Solution Manual - Chemistry-4th Ed. (McMurry)
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Transcript of Solution Manual - Chemistry-4th Ed. (McMurry)
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1
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 _____________________________________________________________________________
2
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
3
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
4
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.
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
5
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
6
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.
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
7
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
8
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.
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
9
(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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
10
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
11
(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
_9
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
13
(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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
14
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
15
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
16
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
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
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
18
The melting point of sodium chloride (NaCl) on the gallium scale is 355oG.
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
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.
Chapter 1 S Chemistry: Matter and Measurement _____________________________________________________________________________
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.
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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 amu 1 C atom x = 2.01 x C atoms10
1.6605 x g 12.011 amu10
1.00 g x 23_ 24
1 amu 1 H atom x = 5.97 x H atoms10
1.6605 x g 1.008 amu10
23
23
C 2.01 x C atoms 1 C10 = = H 5.97 x H atoms 3 H10
A possible formula for ethane is CH3.
For ethylene:
4.29 g x 23_ 24
1 amu 1 C atom x = 2.15 x C atoms10
1.6605 x g 12.011 amu10
0.71 g x 23_ 24
1 amu 1 H atom x = 4.2 x H atoms10
1.6605 x g 1.008 amu10
23
23
C 2.15 x C atoms 1 C10 = = H 4.2 x H atoms 2 H10
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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 amu 1 C atom x = 6.62 x C atoms10
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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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.
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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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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%
Chapter 2 S Atoms, Molecules, and Ions ______________________________________________________________________________
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
32
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
4.50 g C7H6O3 x OHC mol 1OHC g 180.2
x OHC mol 1OHC mol 1
x OHC g 138.1
OHC mol 1
489
489
367
489
367
367 = 5.87 g
C9H8O4
4.50 g C7H6O3 x OHC mol 1OHC g 60.1
x OHC mol 1OHC mol 1
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 %
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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.
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
OFe g 159.7OFe mol 1
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 17.1 = OHC mol 1
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
O mol 1 = 0.28 mol O
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.
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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 7.02 = OHC mol 1
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.
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
0.400 g NaCl x NaCl g 58.4
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
4.30 g NaCl x NaCl g 58.4
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-
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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%
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
H mol 1 = 6.17 mol H
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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 %
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
53
4.48 g H x H g 1.01
H mol 1 = 4.44 mol H
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
H mol 1 = 5.37 mol H
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
Fe mol 1 = 0.5375 mol Fe
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
57
mass C = 0.0507 mol C x = C mol 1
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
mass H = 0.043 mol H x = H mol 1
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
mass C = 0.110 mol C x = C mol 1
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
mass H = 0.164 mol H x = H mol 1
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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 _
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Fe mol 1 = 0.133 mol Fe
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
Rearrange to get X = 10.0 g - Y and then substitute it into the equation above to solve for Y.
= 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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
mass Cl = 0.0117 mol Cl x = Cl mol 1
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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
Chapter 3 - Formulas, Equations, and Moles ______________________________________________________________________________
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.
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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)
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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.
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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:
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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
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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)
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(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.
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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
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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)
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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) → 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)
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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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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)
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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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
2-(aq) + 3 H+(aq) + 2 e-] x 3 (oxidation half reaction)
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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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) → 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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
OFe g 159.69OFe mol 1
+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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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)
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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
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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)
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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.
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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.
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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)
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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
2-(aq) + 16 H+(aq) + 10 e-] x 3 (oxidation half reaction)
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)
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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)
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[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,
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
Rearrange to get X = 10.50 g - Y and then substitute it into the equation above to solve for Y.
= 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
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
Rearrange to get X = 10.0 g - Y and then substitute it into the equation above to solve for Y.
= )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
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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.
