k = 9.7 x 10-6/(M ⋅ s)

(a) t = 6.00 day x min 1

s 60 x

hr 1

min 60 x

day 1

hr 24 = 518,400 s

][HI

1 +kt =

][HI

1

0t

= M 0.100

1 + s) 0s))(518,40 /(M10 x (9.7 6_ •

][HI

1

t

= 15.03/M and [HI] = 0.067 M


304

(b) t =

][HI

1 _

][HI

1

k

1

0t

t =

• M) (0.100

1 _

M) (0.020

1

s) /(M10 x 9.7

16_

= 4,123,711 s

t = 4,123,711 s x hr 24

day 1 x

min 60

hr 1 x

s 60

min 1 = 48 days

12.58 t1/2 = M) s)](0.0200/(Mcdot 10 x [4.0

1 =

]HCk[

12_

o64

= 1250 s = 21 min

t = t1/2 = M) s)](0.0100/(Mcdot 10 x [4.0

1 =

]HCk[

12_

o64

= 2500 s = 42 min

12.59 t1/2 = M) s)](0.100 /(M10 x [9.7

1 =

]k[HI

16_

o • = 1,030,928 s

1,030,928 s x hr 24

day 1 x

min 60

hr 1 x

s 60

min 1 = 12 days

t = t1/2 = M) s)](0.0250 /(M10 x [9.7

1 =

]k[HI

16_

o • = 4,123,711 s

4,123,711 s x hr 24

day 1 x

min 60

hr 1 x

s 60

min 1 = 48 days

12.60 time (min) [N2O] ln[N2O] 1/[N2O] 0 0.250 -1.386 4.00 60 0.218 -1.523 4.59 90 0.204 -1.590 4.90 120 0.190 -1.661 5.26 180 0.166 -1.796 6.02

A plot of ln [N2O] versus time is linear. The reaction is first-order in N2O. k = -slope = -(-2.28 x 10-3/min) = 2.28 x 10-3/min

k = 2.28 x 10-3/min x s 60

min 1 = 3.79 x 10-5/s

12.61 time (s) [NOBr] ln[NOBr] 1/[NOBr]

0 0.0400 -3.219 25.0 10 0.0303 -3.497 33.0


305

20 0.0244 -3.713 41.0 30 0.0204 -3.892 49.0 40 0.0175 -4.046 57.1

A plot of 1/[NOBr] versus time is linear. The reaction is second-order in NOBr. k = slope = 0.80/(M ⋅ s)

12.62 k = s 248

0.693 =

t

0.693

2/1

= 2.79 x 10—3/s

12.63 t1/2 = ]k[A

1

o

; t1/2 = 25 min x min 1

s 60 = 1500 s

k = M) s)(0.036 (1500

1 =

][A t

1

02/1

= 1.8 x 10-2 M-1 s-1

12.64 (a) The units for the rate constant, k, indicate the reaction is zeroth-order.

(b) For a zeroth-order reaction, [A]t - [A] o = -kt

t = 30 min x = min 1

s 60 1800 s

[A] t = -kt + [A]o = - (3.6 x 10-5 M/s)(1800 s) + 0.096 M = 0.031 M (c) Let [A]t = [A]o/2

= M/s 10 x 3.6 _

M 0.096 _ M 2/0.096 =

k _

][A _ 2/][A = t 5_

oo2/1 1333 s

= t 2/1 1333 s x = s 60

min 1 22 min


306

12.65 (a)

A plot of [AB] versus time is linear. The reaction is zeroth-order and k = - slope.

k = = min 0 _min 80.0

M 0.200 _ M 0.140 _ 7.50 x 10-4 M/min

k = 7.50 x 10-4 M/min x = s 60

min 1 1.25 x 10-5 M/s

(b) [A] t - [A]o = -kt [A] t = -kt + [A]o = -(7.50 x 10-4 M/min)(126 min) + 0.200 M = 0.105 M

(c) [A] t - [A]o = -kt

t = = k _

][A _ ][A ot = M/min 10 x 7.50_

M 0.200 _ M 0.1004_

133 min

Reaction Mechanisms 12.66 An elementary reaction is a description of an individual molecular event that involves the

breaking and/or making of chemical bonds. By contrast, the overall reaction describes only the stoichiometry of the overall process but provides no information about how the reaction occurs.

12.67 Molecularity is the number of reactant molecules or atoms for an elementary reaction.

Reaction order is the sum of the exponents of the concentration terms in the rate law. 12.68 There is no relationship between the coefficients in a balanced chemical equation for an

overall reaction and the exponents in the rate law unless the overall reaction occurs in a single elementary step, in which case the coefficients in the balanced equation are the exponents in the rate law.

12.69 The rate-determining step is the slowest step in a multistep reaction. The coefficients in

the balanced equation for the rate-determining step are the exponents in the rate law.


307

12.70 (a) H2(g) + ICl(g) → HI(g) + HCl(g)

HI(g) + ICl(g) → I2(g) + HCl(g) Overall reaction H2(g) + 2 ICl(g) → I2(g) + 2 HCl(g) (b) Because HI(g) is produced in the first step and consumed in the second step, it is a reaction intermediate. (c) In each reaction there are two reactant molecules, so each elementary reaction is bimolecular.

12.71 (a) NO(g) + Cl2(g) → NOCl2(g)

NOCl2(g) + NO(g) → 2 NOCl(g) Overall reaction 2 NO(g) + Cl2(g) → 2 NOCl(g) (b) Because NOCl2 is produced in the first step and consumed in the second step, NOCl2 is a reaction intermediate. (c) Each elementary step is bimolecular.

12.72 (a) bimolecular, Rate = k[O3][Cl] (b) unimolecular, Rate = k[NO2]

(c) bimolecular, Rate = k[ClO][O] (d) termolecular, Rate = k[Cl]2[N2] 12.73 (a) unimolecular, Rate = k[I2] (b) termolecular, Rate = k[NO]2[Br2]

(c) bimolecular, Rate = k[CH3Br][OH—] (d) unimolecular, Rate = k[N2O5] 12.74 (a) NO2Cl(g) → NO2(g) + Cl(g)

Cl(g) + NO2Cl(g) → NO2(g) + Cl2(g) Overall reaction 2 NO2Cl(g) → 2 NO2(g) + Cl2(g) (b) 1. unimolecular; 2. bimolecular (c) Rate = k[NO2Cl]

12.75 (a) Mo(CO)6 → Mo(CO)5 + CO

Mo(CO)5 + L → Mo(CO)5L Overall reaction Mo(CO)6 + L → Mo(CO)5L + CO (b) 1. unimolecular; 2. bimolecular (c) Rate = k[Mo(CO)6]

12.76 NO2(g) + F2(g) → NO2F(g) + F(g) (slow)

F(g) + NO2(g) → NO2F(g) (fast) 12.77 O3(g) + NO(g) → O2(g) + NO2(g) (slow)

NO2(g) + O(g) → O2(g) + NO(g) (fast)


308

The Arrhenius Equation 12.78 Very few collisions involve a collision energy greater than or equal to the activation

energy, and only a fraction of those have the proper orientation for reaction. 12.79 The two reactions have frequency factors that differ by a factor of 10. 12.80 Plot ln k versus 1/T to determine the activation energy, Ea

Slope = -1.25 x 104 K Ea = -R(slope) = -[8.314 x 10-3 kJ/(K ⋅ mol)](-1.25 x 104 K) = 104 kJ/mol 12.81 Plot ln k versus 1/T to determine the activation energy, Ea.

Slope = -1.359 x 104 K Ea = -R(slope) = -[8.314 x 10-3 kJ/(K ⋅ mol)](-1.359 x 104 K) = 113 kJ/mol

12.82 (a)

T

1 _

T

1

RE _

= k

kln 12

a

1

2

k1 = 1.3/(M ⋅ s), T1 = 700 K k2 = 23.0/(M ⋅ s), T2 = 800 K

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 7001

_ K 800

1mol)] kJ/(K 10 x )][8.3141.3(ln _ )23.0([ln

_3_

= 134 kJ/mol

(b) k1 = 1.3/(M ⋅ s), T1 = 700 K solve for k2, T2 = 750 K

ln k2 = kln + T

1 _

T

1

RE _

112

a


309

ln k2 = (1.3)ln + K 700

1 _

K 750

1

mol) kJ/(K 10 x 8.314

kJ/mol 133.8_3_

• = 1.795

k2 = e1.795 = 6.0/(M ⋅ s)

12.83

T

1 _

T

1

RE_

= k

kln 12

a

1

2

(a) Because the rate doubles, k2 = 2k1 k1 = 1.0 x 10-3/s, T1 = 25oC = 298 K k2 = 2.0 x 10-3/s, T2 = 35oC = 308 K

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

K 2981

_ K 308

1mol)] kJ/(Kcdot 10 x )][8.31410 x ln(1.0 _ )10 x [ln(2.0

_3_3_3_

= 53 kJ/mol

(b) Because the rate triples, k2 = 3k1 k1 = 1.0 x 10-3/s, T1 = 25oC = 298 K k2 = 3.0 x 10-3/s, T2 = 35oC = 308 K

Ea =

K 2981

_ K 308


_3_3_3_

= 84 kJ/mol

12.84

T

1 _

T

1

RE _

= k

kln 12

a

1

2

assume k1 = 1.0/(M ⋅ s) at T1 = 25oC = 298 K assume k2 = 15/(M ⋅ s) at T2 = 50oC = 323 K

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 2981

_ K 323

1mol)] kJ/(K 10 x )][8.3141.0(ln _ )15([ln

_3_

= 87 kJ/mo1

12.85

T

1 _

T

1

RE _

= k

kln 12

a

1

2

assume k1 = 1.0/(M ⋅ s) at T1 = 15oC = 288 K assume k2 = 6.37/(M ⋅ s) at T2 = 45oC = 318 K


310

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 2881

_ K 318

1mol)] kJ/(K 10 x )][8.3141.0(ln _ )6.37([ln

_3_

= 47.0 kJ/mo1

12.86

12.87 (a) (b) Catalysis 12.88 A catalyst does participate in the reaction, but it is not consumed because it reacts in one

step of the reaction and is regenerated in a subsequent step. 12.89 A catalyst doesn't appear in the chemical equation for a reaction because a catalyst reacts

in one step of the reaction but is regenerated in a subsequent step. 12.90 A catalyst increases the rate of a reaction by changing the reaction mechanism and

lowering the activation energy.


311

12.91 A homogeneous catalyst is one that exists in the same phase as the reactants.

Example: NO(g) acts as a homogeneous catalyst for the conversion of O2(g) to O3(g). A heterogeneous catalyst is one that exists in a different phase from the reactants. Example: solid Ni, Pd, or Pt for catalytic hydrogenation, C2H4(g) + H2(g) → C2H6(g).

12.92 (a) O3(g) + O(g) → 2 O2(g) (b) Cl acts as a catalyst.

(c) ClO is a reaction intermediate. (d) A catalyst reacts in one step and is regenerated in a subsequent step. A reaction intermediate is produced in one step and consumed in another.

12.93 (a) 2 SO2(g) + 2 NO2(g) → 2 SO3(g) + 2 NO(g) 2 NO(g) + O2(g) → 2 NO2(g)

Overall reaction 2 SO2(g) + O2(g) → 2 SO3(g) (b) NO2(g) acts as a catalyst because it is used in the first step and regenerated in the second. NO(g) is a reaction intermediate because it is produced in the first step and consumed in the second.

12.94 (a) NH2NO2(aq) + OH-(aq) → NHNO2

-(aq) + H2O(l) NHNO2

-(aq) → N2O(g) + OH-(aq) Overall reaction NH2NO2(aq) → N2O(g) + H2O(l) (b) OH- acts as a catalyst because it is used in the first step and regenerated in the second. NHNO2

- is a reaction intermediate because it is produced in the first step and consumed in the second. (c) The rate will decrease because added acid decreases the concentration of OH-, which appears in the rate law since it is a catalyst.

12.95 The reaction in Problem 12.77 involves a catalyst (NO) because NO is used in the first

step and is regenerated in the second step. The reaction also involves an intermediate (NO2) because NO2 is produced in the first step and is used up in the second step.

General Problems 12.96 2 AB2 → A2 + 2 B2

(a) Measure the change in the concentration of AB2 as a function of time. (b) and (c) If a plot of [AB2] versus time is linear, the reaction is zeroth-order and k = -slope. If a plot of ln [AB2] versus time is linear, the reaction is first-order and k = -slope. If a plot of 1/[AB2] versus time is linear, the reaction is second-order and k = slope.

12.97 A → B + C (a) Measure the change in the concentration of A as a function of time at several different temperatures. (b) Plot ln [A] versus time, for each temperature. Straight line graphs will result and k at each temperature equals -slope. Graph ln k versus 1/K, where K is the kelvin temperature. Determine the slope of the line. Ea = -R(slope) where R = 8.314 x 10-3


312

kJ/(K ⋅ mol). 12.98 (a) Rate = k[B2][C]

(b) B2 + C → CB + B (slow) CB + A → AB + C (fast) (c) C is a catalyst. C does not appear in the chemical equation because it is consumed in the first step and regenerated in the second step.

12.99 (a)

(b) Reaction 2 is the fastest (smallest Ea), and reaction 3 is the slowest (largest Ea). (c) Reaction 3 is the most endothermic (positive ∆E), and reaction 1 is the most exothermic (largest negative ∆E).

12.100 The first maximum represents the potential energy of the transition state for the first

step. The second maximum represents the potential energy of the transition state for the second step. The saddle point between the two maxima represents the potential energy of the intermediate products.

12.101 Because 0.060 M is half of 0.120 M, 5.2 h is the half-life.

For a first-order reaction, the half-life is independent of initial concentration. Because 0.015 M is half of 0.030 M, it will take one half-life, 5.2 h.

k = h 5.2

0.693 =

t

0.693

2/1

= 0.133/h

ln kt _ = ]ON[

]ON[

052

t52

t = k _

]ON[]ON[ln052

t52

= /h)(0.133 _

0.4800.015

ln

= 26 h (Note that t is five half-lives.)

12.102 (a) The reaction rate will increase with an increase in temperature at constant volume.

(b) The reaction rate will decrease with an increase in volume at constant temperature because reactant concentrations will decrease.


313

(c) The reaction rate will increase with the addition of a catalyst. (d) Addition of an inert gas at constant volume will not affect the reaction rate.

12.103 As the temperature of a gas is raised by 10oC, even though the collision frequency increases by only ~2%, the reaction rate increases by 100% or more because there is an 12.104 (a) Rate = k[C2H4Br2]

m[I -]n

m =

0.1270.343

ln

10 x 6.4510 x 1.74

ln

=

]BrHC[]BrHC[ln

RateRateln

5_

4_

1242

2242

1

2

= 1

n =

••

0.1020.125

ln

)(0.203)10 x (1.74)(0.343)10 x (1.26

ln

=

]I[]I[ln

]BrHC[ Rate

]BrHC[ Rateln 4_

4_

2_

3_

32422

22423

= 1

Rate = k[C2H4Br2][I

-]

(b) From Experiment 1:

k = M) M)(0.102 (0.127

M/s 10 x 6.45 =

]I][BrHC[

Rate 5_

_242

= 4.98 x 10-3/(M ⋅ s)

(c) Rate = k[C2H4Br2][I-] = [4.98 x 10-3(M ⋅ s)](0.150 M)(0.150 M) = 1.12 x 10-4 M/s

12.105 (a) From the data in the table for Experiment 1, we see that 0.20 mol of A reacts with

0.10 mol of B to produce 0.10 mol of D. The balanced equation for the reaction is: 2 A + B → D

(b) From the data in the table, initial Rates = t

A _

∆∆

have been calculated.

For example, from Experiment 1:

Initial rate = s 60

M) 5.00 _ M (4.80 _ =

t

A _

∆∆

= 3.33 x 10-3 M/s

Initial concentrations and initial rate data have been collected in the table below.

EXPT [A]o (M) [B] o (M) [C]o (M) Initial Rate (M/s) 1 5.00 2.00 1.00 3.33 x 10-3 2 10.00 2.00 1.00 6.66 x 10-3 3 5.00 4.00 1.00 3.33 x 10-3 4 5.00 2.00 2.00 6.66 x 10-3


314

Rate = k[A]m[B] n[C]p From Expts 1 and 2, [A] doubles and the initial rate doubles; therefore m = 1. From Expts 1 and 3, [B] doubles but the initial rate does not change; therefore n = 0. From Expts 1 and 4, [C] doubles and the initial rate doubles; therefore p = 1. The reaction is: first-order in A; zeroth-order in B; first-order in C; second-order overall. (c) Rate = k[A][C] (d) C is a catalyst. C appears in the rate law, but it is not consumed in the reaction. (e) A + C → AC (slow)

AC + B → AB + C (fast) A + AB → D (fast) (f) From data in Expt 1:

k = M) M)(1.00 (5.00

s 60M/ 0.10 =

[A][C]

t/D =

[A][C]

Rate ∆∆ = 3.4 x 10-4/(M ⋅ s)

12.106 For Ea = 50 kJ/mol

f =

• K) mol)](300 kJ/(K 10 x [8.314

kJ/mol 50_exp = e 3_

/RTE _ a = 2.0 x 10-9

For Ea = 100 kJ/mol

f =

• K) mol)](300 kJ/(K 10 x [8.314

kJ/mol 100_exp = e 3_

RT/E_ a = 3.9 x 10-18

12.107 ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

k2 = 2.5k1 k1 = 1.0, T1 = 20oC = 293 K k2 = 2.5, T2 = 30oC = 303 K

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 2931

_ K 303

1mol)] kJ/(K 10 x .314ln(1.0)][8 _ [ln(2.5)

_3_

= 68 kJ/mol

k1 = 1.0, T1 = 120oC = 393 K k2 = ?, T2 = 130oC = 403 K Solve for k2.

ln k2 =

T

1 _

T

1

RE_

12

a + ln k1


315

ln k2 =

• K 393

1 _

K 403

1

mol)] kJ/(K 10 x [8.314

kJ/mol 68 _3_

+ ln(1.0) = 0.516

k2 = e0.516 = 1.7; The rate increases by a factor of 1.7. 12.108 (a) 2 NO(g) + Br2(g) → 2 NOBr(g)

(b) Since NOBr2 is generated in the first step and consumed in the second step, NOBr2 is a reaction intermediate. (c) Rate = k[NO][Br2] (d) It can't be the first step. It must be the second step.

12.109 [A] = -kt + [A]o

[A] o/2 = -kt1/2 + [A]o [A] o/2 - [A]o = -kt1/2 -[A] o/2 = -kt1/2 [A] o/2 = kt1/2

For a zeroth-order reaction, t1/2 = k 2

][A o .

For a zeroth-order reaction, each half-life is half of the previous one. For a first-order reaction, each half-life is the same as the previous one. For a second-order reaction, each half-life is twice the previous one.

12.110 (a) 2 NO(g) _ N2O2(g) (fast) N2O2(g) + H2(g) → N2O(g) + H2O(g) (slow) N2O(g) + H2(g) → N2(g) + H2O(g) (fast)

Overall reaction 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) (b) N2O2 and N2O are reaction intermediates because they are produced in one step of the reaction and used up in a subsequent step. (c) Rate = k2[N2O2][H2] (d) Because the forward and reverse rates in step 1 are equal, k1[NO]2 = k-1[N2O2]. Solving for [N2O2] and substituting into the rate law for the second step gives

Rate = k2[N2O2][H2] = k

k k

1_

21 [NO]2[H2]

Because the rate law for the overall reaction is equal to the rate law for the rate-determining step, the rate law for the overall reaction is

Rate = k[NO]2[H2] where k = k

k k

1_

21

12.111 (a) I-(aq) + OCl-(aq) → Cl-(aq) + OI-(aq)


316

(b) From the data in the table, initial rates = t

]I[ __

∆∆

have been calculated.

For example, from Experiment 1:

Initial rate = s 10

M) 10 x 2.40 _ M 10 x (2.17 _ =

Deltat

]I[ _4_4__∆

= 2.30 x 10-6 M/s

Initial concentrations and initial rate data have been collected in the table below.

EXPT [I-]o (M) [OCl-]o (M) [OH-]o (M) Initial Rate (M/s) 1 2.40 x 10-4 1.60 x 10-4 1.00 2.30 x 10-6 2 1.20 x 10-4 1.60 x 10-4 1.00 1.20 x 10-6 3 2.40 x 10-4 4.00 x 10-5 1.00 6.00 x 10-7 4 1.20 x 10-4 1.60 x 10-4 2.00 6.00 x 10-7

Rate = k[I-]m[OCl-]n[OH-]p

From Expts 1 and 2, [I-] is cut in half and the initial rate is cut in half; therefore m = 1. From Expts 1 and 3, [OCl-] is reduced by a factor of four and the initial rate is reduced by a factor of four; therefore n = 1. From Expts 2 and 4, [OH-] is doubled and the initial rate is cut in half; therefore p = -1.

Rate = k]OH[

]OCl][I[_

__

From data in Expt 1:

k = M) 10 x M)(1.60 10 x (2.40

M) M/s)(1.00 10 x (2.30 =

]OCl][I[

]OH[ Rate4_4_

6_

__

_

= 60/s

(c) The reaction does not occur by a single-step mechanism because OH- appears in the rate law but not in the overall reaction.

(d) OCl-(aq) + H2O(l) _ HOCl(aq) + OH-(aq) (fast) HOCl(aq) + I-(aq) → HOI(aq) + Cl-(aq) (slow) HOI(aq) + OH-(aq) → H2O(l) + OI-(aq) (fast)

Overall reaction I-(aq) + OCl-(aq) → Cl-(aq) + OI-(aq)

Because the forward and reverse rates in step 1 are equal, k1[OCl-][H2O] = k-1[HOCl][OH-]. Solving for [HOCl] and substituting into the rate law for the second step gives

Rate = k2[HOCl][I -] = k

k k

1_

21

]OH[

]IO][H][OCl[_

_2

_

[H2O] is constant and can be combined into k.

Because the rate law for the overall reaction is equal to the rate law for the rate-determining step, the rate law for the overall reaction is


317

Rate = k]OH[

]I][OCl[_

__

where k = k

O]H[ k k

1_

221

12.112 (a) Ratef = kf[A] and Rater = kr[B]

(b)

(c) When Ratef = Rater, kf[A] = kr[B], and )10 x (1.0

)10 x (3.0 =

k

k = [A]

[B]3_

3_

r

f = 3

12.113 (a) 1 → 1/2 → 1/4 → 1/8

After three half-lives, 1/8 of the strontium-90 will remain.

(b) k = y 29

0.693 =

t

0.693

2/1

= 0.0239/y = 0.024/y

(c) t = /y0.0239 _

(1)(0.01)

ln =

k _

)90_(Sr

)90_(Srln

o

t

= 193 y

12.114 k = y 5730

0.693 =

t

0.693

2/1

= 1.21 x 10-4/y

t = /y10 x 1.21_

(15.3)(2.3)

ln =

k _

)C(

)C(ln

4_o

14t

14

= 1.6 x 104 y

12.115 Rate = k [N2O4]; k = ]ON[

Rate

42

At 25oC, k1 = M 0.10

M/s 10 x 5.0 3

= 5.0 x 104 s-1

At 40oC, k2 = M 0.15

M/s 10 x 2.3 4

= 1.5 x 105 s-1


318

25oC = 25 + 273 = 298 K and 40oC = 40 + 273 = 313 K

ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 2981

_ K 313

1mol)] kJ/(K 10 x )][8.31410 x ln(5.0 _ )10 x [ln(1.5

_3_45

= 56.8 kJ/mol

12.116 X → products is a first-order reaction

t = 60 min x = min 1

s 60 3600 s

ln = ][X

][X

o

t - kt; k = t_

][X][X

ln o

t

At 25oC, calculate k1: k1 = s 3600 _M 1.000M 0.600

ln

= 1.42 x 10-4 s-1

At 35oC, calculate k2: k2 = s 3600 _M 0.600M 0.200

ln

= 3.05 x 10-4 s-1

At an unknown temperature calculate k3. k3 = s 3600 _M 0.200M 0.010

ln

= 8.32 x 10-4 s-1

T1 = 25oC = 25 + 273 = 298 K T2 = 35oC = 35 + 273 = 308 K

Calculate Ea using k1 and k2.

ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

K 2981

_ K 308


_3_4_4_

= 58.3

kJ/mol

Use Ea, k1, and k3 to calculate T3.


319

T

1 +

RE _kkln

= T

1

1a

1

3

3

= K 298

1 +

mol) kJ/(K 10 x 8.314kJ/mol 58.3 _

10 x 1.4210 x 8.32

ln

3_

4_

4_

•

= 0.003104/K

T3 = = /K0.003104

1 322 K = 322 - 273 = 49oC

At 3:00 p.m. raise the temperature to 49oC to finish the reaction by 4:00 p.m. 12.117 N2O4(g) → 2 NO2(g)

before (mm Hg) 17.0 0 change (mm Hg) -x +2x after (mm Hg) 17.0 - x 2x

2x = 1.3 mm Hg x = 1.3 mm Hg/2 = 0.65 mm Hg Pt(N2O4) = 17.0 - x = 17.0 - 0.65 = 16.35 mm Hg

k = = t

0.693

2/1

= s 10 x 1.3

0.6935_

5.3 x 104 s-1

ln = P

P

o

t - kt; t = = k _PPln

o

t

= s 10 x 5.3 _

1716.35

ln

1_4 7.4 x 10-7 s

12.118 (a) When equal volumes of two solutions are mixed, both concemtrations are cut in half.

[H3O+]o = [OH-]o = 1.0 M

When 99.999% of the acid is neutralized, [H3O+] = [OH-] = 1.0 M - (1.0 M x 0.99999)

= 1.0 x 10-5 M Using the 2nd order integrated rate law:

kt = ]OH[

1 _

]OH[

1

o+

3t+

3

; t =

]OH[

1 _

]OH[

1

k

1

o+

3t+

3

t =

M) (1.0

1 _

M) 10 x (1.0

1

)s M 10 x (1.3

15 _1_1_11

= 7.7 x 10-7 s

(b) The rate of an acid-base neutralization reaction would be limited by the speed of mixing, which is much slower than the intrinsic rate of the reaction itself.

12.119 )]O([

1 +k t 8 =

)]O([

12

o22

2

)M (0.0100

1 + s) )(100.0s M 8(25 =

)]O([

12

1_2_2

2


320

= )]O([

12

2

30,000 M-2

[O2] = M 30,000

12_

= 0.005 77 M

2 NO(g) + O2(g) → 2 NO2(g)

before (M) 0.0200 0.0100 0 change (M) -2x -x +2x after (M) 0.0200 - 2x 0.0100 - x 2x

[O2] = 0.005 77 M = 0.0100 M - x x = 0.0100 M - 0.005 77 M = 0.004 23 M [NO] = 0.0200 M - 2x = 0.0200 M - 2(0.004 23 M) = 0.0115 M [O2] = 0.005 77 M [NO2] = 2x = 2(0.004 23 M) = 0.008 46 M

12.120 Looking at the two experiments at 600 K, when the NO2 concentration is doubled, the

rate increased by a factor of 4. Therefore, the reaction is 2nd order. Rate = k [NO2]

2 Calculate k1 at 600 K: k1 = Rate/[NO2]

2 = 5.4 x 10-7 M s-1/(0.0010 M)2 = 0.54 M-1 s-1 Calculate k2 at 700 K: k2 = Rate/[NO2]

2 = 5.2 x 10-6 M s-1/(0.0020 M)2 = 13 M-1 s-1 Calculate Ea using k1 and k2.

ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 6001

_ K 700

1mol)] kJ/(K 10 x 8.314ln(0.54)][ _ [ln(13)

_3_

= 111 kJ/mol

Calculate k3 at 650 K using Ea and k1.

Solve for k3.

ln k3 =

T

1 _

T

1

RE_

13

a + ln k1

ln k3 =

• K 600

1 _

K 650

1

mol)] kJ/(K 10 x [8.314

kJ/mol 111 _3_

+ ln(0.54) = 1.0955

k3 = e1.0955 = 3.0 M-1 s-1

k3 t = ]NO[

1 _

]NO[

1

o2t2

; t =

]NO[

1 _

]NO[

1

k

1

o2t23


321

t =

M) (0.0050

1 _

M) (0.0010

1

)s M (3.0

11_1_

= 2.7 x 102 s

12.121 Rate = k [A]x[B] y

Comparing Experiments 1 and 2, the concentration of B does not change, the concentration of A doubles, and the rate doubles. This means the reaction is first-order in A (x = 1). Comparing Experiments 1 and 3, the rate would drop to 0.9 x 10-5 M/s as a result of the concentration of A being cut in half. Then with the the concentration of B doubling, the rate increases by a factor of 4, to 3.6 x 10-5 M/s. This means the reaction is second- order in B (y = 2).

At 600 K, k1 = = ][A][B

Rate2

= )M M)(0.50 (0.50

M/s 10 x 4.32

5_

3.4 x 10-4 M-2 s-1

At 700 K, k2 = = ][A][B

Rate2 =

)M M)(0.10 (0.20

M/s 10 x 1.82

5_

9.0 x 10-3 M-2 s-1

ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

K 6001

_ K 700

1mol)] kJ/(Kcdot 10 x )][8.31410 x (3.4ln _ )10 x (9.0[ln

_3_4_3_

= 114 kJ/mol

12.122 A → C is a first-order reaction.

The reaction is complete at 200 s when the absorbance of C reaches 1.200. Because there is a one to one stoichiometry between A and C, the concentration of A must be proportional to 1.200 - absorbance of C. Any two data points can be used to find k. Let [A]o ∝ 1.200 and at 100 s, [A]t ∝ 1.200 - 1.188 = 0.012

ln = ][A

][A

o

t - kt; k = t_

][A][A

ln o

t

; k = s 100 _M 1.200M 0.012

ln

= 0.0461 s-1

= s 0.0461

0.693 =

k

0.693 = t 1_2/1 15 s

12.123 Rate = k [HI]x

][0.10k

][0.30k =

Rate

Ratex

x

1

2


322

x = = 0.477

0.949 =

(0.10)(0.30)

log

10 x 1.810 x 1.6

log =

(0.10)(0.30)

log

RateRate log

5_

4_

1

2

2

Rate = k [HI]2

At 700 K, k1 = = ][HI

Rate2

= )M (0.10

M/s 10 x 1.82

5_

1.8 x 10-3 M-1 s-1

At 800 K, k2 = = ][HI

Rate2 =

)M (0.20

M/s 10 x 3.92

3_

9.7 x 10-2 M-1 s-1

ln

T

1 _

T

1

RE _

= k

k

12

a

1

2

Ea =

T

1 _

T

1

](R)kln _ k[ln _

12

12

Ea =

•

K 7001

_ K 800

1mol)] kJ/(K 10 x )][8.31410 x (1.8ln _ )10 x (9.7[ln

_3_3_2_

= 186 kJ/mol

Calculate k4 at 650 K using Ea and k1. Solve for k4.

ln k4 =

T

1 _

T

1

RE_

14

a + ln k1

ln k4 =

• K 700

1 _

K 650

1

mol)] kJ/(K 10 x [8.314

kJ/mol 186 _3_

+ ln(1.8 x 10-3) = -8.788

k4 = e-8.788 = 1.5 x 10-4 M-1 s-1

[HI] = = s M 10 x 1.5

M/s 10 x 1.0 =

k

Rate1_1_4_

5_

4

0.26 M

12.124 For radioactive decay, ln = N

N

o

- kt

For 235U, = y 10 x 7.1

0.693 =

t

0.693 = k 8

2/11 9.76 x 10-10 y-1

For 238U, = y 10 x 4.51

0.693 =

t

0.693 = k 9

2/12 1.54 x 10-10 y-1

For 235U, ln = 1N

N

o

1 - k1t and ln 1N

N

o

1 + k1t = 0

For 238U, ln = 2N

N

o

2 - k2t and ln 2N

N

o

2 + k2t = 0


323

Set the two equations that are equal to zero equal to each other and solve for t.

ln 1N

N

o

1 + k1t = ln 2N

N

o

2 + k2t

ln 1N

N

o

1 - ln2N

N

o

2 = k2t - k1t = (k2 - k1)t

2NN

1NN

ln

o

2

o

1

= (k2 - k1)t, now No1 = No2, so N

Nln 2

1 = (k2 - k1)t

N

N

2

1 = 7.25 x 10-3, so ln(7.25 x 10-3) = (1.54 x 10-10 y-1 - 9.76 x 10-10 y-1)t

t = = y 10 x 8.22 _

4.93 _1_10_

6.0 x 109 y

The age of the elements is 6.0 x 109 y (6 billion years).

Multi-Concept Problems

12.125 (a) k = Ae RTE_ a

= (6.0 x 108/(M ⋅ s)) e K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 6.3 _

3_ • = 4.7 x 107/(M ⋅ s)

(b) N has 3 electron clouds, is sp2 hybridized, and the molecule is bent.

(c)

(d) The reaction has such a low activation energy because the F–F bond is very weak and the N–F bond is relatively strong.

12.126 2 HI(g) → H2(g) + I2(g)

(a) mass HI = 1.50 L x mL 1

g 0.0101 x

L 1

mL 1000 = 15.15 g HI

15.15 g HI x HI g 127.91

HI mol 1 = 0.118 mol HI

[HI] = L 1.50

mol 0.118 = 0.0787 mol/L

][HIk = t

[HI] _ 2

∆∆

= (0.031/(M ⋅ min))(0.0787 M)2 = 1.92 x 10-4 M/min

2 HI(g) → H2(g) + I2(g)

∆∆

∆∆

t

[HI] _

2

1 =

t

]I[ 2 = 2

M/min 10 x 1.92 4_

= 9.60 x 10-5 M/min

(9.60 x 10-5 M/min)(1.50 L)(6.022 x 1023 molecules/mol) = 8.7 x 1019 molecules/min


324

(b) Rate = k[HI]2

][HI

1 +kt =

][HI

1

ot

= (0.031/(M ⋅ min))

h 1

min 60.0h x 8.00 +

M 0.0787

1 =

27.59/M

[HI] t = /M27.59

1 = 0.0362 M

From stoichiometry, [H2] t = 1/2 ([HI]o - [HI] t) = 1/2 (0.0787 M - 0.0362 M) = 0.0212 M 410oC = 683 K PV = nRT

PH2 = RT

V

n

= K) (683mol K

atm L 06 0.082mol/L) (0.0212

••

= 1.2 atm

12.127 2 NO2(g) → 2 NO(g) + O2(g)

k = 4.7/(M ⋅ s) (a) The units for k indicate a second-order reaction. (b) 383oC = 656 K PV = nRT

[NO2]o = K) (656

mol K atm L

06 0.082

Hg mm 760atm 1.000

x Hg mm 746

= RT

P =

V

n

••

= 0.01823 mol/L

initial rate = k ]NO[ 2o2 = [4.7/(M ⋅ s)](0.01823 mol/L)2 = 1.56 x 10-3 mol/(L ⋅ s)

initial rate for O2 = 2

s) mol/(L 10 x 1.56 =

2NOfor rate initial 3_

2 • = 7.80 x 10-4 mol/(L ⋅

s) initial rate for O2 = [7.80 x 10-4 mol/(L ⋅ s)](32.00 g/mol) = 0.025 g/(L ⋅ s)

(c) ]NO[

1 +kt =

]NO[

1

02t2

= M 0.01823

1 + s) s)](60 /(M[4.7 •

]NO[

1

t2

= 336.9/M and [NO2] = 0.00297 M

2 NO2(g) → 2 NO(g) + O2(g)

before reaction (M) 0.01823 0 0 change (M) -2x +2x +x after 1.00 min (M) 0.01823 - 2x 2x x

after 1.00 min [NO2] = 0.00297 M = 0.01823 - 2x x = 0.00763 M = [O2] mass O2 = (0.00763 mol/L)(5.00 L)(32.00 g/mol) = 1.22 g O2


325

12.128 (a) N2O5, 108.01 amu

[N2O5]o = L 2.00

ON g 108.01ON mol 1

x ON g 2.7052

5252

= 0.0125 mol/L

ln [N2O5] t = -kt + ln [N2O5]o = -(1.7 x 10-3 s-1)

min 1

s 60.0min x 13.0 + ln (0.0125) = -5.71

[N2O5] t = e-5.71 = 3.31 x 10-3 mol/L After 13.0 min, mol N2O5 = (3.31 x 10-3 mol/L)(2.00 L) = 6.62 x 10-3 mol N2O5

N2O5(g) → 2 NO2(g) + 1/2 O2(g)

before reaction (mol) 0.0250 0 0 change (mol) -x +2x +1/2x after reaction (mol) 0.0250 - x 2x 1/2x

After 13.0 min, mol N2O5 = 6.62 x 10-3 = 0.0250 - x x = 0.0184 mol

After 13.0 min, ntotal = n + n + n ONOON 2252

= (6.62 x 10-3) + 2(0.0184) + 1/2(0.0184)

ntotal = 0.0526 mol 55oC = 328 K PV = nRT

Ptotal = V

nRT =

L 2.00

K) (328mol K atm L

06 0.082mol) (0.0526

••

= 0.71 atm

(b) N2O5(g) → 2 NO2(g) + 1/2 O2(g)

∆Horxn = 2 ∆Ho

f(NO2) - ∆Hof(N2O5)

∆Horxn = (2 mol)(33.2 kJ/mol) - (1 mol)(11 kJ/mol) = 55.4 kJ = 5.54 x 104 J

initial rate = k[N2O5]o = (1.7 x 10-3 s-1)(0.0125 mol/L) = 2.125 x 10-5 mol/(L ⋅ s) initial rate absorbing heat = [2.125 x 10-5 mol/(L ⋅ s)](2.00 L)( 5.54 x 104 J/mol) = 2.4 J/s (c)

ln [N2O5] t = -kt + ln [N2O5]o = -(1.7 x 10-3 s-1)

min 1

s 60.0min x 10.0 + ln (0.0125) = -5.40

[N2O5] t = e-5.40 = 4.52 x 10-3 mol/L After 10.0 min, mol N2O5 = (4.52 x 10-3 mol/L)(2.00 L) = 9.03 x 10-3 mol N2O5

N2O5(g) → 2 NO2(g) + 1/2 O2(g)

before reaction (mol) 0.0250 0 0 change (mol) -x +2x +1/2x after reaction (mol) 0.0250 - x 2x 1/2x

After 10.0 min, mol N2O5 = 9.03 x 10-3 = 0.0250 - x x = 0.0160 mol


326

heat absorbed = (0.0160 mol)(55.4 kJ/mol) = 0.89 kJ 12.129 2 N2O(g) → 2 N2(g) + O2(g)

PO2(in exit gas) = 1.0 mm Hg; Ptotal = 1.50 atm = 1140 mm Hg

From the reaction stoichiometry: PN2

(in exit gas) = 2 PO2 = 2.0 mm Hg

P ON2(in exit gas) = Ptotal - PN2

- PO2= 1140 - 2.0 - 1.0 = 1137 mm Hg

Assume P ON2(initial) = Ptotal = 1140 mm Hg (In assuming a constant total pressure in the

tube, we are neglecting the slight change in pressure due to the reaction.) Volume of tube = πr2l = π(1.25 cm)2(20 cm) = 98.2 cm3 = 0.0982 L

Time, t, gases are in the tube = min 1

s 60 x

L/min 0.75

L 0.0982 x

rate flow

tubeof volume= 7.86 s

At time t, Hg mm 1140

Hg mm 1137 =

(initial) P

gas)exit (in P = ]ON[

]ON[

ON

ON

02

t2

2

2 = 0.997 37

Because k = Ae RTE_ a

and A = 4.2 x 109 s-1, k has units of s-1. Therefore, this is a first-order

reaction and the appropriate integrated rate law is kt _ = ]ON[

]ON[ln 02

t2 .

k = s 7.86

37) (0.997ln _ =

t

]ON[]ON[ln _02

t2

= 3.35 x 10-4 s-1

From the Arrhenius equation, ln k = ln A - RTEa

T = 8.00)] (_ _ 16)mol))[(22. kJ/(K 10 x (8.314

kJ/mol 222 =

k]ln _A (R)[ln E

3_

a

• = 885 K

12.130 H2O2, 34.01 amu

mass H2O2 = (0.500 L)(1000 mL/1 L)(1.00 g/ 1 mL)(0.0300) = 15.0 g H2O2

mol H2O2 = 15.0 g H2O2 x OH g 34.01

OH mol 1

22

22 = 0.441 H2O2

[H2O2]o = L0.500

mol 0.441= 0.882 mol /L

k = t

0.693

2/1

= h 10.7

0.693= 6.48 x 10-2/h

ln [H2O2] t = -kt + ln [H2O2]o ln [H2O2] t = -(6.48 x 10-2/h)(4.02 h) + ln (0.882) ln [H2O2] t = -0.386; [H2O2] t = e -0.386 = 0.680 mol/L


327

mol H2O2 = (0.680 mol/L)(0.500 L) = 0.340 mol

2 H2O2(aq) → 2 H2O(l) + O2(g) before reaction (mol) 0.441 0 0 change (mol) - 2x +2x +x after reaction (mol) 0.441 - 2x 2x x

After 4.02 h, mol H2O2 = 0.340 mol = 0.441 - 2x; solve for x. 2x = 0.101 x = 0.0505 mol = mol O2

P = 738 mm Hg x Hg mm 760

atm 1.00 = 0.971 atm

PV = nRT

V = atm 0.971

K) (293mol K atm L

06 0.082mol) (0.0505 =

P

nRT

••

= 1.25 L

P∆V = (0.971 atm)(1.25 L) = 1.21 L ⋅ atm

w = -P∆V = -1.21 L⋅atm = (-1.21 L ⋅ atm)

• atm L

J 101 = -122 J

12.131 (a) CH3CHO(g) → CH4(g) + CO(g)

before (atm) 0.500 0 0 change (atm) -x +x +x after (atm) 0.500 - x x x

At 605 s, Ptotal = P CHOCH3

+ PCH4 + PCO = (0.500 atm - x) + x + x = 0.808 atm

x = 0.808 atm - 0.500 atm = 0.308 atm

The integrated rate law for a second-order reaction in terms of molar concentrations is

][A

1 +kt =

][A

1

ot

. The ideal gas law, PV = nRT, can be rearranged to show how P is

proportional the the molar concentration of a gas.

P = RT V

n (R and T are constant), so P ∝

V

n = molar concentration

Because of this relationship, the second-order integrated rate law can be rewritten in terms of partial pressures.

P

1 +kt =

P

1

ot

; kt = P

1 _

P

1

ot

; k = t

P

1 _

P

1

ot


328

P is the partial pressure of CH3CHO.

At t = 0, Po = 0.500 and at t = 605 s, Pt = 0.500 atm - 0.308 atm = 0.192 atm

k = = s 605

atm 0.5001

_ atm 0.192

1

5.30 x 10-3 atm-1 s-1

(b) Use the ideal gas law to convert atm-1 to M-1.

P = RT V

n;

V

n =

RT

P; M =

n

V = RT

P

1 1_

So, multiply k by RT to convert atm-1 s-1 to M-1 s-1. k = (5.30 x 10-3 atm-1 s-1)RT

k = (5.30 x 10-3 atm-1 s-1) K) (791mol K

atm L 06 0.082

••

= 0.344 s mol

L

• = 0.344 M-1

s-1

(c) CH3CHO(g) → CH4(g) + CO(g) ∆Ho

rxn = [∆Hof(CH4) + ∆Ho

f(CO)] - ∆Hof(CH3CHO)]

∆Horxn = [(1 mol)(-74.8 kJ/mol) + (1 mol)(-110.5 kJ/mol)] - (1 mol)(-166.2 kJ/mol)

∆Horxn = -19.1 kJ per mole of CH3CHO that decomposes

PV = nRT

mol CH3CHO reacted = K) (791

mol K atm L

06 0.082

L) atm)(1.00 (0.308 =

RT

PV

••

= 0.004 74 mol

q = (0.004 74 mol)(19.1 kJ/mol)(1000 J/kJ) = 90.6 J liberated after a reaction time of 605 s

331

13

Chemical Equilibrium

13.1 (a) Kc = ]O[]SO[

]SO[

22

2

23 (b) Kc =

]SO[

]O[]SO[2

3

22

2

13.2 (a) Kc = )10 x (3.5)10 x (3.0

)10 x (5.0 =

]O[]SO[

]SO[3_23_

22_

22

2

23 = 7.9 x 104

(b) Kc = )10 x (5.0

)10 x (3.5)10 x (3.0 =

]SO[

]O[]SO[22_

3_23_

23

22

2 = 1.3 x 10-5

13.3 (a) ]OHC[

] OHC][H[ = K363

_353

+

c

(b) = 0365)](0.100)(0. _ [0.100].0365)[(0.100)(0

= K2

c 1.38 x 10-4

13.4 From (1), = (1)(2)

(1)(2) =

]B[A][

[AB][B] = K

2c 1

For a mixture to be at equilibrium, ]B[A][

[AB][B]

2

must be equal to 1.

For (2), = (2)(1)

(2)(1) =

]B[A][

[AB][B]

2

1. This mixture is at equilibrium.

For (3), = (4)(2)

(1)(1) =

]B[A][

[AB][B]

2

0.125. This mixture is not at equilibrium.

For (4), 2

2

[AB][B] (2)( ) = =

[A][ ] (4)(1)B 1.0 This mixture is at equilibrium.

13.5 Kp = 0)(1.31)(10.

3)(6.12)(20. =

)P)(P(

)P)(P(

OHCO

HCO

2

22 = 9.48

13.6 2 NO(g) + O2 _ 2 NO2(g); ∆n = 2 - 3 = -1

Kp = Kc(RT)∆n, Kc = Kp(1/RT)∆n at 500 K: Kp = (6.9 x 105)[(0.082 06)(500)]-1 = 1.7 x 104

at 1000 K: Kc = (1.3 x 10-2)

06)(1000) (0.082

11_

= 1.1

13.7 (a) Kc = ]OH[

]H[3

2

32 , Kp =

)P(

)P(3

OH

3H

2

2 , ∆n = (3) - (3) = 0 and Kp = Kc

Chapter 13 - Chemical Equilibrium ______________________________________________________________________________

332

(b) Kc = [H2]2[O2], Kp = )P()P( O

2H 22

, ∆n = (3) - (0) = 3 and Kp = Kc(RT)3

(c) Kc = ]H][SiCl[

][HCl2

24

4

, Kp = )P)(P(

)P(2

HSiCl

4HCl

24

, ∆n = (4) - (3) = 1 and Kp = Kc(RT)

(d) Kc = ]Cl][Hg[

12_+2

2

13.8 Kc = 1.2 x 10-42. Because Kc is very small, the equilibrium mixture contains mostly H2

molecules. H is in periodic group 1A. A very small value of Kc is consistent with strong bonding between 2 H atoms, each with one valence electron.

13.9 The container volume of 5.0 L must be included to calculate molar concentrations.

(a) Qc = L) 5.0mol/ (1.0)L 5.0mol/ (0.060

)L 5.0mol/ (0.80 =

]O[][NO

]NO[2

2

t22t

2t2 = 890

Because Qc < Kc, the reaction is not at equilibrium. The reaction will proceed to the right to reach equilibrium.

(b) Qc = L) 5.0mol/ (0.20)L 5.0mol/ 10 x (5.0

)L 5.0mol/ (4.0 =

]O[][NO

]NO[23_

2

t22t

2t2 = 1.6 x 107

Because Qc > Kc, the reaction is not at equilibrium. The reaction will proceed to the left to reach equilibrium.

13.10 4 = ]B][A[

][AB = K

22

2

c ; For a mixture to be at equilibrium, ]B][A[

][AB

22

2

must be equal to 4.

For (1), (1)(1)

)(6 =

]B][A[][AB

= Q2

22

2

c = 36, Qc > Kc

For (2), (2)(2)

)(4 =

]B][A[][AB

= Q2

22

2

c = 4, Qc = Kc

For (3), (3)(3)

)(2 =

]B][A[][AB

= Q2

22

2

c = 0.44, Qc < Kc

(a) (2) (b) (1), reverse; (3), forward

13.11 Kc = ]H[][H

2

2

= 1.2 x 10-42

(a) [H] = )(0.10)10 x (1.2 = ]H[K 42_2c = 3.5 x 10-22 M

(b) H atoms = (3.5 x 10-22 mol/L)(1.0 L)(6.022 x 1023 atoms/mol) = 210 H atoms H2 molecules = (0.10 mol/L)(1.0 L)(6.022 x 1023 molecules/mol) = 6.0 x 1022 H2 molecules

13.12 CO(g) + H2O(g) _ CO2(g) + H2(g)

initial (M) 0.150 0.150 0 0 change (M) -x -x +x +x


333

equil (M) 0.150 - x 0.150 - x x x

Kc = 4.24 = ) x_ (0.150

x = O]H[CO][

]H][CO[2

2

2

22

Take the square root of both sides and solve for x.

) x_ (0.150x = 4.24

2

2

; 2.06 = x_ 0.150

x; x = 0.101

At equilibrium, [CO2] = [H2] = x = 0.101 M [CO] = [H2O] = 0.150 - x = 0.150 - 0.101 = 0.049 M

13.13 N2O4(g) _ 2 NO2(g)

initial (M) 0.0500 0 change (M) -x +2x equil (M) 0.0500 - x 2x

Kc = 4.64 x 10-3 = x)_ (0.0500

)x(2 =

]ON[]NO[ 2

42

22

4x2 + (4.64 x 10-3)x - (2.32 x 10-4) = 0 Use the quadratic formula to solve for x.

x = 8

0.06110 0.00464 _ =

2(4)

)10 x 2.32 4(4)(_ _ )10 x (4.64 )10 x (4.64 _ 4_23_3_ ±±

x = -0.008 22 and 0.007 06 Discard the negative solution (-0.008 22) because it leads to a negative concentration of NO2 and that is impossible. [N2O4] = 0.0500 - x = 0.0500 - 0.007 06 = 0.0429 M [NO2] = 2x = 2(0.007 06) = 0.0141 M

13.14 N2O4(g) _ 2 NO2(g)

Qc = mol/L) (0.0200

)mol/L (0.0300 =

]ON[

]NO[ 2

t42

2t2 = 0.0450; Qc > Kc

The reaction will approach equilibrium by going from right to left. N2O4(g) _ 2 NO2(g)

initial (M) 0.0200 0.0300 change (M) +x -2x equil (M) 0.0200 + x 0.0300 - 2x

Kc = 4.64 x 10-3 = x)+ (0.0200

)x2 _ (0.0300 =

]ON[]NO[ 2

42

22

4x2 - 0.1246x + (8.072 x 10-4) = 0 Use the quadratic formula to solve for x.

x = 8

0.05109 0.1246 =

2(4)

)10 x 4(4)(8.072 _ )0.1246 (_ 0.1246) (_ _ 4_2 ±±

x = 0.0220 and 0.009 19 Discard the larger solution (0.0220) because it leads to a negative concentration of NO2,


334

and that is impossible. [N2O4] = 0.0200 + x = 0.0200 + 0.009 19 = 0.0292 M [NO2] = 0.0300 - 2x = 0.0300 - 2(0.009 19) = 0.0116 M

13.15 Kp = = )P(

)P)(P(

OH

HCO

2

2 2.44, Qp = = (1.20)

0)(1.00)(1.4 1.17, Qp < Kp and the reaction goes to the

right to reach equilibrium. C(s) + H2O(g) _ CO(g) + H2(g)

initial (atm) 1.20 1.00 1.40 change (atm) -x +x +x equil (atm) 1.20 - x 1.00 + x 1.40 + x

Kp = = )P(

)P)(P(

OH

HCO

2

2 2.44 = x)_ (1.20

x)+ x)(1.40+ (1.00

x2 + 4.84x - 1.53 = 0 Use the quadratic formula to solve for x.

x = 2

5.44 4.84 _ =

2(1)

1.53) 4(1)(_ _ )(4.84 (4.84) _ 2 ±±

x = -5.14 and 0.300 Discard the negative solution (-5.14) because it leads to negative partial pressures and that is impossible. P OH2

= 1.20 - x = 1.20 - 0.300 = 0.90 atm

PCO = 1.00 + x = 1.00 + 0.300 = 1.30 atm

PH2 = 1.40 + x = 1.40 + 0.300 = 1.70 atm

13.16 (a) CO(reactant) added, H2 concentration increases.

(b) CO2 (product) added, H2 concentration decreases. (c) H2O (reactant) removed, H2 concentration decreases. (d) CO2 (product) removed, H2 concentration increases.

At equilibrium, Qc = Kc = O]H[CO][

]H][CO[

2

22 . If some CO2 is removed from the

equilibrium mixture, the numerator in Qc is decreased, which means that Qc < Kc and the reaction will shift to the right, increasing the H2 concentration.

13.17 (a) Because there are 2 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of reaction products remains the same. (b) Because there are 2 mol of gas on the left side and 1 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The number of moles of reaction products increases. (c) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to reactants). The number of moles of reaction product decreases.


335

13.18 13.19 Le Châtelier’s principle predicts that a stress of added heat will be relieved by net

reaction in the direction that absorbs the heat. Since the reaction is endothermic, the equilibrium will shift from left to right (Kc will increase) with an increase in temperature. Therefore, the equilibrium mixture will contain more of the offending NO, the higher the temperature.

13.20 The reaction is exothermic. As the temperature is increased the reaction shifts from right

to left. The amount of ethyl acetate decreases.

Kc = OH]HCH][COCH[

O]H][HCCOCH[

5223

25223

As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in Kc.

13.21 There are more AB(g) molecules at the higher temperature. The equilibrium shifted to the right at the higher temperature, which means the reaction is endothermic.

13.22 (a) A catalyst does not affect the equilibrium composition. The amount of CO remains

the same. (b) The reaction is exothermic. An increase in temperature shifts the reaction toward reactants. The amount of CO increases. (c) Because there are 3 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The amount of CO decreases. (d) An increase in pressure as a result of the addition of an inert gas (with no volume change) does not affect the equilibrium composition. The amount of CO remains the same. (e) Adding O2 increases the O2 concentration and shifts the reaction toward products. The amount of CO decreases.

13.23 (a) Because Kc is so large, kf is larger than kr.

(b) Kc = k

k

r

f ; kr = 10 x 3.4

s M 10 x 8.5 =

K

k34

1_1_6

c

f = 2.5 x 10-28 M-1 s-1

(c) Because the reaction is exothermic, Ea (forward) is less than Ea (reverse). Consequently, as the temperature decreases, kr decreases more than kf decreases, and


336

therefore Kc = k

k

r

f increases.

13.24 Hb + O2 _ Hb(O2)

If CO binds to Hb, Hb is removed from the reaction and the reaction will shift to the left resulting in O2 being released from Hb(O2). This will decrease the effectiveness of Hb for carrying O2.

13.25 The equilibrium shifts to the left because at the higher altitude the concentration of O2 is

decreased. 13.26 There are 26 π electrons. 13.27 The partial pressure of O2 in the atmosphere is 0.2095 atm.

PV = nRT

n = K) (298

mol K atm L

06 0.082

L) atm)(0.500 (0.2095 =

RT

PV

••

= 4.28 x 10-3 mol O2

4.28 x 10-3 mol O2 x O mol 1

molecules O 10 x 6.022

2

223

= 2.58 x 1021 O2 molecules

Understanding Key Concepts 13.28 (a) (1) and (3) because the number of A and B's are the same in the third and fourth

box.

(b) Kc = 4

6 =

[A]

[B] = 1.5

(c) Because the same number of molecules appear on both sides of the equation, the volume terms in Kc all cancel. Therefore, we can calculate Kc without including the volume.

13.29 (a) A2 + C2 _ 2 AC (most product molecules)

(b) A2 + B2 _ 2 AB (fewest product molecules)

13.30 (a) Only reaction (3), (2)(2)

(2)(4) =

][B]A[

[A][AB] = K

2c = 2, is at equilibrium.

(b) (1)(1)

(3)(5) =

][B]A[

[A][AB] = Q

2c = 15 for reaction (1). Because Qc > Kc, the reaction will

go in the reverse direction to reach equilibrium.

(3)(3)

(1)(3) =

][B]A[

[A][AB] = Q

2c = 1/3 for reaction (2). Because Qc < Kc, the reaction will go

in the forward direction to reach equilibrium. 13.31 (a) A2 + 2 B _ 2 AB


337

(b) The number of AB molecules will increase, because as the volume is decreased at constant temperature, the pressure will increase and the reaction will shift to the side of fewer molecules to reduce the pressure.

13.32 When the stopcock is opened, the reaction will go in the reverse direction because there

will be initially an excess of AB molecules. 13.33 As the temperature is raised, the reaction proceeds in the reverse direction. This is

consistent with an exothermic reaction where "heat" can be considered as a product. 13.34 (a) AB → A + B

(b) The reaction is endothermic because a stress of added heat (higher temperature)

shifts the AB _ A + B equilibrium to the right. (c) If the volume is increased, the pressure is decreased. The stress of decreased pressure will be relieved by a shift in the equilibrium from left to right, thus increasing the number of A atoms.

13.35 Heat + BaCO3(s) _ BaO(s) + CO2(g)

(a) (b)

13.36 (a) (b) (c)

13.37 This equilibrium mixture has a Kc ∝ )(3

(2)(2)2

and is less than 1. This means that kf < kr.

Additional Problems Equilibrium Expressions and Equilibrium Constants

13.38 (a) O]H][CH[]H[CO][

= K24

32

c (b) ]Cl[]F[

]ClF[ = K

23

2

23

c (c) Kc = ]F][H[

][HF

22

2

13.39 (a) ]O[]HC[

]CHOCH[ = K

22

42

23

c (b) ][NO

]O][N[ = K 222

c (c) ]O[]NH[

]OH[][NO = K 5

24

3

62

4

c


338

13.40 (a) Kp = )P)(P(

)P)(P(

OHCH

3HCO

24

2 , ∆n = 2 and Kp = Kc(RT)2

(b) Kp = )P()P(

)P(

Cl3

F

2ClF

22

3 , ∆n = -2 and Kp = Kc(RT)-2

(c) Kp = )P)(P(

)P(

FH

2HF

22

, ∆n = 0 and Kp = Kc

13.41 (a) Kp = )P()P(

)P(

O2

HC

2CHOCH

242

3 , ∆n = -1 and Kp = Kc(RT)-1

(b) Kp = )P(

)P)(P(2

NO

ON 22 , ∆n = 0 and Kp = Kc

(c) Kp = )P()P(

)P()P(5

O4

NH

6OH

4NO

23

2 , ∆n = 1 and Kp = Kc(RT)

13.42 ]OHHC[

O]H][HOCHC[ = K 252

25252c

13.43 Kc = O]HO][HC[

OH]CHHOCH[

242

22

13.44 [Citrate]

e][Isocitrat = K c

13.45 acid] oaceticacid][oxal [acetic

acid] [citric = K c

13.46 The two reactions are the reverse of each other.

Kc(reverse) = 10 x 7.5

1 =

(forward)K

19_

c

= 1.3 x 108

13.47 The two reactions are the reverse of each other.

Kp(reverse) = 50.2

1 =

(forward)K

1

p

= 1.99 x 10-2

13.48 Kc = )10 x (8.3

)10 x )(3.210 x (1.5 =

]PCl[

]Cl][PCl[3_

2_2_

5

23 = 0.058

13.49 Kp = (0.608))(0.240

)(1.35 =

)P()P(

)P(2

2

Cl2

NO

2ClNO

2

= 52.0


339


Initial [HI] = 9.30 x 10-3 mol/2.00 L = 4.65 x 10-3 M = 0.004 65 M H2(g) + I2(g) _ 2 HI(g)

initial (M) 0 0 0.004 65 change (M) +x +x -2x equil (M) x x 0.004 65 - 2x x = [H2] = [I2] = 6.29 x 10-4 M = 0.000 629 M [HI] = 0.004 65 - 2x = 0.004 65 - 2(0.000 629) = 0.003 39 M

Kc = )629 (0.000

)39 (0.003 =

]I][H[][HI

2

2

22

2

= 29.0

13.51 (a) Kc = H]COCH[

] COCH][H[

23

_23

+

(b) CH3CO2H(aq) _ H+(aq) + CH3CO2-(aq)

initial (M) 1.0 0 0 change (M) -0.0042 +0.0042 +0.0042 equil (M) 1.0 - 0.0042 0.0042 0.0042

Kc = 0.0042) _ (1.0

.0042)(0.0042)(0 = 1.8 x 10-5

13.52 (a) OH]HCH][COCH[

O]H][HCCOCH[ = K

5223

25223c

(b) CH3CO2H(soln) + C2H5OH(soln) _ CH3CO2C2H5(soln) + H2O(soln) initial (mol) 1.00 1.00 0 0 change (mol) -x -x +x +x equil (mol) 1.00 - x 1.00 - x x x

x = 0.65 mol; 1.00 - x = 0.35 mol; Kc = )(0.35

)(0.652

2

= 3.4

Because there are the same number of molecules on both sides of the equation, the volume terms in Kc cancel. Therefore, we can calculate Kc without including the volume.