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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+
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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) 14 H+(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
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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:
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
102
mol Cl in MCl5 = 0.0337 mol S x = S mol 1
Cl mol 5 0.168 mol Cl
mass Cl = 0.168 mol Cl x = Cl mol 1
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:
mol Cl in MCl5 = 0.0337 mol S x = S mol 2
Cl mol 5 0.0842 mol Cl
mass Cl = 0.0842 mol Cl x = Cl mol 1
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:
mol Cl in MCl5 = 0.0337 mol S x = S mol 3
Cl mol 5 0.0562 mol Cl
mass Cl = 0.0562 mol Cl x = Cl mol 1
Cl g 35.453 1.99 g Cl
mass M = 4.61g - 1.99 g = 2.62 g M
mol M = 0.0337 mol S x = S mol 3
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:
mol Cl in MCl5 = 0.0337 mol S x = S mol 4
Cl mol 5 0.0421 mol Cl
mass Cl = 0.0421 mol Cl x = Cl mol 1
Cl g 35.453 1.49 g Cl
mass M = 4.61g - 1.49 g = 3.12 g M
mol M = 0.0337 mol S x = S mol 4
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)
Chapter 4 - Reactions in Aqueous Solutions ______________________________________________________________________________
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
c 3.00 x m / s10E = h = (6.626 x J s) (6.022 x / mol)10 105.49 x m10
• 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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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.
Understanding Key Concepts
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
c 3.00 x m / s10E = h = (6.626 x J s) (6.022 x / mol)10 103.44 x m10
• λ
E = 3.48 x 107 J/mol = 3.48 x 104 kJ/mol λ = 6.71 Η 10S2 m
8_34 23
_ 2
c 3.00 x m / s10E = h = (6.626 x J s) (6.022 x / mol)10 106.71 x m10
• 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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10486.1 x m 1000 J10
E = 246.1 kJ / mol
λ
For n = 5; λ = 434.0 nm = 434.0 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 10434.0 x m 1000 J10
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10364.6 x m 1000 J10
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 102625 x m 1000 J10
E = 45.60 kJ / mol, IR
λ
5.48 λ = 330 nm = 330 x 10S9 m
8_34 23
_9
c 3.00 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10330 x m 1000 J10
E = 363 kJ / mol
λ
5.49 795 nm = 795 x 10S9 m
8_34 23
_9
c 3.00 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10795 x m 1000 J10
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.
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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]
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 10410.2 x m 1000 J10
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 107458 x m 1000 J10
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
c 2.998 x m / s 1 kJ10E = h = (6.626 x Jcdot s) (6.022 x / mol) 10 104653 x m 1000 J10
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•
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
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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
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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
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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
Chapter 5 S Periodicity and Atomic Structure ______________________________________________________________________________
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
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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
122
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 ______________________________________________________________________________
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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.
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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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.
Understanding Key Concepts
6.27 6.28 (a) shows an extended array, which represents an ionic compound.
(b) shows discrete units, which represent a covalent compound.
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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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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.
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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.
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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.
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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)
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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.
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
molar mass of M = mol 10 x 4.20
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
molar mass of M = mol 10 x 2.80
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.
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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_ •
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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
Chapter 6 - Ionic Bonds and Some Main-Group Chemistry ______________________________________________________________________________
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)
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
143
7.11 (a) (b) (c)
(d)
7.12
7.13 (a)
(b)
(c)
(d)
7.14
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
145
For right Isolated oxygen valence electrons 6 oxygen: Bound oxygen bonding electrons 4
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
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
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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
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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
electrons gantibondin
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.
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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.
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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.
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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)
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(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).
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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
For oxygen: Isolated oxygen valence electrons 6
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
For oxygen: Isolated oxygen valence electrons 6
Bound oxygen bonding electrons 4
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Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
(b) 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 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
(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
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154
Bound oxygen bonding electrons 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
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
Bound oxygen bonding electrons 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For right oxygen: Isolated oxygen valence electrons 6
Bound oxygen bonding electrons 4 Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
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
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
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
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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
Bound hydrogen bonding electrons 2 Bound hydrogen nonbonding electrons 0 Formal charge = 1 - ½(2) - 0 = 0
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
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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
Bound hydrogen bonding electrons 2 Bound hydrogen nonbonding electrons 0 Formal charge = 1 - ½(2) - 0 = 0
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
For carbon: Isolated carbon valence electrons 4
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
Bound oxygen bonding electrons 4 Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
For left carbon: Isolated carbon valence electrons 4
Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0
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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 oxygen: Isolated oxygen valence electrons 6
Bound oxygen bonding electrons 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For left carbon: Isolated carbon valence electrons 4
Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0
For right carbon: Isolated carbon valence electrons 4
Bound carbon bonding electrons 8 Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0
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
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158
7.70 Number of Number of
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
7.71 Number of Number of
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
7.72 Number of Number of
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)
7.75 Number of Number of
Bonded Atoms Lone Pairs Shape (a) XeF3
+ 3 2 T shaped (b) SF3
+ 3 1 trigonal pyramidal (c) ClF2
+ 2 2 bent
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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
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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 ↑↓ ↑↓ ↑↓
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σ*2s ↑↓ ↑↓ ↑↓ σ2s ↑↓ ↑↓ ↑↓
O2+ O2 O2
-
Bond order = 2
electrons gantibondin
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
electrons gantibondin
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)
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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
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For For oxygen: Isolated oxygen valence electrons 6 (top) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For oxygen: Isolated oxygen valence electrons 6 (middle) Bound oxygen bonding electrons 4
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
For oxygen: Isolated oxygen valence electrons 6 (left) Bound oxygen bonding electrons 4
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
For oxygen: Isolated oxygen valence electrons 6 (right) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -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
For For oxygen: Isolated oxygen valence electrons 6 (top) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For oxygen: Isolated oxygen valence electrons 6 (middle) Bound oxygen bonding electrons 4
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
For oxygen: Isolated oxygen valence electrons 6 (left) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6
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164
Formal charge = 6 - ½(2) - 6 = -1
For oxygen: Isolated oxygen valence electrons 6 (right) Bound oxygen bonding electrons 4
Bound oxygen nonbonding electrons 4 Formal charge = 6 - ½(4) - 4 = 0
For sulfur: Isolated sulfur valence electrons 6
Bound sulfur bonding electrons 8 Bound sulfur nonbonding electrons 0 Formal charge = 6 - ½(8) - 0 = +2
For For oxygen: Isolated oxygen valence electrons 6 (top) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For oxygen: Isolated oxygen valence electrons 6 (middle) Bound oxygen bonding electrons 6
Bound oxygen nonbonding electrons 2 Formal charge = 6 - ½(6) - 2 = +1
For oxygen: Isolated oxygen valence electrons 6 (left) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
For oxygen: Isolated oxygen valence electrons 6 (right) Bound oxygen bonding electrons 2
Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -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
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7.100
For 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 carbon: Isolated carbon valence electrons 4 (left) Bound carbon bonding electrons 8
Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 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
For carbon: Isolated carbon valence electrons 4 (right) 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
For 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 carbon: Isolated carbon valence electrons 4 (left) Bound carbon bonding electrons 8
Bound carbon nonbonding electrons 0 Formal charge = 4 - ½(8) - 0 = 0
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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 carbon: Isolated carbon valence electrons 4 (right) 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 2 Bound oxygen nonbonding electrons 6 Formal charge = 6 - ½(2) - 6 = -1
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
For oxygen: Isolated oxygen valence electrons 6
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
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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
electrons gantibondin
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
electrons gantibondin
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
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(“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.
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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
electrons gantibondin
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.
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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
electrons gantibondin
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.
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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
electrons gantibondin
ofnumber _
electrons bonding
ofnumber
2 = 3
6 _ 12 =order bond X2
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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
Chapter 7 - Covalent Bonds and Molecular Structure ______________________________________________________________________________
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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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 = D (Reactant bonds) - D (Product bonds)
∆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)]
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
HCl mol 1 = 3.43 mol HCl
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
q = (specific heat) x m x ∆T
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)
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
HCl mol 1 = 0.219 mol HCl
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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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)]
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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 = D (Reactant bonds) - D (Product bonds)
∆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 = D (Reactant bonds) - D (Product bonds)
∆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 = D (Reactant bonds) - D (Product bonds)
∆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 = D (Reactant bonds) - D (Product bonds)
∆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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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 =
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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)]
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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.