13.53 CH3CO2C2H5(soln) + H2O(soln) _ CH3CO2H(soln) + C2H5OH(soln)

Kc(hydrolysis) = 3.4

1 =

(forward)K

1

c

= 0.29

13.54 ∆n = 1 and Kp = Kc(RT) = (0.575)(0.082 06)(500) = 23.6 13.55 2 SO2(g) + O2(g) _ 2 SO3(g); ∆n = 2 - (2 + 1) = -1 and Kp = 3.30

Kc = Kp

∆

06)(1000) (0.082

1(3.30) =

RT

11_n

= 271


340

13.56 Kp = P OH2

= 0.0313 atm; ∆n = 1

Kc = Kp

06)(298) (0.082

1(0.0313) =

RT

1 = 1.28 x 10-3

13.57 Hg mm 760

atm 1 x Hg mm 0.10 = P HC 810

= 1.3 x 10-4 atm

Kp = P HC 810 = 1.3 x 10-4; ∆n = 1 - 0 = 1, T = 27oC = 300 K

Kc = Kp

∆

06)(300) (0.082

1)10 x (1.3 =

RT

1 4_

n

= 5.3 x 10-6

13.58 (a) ][CO

]CO[ = K 3

32

c , )P(

)P( = K 3CO

3CO

p2 (b)

]O[

1 = K 3

2

c , )P(

1 = K 3

O

p

2

(c) Kc = [SO3], P = K SOp 3 (d) Kc = [Ba2+][SO4

2-]

13.59 (a) Kc = ]H[

]OH[3

2

32 , Kp =

)P(

)P(3

H

3OH

2

2 (b) Kc = ]Cl][Ag[

1_+

(c) Kc = ]OH[

][HCl3

2

6

, Kp = )P(

)P(3

OH

6HCl

2

(d) Kc = [CO2], Kp = PCO2

Using the Equilibrium Constant 13.60 (a) Because Kc is very large, the equilibrium mixture contains mostly product.

(b) Because Kc is very small, the equilibrium mixture contains mostly reactants. 13.61 (a) proceeds hardly at all toward completion

(b) goes almost all the way to completion 13.62 (a) Because Kc is very small, the equilibrium mixture contains mostly reactant.

(b) Because Kc is very large, the equilibrium mixture contains mostly product. (c) Because Kc = 1.8, the equilibrium mixture contains an appreciable concentration of both reactants and products.

13.63 (a) Because Kc is very large, the equilibrium mixture contains mostly product.

(b) Because Kc = 7.5 x 10-3, the equilibrium mixture contains an appreciable concentration of both reactants and products. (c) Because Kc is very small, the equilibrium mixture contains mostly reactant.

13.64 Kc = 1.2 x 1082 is very large. When equilibrium is reached, very little if any ethanol will

remain because the reaction goes to completion. 13.65 Because Kc is very small, pure air will contain very little O3 (ozone) at equilibrium.


341

3 O2(g) _ 2 O3(g); Kc = ]O[

]O[3

2

23 = 1.7 x 10-56; [O2] = 8 x 10-3 M

[O3] = 10 x (1.7)10 x (8 = K x ]O[ 56)_33_c

32 = 9 x 10-32 M


Qc = )L 10.0mol/ L)(4.0 10.0mol/ (2.0

)L 10.0mol/ L)(3.0 10.0mol/ (3.0 =

]SH[]CH[

]H[]CS[2

4

2t2t4

4t2t2 = 7.6 x 10-2; Kc = 2.5 x 10-3

The reaction is not at equilibrium because Qc > Kc. The reaction will proceed from right to left to reach equilibrium.

13.67 Qc = 050)(0.035)(0.

)0(0.15)(0.2 =

]CH[]OH[

]H[][CO 3

t4t2

3t2t = 0.69; Kc = 4.7

The reaction is not at equilibrium because Qc < Kc. The reaction will proceed from left to right to reach equilibrium.

13.68 Kc = ]H][N[

]NH[3

22

23 = 0.29; At equilibrium, [N2] = 0.036 M and [H2] = 0.15 M

[NH3] = (0.29))15(0.036)(0. = K x ]H[ x ]N[3

c3

22 = 5.9 x 10-3 M

13.69 Kc = 2.7 x 102 = ]O[]SO[

]SO[

22

2

23 ; Because [SO3] = [SO2], then 2.7 x 102 =

]O[

1

2

[O2] = 3.7 x 10-3 M 13.70 N2(g) + O2(g) _ 2 NO(g)

initial (M) 1.40 1.40 0 change (M) -x -x +2x equil (M) 1.40 - x 1.40 - x 2x

Kc = 1.7 x 10-3 = ) x_ (1.40

)x(2 =

]O][N[][NO

2

2

22

2

Take the square root of both sides and solve for x.

) x_ (1.40

)x(2 = 10 x 1.7

2

23_ ; 4.1 x 10-2 =

x_ 1.40

x2; x = 2.8 x 10-2

At equilibrium, [NO] = 2x = 2(2.8 x 10-2) = 0.056 M [N2] = [O2] = 1.40 - x = 1.40 - (2.8 x 10-2) = 1.37 M

13.71 N2(g) + O2(g) _ 2 NO(g)

initial (M) 2.24 0.56 0 change (M) -x -x +2x equil (M) 2.24 - x 0.56 - x 2x


342

Kc = ]O][N[

][NO

22

2

= 1.7 x 10-3 = x)_ x)(0.56_ (2.24

)x(2 2

4x2 + (4.8 x 10-3)x - (2.1 x 10-3) = 0 Use the quadratic formula to solve for x.

x = 8

0.1834 0.0048_ =

2(4)

)10 x 2.1 4(4)(_ _ )10 x (4.8 )10 x (4.8 _ 3_23_3_ ±±

x = -0.0235 and 0.0223 Discard the negative solution (-0.0235) because it gives a negative NO concentration and that is impossible. [N2] = 2.24 - x = 2.24 - 0.0223 = 2.22 M [O2] = 0.56 - x = 0.56 - 0.0223 = 0.54 M; [NO] = 2x = 2(0.0223) = 0.045 M

13.72 PCl5(g) _ PCl3(g) + Cl2(g)

initial (M) 0.160 0 0 change (M) -x +x +x equil (M) 0.160 - x x x

Kc = x_ 0.160

x = 10 x 5.8 = ]PCl[

]Cl][PCl[ 22_

5

23

x2 + (5.8 x 10-2)x - 0.00928 = 0 Use the quadratic formula to solve for x.

2

0.20 )10 x 5.8(_ =

2(1)

0.00928)4(1)(_ _ )10 x (5.8 )10 x 5.8(_ =x

2_22_2_ ±±

x = 0.071 and -0.129 Discard the negative solution (-0.129) because it gives negative concentrations of PCl3 and Cl2 and that is impossible. [PCl3] = [Cl2] = x = 0.071 M; [PCl5] = 0.160 - x = 0.160 - 0.071 = 0.089 M

13.73 Qc = (0.200)

040)(0.100)(0. = 0.020, Qc < Kc therefore the reaction proceeds from reactants to

products to reach equilibrium. PCl5(g) _ PCl3(g) + Cl2(g)

initial (M) 0.200 0.100 0.040 change (M) -x +x +x equil (M) 0.200 - x 0.100 + x 0.040 + x

Kc = x_ 0.200

x)+ (0.040 x)+ (0.100 = 10 x 5.8 =

]PCl[

]Cl][PCl[ 2_

5

23

x2 + 0.198x - (7.60 x 10-3) = 0 Use the quadratic formula to solve for x.

2

0.264 0.198)(_ =

2(1)

)10 x 7.604(1)(_ _ )(0.198 0.198)(_ =x

3_2 ±±

x = 0.033 and -0.231


343

Discard the negative solution (-0.231) because it gives negative concentrations of PCl3 and Cl2 and that is impossible. [PCl3] = 0.100 + x = 0.100 + 0.033 = 0.133 M [Cl2] = 0.040 + x = 0.040 + 0.033 = 0.073 M [PCl5] = 0.200 - x = 0.200 - 0.033 = 0.167 M

13.74 (a) Kc = (4.0)(6.0)

(x)(12.0) = 3.4 =

OH]HCH][COCH[

O]H][HCCOCH[

5223

25223 ; x = 6.8 moles CH3CO2C2H5

Note that the volume cancels because the same number of molecules appear on both sides of the chemical equation. (b) CH3CO2H(soln) + C2H5OH(soln) _ CH3CO2C2H5(soln) + H2O(soln) initial (mol) 1.00 10.00 0 0 change (mol) -x -x +x +x equil (mol) 1.00 - x 10.00 - x x x

Kc = 3.4 = x)_ x)(10.00_ (1.00

x2

2.4x2 - 37.4x + 34 = 0 Use the quadratic formula to solve for x.

x = 4.8

32.75 37.4 =

2(2.4)

4(2.4)(34) _ )37.4 (_ 37.4) (_ _ 2 ±±

x = 0.969 and 14.6 Discard the larger solution (14.6) because it leads to negative concentrations and that is impossible. mol CH3CO2H = 1.00 - x = 1.00 - 0.969 = 0.03 mol mol C2H5OH = 10.00 - x = 10.00 - 0.969 = 9.03 mol mol CH3CO2C2H5 = mol H2O = x = 0.97 mol

13.75 When equal volumes of two solutions are mixed together, their concentrations are cut in half.

CH3Cl(aq) + OH-(aq) _ CH3OH(aq) + Cl-(aq) initial (M) 0.05 0.1 0 0 assume complete reaction (M) 0 0.05 0.05 0.05 assume small back reaction (M) +x +x -x -x equil (M) x 0.05 + x 0.05 - x 0.05 - x

Kc = x)+ x(0.05

) x_ (0.05 = 10 =

]OHCl][CH[

]ClOH][CH[ 216

_3

_3 ; Because Kc is very large, x << 0.05.

x(0.05))(0.05

102

16 ≈ ; x = 5 x 10-18

[CH3Cl] = x = 5 x 10-18 M; [OH-] = [CH3OH] = [Cl-] ≈ 0.05 M

13.76 ClF3(g) _ ClF(g) + F2(g) initial (atm) 1.47 0 0 change (atm) -x +x +x


344

equil (atm) 1.47 - x x x

Kp = )P(

)P)(P(

ClF

FClF

3

2 = 0.140 = x_ 1.47

(x)(x); solve for x.


x = 2(1)

0.2058)(4)(1)(_ _ )(0.140 (0.140) _ 2±

x = 2

0.918 0.140 _ ±

x = 0.389 and -0.529 Discard the negative solution (-0.529) because it gives negative partial pressures and that is impossible.

P = P FClF 2 = x = 0.389 atm

PClF3= 1.47 - x = 1.47 - 0.389 = 1.08 atm

13.77 Fe2O3(s) + 3 CO(g) _ 2 Fe(s) + 3 CO2(g) initial (atm) 0.978 0 change (atm) -3x +3x equil (atm) 0.978 - 3x 3x

Kp = )P(

)P(3

CO

3CO2 = 19.9 =

)x3 _ (0.978

)x(33

3

; take the cube root of both sides and solve for x.

3 19.9 = 2.71 = x)3 _ (0.978

x3

2.65 - 8.13x = 3x 2.65 = 11.13x x = 2.65/11.13 = 0.238 atm PCO = 0.978 - 3x = 0.978 - 3(0.238) = 0.264 atm

PCO2= 3x = 3(0.238) = 0.714 atm

Le Châtelier's Principle 13.78 (a) Cl- (reactant) added, AgCl(s) increases

(b) Ag+ (reactant) added, AgCl(s) increases (c) Ag+ (reactant) removed, AgCl(s) decreases (d) Cl- (reactant) removed, AgCl(s) decreases

Disturbing the equilibrium by decreasing [Cl-] increases Qc

]Cl[]Ag[

l = Q

t_

t+c to a

value greater than Kc. To reach a new state of equilibrium, Qc must decrease, which means that the denominator must increase; that is, the reaction must go from right to left, thus decreasing the amount of solid AgCl.


345

13.79 (a) ClNO (product) added, NO2 concentration decreases (b) NO (reactant) added, NO2 concentration increases (c) NO (reactant) removed, NO2 concentration decreases (d) ClNO2 (reactant) added, NO2 concentration increases

Adding ClNO2 decreases the value of Qc

][NO]ClNO[

]NO[ClNO][ = Q

2

2c . To reach a new state of

equilibrium, the reaction must go from left to right, thus increasing the concentration of NO2.

13.80 (a) Because there are 2 mol of gas on the left side and 3 mol of gas on the right side of

the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to reactants). The number of moles of reaction products decreases. (b) Because there are 2 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of reaction product remains the same. (c) Because there are 2 mol of gas on the left side and 1 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The number of moles of reaction products increases.

13.81 As the volume increases, the pressure decreases at constant temperature.

(a) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to products). (b) Because there are 3 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to reactants). (c) Because there are 3 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by an increase in volume (decrease in pressure). There is no net reaction in either direction.

13.82 CO(g) + H2O(g) _ CO2(g) + H2(g) ∆Ho = - 41.2 kJ

The reaction is exothermic. [H2] decreases when the temperature is increased. As the temperature is decreased, the reaction shifts to the right. [CO2] and [H2] increase, [CO] and [H2O] decrease, and Kc increases.

13.83 Because ∆Ho is positive, the reaction is endothermic.

heat + 3 O2(g) _ 2 O3(g)


346

Kc = ]O[

]O[3

2

23

As the temperature increases, heat is added to the reaction, causing a shift to the right. The [O3] increases, and the [O2] decreases. This results in an increase in Kc.

13.84 (a) HCl is a source of Cl- (product), the reaction shifts left, the equilibrium [CoCl4

2-] increases. (b) Co(NO3)2 is a source of Co(H2O)6

2+ (product), the reaction shifts left, the equilibrium [CoCl4

2-] increases. (c) All concentrations will initially decrease and the reaction will shift to the right, the equilibrium [CoCl4

2-] decreases. (d) For an exothermic reaction, the reaction shifts to the left when the temperature is increased, the equilibrium [CoCl4

2-] increases. 13.85 (a) Fe(NO3)3 is a source of Fe3+. Fe3+ (reactant) added; the FeCl2+ concentration increases.

(b) Cl- (reactant) removed; the FeCl2+ concentration decreases. (c) An endothermic reaction shifts to the right as the temperature increases; the FeCl2+ concentration increases. (d) A catalyst does not affect the composition of the equilibrium mixture; no change in FeCl2+ concentration.

13.86 (a) The reaction is exothermic. The amount of CH3OH (product) decreases as the

temperature increases.

(b) When the volume decreases, the reaction shifts to the side with fewer gas molecules. The amount of CH3OH increases. (c) Addition of an inert gas (He) does not affect the equilibrium composition. There is no change. (d) Addition of CO (reactant) shifts the reaction toward product. The amount of CH3OH increases. (e) Addition or removal of a catalyst does not affect the equilibrium composition. There is no change.

13.87 (a) An endothermic reaction shifts to the right as the temperature increases. The amount

of acetone increases. (b) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to products). The amount of acetone increases. (c) The addition of Ar (an inert gas) with no volume change does not affect the composition of the equilibrium mixture. The amount of acetone does not change. (d) H2 (product) added; the amount of acetone decreases. (e) A catalyst does not affect the composition of the equilibrium mixture. The amount of acetone does not change.


347

Chemical Equilibrium and Chemical Kinetics 13.88 A + B _ C

ratef = kf[A][B] and rater = kr[C]; at equilibrium, ratef = rater

kf[A][B] = k r[C]; K = [A][B]

[C] =

k

kc

r

f

13.89 An equilibrium mixture that contains large amounts of reactants and small amounts of

products has a small Kc. A small Kc has kf < kr (c).

13.90 Kc = 10 x 6.2

0.13 =

k

k4_

r

f = 210

13.91 Kc = k

k

r

f ; kr = 10

sM 10 x 6 =

K

k16

1_1_6_

c

f = 6 x 10-22 M-1s-1

13.92 kr increases more than kf, this means that Ea (reverse) is greater than Ea (forward). The

reaction is exothermic when Ea (reverse) > Ea (forward). 13.93 kf increases more than kr, this means that Ea (forward) is greater than Ea (reverse). The

reaction is endothermic when Ea (forward) > Ea (reverse). General Problems

13.94 (a) [N2O4] = L 4.00

mol 0.500 = 0.125 M

N2O4(g) _ 2 NO2(g) initial (M) 0.125 0 change (M) -(0.793)(0.125) +(2)(0.793)(0.125) equil (M) 0.125 - (0.793)(0.125) (2)(0.793)(0.125) At equilibrium, [N2O4] = 0.125 - (0.793)(0.125) = 0.0259 M [NO2] = (2)(0.793)(0.125) = 0.198 M

Kc = (0.0259)

)(0.198 =

]ON[]NO[ 2

42

22 = 1.51

∆n = 2 - 1 = 1 and Kp = Kc(RT)∆n; Kp = Kc(RT) = (1.51)(0.082 06)(400) = 49.6

(b) 13.95 Kc is very large. The reaction goes essentially to completion.

Kc = ]CO[

1

2

; [CO2] = 10 x 1.6

1 =

K

124

c

= 6.2 x 10-25 M


348

13.96 Kc = ]H][N[

]NH[3

22

23 = 0.291

At equilibrium, [N2] = 1.0 x 10-3 M and [H2] = 2.0 x 10-3 M

[NH3] = (0.291))10 x )(2.010 x (1.0 = K x ]H[ x ]N[33_3_

c3

22 = 1.5 x 10-6 M

13.97 (a) Kp = P

)P(

F

2F

2

= 7.83 at 1500 K

PF = 00)(7.83)(0.2 = P K Fp 2 = 1.25 atm

(b) F2(g) _ 2 F(g) initial (atm) x 0 change (atm) -y +2y equil (atm) x - y 2y 2y = 1.25 so y = 0.625 x - y = 0.200; x = 0.200 + y = 0.200 + 0.625 = 0.825

f = 0.825

0.625 = 0.758

(c) The shorter bond in F2 is expected to be stronger. However, because of the small size of F, repulsion between the lone pairs of the two halogen atoms are much greater in F2 than in Cl2.

13.98 2 HI(g) _ H2(g) + I2(g)

Calculate Kc. Kc = )(2.1

0)(0.13)(0.7 =

][HI

]I][H[22

22 = 0.0206

[HI] = L 0.5000

mol 0.20 = 0.40 M

2 HI(g) _ H2(g) + I2(g) initial (M) 0.40 0 0 change (M) -2x +x +x equil (M) 0.40 - 2x x x

Kc = 0.0206 = )x2 _ (0.40

x = ][HI

]I][H[2

2

222

Take the square root of both sides, and solve for x.

)x2 _ (0.40x = 0.0206

2

2

; 0.144 = x2 _ 0.40

x; x = 0.045

At equilibrium, [H2] = [I2] = x = 0.045 M; [HI] = 0.40 - 2x = 0.40 - 2(0.045) = 0.31 M 13.99 Note the container volume is 5.00 L

[H2] = [I2] = 1.00 mol/5.00 L = 0.200 M [HI] = 2.50 mol/5.00 L = 0.500 M

H2(g) + I2(g) _ 2 HI(g) initial (M) 0.200 0.200 0.500 change (M) -x -x +2x


349

equil (M) 0.200 - x 0.200 - x 0.500 + 2x

Kc = ) x_ (0.200

)x2 + (0.500 = 129 =

]I][H[][HI

2

2

22

2

Take the square root of both sides, and solve for x.

) x_ (0.200

)x2 + (0.500 = 129

2

2

; 11.4 = x_ 0.200

x2 + 0.500; x = 0.133

[H2] = [I2] = 0.200 - x = 0.200 - 0.133 = 0.067 M [HI] = 0.500 + 2x = 0.500 + 2(0.133) = 0.766 M

13.100 [H2O] = L 5.00

mol 6.00 = 1.20 M

C(s) + H2O(g) _ CO(g) + H2(g) initial (M) 1.20 0 0 change (M) -x +x +x equil (M) 1.20 - x x x

Kc = x_ 1.20

x = 10 x 3.0 = O]H[

]H[CO][ 22_

2

2

x2 + (3.0 x 10-2)x - 0.036 = 0 Use the quadratic formula to solve for x.

x = 2

0.381 0.030_ =

2(1)

0.036) 4(_ _ )(0.030 (0.030) _ 2 ±±

x = 0.176 and -0.206 Discard the negative solution (-0.206) because it leads to negative concentrations and that is impossible. [CO] = [H2] = x = 0.18 M; [H2O] = 1.20 - x = 1.20 - 0.18 = 1.02 M

13.101 (a) Because Kp is larger at the higher temperature, the reaction has shifted toward

products at the higher temperature, which means the reaction is endothermic. (b) (i) Increasing the volume causes the reaction to shift toward the side with more mol of gas (product side). The equilibrium amounts of PCl3 and Cl2 increase while that of PCl5 decresases. (ii) If there is no volume change, there is no change in equilibrium concentrations. (iii) Addition of a catalyst does not affect the equilibrium concentrations.

13.102 A decrease in volume (a) and the addition of reactants (c) will affect the composition of

the equilibrium mixture, but leave the value of Kc unchanged. A change in temperature (b) affects the value of Kc. Addition of a catalyst (d) or an inert gas (e) affects neither the composition of the equilibrium mixture nor the value of Kc.

13.103 (a) Addition of a solid does not affect the equilibrium composition. There is no change

in the number of moles of CO2. (b) Adding a product causes the reaction to shift toward reactants. The number of moles


350

of CO2 decreases. (c) Decreasing the volume causes the reaction to shift toward the side with fewer mol of gas (reactant side). The number of moles of CO2 decreases. (d) The reaction is endothermic. An increase in temperature shifts the reaction toward products. The number of moles of CO2 increases.

13.104 2 monomer _ dimer

(a) In benzene, Kc = 1.51 x 102 2 monomer _ dimer

initial (M) 0.100 0 change (M) -2x +x equil (M) 0.100 - 2x x

Kc = )x2 _ (0.100

x = 10 x 1.51 =

][monomer

[dimer]2

22

604x2 - 61.4x +1.51 = 0 Use the quadratic formula to solve for x.

x = 1208

11.04 61.4 =

2(604)

.51)(4)(604)(1 _ )61.4(_ 61.4) (_ _ 2 ±±

x = 0.0600 and 0.0417

Discard the larger solution (0.0600) because it gives a negative concentration of the monomer and that is impossible. [monomer] = 0.100 - 2x = 0.100 - 2(0.0417) = 0.017 M; [dimer] = x = 0.0417 M

M 0.017

M 0.0417 =

[monomer]

[dimer] = 2.5

(b) In H2O, Kc = 3.7 x 10-2 2 monomer _ dimer


Kc = )x2 _ (0.100

x = 10 x 3.7 =

][monomer

[dimer]2

2_2

0.148x2 - 1.0148x + 0.000 37 = 0 Use the quadratic formula to solve for x.

x = 0.296

1.0147 1.0148 =

2(0.148)

(0.00037)(4)(0.148) _ )1.0148(_ 1.0148) (_ _ 2 ±±

x = 6.86 and 3.7 x 10-4 Discard the larger solution (6.86) because it gives a negative concentration of the monomer and that is impossible. [monomer] = 0.100 - 2x = 0.100 - 2(3.7 x 10-4) = 0.099 M; [dimer] = x = 3.7 x 10-4 M

M 0.099

M 10 x 3.7 =

[monomer]

[dimer] 4_

= 0.0038


351

(c) Kc for the water solution is so much smaller than Kc for the benzene solution because H2O can hydrogen bond with acetic acid, thus preventing acetic acid dimer formation. Benzene cannot hydrogen bond with acetic acid.

13.105 C(s) + CO2(g) _ 2 CO(g)

initial (M) excess 1.50 mol/20.0 L 0 change (M) -x +2x equil (M) 0.0750 - x 2x [CO] = 2x = 7.00 x 10-2 M; x = 0.0350 M (a) [CO2] = 0.0750 - x = 0.0750 - 0.0350 = 0.0400 M

(b) Kc = (0.0400)

)10 x (7.00 =

]CO[][CO 22_

2

2

= 0.122

13.106 (a) [PCl5] = 1.000 mol/5.000 L = 0.2000 M

PCl5(g) _ PCl3(g) + Cl2(g) initial (M) 0.2000 0 0 change (M) -(0.2000)(0.7850) +(0.2000)(0.7850) +(0.2000)(0.7850) equil (M) 0.0430 0.1570 0.1570

Kc = ]PCl[

]Cl][PCl[

5

23 = (0.0430)

.1570)(0.1570)(0 = 0.573

∆n = 1 and Kp = Kc(RT) = (0.573)(0.082 06)(500) = 23.5

(b) Qc = ]PCl[

]Cl][PCl[

5

23 = (0.500)

600)(0.150)(0.= 0.18

Because Qc < Kc, the reaction proceeds to the right to reach equilibrium.

PCl5(g) _ PCl3(g) + Cl2(g) initial (M) 0.500 0.150 0.600 change (M) - x +x +x equil (M) 0.500 - x 0.150 + x 0.600 + x

Kc = ]PCl[

]Cl][PCl[

5

23 = 0.573 = x)_ (0.500

x)+ x)(0.600+ (0.150; solve for x.

x2 + 1.323x - 0.1965 = 0

x = 2

1.593 1.323_ =

2(1)

0.1965)(4)(1)(_ _ )(1.323 (1.323) _ 2 ±±

x = -1.458 and 0.135 Discard the negative solution (-1.458) because it will lead to negative concentrations and that is impossible. [PCl5] = 0.500 - x = 0.500 - 0.135 = 0.365 M [PCl3] = 0.150 + x = 0.150 + 0.135 = 0.285 M [Cl2] = 0.600 + x = 0.600 + 0.135 = 0.735 M

13.107 Qp = (3.00)

0)(2.00)(1.5 = 1.00, Qp < Kp therefore the reaction proceeds from reactants to


352

products to reach equilibrium. PCl5(g) _ PCl3(g) + Cl2(g)

initial (atm) 3.00 2.00 1.50 change (atm) -x +x +x equil (atm) 3.00 - x 2.00 + x 1.50 + x

Kp = x_ 3.00

x)+ (1.50 x)+ (2.00 = 1.42 =

)P(

)P)(P(

PCl

ClPCl

5

23


2

5.41 4.92) (_ =

2(1)

1.26)4(1)(_ _ )(4.92 4.92) (_ =x

2 ±±

x = 0.245 and -5.165 Discard the negative solution (-5.165) because it gives negative partial pressures and that is impossible. PPCl5 = 3.00 - x = 3.00 - 0.245 = 2.76 atm

PPCl3 = 2.00 + x = 2.00 + 0.245 = 2.24 atm

PCl2 = 1.50 + x = 1.50 + 0.245 = 1.74 atm

Ptotal = PPCl5 + PPCl3 + PCl2 = 2.76 + 2.24 + 1.74 = 6.74 atm

13.108 (a) ]HC[

]HC][HC[ = K104

4262c OVERP)P( )P( = K HCHCHCp 1044262

(b) Kp = 12; ∆n = 1; Kc = Kp

06)(773) (0.082

1(12) =

RT

1 = 0.19

(c) C4H10(g) _ C2H6(g) + C2H4(g) initial (atm) 50 0 0 change (atm) -x +x +x equil (atm) 50 - x x x

Kp = 12 = x_ 50

x2

; x2 + 12x - 600 = 0

Use the quadratic formula to solve for x.

x = 2

50.44 12_ =

2(1)

600) 4(1)(_ _ )(12 12)(_ 2 ±±

x = -31.22 and 19.22 Discard the negative solution (-31.22) because it leads to negative concentrations and that is impossible.

% C4H10 converted = 100% x 50

19.22 = 38%

Ptotal = P + P + P HCHCHC 4262104 = (50 - x) + x + x = (50 - 19) + 19 + 19 = 69 atm

(d) A decrease in volume would decrease the % conversion of C4H10.


353

13.109 (a) Because ∆n = 0, Kp = Kc = 1.0 x 105

(b) Kp = 1.0 x 105 = P

P

necyclopropa

propene ; Pcyclopropane = 10 x 1.0

atm 5.0 =

10 x 1.0P

55

propene = 5.0 x 10-5 atm

(c) The ratio of the two concentrations is equal to Kc. The ratio (Kc) cannot be changed by adding cyclopropane. The individual concentrations can change but the ratio of concentrations can't. Because there is one mole of gas on each side of the balanced equation, the composition of the equilibrium mixture is unaffected by a decrease in volume. The ratio of the two concentrations will not change. (d) Because Kc is large, kf > kr. (e) Because the C–C–C bond angles are 60o and the angles between sp3 hybrid orbitals are 109.5o, the hybrid orbitals are not oriented along the bond directions. Their overlap is therefore poor, and the C–C bonds are correspondingly weak.

13.110 (a) Kp = 3.45; ∆n = 1; Kc = Kp

06)(500) (0.082

1(3.45) =

RT

1 = 0.0840

(b) [(CH3)3CCl] = 1.00 mol/5.00 L = 0.200 M (CH3)3CCl(g) _ (CH3)2C=CH2(g) + HCl(g)

initial (M) 0.200 0 0 change (M) -x +x +x equil (M) 0.200 - x x x

Kc = 0.0840 = x_ 0.200

x2

; x2 + 0.0840x - 0.0168 = 0


x = 2

0.272 0.0840_ =

2(1)

0.0168) 4(1)(_ _ )(0.0840 0.0840)(_ 2 ±±

x = -0.178 and 0.094 Discard the negative solution (-0.178) because it leads to negative concentrations and that is impossible. [(CH3)2C=CCH2] = [HCl] = x = 0.094 M [(CH3)3CCl] = 0.200 - x = 0.200 - 0.094 = 0.106 M (c) Kp = 3.45

(CH3)3CCl(g) _ (CH3)2C=CH2(g) + HCl(g) initial (atm) 0 0.400 0.600 change (atm) +x -x -x equil (atm) x 0.400 - x 0.600 - x

Kp = 3.45 = x

x)_ x)(0.600_ (0.400

x2 - 4.45x + 0.240 = 0 Use the quadratic formula to solve for x.


354

x = 2

4.34 4.45 =

2(1)

)4(1)(0.240 _ )4.45 (_ 4.45) (_ _ 2 ±±

x = 0.055 and 4.40 Discard the larger solution (4.40) because it leads to negative partial pressures and that is impossible. Pt-butyl chloride = x = 0.055 atm; Pisobutylene = 0.400 - x = 0.400 - 0.055 = 0.345 atm PHCl = 0.600 - x = 0.600 - 0.055 = 0.545 atm

13.111 (a) The Arrhenius equation gives for the forward and reverse reactions

e A = k RT / E_ff

fa, and e A = k RT / E_rr

ra, Addition of a catalyst decreases the activation energies by ∆Ea, so

e A = k RT / )E _ E_(ff

afa, ∆ = e x e A RT / ERT / E_f

afa, ∆

and e A = k RT / )E _ E_(rr

ara, ∆ = e x e A RT / ERT / E_r

ara, ∆ Therefore, the rate constants for both the forward and reverse reactions increase by the same factor, e RT / Ea∆ .

(b) The equilibrium constant is given by Kc = e A

A = e A

e A = k

k RT / E _

r

f

RT / E_r

RT / E_f

r

f

ra,

fa,∆

where ∆E = Ea,f - Ea,r. Addition of a catalyst decreases the activation energies by ∆Ea.

So, Kc = e A

A = e x e A

e x e A = k

k RT / E _

r

f

RT / ERT / E_r

RT / ERT / E_f

r

f

ara,

afa,∆

∆

∆

Kc is unchanged because of cancellation of e RT / Ea∆ , the factor by which the two rate constants increase.

13.112 The activation energy (Ea) is positive, and for an exothermic reaction, Ea,r > Ea,f.

kf = Af e RT / E _ fa, , kr = Ar e RT / E _ ra,

Kc = e A

A = e A

e A = k

k RT / ) E _ E(

r

f

RT / E _r

RT / E _f

r

f fa,ra,

ra,

fa,

(Ea,r - Ea,f) is positive, so the exponent is always positive. As the temperature increases,

the exponent, RT / )E _ E( fa,ra, , decreases and the value for Kc decreases as well.

13.113 (a) PV = nRT

PoBr2

= V

nRT =

L 1.00

K) (1000mol K atm L

06 0.082mol) (0.974

••

= 80 atm


355

PoH2

= V

nRT =

L 1.00

K) (1000mol K atm L

06 0.082mol) (1.22

••

= 100 atm

Because Kp is very large, assume first that the reaction goes to completion and then is followed by a small back reaction.

H2(g) + Br2(g) _ 2 HBr(g)

before (atm) 100 80 0 100% reaction (atm) -80 -80 +2(80) after (atm) 20 0 160 back reaction (atm) +x +x -2x after (atm) 20 + x x 160 - 2x

PH2

= 20 + x ≈ 20 atm

PHBr = 160 - 2x ≈ 160 atm

Kp = = )P)(P(

)P(

BrH

2HBr

22

2.1 x 106 = (20)(x)

)(160 2

PBr2 = x =

)10 x (20)(2.1

)(1606

2

= 6.1 x 10-4 atm

(b) (ii) Adding Br2 will cause the greatest increase in the pressure of HBr. The very large value of Kp means that the reaction goes essentially to completion. Therefore, the reaction stops when the limiting reactant, Br2, is essentially consumed. No matter how much H2 is added or how far the the equilibrium is shifted (by lowering the temperature) to favor the formation of HBr, the amount of HBr will ultimately be limited by the amount of Br2 present. So more Br2 must be added in order to produce more HBr.

13.114 (a) PV = nRT, K) (300

mol K atm L

06 0.082

L) atm)(1.00 (0.588 =

RT

PV = ntotal

••

= 0.0239 mol

2 NOBr(g) _ 2 NO(g) + Br2(g) initial (mol) 0.0200 0 0 change (mol) -2x +2x +x equil (mol) 0.0200 - 2x 2x x

ntotal = 0.0239 mol = (0.0200 - 2x) + 2x + x = 0.0200 + x x = 0.0239 - 0.0200 = 0.0039 mol Because the volume is 1.00 L, the molarity equals the number of moles. [NOBr] = 0.0200 - 2x = 0.0200 - 2(0.0039) = 0.0122 M [NO] = 2x = 2(0.0039) = 0.0078 M [Br2] = x = 0.0039 M

= )(0.0122

(0.0039))(0.0078 =

][NOBr

]Br[][NO = K 2

2

22

2

c 1.6 x 10-3


356

(b) ∆n = (3) - (2) = 1, Kp = Kc(RT) = (1.6 x 10-3)(0.082 06)(300) = 0.039 13.115 NO2, 46.01 amu

mol NO2 = 4.60 g NO2 x = NO g 46.01

NO mol 1

2

2 0.100 mol NO2

[NO2] = 0.100 mol/10.0 L = 0.0100 M 2 NO2(g) _ N2O4(g)


Kc = = ]NO[

]ON[2

2

42 4.72 = )x2 _ (0.0100

x2

18.88x2 - 1.189x + 0.000 472 = 0 Use the quadratic formula to solve for x.

x = 37.76

1.1739 1.189 =

2(18.88)

472) .0004(18.88)(0 _ )1.189 (_ 1.189) (_ _ 2 ±±

x = 0.0626 and 4.00 x 10-4 Discard the larger solution (0.0626) because it leads to a negative concentration of NO2 and that is impossible.

ntotal = (0.0100 - 2x + x)(10.0 L) = (0.0100 - x)(10.0 L) = [0.0100 mol/L - (4.00 x 10-4 mol/L)](10.0 L) = 0.0960 mol 100oC = 100 + 273 = 373 K

Ptotal = V

RTntotal = L 10.0

K) (373mol K atm L

06 0.082mol) (0.0960

••

= 0.294 atm

13.116 (a) W(s) + 4 Br(g) _ WBr4(g)

)P(P = K 4

Br

WBrp

4 = 100, PWBr4= )P(

4Br (100) = (0.010 atm)4(100) = 1.0 x 10-6 atm

(b) Because Kp is smaller at the higher temperature, the reaction has shifted toward reactants at the higher temperature, which means the reaction is exothermic.

(c) At 2800 K, )(0.010

)10 x (1.0 = Q 4

6 _

p = 100, Qp > Kp so the reaction will go from products

to reactants, depositing tungsten back onto the filament. 13.117 (a) (NH4)(NH2CO2)(s) _ 2 NH3(g) + CO2(g)

initial (atm) 0 0 change (atm) +2x +x equil (atm) 2x x


357

Ptotal = 2x + x = 3x = 0.116 atm x = 0.116 atm/3 = 0.0387 atm PNH3

= 2x = 2(0.0387 atm) = 0.0774 atm

PCO2= x = 0.0387 atm

= )P( )P( = K CO2

NHp 23(0.0774)2(0.0387) = 2.32 x 10-4

(b) (i) The total quantity of NH3 would decrease. When product, CO2, is added, the equilibrium will shift to the left. (ii) The total quantity of NH3 would remain unchanged. Adding a pure solid, (NH4)(NH2CO2), to a heterogeneous equilibrium will not affect the position of the equilibrium. (iii) The total quantity of NH3 would increase. When product, CO2, is removed, the equilibrium will shift to the right. (iv) The total quantity of NH3 would increase. When the total volume is increased, the reaction will shift to the side with more total moles of gas, which in this case is the product side. (v) The total quantity of NH3 would remain unchanged. Neon is an inert gas which will have no effect on the reaction or on the position of equilibrium. (vi) The total quantity of NH3 would increase. Because the reaction is endothermic, an increase in temperature will shift the equilibrium to to the right.

13.118 2 NO2(g) _ N2O4(g)

∆n = (1) - (2) = -1 and Kp = Kc(RT)-1 = (216)[(0.082 06)(298)]-1 = 8.83

)P(P = K 2

NO

ONp

2

42 = 8.83

Let X = P ON 42 and Y = PNO2

.

Ptotal = 1.50 atm = X + Y and = Y

X2

8.83. Use these two equations to solve for X and Y.

X = 1.50 - Y

= Y

Y _ 1.502

8.83

8.83Y2 + Y - 1.50 = 0 Use the quadratic formula to solve for Y.

Y = 17.7

7.35 1 _ =

2(8.83)

1.50) 4(8.83)(_ _ )(1 (1) _ 2 ±±

Y = -0.472 and 0.359 Discard the negative solution (-0.472) because it leads to a negative partial pressure of NO2 and that is impossible. Y = PNO2

= 0.359 atm

X = P ON 42 = 1.50 atm - Y = 1.50 atm - 0.359 atm = 1.14 atm

13.119 500oC = 500 + 273 = 773 K and 840oC = 840 + 273 = 1113 K


358

Calculate the undissociated pressure of F2 at 1113 K.

T

P = T

P

1

1

2

2 ; = T

T P = P1

212 =

K 773

K) atm)(1113 (0.6000.864 atm

F2(g) _ 2 F(g) initial (atm) 0.864 0 change (atm) -x +2x equil (atm) 0.864 - x 2x

Ptotal = (0.864 atm - x) + 2x = 0.864 atm + x = 0.984 atm x = 0.984 atm - 0.864 atm = 0.120 atm PF2

= (0.864 atm - x) = (0.864 atm - 0.120 atm) = 0.744 atm

PF = 2x = 2(0.120 atm) = 0.240 atm

K p = = P

)P(

F

2F

2

= 0.744

)(0.240 2

0.0774

13.120 N2(g) + 3 H2 _ 2 NH3

initial (mol) 0 0 X change (mol) +y +3y -2y equil (mol) y 3y X - 2y y = 0.200 mol Because the volume is 1.00 L, the molarity equals the number of moles. [N2] = y = 0.200 M; [H2] = 3y = 3(0.200) = 0.600 M

4.20 = )600(0.200)(0.

]NH[ =

]H][N[

]NH[ = K 3

23

322

23

c , solve for [NH3]eq

(4.20))600(0.200)(0. = (4.20)]H][N[ = ]NH[ 3322

2eq3

= (4.20))600(0.200)(0. = (4.20)]H][N[ = ]NH[ 3322eq3 0.426 M

[NH3]eq = 0.426 M = X - 2(0.200) = [NH3]o - 2(0.200) [NH3]o = 0.426 + 2(0.200) = 0.826 M 0.826 mol of NH3 were placed in the 1.00 L reaction vessel.


13.121 (a) H2O, 18.015 amu; 125.4 g H2O x OH g 18.015

OH mol 1

2

2 = 6.96 mol H2O

Given that mol CO = mol H2O = 6.96 mol

L 10.0

K) (700mol K atm L

06 0.082mol) (6.96 =

V

T Rn = P = P OHCO 2

••

= 40.0 atm

CO(g) + H2O(g) _ CO2(g) + H2(g) initial (atm) 40.0 40.0 0 0 equil (atm) 9.80 9.80 40.0 - 9.80 40.0 - 9.80

= 30.2 = 30.2


359

Kp = 0)(9.80)(9.8

2)(30.2)(30. =

)P)(P(

)P)(P(

OHCO

HCO

2

22 = 9.50

(b) 31.4 g H2O x OH g 18.015

OH mol 1

2

2 = 1.743 mol H2O

L 10.0

K) (700mol K atm L

06 0.082mol) (1.743 =

V

T Rn = P OH2

••

= 10.0 atm

P OH2has been increased by 10.0 atm; a new equilibrium will be established.

CO(g) + H2O(g) _ CO2(g) + H2(g) initial (atm) 9.80 9.80 +10.0 30.2 30.2 change (atm) -x -x +x +x equil (atm) 9.80 - x 19.8 - x 30.2 + x 30.2 + x

Kp = x)_ x)(19.80_ (9.80

x)+ x)(30.2+ (30.2 = 9.50 =

)P)(P(

)P)(P(

OHCO

HCO

2

22

8.50x2 - 341.6x + 931.34 = 0 Use the quadratic formula to solve for x.

x = 17.0

291.6 341.6 =

2(8.50)

1.34)4(8.50)(93 _ )341.6 (_ 341.6) (_ _ 2 ±±

x = 37.25 and 2.94 Discard the larger solution (37.25) because it leads to negative partial pressures and that is impossible. PCO = 9.80 - x = 9.80 - 2.94 = 6.86 atm

P OH2= 19.8 - x = 19.8 - 2.94 = 16.9 atm

P = P HCO 22= 30.2 + x = 30.2 + 2.94 = 33.1 atm

K) (700mol K atm L

06 0.082

L) (10.0 atm) (33.1 =

T R

V P = nH2

••

= 5.76 mol H2

H mol 1

molecules H 10 x 6.022 x

cm 1

mL 1 x

mL 1000

L 1 x

L 10.0H mol 5.76

2

223

3

2 = 3.47 x 1020 H2

molecules/cm3 13.122 (a) CO2, 44.01 amu; CO, 28.01 amu

79.2 g CO2 x CO g 44.01

CO mol 1

2

2 = 1.80 mol CO2

CO2(g) + C(s) _ 2 CO(g)

initial (mol) 1.80 0 change (mol) -x +2x equil (mol) 1.80 - x 2x


360

total mass of gas in flask = (16.3 g/L)(5.00 L) = 81.5 g 81.5 = (1.80 - x)(44.01) + (2x)(28.01) 81.5 = 79.22 - 44.01x + 56.02x; 2.28 = 12.01x; x = 2.28/12.01 = 0.19 nCO2

= 1.80 - x = 1.80 - 0.19 = 1.61 mol CO2; nCO= 2x = 2(0.19) = 0.38 mol CO

L 5.0

K) (1000mol K atm L

06 0.082mol) (1.61 =

V

T Rn = PCO2

••

= 26.4 atm

L 5.0

K) (1000mol K atm L

06 0.082mol) (0.38 =

V

T Rn = PCO

••

= 6.24 atm

Kp = (26.4)

)(6.24 =

)P()P(

2

CO

2CO

2

= 1.47

(b) At 1100K, the total mass of gas in flask = (16.9 g/L)(5.00 L) = 84.5 g 84.5 = (1.80 - x)(44.01) + (2x)(28.01) 84.5 = 79.22 - 44.01x + 56.02x; 5.28 = 12.01x; x = 5.28/12.01 = 0.44 nCO2

= 1.80 - x = 1.80 - 0.44 = 1.36 mol CO2; nCO= 2x = 2(0.44) = 0.88 mol CO

L 5.0

K) (1100mol K atm L

06 0.082mol) (1.36 =

V

T Rn = PCO2

••

= 24.6 atm

L 5.0

K) (1100mol K atm L

06 0.082mol) (0.88 =

V

T Rn = PCO

••

= 15.9 atm

Kp = (24.6)

)(15.9 =

)P()P(

2

CO

2CO

2

= 10.3

(c) In agreement with Le Châtelier’s principle, the reaction is endothermic because Kp increases with increasing temperature.

13.123 CO2, 44.01 amu; CO, 28.01 amu; BaCO3, 197.34 amu

1.77 g CO2 x CO g 44.01

CO mol 1

2

2 = 0.0402 mol CO2

CO2(g) + C(s) _ 2 CO(g)

initial (mol) 0.0402 0 change (mol) -x +2x equil (mol) 0.0402 - x 2x

3.41 g BaCO3 x BaCO g 197.34

BaCO mol 1

3

3 x BaCO mol 1CO mol 1

3

2 = 0.0173 mol CO2

mol CO2 = 0.0173 = 0.0402 - x; x = 0.0229 mol CO = 2x = 2(0.0229) = 0.0458 mol CO


361

L 1.000

K) (1100mol K atm L

06 0.082mol) (0.0173 =

V

T Rn = PCO2

••

= 1.562 atm

L 1.000

K) (1100mol K atm L

06 0.082mol) (0.0458 =

V

T Rn = PCO

••

= 4.134 atm

Kp = (1.562)

)(4.134 =

)P()P(

2

CO

2CO

2

= 11.0

13.124 (a) N2O4, 92.01 amu

14.58 g N2O4 x ON g 92.01

ON mol 1

42

42 = 0.1585 mol N2O4

PV = nRT

L 1.000

K) (400mol K atm L

06 0.082mol) (0.1585 =

V

T Rn = P ON 42

••

= 5.20 atm

N2O4(g) _ 2 NO2(g)

initial (atm) 5.20 0 change (atm) -x +2x equil (atm) 5.20 - x 2x

P + P = P NOONtotal 242 = (5.20 - x) + (2x) = 9.15 atm

5.20 + x = 9.15 atm x = 3.95 atm P ON 42

= 5.20 - x = 5.20 - 3.95 = 1.25 atm

PNO2 = 2x = 2(3.95) = 7.90 atm

Kp = (1.25)

)(7.90 =

)P(

)P(2

ON

2NO

42

2 = 49.9

∆n = 1 and Kc = Kp 06)(400) (0.082

(49.9) =

RT

1

= 1.52

(b) ∆Horxn = [2 ∆Ho

f(NO2)] - ∆Hof(N2O4)

∆Horxn = [(2 mol)(33.2 kJ/mol)] - [(1mol)(9.16 kJ/mol)] = 57.2 kJ

PV = nRT

moles N2O4 reacted = n = K) (400

mol K atm L

06 0.082

L) atm)(1.000 (3.95 =

RT

PV

••

= 0.1203 mol N2O4

q = (57.24 kJ/mol N2O4)(0.1203 mol N2O4) = 6.89 kJ


362

13.125 C10H8(s) _ C10H8(g)

(a) Kc = [C10H8] = 5.40 x 10-6 room volume = 8.0 ft x 10.0 ft x 8.0 ft = 640 ft2

room volume = 640 ft2 x

ft 1

in 123

x

in 1

cm 2.543

x

cm 1000

L 1.03

= 18122.8 L

C10H8 molecules = (5.40 x 10-6 mol/L)(18122.8 L)(6.022 x 1023 molecules/mol) = 5.89 x 1022 C10H8 molecules

(b) C10H8, 128.17 amu mol C10H8 = (5.40 x 10-6 mol/L)(18122.8 L) = 0.0979 mol C10H8

mass C10H8 = 0.0979 mol C10H8 x HC mol 1

HC g 128.17

810

810 = 12.55 g C10H8

moth ball: r = 12.0 mm/2 = 6.0 mm = 0.60 cm volume of moth ball = (4/3)πr3 = (4/3)π(0.60cm)3 = 0.905 cm3 mass of moth ball = (0.905 cm3/moth ball)(1.16 g/cm3) = 1.05 g/moth ball

number of moth balls = ball/moth HC g 1.05

HC g 12.55

810

810 = 12 moth balls

13.126 The atmosphere is 21% (0.21) O2; PO2

= (0.21)( )Hg mm 720 = 0.199 atm

2 O3(g) _ 3 O2(g)

Kp = )P(

)P(2

O

3O

3

2 ; 10 x 1.3

)(0.199 =

K

)P( = P 57

3

p

3O

O2

3 = 2.46 x 10-30 atm

vol = 10 x 106 m3 x cm 1000

L 1 x

m 1

cm 1003

3

= 1.0 x 1010 L

K) (298mol K atm L

06 0.082

L) 10 x (1.0 atm) 10 x (2.46 =

T R

V P = n

1030_

O3

••

= 1.0 x 10-21 mol O3

O3 molecules = 1.0 x 10-21 mol O3 x O mol 1

molecules O 10 x 6.022

3

323

= 6.0 x 102 O3 molecules

13.127 250.0 mL = 0.2500 L

[CH3CO2H] = 0.0300 mol/0.2500 L = 0.120 M 2 CH3CO2H _ (CH3CO2H)2


Kc = = ][monomer

[dimer]2

1.51 x 102 = )x2 _ (0.120

x2

604x2 - 73.48x + 2.1744 = 0


363


x = 1208

12.08 73.48 =

2(604)

744)4(604)(2.1 _ )73.48 (_ 73.48) (_ _ 2 ±±

x = 0.0708 and 0.0508 Discard the larger solution (0.0708) because it leads to a negative concentration and that is impossible. [monomer] = 0.120 - 2x = 0.120 - 2(0.0508) = 0.0184 M [dimer] = x = 0.0508 M (b) 25oC = 25 + 273 = 298 K

Π = MRT = (0.0184 M + 0.0508 M) K)(298mol K

atm L 06 0.082

••

= 1.69 atm

13.128 PCl5(g) _ PCl3(g) + Cl2(g)

∆n = (2) - (1) = 1 and at 700 K, Kp = Kc(RT) = (46.9)(0.082 06)(700) = 2694 (a) Because Kp is larger at the higher temperature, the reaction has shifted toward products at the higher temperature, which means the reaction is endothermic. Because the reaction involves breaking two P–Cl bonds and forming just one Cl–Cl bond, it should be endothermic. (b) PCl5, 208.24 amu

mol PCl5 = 1.25 g PCl5 x = PCl g 208.24

PCl mol 1

5

5 6.00 x 10-3 mol

PV = nRT, PPCl5 = V

nRT =

L 0.500

K) (700mol K atm L

06 0.082mol) 10 x (6.00 3_

••

= 0.689

atm Because Kp is so large, first assume the reaction goes to completion and then allow for a small back reaction.

PCl5(g) _ PCl3(g) + Cl2(g) before rxn (atm) 0.689 0 0 100% rxn (atm) -0.689 +0.689 +0.689 after rxn (atm) 0 0.689 0.689 back rxn (atm) +x -x -x equil (atm) x 0.689 - x 0.689 - x

Kp = = P

)P)(P(

PCl

ClPCl

5

23 2694 = x

)(0.689

x

) x_ (0.689 22

≈

x = PPCl5 = 2694

)(0.689 2

= 1.76 x 10-4 atm

P + P + P = P ClPClPCltotal 235

= Ptotal x + (0.689 - x) + (0.689 - x) = 0.689 + 0.689 - 1.76 x 10-4 = 1.38 atm


364

% dissociation = = 100% x 0.689

)10 x (1.76 _ 0.689 = 100% x

)P(

)P( _ )P( 4 _

oPCl

PCloPCl

5

55 99.97%

(c) The molecular geometry is trigonal bipyramidal. There is no dipole moment because of a symmetrical distribution of Cl’s around the central P.

The molecular geometry is trigonal pyramidal. There is a dipole moment because of the lone pair of electrons on the P and an unsymmetrical distribution of Cl’s around the central P.

433

16

Applications of Aqueous Equilibria

16.1 (a) HNO2(aq) + OHS(aq) Ω NO2S(aq) + H2O(l); NO2

S (basic anion), pH > 7.00

(b) H3O+(aq) + NH3(aq) Ω NH4

+(aq) + H2O(l); NH4+ (acidic cation), pH < 7.00

(c) OHS(aq) + H3O+(aq) Ω 2 H2O(l); pH = 7.00

16.2 (a) HF(aq) + OHS(aq) Ω H2O(l) + FS(aq)

Kn = _ 4

a

_14w

3.5 x 10K = 1.0 x 10K

= 3.5 x 1010

(b) H3O+(aq) + OHS(aq) Ω 2 H2O(l)

Kn = _14

w

1 1 =

1.0 x 10K = 1.0 x 1014

(c) HF(aq) + NH3(aq) Ω NH4+(aq) + FS(aq)

Kn = _ 4 _5

a b

_14w

(3.5 x )(1.8 x )10 10K K = 1.0 x 10K

= 6.3 x 105

The tendency to proceed to completion is determined by the magnitude of Kn. The larger the value of Kn, the further does the reaction proceed to completion. The tendency to proceed to completion is: reaction (c) < reaction (a) < reaction (b)

16.3 HCN(aq) + H2O(l) Ω H3O+(aq) + CNS(aq)

initial (M) 0.025 ~0 0.010 change (M) Sx +x +x equil (M) 0.025 S x x 0.010 + x

Ka = + _

3 _10[ ][ ] x(0.010 + x) x(0.010)O CNH = 4.9 x = 10[HCN] 0.025 _ x 0.025

≈

Solve for x. x = 1.23 x 10S9 M = 1.2 x 10S9 M = [H3O+]

pH = Slog[H3O+] = Slog(1.23 x 10S9) = 8.91

[OHS] = _14

w

+ _93

1.0 x 10K = [ ] 1.23 x O 10H

= 8.2 x 10S6 M

[Na+] = [CNS] = 0.010 M; [HCN] = 0.025 M

% dissociation = _9

diss

initial

[HCN ] 1.23 x M10 x 100% = x 100%[HCN 0.025 M]

= 4.9 x 10S6 %

Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________

434

16.4 From NH4Cl(s), [NH4+] initial =

0.10 mol

0.500 L = 0.20 M

NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)

initial (M) 0.40 0.20 ~0 change (M) Sx +x +x equil (M) 0.40 S x 0.20 + x x

Kb = + _4 _5

3

[ ][ ] (0.20 + x)(x) (0.20)(x)NH OH = 1.8 x = 10[ ] (0.40 _ x) (0.40)NH

≈

Solve for x. x = [OHS] = 3.6 x 10S5 M

[H3O+] =

_14w

_ _5

1.0 x 10K = [ ] 3.6 x OH 10

= 2.8 x 10S10 M

pH = Slog[H3O+] = Slog(2.8 x 10S10) = 9.55

16.5 Each solution contains the same number of B molecules. The presence of BH+ from

BHCl lowers the percent dissociation of B. Solution (2) contains no BH+, therefore it has the largest percent dissociation. BH+ is the conjugate acid of B. Solution (1) has the largest amount of BH+ and it would be the most acidic solution and have the lowest pH.

16.6 (a) (1) and (3). Both pictures show equal concentrations of HA and AS.

(b) (3). It contains a higher concentration of HA and AS.

16.7 HF(aq) + H2O(l) Ω H3O+(aq) + FS(aq)


Ka = + _

3 _ 4[ ][ ] x(0.50 + x) x(0.50)OH F = 3.5 x = 10[HF] 0.25 _ x 0.25

≈

Solve for x. x = 1.75 x 10S4 M = [H3O+]

For the buffer, pH = Slog[H3O+] = Slog(1.75 x 10S4) = 3.76

(a) mol HF = 0.025 mol; mol FS = 0.050 mol; vol = 0.100 L 100%

FS(aq) + H3O+(aq) HF(aq) + H2O(l)

before (mol) 0.050 0.002 0.025 change (mol) S0.002 S0.002 +0.002 after (mol) 0.048 0 0.027

[H3O+] = _ 4

a _

[HF] 0.27 = (3.5 x )10K

[ ] 0.48F

= 1.97 x 10S4 M

pH = Slog[H3O+] = Slog(1.97 x 10S4) = 3.71

(b) mol HF = 0.025 mol; mol FS = 0.050 mol; vol = 0.100 L


435

100% HF(aq) + OHS(aq) FS(aq) + H2O(l)

before (mol) 0.025 0.004 0.050 change (mol) S0.004 S0.004 +0.004 after (mol) 0.021 0 0.054

[H3O+] = _ 4

a _

[HF] 0.21 = (3.5 x )10K

[ ] 0.54F

= 1.36 x 10S4 M

pH = Slog[H3O+] = Slog(1.36 x 10S4) = 3.87

16.8 HF(aq) + H2O(l) Ω H3O+(aq) + FS(aq)


Ka = + _

3 _ 4[ ][ ] x(0.100 + x) x(0.100)OH F = 3.5 x = 10[HF] 0.050 _ x 0.050

≈

Solve for x. x = [H3O+] = 1.75 x 10S4 M

pH = Slog[H3O+] = Slog(1.75 x 10S4) = 3.76

mol HF = 0.050 mol/L x 0.100 L = 0.0050 mol HF mol FS = 0.100 mol/L x 0.100 L = 0.0100 mol FS mol HNO3 = mol H3O

+ = 0.002 mol 100%

Neutralization reaction: FS(aq) + H3O+(aq) HF(aq) + H2O(l)

before reaction (mol) 0.0100 0.002 0.0050 change (mol) S0.002 S0.002 +0.002 after reaction (mol) 0.008 0 0.007

[HF] = 0.007 mol

0.100 L = 0.07 M; [FS] =

0.008 mol

0.100 L = 0.08 M

[H3O+] = Ka _ 4

_

[HF] (0.07) = (3.5 x )10

[ ] (0.08)F = 3 x 10S4 M

pH = Slog[H3O+] = Slog(3 x 10S4) = 3.5

This solution has less buffering capacity than the solution in Problem 16.7 because it contains less HF and FS per 100 mL. Note that the change in pH is greater than that in Problem 16.7.

16.9 When equal volumes of two solutions are mixed together, the concentration of each

solution is cut in half.

pH = pKa + log2_3

a _3

[base] [ ]CO = + log pK[acid] [ ]HCO

For HCO3S, Ka = 5.6 x 10S11, pKa = Slog Ka = Slog(5.6 x 10S11) = 10.25

pH = 10.25 + log0.050

0.10

= 10.25 S 0.30 = 9.95

16.10 pH = pKa + log2_3

a _3

[base] [ ]CO = + log pK[acid] [ ]HCO


436

For HCO3S, Ka = 5.6 x 10S11, pKa = Slog Ka = Slog(5.6 x 10S11) = 10.25

10.40 = 10.25 + log2_3

_3

[ ]CO[ ]HCO

; log2_3

_3

[ ]CO[ ]HCO

= 10.40 S 10.25 = 0.15

2_3

_3

[ ]CO[ ]HCO

= 100.15 = 1.4

To obtain a buffer solution with pH 10.40, make the Na2CO3 concentration 1.4 times the concentration of NaHCO3.

16.11 Look for an acid with pKa near the required pH of 7.50.

Ka = 10SpH = 10S7.50 = 3.2 x 10S8 Suggested buffer system: HOCl (Ka = 3.5 x 10S8) and NaOCl.

16.12 (a) serine is 66% dissociated at pH = 9.15 + log 66

34

= 9.44

(b) serine is 5% dissociated at pH = 9.15 + log 5

95

= 7.87

16.13 (a) mol HCl = mol H3O

+ = 0.100 mol/L x 0.0400 L = 0.004 00 mol mol NaOH = mol OHS = 0.100 mol/L x 0.0350 L = 0.003 50 mol Neutralization reaction: H3O

+(aq) + OHS(aq) 2 H2O(l) before reaction (mol) 0.004 00 0.003 50 change (mol) S0.003 50 S0.003 50 after reaction (mol) 0.000 50 0

[H3O+] =

0.000 50 mol

(0.0400 L + 0.0350 L) = 6.7 x 10S3 M

pH = Slog[H3O+] = Slog(6.7 x 10S3) = 2.17

(b) mol HCl = mol H3O+ = 0.100 mol/L x 0.0400 L = 0.004 00 mol

mol NaOH = mol OHS = 0.100 mol/L x 0.0450 L = 0.004 50 mol Neutralization reaction: H3O

+(aq) + OHS(aq) 2 H2O(l) before reaction (mol) 0.004 00 0.004 50 change (mol) S0.004 00 S0.004 00 after reaction (mol) 0 0.000 50

[OHS] = 0.000 50 mol

(0.0400 L + 0.0450 L) = 5.9 x 10S3 M

[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 5.9 x OH 10

= 1.7 x 10S12 M

pH = Slog[H3O+] = Slog(1.7 x 10S12) = 11.77

The results obtained here are consistent with the pH data in Table 16.1 16.14 (a) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol

mol HCl = mol H3O+ = 0.0500 mol/L x 0.0600 L = 0.003 00 mol

Neutralization reaction: H3O+(aq) + OHS(aq) 2 H2O(l)

before reaction (mol) 0.003 00 0.004 00 change (mol) S0.003 00 S0.003 00


437

after reaction (mol) 0 0.001 00

[OHS] = 0.001 00 mol

(0.0400 L + 0.0600 L) = 1.0 x 10S2 M

[H3O+] =

_14w

_ _ 2

1.0 x 10K = [ ] 1.0 x OH 10

= 1.0 x 10S12 M

pH = Slog[H3O+] = Slog(1.0 x 10S12) = 12.00

(b) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol mol HCl = mol H3O

+ = 0.0500 mol/L x 0.0802 L = 0.004 01 mol Neutralization reaction: H3O


[H3O+] =

0.000 01 mol

(0.0400 L + 0.0802 L) = 8.3 x 10S5 M

pH = Slog[H3O+] = Slog(8.3 x 10S5) = 4.08

(c) mol NaOH = mol OHS = 0.100 mol/L x 0.0400 L = 0.004 00 mol mol HCl = mol H3O

+ = 0.0500 mol/L x 0.1000 L = 0.005 00 mol Neutralization reaction: H3O


[H3O+] =

0.001 00 mol

(0.0400 L + 0.1000 L) = 7.1 x 10S3 M

pH = Slog[H3O+] = Slog(7.1 x 10S3) = 2.15

16.15 (a) (3), only HA present (b) (1), HA and AS present

(c) (4), only AS present (d) (2), AS and OHS present

16.16 mol NaOH required = 0.016 mol HOCl 1 mol NaOH

(0.100 L) = 0.0016 molL 1 mol HOCl

vol NaOH required = (0.0016 mol)1 L

= 0.040 L = 40 mL0.0400 mol

40 mL of 0.0400 M NaOH are required to reach the equivalence point. (a) mmol HOCl = 0.016 mmol/mL x 100.0 mL = 1.6 mmol mmol NaOH = mmol OHS = 0.0400 mmol/mL x 10.0 mL = 0.400 mmol Neutralization reaction: HOCl(aq) + OHS(aq) OClS(aq) + H2O(l) before reaction (mmol) 1.6 0.400 0 change (mmol) S0.400 S0.400 +0.400 after reaction (mmol) 1.2 0 0.400

[HOCl] = 1.2 mmol

(100.0 mL + 10.0 mL) = 1.09 x 10S2 M

[OClS] = 0.400 mmol

(100.0 mL + 10.0 mL) = 3.64 x 10S3 M


438

HOCl(aq) + H2O(l) Ω H3O+(aq) + OClS(aq)

initial (M) 0.0109 ~0 0.003 64 change (M) Sx +x +x equil (M) 0.0109 S x x 0.003 64 + x

Ka = + _

3 _8[ ][ ] x(0.003 64 + x) x(0.003 64)O OClH = 3.5 x = 10[HOCl] 0.0109 _ x 0.0109

≈


pH = Slog[H3O+] = Slog(1.05 x 10S7) = 6.98

(b) Halfway to the equivalence point, [OClS] = [HOCl] pH = pKa = Slog Ka = Slog(3.5 x 10S8) = 7.46 (c) At the equivalence point the solution contains the salt, NaOCl. mol NaOCl = initial mol HOCl = 0.0016 mol = 1.6 mmol

[OClS] = 1.6 mmol

(100.0 mL + 40.0 mL) = 1.1 x 10S2 M

For OClS, Kb = _14

w

_8a

1.0 x 10K = for HOCl 3.5 x 10K

= 2.9 x 10S7

OClS(aq) + H2O(l) Ω HOCl(aq) + OHS(aq) initial (M) 0.011 0 ~0 change (M) Sx +x +x equil (M) 0.011 S x x x

Kb = _ 2 2

_7_

[HOCl][ ]OH x x = 2.9 x = 10[ ] 0.011 _ x 0.011OCl

≈


[H3O+] =

_14w

_ _5

1.0 x 10K = [ ] 5.65 x OH 10

= 1.77 x 10S10 = 1.8 x 10S10 M

pH = Slog[H3O+] = Slog(1.77 x 10S10) = 9.75

16.17 From Problem 16.16, pH = 9.75 at the equivalence point.

Use thymolphthalein (pH 9.4 S 10.6). Bromthymol blue is unacceptable because it changes color halfway to the equivalence point.

16.18 (a) mol NaOH required to reach first equivalence point

= 2 3

2 3

0.0800 mol 1 mol NaOHSOH (0.0400 L) = 0.003 20 molL 1 mol SOH

vol NaOH required to reach first equivalence point

= (0.003 20 mol)1 L

= 0.020 L = 20.0 mL0.160 mol

20.0 mL is enough NaOH solution to reach the first equivalence point for the titration of the diprotic acid, H2SO3.


439

For H2SO3, _ 2 _ 2

a1 a1a1 = 1.5 x , = _log = _log(1.5 x ) = 1.82pK10 10K K _8 _8

a 2 a 2a 2 = 6.3 x , = _log = _log(6.3 x ) = 7.20pK10 10K K

At the first equivalence point, pH = a1 a 2 + pK pK 1.82 + 7.20 =

2 2 = 4.51

(b) mol NaOH required to reach second equivalence point

= 2 3

2 3

0.0800 mol 2 mol NaOHSOH (0.0400 L) = 0.006 40 molL 1 mol SOH

vol NaOH required to reach second equivalence point

= (0.006 40 mol)1 L

= 0.040 L = 40.0 mL0.160 mol

30.0 mL is enough NaOH solution to reach halfway to the second equivalent point. Halfway to the second equivalence point pH = p _8

a 2 a 2 = _log = _log(6.3 x )10K K = 7.20 (c) mmol HSO3

S = 0.0800 mmol/mL x 40.0 mL = 3.20 mmol volume NaOH added after first equivalence point = 35.0 mL S 20.0 mL = 15.0 mL mmol NaOH = mmol OHS = 0.160 mmol/L x 15.0 mL = 2.40 mmol

Neutralization reaction: HSO3S(aq) + OHS(aq) Ω SO3

2S(aq) + H2O(l) before reaction (mmol) 3.20 2.40 0 change (mmol) S2.40 S2.40 +2.40 after reaction (mmol) 0.80 0 2.40

[HSO3S] =

0.80 mmol = 0.0107 M

(40.0 mL + 35.0 mL)

[SO32S] =

2.40 mmol

(40.0 mL + 35.0 mL) = 0.0320 M

HSO3S(aq) + H2O(l) Ω H3O

+(aq) + SO32S(aq)


Ka = 2_+

3 3 _8_3

[ ][ ] x(0.0320 + x) x(0.0320)O SOH = 6.3 x = 10[ ] 0.0107 _ x 0.0107HSO

≈


pH = Slog[H3O+] = Slog(2.1 x 10S8) = 7.68

16.19 Let H2A

+ = valine cation (a) mol NaOH required to reach first equivalence point

= +

2

+2

0.0250 mol 1 mol NaOHH A (0.0400 L) = 0.001 00 molL 1 mol H A

vol NaOH required to reach first equivalence point


440

= (0.001 00 mol)1 L

= 0.0100 L = 10.0 mL0.100 mol

10.0 mL is enough NaOH solution to reach the first equivalence point for the titration of the diprotic acid, H2A

+. For H2A

+, _ 3 _ 3

a1 a1a1 = 4.8 x , = _ log = _ log(4.8 x ) = 2.32pK10 10K K _10 _10

a 2 a 2a 2 = 2.4 x , = _ log = _ log(2.4 x ) = 9.62pK10 10K K

At the first equivalence point, pH = a1 a 2 + pK pK 2.32 + 9.62 =

2 2 = 5.97

(b) mol NaOH required to reach second equivalence point

= +

2

+2

0.0250 mol 2 mol NaOHH A (0.0400 L) = 0.002 00 molL 1 mol H A

vol NaOH required to reach second equivalence point

= (0.002 00 mol)1 L

= 0.0200 L = 20.0 mL0.100 mol

15.0 mL is enough NaOH solution to reach halfway to the second equivalent point. Halfway to the second equivalence point pH = p _10

a 2 a 2 = _ log = _ log(2.4 x )10K K = 9.62 (c) 20.0 mL is enough NaOH to reach the second equivalence point. At the second equivalence point

mmol AS = (0.0250 mmol/mL)(40.0 mL) = 1.00 mmol AS solution volume = 40.0 mL + 20.0 mL = 60.0 mL

[A S] = 1.00 mmol

60.0 mL = 0.0167 M

AS(aq) + H2O(l) Ω HA(aq) + OHS(aq) initial (M) 0.0167 0 ~0 change (M) Sx +x +x equil (M) 0.0167 S x x x

Kb = w

a

K for HAK

= w

a 2

K

K =

_14

_10

1.0 x 102.4 x 10

= 4.17 x 10S5

Kb = _ 2

_5_

[HA][ ]OH x = 4.17 x = 10[ ] 0.0167 _ xA

x2 + (4.17 x 10S5)x S (6.964 x 10S7) = 0 Use the quadratic formula to solve for x. x =

2_5 _5 _ 7 _5 _3_ (4.17 x ) (4.17 x _ (4)(1)(_ 6.964 x )) (_ 4.17 x ) (1.67 x )10 10 10 10 10 = 2(1) 2

± ±

x = 8.14 x 10S4 and S8.56 x 10S4 Of the two solutions for x, only the positive value has physical meaning because x is the [OHS].


441

x = [OHS] = 8.14 x 10S4 M

[H3O+] =

_14w

_ _ 4

1.0 x 10K = [ ] 8.14 x OH 10

= 1.23 x 10S11 M

pH = Slog[H3O+] = Slog(1.23 x 10S11) = 10.91

16.20 (a) Ksp = [Ag+][Cl S] (b) Ksp = [Pb2+][I S]2

(c) Ksp = [Ca2+]3[PO43S]2 (d) Ksp = [Cr3+][OHS]3

16.21 Ksp = [Ca2+]3[PO4

3S]2 = (2.01 x 10S8)3(1.6 x 10S5)2 = 2.1 x 10S33 16.22 [Ba2+] = [SO4

2S] = 1.05 x 10S5 M; Ksp = [Ba2+][SO42S] = (1.05 x 10S5)2 = 1.10 x 10S10

16.23 (a) AgCl(s) Ω Ag+(aq) + ClS(aq) equil (M) x x Ksp = [Ag+][Cl S] = 1.8 x 10S10 = (x)(x)

molar solubility = x = spK = 1.3 x 10S5 mol/L

AgCl, 143.32 amu

solubility =

_5 143.32 g1.3 x mol x 10

1 mol1 L

= 0.0019 g/L

(b) Ag2CrO4(s) Ω 2 Ag+(aq) + CrO42S(aq)

equil (M) 2x x Ksp = [Ag+]2[CrO4

2S] = 1.1 x 10S12 = (2x)2(x) = 4x3

molar solubility = x = _12

31.1 x 10

4 = 6.5 x 10S5 mol/L

Ag2CrO4, 331.73 amu

solubility =

_5 331.73 g6.5 x mol x 10

1 mol1 L

= 0.022 g/L

Ag2CrO4 has both the higher molar and gram solubility, despite its smaller value of Ksp. 16.24 Let the number of ions be proportional to its concentration.

For AgX, Ksp = [Ag+][X S] % (4)(4) = 16 For AgY, Ksp = [Ag+][Y S] % (1)(9) = 9 For AgZ, Ksp = [Ag+][Z S] % (3)(6) = 18 (a) AgZ (b) AgY

16.25 [Mg2+]0 is from 0.10 M MgCl2.

MgF2(s) Ω Mg2+(aq) + 2 FS(aq) initial (M) 0.10 0 change (M) +x +2x equil (M) 0.10 + x 2x


442

Ksp = 7.4 x 10S11 = [Mg2+][FS]2 = (0.10 + x)(2x)2 . (0.10)(4x2) x = 1.4 x 10S5, molar solubility = x = 1.4 x 10S5 M

16.26 Compounds that contain basic anions are more soluble in acidic solution than in pure

water. AgCN, Al(OH)3, and ZnS all contain basic anions. 16.27 [Cu2+] = (5.0 x 10S3 mol)/(0.500 L) = 0.010 M

Cu2+(aq) + 4 NH3(aq) Ω Cu(NH3)42+(aq)

before reaction (M) 0.010 0.40 0 assume 100 % reaction (M) S0.010 S 4(0.010) +0.010

after reaction (M) 0 0.36 0.010 assume small back reaction (M) +x +4x Sx

equil (M) x 0.36 + 4x 0.010 S x

Kf = 2+

3 1144 4 42+

3

[Cu( ]) (0.010 _ x) 0.010NH = 5.6 x = 10[ ][ (x)(0.36 + 4x x(0.36] ) )Cu NH

≈

Solve for x. x = [Cu2+] = 1.1 x 10S12 M

16.28 AgBr(s) Ω Ag+(aq) + BrS(aq) Ksp = 5.4 x 10S13 Ag+(aq) + 2 S2O3

2S Ag(S2O3)23S(aq) Kf = 4.7 x 1013

dissolution AgBr(s) + 2 S2O32S(aq) Ω Ag(S2O3)2

3S(aq) + BrS(aq) reaction K = (Ksp)(Kf) = (5.4 x 10S13)(4.7 x 1013) = 25.4

AgBr(s) + 2 S2O32S(aq) Ω Ag(S2O3)2

3S(aq) + BrS(aq) initial (M) 0.10 0 0 change (M) S2x x x equil (M) 0.10 S 2x x x

K = 3_ _ 2

2 3 22 22_

2 3

[Ag( ][ ])S O Br x = 25.4 = [ (0.10 _ 2x] )S O

Take the square root of both sides and solve for x. 2

2

x25.4 = (0.10 _ 2x)

; 5.04 = x

0.10 _ 2 x; x = molar solubility = 0.045 mol/L

16.29 On mixing equal volumes of two solutions, the concentrations of both solutions are cut

in half. For BaCO3, Ksp = 2.6 x 10S9 (a) IP = [Ba2+][CO3

2S] = (1.5 x 10S3)(1.0 x 10S3) = 1.5 x 10S6 IP > Ksp; a precipitate of BaCO3 will form. (b) IP = [Ba2+][CO3

2S] = (5.0 x 10S6)(2.0 x 10S5) = 1.0 x 10S10 IP < Ksp; no precipitate will form.

16.30 pH = pKa + log 3a +

4

[base] [ ]NH = + log pK[acid] [ ]NH


443

For NH4+, Ka = 5.6 x 10S10, pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25

pH = 9.25 + log (0.20)

(0.20) = 9.25; [H3O

+] = 10SpH = 10S9.25 = 5.6 x 10S10 M

[OHS] = _14

w

+ _103

1.0 x 10K = [ ] 5.6 x O 10H

= 1.8 x 10S5 M

[Fe2+] = [Mn2+] = _3(25 mL)(1.0 x M)10

250 mL = 1.0 x 10S4 M

For Mn(OH)2, Ksp = 2.1 x 10S13 IP = [Mn2+][OHS]2 = (1.0 x 10S4)(1.8 x 10S5)2 = 3.2 x 10S14 IP < Ksp; no precipitate will form.

For Fe(OH)2, Ksp = 4.9 x 10S17 IP = [Fe2+][OHS]2 = (1.0 x 10S4)(1.8 x 10S5)2 = 3.2 x 10S14 IP > Ksp; a precipitate of Fe(OH)2 will form.

16.31 MS(s) + 2 H3O+(aq) Ω M2+(aq) + H2S(aq) + 2 H2O(l)

Kspa = 2+

22+

3

[ ][ S]M H[ ]OH

For ZnS, Kspa = 3 x 10S2; for CdS, Kspa = 8 x 10S7 [Cd2+] = [Zn2+] = 0.005 M Because the two cation concentrations are equal, Qc is the same for both.

Qc = 2+

2t t2 2+

3 t

[ [ S] ] (0.005)(0.10)M H = [ (0.3] )OH

= 6 x 10S3

Qc > Kspa for CdS; CdS will precipitate. Qc < Kspa for ZnS; Zn2+ will remain in solution. 16.32 This protein has both acidic and basic sites. H3PO4-H2PO4

S is an acidic buffer. It protonates the basic sites in the protein making them positive and the protein migrates towards the negative electrode. H3BO3-H2BO3

S is a basic buffer. At basic pH's, the acidic sites in the protein are dissociated making them negative and the protein migrates towards the positive electrode.

16.33 To increase the rate at which the proteins migrates toward the negative electrode,

increase the number of basic sites that are protonated by lowering the pH. Decrease the [HPO4

2S]/[H2PO4S] ratio (less HPO4

2S, more H2PO4S) to lower the pH.

Understanding Key Concepts 16.34 A buffer solution contains a conjugate acid-base pair in about equal concentrations.

(a) (1), (3), and (4) (b) (4) because it has the highest buffer concentration.


444

16.35 (a) (2) has the highest pH, [AS] > [HA] (3) has the lowest pH, [HA] > [AS]

(b) (c) 16.36 (4); only AS and water should be present 16.37 (a) (1) corresponds to (iii); (2) to (i); (3) to (iv); and (4) to (ii)

(b) Solution (3) has the highest pH; solution (2) has the lowest pH. 16.38 (a) (i) (1), only B present (ii) (4), equal amounts of B and BH+ present

(iii) (3), only BH+ present (iv) (2), BH+ and H3O+ present

(b) The pH is less than 7 because BH+ is an acidic cation. 16.39 (2) is supersaturated; (3) is unsaturated; (4) is unsaturated 16.40 Let the number of ions be proportional to its concentration.

For Ag2CrO4, Ksp = [Ag+]2[CrO42S] % (4)2(2) = 32

For (2), IP = [Ag+]2[CrO42S] % (2)2(4) = 16

For (3), IP = [Ag+]2[CrO42S] % (6)2(2) = 72

For (4), IP = [Ag+]2[CrO42S] % (2)2(6) = 24

A precipitate will form when IP > Ksp. A precipitate will form only in (3). 16.41 (a) The lower curve represents the titration of a strong acid; the upper curve represents

the titration of a weak acid. (b) pH = 7 for titration of the strong acid; pH = 10 for titration of the weak acid. (c) Halfway to the equivalence point, the pH = pKa ~ 6.3.

Additional Problems Neutralization Reactions 16.42 (a) HI(aq) + NaOH(aq) H2O(l) + NaI(aq)

net ionic equation: H3O+(aq) + OHS(aq) 2 H2O(l)

The solution at neutralization contains a neutral salt (NaI); pH = 7.00.

(b) 2 HOCl(aq) + Ba(OH)2(aq) 2 H2O(l) + Ba(OCl)2(aq)


445

net ionic equation: HOCl(aq) + OHS(aq) H2O(l) + OClS(aq) The solution at neutralization contains a basic anion (OClS); pH > 7.00 (c) HNO3(aq) + C6H5NH2(aq) C6H5NH3NO3(aq) net ionic equation: H3O

+(aq) + C6H5NH2(aq) H2O(l) + C6H5NH3+(aq)

The solution at neutralization contains an acidic cation (C6H5NH3+); pH < 7.00.

(d) C6H5CO2H(aq) + KOH(aq) H2O(l) + C6H5CO2K(aq) net ionic equation: C6H5CO2H(aq) + OHS(aq) H2O(l) + C6H5CO2

S(aq) The solution at neutralization contains a basic anion (C6H5CO2

S); pH > 7.00. 16.43 (a) HNO2(aq) + CsOH(aq) H2O(l) + CsNO2(aq)

net ionic equation: HNO2(aq) + OHS(aq) H2O(l) + NO2S(aq)

The solution at neutralization contains a basic anion (NO2S); pH > 7.00

(b) HBr(aq) + NH3(aq) NH4Br(aq) net ionic equation: H3O

+(aq) + NH3(aq) H2O(l) + NH4+(aq)

The solution at neutralization contains an acidic cation (NH4+); pH < 7.00

(c) HClO4(aq) + KOH(aq) H2O(l) + KClO4(aq) net ionic equation: H3O

+(aq) + OHS(aq) 2 H2O(l) The solution at neutralization contains a neutral salt (KClO4); pH = 7.00 (d) HOBr(aq) + NH3(aq) NH4OBr(aq) net ionic equation: HOBr(aq) + NH3(aq) NH4

+(aq) + OBrS(aq) The solution at neutralization contains the salt NH4OBr. Ka(NH4

+) = 5.6 x 10S10 and Kb(OBrS) = 5.0 x 10S5; Kb(OBrS) > Ka(NH4+); pH > 7.00

16.44 (a) Strong acid - strong base reaction Kn = _14

w

1 1 =

1.0 x 10K = 1.0 x 1014

(b) Weak acid - strong base reaction Kn = _8

a

_14w

3.5 x 10K = 1.0 x 10K

= 3.5 x 106

(c) Strong acid - weak base reaction Kn = _10

b

_14w

4.3 x 10K = 1.0 x 10K

= 4.3 x 104

(d) Weak acid - strong base reaction Kn = _5

a

_14w

6.5 x 10K = 1.0 x 10K

= 6.5 x 109

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

16.45 (d) Weak acid - strong base reaction Kn = _ 4

a

_14w

4.5 x 10K = 1.0 x 10K

= 4.5 x 1010

(c) Strong acid - weak base reaction Kn = _5

b

_14w

1.8 x 10K = 1.0 x 10K

= 1.8 x 109

(a) Strong acid - strong base reaction Kn = _14

w

1 1 =

1.0 x 10K = 1.0 x 1014

(d) Weak acid - weak base reaction Kn = _9 _5

a b

_14w

(2.0 x )(1.8 x )10 10K K = 1.0 x 10K

=

3.6


446

(d) < (b) < (a) < (c) 16.46 (a) After mixing, the solution contains the basic salt, NaF; pH > 7.00

(b) After mixing, the solution contains the neutral salt, NaCl; pH = 7.00 Solution (a) has the higher pH.

16.47 (a) After mixing, the solution contains the neutral salt, NaClO4; pH = 7.00

(b) After mixing, the solution contains the acidic salt, NH4ClO4; pH < 7.00 Solution (b) has the lower pH.

16.48 Weak acid - weak base reaction Kn = _10 _9

a b w _14

(1.3 x )(1.8 x )10 10 = overKK K1.0 x 10

=

2.3 x 10S5 Kn is small so the neutralization reaction does not proceed very far to completion.

16.49 Weak acid - weak base reaction Kn = _5 _10

a b

_14w

(8.0 x )(4.3 x )10 10K K = 1.0 x 10K

= 3.4

Because Kn is close to 1, there will be an appreciable amount of aniline present at equilibrium.

The CommonSIon Effect

16.50 HNO2(aq) + H2O(l) Ω H3O+(aq) + NO2

S(aq) (a) NaNO2 is a source of NO2

S (reaction product). The equilibrium shifts towards reactants, and the percent dissociation of HNO2 decreases. (c) HCl is a source of H3O

+ (reaction product). The equilibrium shifts towards reactants, and the percent dissociation of HNO2 decreases. (d) Ba(NO2)2 is a source of NO2

S (reaction product). The equilibrium shifts towards reactants, and the percent dissociation of HNO2 decreases.

16.51 NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)

(a) KOH is a strong base, and it increases the [OHS]. The pH increases. (b) NH4NO3 is a source of NH4

+ (reaction product). The equilibrium shifts towards reactants, and the [OHS] decreases. The pH decreases. (c) NH4Br is a source of NH4

+ (reaction product). The equilibrium shifts towards reactants, and the [OHS] decreases. The pH decreases. (d) KBr does not affect the pH of the solution.

16.52 (a) HF(aq) + H2O(l) Ω H3O+(aq) + FS(aq)

LiF is a source of FS (reaction product). The equilibrium shifts toward reactants, and the [H3O

+] decreases. The pH increases. (b) Because HI is a strong acid, addition of KI, a neutral salt, does not change the pH.

(c) NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)

NH4Cl is a source of NH4+ (reaction product). The equilibrium shifts toward reactants,


447

and the [OHS] decreases. The pH decreases.

16.53 (a) NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)

NH4Cl is a source of NH4+ (reaction product). The equilibrium shifts toward reactants,

and the [OHS] decreases. The pH decreases.

(b) HCO3S(aq) + H2O(l) Ω H3O

+(aq) + CO32S(aq)

Na2CO3 is a source of CO32S (reaction product). The equilibrium shifts toward reactants,

and the [H3O+] decreases. The pH increases.

(c) Because NaOH is a strong base, addition of NaClO4, a neutral salt, does not change the pH.

16.54 For 0.25 M HF and 0.10 M NaF

HF(aq) + H2O(l) Ω H3O+(aq) + FS(aq)


Ka = + _

3 _ 4[ ][ ] x(0.10 + x) x(0.10)OH F = 3.5 x = 10[HF] 0.25 _ x 0.25

≈


pH = Slog[H3O+] = Slog(8.8 x 10S4) = 3.06

16.55 On mixing equal volumes of two solutions, both concentrations are cut in half.

[CH3NH2] = 0.10 M; [CH3NH3Cl] = 0.30 M

CH3NH2(aq) + H2O(l) Ω CH3NH3+(aq) + OHS(aq)


Kb = + _

3 3 _ 4

3 2

[ ][ ] (0.30 + x) x (0.30) xCH NH OH = 3.7 x = 10[ ] 0.10 _ x 0.10CH NH

≈


[H3O+] =

_14w

_ _ 4

1.0 x 10K = [ ] 1.2 x OH 10

= 8.1 x 10S11 M

pH = Slog[H3O+] = Slog(8.1 x 10S11) = 10.09

16.56 For 0.10 M HN3:

HN3(aq) + H2O(l) Ω H3O+(aq) + N3

S(aq) initial (M) 0.10 ~0 0 change (M) Sx +x +x equil (M) 0.10 S x x x

Ka = _+ 2 2

3 3 _5

3

[ ][ ]O NH x x = 1.9 x = 10[ ] 0.10 _ x 0.10HN

≈

Solve for x. x = 1.4 x 10S3 M


448

% dissociation = _3

3 diss

3 initial

[ ] 1.4 x MHN 10 x 100% = x 100%[ 0.10 M]HN

= 1.4%

For 0.10 M HN3 in 0.10 M HCl:

HN3(aq) + H2O(l) Ω H3O+(aq) + N3

S(aq) initial (M) 0.10 0.10 0 change (M) Sx +x +x equil (M) 0.10 S x 0.10 + x x

Ka = _+

3 3 _5

3

[ ][ ] (0.10 + x)(x) (0.10)(x)O NH = 1.9 x = = x10[ ] 0.10 _ x 0.10HN

≈

Solve for x. x = 1.9 x 10S5 M

% dissociation = _5

3 diss

3 initial

[ ] 1.9 x MHN 10 x 100% = x 100%[ 0.10 M]HN

= 0.019%

The % dissociation is less because of the common ion (H3O+) effect.

16.57 NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)

initial (M) 0.30 0 ~0 change (M) Sx +x +x equil (M) 0.30 S x x x

Kb = + _ 2 24 _5

3

[ ][ ]NH OH x x = 1.8 x = 10[ ] 0.30 _ x 0.30NH

≈


[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 2.3 x OH 10

= 4.3 x 10S12 M

pH = Slog[H3O+] = Slog(4.3 x 10S12) = 11.37

Add 4.0 g of NH4NO3.

NH4NO3, 80.04 amu; [NH4+] = molarity of NH4NO3 =

1 mol4.0 g x

80.04 g

0.100 L

= 0.50 M

NH3(aq) + H2O(l) Ω NH4+(aq) + OHS(aq)


Kb = + _4 _5

3

[ ][ ] (0.50 + x) x (0.50) xNH OH = 1.8 x = 10[ ] 0.30 _ x 0.30NH

≈


[H3O+] =

_14w

_ _5

1.0 x 10K = [ ] 1.1 x OH 10

= 9.1 x 10S10 M

pH = Slog[H3O+] = Slog(9.1 x 10S10) = 9.04

The % dissociation decreases because of the common ion effect. Buffer Solutions


449

16.58 Solutions (a), (c) and (d) are buffer solutions. Neutralization reactions for (c) and (d)

result in solutions with equal concentrations of HF and FS. 16.59 Solutions (b), (c) and (d) are buffer solutions. Neutralization reactions for (b) and (d)

result in solutions with equal concentrations of NH3 and NH4+.

16.60 Both solutions buffer at the same pH because in both cases the [NO2

S]/[HNO2] = 1. Solution (a), however, has a higher concentration of both HNO2 and NO2

S, and therefore it has the greater buffer capacity.

16.61 Both solutions buffer at the same pH because in both cases the [NH3]/[NH4

+] = 1.5. Solution (b), however, has a higher concentration of both NH3 and NH4

+, and therefore it has the greater buffer capacity.

16.62 When blood absorbs acid, the equilibrium shifts to the left, decreasing the pH, but not by

much because the [HCO3S]/[H2CO3] ratio remains nearly constant. When blood absorbs

base, the equilibrium shifts to the right, increasing the pH, but not by much because the [HCO3

S]/[H 2CO3] ratio remains nearly constant.

16.63 H2PO4S(aq) + H2O(l) Ω H3O

+(aq) + HPO42S(aq)

For H2PO4S, _8

a 2 a 2a 2 = 6.2 x , = _log = 7.21pK10K K

pH = 7.4 = p2_ 2_4 4

a 2 _ _2 24 4

[ ] [ ]HPO HPO + log = 7.21 + logK[ ] [ ]PO POH H

To maintain pH near 7.4, need log 2_4

_2 4

[ ]HPO = 0.19[ ]POH

and 2_4 0.19

_2 4

[ ]HPO = = 1.510[ ]POH

The principal buffer reactions are: H3O

+(aq) + HPO42S(aq) H2PO4

S(aq) + H2O(l) OHS(aq) + H2PO4

S(aq) HPO42S(aq) + H2O(l)

16.64 pH = pKa + log_

a

[base] [ ]CN = + log pK[acid] [HCN]

For HCN, Ka = 4.9 x 10S10, pKa = Slog Ka = Slog(4.9 x 10S10) = 9.31

pH = 9.31 + log0.12

0.20

= 9.09

The pH of a buffer solution will not change on dilution because the acid and base concentrations will change by the same amount and their ratio will remain the same.

16.65 NaHCO3, 84.01 amu; Na2CO3, 105.99 amu

[HCO3S] = molarity of NaHCO3 =

1 mol4.2 g x

84.01 g

0.20 L

= 0.25 M


450

[CO32S] = molarity of Na2CO3 =

1 mol5.3 g x

105.99 g

0.20 L

= 0.25 M

pH = pKa + log [base]

[acid] = pKa + log

2_3

_3

[ ]CO[ ]HCO

For HCO3S, a 2K = 5.6 x 10S11, p a 2 a 2 = _log K K = Slog(5.6 x 10S11) = 10.25

pH = 10.25 + log[0.25]

[0.25] = 10.25

The pH of a buffer solution will not change on dilution because the acid and base concentrations will change by the same amount and their ratio will remain the same.

16.66 pH = pKa + log 3a +

4


For NH4+, Ka = 5.6 x 10S10, pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25

For the buffer: pH = 9.25 + log(0.200)

(0.200) = 9.25

(a) add 0.0050 mol NaOH, [OHS] = 0.0050 mol/0.500 L = 0.010 M

NH4+(aq) + OHS(aq) Ω NH3(aq) + H2O(l)

before reaction (M) 0.200 0.010 0.200 change (M) S0.010 S0.010 +0.010

after reaction (M) 0.200 S 0.010 0 0.200 + 0.010

pH = 9.25 + log 3

+4

[ ]NH[ ]NH

= 9.25 + log(0.200 + 0.010)

(0.200 _ 0.010) = 9.29

(b) add 0.020 mol HCl, [H3O+] = 0.020 mol/0.500 L = 0.040 M

NH3(aq) + H3O+(aq) Ω NH4

+(aq) + H2O(l) before reaction (M) 0.200 0.040 0.200

change (M) S0.040 S0.040 +0.040 after reaction (M) 0.200 S 0.040 0 0.200 + 0.040

pH = 9.25 + log 3

+4

[ ]NH[ ]NH

= 9.25 + log(0.200 _ 0.040)

(0.200 + 0.040) = 9.07

16.67 pH = pKa + log2_3

a _3

[base] [ ]SO = + log pK[acid] [ ]HSO

For HSO3S, Ka = 6.3 x 10S8, pKa = Slog Ka = Slog(6.3 x 10S8) = 7.20

For the buffer: pH = 7.20 + log(0.300)

(0.500) = 6.98

(a) add (0.0050 L)(0.20 mol/L) = 0.0010 mol HCl = 0.0010 mol H3O+

mol HSO3S = (0.300 L)(0.500 mol/L) = 0.150 mol

mol SO32S = (0.300 L)(0.300 mol/L) = 0.0900 mol

SO32S(aq) + H3O

+(aq) Ω HSO3S(aq) + H2O(l)

before reaction (mol) 0.0900 0.0010 0.150


451

change (mol) S0.0010 S0.0010 +0.0010 after reaction (mol) 0.0900 S 0.0010 0 0.150 + 0.0010

pH = 7.20 + log2_3

_3

[ ]SO[ ]HSO

= 7.20 + log(0.0900 _ 0.0010)

(0.150 + 0.0010) = 6.97

(b) add (0.0050 L)(0.10 mol/L) = 0.00050 mol NaOH = 0.00050 mol OHS

HSO3S(aq) + OHS(aq) Ω SO3

2S(aq) + H2O(l) before reaction (mol) 0.150 0.00050 0.0900

change (mol) S0.00050 S0.00050 +0.00050 after reaction (mol) 0.150 S 0.00050 0 0.0900 + 0.00050

pH = 7.20 + log2_3

_3

[ ]SO[ ]HSO

= 7.20 + log(0.0900 + 0.00050)

(0.150 _ 0.00050) = 6.98

16.68 Acid Ka pKa = Slog Ka

(a) H3BO3 5.8 x 10S10 9.24 (b) HCO2H 1.8 x 10S4 3.74 (c) HOCl 3.5 x 10S8 7.46 The stronger the acid (the larger the Ka), the smaller is the pKa.

16.69 (a) Ka = a_pK10 = 10S5.00 = 1.0 x 10S5 (b) Ka = a_pK10 = 10S8.70 = 2.0 x 10S9

(b) is the weaker acid

16.70 pH = pKa + log a

[base] = + log pK

[acid]

_2

2

[ ]HCO[ H]HCO

For HCO2H, Ka = 1.8 x 10S4; pKa = Slog Ka = Slog(1.8 x 10S4) = 3.74

pH = 3.74 + log(0.50)

(0.25) = 4.04

16.71 pH = pKa + log[base]

[acid] = pKa + log

_3

2 3

[ ]HCO[ ]COH

For H2CO3, Ka = 4.3 x 10S7; pKa = Slog Ka = Slog(4.3 x 10S7) = 6.37

7.40 = 6.37 + log_3

2 3

[ ]HCO[ ]COH

; 1.03 = log_3

2 3

[ ]HCO[ ]COH

_3

2 3

[ ]HCO[ ]COH

= 101.03 = 10.7; 2 3_3

[ ]COH[ ]HCO

= 0.093


[acid] = pKa + log 3

+4

[ ]NH[ ]NH

For NH4+, Ka = 5.6 x 10S10; pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25

9.80 = 9.25 + log 3

+4

[ ]NH[ ]NH

; 0.550 = log 3

+4

[ ]NH[ ]NH

; 3

+4

[ ]NH[ ]NH

= 100.55 = 3.5

The volume of the 1.0 M NH3 solution should be 3.5 times the volume of the 1.0 M NH4Cl solution so that the mixture will buffer at pH 9.80.


452


[acid] = pKa + log

_3 2

3 2

[ ]CH CO[ H]CH CO

For CH3CO2H, Ka = 1.8 x 10S5; pKa = Slog Ka = Slog(1.8 x 10S5) = 4.74

4.44 = 4.74 + log_

3 2

3 2

[ ]CH CO[ H]CH CO

; S0.30 = log_

3 2

3 2

[ ]CH CO[ H]CH CO

_3 2

3 2

[ ]CH CO[ H]CH CO

= 10S0.30 = 0.50

The solution should have 0.50 mol of CH3CO2S per mole of CH3CO2H. For example,

you could dissolve 41g of CH3CO2Na in 1.00 L of 1.00 M CH3CO2H. 16.74 H3PO4, _3

a1 a1a1 = 7.5 x ; = _log = 2.12pK10K K

H2PO4S, _8

a 2 a 2a 2 = 6.2 x ; = _log = 7.21pK10K K

HPO42S, _13

a3 a3a3 = 4.8 x ; = _log = 12.32pK10K K

The buffer system of choice for pH 7.00 is (b) H2PO4S S HPO4

2S because the pKa for H2PO4

S (7.21) is closest to 7.00. 16.75 HSO4

S, Ka2 = 1.2 x 10S2; pKa2 = Slog Ka2 = 1.92 HOCl, Ka = 3.5 x 10S8; pKa = Slog Ka = 7.56 C6H5CO2H, Ka = 6.5 x 10S5; pKa = Slog Ka = 4.19 The buffer system of choice for pH = 4.50 is (c) C6H5CO2H - C6H5CO2

S because the pKa for C6H5CO2H (4.19) is closest to 4.50.

pH Titration Curves 16.76 (a) (0.060 L)(0.150 mol/L)(1000 mmol/mol) = 9.00 mmol HNO3

(b) vol NaOH = (9.00 mmol HNO3)3

1 mmol NaOH 1 mL NaOH

1 mmol 0.450 mmol NaOHHNO

= 20.0 mL NaOH

(c) At the equivalence point the solution contains the neutral salt NaNO3. The pH is 7.00.

(d)


453

16.77

mmol NaOH = (50.0 mL)(1.0 mmol/mL) = 50 mmol mmol HCl = mmol NaOH = 50 mmol

vol HCl = (50 mmol)1.0 mL

1.0 mmol

= 50 mL

50 mL of 1.0 M HCl is needed to reach the equivalence point. 16.78 mmol OHS = (20.0 mL)(0.150 mmol/mL) = 3.00 mmol

mmol acid present = mmol OHS added = 3.00 mmol

[acid] = 3.00 mmol

= 0.0500 M60.0 mL

16.79 mmol OHS = (60.0 mL)(0.240 mmol/mL) = 14.4 mmol

mmol acid present = 14.4 mmol OHS x _

1 mmol acid

2 mmol OH = 7.20 mmol acid

[acid] = 7.20 mmol

= 0.288 M25.0 mL

16.80 HBr(aq) + NaOH(aq) Na+(aq) + BrS(aq) + H2O(l)

(a) [H3O+] = 0.120 M; pH = Slog[H3O

+] = Slog (0.120) = 0.92 (b) (50.0 mL)(0.120 mmol/mL) = 6.00 mmol HBr (20.0 mL)(0.240 mmol/mL) = 4.80 mmol NaOH 6.00 mmol HBr S 4.80 mmol NaOH = 1.20 mmol HBr after neutralization

[H3O+] =

1.20 mmol

(50.0 mL + 20.0 mL) = 0.0171 M

pH = Slog[H3O+] = Slog(0.0171) = 1.77

(c) (24.9 mL)(0.240 mmol/mL) = 5.98 mmol NaOH 6.00 mmol HBr S 5.98 mmol NaOH = 0.02 mmol HBr after neutralization

[H3O+] =

0.02 mmol

(50.0 mL + 24.9 mL) = 3 x 10S4 M

pH = Slog[H3O+] = Slog(3 x 10S4) = 3.5


454

(d) The titration reaches the equivalence point when 25.0 mL of 0.240 M NaOH is added. At the equivalence point the solution contains the neutral salt NaBr. The pH is 7.00. (e) (25.1 mL)(0.240 mmol/mL) = 6.024 mmol NaOH 6.024 mmol NaOH S 6.00 mmol HBr = 0.024 mmol NaOH after neutralization

[OHS] = 0.024 mmol

(50.0 mL + 25.1 mL) = 3.2 x 10S4 M

[H3O+] =

_14w

_ _ 4

1.0 x 10K = [ ] 3.2 x OH 10

= 3.1 x 10S11 M

pH = Slog[H3O+] = Slog(3.1 x 10S11) = 10.5

(f) (40.0 mL)(0.240 mmol/mL) = 9.60 mmol NaOH 9.60 mmol NaOH S 6.00 mmol HBr = 3.60 mmol NaOH after neutralization

[OHS] = 3.60 mmol

(50.0 mL + 40.0 mL) = 0.040 M

[H3O+] =

_14w

_

1.0 x 10K = [ ] 0.040OH

= 2.5 x 10S13 M

pH = Slog[H3O+] = Slog(2.5 x 10S13) = 12.60

16.81 Ba(OH)2(aq) + 2 HNO3(aq) Ba2+(aq) + 2 NO3S(aq) + 2 H2O(l)

(a) [OHS] = 2(0.150 M) = 0.300 M

[H3O+] =

_14w

_

1.0 x 10K = [ ] 0.300OH

= 3.33 x 10S14 M

pH = Slog[H3O+] = Slog(3.33 x 10S14) = 13.48

(b) (40.0 mL)(0.150 mmol/mL) = 6.00 mmol Ba(OH)2

6.00 mmol Ba(OH)2 x _

2

2 mmol OH1 mmol Ba(OH)

= 12.0 mmol OHS

(10.0 mL)(0.400 mmol/mL) = 4.00 mmol HNO3 12.0 mmol OHS S 4.00 mmol HNO3 = 8.00 mmol OHS after neutralization

[OHS] = 8.00 mmol

(40.0 mL + 10.0 mL) = 0.160 M


455

[H3O+] =

_14w

_

1.0 x 10K = [ ] 0.160OH

= 6.25 x 10S14 M

pH = Slog[H3O+] = Slog(6.25 x 10S14) = 13.20

(c) (20.0 mL)(0.400 mmol/mL) = 8.00 mmol HNO3 12.0 mmol OHS S 8.00 mmol HNO3 = 4.00 mmol OHS after neutralization

[OHS] = 4.00 mmol

(40.0 mL + 20.0 mL) = 0.0667 M

[H3O+] =

_14w

_

1.0 x 10K = [ ] 0.0667OH

= 1.50 x 10S13 M

pH = Slog[H3O+] = Slog(1.50 x 10S13) = 12.82

(d) The titration reaches the equivalence point when 30.0 mL of 0.400 M HNO3 is added. At the equivalence point the solution contains the neutral salt Ba(NO3)2. The pH is 7.00. (e) (40.0 mL)(0.400 mmol/mL) = 16.0 mmol HNO3 16.0 mmol HNO3 S 12.0 mmol OHS = 4.00 mmol H3O

+ after neutralization

[H3O+] =

4.00 mmol

(40.0 mL + 40.0 mL) = 0.0500 M

pH = Slog[H3O+] = Slog(0.0500) = 1.30

16.82 mmol HF = (40.0 mL)(0.250 mmol/mL) = 10.0 mmol

mmol NaOH required = mmol HF = 10.0 mmol

mL NaOH required = (10.0 mmol)1.00 mL

= 50.0 mL0.200 mmol


456

50.0 mL of 0.200 M NaOH is required to reach the equivalence point. For HF, Ka = 3.5 x 10S4; pKa = Slog Ka = Slog(3.5 x 10S4) = 3.46 (a) mmol HF = 10.0 mmol mmol NaOH = (0.200 mmol/mL)(10.0 mL) = 2.00 mmol Neutralization reaction: HF(aq) + OHS(aq) FS(aq) + H2O(l) before reaction (mmol) 10.0 2.00 0 change (mmol) S2.00 S2.00 +2.00 after reaction (mmol) 8.0 0 2.00

[HF] = 8.0 mmol

(40.0 mL + 10.0 mL) = 0.16 M; [FS] =

2.00 mmol

(40.0 mL + 10.0 mL) = 0.0400 M

HF(aq) + H2O(l) Ω H3O+(aq) + FS(aq)


Ka = + _

3 _ 4[ ][ ] x(0.0400 + x) x(0.0400)OH F = 3.5 x = 10[HF] 0.16 _ x 0.16

≈


pH = Slog[H3O+] = Slog(1.4 x 10S3) = 2.85

(b) Halfway to the equivalence point, pH = pKa = Slog Ka = Slog(3.5 x 10S4) = 3.46 (c) At the equivalence point only the salt NaF is in solution.

[FS] = 10.0 mmol

(40.0 mL + 50.0 mL) = 0.111 M

FS(aq) + H2O(l) Ω HF(aq) + OHS(aq) initial (M) 0.111 0 ~0 change (M) Sx +x +x equil (M) 0.111 S x x x

For FS, Kb = _14

w

_ 4a

1.0 x 10K = for HF 3.5 x 10K

= 2.9 x 10S11

Kb = _ 2 2

_11_

[HF][ ]OH x x = 2.9 x = 10[ ] 0.111 _ x 0.111F

≈


[H3O+] =

_14w

_ _ 6

1.0 x 10K = [ ] 1.8 x OH 10

= 5.6 x 10S9 M

pH = Slog[H3O+] = Slog(5.6 x 10S9) = 8.25

(d) mmol HF = 10.0 mmol mol NaOH = (0.200 mmol/mL)(80.0 mL) = 16.0 mmol

Neutralization reaction: HF(aq) + OHS(aq) FS(aq) + H2O(l) before reaction (mmol) 10.0 16.0 0 change (mmol) S10.0 S10.0 +10.0 after reaction (mmol) 0 6.0 10.0 After the equivalence point, the pH of the solution is determined by the [OHS].


457

[OHS] = 6.0 mmol

(40.0 mL + 80.0 mL) = 5.0 x 10S2 M

[H3O+] =

_14w

_ _ 2

1.0 x 10K = [ ] 5.0 x OH 10

= 2.0 x 10S13 M

pH = Slog[H3O+] = Slog(2.0 x 10S13) = 12.70

16.83 mmol CH3NH2 = (100.0 mL)(0.100 mmol/mL) = 10.0 mmol

mmol HNO3 required = mmol CH3NH2 = 10.0 mmol

vol HNO3 required = (10.0 mmol)1.00 mL

= 40.0 mL0.250 mmol

40.0 mL of 0.250 M HNO3 are required to reach the equivalence point.

(a) CH3NH2(aq) + H2O(l) Ω CH3NH3+(aq) + OHS(aq)


Kb = + _ 2

3 3 _ 4

3 2

[ ][ ]CH NH OH x = 3.7 x = 10[ ] 0.100 _ xCH NH

x2 + (3.7 x 10S4)x S (3.7 x 10S5) = 0 Use the quadratic formula to solve for x.

x = 2_ 4 _ 4 _5 _ 4_ (3.7 x ) (3.7 x _ (4)(_ 3.7 x )) _ 3.7 x 0.012210 10 10 10 =

2(1) 2

± ±

x = 0.0059 and S0.0063 Of the two solutions for x, only the positive value of x has physical meaning because x is the [OHS]. [OHS] = x = 0.0059 M

[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 5.9 x OH 10

= 1.7 x 10S12 M

pH = Slog[H3O+] = Slog(1.7 x 10S12) = 11.77

(b) 20.0 mL of HNO3 is halfway to the equivalence point.

For CH3NH3+, Ka =

_14w

_ 4b 3 2

1.0 x 10K = for 3.7 x CH NH 10K

= 2.7 x 10S11

pH = pKa = Slog(2.7 x 10S11) = 10.57 (c) At the equivalence point only the salt CH3NH3NO3 is in solution. mmol CH3NH3NO3 = (0.100 mmol/mL)(100.0 mL) = 10.0 mmol

[CH3NH3+] =

10.0 mmol

(100.0 mL + 40.0 mL) = 0.0714 M

CH3NH3+(aq) + H2O(l) Ω H3O

+(aq) + CH3NH2(aq) initial (M) 0.0714 ~0 0 change (M) Sx +x +x equil (M) 0.0714 S x x x


458

Ka = + 2 2

3 3 2 _11+

3 3

[ ][ ]O CH NHH x x = 2.7 x = 10[ ] 0.0714 _ x 0.0714CH NH

≈


pH = Slog[H3O+] = Slog(1.4 x 10S6) = 5.85

(d) mmol CH3NH2 = (0.100 mmol/mL)(100.0 mL) = 10.0 mmol mmol HNO3 = (0.250 mmol/mL)(60.0 mL) = 15.0 mmol Neutralization reaction: CH3NH2(aq) + H3O

+(aq) CH3NH3+(aq) + H2O(l)

before reaction (mmol) 10.0 15.0 0 change (mmol) S10.0 S10.0 +10.0 after reaction (mmol) 0 5.0 10.0 After the equivalence point the pH of the solution is determined by the [H3O

+].

[H3O+] =

5.0 mmol

(100.0 mL + 60.0 mL) = 3.1 x 10S2 M

pH = Slog[H3O+] = Slog(3.1 x 10S2) = 1.51

16.84 For H2A

+, Ka1 = 4.6 x 10S3 and Ka2 = 2.0 x 10S10 (a) (10.0 mL)(0.100 mmol/mL) = 1.00 mmol NaOH added = 1.00 mmol HA produced. (50.0 mL)(0.100 mmol/mL) = 5.00 mmol H2A

+ 5.00 mmol H2A

+ S 1.00 mmol NaOH = 4.00 mmol H2A+ after neutralization

[H2A+] =

4.00 mmol

(50.0 mL + 10.0 mL) = 6.67 x 10S2 M

[HA] = 1.00 mmol

(50.0 mL + 10.0 mL) = 1.67 x 10S2 M

pH = pKa1 + log+

2

[HA]

[ ]H A = Slog(4.6 x 10S3) + log

_ 2

_ 2

1.67 x 106.67 x 10

= 1.74

(b) Halfway to the first equivalence point, pH = pKa1 = 2.34

(c) At the first equivalence point, pH = a1 a 2 + pK pK

2 = 6.02

(d) Halfway between the first and second equivalence points, pH = pKa2 = 9.70 (e) At the second equivalence point only the basic salt, NaA, is in solution.

Kb = _14

w w

_10a a

1.0 x 10K K = = for HA 2 2.0 x 10K K

= 5.0 x 10S5

mmol AS = (50.0 mL)(0.100 mmol/mL) = 5.00 mmol

[A S] = 5.0 mmol

(50.0 mL + 100.0 mL) = 3.3 x 10S2 M

AS(aq) + H2O(l) Ω HA(aq) + OHS(aq) initial (M) 0.033 0 ~0 change (M) Sx +x +x equil (M) 0.033 S x x x

Kb = _ 2

_5_

[HA][ ] (x)(x)OH x = 5.0 x = 10[ ] 0.033 _ x 0.033A

≈

Solve for x.


459

x = [OHS] = _5(5.0 x )(0.033)10 = 1.3 x 10S3 M

[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 1.3 x OH 10

= 7.7 x 10S12 M

pH = Slog[H3O+] = Slog(7.7 x 10S12) = 11.11

16.85 For H2CO3, Ka1 = 4.3 x 10S7 and Ka2 = 5.6 x 10S11

(a) (25.0 mL)(0.0200 mmol/mL) = 0.500 mmol H2CO3 (10.0 mL)(0.0250 mmol/mL) = 0.250 mmol KOH added 0.500 mmol H2CO3 S 0.250 mmol KOH = 0.250 mmol HCO3

S produced This is halfway to the first equivalence point where pH = pKa1 = Slog(4.3 x 10S7) = 6.37

(b) At the first equivalence point, pH = a1 a 2 + pK pK

2 = 8.31

(c) Halfway between the first and second equivalence points, pH = pKa2 = 10.25 (d) At the second equivalence point only the basic salt, K2CO3, is in solution.

Kb = _14

w w_ _11

a a3

1.0 x 10K K = = for 2 5.6 x HCO 10K K

= 1.8 x 10S4

mmol CO32S = mmol H2CO3 = 0.500 mmol

[CO32S] =

0.500 mmol

(25.0 mL + 40.0 mL) = 0.00769 M

CO32S(aq) + H2O(l) Ω HCO3

S(aq) + OHS(aq) initial (M) 0.00769 0 ~0 change (M) Sx +x +x equil (M) 0.00769 S x x x

Kb = _ _3 _ 4

2_3

[ ][ ] (x)(x)HCO OH = 1.8 x = 10[ ] 0.00769 _ xCO

Use the quadratic formula to solve for x. x =

2_ 4 _ 4 _6 _ 4 _3_ (1.8 x ) (1.8 x _ (4)(1)(_ 1.4 x )) (_1.8 x ) (2.37 x )10 10 10 10 10 = 2(1) 2

± ±

x = S1.27 x 10S3 and 1.09 x 10S3 Of the two solutions for x, only the positive value of x has physical meaning because x is the [OHS]. [OHS] = x = 1.09 x 10S3 M

[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 1.09 x OH 10

= 9.2 x 10S12 M

pH = Slog[H3O+] = Slog(9.2 x 10S12) = 11.04

(e) excess KOH (50.0 mL S 40.0 mL)(0.025 mmol/mL) = 0.250 mmol KOH = 0.250 mmol OHS

[OHS] = 0.250 mmol

(25.0 mL + 50.0 mL) = 3.33 x 10S3 M


460

[H3O+] =

_14w

_ _3

1.0 x 10K = [ ] 3.33 x OH 10

= 3.0 x 10S12 M

pH = Slog[H3O+] = Slog(3.0 x 10S12) = 11.52

16.86 When equal volumes of acid and base react, all concentrations are cut in half.

(a) At the equivalence point only the salt NaNO2 is in solution. [NO2

S] = 0.050 M

For NO2S, Kb =

_14w

_ 4a 2

1.0 x 10K = for 4.5 x HNO 10K

= 2.2 x 10S11

NO2S(aq) + H2O(l) Ω HNO2(aq) + OHS(aq)

Initial (M) 0.050 0 ~0 change (M) Sx +x +x equil (M) 0.050 S x x x

Kb = _ 2

2 _11_2

[ ][ ] (x)(x)HNO OH x = 2.2 x = 10[ ] 0.050 _ x 0.050NO

≈


[H3O+] =

_14w

_ _ 6

1.0 x 10K = [ ] 1.1 x OH 10

= 9.1 x 10S9 M

pH = Slog[H3O+] = Slog(9.1 x 10S9) = 8.04

Phenol red would be a suitable indicator. (see Figure 15.4) (b) The pH is 7.00 at the equivalence point for the titration of a strong acid (HI) with a strong base (NaOH). Bromthymol blue or phenol red would be suitable indicators. (Any indicator that changes color in the pH range 4 S 10 is satisfactory for a strong acid S strong base titration.) (c) At the equivalence point only the salt CH3NH3Cl is in solution. [CH3NH3

+] = 0.050 M

For CH3NH3+, Ka =

_14w

_ 4b 3 2

1.0 x 10K = for 3.7 x CH NH 10K

= 2.7 x 10S11

CH3NH3+(aq) + H2O(l) Ω H3O

+(aq) + CH3NH2(aq) initial (M) 0.050 ~0 0 change (M) Sx +x +x equil (M) 0.050 S x x x

Ka = + 2

3 3 2 _11+

3 3

[ ][ ] (x)(x)O CH NHH x = 2.7 x = 10[ ] 0.050 _ x 0.050CH NH

≈


pH = Slog[H3O+] = Slog(1.2 x 10S6) = 5.92

Chlorphenol red would be a suitable indicator. 16.87 When equal volumes of acid and base react, all concentrations are cut in half.

(a) At the equivalence point only the salt C5H11NHNO3 is in solution. [C5H11NH+] = 0.10 M


461

For C5H11NH+, Ka = _14

w

_3b 115

1.0 x 10K = for N 1.3 x C 10K H

= 7.7 x 10S12

C5H11NH+(aq) + H2O(l) Ω H3O+(aq) + C5H11N(aq)

initial (M) 0.10 ~0 0 change (M) Sx +x +x equil (M) 0.10 S x x x

Ka = + 2

3 115 _12+

115

[ ][ N] (x)(x)O CH H x = 7.7 x = 10[ ] 0.10 _ x 0.10C NHH

≈


pH = Slog[H3O+] = Slog(8.8 x 10S7) = 6.06

Alizarin would be a suitable indicator (see Figure 15.4) (b) At the equivalence point only the salt Na2SO3 is in solution. [SO3

2S] = 0.10 M

For SO32S, Kb =

_14w

_ _8a 3

1.0 x 10K = for 6.3 x HSO 10K

= 1.6 x 10S7

SO32S(aq) + H2O(l) Ω HSO3

S(aq) + OHS(aq) Initial (M) 0.10 0 ~0 change (M) Sx +x +x equil (M) 0.10 S x x x

Kb = _ _ 23 _7

2_3

[ ][ ] (x)(x)HSO OH x = 1.6 x = 10[ ] 0.10 _ x 0.10SO

≈


[H3O+] =

_14w

_ _ 4

1.0 x 10K = [ ] 1.26 x OH 10

= 7.9 x 10S11 M

pH = Slog[H3O+] = Slog(7.9 x 10S11) = 10.10

Thymolphthalein would be a suitable indicator. (c) The pH is 7.00 at the equivalence point for the titration of a strong acid (HBr) with a strong base (Ba(OH)2). Alizarin, bromthymol blue, or phenol red would be suitable indicators. (Any indicator that changes color in the pH range 4 - 10 is satisfactory for a strong acid - strong base titration.)