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
q = (specific heat) x m x ∆T
∆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
Chapter 8 - Thermochemistry: Chemical Energy ______________________________________________________________________________
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
Cl mol 1 1.08 mol Cl
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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)
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Understanding Key Concepts
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 765
= 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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 356
= RT
PV
••
= 0.0290 mol
molar mass = mol 0.0290
g 0.9847 = 34.0 g/mol; molecular mass = 34.0 amu
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
molar mass = mol 0.0446
g 1.342 = 30.1 g/mol; molecular mass = 30.1 amu
(b) Assume 1.000 L gas sample
n = K) (298
mol K atm L
06 0.082
L) (1.000Hg mm 760
atm 1.00 x Hg mm 752
= RT
PV
••
= 0.0405 mol
molar mass = mol 0.0405
g 1.053 = 26.0 g/mol; molecular mass = 26.0 amu
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
K) (273.15mol K atm L
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)
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 350
= 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
atm 1.00 x Hg mm 735
= 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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
211
K) (708mol K atm L
06 0.082
L) (155Hg mm 760
atm 1.00 x Hg mm 795
= 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
P X = P totalHCHC 8383• = (0.015)(1.48 atm) = 0.022 atm
0.0050; = mol 100
mol 0.5 = X HC 104
P X = P totalHCHC 104104• = (0.0050)(1.48 atm) = 0.0074 atm
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
HCl mol 1 = 0.137 mol HCl
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 723
= 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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Square both sides of the equation and solve for T.
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
Square both sides of the equation and solve for T.
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 564
= 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
atm 1.00 x Hg mm 219
= RT
PV
••
= 0.011 78 mol
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
217
mol gas in Bulb B = n = K) (203
mol K atm L
06 0.082
L) (1.000Hg mm 760
atm 1.00 x Hg mm 219
= 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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
219
9.103 (a) ntotal = K) (293
mol K atm L
06 0.082
L) (0.500Hg mm 760
atm 1.00 x Hg mm 258
= 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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
K) (373.1mol K atm L
06 0.082mol) (0.196 =
V
nRT = PCO2
••
= 0.600 atm
mol SO2 = 2x = 2(0.196 mol) = 0.392 mol
L) (10.0
K) (373.1mol K atm L
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
K) (473.1mol K atm L
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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
atm 1.00 x Hg mm 623
= 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
atm 1.00 x Hg mm 260
= 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
atm 1.00 x Hg mm 223
= 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
Chapter 9 - Gases: Their Properties and Behavior ______________________________________________________________________________
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.
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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
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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.
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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)
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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.
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
density = cm 10 x 4.74
g10 x 4.22 =
volume
mass323_
22_
= 8.90 g/cm3
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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.
density = cm 10 x 1.2129
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
density = cm 10 x 2.429
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
density = cm 10 x 6.870
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
density = cm 10 x 3.739
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.
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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
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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.
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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)
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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.
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
density = cm 10 x 1.719
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
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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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
atm 1.00 x Hg mm 751
= 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
atm 1.00 x Hg mm 740
= RT
PV = nH2
••
= 0.160 mol H2
M = Group 3A metal; 2 M(s) + 6 H+(aq) → 2 M3+(aq) + 3 H2(g)
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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:
density = cm 10 x 6.59
atom) g/Al 10 x Al)(4.48 (1 =
volumecellunit
mass cellunit 323_
23_
= 0.680 g/cm3
For body-centered cubic:
density = cm 10 x 6.59
atom) g/Al 10 x Al)(4.48 (2 =
volumecellunit
mass cellunit 323_
23_
= 1.36 g/cm3
For face-centered cubic:
density = cm 10 x 6.59
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
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
atm 1.00 x Hg mm 755
= 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+
Chapter 10 - Liquids, Solids, and Changes of State ______________________________________________________________________________
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
density = cm 10 x 1.531
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
volume of solution = g 1.1094
mL 1 x g 100.0 = 90.14 mL = 0.090 14 L
molarity = L 14 0.090
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
volume of solution = g 1.0514
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
molality = kg 0.9285
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
C mol 1 = 7.45 mol C
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
C mol 1 = 4.10 mol C
9.65% H 9.65 g H x H g 1.008
H mol 1 = 9.57 mol H
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
O mol 1 = 1.37 mol O
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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.
molality = kg 0.0965
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
molality = kg 0.9655
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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%
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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.
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
287
Multi-Concept Problems 11.131 (a) 382.6 mL = 0.3826 L; 20.0oC = 293.2 K
PV = nRT
nH2 =
K) (293.2mol K atm L
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.
molar mass = mol 0.0159
g 2.505 = 157.5 g/mol; molecular mass = 157.5 amu
(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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
molar mass = mol 0.0025
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 11 - Solutions and Their Properties ______________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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)
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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.
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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)
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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.
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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)
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
1mol)] kJ/(Kcdot 10 x )][8.31410 x ln(1.0 _ )10 x [ln(3.0
_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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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.