Solubility Equilibria

16.88 (a) Ag2CO3(s) Ω 2 Ag+(aq) + CO32S(aq) Ksp = [Ag+]2[CO3

2S]

(b) PbCrO4(s) Ω Pb2+(aq) + CrO42S(aq) Ksp = [Pb2+][CrO4

2S]

(c) Al(OH)3(s) Ω Al3+(aq) + 3 OHS(aq) Ksp = [Al3+][OHS]3

(d) Hg2Cl2(s) Ω Hg22+(aq) + 2 ClS(aq) Ksp = [Hg2

2+][Cl S]2 16.89 (a) Ksp = [Ca2+][OHS]2 (b) Ksp = [Ag+]3[PO4

3S] (c) Ksp = [Ba2+][CO3

2S] (d) Ksp = [Ca2+]5[PO43S]3[OHS]

16.90 (a) Ksp = [Pb2+][I S]2 = (5.0 x 10S3)(1.3 x 10S3)2 = 8.4 x 10S9


462

(b) [IS] = _9

sp

2+ _ 4

(8.4 x K 10 = [ ] (2.5 x )Pb 10

= 5.8 x 10S3 M

(c) [Pb2+] = _9

sp

2 2_ _ 4

(8.4 x )K 10 = [ (2.5 x ] )I 10

= 0.13 M

16.91 (a) Ksp = [Ca2+]3[PO4

2S]2 = (2.9 x 10S7)3(2.9 x 10S7)2 = 2.1 x 10S33

(b) [Ca2+] = _33

sp3 3

2 22_4

2.1 x K 10 = [ (0.010] )PO

= 2.8 x 10S10 M

(c) [PO42S] =

_33sp

3 32+

2.1 x K 10 = [ (0.010] )Ca

= 4.6 x 10S14 M

16.92 Ag2CO3(s) Ω 2 Ag+(aq) + CO3

2S(aq) equil (M) 2x x [Ag+] = 2x = 2.56 x 10S4 M; [CO3

2S] = x = (2.56 x 10S4 M)/2 = 1.28 x 10S4 M Ksp = [Ag+]2[CO3

2S] = (2.56 x 10S4)2(1.28 x 10S4) = 8.39 x 10S12 16.93 (a) [Cd2+] = [CO3

2S] = 2.5 x 10S6 M Ksp = [Cd2+][CO3

2S] = (2.5 x 10S6)2 = 6.2 x 10S12 (b) [Ca2+] = 1.06 x 10S2 M [OHS] = 2[Ca2+] = 2(1.06 x 10S2 M) = 2.12 x 10S2 M Ksp = [Ca2+][OHS]2 = (1.06 x 10S2)(2.12 x 10S2)2 = 4.76 x 10S6 (c) PbBr2, 367.01 amu

[Pb2+] = molarity of PbBr2 =

1 mol4.34 g x

367.01 g

1 L

= 1.18 x 10S2 M

[BrS] = 2[Pb2+] = 2(1.18 x 10S2 M) = 2.36 x 10S2 M Ksp = [Pb2+][Br S]2 = (1.18 x 10S2)(2.36 x 10S2)2 = 6.57 x 10S6 (d) BaCrO4, 253.32 amu

[Ba2+] = [CrO42S] = molarity of BaCrO4 =

-3 1 mol2.8 x g x 10

253.32 g

1 L

= 1.1 x 10S5 M

Ksp = [Ba2+][CrO42S] = (1.1 x 10S5)2 = 1.2 x 10S10

16.94 (a) BaCrO4(s) Ω Ba2+(aq) + CrO42S(aq)

equil (M) x x Ksp = [Ba2+][CrO4

2S] = 1.2 x 10S10 = (x)(x)

molar solubility = x = _101.2 x 10 = 1.1 x 10S5 M

(b) Mg(OH)2(s) Ω Mg2+(aq) + 2 OHS(aq) equil (M) x 2x

Ksp = [Mg2+][OHS]2 = 5.6 x 10S12 = x(2x)2 = 4x3


463


35.6 x 10

4 = 1.1 x 10S4 M

(c) Ag2SO3(s) Ω 2 Ag+(aq) + SO32S(aq)

equil (M) 2x x Ksp = [Ag+]2[SO3

2S] = 1.5 x 10S14 = (2x)2x = 4x3


31.5 x 10

4 = 1.6 x 10S5 M

16.95 (a) Ag2CO3(s) Ω 2 Ag+(aq) + CO32S(aq)

equil (M) 2x x Ksp = [Ag+]2[CO3

2S] = 8.4 x 10S12 = (2x)2(x) = 4x3


38.4 x 10

4 = 1.3 x 10S4 M

Ag2CO3, 275.75 amu solubility = (1.3 x 10S4 mol/L)(275.75 g/mol) = 0.036 g/L

(b) CuBr(s) Ω Cu+(aq) + BrS(aq) equil (M) x x

Ksp = [Cu+][Br S] = 6.3 x 10S9 = (x)(x)

molar solubility = x = -96.3 x 10 = 7.9 x 10S5 M CuBr, 143.45 amu solubility = (7.9 x 10S5 mol/L)(143.45 g/mol) = 0.011 g/L

(c) Cu3(PO4)2(s) Ω 3 Cu2+(aq) + 2 PO43S(aq)

equil (M) 3x 2x Ksp = [Cu2+]3[PO4

3S]2 = 1.4 x 10S37 = (3x)3(2x)2 = 108x5


51.4 x 10

108 = 1.7 x 10S8

Cu3(PO4)2 , 380.58 amu solubility = (1.7 x 10S8 mol/L)(380.58 g/mol) = 6.5 x 10S6 g/L

Factors That Affect Solubility 16.96 Ag2CO3(s) Ω 2 Ag+(aq) + CO3

2S(aq) (a) AgNO3, source of Ag+; equilibrium shifts left (b) HNO3, source of H3O

+, removes CO32S; equilibrium shifts right

(c) Na2CO3, source of CO32S; equilibrium shifts left

(d) NH3, forms Ag(NH3)2+; removes Ag+; equilibrium shifts right

16.97 BaF2(s) Ω Ba2+(aq) + 2 FS(aq)

(a) H+ from HCl reacts with FS forming the weak acid HF. The equilibrium shifts to the right increasing the solubility of BaF2. (b) KF, source of FS; equilibrium shifts left, solubility of BaF2 decreases. (c) No change in solubility. (d) Ba(NO3)2, source of Ba2+; equilibrium shifts left, solubility of BaF2 decreases.


464

16.98 (a) PbCrO4(s) Ω Pb2+(aq) + CrO42S(aq)

equil (M) x x Ksp = [Pb2+][CrO4

2S] = 2.8 x 10S13 = (x)(x)


(b) PbCrO4(s) Ω Pb2+(aq) + CrO42S(aq)

initial(M) 0 1.0 x 10S3 equil (M) x 1.0 x 10S3 + x

Ksp = [Pb2+][CrO42S] = 1.2 x 10S10 = (x)(1.0 x 10S3 + x) . (x)(1.0 x 10S3)


_3

2.8 x 101 x 10

= 2.8 x 10S10 M

16.99 (a) SrF2(s) Ω Sr2+(aq) + 2 FS(aq) initial (M) 0.010 0 equil (M) 0.010 + x 2x

Ksp = [Sr2+][FS]2 = 4.3 x 10S9 = (0.010 + x)(2x)2 . (0.010)(2x)2 = 0.040 x2

molar solubility = x = _94.3 x 10

0.040 = 3.3 x 10S4 M

(b) SrF2(s) Ω Sr2+(aq) + 2 FS(aq) initial (M) 0 0.010 equil (M) x 0.010 + 2x

Ksp = [Sr2+][FS]2 = 4.3 x 10S9 = (x)(0.010 + 2x)2 . (x)(0.010)2 = x(0.00010)

molar solubility = x = _94.3 x 10

0.00010 = 4.3 x 10S5 M

16.100 (b), (c), and (d) are more soluble in acidic solution.

(a) AgBr(s) Ω Ag+(aq) + BrS(aq)

(b) CaCO3(s) + H3O+(aq) Ω Ca2+(aq) + HCO3

S(aq) + H2O(l)

(c) Ni(OH)2(s) + 2 H3O+(aq) Ω Ni2+(aq) + 4 H2O(l)

(d) Ca3(PO4)2(s) + 2 H3O+(aq) Ω 3 Ca2+(aq) + 2 HPO4

2S(aq) + 2 H2O(l) 16.101 (a), (b), and (d) are more soluble in acidic solution.

(a) MnS(s) + 2 H3O+(aq) Ω Mn2+(aq) + H2S(aq) + 2 H2O(l)

(b) Fe(OH)3(s) + 3 H3O+(aq) Ω Fe3+(aq) + 6 H2O(l)

(c) AgCl(s) Ω Ag+(aq) + ClS(aq)

(d) BaCO3(s) + H3O+(aq) Ω Ba2+(aq) + HCO3

S(aq) + H2O(l) 16.102 On mixing equal volumes of two solutions, the concentrations of both solutions are cut in half.

Ag+(aq) + 2 CNS(aq) Ω Ag(CN)2S(aq)

before reaction (M) 0.0010 0.10 0 assume 100% reaction S0.0010 S2(0.0010) 0.0010


465

after reaction (M) 0 0.098 0.0010 assume small back rxn +x +2x Sx

equil (M) x 0.098 + 2x 0.0010 S x

Kf = 3.0 x 1020 = _2

+ 2 22_

[Ag(CN ]) (0.0010 _ x) 0.0010 =

[ ][ x(0.098 + 2x x(0.098Ag ) )]CN≈

Solve for x. x = [Ag+] = 3.5 x 10S22 M

16.103 Cr3+(aq) + 4 OHS(aq) Ω Cr(OH)4S(aq)

before reaction (M) 0.0050 1.0 0 assume 100% reaction S0.0050 S(4)(0.0050) +0.0050 after reaction(M) 0 0.98 0.0050 assume small back rxn +x +4x Sx

equil (M) x 0.98 + 4x 0.0050 S x

Kf = _

2944 4 43+ _

[Cr(OH ]) (0.0050 _ x) (0.0050) = 8 x = 10

[ ][ (x)(0.98 + 4 x (x)(0.98] ) )Cr OH≈

Solve for x. x = [Cr3+] = 6.8 x 10S33 M = 7 x 10S33 M

fraction uncomplexed Cr3+ = 3+ _33

_4

[ ] 7 x MCr 10 = 0.0050 M[Cr(OH ])

= 1.4 x 10S30 = 1 x 10S30

16.104 (a) AgI(s) Ω Ag+(aq) + IS(aq) Ksp = 8.5 x 10S17 Ag+(aq) + 2 CNS(aq) Ag(CN)2

S(aq) Kf = 3.0 x 1020

dissolution rxn AgI(s) + 2 CNS(aq) Ω Ag(CN)2S(aq) + IS(aq)

K = (Ksp)(Kf) = (8.5 x 10S17)(3.0 x 1020) = 2.6 x 104

(b) Al(OH)3(s) Ω Al3+(aq) + 3 OHS(aq) Ksp = 1.9 x 10S33 Al3+(aq) + 4 OHS(aq) Al(OH)4

S(aq) Kf = 3 x 1033

dissolution rxn Al(OH)3(s) + OHS(aq) Ω Al(OH)4S(aq)

K = (Ksp)(Kf) = (1.9 x 10S33)(3 x 1033) = 6

(c) Zn(OH)2(s) Ω Zn2+(aq) + 2 OHS(aq) Ksp = 4.1 x 10S17 Zn2+(aq) + 4 NH3(aq) Zn(NH3)4

2+(aq) Kf = 7.8 x 108

dissolution rxn Zn(OH)2(s) + 4 NH3(aq) Ω Zn(NH3)42+ + 2 OHS(aq)

K = (Ksp)(Kf) = (4.1 x 10S17)(7.8 x 108) = 3.2 x 10S8

16.105 (a) Zn(OH)2(s) Ω Zn2+(aq) + 2 OHS(aq) Ksp = 4.1 x 10S17 Zn2+(aq) + 4 OHS(aq) Zn(OH)4

2S(aq) Kf = 3 x 1015

dissolution rxn Zn(OH)2(s) + 2 OHS(aq) Ω Zn(OH)42S(aq)

K = (Ksp)(Kf) = (4.1 x 10S17)(3 x 1015) = 0.1

(b) Cu(OH)2(s) Ω Cu2+(aq) + 2 OHS(aq) Ksp = 1.6 x 10S19 Cu2+(aq) + 4 NH3(aq) Cu(NH3)4

2+(aq) Kf = 5.6 x 1011

dissolution rxn Cu(OH)2(s) + 4 NH3(aq) Ω Cu(NH3)42+ + 2 OHS(aq)


466

K = (Ksp)(Kf) = (1.6 x 10S19)(5.6 x 1011) = 9.0 x 10S8

(c) AgBr(s) Ω Ag+(aq) + BrS(aq) Ksp = 5.4 x 10S13 Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq) Kf = 1.7 x 107

dissolution rxn AgBr(s) + 2 NH3(aq) Ω Ag(NH3)2+(aq) + BrS(aq)

K = (Ksp)(Kf) = (5.4 x 10S13)(1.7 x 107) = 9.2 x 10S6

16.106 (a) AgI(s) Ω Ag+(aq) + IS(aq) equil (M) x x

Ksp = [Ag+][I S] = 8.5 x 10S17 = (x)(x)


(b) AgI(s) + 2 CNS(aq) Ω Ag(CN)2S(aq) + IS(aq)

initial (M) 0.10 0 0 change (M) S2x +x +x equil (M) 0.10 S 2x x x

K = (Ksp)(Kf) = (8.5 x 10S17)(3.0 x 1020) = 2.6 x 104

K = 2.6 x 104 = _ _ 222 2_

[Ag(CN ][ ]) I x = [ (0.10 _ 2 x] )CN

Take the square root of both sides and solve for x. molar solubility = x = 0.050 M

16.107 Cr(OH)3(s) + OHS(aq) Ω Cr(OH)4

S(aq) initial (M) 0.50 0 change (M) Sx +x equil (M) 0.50 S x x

K = (Ksp)(Kf) = (6.7 x 10S31)(8 x 1029) = 0.54

K = 0.54 = _4

_

[Cr(OH ]) x =

[ ] 0.50 _ xOH

0.27 S 0.54x = x 0.27 = 1.54x

molar solubility = x = 0.27

1.54= 0.2 M

Precipitation; Qualitative Analysis 16.108 For BaSO4, Ksp = 1.1 x 10S10

Total volume = 300 mL + 100 mL = 400 mL

[Ba2+] = _3(4.0 x M)(100 mL)10

(400 mL) = 1.0 x 10S3 M

[SO42S] =

_ 4(6.0 x M)(300 mL)10(400 mL)

= 4.5 x 10S4 M

IP = [Ba2+] t[SO42S] t = (1.0 x 10S3)(4.5 x 10S4) = 4.5 x 10S7

IP > Ksp; BaSO4(s) will precipitate.


467

16.109 On mixing equal volumes of two solutions, the concentrations of both solutions are cut in

half. For PbCl2, Ksp = 1.2 x 10S5 = [Pb2+][Cl S]2 IP = (0.0050)(0.0050)2 = 1.2 x 10S7 IP < Ksp; no precipitate will form.

[ClS] = _5

sp

2+ _3

1.2 x K 10 = [ ] 5.0 x Pb 10

= 0.049 M

A [ClS] just greater than 0.049 M will result in precipitation. 16.110 BaSO4, Ksp = 1.1 x 10S10; Fe(OH)3, Ksp = 2.6 x 10S39

Total volume = 80 mL + 20 mL = 100 mL

[Ba2+] = _5(1.0 x M)(80 mL)10

(100 mL) = 8.0 x 10S6 M

[OHS] = 2[Ba2+] = 2(8.0 x 10S6) = 1.6 x 10S5 M

[Fe3+] = _52(1.0 x M)(20 mL)10

(100 mL) = 4.0 x 10S6 M

[SO42S] =

_53(1.0 x M)(20 mL)10(100 mL)

= 6.0 x 10S6 M

For BaSO4, IP = [Ba2+] t[SO42S] t = (8.0 x 10S6)(6.0 x 10S6) = 4.8 x 10S11

IP < Ksp; BaSO4 will not precipitate. For Fe(OH)3, IP = [Fe3+] t[OHS] t

3 = (4.0 x 10S6)(1.6 x 10S5)3 = 1.6 x 10S20 IP > Ksp; Fe(OH)3(s) will precipitate.

16.111 (a) [CO32S] =

_3(2.0 x M)(0.10 mL)10(250 mL)

= 8.0 x 10S7 M

Ksp = 5.0 x 10S9 = [Ca2+][CO32S]

IP = [Ca2+][CO32S] = (8.0 x 10S4)(8.0 x 10S7) = 6.4 x 10S10

IP < Ksp; no precipitate will form. (b) Na2CO3, 106 amu; 10 mg = 0.010 g

[CO32S] =

1 mol0.010 g x

106 g

0.250 L

= 3.8 x 10S4 M

IP = [Ca2+][CO32S] = (8.0 x 10S4)(3.8 x 10S4) = 3.0 x 10S7

IP > Ksp; CaCO3(s) will precipitate. 16.112 pH = 10.80; [H3O

+] = 10SpH = 10S10.80 = 1.6 x 10S11 M

[OHS] = _14

w

+ _113

1.0 x 10K = [ ] 1.6 x O 10H

= 6.2 x 10S4 M

For Mg(OH)2, Ksp = 5.6 x 10S12 IP = [Mg2+] t[OHS] t

2 = (2.5 x 10S4)(6.2 x 10S4)2 = 9.6 x 10S11 IP > Ksp; Mg(OH)2(s) will precipitate


468

16.113 Mg(OH)2, Ksp = 5.6 x 10S12; Al(OH)3, Ksp = 1.9 x 10S33

pH = 8; [H3O+] = 10SpH = 10S8 = 1 x 10S8 M

[OHS] = _14

w

+ _83

1.0 x 10K = [ ] 1 x O 10H

= 1 x 10S6 M

For Mg(OH)2, IP = [Mg2+][OHS]2 = (0.01)(1 x 10S6)2 = 1 x 10S14 IP < Ksp; no Mg(OH)2 will precipitate. For Al(OH)3, IP = [Al3+][OHS]3 = (0.01)(1 x 10S6)3 = 1 x 10S20 IP > Ksp; Al(OH)3 will precipitate.

16.114 Kspa = 2+

22+

3

[ ][ S]M H[ ]OH

; FeS, Kspa = 6 x 102; SnS, Kspa = 1 x 10S5

Fe2+ and Sn2+ can be separated by bubbling H2S through an acidic solution containing the two cations because their Kspa values are so different.

For FeS and SnS, Qc = 2

(0.01)(0.10)

(0.3) = 1.1 x 10S2

For FeS, Qc < Kspa, and no FeS will precipitate. For SnS, Qc > Kspa, and SnS will precipitate.

16.115 CoS, Kspa = 2+

22+

3

[ ][ S]Co H[ ]OH

= 3

(i) In 0.5 M HCl, [H3O+] = 0.5 M

Qc = 2+

2t t2 2+

3 t

[ [ S] ] (0.10)(0.10)Co H = [ (0.5] )OH

= 0.04; Qc < Kspa; CoS will not precipitate

(ii) pH = 8; [H3O+] = 10SpH = 10S8 = 1 x 10S8 M

Qc = 2+

2t t2 2+ _8

3 t

[ [ S] ] (0.10)(0.10)Co H = [ (1 x ] )OH 10

= 1 x 1014; Qc > Kspa; CoS(s) will precipitate

16.116 (a) add ClS to precipitate AgCl

(b) add CO32S to precipitate CaCO3

(c) add H2S to precipitate MnS (d) add NH3 and NH4Cl to precipitate Cr(OH)3 (Need buffer to control [OHS]; excess OHS produces the soluble Cr(OH)4

S.) 16.117 (a) add ClS to precipitate Hg2Cl2

(b) add (NH4)2HPO4 to precipitate MgNH4PO4 (c) add HCl and H2S to precipitate HgS (d) add ClS to precipitate PbCl2

General Problems


469

16.118 Prepare aqueous solutions of the three salts. Add a solution of (NH4)2HPO4. If a white

precipitate forms, the solution contains Mg2+. Perform flame test on the other two solutions. A yellow flame test indicates Na+. A violet flame test indicates K+.

16.119 (a), solution contains HCN and CNS

(c), solution can contain HCN and CNS (e), solution can contain HCN and CNS

16.120 (a), solution contains H2CO3 and HCO3

S (b), solution contains HCO3

S and CO32S

(d), solution contains HCO3S and CO3

2S

16.121

(a) The pH for the weak acid is higher. (b) Initially, the pH rises more quickly for the weak acid, but then the curve becomes more level in the region halfway to the equivalence point. (c) The pH is higher at the equivalence point for the weak acid. (d) Both curves are identical beyond the equivalence point because the pH is determined by the excess [OHS]. (e) If the acid concentrations are the same, the volume of base needed to reach the equilavence point is the same.


470

16.122 (a)

(b) mol NaOH required = 0.010 mol HA 1 mol NaOH

(0.0500 L) = 0.000 50 molL 1 mol HA

vol NaOH required = (0.000 50 mol)1 L

= 0.050 L = 50 mL0.010 mol

(c) A basic salt is present at the equivalence point; pH > 7.00 (d) Halfway to the equivalence point, the pH = pKa = 4.00

16.123 (a) AgBr(s) Ω Ag+(aq) + BrS(aq) (i) HBr is a source of BrS (reaction product). The solubility of AgBr is decreased. (ii) unaffected (iii) AgNO3 is a source of Ag+ (reaction product). The solubility of AgBr is decreased. (iv) NH3 forms a complex with Ag+, removing it from solution. The solubility of AgBr is increased.

(b) BaCO3(s) Ω Ba2+(aq) + CO32S(aq)

(i) HNO3 reacts with CO32S, removing it from the solution. The solubility of BaCO3 is

increased. (ii) Ba(NO3)2 is a source of Ba2+ (reaction product). The solubility of BaCO3 is decreased. (iii) Na2CO3 is a source of CO3

2S (reaction product). The solubility of BaCO3 is decreased. (iv) CH3CO2H reacts with CO3

2S, removing it from the solution. The solubility of BaCO3 is increased.

16.124 For NH4+, Ka =

_14w

_5b 3

1.0 x 10K = for 1.8 x NH 10K

= 5.6 x 10S10

pKa = Slog Ka = Slog(5.6 x 10S10) = 9.25

pH = pKa + log 3

+4

[ ]NH[ ]NH

; 9.40 = 9.25 + log 3

+4

[ ]NH[ ]NH

log 3

+4

[ ]NH[ ]NH

= 9.40 S 9.25 = 0.15; 3

+4

[ ]NH[ ]NH

= 100.15 = 1.41


471

Because the volume is the same for both NH3 and NH4+, 3

+4

mol NHmol NH

= 1.41.

mol NH3 = (0.20 mol/L)(0.250 L) = 0.050 mol NH3

mol NH4+ = 3mol 0.050NH =

1.41 1.41 = 0.035 mol NH4

+

vol NH4+ =

1 L(0.035 mol)

3.0 mol

= 0.012 L = 12 mL

12 mL of 3.0 M NH4Cl must be added to 250 mL of 0.20 M NH3 to obtain a buffer solution having a pH = 9.40.

16.125 H2PO4S(aq) + H2O(l) Ω H3O

+(aq) + HPO42S(aq)

(a) Na2HPO4, source of HPO42S, equilibrium shifts left, pH increases.

(b) Addition of the strong acid, HBr, decreases the pH. (c) Addition of the strong base, KOH, increases the pH. (d) There is no change in the pH with the addition of the neutral salt KI. (e) H3PO4, source of H2PO4

S, equilibrium shifts right, pH decreases. (f) Na3PO4, source of PO4

3S, decreases [H3O+] by forming HPO4

2S, pH increases. 16.126 pH = 10.35; [H3O

+] = 10SpH = 10S10.35 = 4.5 x 10S11 M

[OHS] = _14

w

+ _113

1.0 x 10K = [ ] 4.5 x O 10H

= 2.2 x 10S4 M

[Mg2+] = _ _ 4[ ] 2.2 x OH 10 =

2 2 = 1.1 x 10S4 M

Ksp = [Mg2+][OHS]2 = (1.1 x 10S4)(2.2 x 10S4)2 = 5.3 x 10S12 16.127 mmol Hg2

2+ = (0.010 mmol/mL)(1.0 mL) = 0.010 mmol mmol ClS = (6 mmol/mL)( 0.05 mL) = 0.3 mmol Assume complete reaction.

Hg22+(aq) + 2 ClS(aq) Hg2Cl2(s)

before reaction (mmol) 0.010 0.3 change (mmol) S0.010 S2(0.010) after reaction (mmol) 0 0.28

[ClS] = 0.28 mmol

1.05 mL = 0.27 M

Allow Hg2Cl2 to establish a new equilibrium.

Hg2Cl2(s) Ω Hg22+(aq) + 2 ClS(aq)

initial (M) 0 0.27 equil (M) x 0.27 + 2x

Ksp = [Hg22+][Cl S]2 = 1.4 x 10S18 = x(0.27 + 2x)2 . x(0.27)2

x = [Hg22+] =

_18

2

1.4 x 10(0.27)

= 2 x 10S17 mol/L

Hg22+, 401.18 amu

Hg22+ concentration = (2 x 10S17 mol/L)(401.18 g/mol) = 8 x 10S15 g/L


472

16.128 NaOH, 40.0 amu; 20 g x 1 mol

40.0 g = 0.50 mol NaOH

(0.500 L)(1.5 mol/L) = 0.75 mol NH4Cl

NH4+(aq) + OHS(aq) Ω NH3(aq) + H2O(l)

before reaction (mol) 0.75 0.50 0 change (mol) S0.50 S0.50 +0.50

after reaction (mol) 0.25 0 0.50 This reaction produces a buffer solution. [NH4

+] = 0.25 mol/0.500 L = 0.50 M; [NH3] = 0.50 mol/0.500 L = 1.0 M

pH = pKa + log 3a +

4


For NH4+, Ka =

_14w

_5b 3

1.0 x 10K = for 1.8 x NH 10K

= 5.6 x 10S10; pKa = Slog Ka = 9.25

pH = 9.25 + log1.0

0.5

= 9.55

16.129 (a) AgCl, Ksp = [Ag+][Cl S] = 1.8 x 10S10

[ClS] = _10

sp

+

1.8 x K 10 = 0.030[ ]Ag

= 6.0 x 10S9 M

(b) Hg2Cl2, Ksp = [Hg22+][Cl S]2 = 1.4 x 10S18

[ClS] = _18

sp

2+2

1.4 x K 10 = 0.030[ ]Hg

= 6.8 x 10S9 M

(c) PbCl2, Ksp = [Pb2+][Cl S]2 = 1.2 x 10S5

[ClS] = _5

sp

2+

1.2 x K 10 = [ ] 0.030Pb

= 0.020 M

AgCl(s) will begin to precipitate when the [ClS] just exceeds 6.0 x 10S9 M. At this ClS concentration, IP < Ksp for PbCl2 so all of the Pb2+ will remain in solution.

16.130 For NH4+, Ka =

_14w

_5b 3

1.0 x 10K = for 1.8 x NH 10K

= 5.6 x 10S10; pKa = Slog Ka = 9.25

pH = pKa + log 3

+4

[ ] (0.50)NH = 9.25 + log[ ] (0.30)NH

= 9.47

[H3O+] = 10SpH = 10S9.47 = 3.4 x 10S10 M

For MnS, Kspa = 2+

22+

3

[ ][ S]Mn H[ ]OH

= 3 x 1010

molar solubility = [Mn2+] = 2 2+ 10 _10

spa 3

2

[ ] (3 x )(3.4 x )OK H 10 10 = [ S] (0.10)H

= 3.5 x 10S8 M

MnS, 87.00 amu; solubility = (3.5 x 10S8 mol/L)(87.00 g/mol) = 3 x 10S6 g/L 16.131 pH = 9.00; [H3O

+] = 10SpH = 10S9.00 = 1.0 x 10S9 M


473

[OHS] = _14

w

+ _93

1.0 x 10K = [ ] 1.0 x O 10H

= 1.0 x 10S5 M

Mg(OH)2(s) Ω Mg2+(aq) + 2 OHS(aq) equil (M) x 1.0 x 10S5 (fixed by buffer)

Ksp = [Mg2+][OHS]2 = 5.6 x 10S12 = x(1.0 x 10S5)2


2_5

5.6 x 10 = 0.056 M(1.0 x )10

16.132 60.0 mL = 0.0600 L

mol H3PO4 = 0.0600 L x 3 41.00 mol POH =1.00 L

0.0600 mol H3PO4

mol LiOH = 1.00 L x 0.100 mol LiOH

=1.00 L

0.100 mol LiOH

H3PO4(aq) + OHS(aq) H2PO4

S(aq) + H2O(l) before reaction (mol) 0.0600 0.100 0 change (mol) S0.0600 S0.0600 +0.0600 after reaction (mol) 0 0.040 0.0600

H2PO4

S(aq) + OHS(aq) HPO42S(aq) + H2O(l)

before reaction (mol) 0.0600 0.040 0 change (mol) S0.040 S0.040 +0.040 after reaction (mol) 0.020 0 0.040 The resulting solution is a buffer because it contains the conjugate acid-base pair, H2PO4

S and HPO42S, at acceptable buffer concentrations.

For H2PO4S, Ka2 = 6.2 x 10S8 and pKa2 = S log Ka2 = S log (6.2 x 10S8) = 7.21

pH = pKa2 + log2_4

_2 4

[ ] (0.040 mol / 1.06 L)HPO = 7.21 + log [ ] (0.020 mol / 1.06 L)POH

pH = (0.040)

7.21 + log = 7.21 + 0.30 =(0.020)

7.51

16.133 (a) The mixture of 0.100 mol H3PO4 and 0.150 mol NaOH is a buffer and contains

mainly H2PO4S and HPO4

2S from the reactions:

H3PO4(aq) + OHS(aq) H2PO4S(aq) + H2O(l)

before (mol) 0.100 0.150 0 change (mol) S0.100 S0.100 +0.100 after (mol) 0 0.050 0.100

H2PO4

S(aq) + OHS(aq) HPO42S(aq) + H2O(l)

before (mol) 0.100 0.050 0 change (mol) S0.050 S0.050 +0.050 after (mol) 0.050 0 0.050


474

If water were used to dilute the solution instead of HCl, the pH would be equal to pKa2 because [H2PO4

S] = [HPO42S] = 0.050 mol/1.00 L = 0.050 M

H2PO4S(aq) + H2O(l) Ω H3O

+(aq) + HPO42S(aq) Ka2 = 6.2 x 10S8

pKa2 = Slog K2a = Slog(6.2 x 10S8) = 7.21

pH = pKa2 + log 2_4

_2 4

[ ]HPO[ ]POH

= pKa2 + log(1) = pKa2 = 7.21

The pH is lower (6.73) because the added HCl converts some HPO4

2S to H2PO4S.

HPO42S(aq) + H3O

+(aq) H2PO4S(aq) + H2O(l)

before (M) 0.050 x 0.050 change (M) Sx Sx +x after (M) 0.050 S x 0 0.050 + x [HPO4

2S] + [H2PO4S] = (0.050 S x) + (0.050 + x) = 0.100 M

pH = pKa2 + log 2_4

_2 4

[ ]HPO[ ]POH

[HPO42S] = 0.100 S [H2PO4

S]

6.73 = 7.21 + log _

2 4_

2 4

(0.100 _ [ ])POH[ ]POH

6.73 S 7.21 = S0.48 = log _

2 4_

2 4

(0.100 _ [ ])POH[ ]POH

10S0.48 = 0.331 = _

2 4_

2 4

(0.100 _ [ ])POH[ ]POH

(0.331)[H2PO4S] = 0.100 S [H2PO4

S] (1.331)[H2PO4

S] = 0.100 [H2PO4

S] = 0.100/1.331 = 0.075 M [HPO4

2S] = 0.100 S [H2PO4S] = 0.100 S 0.075 = 0.025 M

H3PO4(aq) + H2O(l) Ω H3O+(aq) + H2PO4

S(aq) Ka1 = 7.5 x 10S3

Ka1 = _+

3 2 4

3 4

[ ][ ]O POH H[ ]POH

[H3PO4] = _+

3 2 4

a1

[ ][ ]O POH H

K

[H3O+] = 10SpH = 10S6.73 = 1.86 x 10S7 M

[H3PO4] = _7

_3

(1.86 x )(0.075)10 =7.5 x 10

1.9 x 10S6 M

(b) If distilled water were used and not HCl, the mole amounts of both H2PO4

S and HPO42S

would be 0.050 mol. The HCl converted some HPO42S to H2PO4

S. HPO4

2S(aq) + H3O+(aq) H2PO4

S(aq) + H2O(l) before (mol) 0.050 x 0.050 change (mol) Sx Sx +x after (mol) 0.050 S x 0 0.050 + x


475

From part (a), [HPO42S] = 0.025 M

mol HPO42S = (0.025 mol/L)(1.00 L) = 0.025 mol = 0.050 S x

x = mol H3O+ = mol HCl inadvertently added = 0.050 S 0.025 = 0.025 mol HCl

16.134 For CH3CO2H, Ka = 1.8 x 10S5 and pKa = Slog Ka = Slog(1.8 x 10S5) = 4.74

The mixture will be a buffer solution containing the conjugate acid-base pair, CH3CO2H and CH3CO2

S, having a pH near the pKa of CH3CO2H.

pH = pKa + log_

3 2

3 2

[ ]CH CO [ H]CH CO

4.85 = 4.74 + log_

3 2

3 2

[ ]CH CO [ H]CH CO

; 4.85 S 4.74 = log_

3 2

3 2

[ ]CH CO [ H]CH CO

0.11 = log_

3 2

3 2

[ ]CH CO [ H]CH CO

; _

3 2

3 2

[ ]CH CO [ H]CH CO

= 100.11 = 1.3

In the Henderson-Hasselbalch equation, moles can be used in place of concentrations because both components are in the same volume so the volume terms cancel. 20.0 mL = 0.0200 L

Let X equal the volume of 0.10 M CH3CO2H and Y equal the volume of 0.15 M CH3CO2

S. Therefore, X + Y = 0.0200 L and _

3 2

3 2

Y x [ ] Y (0.15 mol/L)CH CO = =X x [ H] X (0.10 mol/L)CH CO

1.3

X = 0.0200 S Y Y (0.15 mol/L)

=(0.020 _ Y)(0.10 mol/L)

1.3

0.15Y =

0.0020 _ 0.10Y1.3

0.15Y = 1.3(0.0020 S 0.10Y) 0.15Y = 0.0026 S 0.13Y 0.15Y + 0.13Y = 0.0026 0.28Y = 0.0026 Y = 0.0026/0.28 = 0.0093 L X = 0.0200 S Y = 0.0200 S 0.0093 = 0.0107 L X = 0.0107 L = 10.7 mL and Y = 0.0093 L = 9.3 mL You need to mix together 10.7 mL of 0.10 M CH3CO2H and 9.3 mL of 0.15 M NaCH3CO2 to prepare 20.0 mL of a solution with a pH of 4.85.

16.135 [H3O

+] = 10SpH = 10S2.37 = 0.004 27 M

H3Cit(aq) + H2O(l) Ω H3O+(aq) + H2CitS(aq)

Ka1 = 7.1 x 10S4 = + _

3 2

3

[ ][ ]O CitH H[ Cit]H

(7.1 x 10S4)[H3Cit] = (0.004 27)[H2CitS] [H3Cit] = (0.004 27)[H2CitS]/(7.1 x 10S4) = (6.01)[H2CitS]


476

H2CitS(aq) + H2O(l) Ω H3O+(aq) + HCit2S(aq)

Ka2 = 1.7 x 10S5 = + 2_

3

_2

[ ][ ]O HCitH[ ]CitH

(1.7 x 10S5)[H2CitS] = (0.004 27)[HCit2S] [HCit2S] = (1.7 x 10S5)[H2CitS]/(0.004 27) = (0.003 98)[H2CitS] [H3Cit] + [H2CitS] + [HCit2S] + [Cit3S] = 0.350 M Now assume [Cit3S] . 0, so [H3Cit] + [H2CitS] + [HCit2S] = 0.350 M and then by substitution: (6.01)[H2CitS] + [H2CitS] + (0.003 98)[H2CitS] = 0.350 M (7.01)[H2CitS] = 0.350 M [H2CitS] = 0.350 M/7.01 = 0.050 M [H3Cit] = (6.01)[H2CitS] = (6.01)(0.050 M) = 0.30 M [HCit2S] = (0.003 98)[H2CitS] = (0.003 98)(0.050 M) = 2.0 x 10S4 M

HCit2S(aq) + H2O(l) Ω H3O+(aq) + Cit3S(aq)

Ka3 = 4.1 x 10S7 = + 3_

3

2_

[ ][ ]O CitH[ ]HCit

[Cit3S] = 2_

a3

+3

( )[ ]HCitK[ ]OH

= _ 7 _ 4(4.1 x )(2.0 x )10 10 =

(0.004 27) 1.9 x 10S8 M

16.136 (a) HCl is a strong acid. HCN is a weak acid with Ka = 4.9 x 10S10. Before the titration,

the [H3O+] = 0.100 M. The HCN contributes an insignificant amount of additional

H3O+, so the pH = Slog[H3O

+] = Slog(0.100) = 1.00 (b) 100.0 mL = 0.1000 L

mol H3O+ = 0.1000 L x

0.100 mol HCl =

1.00 L 0.0100 mol H3O

+

add 75.0 mL of 0.100 M NaOH; 75.0 mL = 0.0750 L

mol OHS = 0.0750 L x 0.100 mol NaOH

=1.00 L

0.00750 mol OHS

H3O

+(aq) + OHS(aq) 2 H2O(l) before reaction (mol) 0.0100 0.0075 change (mol) S0.0075 S0.0075 after reaction (mol) 0.0025 0

[H3O+] =

+30.0025 mol OH =

0.1000 L + 0.0750 L 0.0143 M

pH = Slog[H3O+] = Slog(0.0143) = 1.84

(c) 100.0 mL of 0.100 M NaOH will completely neutralize all of the H3O+ from 100.0

mL of 0.100 M HCl. Only NaCl and HCN remain in the solution. NaCl is a neutral salt and does not affect the pH of the solution. [HCN] changes because of dilution. Because the solution volume is doubled, [HCN] is cut in half.


477

[HCN] = 0.100 M/2 = 0.0500 M

HCN(aq) + H2O(l) Ω H3O+(aq) + CNS(aq)

initial (M) 0.0500 ~0 0 change (M) Sx +x +x equil (M) 0.0500 S x x x

Ka = + _

3[ ][ ]O CNH =HCN

4.9 x 10S10 = 2x

0.0500 _ x .

2x0.0500

[H3O+] = x = _10(0.0500)(4.9 x )10 = 4.95 x 10S6 M

pH = Slog[H3O+] = Slog(4.95 x 10S6) = 5.31

(d) Add an additional 25.0 mL of 0.100 M NaOH. 25.0 mL = 0.0250 L

additional mol OHS = 0.0250 L x 0.100 mol NaOH

=1.00 L

0.00250 mol OHS

mol HCN = 0.200 L x 0.0500 mol HCN

=1.00 L

0.0100 mol HCN

HCN(aq) + OHS(aq) CNS(aq) + H2O(l) before reaction (mol) 0.0100 0.00250 0 change (mol) S0.00250 S0.00250 +0.00250 after reaction (mol) 0.0075 0 0.00250 The resulting solution is a buffer because it contains the conjugate acid-base pair, HCN and CNS, at acceptable buffer concentrations. For HCN, Ka = 4.9 x 10S10 and pKa = S log Ka = S log (4.9 x 10S10) = 9.31

pH = pKa + log_[ ] (0.00250 mol / 0.2250 L)CN = 9.31 + log

[HCN] (0.0075 mol / 0.2250 L)

pH = (0.00250)

9.31 + log = 9.31 _ 0.48 =(0.0075)

8.83

16.137 (a) Cd(OH)2(s) Ω Cd2+(aq) + 2 OHS(aq) initial (M) 0 ~0 equil (M) x 2x Ksp = [Cd2+][OHS]2 = 5.3 x 10S15 = (x)(2x)2 = 4x3


35.3 x 10

4 = 1.1 x 10S5 M

[OHS] = 2x = 2(1.1 x 10S5 M) = 2.2 x 10S5 M

[H3O+] =

_14

_5

1.0 x 10 =2.2 x 10

4.5 x 10S10 M

pH = Slog[H3O+] = Slog(4.5 x 10S10) = 9.35

(b) 90.0 mL = 0.0900 L mol HNO3 = (0.100 mol/L)(0.0900 L) = 0.009 00 mol HNO3 The addition of HNO3 dissolves some Cd(OH)2(s).

Cd(OH)2(s) + 2 HNO3(aq) Cd2+(aq) + 2 H2O(l)


478

before (mol) 0.100 0.009 00 1.1 x 10S5 change (mol) S0.0045 S2(0.0045) 1.1 x 10S5 + 0.0045 after (mol) 0.0955 0 ~0.0045 total volume = 100.0 mL + 90.0 mL = 190.0 mL = 0.1900 L [Cd2+] = 0.0045 mol/0.1900 L = 0.024 M Ksp = 5.3 x 10S15 = [Cd2+][OHS]2 = (0.024)[OHS]2

[OHS] = _155.3 x 10 =

0.024 4.7 x 10S7 M

[H3O+] =

_14

_ 7

1.0 x 10 =4.7 x 10

2.1 x 10S8 M

pH = Slog[H3O+] = Slog(2.1 x 10S8) = 7.68

(c) volume HNO3 = 0.0100 mol Cd(OH)2 x 3

2

2 mol HNO1 mol Cd(OH)

x 1.00 L

0.100 mol x

1000 mL =

1.00 L 200

mL

16.138 (a) Zn(OH)2(s) Ω Zn2+(aq) + 2 OHS(aq) initial (M) 0 ~0 equil (M) x 2x Ksp = [Zn2+][OHS]2 = 4.1 x 10S17 = (x)(2x)2 = 4x3


34.1 x 10

4 = 2.2 x 10S6 M

(b) [OHS] = 2x = 2(2.2 x 10S6 M) = 4.4 x 10S6 M

[H3O+] =

_14

_ 6

1.0 x 10 =4.4 x 10

2.3 x 10S9 M

pH = Slog[H3O+] = Slog(2.3 x 10S9) = 8.64

(c) Zn(OH)2(s) Ω Zn2+(aq) + 2 OHS(aq) Ksp = 4.1 x 10S17

Zn2+(aq) + 4 OHS(aq) Ω Zn(OH)42S(aq) Kf = 3 x 1015

Zn(OH)2(s) + 2 OHS(aq) Ω Zn(OH)42S(aq) K = Ksp≅Kf = 0.123

initial (M) 0.10 0 change (M) S2x +x equil (M) 0.10 S 2x x

K = 2_42_

[Zn(OH ]) =

[ ]OH 0.123 =

2

x

(0.10 _ 2x)

0.492x2 S 1.0492x + 0.00123 = 0 Use the quadratic formula to solve for x.

x = 2_ (_ 1.0492) (_ 1.0492 _ (4)(0.492)(0.00123)) 1.0492 1.0480

= 2(0.492) 0.984

± ±

x = 2.1 and 1.2 x 10S3 Of the two solutions for x, only 1.2 x 10S3 has physical meaning because the other solution leads to a negative [OHS]. molar solubility of Zn(OH)4

2S in 0.10 M NaOH = x = 1.2 x 10S3 M


479

16.139 (a) Fe(OH)3(s) Ω Fe3+(aq) + 3 OHS(aq) Ksp = 2.6 x 10S39

H3Cit(aq) + H2O(l) Ω H3O+(aq) + H2CitS(aq) Ka1 = 7.1 x 10S4

H2CitS(aq) + H2O(l) Ω H3O+(aq) + HCit2S(aq) Ka2 = 1.7 x 10S5

HCit2S(aq) + H2O(l) Ω H3O+(aq) + Cit3S(aq) Ka3 = 4.1 x 10S7

Fe3+(aq) + Cit3S(aq) Ω Fe(Cit)(aq) Kf = 6.3 x 1011

3 [H3O+(aq) + OHS(aq) Ω 2 H2O(l)] (1/Kw)3 = 1.0 x 1042

Fe(OH)3(s) + H3Cit(aq) Ω Fe(Cit)(aq) + 3 H2O(l)

K = Ksp Ka1 Ka2 Ka3Kf(1/Kw)3 = 8.1

(b) Fe(OH)3(s) + H3Cit(aq) Ω Fe(Cit)(aq) + 3 H2O(l) initial (M) 0.500 0 change (M) Sx +x equil (M) 0.500 S x x

K = 3

[Fe(Cit)] =

[ Cit]H 8.1 =

x

0.500 _ x

8.1(0.500 S x) = x 4.05 S 8.1x = x 4.05 = 9.1x x = molar solubility = 4.05/9.1 = 0.45 M


16.140 (a) HAS(aq) + H2O(l) Ω H3O+(aq) + A2S(aq) Ka2 = 10S10

HAS(aq) + H2O(l) Ω H2A(aq) + OHS(aq) Kb = w

a

K1K

= 10S10

2 HAS(aq) Ω H2A(aq) + A2S(aq) K = a

a

2K1K

= 10S6

2 H2O(l) Ω H3O+(aq) + OHS(aq) Kw

= 1.0 x 10S14

The principal reaction of the four is the one with the largest K, and that is the third reaction.

(b) Ka1 = + _

3

2

[ ][ ]OH HA[ A]H

and Ka2 = + 2_

3

_

[ ][ ]OH A[ ]HA

[H3O+] = a 2

_

1 [ A]K H[ ]HA

and [H3O+] =

_a

2_

2 [ ]K HA[ ]A

a 2

_

1 [ A]K H[ ]HA

x _

a

2_

2 [ ]K HA[ ]A

= [H3O+]2; a a 2

2_

1 2 [ A]K K H[ ]A

= [H3O+]2

Because the principal reaction is 2 HAS(aq) Ω H2A(aq) + A2S(aq), [H2A] = [A 2S]. Ka1 Ka2 = [H3O

+]2 log Ka1 + log Ka2 = 2 log [H3O

+]


480

a a +3

log 1 + log 2K K = log [ ]OH2

; a a +3

_ log 1 + (_ log 2)K K = _ log [ ]OH2

a ap 1 + p 2K K = pH2

(c) 2 HAS(aq) Ω H2A(aq) + A2S(aq) initial (M) 1.0 0 0 change (M) S2x +x +x equil (M) 1.0 S 2x x x

K = 2_

22_

[ A][ ]H A[ ]HA

= 1 x 10S6 = 2

2

x(1.0 _ 2 x)

Take the square root of both sides and solve for x. x = [A2S] = 1 x 10S3 M mol A2S = (1 x 10S3 mol/L)(0.0500 L) = 5 x 10S5 mol A2S number of A2S ions = (5 x 10S5 mol A2S)(6.022 x 1023 ions/mol) = 3 x 1019 A2S ions

16.141 (a) (i) en(aq) + H2O(l) Ω enH+(aq) + OHS(aq)


Kb = + _

_ 4[ ][ ] (x)(x)enH OH = 5.2 x = 10[en] 0.100 _ x

x2 + (5.2 x 10S4)x S (5.2 x 10S5) = 0 Use the quadratic formula to solve for x.

x = 2_ 4 _ 4 _5 _ 4_ (5.2 x ) (5.2 x _ 4(1)(_ 5.2 x )) _ 5.2 x 0.0144310 10 10 10 =

2(1) 2

± ±

x = S0.0075 and 0.0070 Of the two solutions for x, only the positive value of x has physical meaning because x is the [OHS]. [OHS] = x = 0.0070 M

[H3O+] =

_14w

_

1.0 x 10K = [ ] 0.0070OH

= 1.43 x 10S12 M

pH = Slog[H3O+] = Slog(1.43 x 10S12) = 11.84

(ii) (30.0 mL)(0.100 mmol/mL) = 3.00 mmol en (15.0 mL)(0.100 mmol/mL) = 1.50 mmol HCl Halfway to the first equivalence point, [OHS] = Kb1

[H3O+] =

_14w

_ _ 4

1.0 x 10K = [ ] 5.2 x OH 10

= 1.92 x 10S11 M

pH = Slog[H3O+] = Slog(1.92 x 10S11) = 10.72


481

(iii) At the first equivalence point pH = a ap 1 + p 2K K2

= 9.14

(iv) Halfway between the first and second equivalence points, [OHS] = Kb2 = 3.7 x 10S7 M

[H3O+] =

_14w

_ _ 7

1.0 x 10K = [ ] 3.7 x OH 10

= 2.70 x 10S8 M

pH = Slog[H3O+] = Slog(2.70 x 10S8) = 7.57

(v) At the second equivalence point only the acidic enH2Cl2 is in solution.

For enH22+, Ka =

_14w w

+ _7b b

1.0 x 10K K = = for 2 3.7 x enH 10K K

= 2.70 x 10S8

[enH22+] =

3.00 mmol

(30.0 mL + 60.0 mL) = 0.0333 M

enH22+(aq) + H2O(l) Ω H3O

+(aq) + enH+(aq) initial (M) 0.0333 ~0 0 change (M) Sx +x +x equil (M) 0.0333 S x x x

Ka = + +

3

2+2

[ ][ ]O enHH[ ]enH

= 2.70 x 10S8 = 2(x)(x) x

0.0333 _ x 0.0333≈

Solve for x. x = [H3O+] = _8(2.70 x )(0.0333)10 = 3.00 x 10S5 M

pH = Slog[H3O+] = Slog(3.00 x 10S5) = 4.52

(vi) excess HCl (75.0 mL S 60.0 mL)(0.100 mmol/mL) = 1.50 mmol HCl = 1.50 mmol H3O

+

[H3O+] =

1.50 mmol

(30.0 mL + 75.0 mL) = 0.0143 M

pH = Slog[H3O+] = Slog(0.0143) = 1.84


482

(b) Each of the two nitrogens in ethylenediamine can accept a proton.

(c) Each nitrogen is sp3 hybridized.

16.142 (a) The first equivalence point is reached when all the H3O

+ from the HCl and the H3O+

form the first ionization of H3PO4 is consumed.

At the first equivalence point pH = a1 a 2 + pK pK

2 = 4.66

[H3O+] = 10SpH = 10(S4.66) = 2.2 x 10S5 M

(88.0 mL)(0.100 mmol/mL) = 8.80 mmol NaOH are used to get to the first equivalence point (b) mmol (HCl + H3PO4) = mmol NaOH = 8.8 mmol mmol H3PO4 = (126.4 mL S 88.0 mL)(0.100 mmol/mL) = 3.84 mmol mmol HCl = (8.8 S 3.84) = 4.96 mmol

[HCl] = 4.96 mmol

40.0 mL = 0.124 M; [H3PO4] =

3.84 mmol

40.0 mL = 0.0960 M

(c) 100% of the HCl is neutralized at the first equivalence point.

(d) H3PO4(aq) + H2O(l) Ω H3O+(aq) + H2PO4

S(aq) initial (M) 0.0960 0.124 0 change (M) Sx +x +x equil (M) 0.0960 S x 0.124 + x x

Ka1 = _+

3 2 4

3 4

[ ][ ]O POH H[ ]POH

= 7.5 x 10S3 = (0.124 + x)(x)

0.0960 _ x

x2 + 0.132x S (7.2 x 10S4) = 0 Use the quadratic formula to solve for x.

x = 2 _ 4_ (0.132) (0.132 _ 4(1)(_ 7.2 x )) _ 0.132 0.14210

= 2(1) 2

± ±

x = S0.137 and 0.005 Of the two solutions for x, only the positive value of x has physical meaning because the other solution would give a negative [H3O

+]. [H3O

+] = 0.124 + x = 0.124 + 0.005 = 0.129 M pH = Slog[H3O

+] = Slog(0.129) = 0.89 (e)


483

(f) Bromcresol green or methyl orange are suitable indicators for the first equivalence point. Thymolphthalein is a suitable indicator for the second equivalence point.

16.143 (a) PV = nRT; 25oC = 298 K

HCln =

1.00 atm732 mm Hg x (1.000 L)

760 mm HgPV =

L atmRT 0.082 06 (298 K)K mol

• •

= 0.0394 mol HCl

Na2CO3, 105.99 amu

mol Na2CO3 = 6.954 g Na2CO3 x 2 3

2 3

1 mol Na CO105.99 g Na CO

= 0.0656 mol Na2CO3

CO32S(aq) + H3O

+(aq) HCO3S(aq) + H2O(l)

before reaction (mol) 0.0656 0.0394 0 change (mol) S0.0394 S0.0394 +0.0394

after reaction (mol) 0.0656 S 0.0394 0 0.0394

mol CO32S = 0.0656 S 0.0394 = 0.0262 mol and mol HCO3

S = 0.0394 mol Therefore, we have an HCO3

S/CO32S buffer solution.

pH = pKa2 + log 2_3

_3

[ ]CO[ ]HCO

= S log(5.6 x 10S11) + log 0.0262 mol/V

0.0394 mol/V

pH = 10.25 S 1.77 = 10.08

(b) mol Na+ = 2(0.0656 mol) = 0.1312 mol mol CO3

2S = 0.0262 mol mol HCO3

S = 0.0394 mol mol ClS = 0.0394 mol total ion moles = 0.2362 mol

∆Tf = Kf Α m, ∆Tf = oC kg 0.2362 mol 1.86 mol 0.2500 kg

•

= 1.76oC

Solution freezing point = 0oC S ∆Tf = S1.76oC


484

(c) H2O, 18.02 amu

mol H2O = 250.0 g x 2

2

1 mol OH18.02 g OH

= 13.87 mol H2O

Xsolv = 2

2

mol OHmol O + mol ionsH

= 13.87 mol

13.87 mol + 0.2362 mol = 0.9833

Psoln = Psolv ≅ Xsolv = (23.76 mm Hg)(0.9833) = 23.36 mm Hg 16.144 25oC = 298 K

Π = 2MRT; M =

1.00 atm74.4 mm Hg x

760 mm Hg =

L atm2RT (2) 0.082 06 (298 K)K mol

Π

• •

= 0.00200 M

[M +] = [XS] = 0.00200 M Ksp = [M+][X S] = (0.00200)2 = 4.00 x 10S6

16.145 (a) HCO3S(aq) + OHS(aq) CO3

2S(aq) + H2O(l) (b) mol HCO3

S = (0.560 mol/L)(0.0500 L) = 0.0280 mol HCO3S

mol OHS = (0.400 mol/L)(0.0500 L) = 0.0200 mol OHS

HCO3S(aq) + OHS(aq) CO3

2S(aq) + H2O(l) before reaction (mol) 0.0280 0.0200 0

change (mol) S0.0200 S0.0200 +0.0200 after reaction (mol) 0.0280 S 0.0200 0 0.0200

mol HCO3

S = 0.0280 S 0.0200 = 0.0080 mol

[HCO3S] =

0.0080 mol

0.1000 L = 0.080 M [CO3

2S] = 0.0200 mol

0.1000 L= 0.200 M

HCO3S(aq) + H2O(l) Ω H3O

+(aq) + CO32S(aq)


Ka = 2_+

3 3 _11_3

[ ][ ] x(0.200 + x) x(0.200)O COH = 5.6 x = 10[ ] 0.080 _ x 0.080HCO

≈


pH = Slog[H3O+] = Slog(2.24 x 10S11) = 10.65

Because this solution contains both a weak acid (HCO3S) and its conjugate base, the

solution is a buffer.

(c) HCO3S(aq) + OHS(aq) CO3

2S(aq) + H2O(l) ∆Ho

rxn = [∆Hof(CO3

2S) + ∆Hof(H2O)] S [∆Ho

f(HCO3S) + ∆Ho

f(OHS)] ∆Ho

rxn = [(1 mol)(S677.1 kJ/mol) + (1 mol)(S285.8 kJ/mol)] S [(1 mol)(S692.0 kJ/mol) + (1 mol)(S230 kJ/mol)]

∆Horxn = S 40.9 kJ

0.0200 moles each of HCO3S and OHS reacted.

heat produced = q = (0.0200 mol)(40.9 kJ/mol) = 0.818 kJ = 818 J


485

(d) q = m x specific heat x ∆T

∆T = q

m x specific heat =

o

818 J

(100.0 g)[4.18 J/(g C)]• = 2.0oC

Final temperature = 25oC + 2.0oC = 27oC 16.146 (a) species present initially:

NH4+ CO3

2S H2O acid base acid or base

2H2O(l) Ω H3O+(aq) + OHS(aq)

NH4+(aq) + H2O(l) Ω NH3(aq) + H3O

+(aq)


S(aq) + OHS(aq)

NH3, Kb = 1.8 x 10S5 NH4

+, Ka = 5.6 x 10S10 CO3

2S, Kb = 1.8 x 10S4 HCO3

S, Ka = 5.6 x 10S11

In the mixture, proton transfer takes place from the stronger acid to the stronger base, so

the principal reaction is NH4+(aq) + CO3

2S(aq) Ω HCO3S(aq) + NH3(aq)

(b) NH4+(aq) + OHS(aq) Ω NH3(aq) + H2O(l) K1 = 1/Kb(NH3)


S(aq) + OHS(aq) K2 = Kb(CO32S)

NH4+(aq) + CO3

2S(aq) Ω HCO3S(aq) + NH3(aq) K = K1≅K2

initial (M) 0.16 0.080 0 0.16 change (M) Sx Sx +x +x equil (M) 0.16 S x 0.080 S x x 0.16 + x

K = _

332_+

4 3

[ ][ ]HCO NH =[ ][ ]NH CO

_ 4

_5

1.8 x 10 =1.8 x 10

10 = x(0.16 + x)

(0.16 _ x)(0.080 _ x)

9x2 S 2.56x + 0.128 = 0 Use the quadratic formula to solve for x.

x = 2_ (_ 2.56) (_ 2.56 _ (4)(9)(0.128)) 2.56 1.395

= 2(9) 18

± ±

x = 0.220 and 0.0647 Of the two solutions for x, only 0.00647 has physical meaning because 0.220 leads to negative concentrations. [NH4

+] = 0.16 S x = 0.16 S 0.0647 = 0.0953 M = 0.095 M [NH3] = 0.16 + x = 0.16 + 0.0647 = 0.225 M = 0.22 M [CO3

2S] = 0.080 S x = 0.080 S 0.0647 = 0.0153 M = 0.015 M [HCO3

S] = x = 0.0647 M = 0.065 M The solution is a buffer containing two different sets of conjugate acid-base pairs. Either pair can be used to calculate the pH. For NH4

+, Ka = 5.6 x 10S10 and pKa = 9.25


486

pH = pKa + log 3

+4

[ ]NH [ ]NH

= 9.25 + log(0.225)

(0.0953)

= 9.62

[H3O+] = 10SpH = 10S9.62 = 2.4 x 10S10 M

[OHS] = _14

_10

1.0 x 10 =2.4 x 10

4.2 x 10S5 M

[H2CO3] = _ +

33

a

[ ][ ]HCO OH =K

_10

_7

(0.647)(2.4 x )10 =(4.3 x )10

3.6 x 10S4 M

(c) For MCO3, IP = [M2+][CO32S] = (0.010)(0.0153) = 1.5 x 10S4

Ksp(CaCO3) = 5.0 x 10S9, 103 Ksp = 5.0 x 10S6 Ksp(BaCO3) = 2.6 x 10S9, 103 Ksp = 2.6 x 10S6 Ksp(MgCO3) = 6.8 x 10S6, 103 Ksp = 6.8 x 10S3

IP > 103 Ksp for CaCO3 and BaCO3, but IP < 103 Ksp for MgCO3 so the [CO32S] is large

enough to give observable precipitation of CaCO3 and BaCO3, but not MgCO3. (d) For M(OH)2, IP = [M2+][OHS]2 = (0.010)(4.17 x 10S5)2 = 1.7 x 10S11 Ksp(Ca(OH)2) = 4.7 x 10S6, 103 Ksp = 4.7 x 10S3 Ksp(Ba(OH)2) = 5.0 x 10S3, 103 Ksp = 5.0 Ksp(Mg(OH)2) = 5.6 x 10S12, 103 Ksp = 5.6 x 10S9 IP < 103 Ksp for all three M(OH)2. None precipitate.

(e) CO32S(aq) + H2O(l) Ω HCO3

S(aq) + OHS(aq) initial (M) 0.08 0 ~0

change (M) Sx +x +x equil (M) 0.08 S x x x

Kb = _ _3

2_3

[ ][ ]HCO OH =[ ]CO

1.8 x 10S4 = 2x

(0.08 _ x)

x2 + (1.8 x 10S4)x S (1.44 x 10S5) = 0 Use the quadratic formula to solve for x. x =

2_ 4 _ 4 _5 _ 4 _3_ (1.8 x ) (1.8 x _ (4)(1)(_ 1.44 x )) _ (1.8 x ) 7.59 x 10 10 10 10 10 = 2(1) 2

± ±

x = 0.0037 and S0.0039 Of the two solutions for x, only 0.0037 has physical meaning because S0.0039 leads to negative concentrations. [OHS] = x = 3.7 x 10S3 M For MCO3, IP = [M2+][CO3

2S] = (0.010)(0.08) = 8.0 x 10S4 For M(OH)2, IP = [M2+][OHS]2 = (0.010)(3.7 x 10S3)2 = 1.4 x 10S7 Comparing IP=s here and 103 Ksp=s in (c) and (d) above, Ca2+ and Ba2+ cannot be separated from Mg2+ using 0.08 M Na2CO3. Na2CO3 is more basic than (NH4)2CO3 and Mg(OH)2 would precipitate along with CaCO3 and BaCO3.

16.147 (a) H2SO4, 98.09 amu

Assume 1.00 L = 1000 mL of solution. mass of solution = (1000 mL)(1.836 g/mL) = 1836 g mass H2SO4 = (0.980)(1836 g) = 1799 g H2SO4


487

mol H2SO4 = 1799 g H2SO4 x 2 4

2 4

1 mol SOH =98.09 g SOH

18.3 mol H2SO4

[H2SO4] = 18.3 mol/ 1.00 L = 18.3 M (b) Na2CO3, 105.99 amu; 1 kg = 1000 g = 2.2046 lb H2SO4(aq) + Na2CO3(s) Na2SO4(aq) + H2O(l) + CO2(g)

mass H2SO4 = (0.980)(36 tons) x 2000 lb

1 ton x

1000 g =

2.2046 lb 3.20 x 107 g H2SO4

mol H2SO4 = 3.20 x 107 g H2SO4 x 2 4

2 4

1 mol SOH =98.09 g SOH

3.26 x 105 mol H2SO4

mass Na2CO3 = 3.26 x 105 mol H2SO4 x 2 3

2 4

1 mol Na CO1 mol SOH

x 2 3

2 3

105.99 g Na CO1 mol Na CO

x

1 kg

=1000 g

3.5 x 104 kg Na2CO3

(c) mol CO2 = 3.26 x 105 mol H2SO4 x 2

2 4

1 mol CO =1 mol SOH

3.26 x 105 mol CO2

18oC = 18 + 273 = 291 K PV = nRT

V =

5 L atm(3.26 x mol) 0.082 06 (291 K)10

nRT K mol = =

P 1.00 atm745 mm Hg x

760 mm Hg

• •

7.9 x 106 L

16.148 Pb(CH3CO2)2, 325.29 amu; PbS, 239.27 amu

(a) mass PbS = (2 mL)(1 g/mL)(0.003) x 3 2 2

3 2 2

1 mol Pb( )CH CO x325.29 g Pb( )CH CO

3 2 2

1 mol PbS x

1 mol Pb( )CH CO

239.27 g PbS x (30 /100)

1 mol PbS = 0.0013 g

= 1.3 mg PbS per dye application

(b) [H3O+] = 10SpH = 10S5.50 = 3.16 x 10S6 M

PbS(s) + 2 H3O+(aq) Ω Pb2+(aq) + H2S(aq) + 2 H2O(l)

initial (M) 3.16 x 10S6 0 0 change (M) S2x +x +x equil (M) 3.16 x 10S6 S 2x x x

Kspa = 2+

22+

3

[ ][ S]Pb H =[ ]OH

2

2_ 6

x (3.16 x _ 2x)10

≈2

2_ 6

x =(3.16 x )10

3 x 10S7

x2 = (3.16 x 10S6)2(3 x 10S7) = 3.0 x 10S18 x = 1.7 x 10S9 M = [Pb2+] for a saturated solution. mass of PbS dissolved per washing =


488

(3 gal)(3.7854 L/1 gal)(1.7 x 10S9 mol/L) x 239.27 g PbS

=1 mol PbS

4.7 x 10S6 g PbS/washing

Number of washings required to remove 50% of the PbS from one application =

_ 6

(0.0013 g PbS)(50 /100) =

(4.7 x g PbS/washing)101.4 x 102

washings (c) The number of washings does not look reasonable. It seems too high considering that frequent dye application is recommended. If the PbS is located mainly on the surface of the hair, as is believed to be the case, solid particles of PbS can be lost by abrasion during shampooing.

489

17

Thermodynamics: Entropy, Free Energy, and Equilibrium

17.1 (a) spontaneous; (b), (c), and (d) nonspontaneous 17.2 (a) H2O(g) → H2O(l)

A liquid is more ordered than a gas. Therefore, ∆S is negative. (b) I2(g) → 2 I(g) ∆S is positive because the reaction increases the number of gaseous particles from 1 mol to 2 mol. (c) CaCO3(s) → CaO(s) + CO2(g) ∆S is positive because the reaction increases the number of gaseous molecules. (d) Ag+(aq) + Br-(aq) → AgBr(s) A solid is more ordered than +1 and -1 charged ions in an aqueous solution. Therefore, ∆S is negative.

17.3 (a) A2 + AB3 → 3 AB

(b) ∆S is positive because the reaction increases the number of gaseous molecules. 17.4 (a) disordered N2O

(b) silica glass (amorphous solid, more disorder) (c) 1 mole N2 at STP (larger volume, more disorder) (d) 1 mole N2 at 273 K and 0.25 atm (larger volume, more disorder)

17.5 CaCO3(s) → CaO(s) + CO2(g)

∆So = [So(CaO) + So(CO2)] - So(CaCO3)

∆So = [(1 mol)(39.7 J/(K⋅ mol)) + (1 mol)(213.6 J/(K ⋅ mol))] - (1 mol)(92.9 J/(K ⋅ mol)) = +160.4 J/K

17.6 From Problem 17.5, ∆Ssys = ∆So = 160.4 J/K

CaCO3(s) → CaO(s) + CO2(g) ∆Ho = [∆Ho

f(CaO) + ∆Hof(CO2)] - ∆Ho

f(CaCO3) ∆Ho = [(1 mol)(-635.1 kJ/mol) + (1 mol)(-393.5 kJ/mol)]

- (1 mol)(-1206.9 kJ/mol) = +178.3 kJ

∆Ssurr = K 298

J 178,300 _ =

T H _ o∆

= -598 J/K

∆Stotal = ∆Ssys + ∆Ssurr = 160.4 J/K + (-598 J/K) = -438 J/K Because ∆Stotal is negative, the reaction is not spontaneous under standard-state conditions at 25oC.

17.7 (a) ∆G = ∆H - T∆S = 57.1 kJ - (298 K)(0.1758 kJ/K) = +4.7 kJ

Because ∆G > 0, the reaction is nonspontaneous at 25oC (298 K)

Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________

490

(b) Set ∆G = 0 and solve for T.

0 = ∆H - T∆S; T = kJ/K 0.1758

kJ 57.1 =

S

H

∆∆

= 325 K = 52oC

17.8 (a) ∆G = ∆H - T∆S = 58.5 kJ/mol - (598 K)[0.0929 kJ/(K ⋅ mol)] = +2.9 kJ/mol

Because ∆G > 0, Hg does not boil at 325oC and 1 atm. (b) The boiling point (phase change) is associated with an equilibrium. Set ∆G = 0 and solve for T, the boiling point.

0 = ∆Hvap - T∆Svap; Tbp = mol) kJ/(K 0.0929

kJ/mol 58.5 =

S

H

vap

vap

•∆∆

= 630 K = 357oC

17.9 ∆H < 0 (reaction involves bond making - exothermic)

∆S < 0 (the reaction becomes more ordered in going from reactants (2 atoms) to products (1 molecule)

∆G < 0 (the reaction is spontaneous) 17.10 From Problems 17.5 and 17.6: ∆Ho = 178.3 kJ and ∆So = 160.4 J/K = 0.1604 kJ/K

(a) ∆Go = ∆Ho - T∆So = 178.3 kJ - (298 K)(0.1604 kJ/K) = +130.5 kJ (b) Because ∆G > 0, the reaction is nonspontaneous at 25oC (298 K). (c) Set ∆G = 0 and solve for T, the temperature above which the reaction becomes spontaneous.

0 = ∆H - T∆S; T = kJ/K 0.1604

kJ 178.3 =

S

H

∆∆

= 1112 K = 839oC

17.11 2 AB2 → A2 + 2 B2

(a) ∆So is positive because the reaction increases the number of molecules. (b) ∆Ho is positive because the reaction is endothermic. ∆Go = ∆Ho - T∆So For the reaction to be spontaneous, ∆Go must be negative. This will only occur at high temperature where T∆So is greater than ∆Ho.

17.12 (a) CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(s)

∆Go = [∆Gof(C2H2) +∆Go

f(Ca(OH)2)] - [∆Gof(CaC2) + 2 ∆Go

f(H2O)] ∆Go = [(1 mol)(209.2 kJ/mol) + (1 mol)(-898.6 kJ/mol)]

- [(1 mol)(-64.8 kJ/mol) + (2 mol)(-237.2 kJ/mol)] = -150.2 kJ This reaction can be used for the synthesis of C2H2 because ∆G < 0. (b) It is not possible to synthesize acetylene from solid graphite and gaseous H2 at 25oC and 1 atm because ∆Go

f(C2H2) > 0. 17.13 C(s) + 2 H2(g) → C2H4(g)

Qp = )(100

(0.10) =

)P(P

22H

HC

2

42 = 1.0 x 10-5

∆G = ∆Go + RT ln Qp


491

∆G = 68.1 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln(1.0 x 10-5) = +39.6 kJ/mol Because ∆G > 0, the reaction is spontaneous in the reverse direction.

17.14 ∆G = ∆Go + RT lnQ and ∆Go = 15 kJ

For A2(g) + B2(g) = 2 AB(g), Qp = )P)(P(

)P(

BA

2AB

22

Let the number of molecules be proportional to the partial pressure. (1) Qp = 1.0 (2) Qp = 0.0667 (3) Qp = 18 (a) Reaction (3) has the largest ∆G because Qp is the largest. Reaction (2) has the smallest ∆G because Qp is the smallest. (b) ∆G = ∆Go = 15 kJ because Qp = 1 and ln (1) = 0.

17.15 From Problem 17.10, ∆Go = +130.5 kJ

∆Go = -RT ln Kp

ln Kp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 130.5_ =

RT G_

3_

o

•∆

= -52.7

Kp = e-52.7 = 1 x 10-23 17.16 H2O(l) _ H2O(g)

Kp = P OH2; Kp is equal to the vapor pressure for H2O.

∆Go =∆Gof(H2O(g)) - ∆Go

f(H2O(l)) ∆Go = (1 mol)(-228.6 kJ/mol) - (1 mol)(-237.2 kJ/mol) = +8.6 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 8.6 _ =

RT G_

3_

o

•∆

= -3.5

Kp = P OH2 = e-3.5 = 0.03 atm

17.17 ∆Go = -RT ln K = -[8.314 x 10-3 kJ/(K ⋅ mol)](298 K) ln (1.0 x 10-14) = 80 kJ/mol 17.18 Photosynthetic cells in plants use the sun’s energy to make glucose, which is then used

by animals as their primary source of energy. The energy an animal obtains from glucose is then used to build and organize complex molecules, resulting in a decrease in entropy for the animal. At the same time, however, the entropy of the surroundings increases as the animal releases small, simple waste products such as CO2 and H2O. Furthermore, heat is released by the animal, further increasing the entropy of the surroundings. Thus, an organism pays for its decrease in entropy by increasing the entropy of the rest of the universe.

17.19 You would expect to see violations of the second law if you watched a movie run

backwards. Consider an action-adventure movie with a lot of explosions. An explosion is a spontaneous process that increases the entropy of the universe. You would see an explosion go backwards if you run the the movie backwards but this is impossible because it would decrease the entropy of the universe.


492

Understanding Key Concepts 17.20 (a)

(b) ∆H = 0 (no heat is gained or lost in the mixing of ideal gases) ∆S > 0 (the mixture of the two gases is more disordered) ∆G < 0 (the mixing of the two gases is spontaneous) (c) For an isolated system, ∆Ssurr = 0 and ∆Ssys = ∆STotal > 0 for the spontaneous process. (d) ∆G > 0 and the process is nonspontaneous.

17.21 ∆H > 0 (heat is absorbed during sublimation)

∆S > 0 (gas is more disordered than solid) ∆G < 0 (the reaction is spontaneous)

17.22 ∆H < 0 (heat is lost during condensation)

∆S < 0 (liquid is more ordered than vapor) ∆G < 0 (the reaction is spontaneous)

17.23 ∆H = 0 (system is an ideal gas at constant temperature)

∆S < 0 (there is more order in the smaller volume) ∆G > 0 (compression of a gas is not spontaneous)

17.24 (a) 2 A2 + B2 → 2 A2B

(b) ∆H < 0 (because ∆S is negative, ∆H must also be negative in order for ∆G to be negative)

∆S < 0 (the reaction becomes more ordered in going from reactants (3 molecules) to products (2 molecules))

∆G < 0 (the reaction is spontaneous) 17.25 (a) For initial state 1, Qp < Kp

(more reactant (A2) than product (A) compared to the equilibrium state) For initial state 2, Qp > Kp (more product (A) than reactant (A2) compared to the equilibrium state)

(b) ∆H > 0 (reaction involves bond breaking - endothermic) ∆S > 0 (equilibrium state is more disordered than initial state 1) ∆G < 0 (reaction spontaneously proceeds toward equilibrium)

(c) ∆H < 0 (reaction involves bond making - exothermic) ∆S < 0 (equilibrium state is more ordered than initial state 2)


493

∆G < 0 (reaction spontaneously proceeds toward equilibrium) (d) State 1 lies to the left of the minimum in Figure 17.10. State 2 lies to the right of the minimum.

17.26 (a) ∆Ho > 0 (reaction involves bond breaking - endothermic)

∆So > 0 (2 A's are more disordered than A2) (b) ∆So is for the complete conversion of 1 mole of A2 in its standard state to 2 moles of A in its standard state. (c) There is not enough information to say anything about the sign of ∆Go. ∆Go decreases (becomes less positive or more negative) as the temperature increases. (d) Kp increases as the temperature increases. As the temperature increases there will be more A and less A2. (e) ∆G = 0 at equilibrium.

17.27 (a) Because the free energy decreases as pure reactants form products and also decreases

as pure products form reactants, the free energy curve must go through a minimum somewhere between pure reactants and pure products. At the minimum point, ∆G = 0 and the system is at equilibrium. (b) The minimum in the plot is on the left side of the graph because ∆Go > 0 and the equilibrium composition is rich in reactants.

17.28 ∆Go = -RT ln K where K = [A]

[X] or

[A]

[Y] or

[A]

[Z]

Let the number of molecules be proportional to the concentration. (1) K = 1, ln K = 0, and ∆Go = 0. (2) K > 1, ln K is positive, and ∆Go is negative. (3) K < 1, ln K is negative, and ∆Go is positive.

17.29 The equilibrium mixture is richer in reactant A at the higher temperature. This means the

reaction is exothermic (∆H < 0). At 25oC, ∆Go < 0 because K > 1 and at 45oC, ∆Go > 0 because K < 1. Using the relationship. ∆Go = ∆Ho - T∆So, with ∆Ho < 0, ∆Go will become positive at the higher temperature only if ∆So is negative.

Additional Problems Spontaneous Processes 17.30 A spontaneous process is one that proceeds on its own without any external influence.

For example: H2O(s) → H2O(l) at 25oC A nonspontaneous process takes place only in the presence of some continuous external influence. For example: 2 NaCl(s) → 2 Na(s) + Cl2(g)

17.31 Spontaneous does not mean instantaneous. Even though the decomposition can occur (is

spontaneous), the rate of decomposition is determined by the kinetics of the reaction.


494

17.32 (a) and (d) nonspontaneous; (b) and (c) spontaneous 17.33 (a) and (c) spontaneous; (b) and (d) nonspontaneous. 17.34 (b) and (d) spontaneous (because of the large positive Kp's) 17.35 (a) and (d) nonspontaneous (because of the small K's). Entropy 17.36 Molecular randomness or disorder is called entropy. For the following reaction, the

entropy (disorder) increases: H2O(s) → H2O(l) at 25oC. 17.37 Exothermic reactions can become nonspontaneous at high temperatures if ∆S is negative.

Endothermic reactions can become spontaneous at high temperatures if ∆S is positive. 17.38 (a) + (solid → gas) (b) - (liquid → solid)

(c) - (aqueous ions → solid) (d) + (CO2(aq) → CO2(g)) 17.39 (a) + (increase in moles of gas)

(b) - (decrease in moles of gas and formation of liquid) (c) + (aqueous ions to gas) (d) - (decrease in moles of gas)

17.40 (a) - (liquid → solid)

(b) - (decrease in number of O2 molecules) (c) + (gas is more disordered in larger volume) (d) - (aqueous ions → solid)

17.41 (a) + (solid dissolved in water) (b) + (increase in moles of gas)

(c) + (mixed gases are more disordered) (d) + (liquid to gas) 17.42 S = k ln W, k = 1.38 x 10-23 J/K

(a) S = (1.38 x 10-23 J/K) ln (412) = 2.30 x 10-22 J/K (b) S = (1.38 x 10-23 J/K) ln (4120) = 2.30 x 10-21 J/K (c) S = (1.38 x 10-23 J/K) ln (4 10 x 6.02 23

) = 11.5 J/K If all C–D bonds point in the same direction, S = 0.

17.43 S = k ln W, k = 1.38 x 10-23 J/K

(a) W = 1; S = k ln (1) = 0 (b) W = 32, = 9; S = k ln (32) = 3.03 x 10-23 J/K (c) W = 1; S = k ln (1) = 0 (d) W = 33 = 27; S = k ln (33) = 4.55 x 10-23 J/K (e) W = 1; S = k ln (1) = 0

(f) W = 3 10 x 6.02 23

; S = k ln )3( 10 x 6.02 23

= 9.13 J/K


495

∆S = R ln

V

V

i

f = (8.314 J/K)ln 3 = 9.13 J/K The results are the same.

17.44 (a) H2 at 25oC in 50 L (larger volume)

(b) O2 at 25oC, 1 atm (larger volume) (c) H2 at 100oC, 1 atm (larger volume and higher T) (d) CO2 at 100oC, 0.1 atm (larger volume and higher T)

17.45 (a) ice at 0oC, because of the higher temperature.

(b) N2 at STP, because it has the larger volume. (c) N2 at 0oC and 50 L, because it has the larger volume. (d) water vapor at 150oC and 1 atm, because it has a larger volume and higher temperature.

Standard Molar Entropies and Standard Entropies of Reaction 17.46 The standard molar entropy of a substance is the entropy of 1 mol of the pure substance at

1 atm pressure and 25oC. ∆So = So(products) - So(reactants)

17.47 (a) Units of So = mol K

J

• (b) Units of ∆So = J/K

Standard molar entropies are called absolute entropies because they are measured with respect to an absolute reference point, the entropy of the substance at 0 K.

So = 0 mol K

J

• at T = 0 K.

17.48 (a) C2H6(g); more atoms/molecule

(b) CO2(g); more atoms/molecule (c) I2(g); gas is more disordered than the solid (d) CH3OH(g); gas is more disordered than the liquid.

17.49 (a) NO2(g); more atoms/molecule

(b) CH3CO2H(l); more atoms/molecule (c) Br2(l); liquid is more disordered than the solid (d) SO3(g); gas is more disordered than the solid

17.50 (a) 2 H2O2(l) → 2 H2O(l) + O2(g)

∆So = [2 So(H2O(l)) + So(O2)] - 2 So(H2O2) ∆So = [(2 mol)(69.9 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))]

- (2 mol)(110 J/(K ⋅ mol)) = +125 J/K (+, because moles of gas increase)

(b) 2 Na(s) + Cl2(g) → 2 NaCl(s) ∆So = 2 So(NaCl) - [2 So(Na) + So(Cl2)] ∆So = (2 mol)(72.1 J/(K ⋅ mol)) - [(2 mol)(51.2 J/(K ⋅ mol)) + (1 mol)(223.0 J/(K ⋅ mol))] ∆So = -181.2 J/K (-, because moles of gas decrease)


496

(c) 2 O3(g) → 3 O2(g) ∆So = 3 So(O2) - 2 So(O3) ∆So = (3 mol)(205.0 J/(K ⋅ mol)) - (2 mol)(238.8 J/(K ⋅ mol)) ∆So = +137.4 J/K (+, because moles of gas increase) (d) 4 Al(s) + 3 O2(g) → 2 Al2O3(s) ∆So = 2 So(Al 2O3) - [4 So(Al) + 3 So(O2)] ∆So = (2 mol)(50.9 J/(K ⋅ mol)) - [(4 mol)(28.3 J/(K ⋅ mol)) + (3 mol)(205.0 J/(K ⋅ mol))] ∆So = -626.4 J/K (-, because moles of gas decrease)

17.51 (a) 2 S(s) + 3 O2(g) → 2 SO3(g)

∆So = 2 So(SO3) - [2 So(S) + 3 So(O2)] ∆So = (2 mol)(256.6 J/(K ⋅ mol)) - [(2 mol)(31.8 J/(K ⋅ mol)) + (3 mol)(205.0 J/(K ⋅ mol))] ∆So = -165.4 J/K (-, because moles of gas decrease) (b) SO3(g) + H2O(l) → H2SO4(aq) ∆So = So(H2SO4) - [S

o(SO3) + So(H2O)] ∆So = (1 mol)(20 J/(K ⋅ mol)) - [(1 mol)(256.6 J/(K ⋅ mol)) + (1 mol)(69.9 J/(K ⋅ mol))] ∆So = -306 J/K (-, because of the conversion of a gas and water to an aqueous solution) (c) AgCl(s) → Ag+(aq) + Cl-(aq) ∆So = [So(Ag+) + So(Cl-)] - So(AgCl)] ∆So = [(1 mol)(72.7 J/(K ⋅ mol)) + (1 mol)(56.5 J/(K ⋅ mol))] - (1 mol)(96.2 J/(K ⋅ mol))] ∆So = +33.0 J/K (+, because a solid is converted to ions in aqueous solution) (d) NH4NO3(s) → N2O(g) + 2 H2O(g) ∆So = [So(N2O) + 2 So(H2O)] - So(NH4NO3) ∆So = [(1 mol)(219.7 J/(K ⋅ mol)) + (2 mol)(188.7 J/(K ⋅ mol))] - (1 mol)(151.1 J/(K ⋅ mol)) ∆So = +446.0 J/K (+, because moles of gas increase)

Entropy and the Second Law of Thermodynamics 17.52 In any spontaneous process, the total entropy of a system and its surroundings always

increases. 17.53 For a spontaneous process, ∆Stotal = ∆Ssys + ∆Ssurr > 0. For an isolated system, ∆Ssurr = 0,

and so ∆Ssys > 0 is the criterion for spontaneous change. An example of a spontaneous process in an isolated system is the mixing of two gases.

17.54 ∆Ssurr = T

H _ ∆; the temperature (T) is always positive.

(a) For an exothermic reaction, ∆H is negative and ∆Ssurr is positive. (b) For an endothermic reaction, ∆H is positive and ∆Ssurr is negative.

17.55 ∆Ssurr ∝ T

1

Consider the surroundings as an infinitely large constant-temperature bath to which heat can be added without changing its temperature. If the surroundings have a low temperature, they have only a small amount of disorder, in which case addition of a given


497

quantity of heat results in a substantial increase in the amount of disorder (a relatively large value of ∆Ssurr). If the surroundings have a high temperature, they already have a large amount of disorder, and addition of the same quantity of heat produces only a marginal increase in the amount of disorder (a relatively small value of ∆Ssurr). Thus, we expect ∆Ssurr to vary inversely with temperature.

17.56 N2(g) + 2 O2(g) → N2O4(g)

∆Ho = ∆Hof(N2O4) = 9.16 kJ

∆Ssys = ∆So = So(N2O4) - [So(N2) + 2 So(O2)]

∆Ssys = (1 mol)(304.2 J/(K⋅ mol)) - [(1 mol)(191.5 J/(K ⋅ mol)) + (2 mol)(205.0 J/(K ⋅ mol))] = -297.3 J/K

∆Ssurr = K 298

kJ 9.16 _ =

T H _ o∆

= -0.0307 kJ/K = -30.7 J/K

∆Stotal = ∆Ssys + ∆Ssurr = -297.3 J/K + (-30.7 J/K) = -328.0 J/K Because ∆Stotal < 0, the reaction is nonspontaneous.

17.57 Cu2S(s) + O2(g) → 2 Cu(s) + SO2(g)

∆Ho = ∆Hof(SO2) - ∆Ho

f(Cu2S) ∆Ho = (1 mol)(-296.8 kJ/mol) - (1 mol)(-79.5 kJ/mol) = -217.3 kJ ∆Ssys = ∆So = [2 So(Cu) + So(SO2)] - [S

o(Cu2S) + So(O2)] ∆Ssys = [(2 mol)(33.1 J/(K ⋅ mol)) + (1 mol)(248.1 J/(K ⋅ mol))]

- [(1 mol)(120.9 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] = -11.6 J/K

∆Ssurr = K 298.15

J) 217,300(_ _ = overT H _ o∆ = +728.8 J/K

∆Stotal = ∆Ssys + ∆Ssurr = -11.6 J/K + 728.8 J/K = +717.2 J/K Because ∆Stotal is positive, the reaction is spontaneous under standard-state conditions at 25oC.

17.58 (a) ∆Ssurr = K 343

J/mol 30,700 _ =

TH _ vap∆

= -89.5 J/(K ⋅ mol)

∆Stotal = ∆Svap + ∆Ssurr = 87.0 J/(K ⋅ mol) + (-89.5 J/(K ⋅ mol)) = - 2.5 J/(K ⋅ mol)

(b) ∆Ssurr = K 353

J/mol 30,700 _ =

TH _ vap∆

= -87.0 J/(K ⋅ mol)

∆Stotal = ∆Svap + ∆Ssurr = 87.0 J/(K ⋅ mol) + (- 87.0 J/(K ⋅ mol)) = 0

(c) ∆Ssurr = K 363

J/mol 30,700 _ =

TH _ vap∆

= -84.6 J/(K ⋅ mol)

∆Stotal = ∆Svap + ∆Ssurr = 87.0 J/(K ⋅ mol) + (- 84.6 J/(K ⋅ mol)) = +2.4 J/(K ⋅ mol) Benzene does not boil at 70oC (343 K) because ∆Stotal is negative.


498

The normal boiling point for benzene is 80oC (353 K), where ∆Stotal = 0.

17.59 (a) ∆Ssurr = K 1050

J/mol 30,200 _ =

TH _ fusion∆

= -28.8 J/(K ⋅ mol)

∆Stotal = ∆Ssys + ∆Ssurr = 28.1 J/(K ⋅ mol) + (-28.8 J/(K ⋅ mol)) = -0.7 J/(K ⋅ mol)

(b) ∆Ssurr = K 1075

J/mol 30,200 _ =

TH _ fusion∆

= -28.1 J/(K ⋅ mol)

∆Stotal = ∆Ssys + ∆Ssurr = 28.1 J/(K ⋅ mol) + (-28.1 J/(K ⋅ mol)) = 0

(c) ∆Ssurr = K 1100

J/mol 30,200 _ =

TH _ fusion∆

= -27.5 J/(K ⋅ mol)

∆Stotal = ∆Ssys + ∆Ssurr = 28.1 J/(K ⋅ mol) + (-27.5 J/(K ⋅ mol)) = +0.6 J/(K ⋅ mol) NaCl melts at 1100 K because ∆Stotal > 0. The melting point of NaCl is 1075 K, where ∆Stotal = 0.

Free Energy 17.60 ∆H ∆S ∆G = ∆H - T∆S Reaction Spontaneity

- + - Spontaneous at all temperatures - - - or + Spontaneous at low temperatures

where ∆H > T∆S Nonspontaneous at high temperatures

where ∆H < T∆S + - + Nonspontaneous at all temperatures + + - or + Spontaneous at high temperatures

where T∆S > ∆H Nonspontaneous at low temperature

where T∆S < ∆H 17.61 When ∆H and ∆S are both positive or both negative, the temperature determines the

direction of spontaneous reaction. See Problem 17.60 for an explanation. 17.62 (a) 0oC (temperature is below mp); ∆H > 0, ∆S > 0, ∆G > 0

(b) 15oC (temperature is above mp); ∆H > 0, ∆S > 0, ∆G < 0 17.63 (a) ∆H = 0

∆S = R ln V

V

initial

final = (8.314 J/K) ln 2 = 5.76 J/K

∆G = ∆H - T∆S Because ∆H = 0, ∆G = -T∆S = -(298 K)(5.76 J/K) = -1717 J = -1.72 kJ


499

(b) For a process in an isolated system, ∆Ssurr = 0. Therefore, ∆Stotal = ∆Ssys > 0, and the process is spontaneous.

17.64 ∆Hvap = 30.7 kJ/mol

∆Svap = 87.0 J/(K ⋅ mol) = 87.0 x 10-3 kJ/(K ⋅ mol) ∆Gvap = ∆Hvap - T∆Svap (a) ∆Gvap = 30.7 kJ/mol - (343 K)(87.0 x 10-3 kJ/(K ⋅ mol)) = +0.9 kJ/mol At 70oC (343 K), benzene does not boil because ∆Gvap is positive. (b) ∆Gvap = 30.7 kJ/mol - (353 K)(87.0 x 10-3 kJ/(K ⋅ mol)) = 0 80oC (353 K) is the boiling point for benzene because ∆Gvap = 0 (c) ∆Gvap = 30.7 kJ/mol - (363 K)(87.0 x 10-3 kJ/(K ⋅ mol)) = -0.9 kJ/mol At 90oC (363 K), benzene boils because ∆Gvap is negative.

17.65 ∆Hfusion = 30.2 kJ/mol; ∆Sfusion = 28.1 x 10-3 kJ/(K ⋅ mol)

∆Gfusion = ∆Hfusion - T∆Sfusion (a) ∆Gfusion = 30.2 kJ/mol - (1050 K)(28.1 x 10-3 kJ/(K ⋅ mol)) = +0.7 kJ/mol At 1050 K, NaCl does not melt because ∆Gfusion is positive. (b) ∆Gfusion = 30.2 kJ/mol - (1075 K)(28.1 x 10-3 kJ/(K ⋅ mol)) = 0 1075 K is the melting point for NaCl because ∆Gfusion = 0. (c) ∆Gfusion = 30.2 kJ/mol - (1100 K)(28.1 x 10-3 kJ/(K ⋅ mol)) = -0.7 kJ/mol At 1100 K, NaCl does melt because ∆Gfusion is negative.

17.66 At the melting point (phase change), ∆Gfusion = 0

∆Gfusion = ∆Hfusion - T∆Sfusion

0 = ∆Hfusion - T∆Sfusion; T = mol) kJ/(K 10 x 43.8

kJ/mol 17.3 =

S

H3_

fusion

fusion

•∆∆

= 395 K = 122oC

17.67 128oC = 401 K

At the melting point (phase change), ∆Gfusion = 0 ∆Gfusion = ∆Hfusion - T∆Sfusion 0 = ∆Hfusion - T∆Sfusion ∆Hfusion = T∆Sfusion = (401 K)[47.7 x 10-3 kJ/(K ⋅ mol)] = 19.1 kJ/mol

Standard Free-Energy Changes and Standard Free Energies of Formation 17.68 (a) ∆Go is the change in free energy that occurs when reactants in their standard states are

converted to products in their standard states. (b) ∆Go

f is the free-energy change for formation of one mole of a substance in its standard state from the most stable form of the constituent elements in their standard states.

17.69 The standard state of a substance (solid, liquid, or gas) is the most stable form of a pure

substance at 25oC and 1 atm pressure. For solutes, the condition is 1 M at 25oC.


500

17.70 (a) N2(g) + 2 O2(g) → 2 NO2(g)

∆Ho = 2 ∆Hof(NO2) = (2 mol)(33.2 kJ/mol) = 66.4 kJ

∆So = 2 So(NO2) - [So(N2) + 2 So(O2)]

∆So = (2 mol)(240.0 J/(K ⋅ mol)) - [(1 mol)(191.5 J/(K ⋅ mol)) + (2 mol)(205.0 J/(K ⋅ mol))] ∆So = -121.5 J/K = -121.5 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = 66.4 kJ - (298 K)(-121.5 x 10-3 kJ/K) = +102.6 kJ Because ∆Go is positive, the reaction is nonspontaneous under standard-state conditions at 25oC. (b) 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ∆Ho = 2 ∆Ho

f(KCl) - 2 ∆Hof(KClO3)

∆Ho = (2 mol)(-436.7 kJ/mol) - (2 mol)(-397.7 kJ/mol) = -78.0 kJ ∆So = [2 So(KCl) + 3 So(O2)] - 2 So(KClO3) ∆So = [(2 mol)(82.6 J/(K ⋅ mol)) + (3 mol)(205.0 J/(K ⋅ mol))] - (2 mol)(143 J/(K ⋅ mol)) ∆So = 494.2 J/(K ⋅ mol) = 494.2 x 10-3 kJ/(K ⋅ mol) ∆Go = ∆Ho - T∆So = -78.0 kJ - (298 K)(494.2 x 10-3 kJ/(K ⋅ mol)) = -225.3 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC. (c) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ∆Ho = [∆Ho

f(CH3CO2H) + ∆Hof(H2O)] - ∆Ho

f(CH3CH2OH) ∆Ho = [(1 mol)(-484.5 kJ/mol) + (1 mol)(-285.8 kJ/mol)] - (1 mol)(-277.7 kJ/mol) = - 492.6 kJ ∆So = [So(CH3CO2H) + So(H2O)] - [So(CH3CH2OH) + So(O2)] ∆So = [(1 mol)(160 J/(K ⋅ mol)) + (1 mol)(69.9 J/(K ⋅ mol))]

- [(1 mol)(161 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] ∆So = -136.1 J/(K ⋅ mol) = -136.1 x 10-3 kJ/(K ⋅ mol) ∆Go = ∆Ho - T∆So = -492.6 kJ - (298 K)(-136.1 x 10-3 kJ/(K ⋅ mol)) = - 452.0 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC.

17.71 (a) 2 SO2(g) + O2(g) → 2 SO3(g)

∆Ho = 2 ∆Hof(SO3) - 2 ∆Ho

f SO2) ∆Ho = (2 mol)(-395.7 kJ/mol) - (2 mol)(-296.8 kJ/mol) = -197.8 kJ ∆So = 2 So(SO3) - [2 So(SO2) + So(O2)] ∆So = (2 mol)(256.6 J/(K ⋅ mol)) - [(2 mol)(248.1 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] ∆So = -188.0 J/K = -188.0 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -197.8 kJ - (298 K)(-188.0 x 10-3 kJ/K) = -141.8 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC. (b) N2(g) + 2 H2(g) → N2H4(l) ∆Ho = ∆Ho

f(N2H4) ∆Ho = (1 mol)(50.6 kJ/mol) = 50.6 kJ ∆So = So(N2H4) - [S

o(N2) + 2 So(H2)] ∆So = (1 mol)(121.2 J/(K ⋅ mol)) - [(1 mol)(191.5 J/(K ⋅ mol)) + (2 mol)(130.6 J/(K ⋅ mol))] ∆So = -331.5 J/K = -331.5 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = 50.6 kJ - (298 K)(-331.5 x 10-3 kJ/K) = +149.4 kJ Because ∆Go is positive, the reaction is nonspontaneous under standard-state conditions at 25oC. (c) CH3OH(l) + O2(g) → HCO2H(l) + H2O(l) ∆Ho = [∆Ho

f(HCO2H) + ∆Hof(H2O)] - ∆Ho

f(CH3OH) ∆Ho = [(1 mol)(-424.7 kJ/mol) + (1 mol)(-285.8 kJ/mol)] - (1 mol)(-238.7 kJ/mol) = - 471.8 kJ


501

∆So = [So(HCO2H) + So(H2O)] - [So(CH3OH) + So(O2)]

∆So = [(1 mol)(129.0 J/(K ⋅ mol)) + (1 mol)(69.9 J/(K ⋅ mol))] - [(1 mol)(127 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))]

∆So = -133.1 J/K = -133.1 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -471.8 kJ - (298 K)(-133.1 x 10-3 kJ/K) = - 432.1 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC.

17.72 (a) N2(g) + 2 O2(g) → 2 NO2(g)

∆Go = 2 ∆Gof(NO2) = (2 mol)(51.3 kJ/mol) = +102.6 kJ

(b) 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ∆Go = 2 ∆Go

f(KCl) - 2 ∆Gof(KClO3)

∆Go = (2 mol)(- 409.2 kJ/mol) - (2 mol)(-296.3 kJ/mol) = -225.8 kJ (c) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ∆Go = [∆Go

f(CH3CO2H) +∆Gof(H2O)] -∆Go

f(CH3CH2OH) ∆Go = [(1 mol)(-390 kJ/mol) + (1 mol)(-237.2 kJ/mol)] - (1 mol)(-174.9 kJ/mol) = - 452 kJ

17.73 (a) 2 SO2(g) + O2(g) → 2 SO3(g)

∆Go = 2 ∆Gof(SO3) - 2 ∆Go

f(SO2) ∆Go = (2 mol)(-371.1 kJ/mol) - (2 mol)(-300.2 kJ/mol) = -141.8 kJ (b) N2(g) + 2 H2(g) → N2H4(l) ∆Go = ∆Go

f(N2H4) = (1 mol)(149.2 kJ/mol) = 149.2 kJ (c) CH3OH(l) + O2(g) → HCO2H(l) + H2O(l) ∆Go = [∆Go

f(HCO2H) +∆Gof(H2O)] - ∆Go

f(CH3OH) ∆Go = [(1 mol)(-361.4 kJ/mol) + (1 mol)(-237.2 kJ/mol)] - (1 mol)(-166.4 kJ/mol) ∆Go = - 432.2 kJ

17.74 A compound is thermodynamically stable with respect to its constituent elements at 25oC

if ∆Gof is negative.

∆Gof (kJ/mol) Stable

(a) BaCO3(s) -1138 yes (b) HBr(g) -53.4 yes (c) N2O(g) +104.2 no (d) C2H4(g) +68.1 no

17.75 A compound is thermodynamically stable with respect to its constituent elements at 25oC

if ∆Gof is negative.

∆Gof (kJ/mol) Stable

(a) C6H6(l) +124.5 no (b) NO(g) +86.6 no (c) PH3(g) +13 no (d) FeO(s) -255 yes

17.76 CH2=CH2(g) + H2O(l) → CH3CH2OH(l)

∆Ho =∆Hof(CH2CH2OH) - [∆Ho

f(CH2=CH2) + ∆Hof(H2O)]


502

∆Ho = (1 mol)(-277.7 kJ/mol) - [(1 mol)(52.3 kJ/mol) + (1 mol)(-285.8 kJ/mol)] ∆Ho = - 44.2 kJ ∆So = So(CH3CH2OH) - [So(CH2=CH2) + So(H2O)] ∆So = (1 mol)(161 J/(K ⋅ mol)) - [(1 mol)(219.5 J/(K ⋅ mol)) + (1 mol)(69.9 J/(K ⋅ mol))] ∆So = -128 J/(K ⋅ mol) = -128 x 10-3 kJ/(K ⋅ mol) ∆Go = ∆Ho - T∆So = - 44.2 kJ - (298 K)(-128 x 10-3 kJ/K) = -6.1 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC. The reaction becomes nonspontaneous at high temperatures because ∆So is negative. To find the crossover temperature, set ∆G = 0 and solve for T.

T = J/K 128_

J 44,200 _ =

S Ho

o

∆∆

= 345 K = 72oC

The reaction becomes nonspontaneous at 72oC. 17.77 2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g)

∆Ho = 2 ∆Hof(H2O) - [2 ∆Ho

f(H2S) + ∆Hof(SO2)]

∆Ho = (2 mol)(-241.8 kJ/mol) - [(2 mol)(-20.6 kJ/mol) + (1 mol)(-296.8 kJ/mol) = -145.6 kJ ∆So = [3 So(S) + 2 So(H2O)] - [2 So(H2S) + So(SO2)] ∆So = [(3 mol)(31.8 J/(K ⋅ mol)) + (2 mol)(188.7 J/(K ⋅ mol))]

- [(2 mol)(205.7 J/(K ⋅ mol)) + (1 mol)(248.1 J/(K ⋅ mol))] ∆So = -186.7 J/K = -186.7 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -145.6 kJ - (298 K)(-186.7 x 10-3 kJ/K) = -90.0 kJ Because ∆Go is negative, the reaction is spontaneous under standard-state conditions at 25oC. The reaction becomes nonspontaneous at high temperatures because ∆So is negative. To find the crossover temperature set ∆G = 0 and solve for T.

T = J/K 186.7_

J 145,600_ =

S Ho

o

∆∆

= 780 K = 507oC

The reaction becomes nonspontaneous at 507oC. 17.78 3 C2H2(g) → C6H6(l)

∆Go =∆Gof(C6H6) - 3 ∆Go

f(C2H2) ∆Go = (1 mol)(124.5 kJ/mol) - (3 mol)(209.2 kJ/mol) = -503.1 kJ Because ∆Go is negative, the reaction is possible. Look for a catalyst. Because ∆Go

f for benzene is positive (+124.5 kJ/mol), the synthesis of benzene from graphite and gaseous H2 at 25oC and 1 atm pressure is not possible.

17.79 CH2ClCH2Cl(l) → CH2=CHCl(g) + HCl(g)

∆Go = [∆Gof(CH2=CHCl) +∆Go

f(HCl)] - ∆Gof(CH2ClCH2Cl)

∆Go = [(1 mol)(51.9 kJ/mol) + (1 mol)(-95.3 kJ/mol)] - (1 mol)(-79.6 kJ/mol) = +36.2 kJ Because ∆Go is positive, the reaction is nonspontaneous under standard-state conditions at 25oC.

CH2ClCH2Cl(l) → CH2=CHCl(g) + HCl(g)

Sum: NaOH(aq) + HCl(g) → Na+(aq) + Cl-(aq) + H2O(l) CH2ClCH2Cl(l) + NaOH(aq) → CH2=CHCl(g) + Na+(aq) + Cl-(aq) + H2O(l) ∆Go = [∆Go

f(CH2=CHCl) +∆Gof(Na+) + ∆Go

f(Cl-) + ∆Gof(H2O)]


503

- [∆Gof(CH2ClCH2Cl) + ∆Go

f(NaOH)] ∆Go = [(1 mol)(51.9 kJ/mol) + (1 mol)(-261.9 kJ/mol)

+ (1 mol)(-131.3 kJ/mol) + (1 mol)(-237.2 kJ/mol)] - [(1 mol)(-79.6 kJ/mol) + (1 mol)(-419.2 kJ/mol)] = -79.7 kJ

Using NaOH(aq), ∆Go = -79.7 kJ and the reaction is spontaneous. (More generally, base removes HCl, driving the reaction to the right.) The synthesis of a compound from its constituent elements is thermodynamically feasible at 25oC and 1 atm pressure if ∆Go

f is negative. Because ∆Go

f(CH2=CHCl) = +51.9 kJ, the synthesis of vinyl chloride from its elements is not possible at 25oC and 1 atm pressure.

Free Energy, Composition, and Chemical Equilibrium 17.80 ∆G = ∆Go + RT ln Q 17.81 ∆G = ∆Go + RT ln Q

(a) If Q < 1, then RT ln Q is negative and ∆G < ∆Go. (b) If Q = 1, then RT ln Q = 0 and ∆G = ∆Go. (c) If Q > 1, then RT ln Q is positive and ∆G > ∆Go. As Q increases the thermodynamic driving force decreases.

17.82 ∆G = ∆Go + RT ln

)P()P(

)P(

O2

SO

2SO

22

3

(a) ∆G = (-141.8 kJ/mol) + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln

(100))(100

)(1.02

2

= -176.0 kJ/mol

(b) ∆G = (-141.8 kJ/mol) + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln

(1.0))(2.0

)(102

2

= -133.8 kJ/mol

(c) Q = 1, ln Q = 0, ∆G = ∆Go = -141.8 kJ/mol

17.83 ∆G = ∆Go + RT ln

)P()P(

]CONHNH[

CO2

NH

22

23

(a) ∆G = -13.6 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln

(10))(10

1.02

= -30.7 kJ/mol

Because ∆G is negative, the reaction is spontaneous.

(b) ∆G = -13.6 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln

(0.10))(0.10

1.02

= +3.5 kJ/mol

Because ∆G is positive, the reaction is nonspontaneous. 17.84 ∆Go = -RT ln K

(a) If K > 1, ∆Go is negative. (b) If K = 1, ∆Go = 0.


504

(c) If K < 1, ∆Go is positive.

17.85 K = e RT G _ o∆

(a) If ∆Go is positive, K is small. (b) If ∆Go is negative, K is large. 17.86 ∆Go = -RT ln Kp = -141.8 kJ

ln Kp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 141.8(_ _ =

RT G _

3_

o

•∆

= 57.23

Kp = e57.23 = 7.1 x 1024 17.87 ∆Go = - RT ln K = -13.6 kJ

ln K = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 13.6(_ _ =

RT G _

3_

o

•∆

= 5.49

K = e5.49 = 2.4 x 102 17.88 C2H5OH(l) _ C2H5OH(g)

∆Go = ∆Gof(C2H5OH(g)) - ∆Go

f(C2H5OH(l)) ∆Go = (1 mol)(-168.6 kJ/mol) - (1 mol)(-174.9 kJ/mol) = +6.3 kJ ∆Go = -RT ln K

ln K = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) (6.3 _ =

RT G _

3_

o

•∆

= -2.54

K = e-2.54 = 0.079; K = Kp = P OHHC 52 = 0.079 atm

17.89 ∆Go = -RT ln Ka

∆Go = -[8.314 x 10-3 kJ/(K ⋅ mol)](298 K)ln(3.0 x 10-4) = +20.1 kJ/mol 17.90 2 CH2=CH2(g) + O2(g) → 2 C2H4O(g)

∆Go = 2 ∆Gof(C2H4O) - 2 ∆Go

f(CH2=CH2) ∆Go = (2 mol)(-13.1 kJ/mol) - (2 mol)(68.1 kJ/mol) = -162.4 kJ ∆Go = -RT ln K

ln K = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 162.4(_ _ =

RT G _

3_

o

•∆

= 65.55

K = Kp = e65.55 = 2.9 x 1028 17.91 CO(g) + 2 H2(g) _ CH3OH(g)

∆Go = ∆Gof(CH3OH) - ∆Go

f(CO) ∆Go = (1 mol)(-161.9 kJ/mol) - (1 mol)(-137.2 kJ/mol) = -24.7 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 24.7(_ _ =

RT G _

3_

o

•∆

= 9.97

Kp = e9.97 = 2.1 x 104 ∆G = ∆Go + RT ln Q

∆G = -24.7 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K) ln

)(20)(20

202

= -39.5 kJ/mol


505

General Problems 17.92 The kinetic parameters [(a), (b), and (h)] are affected by a catalyst. The thermodynamic

and equilibrium parameters [(c), (d), (e), (f), and (g)] are not affected by a catalyst. 17.93 (a), (c), and (d) are nonspontaneous; (b) is spontaneous. 17.94 (a) Spontaneous does not mean fast, just possible.

(b) For a spontaneous reaction ∆Stotal > 0. ∆Ssys can be positive or negative. (c) An endothermic reaction can be spontaneous if ∆Ssys > 0. (d) This statement is true because the sign of ∆G changes when the direction of a reaction is reversed.

17.95 Point Total Possible Ways Number of Ways

2 (1+1) 1 3 (2+1)(1+2) 2 4 (1+3)(2+2)(3+1) 3 5 (1+4)(2+3)(3+2)(4+1) 4 6 (1+5)(2+4)(3+3)(4+2)(5+1) 5 7 (1+6)(2+5)(3+4)(4+3)(5+2)(6+1) 6 8 (2+6)(3+5)(4+4)(5+3)(6+2) 5 9 (3+6)(4+5)(5+4)(6+3) 4

10 (4+6)(5+5)(6+4) 3 11 (6+5)(5+6) 2

12 (6+6) 1 Because a point total of 7 can be rolled in the most ways, it is the most probable point total.

17.96 17.97 (a) Q = 1, ln Q = 0, ∆G = ∆Go = +79.9 kJ

Because ∆G is positive, the reaction is spontaneous in the reverse direction. (b) ∆G = ∆Go + RT ln Q; Q = [H3O

+][OH-] = (1.0 x 10-7)2 = 1.0 x 10-14 ∆G = 79.9 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K) ln(1.0 x 10-14) = 0 Because ∆G = 0, the reaction is at equilibrium. (c) ∆G = ∆Go + RT ln Q


506

Q = [H3O+][OH-] = (1.0 x 10-7)(1.0 x 10-10) = 1.0 x 10-17

∆G = 79.9 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K) ln(1.0 x 10-17) = -17.1 kJ/mol Because ∆G is negative, the reaction is spontaneous in the forward direction. The results are consistent with Le Châtelier's principle. When the [H3O

+] and [OH-] are larger than the equilibrium concentrations (a), the reverse reaction takes place. When the product of [H3O

+] and the [OH-] is less than the equilibrium value, the forward reaction is spontaneous. ∆Go = -RT ln K

ln K = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 79.9_ =

RT G _

3_

o

•∆

= -32.25

K = Ka = e-32.25 = 9.9 x 10-15 17.98 At the normal boiling point, ∆G = 0.

∆Gvap = ∆Hvap - T∆Svap; T = J/K 110

J 38,600 =

S

H

vap

vap

∆∆

= 351 K = 78oC

17.99 At the normal boiling point, ∆Gvap = 0. 61oC = 334 K

∆Gvap = ∆Hvap - T∆Svap; ∆Svap = K 334

J 29,240 =

THvap∆

= 87.5 J/K

17.100 ∆G = ∆H - T∆S

(a) ∆H must be positive (endothermic) and greater than T∆S in order for ∆G to be positive (nonspontaneous reaction). (b) Set ∆G = 0 and solve for ∆H. ∆G = 0 = ∆H - T∆S = ∆H - (323 K)(104 J/K) = ∆H - (33592 J) = ∆H - (33.6 kJ) ∆H = 33.6 kJ ∆H must be greater than 33.6 kJ.

17.101 NH4NO3(s) → N2O(g) + 2 H2O(g)

(a) ∆Go = [∆Gof(N2O) + 2 ∆Go

f(H2O)] - ∆Gof(NH4NO3)

∆Go = [(1 mol)(104.2 kJ/mol) + (2 mol)(-228.6 kJ/mol)] - (1 mol)(-184.0 kJ/mol) ∆Go = -169.0 kJ Because ∆Go is negative, the reaction is spontaneous. (b) Because the reaction increases the number of moles of gas, ∆So is positive. ∆Go = ∆Ho - T∆So As the temperature is raised, ∆Go becomes more negative. (c) ∆Go = -RT ln K

ln K = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 169.0(_ _ =

RT G _

3_

o

•∆

= 68.21

K = Kp = e68.21 = 4.2 x 1029

(d) Q = )P)(P(2

OHON 22 = (30)(30)2 = (30)3

∆G = ∆Go + RT ln Q ∆G = -169.0 kJ/mol + [8.314 x 10-3kJ/(K ⋅ mol)](298 K) ln[(30)3] = -143.7 kJ/mol


507

17.102 (a) 2 Mg(s) + O2(g) → 2 MgO(s)

∆Ho = 2 ∆Hof(MgO) = (2 mol)(-601.7 kJ/mol) = -1203.4 kJ

∆So = 2 So(MgO) - [2 So(Mg) + So(O2)] ∆So = (2 mol)(26.9 J/(K ⋅ mol)) - [(2 mol)(32.7 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] ∆So = -216.6 J/K = -216.6 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -1203.4 kJ - (298 K)(-216.6 x 10-3 kJ/K) = -1138.8 kJ Because ∆Go is negative, the reaction is spontaneous at 25oC. ∆Go becomes less negative as the temperature is raised. (b) MgCO3(s) → MgO(s) + CO2(g) ∆Ho = [∆Ho

f(MgO) + ∆Hof(CO2)] - ∆Ho

f(MgCO3) ∆Ho = [(1 mol)(-601.1 kJ/mol) + (1 mol)(-393.5 kJ/mol)] - (1 mol)(-1096 kJ/mol) = +101 kJ ∆So = [So(MgO) + So(CO2)] - S

o(MgCO3) ∆So = [(1 mol)(26.9 J/(K ⋅ mol)) + (1 mol)(213.6 J/(K ⋅ mol))] - (1 mol)(65.7 J/(K ⋅ mol)) ∆So = 174.8 J/K = 174.8 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = 101 kJ - (298 K)(174.8 x 10-3 kJ/K) = +49 kJ Because ∆Go is positive, the reaction is not spontaneous at 25oC. ∆Go becomes less positive as the temperature is raised. (c) Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ∆Ho = ∆Ho

f(Al 2O3) - ∆Hof(Fe2O3)

∆Ho = (1 mol)(-1676 kJ/mol) - (1 mol)(-824.2 kJ/mol) = -852 kJ ∆So = [So(Al 2O3) + 2 So(Fe)] - [So(Fe2O3) + 2 So(Al)] ∆So = [(1 mol)(50.9 J/(K ⋅ mol)) + (2 mol)(27.3 J/(K ⋅ mol))]

- [(1 mol)(87.4 J/(K ⋅ mol)) + (2 mol)(28.3 J/(K ⋅ mol))] ∆So = -38.5 J/K = -38.5 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -852 kJ - (298 K)(-38.5 x 10-3 kJ/K) = -840 kJ Because ∆Go is negative, the reaction is spontaneous at 25oC. ∆Go becomes less negative as the temperature is raised. (d) 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) ∆Ho = [∆Ho

f(Na2CO3) + ∆Hof(CO2) + ∆Ho

f(H2O)] - 2 ∆Hof(NaHCO3)

∆Ho = [(1 mol)(-1130.7 kJ/mol) + (1 mol)(-393.5 kJ/mol) + (1 mol)(-241.8 kJ/mol)] - (2 mol)(-950.8 kJ/mol) = +135.6 kJ

∆So = [So(Na2CO3) + So(CO2) + So(H2O)] - 2 So(NaHCO3) ∆So = [(1 mol)(135.0 J/(K ⋅ mol)) + (1 mol)(213.6 J/(K ⋅ mol))

+ (1 mol)(188.7 J/(K ⋅ mol))] - (2 mol)(102 J/(K ⋅ mol)) ∆So = +333 J/K = +333 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = +135.6 kJ - (298 K)(+333 x 10-3 kJ/K) = +36.4 kJ Because ∆Go is positive, the reaction is not spontaneous at 25oC. ∆Go becomes less positive as the temperature is raised.

17.103 (a) ∆Hvap/Tbp

ammonia 120 J/K benzene 87 J/K carbon tetrachloride 85 J/K


508

chloroform 87 J/K mercury 90 J/K

(b) All processes are the conversion of a liquid to a gas at the boiling point. They should should all have similar ∆S values. ∆Hvap/Tbp is equal to ∆Svap. (c) NH3 deviates from Trouton's rule because of hydrogen bonding. Because NH3(l) is more ordered than the other liquids, ∆Svap is larger.

17.104 (a) 6 C(s) + 3 H2(g) → C6H6(l)

∆Sof = So(C6H6) - [6 So(C) + 3 So(H2)]

∆Sof = (1 mol)(172.8 J/(K ⋅ mol)) - [(6 mol)(5.7 J/(K ⋅ mol)) + (3 mol)(130.6 J/(K ⋅ mol))]

∆Sof = -253 J/K = -253 J/(K ⋅ mol)

∆Gof = ∆Ho

f - T∆Sof

∆Sof =

K 298

kJ/mol 124.5 _ kJ/mol 49.0 =

T G _ H o

fof ∆∆

= -0.253 kJ/(K ⋅ mol)

∆Sof = -253 J/(K ⋅ mol)

Both calculations lead to the same value of ∆Sof.

(b) Ca(s) + S(s) + 2 O2(g) → CaSO4(s) ∆So

f = So(CaSO4) - [So(Ca) + So(S) + 2 So(O2)]

∆Sof = (1 mol)(107 J/(K ⋅ mol))

- [(1 mol)(41.4 J/(K ⋅ mol)) + (1 mol)(31.8 J/(K ⋅ mol)) + (2 mol)(205.0 J/(K ⋅ mol))] ∆So

f = -376 J/K = -376 J/(K ⋅ mol)

∆Gof = ∆Ho

f - T∆Sof

∆Sof =

K 298

kJ/mol) 1321.9(_ _ kJ/mol 1434.1_ =

T G _ H o

fof ∆∆

= -0.376 kJ/(K ⋅ mol)

∆Sof = -376 J/(K ⋅ mol)


(c) 2 C(s) + 3 H2(g) + 1/2 O2(g) → C2H5OH(l) ∆So

f = So(C2H5OH) - [So(C) + So(H2) + 1/2 So(O2)]

∆Sof = (1 mol)(161 J/(K ⋅ mol))

- [(2 mol)(5.7 J/(K ⋅ mol)) + (3 mol)(130.6 J/(K ⋅ mol)) + (0.5 mol)(205.0 J/(K ⋅ mol))] ∆So

f = -345 J/K = -345 J/(K ⋅ mol)

∆Gof = ∆Ho

f - T∆Sof

∆Sof =

K 298

kJ/mol) 174.9(_ _ kJ/mol 277.7_ =

T G _ H o

fof ∆∆

= -0.345 kJ/(K ⋅ mol)

∆Sof = -345 J/(K ⋅ mol)


17.105 MgCO3(s) → MgO(s) + CO2(g)

From Problem 17.102(b) ∆Ho = +101 kJ; ∆So = 174.8 J/K = 174.8 x 10-3 kJ/K The equilibrium pressure of CO2 is equal to Kp = PCO2

. Kp is not affected by the

quantities of MgCO3 and MgO present. Kp can be calculated from ∆Go. ∆Go = ∆Ho - T∆So ∆Go = -RT ln Kp


509

(a) ∆Go = 101 kJ - (298 K)(174.8 x 10-3 kJ/K) = +49 kJ

ln Kp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 49 _ =

RT G _

3_

o

•∆

= - 19.8

Kp = PCO2 = e-19.8 = 3 x 10-9 atm

(b) ∆Go = 101 kJ - (553 K)(174.8 x 10-3 kJ/K) = 4.3 kJ

ln Kp = K) mol)](553 kJ/(K 10 x [8.314

kJ/mol 4.3 _ =

RT G _

3_

o

•∆

= -0.94

Kp = PCO2 = e-0.94 = 0.39 atm

(c) PCO2 = 0.39 atm because the temperature is the same as in (b).