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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.
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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)
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
1mol)] kJ/(Kcdot 10 x )][8.31410 x ln(1.42 _ )10 x [ln(3.05
_3_4_4_
= 58.3
kJ/mol
Use Ea, k1, and k3 to calculate T3.
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 12 - Chemical Kinetics _____________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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,
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
339
13.50 The container volume of 2.00 L must be included to calculate molar concentrations.
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
Chapter 13 - Chemical 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
13.66 The container volume of 10.0 L must be included to calculate molar concentrations.
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
x2 + 0.140x - 0.2058 = 0 Use the quadratic formula to 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.
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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)
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
initial (M) 0.100 0 change (M) -2x +x equil (M) 0.100 - 2x x
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
x2 + 4.92x - 1.26 = 0 Use the quadratic formula to solve for x.
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.
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Use the quadratic formula to solve for x.
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.
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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)
initial (M) 0.0100 0 change (M) -2x +x equil (M) 0.0100 - 2x x
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
Multi-Concept Problems
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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
initial (M) 0.120 0 change (M) -2x +x equil (M) 0.120 - 2x x
Kc = = ][monomer
[dimer]2
1.51 x 102 = )x2 _ (0.120
x2
604x2 - 73.48x + 2.1744 = 0
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
363
Use the quadratic formula to solve for x.
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
Chapter 13 - Chemical Equilibrium ______________________________________________________________________________
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.
365
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)
initial (M) 0.25 ~0 0.50 change (M) Sx +x +x equil (M) 0.25 S x x 0.50 + x
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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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)
initial (M) 0.050 ~0 0.100 change (M) Sx +x +x equil (M) 0.050 S x x 0.100 + x
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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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
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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
+(aq) + OHS(aq) 2 H2O(l) before reaction (mol) 0.004 01 0.004 00 change (mol) S0.004 00 S0.004 00 after reaction (mol) 0.000 01 0
[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
+(aq) + OHS(aq) 2 H2O(l) before reaction (mol) 0.005 00 0.004 00 change (mol) S0.004 00 S0.004 00 after reaction (mol) 0.001 00 0
[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
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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
≈
Solve for x. x = [H3O+] = 1.05 x 10S7 M
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
≈
Solve for x. x = [OHS] = 5.65 x 10S5 M
[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.
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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)
initial (M) 0.0107 ~0 0.0320 change (M) Sx +x +x equil (M) 0.0107 S x x 0.0320 + x
Ka = 2_+
3 3 _8_3
[ ][ ] x(0.0320 + x) x(0.0320)O SOH = 6.3 x = 10[ ] 0.0107 _ x 0.0107HSO
≈
Solve for x. x = [H3O+] = 2.1 x 10S8 M
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
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= (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].
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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
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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
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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.
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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)
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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
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(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,
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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)
initial (M) 0.25 ~0 0.10 change (M) Sx +x +x equil (M) 0.25 S x x 0.10 + x
Ka = + _
3 _ 4[ ][ ] x(0.10 + x) x(0.10)OH F = 3.5 x = 10[HF] 0.25 _ x 0.25
≈
Solve for x. x = [H3O+] = 8.8 x 10S4 M
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)
initial (M) 0.10 0.30 ~0 change (M) Sx +x +x equil (M) 0.10 S x 0.30 + x x
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
≈
Solve for x. x = [OHS] = 1.2 x 10S4 M
[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
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% 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
≈
Solve for x. x = [OHS] = 2.3 x 10S3 M
[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)
initial (M) 0.30 0.50 ~0 change (M) Sx +x +x equil (M) 0.30 S x 0.50 + x x
Kb = + _4 _5
3
[ ][ ] (0.50 + x) x (0.50) xNH OH = 1.8 x = 10[ ] 0.30 _ x 0.30NH
≈
Solve for x. x = [OHS] = 1.1 x 10S5 M
[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
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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
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[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
[base] [ ]NH = + log pK[acid] [ ]NH
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
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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
16.72 pH = pKa + log[base]
[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.