17.106 ∆Go = -RT ln Kb

At 20 oC: ∆Go = -[8.314 x 10-3 kJ/(K ⋅ mol)](293 K) ln(1.710 x 10-5) = +26.74 kJ/mol At 50 oC: ∆Go = -[8.314 x 10-3 kJ/(K ⋅ mol)](323 K) ln(1.892 x 10-5) = +29.20 kJ/mol ∆Go = ∆Ho - T∆So

26.74 = ∆Ho - 293∆So 29.20 = ∆Ho - 323∆So Solve these two equations simultaneously for ∆Ho and ∆So.

26.74 + 293∆So = ∆Ho 29.20 + 323∆So = ∆Ho Set these two equations equal to each other.

26.74 + 293∆So = 29.20 + 323∆So 26.74 - 29.20 = 323∆So - 293 ∆So -2.46 = 30∆So ∆So = -2.46/30 = -0.0820 = -0.0820 kJ/K = -82.0 J/K 26.74 + 293∆So = 26.74 + 293(-0.0820) = ∆Ho = +2.71 kJ

17.107 (a) ∆Ho = 2 ∆Ho

f(NH3) = (2 mol)(- 46.1 kJ/mol) = -92.2 kJ ∆Go = 2 ∆Go

f(NH3) = (2 mol)(-16.5 kJ/mol) = -33.0 kJ ∆Go = ∆Ho - T∆So ∆Ho - ∆Go = T∆So

∆So = = T

G _ H oo ∆∆ =

K 298

kJ) 33.0(_ _ kJ 92.2_ -0.199 kJ/K = -199 J/K

(b) ∆So is negative because the number of mol of gas molecules decreases from 4 mol to 2 mol on going from reactants to products. (c) The reaction is spontaneous because ∆Go is negative. (d) ∆Go = ∆Ho - T∆So = -92.2 kJ - (350 K)(-0.199 kJ/K) = -22.55 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](350 kJ/(K 10 x [8.314

kJ/mol) 22.55 (_ _ =

RT G _

3_

o

•∆

= 7.749

Kp = e7.749 = 2.3 x 103 ∆n = 2 - (1 + 3) = -2


510

Kc = = RT

1 K

n

p

∆

= RT

1 )10 x (2.3

2_

3

(2.3 x 103)(RT)2

Kc = (2.3 x 103)[(0.082 06)(350)]2 = 1.9 x 106 17.108 (a) ∆Ho = [∆Ho

f(Ag+(aq)) + ∆Hof(Br-(aq))] - ∆Ho

f(AgBr(s)) ∆Ho = [(1 mol)(105.6 kJ/mol) + (1 mol)(-121.5 kJ/mol)] - (1 mol)(-100.4 kJ/mol) = +84.5 kJ ∆So = [So(Ag+(aq)) + So(Br-(aq))] - So(AgBr(s)) ∆So = [(1 mol)(72.7 J/(K ⋅ mol)) + (1 mol)(82.4 J/(K ⋅ mol))]

- (1 mol)(107 J/(K ⋅ mol)) = +48.1 J ∆Go = ∆Ho - T∆So = 84.5 kJ - (298 K)(48.1 x 10-3 kJ/K) = +70.2 kJ (b) ∆Go = -RT ln Ksp

ln Ksp = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol 70.2 _ =

RT G _

3_

o

•∆

= -28.3

Ksp = e-28.3 = 5 x 10-13 (c) Q = [Ag+][Br -] = (1.00 x 10-5)(1.00 x 10-5) = 1.00 x 10-10 ∆G = ∆Go + RTlnQ ∆G = 70.2 kJ/mol + [8.314 x 10-3 kJ/(K ⋅ mol)](298 K) ln(1.00 x 10-10) = 13.2 kJ/mol A positive value of ∆G means that the forward reaction is nonspontaneous under these conditions. The reverse reaction is therefore spontaneous, which is consistent with the fact that Q > Ksp.

17.109 (a) ∆Go = ∆Ho - T∆So and ∆Go = -RT ln K

Set the two equations equal to each other. -RT ln K = ∆Ho - T∆So

ln K = _RT

ST _ H oo ∆∆; ln K =

RT ST

+ RT

H _ oo ∆∆; ln K =

R S

+ RT

H _ oo ∆∆

ln K = R

S +

T

1

R H _ oo ∆

∆ This is the equation for a straight line (y = mx + b).

y = ln K; m = -R

H o∆ = slope; x =

T

1; b =

R S o∆

= intercept

(b) Plot ln K versus 1/T ∆Ho = -R(slope) ∆So = R(intercept) (c) For a reaction where K increases with increasing temperature, the following plot would be obtained:

The slope is negative.


511

Because ∆Ho = -R(slope), ∆Ho is positive, and the reaction is endothermic.

This prediction is in accord with LeChâtelier's principle because when you add heat (raise the temperature) for an endothermic reaction, the reaction in the forward direction takes place, the product concentrations increase and the reactant concentrations decrease. This results in an increase in K.

17.110 Br2(l) _ Br2(g)

∆So = So(Br2(g)) - So(Br2(l)) ∆So = (1 mol)(245.4 J/(K ⋅ mol)) - (1 mol)(152.2 J/(K ⋅ mol)) = 93.2 J/K = 93.2 x 10-3 kJ/K ∆G = ∆Ho - T∆So At the boiling point, ∆G = 0. 0 = ∆Ho - Tbp∆So

Tbp = S Ho

o

∆∆

∆Ho = Tbp ∆So = (332 K)(93.2 x 10-3 kJ/K) = 30.9 kJ

Kp = PBr2 =

Hg mm 760

atm 1 x Hg mm 227 = 0.299 atm

∆Go = -RT ln Kp and ∆Go = ∆Ho - T∆So (set equations equal to each other) ∆Ho - T∆So = -RT ln Kp (rearrange)

R S

+ T

1

R H _

= Kln oo

p∆∆

(solve for T)

T =

••

•

∆

∆

mol) kJ/(K 10 x 8.314mol) kJ/(K 10 x 93.2

_ (0.299)ln

mol) kJ/(K 10 x 8.314kJ/mol 30.9_

=

R S

_ Kln

R H _

3_

3_

3_

o

p

o

= 299 K = 26oC

Br2(l) has a vapor pressure of 227 mm Hg at 26oC. 17.111 For PbI2, Ksp = [Pb2+][I -]2

PbI2(s) _ Pb2+(aq) + 2 I-(aq) initial (M) 0 0 equil (M) x 2x Ksp = x(2x)2 = 4x3, where x = molar solubility At 20oC = 20 + 273 = 293 K, Ksp = 4(1.45 x 10-3)3 = 1.22 x 10-8 At 80oC = 80 + 273 = 353 K, Ksp = 4(6.85 x 10-3)3 = 1.29 x 10-6

From problem 17.109, ln K = R

S +

RT H _ oo ∆∆


512

ln K1 - ln K2 = R

S +

RT

H _ o

1

o ∆∆ -

∆∆R

S +

RT

H _ o

2

o

= K

Kln 2

1

RT

H _

1

o∆ -

RT

H _

2

o∆ =

∆T

1 _

T

1

R H _

21

o

=

∆T

1 _

T

1

R H

12

o

∆Ho =

T

1 _

T

1

R ]Kln _ K[ln

12

21

∆Ho =

•

K 2931

_ K 353

1mol)] kJ/(K 10 x )][8.31410 x ln(1.29 _ )10 x [ln(1.22 3_6_8_

= 66.8 kJ/mol

∆Go = -RT ln Ksp = - [8.314 x 10-3 kJ/(K ⋅ mol)](293 K) ln(1.22 x 10-8) = 44.4 kJ/mol ∆Go = ∆Ho - T∆So ∆Ho - ∆Go = T∆So

∆So = T

G _ H oo ∆∆

∆So = = K 293

kJ/mol 44.4 _ kJ/mol 66.8 0.0765 kJ/(K ⋅ mol) = 76.5 J/(K ⋅ mol)

17.112 ∆Ho = [2 ∆Ho

f(Cl-(aq))] - [2 ∆Hof(Br-(aq))]

∆Ho = [(2 mol)(-167.2 kJ/mol)] - [(2 mol)(-121.5 kJ/mol)] = -91.4 kJ ∆So = [So(Br2(l)) + 2 So(Cl-(aq))] - [2 So(Br-(aq)) + So(Cl2(g))] ∆So = [(1 mol)(152.2 J/(K ⋅ mol)) + (2 mol)(56.5 J/(K ⋅ mol))]

- [(2 mol)(82.4 J/(K ⋅ mol)) + (1 mol)(223.0 J/(K ⋅ mol))] = -122.6 J/K 80oC = 80 + 273 = 353 K ∆Go = ∆Ho - T∆So = -91.4 kJ - (353 K)(-122.6 x 10-3 kJ/K) = - 48.1 kJ ∆Go = -RT ln K

ln K = K) mol)](353 kJ/(K 10 x [8.314

kJmol) 48.1 (_ _ =

RT G _

3_

o

•∆

= 16.4

K = e16.4 = 1.3 x 107 17.113 CS2(l) _ CS2(g)

∆Ho = ∆Hof(CS2(g)) - ∆Ho

f(CS2(l)) ∆Ho = [(1 mol)(116.7 kJ/mol)] - [(1 mol)(89.0 kJ/mol)] = 27.7 kJ ∆So = So(CS2(g)) - So(CS2(l)) ∆So = [(1 mol)(237.7 J/(K ⋅ mol))] - [(1 mol)(151.3 J/(K ⋅ mol))] = 86.4 J/K ∆G = ∆Ho - T∆So At the boiling point, ∆G = 0. 0 = ∆Ho - Tbp∆So

Tbp = = S Ho

o

∆∆

= kJ/K 10 x 86.4

kJ 27.73_

321 K

Tbp = 321 K = 321 - 273 = 48oC


513

17.114 35oC = 35 + 273 = 308 K ∆Go = ∆Ho - T∆So = -352 kJ - (308 K)(-899 x 10-3 kJ/K) = -75.1 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](308 kJ/(K 10 x [8.314

kJ/mol) 75.1 (_ _ =

RT G _

3_

o

•∆

= 29.33

Kp = e29.33 = 5.5 x 1012

Kp = = )P(

16

OH2

5.5 x 1012

= P OH2=

10 x 5.5

16

120.0075 atm

= P OH2 0.0075 atm x =

atm 1

Hg mm 7605.7 mm Hg

17.115 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

∆Ho = 2 ∆Hof(KCl) - 2 ∆Ho

f(KClO3) ∆Ho = (2 mol)(- 436.7 kJ) - (2 mol)(-397.7 kJ) = -78.0 kJ 25oC = 25 + 273 = 298 K ∆Go = ∆Ho - T∆So ∆Ho - ∆Go = T∆So

∆So = = T

G _ H oo ∆∆ =

K 298

kJ) 225.8(_ _ kJ 78.0_ 0.496 kJ/K = 496 J/K

∆So = [2 So(KCl) + 3 So(O2)] - 2 So(KClO3) 496 J/K = [(2 mol)(82.6 J/(K ⋅ mol)) + (3 mol)So(O2)] - (2 mol)(143 J/(K ⋅ mol)) (3 mol)So(O2) = 496 J/K - (2 mol)(82.6 J/(K ⋅ mol)) + (2 mol)(143 J/(K ⋅ mol)) (3 mol)So(O2) = 616.8 J/K So(O2) = (616.8 J/K)/(3 mol) = 205.6 J/(K ⋅ mol) = 206 J/(K ⋅ mol)

17.116 N2O4(g) _ 2 NO2(g)

∆Ho = 2 ∆Hof(NO2) - ∆Ho

f(N2O4) = (2 mol)(33.2 kJ) - (1 mol)(9.16 kJ) = 57.2 kJ ∆So = 2 So(NO2) - S

o(N2O4) = (2 mol)(240.0 J/(K ⋅ mol)) - (1 mol)(304.2 J/(K ⋅ mol)) ∆So = 175.8 J/K = 175.8 x 10-3 kJ/K ∆Go = ∆Ho - T∆So and ∆Go = -RT ln Kp; Set these two equations equal to each other and solve for T. ∆Ho - T∆So = -RT ln Kp ∆Ho = T∆So - RT ln Kp = T(∆So - R ln Kp)

T = Kln R _ S

H

po

o

∆∆

(a) = P + P NOON 242 1.00 atm and P 2 = P ONNO 422

= P 2 + P ONON 4242= P 3 ON 42

1.00 atm

= P ON 42 1.00 atm/3 = 0.333 atm

= P _ atm 1.00 = P ONNO 4221.00 - 0.333 = 0.667 atm


514

Kp = P

)P(

ON

2NO

42

2 = = (0.333)

)(0.667 2

1.34

T = Kln R _ S

H

po

o

∆∆

T = (1.34)ln mol)] kJ/(K 10 x [8.314 _ mol)] kJ/(K 10 x [175.8

kJ/mol 57.23_3_ ••

= 330 K

T = 330 K = 330 - 273 = 57oC (b) = P + P NOON 242

1.00 atm and P = P ONNO 422 so P = P ONNO 422

= 0.50 atm

Kp = P

)P(

ON

2NO

42

2 = = (0.500)

)(0.500 2

0.500

T = Kln R _ S

H

po

o

∆∆

T = (0.500)ln mol)] kJ/(K 10 x [8.314 _ mol)] kJ/(K 10 x [175.8

kJ/mol 57.23_3_ ••

= 315 K

T = 315 K = 315 - 273 = 42oC Multi-Concept Problems 17.117 N2O4(g) _ 2 NO2(g)

∆Ho = 2 ∆Hof(NO2) - ∆Ho

f(N2O4) = (2 mol)(33.2 kJ) - (1 mol)(9.16 kJ) = 57.2 kJ ∆So = 2 So(NO2) - S

o(N2O4) = (2 mol)(240.0 J/(K ⋅ mol)) - (1 mol)(304.2 J/(K ⋅ mol)) ∆So = 175.8 J/K = 175.8 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = 57.2 kJ - (373 K)(175.8 x 10-3 kJ/K) = -8.4 kJ

Kp = P

)P(

ON

2NO

42

2

∆Go = -RT ln Kp

ln Kp = K) mol)](373 kJ/(K 10 x [8.314

kJ/mol) 8.4(_ _ =

RT G_

3_

o

•∆

= 2.71

Kp = e2.71 = 15 N2O4(g) _ 2 NO2(g)

initial (atm) 1.00 1.00 change (atm) -x +2x equil (atm) 1.00 - x 1.00 + 2x

Kp = x)_ (1.00

)x2 + (1.00 = 15 =

P

)P(2

ON

2NO

42

2

4x2 + 19x - 14 = 0 Use the quadratic formula to solve for x.

x = 8

24.2 19_ =

2(4)

14) (4)(4)(_ _ )(19 (19) _ 2 ±±


515

x = 0.65 and -5.4 Of the two solutions for x, only 0.65 has physical meaning because -5.4 would lead to a negative partial pressure for NO2. P ON 42

= 1.00 - x = 1.00 - 0.65 = 0.35 atm; PNO2 = 1.00 + 2x = 1.00 + 2(0.65) = 2.30 atm

17.118 N2(g) + 3 H2(g) _ 2 NH3(g)

∆Ho = 2 ∆Hof(NH3) - [∆Ho

f(N2) + 3 ∆Hof(H2)] = (2 mol)(- 46.1 kJ) - [0] = -92.2 kJ

∆So = 2 So(NH3) - [So(N2) + 3 So(H2)]

∆So = (2 mol)(192.3 J/(K ⋅ mol)) - [(1 mol)(191.5 J/(K ⋅ mol)) + (3 mol)(130.6 J/(K ⋅ mol))] = -198.7 J/K

∆Go = ∆Ho - T∆So = -92.2 kJ - (673 K)(-198.7 x 10-3 kJ/K) = 41.5 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](673 kJ/(K 10 x [8.314

kJ/mol 41.5 _ =

RT G_

3_

o

•∆

= -7.42

Kp = e-7.42 = 6.0 x 10-4 Because Kp = Kc(RT)∆n, Kc = Kp(RT)-∆n Kc = Kp(RT)2 = (6.0 x 10-4)[(0.082 06)(673)]2 = 1.83 N2, 28.01 amu; H2, 2.016 amu Initial concentrations:

[N2] = L 5.00

g 28.01mol 1

g) (14.0

= 0.100 M and [H2] = L 5.00

g 2.016mol 1

g) (3.024

= 0.300 M

N2(g) + 3 H2(g) _ 2 NH3(g)

initial (M) 0.100 0.300 0 change (M) -x -3x +2x equil (M) 0.100 - x 0.300 - 3x 2x

Kc = ]H[ ]N[

]NH[3

22

23 =

)x3 _ x)(0.300_ (0.100

)x(23

2

= ) x_ 27(0.100

x44

2

= 1.83

) x_ (0.100

x 2

2

= 4

(27)(1.83) = 12.35;

) x_ (0.100

x2 = 12.35 = 3.515

3.515x2 - 1.703x + 0.03515 = 0 Use the quadratic formula to solve for x.

x = 7.030

1.551 1.703 =

2(3.515)

(0.03515)(4)(3.515) _ )1.703 (_ 1.703) (_ _ 2 ±±

x = 0.463 and 0.0216 Of the two solutions for x, only 0.0216 has physical meaning because 0.463 would lead to negative concentrations of N2 and H2. [N2] = 0.100 - x = 0.100 - 0.0216 = 0.078 M [H2] = 0.300 - 3x = 0.300 - 3(0.0216) = 0.235 M


516

[NH3] = 2x = 2(0.0216) = 0.043 M 17.119 2 SO2(g) + O2(g) _ 2 SO3(g)

∆Ho = 2 ∆Hof(SO3) - 2 ∆Ho

f(SO2) ∆Ho = (2 mol)(-395.7 kJ/mol) - (2 mol)(-296.8 kJ/mol) = -197.8 kJ ∆So = 2 So(SO3) - [2 So(SO2) + So(O2)] ∆So = (2 mol)(256.6 J/(K ⋅ mol)) - [(2 mol)(248.1 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] ∆So = -188.0 J/K = -188.0 x 10-3 kJ/K ∆Go = ∆Ho - T∆So = -197.8 kJ - (800 K)(-188.0 x 10-3 kJ/K) = - 47.4 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](800 kJ/(K 10 x [8.314

kJ/mol) 47.4 (_ _ =

RT G _

3_

o

•∆

= 7.13

Kp = e7.13 = 1249 SO2, 64.06 amu; O2, 32.00 amu At 800 K:

L 15.0

K) (800mol K atm L

06 0.082g 64.06 mol 1

x g 192

= V

T Rn = PSO2

••

= 13.1 atm

L 15.0

K) (800mol K atm L

0.08206g 32.00 mol 1

x g 48.0

= V

T Rn = PO2

••

= 6.57 atm

2 SO2(g) + O2(g) _ 2 SO3(g)

initial (atm) 13.1 6.57 0 assume complete rxn (atm) 0 0 13.1 assume a small back rxn +2x +x -2x

equil (atm) 2x x 13.1 - 2x

Kp = 1249 = (x))x(2

)(13.1

(x))x(2

)x2 _ (13.1 =

]O[]SO[

]SO[2

2

2

2

22

2

23 ≈

Solve for x. x3 = 0.0343; x = 0.325 Use successive approximations to solve for x because 2x is not negligible compared with 13.1.

Second approximation:

(x))x(2

](2)(0.325) _ [13.1 = 1249

2

2

; Solve for x. x3 = 0.0310; x = 0.314

Third approximation:

(x))x(2

](2)(0.314) _ [13.1 = 1249

2

2

; Solve for x. x3 = 0.0311; x = 0.315 (x has converged)


517

PSO2 = 2x = 2(0.315) = 0.63 atm

PO2 = x = 0.32 atm

PSO3 = 13.1 - 2x = 13.1 - 2(0.315) = 12.5 atm

(b) The % yield of SO3 decreases with increasing temperature because ∆So is negative. ∆Go becomes less negative and Kp gets smaller as the temperature increases. (c) At 1000 K: ∆Go = ∆Ho - T∆So = -197.8 kJ - (1000 K)(-188.0 x 10-3 kJ/K) = -9.8 kJ ∆Go = -RT ln Kp

ln Kp = K) mol)](1000 kJ/(K 10 x [8.314

kJ/mol) 9.8(_ _ =

RT G _

3_

o

•∆

= 1.179

Kp = e1.179 = 3.25

L 15.0

K) (1000mol K atm L

06 0.082g 64.06 mol 1

x g 192

= V

T Rn = PSO2

••

= 16.4 atm

L 15.0

K) (1000mol K atm L

06 0.082g 32.00 mol 1

x g 48.0

= V

T Rn = PO2

••

= 8.2 atm

2 SO2(g) + O2(g) _ 2 SO3(g)

initial (atm) 16.4 8.2 0 assume complete rxn (atm) 0 0 16.4 assume a small back rxn +2x +x -2x

equil (atm) 2x x 16.4 - 2x

Kp = 3.25 = (x))x(2

)(16.4

(x))x(2

)x2 _ (16.4 =

]O[]SO[

]SO[2

2

2

2

22

2

23 ≈

Solve for x. x3 = 20.7; x = 2.7 Use successive approximations to solve for x because 2x is not negligible compared with 16.4.

Second approximation:

(x))x(2

](2)(2.7) _ [16.4 = 3.25

2

2

; Solve for x. x3 = 9.31; x = 2.1

Third approximation:

(x))x(2

](2)(2.1) _ [16.4 = 3.25

2

2

; Solve for x. x3 = 11.4; x = 2.3

Fourth approximation:

(x))x(2

](2)(2.3) _ [16.4 = 3.25

2

2

; Solve for x. x3 = 10.7; x = 2.2 (x has converged)


518

PSO2 = 2x = 2(2.2) = 4.4 atm

PO2 = x = 2.2 atm

PSO3 = 16.4 - 2x = 16.4 - 2(2.2) = 12.0 atm

Ptotal = PSO2 + PO2

+ PSO3 = 4.4 + 2.2 + 12.0 = 18.6 atm

On going from 800 K to 1000 K, Ptotal increases to 18.6 atm (because Kp decreases, but P increases with temperature at constant volume).

17.120 Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4

-(aq) → 2 PbSO4(s) + 2 H2O(l) (a) ∆Go = [2 ∆Go

f(PbSO4) + 2 ∆Gof(H2O)] - [∆Go

f(PbO2) + 2 ∆Gof(HSO4

-)] ∆Go = (2 mol)(-813.2 kJ/mol) + (2 mol)(-237.2 kJ/mol)]

- [(1 mol)(-217.4 kJ/mol) + (2 mol)(-756.0 kJ/mol)] = -371.4 kJ (b) oC = 5/9(oF - 32) = 5/9(10 - 32) = -12.2oC; -12.2oC = 261 K ∆Ho = [2 ∆Ho

f(PbSO4) + 2 ∆Hof(H2O)] - [∆Ho

f(PbO2) + 2 ∆Hof(HSO4

-)] ∆Ho = [(2 mol)(-919.9 kJ/mol) + (2 mol)(-285.8 kJ/mol)]

- [(1 mol)(-277 kJ/mol) + (2 mol)(-887.3 kJ/mol)] = -359.8 kJ ∆So = [2 So(PbSO4) + 2 So(H2O)] - [So(Pb) + So(PbO2) + 2 So(H+) + 2 So(HSO4

-)] ∆So = [(2 mol)(148.6 J/(K ⋅ mol)) + (2 mol)(69.9 J/(K ⋅ mol))]

- [(1 mol)(64.8 J/(K ⋅ mol)) + (1 mol)(68.6 J/(K ⋅ mol)) + (2 mol)(132 J/(K ⋅ mol))] = 39.6 J/K = 39.6 x 10-3 kJ/K

∆Go = ∆Ho - T∆So = -359.8 kJ -(261 K)(39.6 x 10-3 kJ/K) = -370.1 kJ at 261 K

HSO4-(aq) + H2O(l) _ H3O

+(aq) + SO42-(aq)

initial (M) 0.100 0.100 0 change (M) -x +x +x equil (M) 0.100 - x 0.100 + x x

Ka2 = ] HSO[

] SO][OH[_4

_24

+3 = 1.2 x 10-2 =

x_ 0.100

x x)+ (0.100

x2 + 0.112x - (1.2 x 10-3) = 0 Use the quadratic formula to solve for x.

x = 2

0.132 0.112_ =

2(1)

)10 x 1.2 (4)(1)(_ _ )(0.112 (0.112) _ 3_2 ±±

x = -0.122 and 0.010 Of the two solutions for x, only 0.010 has physical meaning because -0.122 would lead to negative concentrations of H3O

+ and SO42-.

[H+] = 0.100 + x = 0.100 + 0.010 = 0.110 M [HSO4

-] = 0.100 - x = 0.100 - 0.010 = 0.090 M

∆G = ∆Go + RT ln] HSO[ ]H[

12_

42+

∆G = (-370.1 kJ/mol) + [8.314 x 10-3 kJ/(K ⋅ mol)](261 K) ln )(0.090 )(0.110

122

∆G = -350.1 kJ/mol


519

17.121 CaCO3(s) _ Ca2+(aq) + CO3

2-(aq) ∆Ho = [∆Ho

f(Ca2+) + ∆Hof(CO3

2-)] - ∆Hof(CaCO3)

∆Ho = [(1 mol)(-542.8 kJ/mol) + (1 mol)(-677.1 kJ/mol)] - (1 mol)(-1206.9 kJ/mol) ∆Ho = -13.0 kJ ∆So = [So(Ca2+) + So(CO3

2-)] - So(CaCO3) ∆So = [(1 mol)(-53.1 J/(K ⋅ mol)) + (1 mol)(-56.9 J/(K ⋅ mol))] - (1 mol)(92.9 J/(K ⋅ mol)) ∆So = -202.9 J/K = -202.9 x 10-3 kJ/K 50oC = 50 + 273 = 323 K ∆G = ∆Ho - T∆So = -13.0 kJ - (323 K)(-202.9 x 10-3 kJ/K) = +52.54 kJ ∆G = -RT ln Ksp

ln Ksp = K) mol)](323 kJ/(K 10 x [8.314

kJ/mol 52.54 _ =

RT

G_3_ •

∆ = -19.56

Ksp = e-19.56 = 3.2 x 10-9

20oC = 20 +273 = 293 K

nCO2 =

K) (293mol K atm L

06 0.082

L) (1.000Hg mm 760


= RT

PV

••

= 0.0400 mol CO2

Ca(OH)2, 74.09 amu

mol Ca(OH)2 = 3.335 g Ca(OH)2 x )Ca(OH g 74.09

)Ca(OH mol 1

2

2 = 0.0450 mol Ca(OH)2

CO2(g) + H2O(l) → H2CO3(aq)

Ca(OH)2(aq) + H2CO3(aq) → CaCO3(s) + 2 H2O(l) before (mol) 0.0450 0.0400 0 change (mol) -0.0400 -0.0400 +0.0400 after (mol) 0.0050 0 0.0400

500.0 mL = 0.5000 L [Ca(OH)2] = [Ca2+] = 0.0050 mol/0.5000 L = 0.010 M

CaCO3(s) _ Ca2+(aq) + CO3

2-(aq) initial (M) 0.010 0 change (M) +x +x equil (M) 0.010 + x x Ksp = [Ca2+][CO3

2-] = 3.2 x 10-9 = (0.010 + x)x ≈0.010x x = molar solubility = 3.2 x 10-9/0.010 = 3.2 x 10-7 M Because ∆Ho is negative (exothermic), the solubility of CaCO3 is lower at 50oC.

17.122 PV = nRT


520

nNH3 =


06 0.082

L) (1.00Hg mm 760


= RT

PV

••

= 0.0400 mol NH3

500.0 mL = 0.5000 L [NH3] = 0.0400 mol/0.5000 L = 0.0800 M NH3(aq) + H2O(l) _ NH4

+(aq) + OH-(aq) ∆Ho = [∆Ho

f(NH4+) + ∆Ho

f(OH-)] - [∆Hof(NH3) + ∆Ho

f(H2O)] ∆Ho = [(1 mol)(-132.5 kJ/mol) + (1 mol)(-230.0 kJ/mol)]

- [(1 mol)(-80.3 kJ/mol) + (1 mol)(-285.8 kJ/mol)] = +3.6 kJ ∆So = [So(NH4

+) + So(OH-)] - [So(NH3) + So(H2O)] ∆So = [(1 mol)(113 J/(K ⋅ mol)) + (1 mol)(-10.8 J/(K ⋅ mol))]

- [(1 mol)(111 J/(K ⋅ mol)) + (1 mol)(69.9 J/(K ⋅ mol))] = -78.7 J/K T = 2.0oC = 2.0 + 273.1 = 275.1 K ∆Go = ∆Ho - T∆So = 3.6 kJ - (275.1 K)(-78.7 x 10-3 kJ/K) = 25.3 kJ ∆Go = -RT ln Kb

ln Kb = K) 1mol)](275. kJ/(K 10 x [8.314

kJ/mol 25.3 _ =

RT G _

3_

o

•∆

= -11.06

Kb = e-11.06 = 1.6 x 10-5

NH3(aq) + H2O(l) _ NH4+(aq) + OH-(aq)

initial (M) 0.0800 0 ~0 change (M) -x +x +x equil (M) 0.0800 - x x x

at 2oC, Kb = = ]NH[

]OH][NH[

3

_+4 1.6 x 10-5 = ≈

x_ 0.0800x2

0.0800x2

x2 = (1.6 x 10-5)(0.0800)

x = [OH-] = )(0.0800)10 x (1.6 5_ = 1.13 x 10-3 M

[H3O+] = =

10 x 1.1310 x 1.0

3_

14_

8.85 x 10-12 M

pH = -log[H3O+] = -log(8.85 x 10-12) = 11.05

17.123 (a) I2(s) → 2 I-(aq)

[I2(s) + 2 e- → 2 I-(aq)] x 5 reduction half reaction

I2(s) → 2 IO3-(aq)

I2(s) + 6 H2O(l) → 2 IO3-(aq)

I2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq)

I2(s) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 10 e- oxidation half reaction


521

Combine the two half reactions. 6 I2(s) + 6 H2O(l) → 10 I-(aq) + 2 IO3

-(aq) + 12 H+(aq) Divide all coefficients by 2. 3 I2(s) + 3 H2O(l) → 5 I-(aq) + IO3

-(aq) + 6 H+(aq) 3 I2(s) + 3 H2O(l) + 6 OH-(aq) → 5 I-(aq) + IO3

-(aq) + 6 H+(aq) + 6 OH-(aq) 3 I2(s) + 3 H2O(l) + 6 OH-(aq) → 5 I-(aq) + IO3

-(aq) + 6 H2O(l) 3 I2(s) + 6 OH-(aq) → 5 I-(aq) + IO3

-(aq) + 3 H2O(l) (b) ∆Go = [5 ∆Go

f(I-) + ∆Go

f(IO3-) + 3 ∆Go

f(H2O(l))] - 6 ∆Gof(OH-)

∆Go = [(5 mol)(-51.6 kJ/mol) + (1 mol)(-128.0 kJ/mol) + (3 mol)(-237.2 kJ/mol)] - (6 mol)(-157.3 kJ/mol) = -153.8 kJ

(c) The reaction is spontaneous because ∆Go is negative. (d) 25oC = 25 + 273 = 298 K ∆Go = -RT ln Kc

ln Kc = K) mol)](298 kJ/(K 10 x [8.314

kJ/mol) 153.8(_ _ =

RT G _

3_

o

•∆

= 62.077

Kc = e62.077 = 9.1 x 1026

Kc = = ] OH[

] IO[] I[6_

_3

5_

9.1 x 1026 = ] OH[

(0.50))(0.106_

5

[OH-] = = 10 x 9.1

(0.50))(0.106

26

5

4.2 x 10-6 M

[H3O+] = =

10 x 4.210 x 1.0

6_

14_

2.38 x 10-9 M

pH = -log[H3O+] = -log(2.38 x 10-9) = 8.62

523

18

Electrochemistry 18.1 2 Ag+(aq) + Ni(s) → 2 Ag(s) + Ni2+(aq)

There is a Ni anode in an aqueous solution of Ni2+, and a Ag cathode in an aqueous solution of Ag+. A salt bridge connects the anode and cathode compartment. The electrodes are connected through an external circuit.

18.2 Fe(s)Fe2+(aq)Sn2+(aq)Sn(s) 18.3 Pb(s) + Br2(l) → Pb2+(aq) + 2 Br-(aq)

There is a Pb anode in an aqueous solution of Pb2+. The cathode is a Pt wire that dips into a pool of liquid Br2 and an aqueous solution that is saturated with Br2. A salt bridge connects the anode and cathode compartment. The electrodes are connected through an external circuit.

18.4 (a) and (b)

(c) 2 Al(s) + 3 Co2+(aq) → 2 Al3+(aq) + 3 Co(s)

524

(d) Al(s)Al 3+(aq)Co2+(aq)Co(s) 18.5 Al(s) + Cr3+(aq) → Al3+(aq) + Cr(s)

∆Go = -nFEo = -(3 mol e-)

•

V C 1

J 1V) (0.92

e mol 1

C 96,500_

= -266,340 J = -270 kJ

18.6 oxidation: Al(s) → Al3+(aq) + 3 e- Eo = 1.66 V

reduction: Cr3+(aq) + 3 e- → Cr(s) Eo = ? overall Al(s) + Cr3+(aq) → Al3+(aq) + Cr(s) Eo = 0.92 V The standard reduction potential for the Cr3+/Cr half cell is: Eo = 0.92 - 1.66 = -0.74 V

18.7 (a) Cl2(g) + 2 e- → 2 Cl-(aq) Eo = 1.36 V

Ag+(aq) + e- → Ag(s) Eo = 0.80 V Cl2 has the greater tendency to be reduced (larger Eo). The species that has the greater tendency to be reduced is the stronger oxidizing agent. Cl2 is the stronger oxidizing agent.

(b) Fe2+(aq) + 2 e- → Fe(s) Eo = -0.45 V Mg2+(aq) + 2 e- → Mg(s) Eo = -2.37 V The second half-reaction has the lesser tendency to occur in the forward direction (more negative Eo) and the greater tendency to occur in the reverse direction. Therefore, Mg is the stronger reducing agent.

18.8 (a) 2 Fe3+(aq) + 2 I-(aq) → 2 Fe2+(aq) + I2(s)

reduction: Fe3+(aq) + e- → Fe2+(aq) Eo = 0.77 V oxidation: 2 I-(aq) → I2(s) + 2 e- Eo = -0.54 V

overall Eo = 0.23 V Because Eo for the overall reaction is positive, this reaction can occur under standard-state conditions.

(b) 3 Ni(s) + 2 Al3+(aq) → 3 Ni2+(aq) + 2 Al(s) oxidation: Ni(s) → Ni2+(aq) + 2 e- Eo = 0.26 V reduction: Al3+(aq) + 3 e- → Al(s) Eo = -1.66 V

overall Eo = -1.40 V Because Eo for the overall reaction is negative, this reaction cannot occur under standard-state conditions. This reaction can occur in the reverse direction.

18.9 (a) D is the strongest reducing agent. D+ has the most negative standard reduction

potential. A3+ is the strongest oxidizing agent. It has the most positive standard reduction potential. (b) An oxidizing agent can oxidize any reducing agent that is below it in the table. B2+ can oxidize C and D. A reducing agent can reduce any oxidizing agent that is above it in the table. C can reduce A3+ and B2+.

(c) Use the two half-reactions that have the most positive and the most negative standard

Chapter 18 - Electrochemistry _____________________________________________________________________________

525

reduction potentials, respectively. A3+ + 2 e- → A+ 1.47 V 2 x (D → D+ + e-) 1.38 V A3+ + 2 D → A+ + 2 D+ 2.85 V

18.10 Cu(s) + 2 Fe3+(aq) → Cu2+(aq) + 2 Fe2+(aq)

Eo = E + E oFe _ Fe

oCu _Cu +2+3+2 = -0.34 V + 0.77 V = 0.43 V; n = 2 mol e-

E = Eo - ]Fe[

]Fe][Cu[ log

n

V 0.05922+3

2+2+2

= 0.43 V - )10 x (1.0

)0(0.25)(0.2 log

2

V) (0.059224_

2

= 0.25 V

18.11 5 [Cu(s) → Cu2+(aq) + 2 e-] oxidation half reaction

2 [5 e- + 8 H+(aq) + MnO4-(aq) → Mn2+(aq) + 4 H2O(l)] reduction half reaction

5 Cu(s) + 16 H+(aq) + 2 MnO4

-(aq) → 5 Cu2+(aq) + 2 Mn2+(aq) + 8 H2O(l)

∆E = -]H[]MnO[

]Mn[]Cu[ log

n

V 0.059216+2 _

4

2+25+2

(a) The anode compartment contains Cu2+.

∆E = -)(1 )(1

)(1 )(0.01 log

10

V 0.0592162

25

= +0.059 V

(b) The cathode compartment contains Mn2+, MnO4-, and H+.

∆E = -)(0.01 )(0.01

)(0.01 )(1 log

10

V 0.0592162

25

= -0.19 V

18.12 H2(g) + Pb2+(aq) → 2 H+(aq) + Pb(s)

Eo = E + E oPb _ Pb

oH _ H +2+

2 = 0 V + (-0.13 V) = -0.13 V; n = 2 mol e-

E = Eo - )P](Pb[

]OH[ log n

V 0.0592

H+2

2+3

2

0.28 V = -0.13 V - (1)(1)

]OH[ log 2

V) (0.0592 2+3 = -0.13 V - (0.0592 V) log [H3O

+]

pH = -log[H3O+] therefore 0.28 V = -0.13 V + (0.0592 V) pH

pH = V 0.0592

V) 0.13 + V (0.28 = 6.9

18.13 4 Fe2+(aq) + O2(g) + 4 H+(aq) → 4 Fe3+(aq) + 2 H2O(l)

Eo = E + E oOH _ O

oFe _ Fe 22+3+2 = -0.77 V + 1.23 V = 0.46 V; n = 4 mol e-

Eo = n

V 0.0592 log K

log K = V 0.0592

V) (4)(0.46 =

V 0.0592 nE o

= 31; K = 1031 at 25oC


526

18.14 Eo = )10 x (1.8 log 2

V 0.0592 =K log

n

V 0.0592 5_ = -0.140 V

18.15 (a) Zn(s) + 2 MnO2(s) + 2 NH4

+(aq) → Zn2+(aq) + Mn2O3(s) + 2 NH3(aq) + H2O(l) (b) Zn(s) + 2 MnO2(s) → ZnO(s) + Mn2O3(s) (c) Zn(s) + HgO(s) → ZnO(s) + Hg(l) (d) Cd(s) + 2 NiO(OH)(s) + 2 H2O(l) → Cd(OH)2(s) + 2 Ni(OH)2(s)

18.16 (a) [Mg(s) → Mg2+(aq) + 2 e-] x 2

O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) 2 Mg(s) + O2(g) + 4 H+(aq) → 2 Mg2+(aq) + 2 H2O(l) (b) [Fe(s) → Fe2+(aq) + 2 e-] x 4 [O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l)] x 2 4 Fe2+(aq) + O2(g) + 4 H+(aq) → 4 Fe3+(aq) + 2 H2O(l) [2 Fe3+(aq) + 4 H2O(l) → Fe2O3 ⋅ H2O(s) + 6 H+(aq)] x 2 4 Fe(s) + 3 O2(g) + 2 H2O(l) → 2 Fe2O3 ⋅ H2O(s)

18.17 (a)

(b) anode reaction 4 OH-(l) → O2(g) + 2 H2O(l) + 4 e- cathode reaction 4 K+(l) + 4 e- → 4 K(l) overall reaction 4 K+(l) + 4 OH-(l) → 4 K(l) + O2(g) + 2 H2O(l)

18.18 (a) anode reaction 2 Cl-(aq) → Cl2(g) + 2 e-

cathode reaction 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) overall reaction 2 Cl-(aq) + 2 H2O(l) → Cl2(g) + H2(g) + 2 OH-(aq)

(b) anode reaction 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e-

cathode reaction 2 Cu2+(aq) + 4 e- → 2 Cu(s) overall reaction 2 Cu2+(aq) + 2 H2O(l) → 2 Cu(s) + O2(g) + 4 H+(aq)


527

18.19

anode reaction Ag(s) → Ag+(aq) + e- cathode reaction Ag+(aq) + e- → Ag(s) The overall reaction is transfer of silver metal from the silver anode to the spoon.

18.20 Charge =

min

s 60

h

min 60h) (8.00

s

C 10 x 1.00 5 = 2.88 x 109 C

Moles of e- = (2.88 x 109 C)

C 96,500e mol 1 _

= 2.98 x 104 mol e-

cathode reaction: Al3+ + 3 e- → Al

mass Al = (2.98 x 104 mol e-) x g 1000

kg 1 x

Al mol 1

Al g 26.98 x

e mol 3

Al mol 1_

= 268 kg Al

18.21 3.00 g Ag x Ag g 107.9

Ag mol 1 = 0.0278 mol Ag

cathode reaction: Ag+(aq) + e- → Ag(s)

Charge = (0.0278 mol Ag)

e mol 1

C 96,500

Ag mol 1e mol 1

_

_

= 2682.7 C

Time =

s 3600

h 1 x

C/s 0.100

C 2682.7 =

A

C = 7.45 h

18.22 When a beam of white light strikes the anodized surface, part of the light is reflected

from the outer TiO2, while part penetrates through the semitransparent TiO2 and is reflected from the inner metal. If the two reflections of a particular wavelength are out of phase, they interfere destructively and that wavelength is canceled from the reflected light. Because nλ = 2d x sin θ, the canceled wavelength depends on the thickness of the TiO2 layer.


528

18.23 volume = )cm (10.0mm 10

cm 1 x mm 0.0100 2

= 0.100 cm3

mol Al2O3 = (0.100 cm3)(3.97 g/cm3)OAl g 102.0

OAl mol 1

32

32 = 3.892 x 10-3 mol Al2O3

mole e- = 3.892 x 10-3 mol Al2O3 x OAl mol 1

e mol 6

32

_

= 0.02335 mol e-

coulombs = 0.02335 mol e- x e mol 1

C 96,500_

= 2253 C

time = A

C =

s 60

min 1 x

C/s 0.600

C 2253 = 62.6 min


18.24 (a) - (d)

(e) anode reaction Zn(s) → Zn2+(aq) + 2 e- cathode reaction Pb2+(aq) + 2 e- → Pb(s) overall reaction Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

18.25 (a) anode is Ni; cathode is Pt

(b) anode reaction 3 Ni(s) → 3 Ni2+(aq) + 6 e- cathode reaction Cr2O7

2-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) overall reaction Cr2O7

2-(aq) + 3 Ni(s) + 14 H+(aq) → 2 Cr3+(aq) + 3 Ni2+(aq) + 7 H2O(l)

(c) Ni(s)Ni2+(aq) Cr2O72-(aq), Cr3+Pt(s)

18.26 (a) The three cell reactions are the same except for cation concentrations.

anode reaction Cu(s) → Cu2+(aq) + 2 e- Eo = -0.34 V cathode reaction 2 Fe3+(aq) + 2 e- → 2 Fe2+(aq) Eo = 0.77 V overall reaction Cu(s) + 2 Fe3+(aq) → Cu2+(aq) + 2 Fe2+(aq) Eo = 0.43 V


529

(b)

(c) E = Eo - ]Fe[

]Fe][Cu[ log

n

V 0.05922+2

2+2+2

; n = 2 mol e-

(1) E = Eo = 0.43 V because all cation concentrations are 1 M.

(2) E = Eo - )(1

)(1)(5 log

2

V 0.05922

2

= 0.39 V

(3) E = Eo - )(0.1

)(0.1)(0.1 log

2

V 0.05922

2

= 0.46 V

Cell (3) has the largest potential, while cell (2) has the smallest as calculated from the Nernst equation.

18.27 (a) - (b)

(c) anode reaction 2 Br-(aq) → Br2(aq) + 2 e- cathode reaction Cu2+(aq) + 2 e- → Cu(s) overall reaction Cu2+(aq) + 2 Br-(aq) → Cu(s) + Br2(aq)


530

18.28 (a) This is an electrolytic cell that has a battery connected between two inert electrodes.

(b)

(c) anode reaction 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e- cathode reaction Ni2+(aq) + 2 e- → Ni(s) overall reaction 2 Ni2+(aq) + 2 H2O(l) → 2 Ni(s) + O2(g) + 4 H+(aq)

18.29 (a) & (b)

(c) anode reaction 2 O2- → O2(g) + 4 e-

cathode reaction TiO2(s) + 4 e- → Ti(s) + 2 O2- overall reaction TiO2(s) → Ti(s) + O2(g)

18.30 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s); E = Eo - ]Cu[

]Zn[ log

2

V 0.0592+2

+2

(a) E increases because increasing [Cu2+] decreases ]Cu[

]Zn[ log

+2

+2

.

(b) E will decrease because addition of H2SO4 increases the volume which decreases


531

[Cu2+] and increases ]Cu[

]Zn[ log

+2

+2

.

(c) E decreases because increasing [Zn2+] increases ]Cu[

]Zn[ log

+2

+2

.

(d) Because there is no change in [Zn2+], there is no change in E.

18.31 Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s); E = Eo - ]Ag[

]Cu[ log

2

V 0.05922+

+2

(a) E decreases because addition of NaCl precipitates AgCl which decreases [Ag+] and

increases ]Ag[

]Cu[ log

2+

+2

.

(b) E increases because addition of NaCl increases the volume which decreases [Cu2+]

and decreases ]Ag[

]Cu[ log

2+

+2

.

(c) E decreases because addition of NH3 complexes Ag+, yielding Ag(NH3)2+, which

decreases [Ag+] and increases ]Ag[

]Cu[ log

2+

+2

.

(d) E increases because addition of NH3 complexes Cu2+, yielding Cu(NH3)42+, which

decreases [Cu2+] and decreases ]Ag[

]Cu[ log

2+

+2

.

Additional Problems Galvanic Cells 18.32 The electrode where oxidation takes place is called the anode. For example, the lead

electrode in the lead storage battery. The electrode where reduction takes place is called the cathode. For example, the PbO2 electrode in the lead storage battery.

18.33 The oxidizing agent gets reduced and reduction takes place at the cathode. 18.34 The cathode of a galvanic cell is considered to be the positive electrode because

electrons flow through the external circuit toward the positive electrode (the cathode). 18.35 The salt bridge maintains charge neutrality in both the anode and cathode compartments

of a galvanic cell.


532

18.36 (a) Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s)

(b) 2 Al(s) + 3 Cd2+(aq) → 2 Al3+(aq) + 3 Cd(s)

(c) 6 Fe2+(aq) + Cr2O72-(aq) + 14 H+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)


533

18.37 (a) 3 Cu2+(aq) + 2 Cr(s) → 3 Cu(s) + 2 Cr3+(aq)

(b) Pb(s) + 2 H+(aq) → Pb2+(aq) + H2(g)

(c) Cl2(g) + Sn2+(aq) → Sn4+(aq) + 2 Cl-(aq)


534

18.38 (a) Cd(s)Cd2+(aq)Sn2+(aq)Sn(s)

(b) Al(s)Al 3+(aq)Cd2+(aq)Cd(s) (c) Pt(s)Fe2+(aq), Fe3+(aq)Cr2O7

2-(aq), Cr3+(aq)Pt(s) 18.39 (a) Cr(s)Cr3+(aq)Cu2+(aq)Cu(s)

(b) Pb(s)Pb2+(aq)H+(aq)H2(g)Pt(s) (c) Pt(s)Sn2+(aq), Sn4+(aq)Cl2(g)Cl-(aq)Pt(s)

18.40 (a)

(b) anode reaction H2(g) → 2 H+(aq) + 2 e- cathode reaction 2 Ag+(aq) + 2 e- → 2 Ag(s) overall reaction H2(g) + 2 Ag+(aq) → 2 H+(aq) + 2 Ag(s)

(c) Pt(s)H2(g)H+(aq)Ag+(aq)Ag(s)


535

18.41 (a)

(b) anode reaction Zn(s) → Zn2+(aq) + 2 e- cathode reaction Cl2(g) + 2 e- → 2 Cl-(aq) overall reaction Zn(s) + Cl2(g) → Zn2+(aq) + 2 Cl-(aq) (c) Zn(s)Zn2+(aq)Cl2(g)Cl-(aq)C(s)

18.42 (a) anode reaction Co(s) → Co2+(aq) + 2 e-

cathode reaction Cu2+(aq) + 2 e- → Cu(s) overall reaction Co(s) + Cu2+(aq) → Co2+(aq) + Cu(s)

(b) anode reaction 2 Fe(s) → 2 Fe2+(aq) + 4 e- cathode reaction O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) overall reaction 2 Fe(s) + O2(g) + 4 H+(aq) → 2 Fe2+(aq) + 2 H2O(l)


536

18.43 (a) anode reaction Mn(s) → Mn2+(aq) + 2 e-

cathode reaction Pb2+(aq) + 2 e- → Pb(s) overall reaction Mn(s) + Pb2+(aq) → Mn2+(aq) + Pb(s)

(b) anode reaction H2(g) → 2 H+(aq) + 2 e- cathode reaction 2 AgCl(s) + 2 e- → 2 Ag(s) + 2 Cl-(aq) overall reaction H2(g) + 2 AgCl(s) → 2 Ag(s) + 2 H+(aq) + 2 Cl-(aq)


537

Cell Potentials and Free-Energy Changes; Standard Reduction potentials 18.44 The SI unit of electrical potential is the volt (V).

The SI unit of charge is the coulomb (C). The SI unit of energy is the joule (J). 1 J = 1 C ⋅ 1 V

18.45 ∆G = -nFE; ∆G is the free energy change for the cell reaction

n is the number of moles of e- F is the Faraday (96,500 C/mol e-) E is the galvanic cell potential

18.46 E is the standard cell potential (Eo) when all reactants and products are in their standard

states--solutes at 1 M concentrations, gases at a partial pressure of 1 atm, solids and liquids in pure form, all at 25oC.

18.47 The standard reduction potential is the potential of the reduction half reaction in a

galvanic cell where the other electrode is the standard hydrogen electrode. 18.48 Zn(s) + Ag2O(s) → ZnO(s) + 2 Ag(s); n = 2 mol e-

∆G = -nFE = -(2 mol e-)

e mol 1

C 96,500_

(1.60 V)

• V C 1

J 1 = -308,800 J = -309 kJ

18.49 Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4

-(aq) → 2 PbSO4(s) + 2 H2O(l)


538

n = 2 mol e-

∆Go = -nFEo = -(2 mol e-) V) (1.924e mol 1

C 96,500_

• V C 1

J 1 = -371,300 J = -371 kJ

18.50 2 H2(g) + O2(g) → 2 H2O(l); n = 4 mol e- and 1 V = 1 J/C

∆Go = 2 ∆Gof(H2O(l)) = (2 mol)(-237.2 kJ/mol) = -474.4 kJ

∆Go = -nFEo

Eo =

∆

e mol 1C 96,500

)e mol (4

J) 474,400 (_ _ =

nF G _

__

o

= +1.23 J/C = +1.23 V

18.51 CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l); n = 8 mol e- and 1 V = 1 J/C

∆Go = [∆Gof(CO2) + 2 ∆Go

f(H2O(l))] - ∆Gof(CH4)

∆Go = [(1 mol)(-394.4 kJ/mol) + (2 mol)(- 237.2 kJ/mol)] - (1 mol)(-50.8 kJ/mol) = -818.0 kJ

∆Go = -nFEo Eo =

∆

e mol 1C 96,500

)e mol (8

J) 818,000_(_ =

nF G _

__

o

= +1.06 J/C = +1.06 V

18.52 oxidation: Zn(s) → Zn2+(aq) + 2 e- Eo = 0.76 V

reduction: Eu3+(aq) + e- → Eu2+(aq) Eo = ? overall Zn(s) + 2 Eu3+(aq) → Zn2+(aq) + 2 Eu2+(aq) Eo = 0.40 V The standard reduction potential for the Eu3+/Eu2+ half cell is: Eo = 0.40 - 0.76 = -0.36 V

18.53 oxidation: 2 Ag(s) + 2 Br-(aq) → 2 AgBr(s) + 2 e- Eo = ?

reduction: Cu2+(aq) + 2 e- → Cu(s) Eo = 0.34 V overall Cu2+(aq) + 2 Ag(s) + 2 Br-(aq) → Cu(s) + 2 AgBr(s) Eo = 0.27 V Eo for the oxidation half reaction = 0.27 - 0.34 = -0.07 V For AgBr(s) + e- → Ag(s) + Br-(aq), Eo = -(-0.07 V) = +0.07 V

18.54 Sn4+(aq) < Br2(l) < MnO4

- 18.55 Pb(s) < Fe(s) < Al(s) 18.56 Cr2O7

2-(aq) is highest in the table of standard reduction potentials, therefore it is the strongest oxidizing agent. Fe2+(aq) is lowest in the table of standard reduction potentials, therefore it is the weakest oxidizing agent.

18.57 From Table 18.1:

Sn2+ is the strongest reducing agent and Fe2+ is the weakest reducing agent.


539

18.58 (a) Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s) oxidation: Cd(s) → Cd2+(aq) + 2 e- Eo = 0.40 V reduction: Sn2+(aq) + 2 e- → Sn(s) Eo = -0.14 V

overall Eo = 0.26 V

n = 2 mol e-


e mol 1

C 96,500_

(0.26 V)

• V C 1

J 1 = -50,180 J = -50 kJ

(b) 2 Al(s) + 3 Cd2+(aq) → 2 Al3+(aq) + 3 Cd(s) oxidation: 2 Al(s) → 2 Al3+(aq) + 6 e- Eo = 1.66 V reduction: 3 Cd2+(aq) + 6 e- → 3 Cd(s) Eo = -0.40 V

overall Eo = 1.26 V n = 6 mol e-


e mol 1

C 96,500_

(1.26 V)

• V C 1

J 1 = -729,540 J = -730 kJ


oxidation: 6 Fe2+(aq) → 6 Fe3+(aq) + 6 e- Eo = -0.77 V reduction: Cr2O7

2-(aq) + 14 H+(aq) + 6 e- + → 2 Cr3+(aq) + 7 H2O(l) Eo = 1.33 V overall Eo = 0.56 V

n = 6 mol e-


e mol 1

C 96,500_

(0.56 V)

• V C 1

J 1 = -324,240 J = -324 kJ

18.59 (a) 3 Cu2+(aq) + 2 Cr(s) → 3 Cu(s) + 2 Cr3+(aq)

oxidation 2 Cr(s) → 2 Cr3+(aq) + 6 e- Eo = 0.74 V reduction 3 Cu2+(aq) + 6 e- → 3 Cu(s) Eo = 0.34 V



e mol 1

C 96,500_

(1.08 V)

• V C 1

J 1 = -625,320 J = -625 kJ

(b) Pb(s) + 2 H+(aq) → Pb2+(aq) + H2(g) oxidation: Pb(s) → Pb2+(aq) + 2 e- Eo = 0.13 V reduction: 2 H+(aq) + 2 e- → H2(g) Eo = 0.00 V



e mol 1

C 96,500_

(0.13 V)

• V C 1

J 1 = -25,090 J = -25 kJ

(c) Cl2(g) + Sn2+(aq) → Sn4+(aq) + 2 Cl-(aq) oxidation Sn2+(aq) → Sn4+(aq) + 2 e- Eo = -0.15 V reduction Cl2(g) + 2 e- → 2 Cl-(aq) Eo = 1.36 V

overall Eo = 1.21 V


540

n = 2 mol e-


e mol 1

C 96,500_

(1.21 V)

• V C 1

J 1 = -233,530 J = -234 kJ

18.60 (a) 2 Fe2+(aq) + Pb2+(aq) → 2 Fe3+(aq) + Pb(s)

oxidation: 2 Fe2+(aq) → 2 Fe3+(aq) + 2 e- Eo = -0.77 V reduction: Pb2+(aq) + 2 e- → Pb(s) Eo = -0.13 V

overall Eo = -0.90 V Because Eo is negative, this reaction is nonspontaneous. (b) Mg(s) + Ni2+(aq) → Mg2+(aq) + Ni(s) oxidation: Mg(s) → Mg2+(aq) + 2 e- Eo = 2.37 V reduction: Ni2+(aq) + 2 e- → Ni(s) Eo = -0.26 V

overall Eo = 2.11 V Because Eo is positive, this reaction is spontaneous.

18.61 (a) 5 Ag+(aq) + Mn2+(aq) + 4 H2O(l) → 5 Ag(s) + MnO4

-(aq) + 8 H+(aq) oxidation: Mn2+(aq) + 4 H2O(l) → MnO4

-(aq) + 8 H+(aq) + 5 e- Eo = -1.51 V reduction: 5 Ag+(aq) + 5 e- → 5 Ag(s) Eo = 0.80 V

overall Eo = -0.71 V Because Eo is negative, this reaction is nonspontaneous. (b) 2 H2O2(aq) → O2(g) + 2 H2O(l) oxidation: H2O2(aq) → O2(g) + 2 H+(aq) + 2 e- Eo = -0.70 V reduction: H2O2(aq) + 2 H+(aq) + 2 e- → 2 H2O(l) Eo = 1.78 V

overall Eo = 1.08 V Because Eo is positive, this reaction is spontaneous.

18.62 (a) oxidation: Sn2+(aq) → Sn4+(aq) + 2 e- Eo = -0.15 V

reduction: Br2(l) + 2 e- → 2 Br-(aq) Eo = 1.09 V overall Eo = +0.94 V

Because the overall Eo is positive, Sn2+(aq) can be oxidized by Br2(l). (b) oxidation: Sn2+(aq) → Sn4+(aq) + 2 e- Eo = -0.15 V reduction: Ni2+(aq) + 2 e- → Ni(s) Eo = -0.26 V

overall Eo = -0.41 V Because the overall Eo is negative, Ni2+(aq) cannot be reduced by Sn2+(aq). (c) oxidation: 2 Ag(s) → 2 Ag+(aq) + 2 e- Eo = -0.80 V reduction: Pb2+(aq) + 2 e- → Pb(s) Eo = -0.13 V

overall Eo = -0.93 V Because the overall Eo is negative, Ag(s) cannot be oxidized by Pb2+(aq). (d) oxidation: H2SO3(aq) + H2O(l) → SO4

2-(aq) + 4 H+(aq) + 2 e- Eo = -0.17 V reduction: I2(s) + 2 e- → 2 I-(aq) Eo = 0.54 V

overall Eo = +0.37 V Because the overall Eo positive, I2(s) can be reduced by H2SO3.


541

18.63 (a) oxidation: Zn(s) → Zn2+(aq) + 2 e- Eo = 0.76 V reduction: Pb2+(aq) + 2 e- → Pb(s) Eo = -0.13 V

overall Eo = 0.63 V Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s) The reaction is spontaneous because Eo is positive.

(b) oxidation: 4 Fe2+(aq) → 4 Fe3+(aq) + 4 e- Eo = -0.77 V reduction: O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) Eo = 1.23 V

overall Eo = 0.46 V 4 Fe2+(aq) + O2(g) + 4 H+(aq) → 4 Fe3+(aq) + 2 H2O(l) The reaction is spontaneous because Eo is positive. (c) oxidation: 2 Ag(s) → 2 Ag+(aq) + 2 e- Eo = -0.80 V

reduction: Ni2+(aq) + 2 e- → Ni(s) Eo = -0.26 V overall Eo = -1.06 V

There is no reaction because Eo is negative. (d) oxidation: H2(g) → 2 H+(aq) + 2 e- Eo = 0.00 V

reduction: Cd2+(aq) + 2 e- → Cd(s) Eo = -0.40 V overall Eo = -0.40 V

There is no reaction because Eo is negative. The Nernst Equation 18.64 2 Ag+(aq) + Sn(s) → 2 Ag(s) + Sn2+(aq)

oxidation: Sn(s) → Sn2+(aq) + 2 e- Eo = 0.14 V reduction: 2 Ag+(aq) + 2 e- → 2 Ag(s) Eo = 0.80 V

overall Eo = 0.94 V

E = Eo - )(0.010

(0.020) log

2

V) (0.0592 _ V 0.94 =

]Ag[

]Sn[ log

n

V 0.059222+

+2

= 0.87 V

18.65 2 Fe2+(aq) + Cl2(g) → 2 Fe3+(aq) + 2 Cl-(aq)

oxidation: 2 Fe2+(aq) → 2 Fe3+(aq) + 2 e- Eo = -0.77 V reduction: Cl2(g) + 2 e- → 2 Cl-(aq) Eo = 1.36 V

overall Eo = 0.59 V

E = Eo - (0.50))(1.0

)(0.0030)(0.0010 log

2

V) (0.0592 _ V 0.59 =

P ]Fe[

]Cl[]Fe[ log

n

V 0.05922

22

Cl2+2

2_2+3

2

= 0.91 V

18.66 Pb(s) + Cu2+(aq) → Pb2+(aq) + Cu(s)

oxidation: Pb(s) → Pb2+(aq) + 2 e- Eo = 0.13 V reduction: Cu2+(aq) + 2 e- → Cu(s) Eo = 0.34 V

overall Eo = 0.47 V

E = Eo - )10 x (1.0

1.0 log

2

V) (0.0592 _ V 0.47 =

]Cu[

]Pb[ log

n

V 0.05924-+2

+2

= 0.35 V


542

When E = 0, 0 = Eo - ]Cu[

1.0 log

2

V) (0.0592 _ V 0.47 =

]Cu[

]Pb[ log

n

V 0.0592+2+2

+2

0 = 0.47 V + 2

V) (0.0592 log [Cu2+]

log [Cu2+] = (-0.47 V)

V 0.0592

2 = -15.88; [Cu2+] = 10-15.88 = 1 x 10-16 M

18.67 Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

oxidation: Fe(s) → Fe2+(aq) + 2 e- Eo = 0.45 V reduction: Cu2+(aq) + 2 e- → Cu(s) Eo = 0.34 V

overall Eo = 0.79 V

E = 0.67 V = Eo -

]Cu[

0.10 log

2

V) (0.0592 _ V 0.79 =

]Cu[

]Fe[ log

n

V 0.0592+2+2

+2

0.67 V = ])Cu[ log _ (0.10) (log 2

V) (0.0592 _ V 0.79 +2

log [Cu2+] = -5.05; [Cu2+] = 10-5.05 = 8.9 x 10-6 M

18.68 (a) E = Eo - )(0.020 log 2

V) (0.0592 _ V 0.54 = ]I[ log

n

V 0.0592 22_ = 0.64 V

(b) E = Eo -

0.10

0.10 log

1

V) (0.0592 _ V 0.77 =

]Fe[

]Fe[ log

n

V 0.0592+3

+2

= 0.77 V

(c) E = Eo -

0.0010

0.40 log

2

V) (0.0592 _ V 0.15_ =

]Sn[

]Sn[ log

n

V 0.0592+2

+4

= -0.23 V

(d) E = Eo -

1.0)0(1.0)(0.01

log 6

V) (0.0592 _ V 1.33_ =

]Cr[

]H][ OCr[ log

n

V 0.0592 14

2+3

14+_272

E = (0.010) log (14) 6

V) (0.0592 _ V 1.33_ = -1.05 V

18.69 E = Eo - ]OH[

P log n

V 0.05922+

3

H2 ; Eo = 0, n = 2 mol e-, and PH2 = 1 atm

(a) [H3O+] = 1.0 M; E = -

)(1.0

1 log

2

V 0.05922

= 0

(b) pH = 4.00, [H3O+] = 10-4.00 = 1.0 x 10-4 M

E = - )10 x (1.0

1 log

2

V 0.059224_ = -0.24 V

(c) [H3O+] = 1.0 x 10-7 M; E = -

)10 x (1.0

1 log

2

V 0.059227_ = -0.41 V


543

(d) [OH-] = 1.0 M; [H3O+] =

1.010 x 1.0

= ]OH[

K 14_

_

w = 1.0 x 10-14 M

E = -)10 x (1.0

1 log

2

V 0.0592214_ = -0.83 V

18.70 H2(g) + Ni2+(aq) → 2 H+(aq) + Ni(s)

Eo = E + E oNi _ Ni

oH _ H +2+

2 = 0 V + (-0.26 V) = -0.26 V

E = Eo - )P](Ni[

]OH[ log n

V 0.0592

H+2

2+3

2

0.27 V = -0.26 V - (1)(1)

]OH[ log 2

V) (0.0592 2+3

0.27 V = -0.26 V - (0.0592 V) log [H3O+]

pH = - log [H3O

+] therefore 0.27 V = -0.26 V + (0.0592 V) pH

pH = V 0.0592

V) 0.26 + V (0.27 = 9.0

18.71 Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g)

Eo = E + E oZn _Zn

oH _ H +2

2+ = 0 V + 0.76 V = 0.76 V

E = Eo - ]OH[

)P](Zn[ log

n

V 0.05922+

3

H+2

2

0.58 V = 0.76 V - ]OH[

(1)(1) log

2

V) (0.05922+

3

0.58 V = 0.76 V + (0.0592 V) log [H3O+]

pH = - log [H3O+] therefore 0.58 V = 0.76 V - (0.0592 V) pH

pH = V 0.0592

V) 0.76 _ V _(0.58 = 3.0

Standard Cell Potentials and Equilibrium Constants 18.72 ∆Go = -nFEo

Because n and F are always positive, ∆Go is negative when Eo is positive because of the negative sign in the equation.

Eo = n

V 0.0592 log K; log K =

V 0.0592 En o

; K = 100.0592 En o

If Eo is positive, the exponent is positive (because n is positive), and K is greater than 1. 18.73 If K < 1, Eo < 0. When Eo = 0, K = 1. 18.74 Ni(s) + 2 Ag+(aq) → Ni2+(aq) + 2 Ag(s)

oxidation: Ni(s) → Ni2+(aq) + 2 e- Eo = 0.26 V


544

reduction: 2 Ag+(aq) + 2 e- → 2 Ag(s) Eo = 0.80 V overall Eo = 1.06 V

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (2)(1.06 =

V 0.0592 En o

= 35.8; K = 1035.8 = 6 x 1035

18.75 2 MnO4

-(aq) + 10 Cl-(aq) + 16 H+(aq) → 2 Mn2+(aq) + 5 Cl2(g) + 8 H2O(l) oxidation: 10 Cl-(aq) → 5 Cl2(g) + 10 e- Eo = -1.36 V reduction: 2 MnO4

-(aq) + 16 H+(aq) + 10 e- → 2 Mn2+(aq) + 8 H2O(l) Eo = 1.51 V overall Eo = 0.15 V

Eo = K log n

V 0.0592; log K =

V 0.0592

V) (10)(0.15 =

V 0.0592 En o

= 25.3; K = 1025.3 = 2 x

1025 18.76 Eo and n are from Problem 18.58.

Eo = n

V 0.0592 log K; log K =

V 0.0592 En o

(a) Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s); Eo = 0.26 V and n = 2 mol e-

log K = V 0.0592

V) (2)(0.26 = 8.8; K = 108.8 = 6 x 108

(b) 2 Al(s) + 3 Cd2+(aq) → 2 Al3+(aq) + 3 Cd(s); Eo = 1.26 V and n = 6 mol e-

log K = V 0.0592

V) (6)(1.26 = 128; K = 10128


Eo = 0.56 V and n = 6 mol e-

log K = V 0.0592

V) (6)(0.56 = 57; K = 1057

18.77 Eo and n are from Problem 18.59.

Eo = n

V 0.0592 log K; log K =

V 0.0592 En o

(a) 3 Cu2+(aq) + 2 Cr(s) → 3 Cu(s) + 2 Cr3+(aq); Eo = 1.08 V and n = 6 mol e-

log K = V 0.0592

V) (6)(1.08 = 109; K = 10109

(b) Pb(s) + 2 H+(aq) → Pb2+(aq) + H2(g); Eo = 0.13 V and n = 2 mol e-

log K = V 0.0592

V) (2)(0.13 = 4.4; K = 104.4 = 3 x 104

(c) Cl2(g) + Sn2+(aq) → Sn4+(aq) + 2 Cl-(aq); Eo = 1.21 V and n = 2 mol e-

log K = V 0.0592

V) (2)(1.21 = 40.9; K = 1040.9 = 8 x 1040

18.78 Hg2

2+(aq) → Hg(l) + Hg2+(aq) oxidation: ½[Hg2

2+(aq) → 2 Hg2+(aq) + 2 e-] Eo = -0.92 V


545

reduction: ½[Hg22+(aq) + 2 e- → 2 Hg(l)] Eo = 0.80 V

overall Eo = -0.12 V

Eo = K log n

V 0.0592

log K = V 0.0592

V) 0.12 (1)(_ =

V 0.0592 En o

= -2.027; K = 10-2.027 = 9 x 10-3

18.79 2 H2O2(aq) → 2 H2O(l) + O2(g) oxidation: H2O2(aq) → O2(g) + 2 H+(aq) + 2e- Eo = -0.70 V reduction: H2O2(aq) + 2 H+(aq) + 2e- → 2 H2O(l) Eo = 1.78 V

overal Eo = 1.08 V

Eo = K log n

V 0.0592; log K =

V 0.0592

V) (2)(1.08 =

V 0.0592 En o

= 36.5; K = 1036.5 = 3 x

1036 Batteries; Corrosion 18.80 Rust is a hydrated form of iron(III) oxide (Fe2O3 ⋅ H2O). Rust forms from the oxidation

of Fe in the presence of O2 and H2O. Rust can be prevented by coating Fe with Zn (galvanizing).

18.81 Cr forms a protective oxide coating similar to Al. 18.82 Cathodic protection is the attachment of a more easily oxidized metal to the metal you

want to protect. This forces the metal you want to protect to be the cathode, hence the name, cathodic protection. Zn and Al can offer cathodic protection to Fe (Ni and Sn cannot).

18.83 A sacrificial anode is a metal used for cathodic protection. It behaves as an anode and is

more easily oxidized than the metal it is protecting. An example of a sacrificial anode is Zn for protecting Fe (galvanizing).

18.84 (a) (b) Anode: Pb(s) + HSO4

-(aq) → PbSO4(s) + H+(aq) + 2 e- Eo = 0.296 V


546

Cathode: PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- → PbSO4(s) + 2 H2O(l) Eo = 1.628 V

Overall Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) → 2 PbSO4(s) + 2 H2O(l) Eo = 1.924 V

(c) Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (2)(1.924 =

V 0.0592 En o

= 65.0; K = 1 x 1065

(d) When the cell reaction reaches equilibrium the cell voltage = 0. 18.85 oxidation: 2 H2(g) + 4 OH-(aq) → 4 H2O(l) + 4 e- Eo = 0.83 V

reduction: O2(g) + 2 H2O(l) + 4 e- → 4 OH-(aq) Eo = 0.43 V 2 H2(g) + O2(g) → 2 H2O(l) Eo = 1.23 V

n = 4 mol e- and 1 J = 1 C x 1 V


e mol 1

C 96,500_

(1.23 V)

• V C 1

J 1 = - 474,780 J = - 475 kJ

Eo = K log n

V 0.0592; log K =

V 0.0592

V) (4)(1.23 =

V 0.0592 En o

= 83.1; K = 1083.1 = 1 x

1083

E = Eo - )P( )P(

1 log

n

V 0.0592

O2

H 22

= 1.23 V - (25) )(25

1 log

4

V 0.05922

= 1.29 V

18.86 Zn(s) + HgO(s) → ZnO(s) + Hg(l); Zn, 65.39 amu; HgO, 216.59 amu

mass HgO = 2.00 g Zn x HgO mol 1

HgO g 216.59 x

Znmol 1

HgO mol 1 x

Zng 65.39

Znmol 1 = 6.62 g HgO

18.87 Cd(OH)2(s) + 2 Ni(OH)2(s) → Cd(s) + 2 NiO(OH)(s) + 2 H2O(l)

Ni(OH)2, 92.71 amu; Cd, 112.41 amu

mass Cd = 10.0 g Ni(OH)2 x Cd mol 1

Cd g 112.41 x

)Ni(OH mol 2

Cd mol 1 x

)Ni(OH g 92.71

)Ni(OH mol 1

22

2 = 6.06 g

Cd Electrolysis


547

18.88 (a)

(b) anode: 2 Cl-(l) → Cl2(g) + 2 e- cathode: Mg2+(l) + 2 e- → Mg(l) overall: Mg2+(l) + 2 Cl-(l) → Mg(l) + Cl2(g)

18.89 (a)

(b) anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e- cathode: 4 H+(aq) + 4 e- → 2 H2(g) overall: 2 H2O(l) → O2(g) + 2 H2(g)

18.90 possible anode reactions:

2 Cl-(aq) → Cl2(g) + 2 e- 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e-

possible cathode reactions: 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) Mg2+(aq) + 2 e- → Mg(s)

actual reactions: anode: 2 Cl-(aq) → Cl2(g) + 2 e- cathode: 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) This anode reaction takes place instead of 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e- because of a high overvoltage for formation of gaseous O2. This cathode reaction takes place instead of Mg2+(aq) + 2 e- → Mg(s) because H2O is easier to reduce than Mg2+.


548

18.91 (a) K(l) and Cl2(g) (b) H2(g) and Cl2(g). Solvent H2O is reduced in preference to K+. 18.92 (a) NaBr

anode: 2 Br-(aq) → Br2(l) + 2 e- cathode: 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) overall: 2 H2O(l) + 2 Br-(aq) → Br2(l) + H2(g) + 2 OH-(aq) (b) CuCl2 anode: 2 Cl-(aq) → Cl2(g) + 2 e- cathode: Cu2+(aq) + 2 e- → Cu(s) overall: Cu2+(aq) + 2 Cl-(aq) → Cu(s) + Cl2(g) (c) LiOH anode: 4 OH-(aq) → O2(g) + 2 H2O(l) + 4 e- cathode: 4 H2O(l) + 4 e- → 2 H2(g) + 4 OH-(aq) overall: 2 H2O(l) → O2(g) + 2 H2(g)

18.93 (a) Ag2SO4

anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e- cathode: 4 Ag+(aq) + 4 e- → 4 Ag(s) overall: 4 Ag+(aq) + 2 H2O(l) → O2(g) + 4 H+(aq) + 4 Ag(s) (b) Ca(OH)2 anode: 4 OH-(aq) → O2(g) + 2 H2O(l) + 4 e- cathode: 4 H2O(l) + 4 e- → 2 H2(g) + 4 OH-(aq) overall: 2 H2O(l) → O2(g) + 2 H2(g) (c) KI anode: 2 I-(aq) → I2(s) + 2 e- cathode: 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq) overall: 2 I-(aq) + 2 H2O(l) → I2(s) + H2(g) + 2 OH-(aq)

18.94 Ag+(aq) + e- → Ag(s); 1 A = 1 C/s

mass Ag = Ag mol 1

Ag g 107.87 x

e mol 1

Ag mol 1 x

C 96,500e mol 1

x min 1

s 60min x 20.0 x

s

C 2.40

_

_

=

3.22 g 18.95 Cu2+(aq) + 2 e- → Cu(s)

mol e- = 100.0 C 96,500

e mol 1 x

min

s 60 x

h

min 60h x 24.0 x

s

C _

= 89.5 mol e-

mass Cu = 89.5 mol e- x g 1000

kg 1 x

Cu mol 1

Cu g 63.54 x

e mol 2

Cu mol 1_

= 2.84 kg Cu

18.96 2 Na+(l) + 2 Cl-(l) → 2 Na(l) + Cl2(g)

Na+(l) + e- → Na(l); 1 A = 1 C/s; 1.00 x 103 kg = 1.00 x 106 g


549

Charge = 1.00 x 106 g Na x e mol 1

C 96,500 x

Na mol 1e mol 1

x Na g 22.99

Na mol 1_

_

= 4.20 x 109 C

Time = s 3600

h 1 x

C/s 30,000

C 10 x 4.20 9

= 38.9 h

1.00 x 106 g Na x Na mol 2Cl mol 1

x Na g 22.99

Na mol 1 2 = 21,748.6 mol Cl2

PV = nRT

V = atm 1.00


06 0.082mol) (21,748.6 =

P

nRT

••

= 4.87 x 105 L Cl2

18.97 Al3+ + 3 e- → Al; 40.0 kg = 40,000 g; 1 h = 3600 s

Charge = 40,000 g Al x e mol 1

C 96,500 x

Al mol 1e mol 3

x Al g 26.98

Al mol 1_

_

= 4.29 x 108 C

Current = s 3600

C 10 x 4.29 8

= 1.19 x 105 A

18.98 PbSO4(s) + H+(aq) + 2 e- → Pb(s) + HSO4-(aq)

mass PbSO4

=PbSO mol 1PbSO g 303.3

x e mol 2

PbSO mol 1 x

C 96,500e mol 1

x h 1

s 3600h x 1.50 x

s

C 10.0

4

4

_

4_

mass PbSO4 = 84.9 g PbSO4 18.99 Cr3+(aq) + 3 e- → Cr(s)

Charge = 125 g Cr x e mol 1

C 96,500 x

Cr mol 1e mol 3

x Cr g 52.00

Cr mol 1_

_

= 6.96 x 105 C

Time = C/s 200.0

C 10 x 6.96 5

x s 60

min 1 = 58.0 min

General Problems 18.100 (a) 2 MnO4

-(aq) + 16 H+(aq) + 5 Sn2+(aq) → 2 Mn2+(aq) + 5 Sn4+(aq) + 8 H2O(l) (b) MnO4

- is the oxidizing agent; Sn2+ is the reducing agent. (c) Eo = 1.51 V + (-0.15 V) = 1.36 V

18.101 2 Mn3+(aq) + 2 H2O(l) → Mn2+(aq) + MnO2(s) + 4 H+(aq)

Eo = 1.51 V + (-0.95 V) = +0.56 V Because Eo is positive, the disproportionation is spontaneous under standard-state conditions.

18.102 (a) Ag+ is the strongest oxidizing agent because Ag+ has the most positive standard

reduction potential.


550

Pb is the strongest reducing agent because Pb2+ has the most negative standard reduction potential.

(b)

(c) Pb(s) + 2 Ag+(aq) → Pb2+(aq) + 2 Ag(s); n = 2 mol e- Eo = Eo

ox + Eored = 0.13 V + 0.80 V = 0.93 V


e mol 1

C 96,500_

(0.93 V)

• V C 1

J 1 = -179,490 J = -180 kJ

Eo = K log n

V 0.0592; log K =

V 0.0592

V) (2)(0.93 =

V 0.0592 En o

= 31; K = 1031

(d) E = Eo -

)(0.01

0.01 log

2

V 0.0592 _ V 0.93 =

]Ag[

]Pb[ log

n

V 0.059222+

+2

= 0.87 V

18.103 For Pb2+, E = -0.13 - 2

V 0.0592 log

]Pb[

1+2

For Cd2+, E = -0.40 - 2

V 0.0592 log

]Cd[

1+2

Set these two equations for E equal to each other and solve for [Cd2+]/[Pb2+].

-0.13 - 2

V 0.0592 log

]Pb[

1+2

= -0.40 - 2

V 0.0592 log

]Cd[

1+2


551

0.27 = 2

V 0.0592(log[Cd2+] - log[Pb2+]) =

]Pb[

]Cd[ log

2

V 0.0592+2

+2

log0.0592

(0.27)(2) =

]Pb[

]Cd[+2

+2

= 9.1; ]Pb[

]Cd[+2

+2

= 109.1 = 1 x 109

18.104 (a) (b) 2 Al(s) + 6 H+(aq) → 2 Al3+(aq) + 3 H2(g) Eo = Eo

ox + Eored = 1.66 V + 0.00 V = 1.66 V

(c)

E = Eo -

)(0.10

)(10.0)(0.10 log

6

V) (0.0592 _ V 1.66 =

]H[

)P(]Al[ log

n

V 0.05926

32

6+

3H

2+32 = 1.59 V

(d) ∆Go = -nFEo = -(6 mol e-)

e mol 1

C 96,500_

(1.66 V)

• V C 1

J 1 = -961,140 J = -961 kJ

Eo = K log n

V 0.0592; log K =

V 0.0592

V) (6)(1.66 =

V 0.0592 En o

= 168; K = 10168

(e) mass Al =Al mol 1

Al g 26.98 x

e mol 3

Al mol 1 x

C 96,500e mol 1

x min 1

s 60min x 25.0 x

s

C 10.0

_

_

=

1.40 g 18.105 Zn(s) → Zn2+(aq) + 2 e-

mass Zn = 0.100 Znmol 1

Zng 65.39 x

e mol 2

Znmol 1 x

C 96,500e mol 1

x min

s 60 x

h

min 60h x 200.0 x

s

C_

_

mass Zn = 24.4 g 18.106 2 Cl-(aq) → Cl2(g) + 2 e-

13 million tons = 13 x 106 tons; Cl2, 70.91 amu

13 x 106 tons x Cl g 70.91

Cl mol 1 x

ton1

g 907,200

2

2 = 1.66 x 1011 mol Cl2

Charge = 1.66 x 1011 mol Cl2 x e mol 1

C 96,500 x

Cl mol 1e mol 2

_2

_

= 3.20 x 1016 C

1 J = 1 C x 1 V; Energy = (3.20 x 1016 C)(4.5 V) = 1.44 x 1017 J


552

kWh = (1.44 x 1017 J)

J 10 x 3.6

kWh 16

= 4.0 x 1010 kWh

18.107 (a) From: B + A+ → B+ + A, A+ is reduced more easily than B+

From: C + A+ → C+ + A, A+ is reduced more easily than C+ From: B + C+ → B+ + C, C+ is reduced more easily than B+ A+ + e- → A C+ + e- → C B+ + e- → B (b) A+ is the strongest oxidizing agent; B is the strongest reducing agent (c) A+ + B → B+ + A

18.108 (a) oxidizing agents: PbO2, H

+, Cr2O72-; reducing agents: Al, Fe, Ag

(b) PbO2 is the strongest oxidizing agent. H+ is the weakest oxidizing agent. (c) Al is the strongest reducing agent. Ag is the weakest reducing agent. (d) oxidized by Cu2+: Fe and Al; reduced by H2O2: PbO2 and Cr2O7

2- 18.109 From Appendix D:

AgBr(s) + e- → Ag(s) + Br-(aq) Eo = 0.07 V (a) oxidation: C6H4(OH)2(aq) → C6H4O2(aq) + 2 H+(aq) + 2 e- Eo = -0.699 V reduction: 2[AgBr(s) + e- → Ag(s) + Br-(aq)] Eo = 0.07 V

overall: 2 AgBr(s) + C6H4(OH)2(aq) → 2 Ag(s) + 2 Br-(aq) + C6H4O2(aq) + 2 H+(aq)

overall Eo = -0.699 V + 0.07 V = -0.63 V Because the overall Eo is negative, the reaction is nonspontaneous when [H+] = 1.0 M.

(b) Eo(in 1.0 M OH-) = Eo(in 1.0 M H+) - ])(OHHC[

]OHC[]H[]Br[ log

n

V 0.0592

246

2462+2_

[H+] = 1.0

10 x 1.0 =

]OH[K 14_

_

w = 1.0 x 10-14 M

Eo(in 1.0 M OH-) = -0.63 V - (1)

(1))10()(1 log

2

V) (0.0592 214_2

= +0.20 V

18.110 (a) 3 CH3CH2OH(aq) + 2 Cr2O7

2-(aq) + 16 H+(aq) → 3 CH3CO2H(aq) + 4 Cr3+(aq) + 11 H2O(l)

oxidation: 3 CH3CH2OH(aq) + 3 H2O(l) → 3 CH3CO2H(aq) + 12 H+(aq) + 12 e- Eo = -0.058V reduction: 2 Cr2O7

2-(aq) + 28 H+(aq) + 12 e- + → 4 Cr3+(aq) + 14 H2O(l) Eo = 1.33 V overall Eo = 1.27 V


553

(b) E = Eo - ]H[]OCr[]OHCHCH[

]Cr[]HCOCH[ log

n

V 0.059216+2_2

723

23

4+3323

pH = 4.00, [H+] = 0.000 10 M

E = 1.27 V -

)10 (0.000)(1.0)(1.0

)(1.0)(1.0 log

12

V) (0.05921623

43

E = 1.27 V - )10 (0.000

1 log

12

V) (0.059216

= 0.95 V

18.111 (a) ∆Go = -nFEo

∆Go3 = ∆Go

1 + ∆Go2 therefore -n3FEo

3 = -n1FEo1 + (-n2FEo

2) n3E

o3 = n1E

o1 + n2E

o2

Eo3 =

nE n + E n

3

o22

o11

(b) Eo3 =

1

V) (2)(0.45 + V) 0.04 (3)(_ = 0.78 V

(c) Eo values would be additive (Eo3 = Eo

1 + Eo2) if reaction (3) is an overall cell

reaction because the electrons in the two half reactions, (1) and (2), cancel. That is, n1 = n2 = n3 in the equation for Eo3.

18.112 anode: Ag(s) + Cl-(aq) → AgCl(s) + e-

cathode: Ag+(aq) + e- → Ag(s) overall: Ag+(aq) + Cl-(aq) → AgCl(s) Eo = 0.578 V

For AgCl(s) _ Ag+(aq) + Cl-(aq) Eo = -0.578 V

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) 0.578 (1)(_ =

V 0.0592 En o

= -9.76

K = Ksp = 10-9.76 = 1.7 x 10-10 18.113 (a) anode: Cu(s) → Cu2+(aq) + 2 e- Eo = -0.34 V

cathode: 2 Ag+(aq) + 2 e- → 2 Ag(s) Eo = 0.80 V overall: 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) Eo = 0.46 V

E = Eo -

)(0.050

1.0 log

2

V) (0.0592 _ V 0.46 =

]Ag[

]Cu[ log

n

V 0.059222+

+2

= 0.38 V

(b) [Ag+] = M 1.010 x 5.4

= ]Br[

K 13_

_

sp = 5.4 x 10-13 M

E = Eo -

)10 x (5.4

1.0 log

2

V) (0.0592 _ V 0.46 =

]Ag[

]Cu[ log

n

V 0.0592213_2+

+2

= -0.27 V

The cell potential for the spontaneous reaction is E = 0.27 V. The spontaneous reaction is: Cu2+(aq) + 2 Ag(s) + 2 Br-(aq) → 2 AgBr(s) + Cu(s)


554

(c) Cu2+(aq) + 2 e- → Cu(s) Eo = 0.34 V

2 Ag(s) + 2 Br-(aq) → 2 AgBr(s) + 2 e- Eo = ? Cu2+(aq) + 2 Ag(s) + 2 Br-(aq) → 2 AgBr(s) + Cu(s) Eo = 0.27 V

Eo = ? = 0.27 V - 0.34 V = -0.07 V For: AgBr(s) + e- → Ag(s) + Br-(aq) the standard reduction potential is Eo = 0.07 V

18.114 4 Fe2+(aq) + O2(g) + 4 H+(aq) → 4 Fe3+(aq) + 2 H2O(l)

oxidation 4 Fe2+(aq) → 4 Fe3+(aq) + 4 e- Eo = - 0.77 V reduction O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) Eo = 1.23 V

overall Eo = 0.46 V

= Hg mm 760

atm 1.00 x Hg mm 160 = PO2

0.211 atm

E = Eo - )P(]H[]Fe[

]Fe[ log

n

V 0.0592

O4+4+2

4+3

2

E = 0.46 V - (0.211))10 x (1)10 x (1

)10 x (1 log

4

V 0.059247_47_

47_

E = 0.46 V - 0.42 V = 0.04 V Because E is positive, the reaction is spontaneous.

18.115 H2MoO4(aq) + As(s) → Mo3+(aq) + H3AsO4(aq)

H2MoO4(aq) → Mo3+(aq) H2MoO4(aq) → Mo3+(aq) + 4 H2O(l) 6 H+(aq) + H2MoO4(aq) → Mo3+(aq) + 4 H2O(l) [3 e- + 6 H+(aq) + H2MoO4(aq) → Mo3+(aq) + 4 H2O(l)] x 5

(reduction half reaction) As(s) → H3AsO4(aq) As(s) + 4 H2O(l) → H3AsO4(aq) As(s) + 4 H2O(l) → H3AsO4(aq) + 5 H+(aq) [As(s) + 4 H2O(l) → H3AsO4(aq) + 5 H+(aq) + 5 e-] x 3 (oxidation half reaction)

Combine the two half reactions. 30 H+(aq) + 5 H2MoO4(aq) + 3 As(s) + 12 H2O(l) →

5 Mo3+(aq) + 3 H3AsO4(aq) + 15 H+(aq) + 20 H2O(l) 15 H+(aq) + 5 H2MoO4(aq) + 3 As(s) → 5 Mo3+(aq) + 3 H3AsO4(aq) + 8 H2O(l)

5 x [H2MoO4(aq) + 2 H+(aq) + 2 e- → MoO2(s) + 2 H2O(l)] Eo = +0.646 V 5 x [MoO2(s) + 4 H+(aq) + e- → Mo3+(aq) + 2 H2O(l)] Eo = -0.008 V


555

3 x [As(s) + 3 H2O(l) → H3AsO3(aq) + 3 H+(aq) + 3 e-] Eo = -0.240 V 3 x [H3AsO3(aq) + H2O(l) → H3AsO4(aq) + 2 H+(aq) + 2 e-] Eo = -0.560 V

15 H+(aq) + 5 H2MoO4(aq) + 3 As(s) → 5 Mo3+(aq) + 3 H3AsO4(aq) + 8 H2O(l)

∆Go = -nFEo = -(10 mol e-)

e mol 1

C 96,500_

(0.646 V)

• V C 1

J 1 = -623,390 J = -623.4 kJ


e mol 1

C 96,500_

(-0.008 V)

• V C 1

J 1 = 3,860 J = +3.9 kJ


e mol 1

C 96,500_

(-0.240 V)

• V C 1

J 1 = 208,440 J = +208.4 kJ


e mol 1

C 96,500_

(-0.560 V)

• V C 1

J 1 = 324,240 J = +324.2 kJ

∆Go(total) = -623.4 kJ + 3.9 kJ + 208.4 kJ + 324.2 kJ = -86.9 kJ = -86,900 J 1 V = 1 J/C

∆Go = -nFEo; Eo =

∆

e mol 1C 96,500

)e mol (15

J) 86,900 (_ _ =

nF G _

__

o

= +0.060 J/C = +0.060 V

18.116 First calculate Eo for the galvanic cell in order to determine Eo

1. anode: 5 [2 Hg(l) + 2 Br-(aq) → Hg2Br2(s) + 2 e- ] Eo

1 = ? cathode: 2 [MnO4

-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)] Eo2 = 1.51 V

overall: 2 MnO4-(aq) + 10 Hg(l) + 10 Br-(aq) + 16 H+(aq) →

2 Mn2+(aq) + 5 Hg2Br2(s) + 8 H2O(l) n = 10 mol e-

E = Eo - ]H[]MnO[]Br[

]Mn[ log

n

V 0.059216+2_

410_

2+2

1.214 V = Eo -

)(0.10)(0.10)(0.10

)(0.10 log

10

V) (0.059216210

2

1.214 V = Eo - )(0.10

1 log

10

V) (0.059226

= Eo - 0.154 V

Eo = 1.214 + 0.154 = 1.368 V Eo

1 + Eo2 = 1.368 V; Eo1 +1.51 V = 1.368 V; Eo1 = 1.368 V - 1.51 V = -0.142 V

oxidation: 2 Hg(l) → Hg22+(aq) + 2 e- Eo = -0.80 V (Appendix D)

reduction: Hg2Br2(s) + 2 e- → 2 Hg(l) + 2 Br-(aq) Eo = -0.142 V (from Eo1) overall: Hg2Br2(s) → Hg2

2+(aq) + 2 Br-(aq) Eo = -0.658 V

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) 0.658 (2)(_ =

V 0.0592 En o

= -22.2

K = Ksp = 10-22.2 = 6 x 10-23 18.117 oxidation: Cu+(aq) → Cu2+(aq) + e- Eo = -0.15 V


556

reduction: Cu2+(aq) + 2 CN-(aq) + e- → Cu(CN)2-(aq) Eo = 1.103 V

overall: Cu+(aq) + 2 CN-(aq) → Cu(CN)2-(aq) Eo = 0.953 V

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (1)(0.953 =

V 0.0592 En o

= 16.1

K = Kf = 1016.1 = 1 x 1016 18.118 (a) anode: 4[Al(s) → Al3+(aq) + 3 e-] Eo = 1.66 V

cathode: 3[O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l)] Eo = 1.23 V overall: 4 Al(s) + 3 O2(g) + 12 H+(aq) → 4 Al3+(aq) + 6 H2O(l) Eo = 2.89 V

(b) & (c) E = Eo - ]H[ )P(

]Al[ log

Fn

T R 2.30312+3

O

4+3

2

E = 2.89 V -

•)10 x (1.0 )(0.20

)10 x (1.0 log

) e C/mol )(96,500 e mol (12

K) (310mol K

kJ 10 x 8.314(2.303)

127_3

49_

__

3_

E = 2.89 V - 0.257 V = 2.63 V

18.119 E1 = Eo1 -

6

V 0.0592log

]H[]NO[

)P(]Cu[8+2_

3

2NO

3+2

and E2 = Eo2 -

2

V 0.0592log

]H[] NO[

)P](Cu[4+2_

3

2NO

+22

(a) E1 = 0.62 V - 6

V 0.0592log

)(1.0)(1.0

)10 x (1.0)(0.1082

23_3

= 0.71 V

E2 = 0.45 V - 2

V 0.0592log

)(1.0)(1.0

)10 x (0.10)(1.042

23_

= 0.66 V

Reaction (1) has the greater thermodynamic tendency to occur because of the larger positive potential.

(b) E1 = 0.62 V - 6

V 0.0592log

)(10.0)(10.0

)10 x (1.0)(0.1082

23_3

= 0.81 V

E2 = 0.45 V - 2

V 0.0592log

)(10.0)(10.0

)10 x (0.10)(1.042

23_

= 0.83 V

Reaction (2) has the greater thermodynamic tendency to occur because of the larger positive potential. (c) Set the two equations equal to each other and solve for x.

0.62 V - 6

V 0.0592log

)(x)(x

)10 x (1.0)(0.1082

23_3

= 0.45 V - 2

V 0.0592log

)(x)(x

)10 x (0.10)(1.042

23_

0.17 - 6

V 0.0592[(-9) - 10 log x] = -

2

V 0.0592[(-7) - 6 log x]

0.0516 = 0.0789 log x; 0.0789

0.0516 = log x; 0.654 = log x

[HNO3] = x = 100.654 = 4.5 M


557

Multi-Concept Problems 18.120 (a) 4 CH2=CHCN + 2 H2O → 2 NC(CH2)4CN + O2

(b) mol e- = 3000 C/s x 10.0 h x C 96,500

e mol 1 x

h 1

s 3600 _

= 1119.2 mol e-

mass adiponitrile =

1119.2 mol e- x g 1000

kg 1.0 x

leadiponitri mol 1

leadiponitri g 108.14 x

e mol 2

leadiponitri mol 1_

= 60.5 kg

(c) 1119.2 mol e- x e mol 4O mol 1

_

2 = 279.8 mol O2

PV = nRT

V =

••

Hg mm 760atm 1

x Hg mm 740

K) (298mol Katm L

06 0.082mol) (279.8 =

P

nRT = 7030 L O2

18.121 (a) 2 MnO4

-(aq) + 5 H2C2O4(aq) + 6 H+(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)

(b) oxidation: 5[H2C2O4(aq) → 2 CO2(g) + 2 H+(aq) + 2 e-] Eo = 0.49 V reduction: 2[MnO4

-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)] Eo = 1.51 V overall Eo = 2.00 V

(c)


C 96,500_

• V C 1

J 1 = -1,930,000 J = -1,930 kJ

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (10)(2.00 =

V 0.0592 En o

= 338; K = 10338

(d) Na2C2O4, 134.0 amu

1.200 g Na2C2O4 x OCNa mol 5

KMnO mol 2 x

OCNa g 134.0OCNa mol 1

422

4

422

422 = 3.582 x 10-3 mol KMnO4


mol 10 x 3.582 3_

= 0.1102 M

18.122 (a) Cr2O7

2-(aq) + 6 Fe2+(aq) + 14 H+(aq) → 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(l) (b) The two half reactions are: oxidation: Fe2+(aq) → Fe3+(aq) + e- Eo = -0.77 V reduction: Cr2O7

2-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O(l) Eo = 1.33 V At the equivalence point the potential is given by either of the following expressions:


558

(1) E = 1.33 V - 6

V 0.0592log

]H][ OCr[

]Cr[14+_2

72

2+3

(2) E = 0.77 V - 1

V 0.0592log

]Fe[

]Fe[+3

+2

where E is the same in both because equilibrium is reached and the solution can have only one potential. Multiplying (1) by 6, adding it to (2), and using some stoichiometric relationships at the equivalence point will simplify the log term.

7E = [(6 x 1.33 V) + 0.77 V] - (0.0592 V)log]H][ OCr][Fe[

]Cr][Fe[14+_2

72+3

2+3+2

At the equivalence point, [Fe2+] = 6[Cr2O7

2-] and [Fe3+] = 3[Cr3+]. Substitute these equalities into the previous equation.

7E = [(6 x 1.33 V) + 0.77 V] - (0.0592 V)log]H][ OCr][Cr3[

]Cr][ OCr6[14+_2

72+3

2+3_272

Cancel identical terms.

7E = [(6 x 1.33 V) + 0.77 V] - (0.0592 V)log]H3[

]Cr6[14+

+3

mol Fe2+ = (0.120 L)(0.100 mol/L) = 0.0120 mol Fe2+

mol Cr2O72- = 0.0120 mol Fe2+ x

Fe mol 6OCr mol 1

+2

_272 = 0.00200 mol Cr2O7

2-

volume Cr2O72- = 0.00200 mol x

mol 0.120

L 1 = 0.0167 L

At the equivalence point assume mol Fe3+ = initial mol Fe2+ = 0.0120 mol Total volume at the equivalence point is 0.120 L + 0.0167 L = 0.1367 L

[Fe3+] = L 0.1367

mol 0.0120 = 0.0878 M; [Cr3+] = [Fe3+]/3 = (0.0878 M)/3 = 0.0293 M

[H+] = 10-pH = 10-2.00 = 0.010 M

7E = [(6 x 1.33 V) + 0.77 V] - (0.0592 V)log)3(0.010

6(0.0293)14

= 8.75 - 1.585 = 7.165 V

E = 7

V 7.165 = 1.02 V at the equivalence point.

18.123 2 H2(g) + O2(g) → 2 H2O(l)

(a) ∆Ho = 2 ∆Hof(H2O) = (2 mol)(-285.8 kJ/mol) = -571.6 kJ

∆So = 2 So(H2O) - [2 So(H2) + So(O2)] ∆So = (2 mol)(69.9 J/(K ⋅ mol)) - [(2 mol)(130.6 J/(K ⋅ mol)) + (1 mol)(205.0 J/(K ⋅ mol))] ∆So = -326.4 J/K = -0.3264 kJ/K 95oC = 368 K ∆Go = ∆Ho - T∆So = -571.6 kJ -(368 K)(-0.3264 kJ/K) = - 451.5 kJ 1 V = 1 J/C


559

∆Go = -nFEo; Eo = C) (4)(96,500

J 10 x 451.5 _ _ =

nF G

_3o∆

= 1.17 J/C = 1.17 V

(b) E = Eo - )P()P(

1 log

Fn

T R 2.303

O2

H 22

E = 1.17 V -

•(25) )(25

1 log

) e C/mol )(96,500 e mol (4

K) (368K mol

kJ 10 x 8.314(2.303)

2__

3_

E = 1.17 V + 0.077 V = 1.25 V 18.124 (a) Zn(s) + 2 Ag+(aq) + H2O(l) → ZnO(s) + 2 Ag(s) + 2 H+(aq)

∆Horxn = ∆Ho

f(ZnO) - [2 ∆Hof(Ag+) + ∆Ho

f(H2O)] ∆Ho

rxn = [(1 mol)(-348.3 kJ/mol)] - [(2 mol)(105.6 kJ/mol) + (1 mol)(-285.8 kJ/mol)] ∆Ho

rxn = -273.7 kJ ∆So = [So(ZnO) + 2 So(Ag)] - [So(Zn) + 2 So(Ag+) + So(H2O)] ∆So = [(1 mol)(43.6 J/(K⋅ mol)) + (2 mol)(42.6 J/(K⋅ mol))] - [(1 mol)(41.6 J/(K⋅ mol)) + (2 mol)(72.7 J/(K⋅ mol)) + (1 mol)(69.9 J/(K⋅ mol)) ∆So = - 128.1 J/K ∆Go = ∆Ho - T∆So = - 273.7 kJ - (298 K)(- 128.1 x 10-3 kJ/K) = - 235.5 kJ (b) 1 V = 1 J/C

∆Go = -nFEo Eo =

∆

e mol 1C 96,500

)e mol (2

J) 10 x 235.5_(_ =

nF G _

__

3o

= 1.220 J/C = 1.220 V

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (2)(1.220 =

V 0.0592 En o

= 41.22

K = 1041.22 = 2 x 1041

(c) E = Eo - ] Ag[

] H[ log n

V 0.05922+

2+

The addition of NH3 to the cathode compartment would result in the formation of the Ag(NH3)2

+ complex ion which results in a decrease in Ag+ concentration. The log term in the Nernst equation becomes larger and the cell voltage decreases.

On mixing equal volumes of two solutions, the concentrations of both solutions are cut in half.

Ag+(aq) + 2 NH3(aq) _ Ag(NH3)2+(aq)

before reaction (M) 0.0500 2.00 0 assume 100% reaction -0.0500 -2(0.0500) +0.0500 after reaction (M) 0 1.90 0.0500 assume small back rxn +x +2x -x

equil (M) x 1.90 + 2x 0.0500 - x

Kf = 1.7 x 107 = )(x)(1.90

0.0500

)x2 + (x)(1.90

x)_ (0.0500 =

]NH][Ag[

])NH[Ag(222

3+

+23 ≈


560

Solve for x. x = [Ag+] = 8.15 x 10-10 M

E = Eo - ]Ag[

]H[ log n

V 0.05922+

2+

= 1.220 V - )M 10 x (8.15

)M (1.00 log

2

V 0.0592210_

2

= 0.682

V

(d) Calculate new initial concentrations because of dilution to 110.0 mL.

M i x Vi = Mf x Vf; Mf = [Cl-] = mL 110.0

mL 10.0 x M 0.200 =

V

V x M

f

ii = 0.0182 M

M i x Vi = Mf x Vf; Mf = [Ag+] = mL 110.0

mL 100.0 x M 0.0500 =

V

V x M

f

ii = 0.0455 M

M i x Vi = Mf x Vf; Mf = [NH3] = mL 110.0

mL 100.0 x M 2.00 =

V

V x M

f

ii = 1.82 M

Now calculate the [Ag+] as a result of the following equilibrium: Ag+(aq) + 2 NH3(aq) _ Ag(NH3)2

+(aq) before reaction (M) 0.0455 1.82 0 assume 100% reaction -0.0455 -2(0.0455) +0.0455 after reaction (M) 0 1.73 0.0455 assume small back rxn +x +2x -x

equil (M) x 1.73 + 2x 0.0455 - x

Kf = 1.7 x 107 = )(x)(1.73

0.0455

)x2 + (x)(1.73

x)_ (0.0455 =

]NH][Ag[

])NH[Ag(222

3+

+23 ≈

Solve for x. x = [Ag+] = 8.94 x 10-10 M For AgCl, Ksp = 1.8 x 10-10 IP = [Ag+][Cl -] = (8.94 x 10-10 M)(0.0182 M) = 1.6 x 10-11 IP < Ksp, AgCl will not precipitate. Now calculate new initial concentrations because of dilution to 120.0 mL.

M i x Vi = Mf x Vf; Mf = [Br-] = mL 120.0

mL 10.0 x M 0.200 =

V

V x M

f

ii = 0.0167 M

M i x Vi = Mf x Vf; Mf = [Ag+] = mL 120.0

mL 100.0 x M 0.0500 =

V

V x M

f

ii = 0.0417 M

M i x Vi = Mf x Vf; Mf = [NH3] = mL 120.0

mL 100.0 x M 2.00 =

V

V x M

f

ii = 1.67 M

Now calculate the [Ag+] as a result of the following equilibrium: Ag+(aq) + 2 NH3(aq) _ Ag(NH3)2

+(aq) before reaction (M) 0.0417 1.67 0 assume 100% reaction -0.0417 -2(0.0417) +0.0417 after reaction (M) 0 1.59 0.0417 assume small back rxn +x +2x -x

equil (M) x 1.59 + 2x 0.0417 - x

Kf = 1.7 x 107 = )(x)(1.59

0.0417

)x2 + (x)(1.59

x)_ (0.0417 =

]NH][Ag[

])NH[Ag(222

3+

+23 ≈

Solve for x. x = [Ag+] = 9.70 x 10-10 M


561

For AgBr, Ksp = 5.4 x 10-13 IP = [Ag+][Br -] = (9.70 x 10-10 M)(0.0167 M) = 1.6 x 10-11 IP > Ksp, AgBr will precipitate.

18.125 (a) anode: Fe(s) + 2 OH-(aq) → Fe(OH)2(s) + 2 e-

cathode: 2 x [NiO(OH)(s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)] overall: Fe(s) + 2 NiO(OH)(s) + 2 H2O(l) → Fe(OH)2(s) + 2 Ni(OH)2(s) (b)


e mol 1

C 96,500_

(1.37 V)

• V C 1

J 1 = -264,410 J = -264 kJ

Eo = n

V 0.0592 log K; log K =

V 0.0592

V) (2)(1.37 =

V 0.0592 En o

= 46.3

K = 1046.3 = 2 x 1046 (c) It would still be 1.37 V because OH- does not appear in the overall cell reaction. The overall cell reaction contains only solids and one liquid, therefore the cell voltage does not change because there are no concentration changes. (d) Fe(OH)2, 89.86 amu; 1 A = 1 C/s

mol e- = (0.250 C/s)(40.0 min)

C 96,500e mol 1

min 1

s 60 _

= 6.22 x 10-3 mol e-

mass Fe(OH)2 = (6.22 x 10-3 mol e-) x xe mol 2

)Fe(OH mol 1_

2

)Fe(OH mol 1

)Fe(OH g 89.86

2

2 = 0.279 g

H2O molecules consumed =

(6.22 x 10-3 mol e-) x xe mol 2

OH mol 2_

2 = OH mol 1

molecules OH 10 x 6.022

2

223

3.75 x 1021 H2O

molecules 18.126 (a) Oxidation half reaction: 2 [C4H10(g) + 13 O2-(s) → 4 CO2(g) + 5 H2O(l) + 26 e-]

Reduction half reaction: 13 [O2(g) + 4 e- → 2 O2-(s)] Cell reaction: 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) (b) ∆Ho = [8 ∆Ho

f(CO2)) + 10 ∆Hof(H2O)] - [2 ∆Ho

f(C4H10)] ∆Ho = [(8 mol)(-393.5 kJ/mol) + (10 mol)(-285.8 kJ/mol)]

- [(2 mol)(-126 kJ/mol)] = -5754 kJ ∆So = [8 So(CO2) + 10 So(H2O)] - [2 So(C4H10) + 13 So(O2)] ∆So = [(8 mol)(213.6 J/(K ⋅ mol)) + (10 mol)(69.9 J/(K ⋅ mol))]

- [(2 mol)(310 J/(K ⋅ mol)) + (13 mol)(205 J/(K ⋅ mol))] = -877.2 J/K ∆Go = ∆Ho - T∆So = -5754 kJ - (298 K)(-877.2 x 10-3 kJ/K) = -5493 kJ 1 V = 1 J/C

∆Go = - nFEo; Eo = C) 0(52)(96,50

J 10 x 5493 _ _ =

nF G

_3o∆

= 1.09 J/C = 1.09 V

∆Go = -RT ln K


562

ln K = K) kJ/K)(298 10 x (8.314

kJ) 5493 (_ _ =

RT G _

3_

o∆ = 2217

K = e2217 = 7 x 10962

On raising the temperature, both K and Eo will decrease because the reaction is exothermic (∆Ho < 0). (c) C4H10, 58.12 amu; 10.5 A = 10.5 C/s

mass C4H10 = 10.5 C/s x 8 hr x xmin 1

s 60 x

hr 1

min 60 x

C 96,500e mol 1 _

xe mol 52HC mol 2

_

104

= HC mol 1HC g 58.12

104

104 7.00 g C4H10

n = 7.00 g C4H10 x = HC g 58.12

HC mol 1

104

104 0.120 mol C4H10

20oC = 20 + 273 = 293 K PV = nRT

V =

••

Hg mm 760atm 1.00

x Hg mm 815

K) (293mol K atm L

06 0.082mol) (0.120 =

P

nRT = 2.69 L

18.127 cathode:

(1) MnO2(s) + 4 H+(aq) + 2 e- → Mn2+(aq) + 2 H2O(l) Eo = +1.22 V (2) Mn(OH)2(s) + OH-(aq) → MnO(OH)(s) + H2O(l) + e- Eo = +0.380 V (3) Mn2+(aq) + 2 OH-(aq) → Mn(OH)2(s) K = 1/Ksp = 1/(2.1 x 10-13) = 4.8 x 1012 (4) 4 x [H2O(l) → H+(aq) + OH-(aq)] K = (Kw)4 = (1.0 x 10-14)4 = 1.0 x 10-56 MnO2(s) + H2O(l) + e- → MnO(OH)(s) + OH-(aq)

∆Go1 = -nFEo = -(2 mol e-)

e mol 1

C 96,500_

(1.22 V)

• V C 1

J 1 = -235,460 J = -235.5 kJ

∆Go2 = -nFEo = -(1 mol e-)

e mol 1

C 96,500_

(0.380 V)

• V C 1

J 1 = -36,670 J = -36.7 kJ

∆Go3 = - RT ln K = -(8.314 x 10-3 kJ/K)(298 K) ln (4.8 x 1012) = -72.3 kJ

∆Go4 = - RT ln K = -(8.314 x 10-3 kJ/K)(298 K) ln (1.0 x 10-56) = +319.5 kJ

∆Go(total) = -235.5 kJ - 36.7 kJ - 72.3 kJ + 319.5 kJ = -25.0 kJ = -25,000 J 1 V = 1 J/C

∆Go = -nFEo; Eo =

∆

e mol 1C 96,500

)e mol (1

J) 25,000 (_ _ =

nF G _

__

o

= +0.259 J/C = +0.259 V

Eocathode = +0.259 V


563

anode: (1) Zn(s) → Zn2+(aq) + 2 e- Eo = +0.76 V (2) Zn2+(aq) + 2 OH-(aq) → Zn(OH)2(s) K = 1/Ksp = 1/(4.1 x 10-17) = 2.4 x 1016 Zn(s) + 2 OH-(aq) → Zn(OH)2(s) + 2 e-

∆Go1 = -nFEo = -(2 mol e-)

e mol 1

C 96,500_

(0.76 V)

• V C 1

J 1 = -146,680 J = -146.7 kJ

∆Go2 = -RT ln K = -(8.314 x 10-3 kJ/K)(298 K) ln (2.4 x 1016) = -93.4 kJ

∆Go(total) = -146.7 kJ - 93.4 kJ = -240.1 kJ = -240,100 J 1 V = 1 J/C

∆Go = -nFEo; Eo =

∆

e mol 1C 96,500

)e mol (2

J) 240,100 (_ _ =

nF G _

__

o

= +1.24 J/C = +1.24 V

Eoanode = +1.24 V

Eocell = Eo

cathode + Eoanode = 0.259 V + 1.24 V = 1.50 V

(b) FeO4

2-(aq) → Fe(OH)3(s) FeO4

2-(aq) → Fe(OH)3(s) + H2O(l) FeO4

2-(aq) + 5 H+(aq) → Fe(OH)3(s) + H2O(l) FeO4

2-(aq) + 5 H+(aq) + 3 e- → Fe(OH)3(s) + H2O(l) FeO4

2-(aq) + 5 H+(aq) + 5 OH-(aq) + 3 e- → Fe(OH)3(s) + H2O(l) + 5 OH-(aq) FeO4

2-(aq) + 5 H2O(l) + 3 e- → Fe(OH)3(s) + H2O(l) + 5 OH-(aq) FeO4

2-(aq) + 4 H2O(l) + 3 e- → Fe(OH)3(s) + 5 OH-(aq)

(c) K2FeO4, 198.04 amu; MnO2, 86.94 amu

coulombs = 10.00 g K2FeO4 x xFeOK g 198.04

FeOK mol 1

42

42 xFeOK mol 1e mol 3

42

_

= e mol 1

C 96,500_

1.46 x 104 C from 10.00 g K2FeO4

coulombs = 10.00 g MnO2 x xMnO g 86.94

MnO mol 1

2

2 xMnO mol 1

e mol 1

2

_

= e mol 1

C 96,500_

1.11 x 104 C from 10.00 g MnO2 18.128 (a) 4 [Au(s) + 2 CN-(aq) → Au(CN)2

-(aq) + e-] oxidation half reaction

O2(g) → 2 H2O(l) O2(g) + 4 H+(aq) → 2 H2O(l) 4 e- + O2(g) + 4 H+(aq) → 2 H2O(l) reduction half reaction

Combine the two half reactions. 4 Au(s) + 8 CN-(aq) + O2(g) + 4 H+(aq) → 4 Au(CN)2

-(aq) + 2 H2O(l) 4 Au(s) + 8 CN-(aq) + O2(g) + 4 H+(aq) + 4 OH-(aq)

→ 4 Au(CN)2-(aq) + 2 H2O(l) + 4 OH-(aq)


564

4 Au(s) + 8 CN-(aq) + O2(g) + 4 H2O(l) → 4 Au(CN)2

-(aq) + 2 H2O(l) + 4 OH-(aq) 4 Au(s) + 8 CN-(aq) + O2(g) + 2 H2O(l) → 4 Au(CN)2

-(aq) + 4 OH-(aq) (b) Add the following five reactions together. ∆Go is calculated below each reaction. 4 [Au+(aq) + 2 CN-(aq) → Au(CN)2

-(aq)] K = (Kf)4

∆Go = -RT ln K = -(8.314 x 10-3 kJ/K)(298 K) ln (6.2 x 1038)4 = -885.2 kJ

O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) Eo = 1.229 V


C 96,500_

• V C 1

J 1 = - 474,394 J = - 474.4 kJ

4 [H2O(l) _ H+(aq) + OH-(aq)] K = (Kw)4 ∆Go = -RT ln K = -(8.314 x 10-3 kJ/K)(298 K) ln (1.0 x 10-14)4 = +319.5 kJ

4 [Au(s) → Au3+(aq) + 3 e-] Eo = -1.498 V

∆Go = -nFEo = -(12 mol e-) V) 1.498 (_e mol 1

C 96,500_

• V C 1

J 1 = +1,734,684 J = +1,734.7 kJ

4 [Au3+(aq) + 2 e- → Au+(aq)] Eo = 1.401 V

∆Go = - nFEo = -(8 mol e-) V) (1.401e mol 1

C 96,500_

• V C 1

J 1 = -1,081,572 J = -1,081.6 kJ

Overall reaction: 4 Au(s) + 8 CN-(aq) + O2(g) + 2 H2O(l) → 4 Au(CN)2

-(aq) + 4 OH-(aq) ∆Go = - 885.2 kJ - 474.4 kJ + 319.5 kJ + 1,734.7 kJ - 1,081.6 kJ = - 387.0 kJ

18.129 The overall cell reaction is: 2 Fe3+(aq) + 2 Hg(l) + 2 Cl-(aq) → 2 Fe2+(aq) + Hg2Cl2(s) The Nernst equation can be applied to separate half reactions. One half reaction is for the calomel reference electrode. 2 Hg(l) + 2 Cl-(aq) → Hg2Cl2(s) + 2 e- Eo = -0.28 V When [Cl-] = 2.9 M,

Ecalomel = Eo - ]Cl[

1 log

n

V 0.05922_ = -0.28 V -

)(2.9

1 log

2

V 0.05922 = -0.25 V

Balance the titration redox reaction: MnO4

-(aq) + Fe2+(aq) → Mn2+(aq) + Fe3+(aq) [Fe2+(aq) → Fe3+(aq) + e-] x 5

MnO4

-(aq) → Mn2+(aq) MnO4

-(aq) → Mn2+(aq) + 4 H2O(l) MnO4

-(aq) + 8 H+(aq) → Mn2+(aq) + 4 H2O(l) MnO4

-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l)

Combine the two half reactions. MnO4

-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)


565

initial mol Fe2+ = (0.010 mol/L)(0.1000 L) = 0.0010 mol Fe2+ mL MnO4

- needed to reach endpoint =

0.0010 mol Fe2+ x xFe mol 5

MnO mol 1+2

_4 x

MnO mol 0.010

L 1.00_4

= L 1.00

mL 100020.0 mL

(a) initial mol Fe2+ = (0.010 mol/L)(0.1000 L) = 0.0010 mol Fe2+ mol MnO4

- in 5.0 mL = (0.010 mol/L)(0.0050 L) = 0.000 050 mol MnO4-

MnO4

-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) before (mol) 0.000 050 0.0010 0 change (mol) -0.000 050 -5(0.000 050) (0.000 050) after (mol) 0 0.000 75 0.000 25 Again, the Nernst equation can be applied to separate half reactions. The other half reaction is: Fe3+(aq) + e- → 2 Fe2+(aq) Eo = +0.77 V E for the half reaction after adding 5.0 mL of MnO4

- is

E Fe/Fe +2+3 = Eo - ]Fe[

]Fe[ log

n

V 0.0592+3

+2

= 0.77 V - 25) (0.000

75) (0.000 log

1

V 0.0592= 0.74 V

(Note in the Nernst equation above, we are taking a ratio of Fe2+ to Fe3+ so we can ignore volumes and just use moles instead of molarity.) Ecell = E Fe/Fe +2+3 + Ecalomel = 0.74 V + (-0.25 V) = 0.49 V

(b) initial mol Fe2+ = (0.010 mol/L)(0.1000 L) = 0.0010 mol Fe2+ mol MnO4


MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)

before (mol) 0.000 10 0.0010 0 change (mol) -0.000 10 -5(0.000 10) +5(0.000 10) after (mol) 0 0.000 50 0.000 50 E for the half reaction after adding 10.0 mL of MnO4

- is

E Fe/Fe +2+3 = Eo - ]Fe[

]Fe[ log

n

V 0.0592+3

+2

= 0.77 V - 50) (0.000

50) (0.000 log

1

V 0.0592= 0.77 V

Ecell = E Fe/Fe +2+3 + Ecalomel = 0.77 V + (-0.25 V) = 0.52 V

(c) initial mol Fe2+ = (0.010 mol/L)(0.1000 L) = 0.0010 mol Fe2+ mol MnO4


MnO4

-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) before (mol) 0.000 19 0.0010 0 change (mol) -0.000 19 -5(0.000 19) +5(0.000 19) after (mol) 0 0.000 05 0.000 95 E for the half reaction after adding 19.0 mL of MnO4

- is

E Fe/Fe +2+3 = Eo - ]Fe[

]Fe[ log

n

V 0.0592+3

+2

= 0.77 V - 95) (0.000

05) (0.000 log

1

V 0.0592= 0.85 V

Ecell = E Fe/Fe +2+3 + Ecalomel = 0.85 V + (-0.25 V) = 0.60 V


566

(d) 21.0 mL is past the endpoint so the MnO4

- is in excess and all of the Fe2+ is consumed. initial mol Fe2+ = (0.010 mol/L)(0.1000 L) = 0.0010 mol Fe2+ mol MnO4


MnO4

-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) before (mol) 0.000 21 0.0010 0 0 change (mol) -0.000 20 -5(0.000 20) +5(0.000 20) after (mol) 0.000 01 0 0.000 20 0.0010

Because the Fe2+ is totally consumed, there is a new half reaction: MnO4

-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O(l) Eo = 1.51 V The total volume = 100.0 mL + 21.0 mL = 121.0 mL = 0.1210 L [MnO4

-] = 0.000 01 mol/0.1210 L = 0.000 083 M [Mn2+] = 0.000 20 mol/0.1210 L = 0.001 65 M We need to determine [H+] in order to determine the half reaction potential. [H2SO4]dil ⋅ 121.0 mL = [H2SO4]conc ⋅ 100.0 mL [H2SO4]dil = [(1.50 M)(100.0 mL)]/121.0 mL = 1.24 M We can ignore the small amount of H+ consumed by the titration itself, because the H2SO4 concentration is so large. Consider the dissociation of H2SO4. From the complete dissociation of the first proton, [H+] = [HSO4

-] = 1.24 M. For the dissociation of the second proton, the following equilibrium must be considered:

HSO4-(aq) _ H+(aq) + SO4

2-(aq) initial (M) 1.24 1.24 0 change (M) -x +x +x equil (M) 1.24 - x 1.24 + x x

Ka2 = x_ 1.24

x)(x)+ (1.24 = 10 x 1.2 =

] HSO[

] SO][H[ 2__4

_24

+


x = 2

1.276 1.252 _ =

2(1)

0.0149) 4(1)(_ _ )(1.252 (1.252) _ 2 ±±

x = -1.264 and 0.012 Of the two solutions for x, only the positive value of x has physical meaning, since x is the [SO4

2-]. [H+] = 1.24 + x = 1.24 + 0.012 = 1.25 M

E Mn/MnO +2_4

= Eo - ]H][ MnO[

]Mn[ log

n

V 0.05928+_

4

+2

= 1.51 V - )083)(1.25 (0.000

65) (0.001 log

5

V 0.05928 = 1.50 V

Ecell = E Mn/MnO +2_4

+ Ecalomel = 1.50 V + (-0.25 V) = 1.25 V


567

Notice that there is a dramatic change in the potential at the equivalence point.