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16.73 pH = pKa + log[base]
[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)
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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
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(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
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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
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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)
initial (M) 0.16 ~0 0.0400 change (M) Sx +x +x equil (M) 0.16 S x x 0.0400 + x
Ka = + _
3 _ 4[ ][ ] x(0.0400 + x) x(0.0400)OH F = 3.5 x = 10[HF] 0.16 _ x 0.16
≈
Solve for x. x = [H3O+] = 1.4 x 10S3 M
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
≈
Solve for x. x = [OHS] = 1.8 x 10S6 M
[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].
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[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)
initial (M) 0.100 0 ~0 change (M) Sx +x +x equil (M) 0.100 S x x x
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
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Ka = + 2 2
3 3 2 _11+
3 3
[ ][ ]O CH NHH x x = 2.7 x = 10[ ] 0.0714 _ x 0.0714CH NH
≈
Solve for x. x = [H3O+] = 1.4 x 10S6 M
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.
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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
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[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
≈
Solve for x. x = [OHS] = 1.1 x 10S6 M
[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
≈
Solve for x. x = [H3O+] = 1.2 x 10S6 M
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
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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
≈
Solve for x. x = [H3O+] = 8.8 x 10S7 M
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
≈
Solve for x. x = [OHS] = 1.26 x 10S4 M
[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
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(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
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molar solubility = x = _12
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
molar solubility = x = _14
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
molar solubility = x = _12
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
molar solubility = x = _37
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.
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16.98 (a) PbCrO4(s) Ω Pb2+(aq) + CrO42S(aq)
equil (M) x x Ksp = [Pb2+][CrO4
2S] = 2.8 x 10S13 = (x)(x)
molar solubility = x = _132.8 x 10 = 5.3 x 10S7 M
(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)
molar solubility = x = _13
_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
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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)
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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)
molar solubility = x = _178.5 x 10 = 9.2 x 10S9 M
(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.
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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
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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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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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.
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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
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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
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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
[base] [ ]NH = + log pK[acid] [ ]NH
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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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
molar solubility = x = _12
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
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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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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]
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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.
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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
molar solubility = x = _15
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)
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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
molar solubility = x = _17
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
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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
Multi-Concept Problems
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
+]
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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)
initial (M) 0.100 0 ~0 change (M) Sx +x +x equil (M) 0.100 S x x x
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
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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
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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)
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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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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)
initial (M) 0.080 ~0 0.200 change (M) Sx +x +x equil (M) 0.080 S x x 0.200 + x
Ka = 2_+
3 3 _11_3
[ ][ ] x(0.200 + x) x(0.200)O COH = 5.6 x = 10[ ] 0.080 _ x 0.080HCO
≈
Solve for x. x = [H3O+] = 2.24 x 10S11 M
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
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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)
CO32S(aq) + H2O(l) Ω HCO3
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)
CO32S(aq) + H2O(l) Ω HCO3
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
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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
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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 =
Chapter 16 S Applications of Aqueous Equilibria ______________________________________________________________________________
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(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)
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(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
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∆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.
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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)
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∆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.
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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
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∆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)
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(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
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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.
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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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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.
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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)]
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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)]
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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.
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
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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)
Both calculations lead to the same value of ∆Sof.
(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)
Both calculations lead to the same value of ∆Sof.
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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.
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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 ∆∆
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
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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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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 ±±
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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)
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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)
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
atm 1.00 x Hg mm 731
= 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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
520
nNH3 =
K) (298.1mol K atm L
06 0.082
L) (1.00Hg mm 760
atm 1.00 x Hg mm 744
= 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
Chapter 17 - Thermodynamics: Entropy, Free Energy, and Equilibrium ______________________________________________________________________________
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
522
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Understanding Key Concepts
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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-
∆Go = -nFEo = -(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-
∆Go = -nFEo = -(6 mol e-)
e mol 1
C 96,500_
(1.26 V)
• V C 1
J 1 = -729,540 J = -730 kJ
(c) 6 Fe2+(aq) + Cr2O72-(aq) + 14 H+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
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-
∆Go = -nFEo = -(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
overall Eo = 1.08 V n = 6 mol e-
∆Go = -nFEo = -(6 mol e-)
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
overall Eo = 0.13 V n = 2 mol e-
∆Go = -nFEo = -(2 mol e-)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
540
n = 2 mol e-
∆Go = -nFEo = -(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.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
(c) 6 Fe2+(aq) + Cr2O72-(aq) + 14 H+(aq) → 6 Fe3+(aq) + 2 Cr3+(aq) + 7 H2O(l)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
∆Go = -nFEo = -(4 mol e-)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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+.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
K) (273.15mol K atm L
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.