657

22

Nuclear Chemistry 22.1 (a) In beta emission, the mass number is unchanged, and the atomic number increases

by one. Rh + e _Ru 10645

0 1_

10644

(b) In alpha emission, the mass number decreases by four, and the atomic number decreases by two. Tl + He _ Bi 185

81 42

18983

(c) In electron capture, the mass number is unchanged, and the atomic number decreases by one. Bi _ e + Po 204

83 0 1_

20484

22.2 The mass number decreases by four, and the atomic number decreases by two. This is

characteristic of alpha emission. He + Ra _Th 42

21088

21490

22.3 t1/2 = h 10 x 1.08

0.693 =

k

0.6931_2_ = 64.2 h

22.4 k = y 10 x 1.21 = y 5730

0.693 =

t

0.693 1_4-

2/1

22.5

y 5730

y 16,2300.693_ =

t

t0.693_ =

N

Nln

2/10

= -1.963

N

N

0

= e-1.963 = 0.140; 100%

N = 0.140; N = 14.0%

22.6 ln

t

t0.693)(_ =

N

N

2/10

; 0 =at t rateDecay

tat time rateDecay =

N

N

0

ln

t

d 28.00.693)(_ =

16,800

10,860

2/1

; -0.436 = (-0.693)

t

d 28.0

2/1

0.436)(_

d) 00.693)(28.(_ = t 2/1 = 44.5 d

Chapter 22 - Nuclear Chemistry ______________________________________________________________________________

658

22.7 ln

t

t0.693)(_ =

N

N

2/10

ln

t

d 100.693)(_ =

100

10

2/1

; -2.303 = (-0.693)

t

d 10

2/1

2.303)(_

d) 0.693)(10(_ = t 2/1 = 3.0 d

22.8 Ni + e _Cu 59

2801

5929

16 59Cu → 8 59Cu → 4 59Cu → 2 59Cu; three half-lives have passed. 22.9 199Au has a higher neutron/proton ratio and decays by beta emission. 173Au has a

lower neutron/proton ratio and decays by alpha emission. 22.10 The shorter arrow pointing right is for beta emission. The longer arrow pointing left is

for alpha emission. A = X 97 +153

97 = Bk 25097

B = X 98 +15298 = Cf 250

98

C = X 96 +15096 = Cm 246

96

D = X 94 +14894 = Pu 242

94

E = X 92 +14692 = U 238

92 22.11 For O 16

8 : First, calculate the total mass of the nucleons (8 n + 8 p) Mass of 8 neutrons = (8)(1.008 66 amu) = 8.069 28 amu Mass of 8 protons = (8)(1.007 28 amu) = 8.058 24 amu Mass of 8 n + 8 p = 16.127 52 amu

Next, calculate the mass of a 16O nucleus by subtracting the mass of 8 electrons from the mass of a 16O atom. Mass of 16O atom = 15.994 92 amu -Mass of 8 electrons = -(8)(5.486 x 10-4 amu) = -0.004 39 amu Mass of 16O nucleus = 15.990 53 amu Then subtract the mass of the 16O nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (16.127 52 amu) - (15.990 53 amu) = 0.136 99 amu Mass defect in g/mol: (0.136 99 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.136 99 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.136 99 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.233 x 1013 J/mol = 1.233 x 1010 kJ/mol


659

∆E = nucleons 16

nucleus 1 x

J 10 x 1.60

MeV 1 x

nuclei/mol 10 x 6.022

J/mol 10 x 1.23313_23

13

= 8.00 nucleon

MeV

22.12 ∆E = -852 kJ/mol = -852 x 103 J/mol; 1 J = 1 kg ⋅ m2/s2

∆E = ∆mc2; ∆m = ( )m/s 10 x 3.00

mol sm kg

10 x 852_

= c

E8 2

2

23

2

••

∆ = -9.47 x 10-12 kg/mol

∆m = -9.47 x 10-12 kg 1

g 1000 x

mol

kg = -9.47 x 10-9 g/mol

22.13 n 2 +Zr + Te _ U +n 1

09740

13752

23592

10

mass U 23592 235.0439 amu

mass n 10 1.008 66 amu

-mass Te 13752 -136.9254 amu

-mass Zr 9740 -96.9110 amu

-mass n 2 10 -(2)(1.008 66) amu

mass change 0.1988 amu

(0.1988 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.1988 g/mol ∆E = ∆mc2 = (0.1988 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.79 x 1013 J/mol = 1.79 x 1010 kJ/mol

22.14 He _ H + H 3

221

11

mass 1H 1.007 83 amu mass 2H 2.014 10 amu -mass 3He -3.016 03 amu mass change 0.005 90 amu

(0.005 90 amu)(1.660 54 x 10-24 g/amu)(6.022 x1023 mol-1) = 0.005 90 g/mol ∆E = ∆mc2 = (0.005 90 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 5.31 x 1011 J/mol = 5.31 x 108 kJ/mol

22.15 n +K _ p +Ar 1

04019

11

4018

22.16 n 2 + Np _ H + U 1

023893

21

23892

22.17 ln

t

t0.693)(_ =

N

N

2/10

; 0 = t at time rateDecay


N

N

0


660

ln

y 5730

t0.693)(_ =

15.3

2.4; t = 1.53 x 104 y

22.18 Elements heavier than iron arise from nuclear reactions occuring as a result of

supernova explosions. Understanding Key Concepts 22.19 16 40K → 8 40K → 4 40K; two half-lives have passed. 22.20 The isotope contains 8 neutrons and 6 protons. The isotope symbol is C 14

6 .

C 146 would decay by beta emission because the n/p ratio is high.

22.21 Tm 148

69 decays to Er 14868 by either positron emission or electron capture.

22.22 The shorter arrow pointing right is for beta emission. The longer arrow pointing left is

for alpha emission. A = X 94 +147

94 = Pu 24194

B = X 95 +14695 = Am 241

95

C = X 93 +14493 = Np 237

93

D = X 91 +14291 = Pa 233

91

E = X 92 +14192 = U 233

92 22.23 The half-life is approximately 3 years. Additional Problems Nuclear Reactions and Radioactivity 22.24 Positron emission is the conversion of a proton in the nucleus into a neutron plus an

ejected positron. Electron capture is the process in which a proton in the nucleus captures an inner-shell electron and is thereby converted into a neutron.

22.25 An alpha particle ( )He +24

2 is a helium nucleus. The He atom has two electrons and is neutral.

22.26 Alpha particles move relatively slowly and can be stopped by the skin. However,

inside the body, alpha particles give up their energy to the immediately surrounding tissue. Gamma rays move at the speed of light and are very penetrating. Therefore they are equally hazardous internally and externally.

22.27 "Neutron rich" nuclides emit beta particles to decrease the number of neutrons and


661

increase the number of protons in the nucleus. "Neutron poor" nuclides decrease the number of protons and increase the n/p ratio by either alpha emission, positron emission, or electron capture.

22.28 There is no radioactive "neutralization" reaction like there is an acid-base

neutralization reaction. 22.29 The nuclei of 24Na and 24Na+ are identical so their nuclear reactions must be the same

even though their chemical reactions are completely different. 22.30 (a) Sb + e _Sn 126

51 0 1_

12650 (b) Rn + He _ Ra 206

86 42

21088

(c) Kr + e _ Rb 7736

01

7737 (d) Br _ e +Kr 76

350 1_

7636

22.31 (a) Y + e _Sr 90

390 1_

9038 (b) Cf + He _ Fm 243

98 42

247100

(c) Cr + e _Mn 4924

01

4925 (d) Cl _ e +Ar 37

170 1_

3718

22.32 (a) e +Au _ Hg 0

118879

18880 (b) He + Bi _At 4

221483

21885

(c) e + Pa _Th 0 1_

23491

23490

22.33 (a) e + Mg _ Na 0

1_2412

2411 (b) e +Pr _ Nd 0

113559

13560

(c) He + Os _Pt 42

16676

17078

22.34 (a) He + Ta _ Re 4

215873

16275 (b) Pm _ e + Sm 138

61 0 1_

13862

(c) e + Re _W 0 1_

18875

18874 (d) e + Hf _ Ta 0

116572

16573

22.35 (a) e + Gd _Eu 0

1_15764

15763 (b) Cs _ e + Ba 126

55 0 1_

12656

(c) He + Nd _ Sm 42

14260

14662 (d) e + Cs _ Ba 0

112555

12556

22.36 160W is neutron poor and decays by alpha emission. 185W is neutron rich and decays

by beta emission. 22.37 I 136

53 is neutron rich and decays by beta emission.

I 12253 is neutron poor and decays by positron emission.

22.38 He + Np _ Am 4

223793

24195

He + Pa _ Np 42

23391

23793

e + U _ Pa 0 1_

23392

23391

He +Th _ U 42

22990

23392

He + Ra _Th 42

22588

22990

e + Ac _ Ra 0 1_

22589

22588

He +Fr _ Ac 42

22187

22589


662

He +At _Fr 42

21785

22187

He + Bi _At 42

21383

21785

e + Po _ Bi 0 1_

21384

21383

He + Pb _ Po 42

20982

21384

e + Bi _ Pb 0 1_

20983

20982

22.39 He + Po _Rn 4

221884

22286

He + Pb _ Po 42

21482

21884

He + Hg _ Pb 42

21080

21482

e + Tl _ Hg 0 1_

21081

21080

e + Pb _ Tl 0 1_

21082

21081

22.40 Each alpha emission decreases the mass number by four and the atomic number by

two. Each beta emission increases the atomic number by one. Pb _Th 208

82 23290

Number of α emissions = 4

number mass Pb _number massTh

= 4

208 _ 232 = 6 α emissions

The atomic number decreases by 12 as a result of 6 alpha emissions. The resulting atomic number is (90 - 12) = 78. Number of β emissions = Pb atomic number - 78 = 82 - 78 = 4 β emissions

22.41 Each alpha emission decreases the mass number by four and the atomic number by

two. Each beta emission increases the atomic number by one. Pb _ U 207

82 23592


number mass Pb _number mass U

= 4


The atomic number decreases by 14 as a result of 7 alpha emissions. The resulting atomic number is (92 - 14) = 78. Number of β emissions = Pb atomic number - 78 = 82 - 78 = 4 β emissions

Radioactive Decay Rates 22.42 If the half-life of 59Fe is 44.5 d, it takes 44.5 days for half of the original amount of

59Fe to decay. 22.43 The half-life is the time it takes for one-half of a radioactive sample to decay.


663

The decay constant is the rate constant for the first order radioactive decay.

k = t

0.693

2/1

22.44 k = d 2.805

0.693 =

t

0.693

2/1

= 0.247 d-1

22.45 k = h 78.25

0.693 =

t

0.693

2/1

= 8.86 x 10-3 h-1

22.46 t 2/1 = d 0.228

0.693 =

k

0.6931_ = 3.04 d

22.47 y 10 x 2.88

0.693 =

k

0.693 = t 1_5_2/1 = 2.41 x 104 y

22.48 After 65 d:

y 432.2

d/y 365d 65

0.693_ = t

t0.693_ =

N

Nln

2/10

= -0.000 285 5

N

N

0

= e-0.0002855 = 0.9997; 100%

N = 0.9997; N = 99.97%

After 65 y:

y 432.2

y 650.693_ =

t

t0.693_ =

N

Nln

2/10

= -0.1042

N

N

0

= e-0.1042 = 0.9010; 100%

N = 0.9010; N = 90.10%

After 650 y:

y 432.2

y 6500.693_ =

t

t0.693_ =

N

Nln

2/10

= -1.042

N

N

0

= e-1.042 = 0.3527; 100%

N = 0.3527; N = 35.27%

22.49 After 24 min: ln

min 109.8

min 240.693)(_ =

t

t0.693)(_ =

N

N

2/10

= -0.1515

N

N

0

= e-0.1515 = 0.8594; 100%

N = 0.8594; N = 85.94%


664

After 24 h: ln

min 109.8h 1min 60

h x 240.693)(_ =

t

t0.693)(_ =

N

N

2/10

= -9.089

N

N

0

= e-9.089 = 0.000 113 0; 100%

N = 0.000 113 0; N = 0.011 30%

After 24 d: ln

min 109.8h 1min 60

x d 1h 24

x d 240.693)(_ =

t

t0.693)(_ =

N

N

2/10

= -218.1

N

N

0

= e-218.1 = 1.861 x 10-95; 100%

N = 1.861 x 10-95; N = 1.861 x 10-93%

22.50

t

t0.693)(_ =

N

Nln

2/10

;

y 5730

t0.693)(_ = )0.43ln( ; t = 6980 y

22.51 Assume a sample of K 40

19 containing 100 atoms.

Ar _ e +K 4018

0 1_

4019

before decay (atoms) 100 0 after decay (atoms) 100 - x x

x_ 100

x =

K

Ar 40

40

= 1.15; Solve for x. x = 53.5

ln

t

t0.693)(_ =

N

N

2/10

N = 100 - x = 100 - 53.5 = 46.5, the amount of 40K at time t. N0 = 100, the original amount of 40K.

ln

y 10 x 1.28

t0.693) (_ =

100

46.59

; t = 1.41 x 109 y

22.52 t 2/1 = d 10 x 7.89

0.693 =

k

0.6931_3_ = 87.83 d

t

t0.693)(_ =

N

Nln

2/10

d 87.83

d 1850.693)(_ = = -1.4597

N

N

0

= e-1.4597 = 0.2323; 100%

N = 0.2323; N = 23.2%

22.53 y 10 x 2.88

0.693 =

k

0.693 = t 1_5_2/1 = 2.41 x 104 y


665

After 1000 y: ln

y 10 x 2.41

y 10000.693)(_ =

t

t0.693)(_ =

N

N4

2/10

= -0.028 76

N

N

0

= e-0.02876 = 0.9717; 100%

N = 0.9717; N = 97.17%

After 25,000 y: ln

y 10 x 2.41

y 25,0000.693)(_ =

t

t0.693)(_ =

N

N4

2/10

= -0.7189

N

N

0

= e-0.7189 = 0.4873; 100%

N = 0.4873; N = 48.73%

After 100,000 y: ln

y 10 x 2.41

y 100,0000.693)(_ =

t

t0.693)(_ =

N

N4

2/10

= -2.876

N

N

0

= e-2.876 = 0.0564; 100%

N = 0.0564; N = 5.64%

22.54 t1/2 = (102 y)(365 d/y)(24 h/d)(3600 s/h) = 3.2167 x 109 s

k = s 10 x 3.2167

0.693 =

t

0.6939

2/1

= 2.1544 x 10-10 s-1

N = (1.0 x 10-9 g)

Po g 209

Po mol 1(6.022 x 1023 atoms/mol) = 2.881 x 1012 atoms

Decay rate = kN = (2.1544 x 10-10 s-1)(2.881 x 1012 atoms) = 6.21 x 102 s-1 621 α particles are emitted in 1.0 s.

22.55 t1/2 = (3.0 x 105 y)(365 d/y)(24 h/d)(60 min/h) = 1.6 x 1011 min

k = min 10 x 1.6

0.693 =

t

0.69311

2/1

= 4.3 x 10-12 min-1

N = (5.0 x 10-3 g)

g 36

Cl mol 1 36

(6.022 x 1023 atoms/mol) = 8.4 x 1019 atoms

Decay rate = kN = (4.3 x 10-12 min-1)(8.4 x 1019 atoms) = 3.6 x 108 min-1

Curies = (3.6 x 108/min)

/s10 x 3.7

Ci 1

s 60

min 110

= 1.6 x 10-4 Ci

22.56 Decay rate = kN

N = (1.0 x 10-3 g)

g 79

Se mol 1 79

(6.022 x 1023 atoms/mol) = 7.6 x 1018 atoms

k = 10 x 7.6

/s10 x 1.5 =

N

rateDecay 18

5

= 2.0 x 10-14 s-1

s 10 x 2.0

0.693 =

k

0.693 = t 1_14_2/1 = 3.5 x 1013 s


666

d 365

y 1

h 24

d 1

s 3600

h 1s) 10 x (3.5 = t 13

2/1 = 1.1 x 106 y


N = (1.0 x 10-9 g)

Ti g 44

Ti mol 1(6.022 x 1023 atoms/mol) = 1.37 x 1013 atoms

k = 10 x 1.37

s 10 x 4.8 =

N

rateDecay 13

1_3

= 3.50 x 10-10 s-1

k = (3.50 x 10-10 s-1)(3600 s/h)(24 h/d)(365 d/y) = 1.10 x 10-2 y-1

t1/2 = y 10 x 1.10

0.693 =

k

0.6931_2_ = 63 y

22.58 ln

t

t0.693)(_ =

N

N

2/10



N

N

0

ln

t

d 10.00.693)(_ =

8540

6990

2/1

; t 2/1 = 34.6 d

22.59

t

t0.693)(_ =

N

Nln

2/10



N

N

0

t

h 48.00.693)(_ =

53,500

10,980ln

2/1

; t1/2 = 21.0 h

Energy Changes During Nuclear Reactions 22.60 The loss in mass that occurs when protons and neutrons combine to form a nucleus is

called the mass defect. The lost mass is converted into the binding energy that is used to hold the nucleons together.

22.61 Energy (heat) is absorbed in an endothermic reaction. The energy is converted to

mass. The mass of the products is slightly larger than the mass of the reactants.

22.62 E = (1.50 MeV)

MeV 1

J 10 x 1.60 13_

= 2.40 x 10-13 J

λ = J 10 x 2.40

m/s) 10 x s)(3.00Jcdot 10 x (6.626 =

E

hc13_

834_

= 8.28 x 10-13 m = 0.000 828 nm

22.63 E = (6.82 keV)

MeV 1

J10 x 1.60

keV 10

MeV 1 13_

3 = 1.09 x 10-15 J


667

s J10 x 6.626

J10 x 1.09 =

h

E =

34_

15_

•ν = 1.65 x 1018/s = 1.65 x 1018 Hz

22.64 (a) For Fe 52

26 : First, calculate the total mass of the nucleons (26 n + 26 p) Mass of 26 neutrons = (26)(1.008 66 amu) = 26.225 16 amu Mass of 26 protons = (26)(1.007 28 amu) = 26.189 28 amu Mass of 26 n + 26 p = 52.414 44 amu Next, calculate the mass of a 52Fe nucleus by subtracting the mass of 26 electrons from the mass of a 52Fe atom. Mass of 52Fe atom = 51.948 11 amu -Mass of 26 electrons = -(26)(5.486 x 10-4 amu) = -0.014 26 amu Mass of 52Fe nucleus = 51.933 85 amu Then subtract the mass of the 52Fe nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (52.414 44 amu) - (51.933 85 amu) = 0.480 59 amu Mass defect in g/mol: (0.480 59 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.480 59 g/mol

(b) For Mo 92

42 : First, calculate the total mass of the nucleons (50 n + 42 p) Mass of 50 neutrons = (50)(1.008 66 amu) = 50.433 00 amu Mass of 42 protons = (42)(1.007 28 amu) = 42.305 76 amu Mass of 50 n + 42 p = 92.738 76 amu Next, calculate the mass of a 92Mo nucleus by subtracting the mass of 42 electrons from the mass of a 92Mo atom. Mass of 92Mo atom = 91.906 81 amu -Mass of 42 electrons = -(42)(5.486 x 10-4 amu) = -0.023 04 amu Mass of 92Mo nucleus = 91.883 77 amu Then subtract the mass of the 92Mo nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (92.738 76 amu) - (91.883 77 amu) = 0.854 99 amu Mass defect in g/mol: (0.854 99 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.854 99 g/mol

22.65 (a) For S 32

16 : First, calculate the total mass of the nucleons (16 n + 16 p) Mass of 16 neutrons = (16)(1.008 66 amu) = 16.138 56 amu Mass of 16 protons = (16)(1.007 28 amu) = 16.116 48 amu Mass of 16 n + 16 p = 32.255 04 amu Next, calculate the mass of a 32S nucleus by subtracting the mass of 16 electrons from the mass of a 32S atom.


668

Mass of 32S = 31.972 07 amu Mass of 16 electrons = -(16)(5.486 x 10-4 amu) = -0.008 78 amu Mass of 32S nucleus = 31.963 29 amu Then subtract the mass of the 32S nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (32.255 04 amu) - (31.963 29 amu) = 0.291 75 amu Mass defect in g/mol: (0.291 75 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.291 74 g/mol

(b) For Ca 4020 :

First, calculate the total mass of the nucleons (20 n + 20 p) Mass of 20 neutrons = (20)(1.008 66 amu) = 20.173 20 amu Mass of 20 protons = (20)(1.007 28 amu) = 20.145 60 amu Mass of 20 n + 20 p = 40.318 80 amu Next, calculate the mass of a 40Ca nucleus by subtracting the mass of 20 electrons from the mass of a 40Ca atom. Mass of 40Ca = 39.962 59 amu -Mass of 20 electrons = -(20)(5.486 x 10-4 amu) = -0.010 97 amu Mass of 40Ca nucleus = 39.951 62 amu Then substract the mass of the 40Ca nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus

= (40.318 80 amu) - (39.951 62 amu) = 0.367 18 amu Mass defect in g/mol: (0.367 18 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.367 17 g/mol

22.66 (a) For Ni 58

28 : First, calculate the total mass of the nucleons (30 n + 28 p) Mass of 30 neutrons = (30)(1.008 66 amu) = 30.259 80 amu Mass of 28 protons = (28)(1.007 28 amu) = 28.203 84 amu Mass of 30 n + 28 p = 58.463 64 amu Next, calculate the mass of a 58Ni nucleus by subtracting the mass of 28 electrons from the mass of a 58Ni atom. Mass of 58Ni atom = 57.935 35 amu -Mass of 28 electrons = -(28)(5.486 x 10-4 amu) = -0.015 36 amu Mass of 58Ni nucleus = 57.919 99 amu Then subtract the mass of the 58Ni nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (58.463 64 amu) - (57.919 99 amu) = 0.543 65 amu


669

Mass defect in g/mol: (0.543 65 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.543 65 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.543 65 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 4.893 x 1013 J/mol = 4.893 x 1010 kJ/mol

∆E = nucleons 58

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 4.89313_23

13

= 8.76 nucleon

MeV

(b) For Kr 84

36 : First, calculate the total mass of the nucleons (48 n + 36 p) Mass of 48 neutrons = (48)(1.008 66 amu) = 48.415 68 amu Mass of 36 protons = (36)(1.007 28 amu) = 36.262 08 amu Mass of 48 n + 36 p = 84.677 76 amu Next, calculate the mass of a 84Kr nucleus by subtracting the mass of 36 electrons from the mass of a 84Kr atom. Mass of 84Kr atom = 83.911 51 amu -Mass of 36 electrons = -(36)(5.486 x 10-4 amu) = -0.019 75 amu Mass of 84Kr nucleus = 83.891 76 amu Then subtract the mass of the 84Kr nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (84.677 76 amu) - (83.891 76 amu) = 0.786 00 amu Mass defect in g/mol: (0.786 00 amu)(1.660 54 x 10-24 g/mol)(6.022 x 1023 mol-1) = 0.786 00 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.786 00 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 7.074 x 1013 J/mol = 7.074 x 1010 kJ/mol

∆E = nucleons 84

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 7.07413_23

13

= 8.74 nucleon

MeV

22.67 (a) For Cu 63

29 : First, calculate the total mass of the nucleons (34 n + 29 p) Mass of 34 neutrons = (34)(1.008 66 amu) = 34.294 44 amu Mass of 29 protons = (29)(1.007 28 amu) = 29.211 12 amu Mass of 34 n + 29 p = 63.505 56 amu Next calculate the mass of a 63Cu nucleus by subtracting the mass of 29 electrons from the mass of a 63Cu atom. Mass of 63Cu atom = 62.939 60 amu -Mass of 29 electrons = -(29)(5.486 x 10-4amu) = -0.015 91 amu Mass of 63Cu nucleus = 62.923 69 amu Then subtract the mass of the 63Cu nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus

= (63.505 56 amu) - (62.923 69 amu) = 0.581 87 amu


670

Mass defect in g/mol: (0.581 87 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.581 86 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.581 86 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 5.237 x 1013 J/mol = 5.237 x 1010 kJ/mol

∆E = nucleon

MeV 8.63 =

nucleons 63

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 5.23713_23

13

(b) For Sr 84

38 : First, calculate the total mass of the nucleons (46 n + 38 p) Mass of 46 neutrons = (46)(1.008 66 amu) = 46.398 36 amu Mass of 38 protons = (38)(1.007 28 amu) = 38.276 64 amu Mass of 46 n + 38 p = 84.675 00 amu Next, calculate the mass of a 84Sr nucleus by subtracting the mass of 38 electrons from the mass of a 84Sr atom. Mass of 84Sr atom = 83.913 43 amu -Mass of 38 electrons = -(38)(5.486 x 10-4 amu) = -0.020 85 amu Mass of 84Sr nucleus = 83.892 58 amu Then subtract the mass of the 84Sr nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus

= (84.675 00 amu) - (83.892 58 amu) = 0.782 42 amu Mass defect in g/mol: (0.782 42 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.782 40 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.782 40 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 7.042 x 1013 J/mol = 7.042 x 1010 kJ/mol

∆E = nucleon

MeV 8.70 =

nucleons 84

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 7.04213_23

13

22.68 He + Re _Ir 4

217075

17477

mass Ir 17477 173.966 66 amu

-mass Re 17075 -169.958 04 amu

-mass He 42 - 4.002 60 amu

mass change 0.006 02 amu

(0.006 02 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.006 02 g/mol ∆E = ∆mc2 = (0.006 02 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 5.42 x 1011 J/mol = 5.42 x 108 kJ/mol

22.69 e + Al _ Mg 0

1_2813

2812


671

Reactant: Mg 2812 nucleus = Mg 28

12 atom - 12 e-

Product: Al 2813 nucleus + e- = ( Al 28

13 nucleus - 13 e-) + e- = Al 2813 nucleus - 12 e-

Change : ( Mg 28

12 atom - 12 e-) - ( Al 2813 nucleus - 12 e-) = Mg 28

12 atom - Al 2813 atom

(electrons cancel) Mass change = 27.983 88 amu - 27.981 91 amu = 0.001 97 amu


22.70 ∆m = )m/s 10 x (3.00

s/m kg 10 x 92.2 =

)m/s 10 x (3.00

J 10 x 92.2 =

c

E28

223

28

3

2

•∆ = 1.02 x 10-12 kg

∆m = 1.02 x 10-9 g

22.71 ∆m = )m/s 10 x (3.00s/m kg 10 x 131

= )m/s 10 x (3.00

J 10 x 131 =

c

E28

223

28

3

2

•∆ = 1.46 x 10-12 kg

∆m = 1.46 x 10-9 g 22.72 Mass of positron and electron

= 2(9.109 x 10-31 kg)(6.022 x 1023 mol-1) = 1.097 x 10-6 kg/mol ∆E = ∆mc2 = (1.097 x 10-6 kg/mol)(3.00 x 108 m/s)2 ∆E = 9.87 x 1010 J/mol = 9.87 x 107 kJ/mol

22.73 n + He _ H 2 1

032

21

mass H 2 21 2(2.0141) amu

-mass He 32 -3.0160 amu

-mass n 10 -1.008 66 amu mass change 0.003 54 amu

(0.003 54 amu)(1.660 54 x 10-24 g/mol)(6.022 x 1023 mol-1) = 0.003 54 g/mol ∆E = ∆mc2 = (0.003 54 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 3.2 x 1011 J/mol = 3.2 x 108 kJ/mol

Nuclear Transmutation 22.74 (a) In _ He + Ag 113

49 42

10947 (b) n + N _ He + B 1

0137

42

105


672

22.75 (a) n 3 +Zn + Sm _ U 10

7230

16062

23592

(b) n 2 + La +Br _ U 1

014657

8735

23592

22.76 n +Mt _ Fe + Bi 1

0266109

5826

20983

22.77 Mo _n + Mo 99

4210

9842

22.78 n 4 + Cf _ C + U 1

024698

126

23892

22.79 (a) n 4 + No _ C + Cm 1

0254102

126

24696

(b) n + Md _ He + Es 1

0256101

42

25399

(c) n 4 +Lr _ B + Cf 1

0257103

115

25098

General Problems 22.80 e 4 + He 6 + Pb _Th 0

1_42

20882

23290

Reactant: Th 23290 nucleus = Th 232

90 atom - 90 e-

Product: Pb 20882 nucleus + (6) nucleus) He ( 4

2 + 4 e-

= ( Pb 20882 atom - 82 e-) + (6)( He 4

2 atom - 2 e-) + 4 e-

= Pb 20882 atom + (6)( He 4

2 atom) - 90 e- Change: ( Th 232

90 atom - 90 e-) - [ Pb 20882 atom + (6)( He 4

2 atom) - 90 e-]

= Th 23290 atom - [ Pb 208

82 atom + (6)( He 42 atom)] (electrons cancel)

Mass change = 232.038 054 amu - [207.976 627 amu + (6)(4.002 603 amu)] = 0.045 809 amu



k = y 10 x 1.28

0.693 =

t

0.6939

2/1

= 5.41 x 10-10 y-1

KCl, 74.55 amu N = number of 40K+ ions in a 1.00 g sample of KCl

N = (0.000 117)(1.00 g)

KCl mol 1K mol 1

g 74.55

KCl mol 1 +

(6.022 x 1023 mol-1)


673

N = 9.45 x 1017 40K+ ions Decay rate = kN = (5.41 x 10-10 y-1)(9.45 x 1017) = 5.11 x 108/y

Disintegration/s = (5.11 x 108/y)

s 3600

h 1

h 24

d 1

d 365

y 1 = 16.2/s

22.82 e 2 + He 2 + U _Pu 0

1_42

23392

24194

Reactant: Pu 24194 nucleus = Pu 241

94 atom - 94 e-

Product: U 23392 nucleus + (2)( He 4

2 nucleus) + 2 e-

= ( U 23392 atom - 92 e-) + (2)( He 4

2 atom - 2 e-) + 2 e-

= U 23392 atom + (2)( He 4

2 atom) - 94 e- Change: ( Pu 241

94 atom - 94 e-) - [ U 23392 atom + (2)( He 4

2 atom) - 94 e-]

= Pu 24194 atom - [ U 233

92 atom + (2)( He 42 atom)]

Mass change = 241.056 845 amu - [233.039 628 amu + (2)(4.002 603 amu)]

= 0.012 011 amu (0.012 011 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.012 011 g/mol ∆E = ∆mc2 = (0.012 011 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.08 x 1012 J/mol = 1.08 x 109 kJ/mol

22.83 X 293

118 , Y 289116 , and Z 285

114

22.84

t

t0.693)(_ =

N

Nln

2/10



N

N

0

( )s 1.53tover0.693)(_ = 100

99.99 _ 100ln

; t = 20.3 s

22.85 s 0.063

0.693 =

k

0.693 = t 1_2/1 = 11 s

ln

t

t 0.693)(_ =

N

N

2/10



N

N

0

ln

s 11

t0.693)(_ =

100

99.99 _ 100; t = 150 s

22.86 (a) For Cr 50

24 : First, calculate the total mass of the nucleons (26 n + 24 p) Mass of 26 neutrons = (26)(1.008 66 amu) = 26.225 16 amu Mass of 24 protons = (24)(1.007 28 amu) = 24.174 72 amu Mass of 26 n + 24 p = 50.399 88 amu Next, calculate the mass of a 50Cr nucleus by subtracting the mass of 24 electrons from


674

the mass of a 50Cr atom. Mass of 50Cr atom = 49.946 05 amu -Mass of 24 electrons = -(24)(5.486 x 10-4 amu) = -0.013 17 amu Mass of 50Cr nucleus = 49.932 88 amu Then subtract the mass of the 50Cr nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (50.399 88 amu) - (49.932 88 amu) = 0.467 00 amu Mass defect in g/mol: (0.467 00 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.467 00 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.467 00 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 4.203 x 1013 J/mol = 4.203 x 1010 kJ/mol

∆E = nucleons 50

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 4.20313_23

13

= 8.72 nucleon

MeV

(b) For Zn 6430 :

First, calculate the total mass of the nucleons (34 n + 30 p) Mass of 34 neutrons = (34)(1.008 66 amu) = 34.294 44 amu Mass of 30 protons = (30)(1.007 28 amu) = 30.218 40 amu Mass of 34 n + 30 p = 64.512 84 amu Next, calculate the mass of a 64Zn nucleus by subtracting the mass of 30 electrons from the mass of a 64Zn atom. Mass of 64Zn atom = 63.929 15 amu -Mass of 30 electrons = -(30)(5.486 x 10-4 amu) = -0.016 46 amu Mass of 64Zn nucleus = 63.912 69 amu Then subtract the mass of the 64Zn nucleus from the mass of the nucleons to find the mass defect: Mass defect = mass of nucleons - mass of nucleus = (64.512 84 amu) - (63.912 69 amu) = 0.600 15 amu Mass defect in g/mol: (0.600 15 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.600 15 g/mol Now, use the Einstein equation to convert the mass defect into the binding energy. ∆E = ∆mc2 = (0.600 15 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 5.401 x 1013 J/mol = 5.401 x 1010 kJ/mol

∆E = nucleons 64

nucleus 1 x

J 10 x 1.60

MeV 1 x


J/mol 10 x 5.40113_23

13

= 8.76 nucleon

MeV

The 64Zn is more stable.


675

22.87 ln

t

t0.693)(_ =

N

N

2/10



N

N

0

ln

y 5730

t0.693)(_ =

15.3

2.9; t = 1.38 x 104 y

22.88 H + He _ He + H 1

142

32

21

mass H 21 2.0141 amu

mass He 32 3.0160 amu

-mass He 42 - 4.0026 amu

-mass H 11 -1.0078 amu mass change 0.0197 amu

(0.0197 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.0197 g/mol ∆E = ∆mc2 = (0.0197 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.77 x 1012 J/mol = 1.77 x 109 kJ/mol

22.89 t1/2 = 1.1 x 1020 y = (1.1 x 1020 y)(365 d/y) = 4.0 x 1022 d

k = d 10 x 4.0

0.693 =

t

0.69322

2/1

= 1.7 x 10-23 d-1

N = 6.02 x 1023 atoms Decay rate = kN = (1.7 x 10-23 d-1)(6.02 x 1023 atoms) = 10/d There are 10 disintegrations per day.

22.90 e 2 +Pu _n + U 0

1_23994

10

23892

22.91 3.9 x 1023 kJ = 3.9 x 1026 J = 3.9 x 1026 kg ⋅ m2/s2

∆E = ∆mc2; ∆m = )m/s 10 x (3.00

s/m kg 10 x 3.9 =

c

E28

2226

2

•∆ = 4.3 x 109 kg

The sun loses mass at a rate of 4.3 x 109 kg/s. 22.92 10B + 1n → 4He + 7Li + γ

mass 10B 10.012 937 amu mass 1n 1.008 665 amu -mass 4He - 4.002 603 amu -mass 7Li -7.016 004 amu mass change 0.002 995 amu

(0.002 995 amu)(1.660 54 x 10-24 g/amu) = 4.973 x 10-27 g ∆E = ∆mc2 = (4.973 x 10-27 g)(10-3 kg/g)(3.00 x 108 m/s)2 = 4.476 x 10-13 J


676

Kinetic energy = 2.31 MeV x = MeV 1

J 10 x 1.60 13_

3.696 x 10-13 J

γ photon energy = ∆E - KE = 4.476 x 10-13 J - 3.696 x 10-13 J = 7.80 x 10-14 J 22.93 Each alpha emission decreases the mass number by four and the atomic number by

two. Each beta emission increases the atomic number by one. Bi _ Np 209

83 23793


number mass Bi _number mass Np

= 4


The atomic number decreases by 14 as a result of 7 alpha emissions. The resulting atomic number is (93 - 14) = 79. Number of β emissions = Bi atomic number - 79 = 83 - 79 = 4 β emissions

22.94 (a) Mo + e _ Tc 100

42 01

10043 (positron emission)

Mo _ e + Tc 10042

0 1_

10043 (electron capture)

(b) Positron emission Reactant: Tc 100

43 nucleus = Tc 10043 atom - 43 e-

Product: Mo 10042 nucleus + e+ = Mo 100

42 atom - 42 e- + 1 e+ Change: ( Tc 100

43 atom - 43 e-) - ( Mo 10042 atom - 42 e- + 1 e+)

= Tc 10043 atom - Mo 100

42 atom - 2 e-

Mass change = 99.907 657 amu - 99.907 48 amu - (2)(0.000 5486 amu) = -0.000 92 amu

(-0.000 92 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = -0.000 92 g/mol ∆E = ∆mc2 = (-0.000 92 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = -8.3 x 1010 J/mol = -8.3 x 107 kJ/mol

Electron Capture Reactant: Tc 100

43 nucleus + e- = Tc 10043 atom - 42 e-

Product: Mo 10042 nucleus = Mo 100

42 atom - 42 e- Change: ( Tc 100

43 atom - 42 e-) - ( Mo 10042 atom - 42 e-)

= Tc 10043 atom - Mo 100

42 atom (electrons cancel)

Mass change = 99.907 657 amu - 99.907 48 amu = 0.000 177 amu


677


Only electron capture is observed because there is a mass decrease and a release of energy.

22.95 (a) α emission: He +Fr _ Ac 4

222287

22689

β emission: e +Th _ Ac 0 1_

22690

22689

electron capture: Ra _ e + Ac 22688

0 1_

22689

(b) t 2/1 = d 0.556

0.693 =

k

0.6931_ = 1.25 d

If 80% reacts, then 20% is left.

t

t0.693)(_ =

N

Nln

2/10

;

d 1.25

t0.693)(_ =

100

20ln

t = 0.693) (_

d) (1.2510020

ln

= 2.90 d

Multi-Concept Problems 22.96 BaCO3, 197.34 amu

1.000 g BaCO3 x BaCO g 197.34

BaCO mol 1

3

3 x BaCO mol 1

C mol 1

3

x C mol 1

C g 12.011 = 0.060 86 g C

4.0 x 10-3 Bq = 4.0 x 10-3 disintegrations/s (4.0 x 10-3 Bq = 4.0 x 10-3 disintegrations/s)(60 s/min) = 0.24 disintegrations/min

sample radioactivity = C g 86 0.060

tions/mindisintegra 0.24= 3.94 disintegrations/min per gram of

C

t

t0.693)(_ =

N

Nln

2/10



N

N

0

y 5730

t0.693)(_ =

15.3

3.94ln ; t = 11,000 y

22.97 t1/2 = 138 d = 138 d x d 365

y 1= 0.378 y

k = y 0.378

0.693 =

t

0.693

2/1

= 1.83 y-1


678

0.700 mg x mg 1

g 10 x 1 3_

= 7.00 x 10-4 g

No = (7.00 x 10-4 g)

Po g 210

Po mol 1(6.022 x 1023 atoms/mol) = 2.01 x 1018 atoms

N

Nln

o

= -kt = - (1.83 y-1)(1 y) = -1.83; N

N

o

= e-1.83 = 0.160

N = 0.160 No = (0.160)(2.01 x 1018 atoms) = 0.322 x 1018 atoms atoms He = atoms Po decayed atoms He = 2.01 x 1018 atoms - 0.322 x 1018 atoms = 1.688 x 1018 atoms

mol He = atoms/mol 10 x 6.022

atoms He 10 x 1.68823

18

= 2.80 x 10-6 mol He

20oC = 293 K

P = V

nRT =

L 0.2500

K) (293mol K atm L

06 0.082mol) 10 x (2.80 6_

••

= 2.69 x 10-4 atm

P = 2.69 x 10-4 atm x atm 1.00

Hg mm 760= 2.05 mm Hg

22.98 First find the activity of the 51Cr after 17.0 days.

t

t0.693) (_ =

N

Nln

2/10



N

N

0

d 27.7

d 17.00.693) (_ =

4.10

Nln

ln N - ln(4.10) = -0.4253 ln N = -0.4253 + ln(4.10) = 0.9857 N = e0.9857 = 2.68 µCi/mL

(20.0 mL)(2.68 µCi/mL) = (total blood volume)(0.009 35 µCi/mL) total blood volume = 5732 mL = 5.73 L

22.99 First find the activity (N) that the 28Mg would have after 2.4 hours assuming that none

of it was removed by precipitation as MgCO3.

t

t0.693) (_ =

N

Nln

2/10



N

N

0

h 20.91

h 2.400.693) (_ =

0.112

Nln

ln N - ln(0.112) = -0.0795


679

ln N = - 0.0795 + ln(0.112) = -2.27 N = e-2.27 = 0.103 µCi/mL 20.00 mL = 0.020 00 L and 15.00 mL = 0.015 00 L mol MgCl2 = (0.007 50 mol/L)(0.020 00 L) = 1.50 x 10-4 mol MgCl2 mol Na2CO3 = (0.012 50 mol/L)(0.015 00 L) = 1.87 x 10-4 mol Na2CO3 The mol of CO3

2- are in excess, so assume that all of the Mg2+ precipitates as MgCO3 according to the reaction:

Mg2+(aq) + CO32-(aq) → MgCO3(s)

initial (mol) 0.000 150 0.000 187 0 change (mol) - 0.000 150 - 0.000 150 + 0.000 150

final (mol) 0 0.000 037 0.000 150 [CO3

2-] = 0.000 037 mol/(0.020 00 L + 0.015 00 L) = 0.001 06 M

Now consider the dissolution of MgCO3 in the presence of CO32-.

MgCO3(s) _ Mg2+(aq) + CO32-(aq)

initial (M) 0 0.001 06 change (M) +x +x equil (M) x 0.001 06 + x

The [Mg2+] in the filtrate is proportional to its activity after 2.40 h.

x = [Mg2+] = 0.029 µCi/mL x = Ci/mL 0.103

M 50 0.007

µ 0.002 11 M

[CO32-] = 0.001 06 + x = 0.001 06 + 0.002 11 = 0.003 17 M

Ksp = [Mg2+][CO32-] = (0.002 11)(0.003 17) = 6.7 x 10-6

681

23

Organic Chemistry

23.1

23.2

23.3

23.4 C7H16 23.5 Structures (a) and (c) are identical. They both contain a chain of six carbons with two –

CH3 branches at the fourth carbon and one –CH3 branch at the second carbon. Structure (b) is different, having a chain of seven carbons.

682

23.6 The two structures are identical. The compound is

23.7 (a) pentane

2-methylbutane

2,2-dimethylpropane (b) 3,4-dimethylhexane (c) 2,4-dimethylpentane (d) 2,2,5-trimethylheptane

23.8 (a) (b)

(c) (d) 23.9 2,3-dimethylhexane 23.10 (a) 1,4-dimethylcyclohexane (b) 1-ethyl-3-methylcyclopentane

(c) isopropylcyclobutane

23.11 (a) (b) (c)

23.12

Chapter 23 - Organic Chemistry ______________________________________________________________________________

683

23.13 (a) (b)

23.14 (a) (b) 23.15 (a) 3-methyl-1-butene (b) 4-methyl-3-heptene (c) 3-ethyl-1-hexyne

23.16 (a) (b)

(c)

23.17 (a) (b) (c)

23.18

23.19

23.20 (a) (b) (c)


684

23.21 (a) (b) (c)

23.22

23.23 (a) (b)

23.24 (a) (b)

(c)

23.25 (a) (b)

23.26

23.27 (a) (b)


685

23.28 Understanding Key Concepts

23.29

23.30 (a) (b)

23.31 (a) (b) 23.32 (a) alkene, ketone, ether (b) alkene, amine, carboxylic acid 23.33 (a) 2,3-dimethylpentane (b) 2-methyl-2-hexene

23.34

23.35


686

23.36 There are many possibilities. Here are two:

23.37

Additional Problems Functional Groups and Isomers 23.38 A functional group is a part of a larger molecule and is composed of an atom or group of

atoms that has a characteristic chemical behavior. They are important because their chemistry controls the chemistry in molecules that contain them.

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

23.40 (a) (b) (c)

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

23.42

23.43 (a)

(b)


687

(c) (d)

23.44 (a) alkene and aldehyde (b) aromatic ring, alcohol, and ketone 23.45 ester, aromatic ring, and amine Alkanes 23.46 In a straight-chain alkane, all the carbons are connected in a row. In a branched-chain

alkane, there are branching connections of carbons along the carbon chain. 23.47 An alkane is a compound that contains only carbon and hydrogen and has only single

bonds. An alkyl group is the part of an alkane that remains when a hydrogen is removed. 23.48 In forming alkanes, carbon uses sp3 hybrid orbitals. 23.49 Because each carbon is bonded to its maximum number of atoms and cannot bond to

additional atoms, an alkane is said to be saturated. 23.50 C3H9 contains one more H than needed for an alkane.

23.51 (a) Underlined carbon has five bonds.

(b) Underlined carbon has five bonds.

(c) Underlined carbon has six bonds. 23.52 (a) 4-ethyl-3-methyloctane (b) 4-isopropyl-2-methylheptane

(c) 2,2,6-trimethylheptane (d) 4-ethyl-4-methyloctane 23.53 2,2,4-trimethylpentane


688

23.54 (a) (b)

(c) (d)

23.55 (a) (b) (c) (d) 23.56 (a) 1,1-dimethylcyclopentane (b) 1-isopropyl-2-methylcyclohexane

(c) 1,2,4-trimethylcyclooctane 23.57 (a) The longest chain contains six carbons and the molecule should be named from a

hexane root; the correct name is 3,3-dimethylhexane. (b) The longest chain contains seven carbons and the molecule should be named from a heptane root; the correct name is 3,5-dimethylheptane. (c) The ring is a cycloheptane ring and the methyl groups are in the 1 and 3 position; the correct name is 1,3-dimethylcycloheptane.

23.58 The structures are shown in Problem 23.2.

hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane

23.59 heptane 2-methylhexane

3-methyhexane 2,2-dimethylpentane

3,3-dimethylpentane 2,3-dimethylpentane 2,4-dimethypentane


689

2,2,3-trimethylbutane 3-ethylpentane

23.60 (a)

(b)

(c) 23.61 Reaction (a) is likely to have a higher yield because there is only one possible

monochlorinated substitution product. Reaction (b) has four possible monochlorinated substitution products, which would result in a lower yield of the one product shown.

Alkenes, Alkynes, and Aromatic Compounds 23.62 (a) sp2 (b) sp (c) sp2 23.63 Alkenes, alkynes, and aromatic compounds are said to be unsaturated because they do

not contain as many hydrogens as their alkane analogs. 23.64 Today the term "aromatic" refers to the class of compounds containing a six-membered

ring with three double bonds, not to the fragrance of a compound. 23.65 An addition reaction is the reaction of an XY molecule with an alkene or alkyne.


690

23.66 (a) (b) (c)

23.67

23.68 (a) 4-methyl-2-pentene (b) 3-methyl-1-pentene

(c) 1,2-dichlorobenzene, or o-dichlorobenzene (d) 2-methyl-2-butene (e) 7-methyl-3-octyne

23.69 (a) (b) (c)

23.70 1-pentene 2-pentene

2-methyl-1-butene

Only 2-pentene can exist as cis-trans isomers.

2-methyl-2-butene 3-methyl-1-butene

23.71 1-pentyne 2-pentyne

3-methyl-1-butyne

23.72 (a) CH2=CHCH2CH2CH2CH3 This compound cannot form cis-trans isomers.

(b) CH3CH=CHCH2CH2CH3 This compound can form cis-trans isomers because of the different groups on each double bond C.

(c) CH3CH2CH=CHCH2CH3 This compound can form cis-trans isomers because


691

of the different groups on each double bond C.

23.73 (a) This compound can form cis-trans isomers because of the different groups on each double bond C.

(b) This compound cannot form cis-trans isomers.

(c) This compound can form cis-trans isomers because of the different groups on each double bond C.

23.74 (a) (b)

(c)

23.75 (a) (b) 5-methyl-2-hexene

2,2-dimethyl-3-hexyne

(c) (d) 2-methyl-1-hexene

1,3-diethylbenzene


692

23.76 Cis-trans isomers are possible for substituted alkenes because of the lack of rotation about the carbon-carbon double bond. Alkanes and alkynes cannot form cis-trans isomers because alkanes have free rotation about carbon-carbon single bonds and alkynes are linear about the carbon-carbon triple bond.

23.77 Small-ring cycloalkenes don't exist as cis-trans isomers because the trans isomer could

not close the carbon-carbon chain back on itself to form a ring.

23.78 (a)

(b)

(c)

23.79 (a)

(b)

(c)

23.80 (a)


693

(b)

(c)

(d)

23.81 cyclohexane

Alcohols, Amines, and Carbonyl Compounds

23.82 (a) (b)

(c) (d) 23.83 (a) CH3CH2CH2NH2 (b) (CH3CH2)2NH (c) CH3CH2CH2NHCH3


694

23.84 Quinine, a base will dissolve in aqueous acid, but menthol is insoluble. 23.85 Pentanoic acid will react with aqueous NaHCO3 to yield CO2, but methyl butanoate will

not. 23.86 An aldehyde has a terminal carbonyl group. A ketone has the carbonyl group located

between two carbon atoms. 23.87 In aldehydes and ketones, the carbonyl-group carbon is bonded to atoms (H and C) that

don't attract electrons strongly. In carboxylic acids, esters, and amides, the carbonyl-group carbon is bonded to an atom (O or N) that does attract electrons strongly.

23.88 The industrial preparation of ketones and aldehydes involves the oxidation of the related

alcohol. 23.89 Carboxylic acids, esters, and amides undergo carbonyl-group substitution reactions, in

which a group –Y substitutes for the –OH, –OC, or –N group of the starting material. 23.90 (a) ketone (b) aldehyde (c) ketone (d) amide (e) ester

23.91 (a) N,N-dimethylpropanamide pentanamide

N-methylbutanamide

(b) methyl pentanoate ethyl butanoate

propyl propanoate

23.92 C6H5CO2H(aq) + H2O(l) _ H3O

+(aq) + C6H5CO2-(aq)

initial (M) 1.0 ~0 0 change (M) -x +x +x equil (M) 1.0 - x x x

Ka = 1.0x

x_ 1.0x = 10 x 6.5 =

H]COHC[

] COHC][OH[ 225_

256

_256

+3 ≈

x = [H3O+] = [C6H5CO2H]diss = 0.0081 M

% dissociation = 100% x M 1.0

M 0.0081 = 100% x

]HCOHC[

]HCOHC[

initial256

diss256 = 0.81%


695

23.93 (a) (b) (c) 23.94 (a) methyl 4-methylpentanoate (b) 4,4-dimethylpentanoic acid

(c) 2-methylpentanamide 23.95 (a) N,N-dimethyl-4-methylhexanamide (b) isopropyl 2-methylpropanoate

(c) N-ethyl-p-chlorobenzamide

23.96 (a) (b)

(c)

23.97 (a) (b)

(c) 23.98 (a)

(b)

(c)


696

23.99 (a)

(b)

(c)

23.100 amine, aromatic ring, and ester

carboxylic acid alcohol

23.101

Polymers 23.102 Polymers are large molecules formed by the repetitive bonding together of many smaller

molecules, called monomers. 23.103 Polyethylene results from the polymerization of a simple alkene by an addition reaction

to the double bond. Nylon results from the sequential reaction of two difunctional molecules.

23.104


697

23.105 (a)

(b)

23.106 (a) (b) (c)

23.107

23.108

23.109 General Problems

23.110 (a) (b)

(c) (d)


698

(e) (f) 23.111 (a) 2,3-dimethylhexane (b) 4-isopropyloctane

(c) 4-ethyl-2,4-dimethylhexane (d) 3,3-diethylpentane 23.112 Cyclohexene will react with Br2 and decolorize it. Cyclohexane will not react. 23.113 Cyclohexene will react with Br2 and decolorize it. Benzene will not react with Br2

without a catalyst.

23.114 (a) (b) (c)

23.115 Multi-Concept Problems 23.116 (a) Calculate the empirical formula. Assume a 100.0 g sample of fumaric acid.

41.4 g C x C g 12.01

C mol 1 = 3.45 mol C; 3.5 g H x

H g 1.008


55.1 g O x O g 16.00


Because the mol amounts for the three elements are essentially the same, the empirical formula is CHO (29 amu).

(b) Calculate the molar mass from the osmotic pressure.

Π = MRT; M = T R

Π =

K) (298mol K atm L

06 0.082

Hg mm 760atm 1.00

x Hg mm 240.3

••

= 0.0129 M

(0.1000 L)(0.0129 mol/L) = 1.29 x 10-3 mol fumaric acid


699

fumaric acid molar mass = mol 10 x 1.29

g 0.15003_

= 116 g/mol

molecular mass = 116 amu

(c) Determine the molecular formula. 29

116 =

mass formula empirical

massmolar = 4

molecular formula = C(1 x 4)H(1 x 4)O(1 x 4) = C4H4O4 From the titration, the number of carboxylic acid groups can be determined.

mol C4H4O4 = 0.573 g x g 116

OHC mol 1 444 = 0.004 94 mol C4H4O4

mol NaOH used = (0.0941 L)(0.105 mol/L) = 0.0099 mol NaOH

mol 94 0.004

mol 0.0099 =

OHC mol

NaOH mol

444

= 2

Because 2 mol of NaOH are required to titrate 1 mol C4H4O4, C4H4O4 is a diprotic acid. Because C4H4O4 gives an addition product with HCl and a reduction product with H2, it contains a double bond.

(d) The correct structure is 23.117 (a) CO2, 44.01 amu; H2O, 18.02 amu

mol CO2 = 0.1213 g CO2 x CO g 44.01

CO mol 1

2

2 = 0.00276 mol CO2

mol H2O = 0.0661 g H2O x OH g 18.02

OH mol 1

2

2 = 0.00367 mol H2O

mass C = 0.00276 mol CO2 x C mol 1

C g 12.011 x

CO mol 1

C mol 1

2

= 0.0332 g C

mass H = 0.00367 mol H2O x H mol 1

H g 1.008 x

OH mol 1

H mol 2

2

= 0.00740 g H

mass O = 0.0552 g sample - 0.0332 g C - 0.00740 g H = 0.0146 g O

mol C = 0.00276 mol CO2 x CO mol 1

C mol 1

2

= 0.00276 mol C


700

mol H = 0.00367 mol H2O x OH mol 1

H mol 2

2

= 0.00734 mol H

mol O = 0.0146 g O x O g 16.00

O mol 1= 0.000913 mol O

C0.00276 H0.00734 O0.000913 (divide each subscript by the smallest) C0.00276 / 0.000913 H0.00734 / 0.000913 O0.000913 / 0.000913 C3.023 H8.039 O C3H8O

2 C3H8O(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(l)

(b) C3H8O is a molecular formula because a multiple such as C6H16O2 is not possible. (c)

(d) Acetone is . The most likely structure for C3H8O is .

(e) C3H8O, 60.10 amu

mol C3H8O = 5.000 g C3H8O x OHC g 60.10

OHC mol 1

83

83 = 0.08320 mol C3H8O

∆Hocombustion =

mol 0.08320

kJ 166.9 _= -2006 kJ/mol = -2006 kJ

C3H8O(l) + 9/2 O2(g) → 3 CO2(g) + 4 H2O(l) ∆Ho

combustion = [ 3 ∆Hof(CO2) + 4 ∆Ho

f(H2O)] - ∆Hof(C3H8O)

∆Hof(C3H8O) = [ 3 ∆Ho

f(CO2) + 4 ∆Hof(H2O)] - ∆Ho

combustion = [(3 mol)(-393.5 kJ/mol) + (4 mol)(-285.8 kJ/mol)] - (-2006 kJ) = -317.7 kJ

∆Hof = -317.7 kJ/mol

23.118 (a) propanamide

(b)


701

(c) (d) An observed trigonal planar N does not agree with the VSEPR prediction. The

second resonance structure is consistent with a trigonal planar N.

solution manual - chemistry-4th ed. (mcmurry)

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