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
∆Go = -nFEo = -(2 mol e-)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
∆Go = -nFEo = -(5 mol e-)
e mol 1
C 96,500_
(-0.008 V)
• V C 1
J 1 = 3,860 J = +3.9 kJ
∆Go = -nFEo = -(9 mol e-)
e mol 1
C 96,500_
(-0.240 V)
• V C 1
J 1 = 208,440 J = +208.4 kJ
∆Go = -nFEo = -(6 mol e-)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
∆Go = -nFEo = -(10 mol e-) V) (2.00e mol 1
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
molarity = L 50 0.032
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:
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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 ≈
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
∆Go = -nFEo = -(2 mol e-)
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
∆Go = -nFEo = -(4 mol e-) V) (1.229e mol 1
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)
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
- in 10.0 mL = (0.010 mol/L)(0.0100 L) = 0.000 10 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
- in 19.0 mL = (0.010 mol/L)(0.0190 L) = 0.000 19 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
Chapter 18 - Electrochemistry _____________________________________________________________________________
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
- in 21.0 mL = (0.010 mol/L)(0.0210 L) = 0.000 21 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
+
x2 + 1.252x - 0.0149 = 0 Use the quadratic formula to solve for x.
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
Chapter 18 - Electrochemistry _____________________________________________________________________________
567
Notice that there is a dramatic change in the potential at the equivalence point.
568
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
tat time rateDecay =
N
N
0
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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 of α emissions = 4
number mass Pb _number mass U
= 4
207 _ 235 = 7 α emissions
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.
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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%
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
22.57 Decay rate = kN
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
; 0 = t at time rateDecay
tat time rateDecay =
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
; 0 = t at time rateDecay
tat time rateDecay =
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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.
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
nuclei/mol 10 x 6.022
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
nuclei/mol 10 x 6.022
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
nuclei/mol 10 x 6.022
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
nuclei/mol 10 x 6.022
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
(0.001 97 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.001 97 g/mol ∆E = ∆mc2 = (0.001 97 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.77 x 1011 J/mol = 1.77 x 108 kJ/mol
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
(0.045 809 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.045 809 g/mol ∆E = ∆mc2 = (0.045 809 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 4.12 x 1012 J/mol = 4.12 x 109 kJ/mol
22.81 Decay rate = kN
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)
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
; 0 = t at time rateDecay
tat time rateDecay =
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
; 0 = t at time rateDecay
tat time rateDecay =
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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
nuclei/mol 10 x 6.022
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
nuclei/mol 10 x 6.022
J/mol 10 x 5.40113_23
13
= 8.76 nucleon
MeV
The 64Zn is more stable.
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
675
22.87 ln
t
t0.693)(_ =
N
N
2/10
; 0 = t at time rateDecay
tat time rateDecay =
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
Chapter 22 - Nuclear Chemistry ______________________________________________________________________________
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 of α emissions = 4
number mass Bi _number mass Np
= 4
209 _ 237 = 7 α emissions
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
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(0.000 177 amu)(1.660 54 x 10-24 g/amu)(6.022 x 1023 mol-1) = 0.000 177 g/mol ∆E = ∆mc2 = (0.000 177 g/mol)(10-3 kg/g)(3.00 x 108 m/s)2 ∆E = 1.6 x 1010 J/mol = 1.6 x 107 kJ/mol
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
; 0 = t at time rateDecay
tat time rateDecay =
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
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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
; 0 = t at time rateDecay
tat time rateDecay =
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
; 0 = t at time rateDecay
tat time rateDecay =
N
N
0
h 20.91
h 2.400.693) (_ =
0.112
Nln
ln N - ln(0.112) = -0.0795
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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
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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)
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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)
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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
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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)
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(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
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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
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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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.
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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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
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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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)
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(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
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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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%
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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)
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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
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23.105 (a)
(b)
23.106 (a) (b) (c)
23.107
23.108
23.109 General Problems
23.110 (a) (b)
(c) (d)
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(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
H mol 1 = 3.47 mol H
55.1 g O x O g 16.00
O mol 1 = 3.44 mol O
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
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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
Chapter 23 - Organic Chemistry ______________________________________________________________________________
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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)
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(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.